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General Physics Particle Kinematics S.C. Pandey Dynamics of Particle Circular Motion Energy and Momentum
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(Volume 3) er s
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Waves, Optics and Thermodynamics
Chandigarh • Delhi • Chennai https://t.me/universitarios_infohttps://www.jamarana.com
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Production Editor: Nitkiran Bedi Composition: Tantla Composition Services, Chandigarh Printer: Sanat Printers
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The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertant omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book.
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Copyright © 2010 Dorling Kindersley (India) Pvt. Ltd
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This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated, without the publisher’s prior written consent, in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise) without the prior written permission of both the copyright owner and the abovementioned publisher of this book. ISBN 978-81-317-3411-7 10 9 8 7 6 5 4 3 2 1 Published by Dorling Kindersley (India) Pvt. Ltd, licensees of Pearson Education in South Asia.
Head Office: 7th Floor, Knowledge Boulevard, A-8-(A) Noida-201309, India. Registered Office: 11 Community Centre, Panchsheel Park, New Delhi 110 017, India.
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Preface
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Contents Chapter 2
Kinetic Theory of Gases and Thermodynamics
Chapter 3
Wave Mechanics
1.1–1.51 2.1–2.114 3.1–3.69 4A.1–4A.123 4B.1–4B.95
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Chapter 4B Light (Wave Optics)
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Chapter 4A Light (Geometrical Optics)
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Heat
Chapter 1
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Preface er si ta rio s
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Salient feature
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For a science student, physics is the most important subject as it requires logical reasoning and high imagination. Without improving the level of physics, it is difficult to achieve a goal with the kind of competition that exists today. This five part volume covers all parts of general physics—Mechanics, Heat, Wave, Light, Electromagnetism and Modern Physics—which is written in accordance with the latest syllabus of the IIT-JEE and AIEEE. There is no single book that is available in the market that contains a large amount of solved examples.
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■ Entire syllabus is covered in five volumes. ■ Content of each chapter is well defined and builds new concepts from the scratch. ■ Each chapter describes the theory in a simple and lucid style. ■ Covers a wide spectrum of questions to enable the student to develop enough expertise to tackle any problem. ■ Helps students in building analytical and quantitative skills, which, in turn, develop confidence in problem solving. ■ Practice exercises are given at the end of each chapter. ■ Numerous diagrams in every chapter. After studying the entire chapter, students will be able to learn different tricks and techniques of problem solving with suitable level of analytical ability. Suggestions for improving the book are always welcome.
All the best!
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S. C. PANDEY
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c h a p t e r
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Heat
1
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Zeroth Law of Thermodynamics
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It is a form of energy which determines the change in thermal state of the body and is defined as the flow of energy from one body to the other body due to difference in the degree of hotness of two bodies (temperature). It flows from the body which is at a higher temperature to the other at low temperature.
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If two bodies A and B are in thermal equilibrium and A and C are also in thermal equilibrium then B and C are also in thermal equilibrium. Calorimetry It is the branch of heat transfer that deals with the measurement of heat. The heat is usually measured in calories or kilocalories. One Calorie 1 calorie is the quantity of heat required to raise the temperature of 1 gram of water by 1ºC. Specific heat The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. s=
∆θ m∆T
Its unit is cal/gm/ºC. The heat lost by the body or gained from the body depends upon the difference in the temperature. Heat lost or gained by a body = ms θ
where, m = Mass of the body.
s = Specific heat. θ = Rise or fall in the temperature of body.
1.1 HEA
a
Heat
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Heat
Heat Capacity The heat capacity of the body is the quantity of heat required by the body to raise its temperature by 1ºC. It is also known as thermal capacity. Heat capacity = ms (mass × specific heat)
Water Equivalent It is the quantity of water whose thermal capacity is same as the heat capacity of the body. It is denoted by W.
rinciple of Calorimetry
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W = ms = Heat capacity of the body.
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m1s1 (θ2 – θ) = m2s2 (θ – θ1)
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If two bodies at different temperatures are made to come in contact with each other in the calorimeter. As a result, heat is exchanged between the objects as well as with the calorimeter. The principle of calorimetry states that the total heat given by the hot bodies equals the total heat received by the cold bodies. Suppose, two bodies are in the contact. The temperature of one body is θ1 and the other body temperature is θ2 (θ2 > θ1). Let the temperature of equilibrium is θ. Let also the mass of bodies is m1 and m2 respectively and specific heat s1 and s2. Then, from principle of calorimetry, Heat lost by hot body = Heat gained by cold body
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Latent heat During this process of melting or vaporization, the temperature remains constant. The amount of heat needed to melt a solid of mass m
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Latent Heat of Fusion
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Q = mL where, L = Latent heat of fusion or vaporization.
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It is the quantity of heat required to change the unit mass of a solid substance to the liquid state at its melting point. For ice, latent heat of fusion is 80 cal/gram.
Latent Heat of Vaporization It is the quantity of heat required to change the unit mass of a liquid to gaseous state at the boiling point of the liquid. For water, latent heat of vaporization is 540 cal/gm. Example The temperature of 100 gm of water is to be raised from 24ºC to 90ºC by adding steam to it. Calculate the mass of steam required for this purpose. )
) Solution
Let the mass required of stream is m gm Heat lost by steam = m × 540 + ms (100 – 90) = (m × 540 + m × 10) calorie
1.2
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Heat
1.3
Specific heat of water is 1 calorie/gm/ºC.
Heat lost = 550 m calorie Heat gained by water = 100 × 1 × (90 – 24) calories = 6600 calories By the principle of calorimetry Heat lost = heat gained 550 m = 6600 m=
∴
6600 = 12 gm. 550
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Example Three liquids A, B and C are given. The temperature of A, B and C are 60ºC, 55ºC and 50ºC respectively. When 4g of A and 3g of C are mixed the temperature is 55ºC, while when 2g of A and 3g of B are mixed, the temperature is 57ºC. What would be the temperature of the mixture when equal masses of B and C are mixed? )
m
) Solution Let the specific heat capacity of the liquid is sA, sB and sC respectively.
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Thus, when the liquids A and C are mixed, Heat lost by A = 4 × sA (60 – 55) = 4 × sA × 5
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sA 3 = sC 4
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4 × sA × 5 = 3 × sC × 5
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∴
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Heat gained by C = 3 × sC (55 – 50) = 3 × sC × 5
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When, the liquids A and B are mixed. Heat lost by A = 2 × sA (60 – 57) = 2 × sA × 3
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sA = sB
...(ii)
or
2 × sA × 3 = 3 × sB × 2
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∴
Heat gained by B = 3 × sB (57 – 55) = 3 × sB × 2
...(iii)
When the liquids B and C are mixed Heat lost by B = m × sB (55 – θ) Heat gained by C = m × sC (θ – 50) ∴ sB (55 – θ) = sC × (θ – 50) From equations (i), (ii) and (iii), we get
4 sA (θ − 50) 3 165 – 3θ = 4θ – 200
sA × (55 − θ) =
or
365 = 7θ
∴
θ=
or
365 = 52.14°C 7
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Heat
Example When a block of metal of specific heat 0.1 cal/gm/ºC and weighing 110 gm is heated to 100ºC and then quickly transferred to a calorimeter containing 200 gm of a liquid at 10ºC, the resulting temperature is 18ºC. On repeating the experiement with 400 gm of same liquid in the same calorimeter at some initial temperature, the resulting temperature is 14.5ºC. Find (i) Specific heat of liquid. (ii) The water equivalent at calorimeter. )
) Solution
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Let the specific heat capacity of liquid is s and water equivalent of calorimeter is W. Then, the heat lost by block in first case 110 × 0.1 × (100 – 18) = (110 × 0.1 × 82) calorie Heat gained by (liquid + calorimeter) = 200 × s × (18 – 10) + W (18 – 10) = 200 × s × 8 + 8 W From the principle of calorimetry Heat lost = Heat gained or 110 × 0.1 × 82 = 200 × s × 8 + 8 W ...(i) From second case 110 × 0.1 × 85.5 = 400 × s × 4.5 + 4.5 W ...(ii) Solving equations (i) and (ii), we get s = 0.48 cal/g/ºC and W = 16.6 gm.
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Example A lead ball at 30ºC is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. Calculate the latent heat of fusion of lead. Specific heat capacity of lead = 126 J/kg.ºC and melting point of lead = 330ºC. Assume that any mechanical energy lost is using heat for of ball. Use, g = 10 m/s2. )
) Solution
Initial gravitational potential energy of the ball = mgh = m × 10 × 6.2 × 103 = m × (6.2 × 104 J/kg) All this energy is used to heat the ball as it reaches the ground with a small velocity. Energy required to take the ball from 30ºC to 330ºC is = m × (126 J/kg – ºC) × 300ºC = m × 37800 J/kg And energy required to melt the ball at 330ºC = mL where, L = latent heat of fusion of lead. Thus, m × (6.2 × 104) = m × 37800 + mL or L = 2.4 × 104 J/kg.
1.4
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Heat
1.5
Example An aluminium container of mass 100 g contains 200 g of ice at – 20ºC. Heat is added to the system at a rate of 100 cal/s. What is the temperature of the system after 4 minutes? Draw a rough sketch showing the variation in the temperature of the system as a function of time. Specific heat capacity of ice = 0.5 cal/gm-ºC. Specific heat capacity of aluminium = 0.2 cal/gm-ºC. Specific heat capacity of water = 1 cal/gmºC and latent heat of fusion of ice = 80 cal/gm.
) Solution
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Total heat supplied to the system in 4 minutes Q = 100 cal/s × 240 s = 2.4 × 104 calorie. Heat required to take the system from – 20ºC to 0ºC. = (100 g) × (0.2 cal/gm–ºC) × (20ºC) + (200 g) × (0.5 cal/gm–ºC) × (20ºC)
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Thus, 2.4 × 104 cal = 2400 cal. + 16000 cal. + (220 cal/ºC) θ 5600 or θ= = 25.5°C. 220 Variation in the temperature as a function of time is sketched in the figure below.
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= 400 cal + 2000 cal = 2400 cal. 2400 Time taken in this process = = 24sec. 100 Heat required to melt the ice at 0ºC = 200 × 80 = 16000 cal. 16000 Time taken in this process = sec. = 160 sec. 100 If the final temperature is θ, the heat required to take the system to the final temperature = (100 g) (0.2 cal/gm-ºC) θ + (200 g) (1 cal/gm-ºC) θ
±
7LPH
±
Example 300 gm of ice at a temperature of – 10ºC is dropped into a calorimeter containing 200 gm of water at 30ºC. What will be the temperature and its contents, after thermal equilibrium is established? The specific heat of ice is 0.48 cal gm–1 (ºC)–1 and the latent heat of fusion of ice is 80 cal gm–1. The heat capacity of the calorimeter is negligible. https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
) Solution
Final temperature of the mixture can be greater than 0ºC or less than 0ºC. Let us analyse each of these cases:
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(i) Final temperature greater than 0ºC: All the ice in the calorimeter absorbs heat and rises to 0ºC. Ice at 0ºC absorbs latent heat and becomes water at 0ºC. Now, the temperature of this water is raised to θºC. Heat required for the ice at –10º to become water at θºC = (300 × 0.8 × 10) + (300 × 80) + 300 × 1 × θ = (26880 + 300 θ) calorie Heat lost by water at 30ºC = 200 (30 – θ) = (6000 – 200 θ) calorie Since, the heat required is greater than the heat lost by water for any temperature between 0 and 30ºC, the system cannot attain a temperature greater than 0ºC. (ii) Final temperature less than 0ºC: For the final temperature to be less than 0ºC, the entire water in the calorimeter should condense and the temperature becomes – θºC. Hence, to be released by water attain the temperature – θºC. = (200 × 1 × 30) + (200 × 80) + (200 × 0.4 × θ) = (22000 + 96 θ) calorie Heat absorbed by ice = 300 × 0.4 × (10 – θ) = (1440 – 144 θ) calorie Since, the heat which can be absorbed by ice is less than the heat to be released by water for all temperatures between 0 and – 10ºC, the system cannot attain a temperature less than 0ºC. (iii) Thus, the final temperature of the mixture can be neither greater than 0ºC nor less than 0ºC. Therefore, the calorimeter will contain both ice and water the final temperature of the mixture will be 0ºC.
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Example A solid material is supplied with heat at a constant rate. The temperature of the material is changing with the heat input as shown in the graph. Study the graph carefully and answer the following questions.
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Heat
(i) (ii) (iii) (iv)
1.7
What do the horizontal regions AB and CD represent? If CD = 2 AB, what do you infer? What does the slope DE represent? The slope of OA > slope of BC. What does this indicate?
)
) Solution
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(i) AB represents change of state from solid to liquid (at constant temperature). CD represents the change of state from liquid to gas (at constant temperature). (ii) For the given mass, heat required to change liquid state into gaseous state is double that required to change the solid state into the liquid state. Hence, the latent heat of vaporization of the material is double the latent heat of fusion. dT (for gaseous state) (iii) Slope DE = dθ dθ is the heat required to change the temperature of the given mass Now dT through 1ºC, assuming the gas is enclosed at constant volume or at constant pressure. 1 = Reciprocal of the heat capacity under the given ∴ Slope = heat capacity
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AL EX A SIO P
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Solid have length, breadth and thickness. When heated, in case of a thin rod of larger length, linear expansion is more important. In case of a thin plate having length and breadth with negligible thickness, area expansion (superficial expansion) is important. When there is a change in length breadth and thickness of a solid, when heated, volume expansion is considered. Accordingly, a solid can have linear expansion, superficial expansion, and volume (cubical) expansion. Linear expansion (expansion in length of a solid) Let l1 and l2 be the length of rod at temperature θ1ºC and θ2 ºC respectively. Then
l2 = l1 [1 + α (θ2 – θ1)] = l1 (1 + αθ)
where, α = Coefficient of linear expansion θ = (θ2 – θ1) If l0 is the length of the rod at 0ºC, then the length of the rod at θºC is
lt = l0 (1 + αθ)
or
1.2 THE
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assumption. (iv) Slope OA > slope BC Heat capacity for the solid state < Heat capacity for the liquid state. But since the mass is the same (heat capacity = ms), specific heat of the liquid state is more than that of the solid state.
α=
lt − l0 l0 θ
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Heat
Coefficient of linear expansion is defined as the change in length per unit length per unit degree change in temperature. Superficial expansion (expansion in surface area) If A1 is the area of solid at θ1 ºC and A2 is the area at θ2 ºC, then A2 = A1 (1 + βθ) where, θ = θ2 – θ1 β = coefficient of superficial (areal) expansion. Volume expansion (expansion in volume) If V1 is the volume of solid at θ1ºC and V2 is the volume at θ2ºC, then
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V2 = V1 (1 + γθ) where, γ = Coefficient of cubical (volume expansion)
d1 (1 + γ θ)
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d2 =
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note: For isotropic solids: β = 2α, γ = 3α, [6α = 3β = 2γ] As temperature increases, density of solid decreases. If d1 is the density at θ1ºC, d2 is the density at θ2ºC, then
s:
Heating a metallic Scale
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A metallic scale (linear) expands in length when heated. As a result all the marking are displaced from their usual (correct) positions. A reading of 1 unit on a heated scale is equivalent to an actual length of 1 (1 + αθ), where, α is a coefficient of linear expansion of material of scale, and θ is rise in the temperature of the scale.
ht
1.8
If the reading is x, actual length = x (1 + αθ) and actual length = reading (1 + αθ) https://t.me/universitarios_infohttps://www.jamarana.com
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1.9
ifference in Length of Two ods R
D
Heat
Consider two rods 1 and 2 of lengths l1 and l2. Let, they be heated through a temperature θ. If l1′ and l2′ are their expanded lengths, then and
l2′ = l2 (1 + α2 θ) l1′ = l1 (1 + α1 θ)
where α2 = Coefficient of linear expansion of rod 2, and α1 = Coefficient of linear expansion of rod 1. Since, the difference of lengths of two rods is constant, therefore l1 α1 = l2 α2
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P
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Time eriod of endulum
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and
l2′ – l1′ = l2 – l1
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Time period (T) of a simple pendulum of length l is given by
rs
l0 g
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If there is rise in temperature by θ, length of the pendulum increases and, hence time period increases. As the result, the clock supporting such a simple pendulum slows down. If l0 be the length of pendulum and corresponding time period be T0, then
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If the pendulum heated by θ (rise in temperature), the new time period Tt will be given by Tt = 2π
where lt = Expanded length
lt g
⇒
Tt 1 = 1 + αθ = 1 + αθ (as α is very small) T0 2
⇒
⇒
Tt − T0 1 = αθ T0 2
⇒
⇒
Tt l l (1 + αθ) = t = 0 T0 l0 l0
⇒
∆T 1 = αθ T0 2
∆T = fractional increases in the time period = time lost in 1 second. T0 1 Time lost in one day = αθ × 86400 s. 2
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Heat
Example A steel tube of coefficient of linear expansion 18 × 10–6/ºC contains mercury, whose coefficient of absolute volume expansion is 180 × 10–6/ºC. The volume of mercury contained in the tube is 10–5m3 at 0ºC and it is desired that the length of the mercury column should remain constant at all normal temperature. This is achieved by fixing a rod of silica, whose thermal expansion is negligible at the bottom of the tube, to be immersed fully in mercury. Calculate the volume of the silica rod. )
) Solution At 0ºC:
m
At θº C:
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Let, V0 = Volume of mercury. V = Volume of silica rod. l0 = Length of mercury column. A0 = Length of mercury column. A0 = Area of cross-section of the steel tube. A0l0 = V0 + V
...(i)
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V0 ( γ Hg − 2.α) 2α
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180 × 10−16 − 2 × 18 × 10−6 = 10−5 2 × 18 × 10−16 –5 3 = 4 × 10 m
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∴
(V0 + V) (1 + 2 · α · θ) = (V0 + V) + (V0 · γHg · θ)
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...(ii)
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A0 (1 + 2 · α θ) l0 = Vθ + V = V0 (1 + γHg · θ) + V From equation (i), A0 · l0 = V0 + V or
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Aθ = A0 (1 + 2 · α θ) Since, l0 should remain constant, therefore Aθ (l0) = Vθ + V
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Example A glass bottle of volume 50 cm3 at 0ºC, is filled with paraffin at 15ºC. What is the mass of paraffin? Given d0 = 0.82 g cm–3 and γparaffin = 9 × 10–4/ºC. Coefficient of linear expansion of glass = 9 × 10–6/ºC.
) Solution )
1.10
V0 = Volume of glass bottle at 0ºC = 50 cm3 V15 = Volume of the bottle at 15ºC = 50 (1 + γg θ)
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Heat
1.11
= 50 (1 + 3 × 9 × 10–6 × 15) = 50 × 1.000405 cm3 d0 = Density of paraffin at 0ºC = 0.82 g cm–3 d15 = Density of paraffin at 15ºC d0 1 + γ ρθ
=
0.82 1 + 9 × 10−4 × 15
=
0.82 1.0135
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=
ar a
= 40.47 g
rio s_ in fo
∴ Mass of the paraffin filling the bottle at 15ºC = V15 d15
iv
The length of iron rod at 100ºC is l1 = (50) [1 + (12 × 10–6) (100 – 20)] = 50.048 cm The length of the aluminium rod at 100ºC is l2 = (100) [1 + (24 × 10–6) (100 – 20)] = 100.192 cm The length of the composite system at 100ºC is = (50.048 + 100.192) cm = 150.24 cm Since length of the composite system at 20ºC is 150 cm, hence, the average coefficient of linear expansion of the composite rod is,
un
s: //
t.m e/
ht tp s:
ht tp
//t
ht
)
.m
) Solution
er si ta rio s
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
Example A iron rod of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurement refer to 20ºC. Find the length of the composite system at 100ºC and its average coefficient of linear expansion. The coefficient of linear expansion of iron and aluminium are 12 × 10–6/ºC and 24 × 10–6/ºC respectively.
α=
0.24 150 × (100 − 20)
= 20 × 10–6/ºC.
Example A glass vessel of volume 100 cm3 is filled with mercury and is heated from 25ºC to 75ºC. What volume of mercury will over flow? Coefficient of linear expansion of glass = 1.8 × 10–6/ºC and the coefficient of volume expansion of mercury is 1.8 × 10–4/ºC. https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
)
) Solution
m
rio s_ in fo
ar a
na .c om
Volume of mercury at 25ºC is V0 = 100 cm3 Coefficient of volume expansion of mercury γL = 1.8 × 10–4/ºC Coefficient of volume expansion of glass γS = 3 × 1.8 × 10–6/ºC = 5.4 × 10–6/ºC Thus, the volume of mercury at 75ºC is VLθ = V0 (1 + γL θ) and the volume of vessel at 75ºC is VSθ = V0 (1 + γS θ) Volume of mercury overflow = VLθ – VSθ = V0 (γL – γS) θ = 100 cm3 (1.8 × 10–4 – 5.4 × 10–6)/ºC × 50ºC = 0.87 cm3
un
//t
iv
er si ta rio s
un iv
.m
tp
e/
s:
//w
w
er s
w
ita
.ja
N
ote: γa = (γL – γS) acts as the effective coefficient of expansion of the liquid with respect to the solid. The expansion of mercury as seen from the glass can be written as Vθ – V0 = V0 γa θ or Vθ = V0 (1 + γa θ) The constant γa is called the apparent coefficient of expansion of the liquid with respect to the solid.
ht
ht tp
s: //
t.m e/
ht tp s:
Example Find the difference in levels of mercury in two identical vertical communicating open glass tubes with horizontal base, if the left hand tube is maintained at 0ºC and the right hand tube is heated to 100ºC. The height of mercury column in the left hand tube is 90 cm. g = 1.82 × 10–4/ºC. Neglect the expansion of glass. )
) Solution
Given, h0 = 90 cm to find (h100 – h0). Let d0 and d100 be the corresponding densities. Therefore P0 = Pressure due to the left hand column = h0d0g P100 = Pressure due to the right hand column = h100d100g Since, the column are balanced, therefore P0 = P100 or h0d0g = h100d100g
or
d0 h = 100 d100 h0
But
1.12
d0 = 1 + γθ d100
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Heat
1.13
h100 = 1 + γθ h0
∴
h100 – h0 = h0 γθ = 90 × 1.82 × 10–4 × 100 = 1.638 cm
∴
Example A pendulum clock loses 12 sec in a day if the temperature is 40ºC and goes fast by 4 sec a day if the temperature is 20ºC. Find the temperature at which the clock will show correct time and the coefficient of linear expansion of the metal of the pendulum shaft.
) Solution
rio s_ in fo
er si ta rio s
un iv
e/
32 86400 × 20
iv
α=
.m
∴
32 = 86400 α (40 – 20)
un
//t
ht
tp
s:
//w
w
er s
w
ita
.ja
m
ar a
na .c om
Let θ be the temperature at which the clock is correct. 1 Time lost per day = α (rise in temperature) × 86400 2 1 ⇒ 12 = α (40 – θ) × 86400 2 1 Time gained by day = α (drop in temperature) × 86400 2 1 ⇒ 4 = α (θ – 20) × 86400 2 Adding equations (i) and (ii), we get
ht tp
s: //
t.m e/
ht tp s:
= 1.85 × 10–5/ºC
G
5
I
Dividing equations (i) by equation (ii), we get 12 (θ – 20) = 4 (40 – θ) or 12θ – 240 = 160 – 4θ or 16θ = 400 ∴ θ = 25ºC Hence, clock shows correct time at 25ºC. https://t.me/universitarios_infohttps://www.jamarana.com
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...(i)
...(ii)
Heat
Example A copper and a tungsten plate having a thickness δ = 2 mm each are riveted together so that at 0ºC they form a flat bimetallic plate. Find the average radius of the curvature of this plate at θ = 200ºC. The coefficient of linear expansion for copper and tungsten are αcu = 1.7 × 10–5/k and αt = 0.4 × 10–5/k, respectively. )
) Solution
fo
…(i)
rio
m
δ φ R + = l0 (1 + α c θ) 2
s_ in
ar an a
.c om
The average length of copper plate at a temperature θ = 200ºC is lc = l0 (1 + αc θ) where, l0 is the length of copper plate at 0ºC. Length of the tungsten plate is lt = l0 (1 + αtθ) From figure we have, lc = φ (R + δ/2) and lt = φ (R – δ/2) Consequently
rio s
ita
rs ve
.m
e/ un i
s: //
δ [Neglecting (αc + αt) in numerator.] (α c − α t )θ
R=
t.m
2 + (α c + α t )θ R = δ [2(α c − αt )θ]
ht tp
ht
tp
s:
or
or
si ta
iv er
un
e/
d 2 = (1 + α c θ) δ (1 + α t θ) R− 2 R
//t
tp s:
//w
w
w .ja
δ φ R − = l0 (1 + α t θ) …(ii) 2 We shall assume that the edges of plates are not displaced during deformation and that an increase in the plate thickness due to heating can be neglected. From equations (i) and (ii), we have
ht
Substituting the values in above relation, we get R = 0.769 m.
1.14
Example A mercury thermometer is to be made with glass tubing of internal bore 0.5 mm diameter and the distance between the fixed points is to be 20 cm. Estimate the internal volume of the bulb and stem below the lower fixed point. Coefficient of cubical expansion of mercury is 18 × 10–5/ ºC and coefficient of linear expansion of glass is 9 × 10–6/ºC.
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Heat
1.15
)
) Solution
Let, V0 = Volume of mercury upto the lower fixed mark (0ºC). When heated to 100ºC, the increase in volume is given as Expansion = πr2 · L 2
=
39.3 × 10−3 V0 × 100
=
39.3 × 10−5 V0
rio s_ in fo
Vθ − V0 V0 θ
ar a
er s
w
.m
iv un
t.m e/
V0 = 2.57 cm3.
s: //
ht tp s:
R
N
39.3 V0
//t
15.3 =
1.3 HEAT T A SFE R
39.3 × 10−5 + 2.7 × 10−5 (when there is no over flow.) V0
e/
tp or
ht
or
un iv
//w
w
18 × 10−5 =
s:
or
Heat can be transferred from one place to another by three different methods, namely, conduction, convection and radiation.
ht tp
ita
.ja
m
Coefficient of cubical expansion of glass γg = 3α = 3 × 9 × 10–6 = 27 × 10–6/ºC ∴ γ = γa + γg
er si ta rio s
γa =
na .c om
∴
0.5 = π × 20 20 = 39.3 × 10–3cm3 Coefficient of apparent expansion
Conduction It is the process by which the heat is transferred in solids. In conduction, molecules vibrate about a fixed location and transfer the heat by collision. Convection It is the process by which heat is transferred in fluids (liquid and gases). In convection, transfer of heat takes place by transporting of matter (in form of motion of particles). When a liquid in container is heated, the molecules at the lower layers are heated up as their densities decreases. As a result the molecules rise up and heavier ones come down, hence a continuous movement of molecules takes place giving rise to convection currents. In this manner the whole of liquid gets heated.
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1.16
Heat
radiation It is a process in which the heat is transferred in the form of electromagnetic waves without the aid of any material medium. Heat is radiated from a body due to its own temperature (greater than) is called as radiant energy.
1.4 THErmAL COnduCTIVITy
ar a er s
w w
;
un iv
//w s:
$
7
ita
.ja
m
$
er si ta rio s
7
rio s_ in fo
na .c om
The ability of a material to conduct heat is measured by thermal conductivity. Consider a slab of uniform cross-section A and length x. Let one face of the slab be maintained at temperature T1 and other at T2. Also suppose the remaining surface is covered with a non-conducting material so that no heat is transferred through the sides. After sufficient time, steady state is reached and the temperature at any point remains unchanged as time passes. In such a case, the amount of heat crossing per unit time through any cross-section of the slab is equal.
e/
.m
ht
tp
If ∆Q amount of heat crosses through any cross-section in time ∆t, then
∆Q is ∆t
…(i)
s: //
A(T1 − T2 ) ∆Q =k ∆t x
ht tp
Thus,
t.m e/
ht tp s:
un
//t
iv
called the heat current. It is found that in steady-state the heat current is proportional to the area of crosssection A, proportional to the temperature difference (T1 – T2) between the ends and inversely proportional to the length x.
where, k is a constant for the material of the slab and is called the thermal conductivity of the material. If area is non-uniform ∆Q dT = − kA …(ii) ∆t dx dT The quantity is called the temperature gradient. The minus sign indicates that dx dT is negative along the direction of heat flow. dx The unit of thermal conductivity can be easily worked using equations (i) and (ii). The SI unit is J/s-m-K or W/m-K.
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Heat
1.17
Heat Conduction Through a Composite Slab (rod) (A) Two rods connected in series. (B) Three rods connected in series. (A) Consider a composite rod made up of two rods of lengths d1 and d2 and each of cross-section A, joined end-to end (as shown). Let, k1 and k2 be the coefficients of thermal conductivities of two rods and T1 and T2 be the temperature of two ends of the composite rod. 7
N
N
G
ita
G
un iv
//w
Let us assume T as the temperature of the junction of two rods and T1 > T2. Heat flows from left to the right. In steady-state, heat flow per second is ∆Q same through each rod. It is given by H = ∆t k For Ist rod: H = 1 A(T1 − T) …(i) d1
un
k2 A (T − T2 ) d2
…(ii)
s: //
For IInd rod: H =
t.m e/
ht tp s:
//t
iv
.m
e/
s: tp ht
$
er si ta rio s
w
er s
w
.ja
m
ar a
$
7
rio s_ in fo
na .c om
7
ht tp
Eliminating T from equations (i) and (ii), we get H=
A (T1 − T2 ) d1 d 2 + k1 k2
Equating the equations (i) and (ii), we get k1 k A (T1 − T1 ) = 2 A (T − T2 ) d1 d2
⇒
k1 k T1 + 2 T2 d1 d2 T= k1 k2 + d1 d 2
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Heat
(B) Consider a composite slab made up of three rods of lengths d1, d2 and d3 joined end to end. Let area of cross-section of each rod be A. The left side of the composite rod is maintained at T1 and right side at T2 (T1 > T2). Let the coefficient of thermal conductivities of three rods be k1, k2 and k3 and junction temperature of rods 1 and 2 be TA and that of 2 and 3 be TB. 7
7$
+
7%
N
N
7
+
N
$
ar a
G
G
rio s_ in fo
na .c om
$
G
iv
un
//t
ht tp s:
k3 A (TB − T2 ) d3
t.m e/
.m
tp
For IIIrd rod: H =
er si ta rio s
ita
k2 A (TA − TB ) d2
e/
s:
For IInd rod: H =
k1 A (T1 − TA ) d1
er s
H=
//w
w
w
For Ist rod:
un iv
.ja
m
As T1 > T2, heat flows from left to the right side. At steady-state let H be the heat flowing per second.
ht
H d1 A k1
s: //
From equation (i) T1 − TA =
…(i) …(ii) …(iii) …(iv)
From equation (ii) TA − TB =
H d2 A k2
…(v)
From equation (iii) TB − T2 =
H d3 A k3
…(vi)
ht tp
1.18
Since, T1 and T2 is known, so TA and TB can be determined. To eliminate TA and TB, adding the equations (iv), (v) and (vi), we get T1 − T2 =
⇒
H=
H d1 d 2 d3 + + A k1 k2 k3 (T1 − T2 ) A d1 d 2 d3 + + k1 k2 k3
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Heat
Conduction of Heat Through a Cylindrical Shell U
U
rio s_ in fo er si ta rio s
er s
w
U
un iv
w +
iv
.m
+
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
+
U
e/
//w s: tp
+
ita
.ja
m
ar a
na .c om
/
Let, T1 = Temperature of the inner shell (T1 > T2). T2 = Temperature of the outer surface. L = Length of the cylinder. k = Thermal conductivity. H = −kA
∴ or
H
dT dT = −k (2π rL) dr dr
dr = −2π kL dT r
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1.19
Heat
integrating
r2
∫H r1
or
H In
T
2 dr = −2π kL ∫ dT r T1
r2 = −2π kL (T2 − T1 ) r1 = 2π kL (T1–T2)
or
H=
2π L (T1 − T2 ) r 1 log 2 k r1
un iv e/
iv
.m t.m e/
un
//t
2π L (T1 − T2 ) r r 1 1 log 2 + log 3 k1 r1 k2 r1
ht tp
s: //
ht tp s:
U
er si ta rio s
er s
w w //w s: tp
H=
U
ita
.ja
m
ar a
U
rio s_ in fo
na .c om
In case of two cylindrical layers:
ht
1.20
Conduction of Heat Through a Spherical Shell H=
4π (T1 − T2 ) 1 1 1 − k r1 r2
In case of multiple shells: H=
4π (T1 − T2 ) 11 1 1 1 1 − + − k1 r1 r2 k2 r2 r3
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Heat
1.21
U 7
U 7
na .c om
1.5 GrOWTH OF ICE On THE SurFACE OF A LAKE
rio s_ in fo
er si ta rio s
un iv
e/
∆Q kA (θ1 − θ2 ) = ∆t d
iv
ht
We know,
.m
tp
s:
//w
w
er s
w
ita
.ja
m
ar a
A layer of ice of thickness x is formed on the surface of a lake. The water below the ice layer is at a temperature of 0ºC. The air above the ice layer is at temperature of – θºC (below 0ºC). Heat flows from water to air through ice. As a result of this heat loss, the layer increase in thickness from below. Let dx be the increase in thickness in time dt. Heat loss from water to air = mass × latent heat = (A dx) ρ L where, ρ = Density of ice. L = Latent heat of ice. A = Area of cross-section of ice slab.
t.m e/
ht tp s:
un
//t
dx kAθ − ⇒ Heat flow per second = ΑρL = x dt
s: //
Rate of growth of thickness =
dx k θ = dt ρLx
⇒
2 2
2 1
ht tp
If the thickness changes from x1 to x2 in t seconds, then
x2
kθ
t
∫ x dx = ρL ∫ dt
x1
0
x −x kθ t = 2 ρL
DLU
Tq& LFHOD\HU
:DWHU&
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Heat
Example Water is boiled in a rectangular steel tank, the bottom of which is 5 mm thick. If the water level falls steady at the rate of 1 cm in 5 minutes, what is the temperature of the lower surface of the tank. Thermal conductivity of steel = 0.12 CGS units. Latent heat of steam = 540 cal g–1. )
) Solution
Let area of the bottom plate of the tank be A cm2 and the temperature of the lower surface be θºC. Since, the upper surface is in contact with the boiling water, its temperature may be assumed to be 100ºC. θ − 100 °C cm −1. 0.5 The heat passing by conduction through the plate is sufficient to vapourise A cm3 of water in 5 minutes or A gram in 5 minutes. This requires 540 A calories in 5 minutes. The rate of flow of heat through the plate is
ar an
in
a.
fo
co
m
Temperature gradient of steel plate =
m
rio
s_
θ 540 A = cal s −1 t 5 × 60
si
ta
.ja
Substituting in the conductivity equation
rio
s
er iv
s:
un
//t
iv
er
si
ta
un
e/
θ = 107.5ºC
⇒
540 A 0.12 × A(θ − 100) = 5 × 60 0.5
.m
tp
s:
we have
//w w
w
θ kA (θ1 − θ2 ) = t d
ht
ht
tp
s:
//t
ht
.m
tp
e/
Example A closed cubical box is made of perfectly insulating material and the only way for heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross-sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface of one plug is kept at a temperature of 100ºC while the outer surface of the other plug is maintained at a temperature of 4ºC. the thermal conductivity of the material of the plug is 2.0 W/m–ºC A source of energy generated 13 W is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is same at all points on the inner surface.
) Solution )
1.22
The situation is shown in the figure. Let the temperature inside the box be θ. The rate at which heat enters the box through the left plug is ∆Q1 kA (θ1 − θ) = ∆t x Rate of heat generation in the box = 13 W.
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Heat
1.23
The rate at which heat flows out of the box through the right plug is ∆Q 2 kA (θ − θ2 ) = ∆t x In the steady-state
T &
6
T &
na .c om
∆Q1 ∆Q 2 + 13 = ∆t ∆t
or
2 kA kA θ= (θ1 + θ2 ) + 13W x x
or
θ=
m
un
//t
iv
.m
= 52º + 216.67ºC ≅ 269ºC
er si ta rio s
er s
un iv
100 + 4 13 × 0.08 + 2 2 × 2.0 × 12 × 10−4
e/
=
ita
.ja
w
w //w s: tp ht
θ1 + θ2 (13 W) x + 2 2 kA
rio s_ in fo
kA (θ1 − θ) kA + 13 W = (θ − θ2 ) x x
ar a
or
ht tp
s: //
t.m e/
ht tp s:
Example Two parallel plates A and B are joined together to form a compound plate. The thickness of the plates are 4.0 cm and 2.5 cm respectively and the area of crosssection is 100 cm2 for each plate. A thermal conductivities are kA = 200 W/mºC for the plate A and kB = 400 W/m-ºC for the plate B. The outer surface of the plate A is maintained at 100ºC and the outer surface of the plate B is maintained at 0ºC. Find
&
$
%
&
(i) The rate of heat flow through any cross-section. (ii) The temperature at the interface. (iii) The equivalent thermal conductivity of the compound plate.
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Heat
)
) Solution
(i) Let the temperature of the interface be θ. The area of cross-section of each plate is A = 100 cm2 = 0.01 m2. The thickness are xA = 0.04 m and xB = 0.025 m. 1 xA The thermal resistance of the plate A is R1 = kA A 1 xB and that of the plate B is R 2 = kB A
The equivalent thermal resistance is
m
fo si ta er iv
un
t.m e/
s: tp
ht
3810 W =
s
rio
iv
un
.m e/
∆Q A (θ − θ2 ) = xB ∆t kB
//t
(0.01m 2 )(θ − 0°C) (0.025m) (400 W/m − °C)
s: //
in
si ta
er
w //w
s:
(0.01 m 2 ) (100°C) (0.04) (0.025m) + (200 W/m − °C) (400 W/m − °C)
=3810 W.
tp
s_
an
.ja
m
ar
w
=
rio
co
∆Q θ1 − θ2 A (θ1 − θ2 ) = = xA xB ∆t R + kA kB
Thus
or
…(i)
a.
1 xA xB + A kA kB
=
R= R1 + R2
(ii) We have,
ht
tp
xA + xB xA xB + kA kB Substituting the value of kA, kB, xA and xB, we get k = 248 W/m-ºC.
or
ht
or θ = 24ºC (iii) If k is the equivalent thermal conductivity of the compound plate, its thermal resistance is 1 xA + xB R= A k Comparing with equation (i), we get xA + xB xA xB = + k kA kB
1.24
k=
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Heat
1.25
Example One end of a uniform brass rod 15 cm long and 20 cm2 cross-sectional and is kept at 100ºC. The other end is in perfect thermal contact with an iron rod of identical cross-section and length 8 cm. The free end of iron rod is kept in melting ice and when the composite rod is well lagged and steady-state has been reached it is found that 684 g of ice melt per hour. Calculate the thermal conductivity of iron, given that the thermal conductivity of brass is 0.25 cal s–1cm–1/ºC. L = 80 cal/gm.
) Solution
G
G
For the iron rod, ∴ or
er si ta rio s
un
t.m e/
0.25 × 20(100 − θ) 15
s: //
or
15.2 =
θ = 54.4ºC Q k2 A(θ − θ2 ) = t d2
ht tp
∴
ht tp s:
//t
iv
.m
e/
s:
un iv
//w
w
er s
w
Here, Q = Heat required to melt 684 g of ice = 684 × 80 cal t = 1 hour = 3600 sec. θ 684 × 80 ∴ = = 15.2 cal s −1 t 3600 Since, the rods are well lagged, there is no loss of the heat from the surface, thereθ fore is the same for both. t Q k1 A (θ1 − θ) For the brass rod, = t d1
tp ht
T &
ita
.ja
m
ar a
T &
rio s_ in fo
na .c om
%UDVVN ,URQN "
15.2 =
k2 × 20(54.4 − 0) 8
k2 = 0.112 cal s–1cm–1/ºC.
Example On a cold winter day, the atmospheric temperature is –θ (on celsius scale) which is below 0ºC. A cylindrical drum of height h made of a bad conductor is completely filled with water at 0ºC and is kept outside without any lid. Calculate the time taken for the whole mass of water to freeze. Thermal conductivity of ice is k and its latent heat of fusion is L. Neglect expansion of water on freezing. https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
) Solution Suppose, the ice starts forming at time t = 0 and a thickness is formed
at time t. The amount of heat flown from the water to the surrounding in the time interval t to t + dt is kAθ dt x The Mass of ice formed due to the loss of this amount of heat is ∆Q=
dm =
∆Q kAθ = dt L xL
The thickness dx of ice formed in time dt is
m
±T
ita
.ja
er s
w
un iv
w //w
K
.m
G[
t.m e/
ht tp s:
un
//t
&
[
iv
tp
e/
s:
rio s_ in fo
ρL x dx kθ
kθ dt ρx L
er si ta rio s
dt =
=
na .c om
dm Aρ
ar a
or
dx =
ht T
∫ dt = 0
or
T=
h
s: //
Thus, the time T taken for the whole mass of the water to freeze is given by ρL x dx k θ ∫0
ht tp
1.26
ρ Lh 2 . 2k θ
Example Figure shows a large tank of water at a constant temperature θ0 and a small vessel containing a mass m of water at an initial temperature θ1 (< θ0). A metal rod of length L, area of cross-section A and thermal conductivity k connects the two vessels. Find the time taken for the temperature of the water in the smaller vessel to become θ2 (θ1 < θ2 < θ0). Specific heat capacity of water is s and all other heat capacities are negligible.
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Heat
1.27
T /
T
) Solution
kA (θ0 − θ) dt L
rio s_ in fo
∆Q=
na .c om
Suppose, temperature of the water in the smaller vessel is θ at time t. In the next time interval dt, a heat ∆Q is transferred to it where …(i)
er si ta rio s
un iv
ht tp s:
0
θ
Lms 2 d θ kA θ∫1 θ0 − θ
iv
∫ dt =
un
T
or
Lms d θ kA θ0 − θ
e/
dt =
.m
ht
tp
s:
or
//t
//w
w
er s
w
ita
.ja
m
ar a
This heat increases the temperature of the water of mass m to θ + dθ, where ∆Q = ms dθ ...(ii) From equations (i) and (ii), we have kA (θ0 − θ) dt = ms d θ L
Lms θ0 − θ1 ln kA θ0 − θ2
s: //
T=
ht tp
Thus,
t.m e/
where, T is the time required for the temperature of the water to become θ2.
1.6 HEAT TrAnSFEr THrOuGH rAdIATIOn In radiation, heat is transferred from one body to other or to the surroundings even in the absence of any medium in the intervening space. Heat energy of the sun is transmitted to earth through radiation. Radiation is only a mode of transfer of energy by transverse electromagnetic waves. While studying heat radiation (Radiant energy), we are concerned with thermal radiations which form the infra-red region of electromagnetic waves. All bodies emit heat to the surrounding at all temperatures and at all times. When the temperature of a body remains constant, it emits as much heat to the surrounding as it gains from them. The body is then in a state of dynamic (thermal) equilibrium.
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Heat
Black Body A black body may be described as one that completely absorbs the radiations of all wave-lengths falling upon it. There is no substance which exactly behaves like a black body as described.
Hollow Enclosure
Emissive power
rio s_ in fo
na .c om
A hollow enclosure, kept in constant temperature behaves like a black body. If a small hole is made such a enclosure, will acts as perfectly black body towards the incident radiations. Any ray getting into the enclosure through the hole will be internally reflected not be able to escape. Any radiation coming out of a narrow hole made in a hollow enclosure is generally returned to as black body radiation or cavity radiation.
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
er si ta rio s
'Z
e/
s:
un iv
//w
¨$
er s
w
w
ita
.ja
m
ar a
It is the total amount of energy radiated by a body (at a particular temperature) per second per unit area of the surface.
ht
Consider a small area ∆A of the body emitting the thermal radiation. Consider a small angle ∆ω about the normal to the radiating surface. Energy radiated per second per unit area of the surface.
ht tp
1.28
E= where,
1 ∆Q A ∆t
A = Surface area ∆Q = Heat radiated per sec. from the surface. ∆t
Absorptive power Absorption power of a body is defined as the fraction of the incident radiation that is absorbed by the body. If we denote the absorption power by a, then a=
energy absorbed energy incident
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Heat
1.29
As all the radiation incident on a black body is absorbed, the absorptive power of a black body is unity. Note that the absorptive power is a dimensionless quantity but the emissive power is not.
Kirchhoff ’s Law The ratio of emissive power to absorptive power is the same for all bodies at a given temperature and is equal to the emissive power of a black body at that temperature. E (body) = E (black body) a (body)
Thus,
Stefan’s Law
er si ta rio s
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
rio s_ in fo
na .c om
The energy emitted per second per unit area of a black body is proportional to the fourth power of the absolute temperature. i.e., E = σ T4 where, σ = Stefan’s constant = 5.67 × 10–8 J/m2/s/K4 For any other body: E = ∈ σ T4 where, ∈ = Emissivity of body (∈ = 1 for a black body)
e/
.m
ht
tp
net Heat Loss From the Surface of the Body
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
The rate at which body radiates energy is determined by the temperature of the body, but the rate at which it absorbs energy by radiation depends on the temperature of surrounding. When a body is hotter than its surroundings, the rate of absorption is greater than the rate of emission and its temperature rises. At thermal equilibrium the two rates are equal. :DOOV 7
7
Hence, for a body at a temperature of T1, surrounded by walls at a temperature T2 (as in the figure), the net rate of loss (or gain) of energy per sec by radiation is Hnet = A ∈ σ T14 – A ∈ σ T24 = A ∈ σ (T14 – T24) https://t.me/universitarios_infohttps://www.jamarana.com
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1.30
Heat
1.7 nEWTOn’S LAW OF COOLInG dQ It states that the rate of heat loss from a hot body to the surroundings is given dt dQ by = kA (θ − θ0 ), if (θ – θ0) << absolute temperature of surroundings dt where,
A = Surface area. k = Constant. θ = Absolute temperature of the body. θ0 = Temperature of surroundings. θ – θ0 = Difference in temperature.
na .c om
dθ dQ = ms dt dt
s:
Wien’s Law
er si ta rio s
er s
un iv
//w
ita
.ja
dθ ∝ A(θ − θ0 ) dt
w
ms
w
⇒
rio s_ in fo
m
ar a
where, m = Mass of the body. s = Specific heat of the body. dθ = Rate of fall of temperature (rate of cooling). dt
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
The distribution of energy among the wavelengths of radiations emitted from a “Black Body” was studied by Lummer and Pringsheion. The energy contained over a wide range of wavelengths of radiations emitted by black body is not equally divided. It was found that the relative intensities of different wavelengths depend upon the absolute temperature of the radiator. The energy distributed (E) in the spectrum for the temperature range 893 K to 1893 K was found and plotted against the wavelengths (λ). The general shape of the curve is same for all temperature. From the graph it was analysed that: 7!7!7!7 7 (
7 7 7 OP
O
■ For a particular temperature, the graph is continuous on spreading from a minimum to a maximum at the other end and has a wavelength where the emission is maximum (λm). https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
1.31
■ As the temperature increases, the minimum cut-off wavelength shifts to a lower value and the wavelength where emission is maximum (λm) also shift to a lower value. Then deduced that, λmT = constant where, λm = Wavelength corresponding to maximum energy at a particular temperature. The value of constant is 2.9 × 10–3 mK = 0.29 cm K. This statement is known as “Wien’s Displacement Law”. If E (λm) is the energy associated with λm at a given temperature, then E (λm) ∝ T5
na .c om
ote: Wien’s Law is fairly true for shorter wavelengths.
rio s_ in fo
N
This is known as Wien’s Fifth Power Law or simply Wien’s Displacement Law (II)
ita
er s
w
)
) Solution
.ja
m
ar a
Example A copper ball cools from 62ºC to 50ºC in 10 minutes and to 42ºC in the next 10 minutes. Calculate its temperature at the end of next 10 minutes.
θ1 + θ2 62 + 50 = = 56°C 2 2
iv
.m
The rate of fall temperature
d θ 62 − 50 = = 1.2°C/min dt 10 dθ ∝ (θ − θ0 ) dt where, θ0 = temperature of surroundings ∴ 1.2 = kA (56 – θ0) ...(i) Since, the ball cools from θ2 = 50ºC to θ3 = 42ºC in time t2 = 10 minutes, therefore
⇒
θ′ =
t.m e/
ht tp
s: //
ht tp s:
−
un
//t
ht
tp
e/
s:
θ=
er si ta rio s
un iv
//w
w
Average temperature of the liquid in the first case
θ2 + θ3 = 46°C 2
d θ′ θ2 − θ3 50 − 42 = = = 0.8 °C/min dt t2 10
d θ′ = 0.8 = kA (46 − θ0 ) dt Dividing the equation (i) by equation (ii), we get
∴
1.2 56 − θ0 = 0.8 46 − θ0 θ0 = 26ºC
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…(ii)
Heat
Since, the ball cools from θ3 = 42º to θ4 in t3 = 10 minutes, therefore θ3 + θ4 42 + θ4 = 2 2
θ″ =
d θ″ θ3 − θ4 42 − θ4 = = dt t3 10 42 + θ4 42 + θ4 = kA − 26 2 2
…(iii)
Now,
⇒
6 θ4 = 220 θ4 = 36.7ºC
in s_
m
rio
ar
an
a.
or ∴
fo
co
m
42 − θ4 42 + θ4 − 26 Dividing the equation (iii) by equation (ii), we have 10 2 = 0.8 46 − 26 42 − θ4 42 + θ4 − 52 = or 8 2 × 20 or 5 (42 – θ4) = θ4 – 10
s
ta
e/
2
//t 2
s:
ht
.m
tp
A1 r1 = A 2 r2
∴
si
er
un
//t
iv
A1 = Surface area of the first sphere = 4πr12 A2 = Surface area of the second sphere = 4πr22
s:
)
) Solution
.m
rio
iv
un
e/
s:
(i) The initial rates of loss of heat. (ii) The initial rates of fall of temperature in the two cases.
tp
//w
w
er
w
si
ta
.ja
Example Two solid copper spheres of radii 6 cm and 12 cm are suspended in two evacuated chambers, each being at a temperature 15ºC above its surroundings. Compare:
ht
ht
tp
6 1 = = 4 12
4 Mass of the first sphere, M1 = volume × density = π r13 d 3
4 M 2 = π r23 d 3 3
(i)
3
M1 r1 6 1 = = = M 2 r2 12 8
∴
1.32
dQ1 = kA1 (θ − θ0 ) dt dQ 2 = kA 2 (θ − θ0 ) dt
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Heat
1.33
dQ1 dt = A1 = 1 . dQ 2 A 2 4 dt
dθ (ii) M1s 1 ∝ A1 (θ − θ0 ) dt
dθ M 2 s 2 ∝ A 2 (θ − θ0 ) dt d θ1 M1 dt A ∴ ⋅ = 1 d θ M2 A2 2 dt
o
co
nf
a.
s_ i
an
ta r
m
io
ar
tp
e/
s er si
ta
un
iv
w
s:
//w
w
or
er si
.ja
d θ1 dt = 2 . d θ2 1 dt
rio
d θ1 1 dt 1 or ⋅ = 8 d θ2 4 dt
m
iv
//t
.m
ht tp
e/
un
s: //
ht
t.m
Example A blackened metal sphere of thermal capacity 1000 cal/ºC and surface area 500 cm2 is kept inside an evacuated enclosure with blackened walls maintained at 27ºC. Find the rate of cooling of the sphere when its temperature is at 127ºC, σ = 1.36 × 10–12 cal. cm–2s–1K–4. )
s:
) Solution
ht
tp
Given: Thermal capacity = ms = 1000 cal/ºC A = 500 cm2 T = 273 + 127 = 400 K T0 = 27 + 273 = 300 K It is required to find d θ = Rate of cooling
dt d θ ∈σA (T4 – T 4) ...(where, ∈ = 1 for black body) ms = 0 dt
= 1.36 × 10–12 × 500 × 108 (256 – 81)
∴
dt
or 1000 d θ = 1 × 1.36 × 10–12 × 500 [(400)4 – (300)4] d θ 1.36 × 5 × 10–5 × 175 = dt
= 0.0119ºC/sec.
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Heat
Example Estimate the temperature of the earth, assuming it to be in radioactive equilibrium with the sun. The radius of the sun = 7 × 108 m, temperature of the solar surface = 6000 K, distance of earth from sun = 1.5 × 1011 m.
) Solution Assuming both the sun and the earth as black bodies.
where,
K=
π rE2 4π R 2
rio s_ in fo
Q2 = K × Q1
∴
na .c om
Energy radiated from the sun in unit time Q1 = σ AT4 where, A is the surface of the sun Q1 = σ × 4π rs2 × Ts4 At a distance R from the sun, where the earth is situated, the entire energy radiated by the sun is spread into an area of 4π R2. Out of this area, the earth blocks an effective area of πrE2 (cross-sectional area of the earth) and the radiation spread in this area is incident upon the earth. Let Q2 be the energy absorbed by the earth in unit time
ar a
The fraction of the total radiated energy absorbed by the earth, π rE2 × σ 4π rs2 × Ts4 4π R 2
6XQ¶VUD\V
t.m e/
ht tp s:
un
//t
iv
.m
e/
s: tp
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
Q2 =
m
∴
ht
s: //
(DUWK
Energy radiated by the earth = σ 4π rE2 × TE4 When, there is a radioactive equilibrium, then Energy radiated by the earth = Energy absorbed by the earth
ht tp
1.34
or or or
σ 4π rE 2 × TE 4 =
π rE2 σ 4π rS2 × TS4 4π R 2
r2 TE 4 = TS4 s 2 4R 1/2
r TE = TS s 2R
1/2
7 × 108 = 6000 × 11 2 × 1.5 × 10
∴
TE = 290 K.
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)
) Solution We have, dθ = − k (θ − θ0 ) dt where, θ0 is the temperature of the surrounding and θ is the temperature of the body at time t. Suppose, θ = θ1 at t = 0, then
ita
∴ 10800 + 2000 – 20 θ = 8000 + 500 + 110 θ
= 33ºC.
un
3. An earthen pitcher loses one gram of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water. Calculate the time required for the water in the pitcher cool to 28ºC from its original temperature of 30ºC. Neglect radiation effects. Average latent heat of vaporization of water in this range of temperature is 580 cal g–1.
iv
er si ta rio s
ht tp
s: //
∆Qms θ1 − θ0 = 2ms 2 If the body loses this heat in time t1, the temperature at t1 will be
θ1 − θ0 θ1 + θ0 = 2 2 Putting these values of time and temperature in equation (i), we get
θ1 −
130 θ = 4300 4300 θ= 130
t.m e/
ht tp s:
//t
.m
e/
or
or
un iv
//w
s:
tp
rio s_ in fo
ar a m
w
...(i)
The body continues to lose heat till its temperature becomes equal to the surrounding. The loss of heat in this entire period is ∆Qm = ms (θ1 – θ0) It is the maximum heat, that a body can lose. If the body loses half this heat, the decrease in its temperature will be
ht
xa
)
Heat lost by steam = mL + ms (100 – θ) where, θ is the equilibrium temperature ∴ Heat lost by steam = 20 × 540 + 20 × 1 (100 – θ) = 10800 + 2000 – 20 θ Heat gained by (ice + calorimeter) = 100 × 80 + 100 × 0.5 × 10 + 110 × θ Now Heat lost = Heat gained
er s
θ – θ0 = (θ1 – θ0) e–kt
w
or
ln
) Solution
.ja
θ − θ0 = −kt θ1 − θ0
or
dθ ∫θ θ − θ0 = −k ∫0 dt 1
S
2. 20 gm steam at 100ºC is let into a closed calorimeter of water equivalent 10 gm containing 100 gm ice at – 10ºC. Find the final temperature of the calorimeter and its contents. Latent heat of steam is 540 cal/gm, latent heat of fusion of ice = 80 cal/gm, specific heat of ice = 0.5 cal/ºC gm.
na .c om
t
θ
mp
1.35
U
D
1. A hot body placed in air is cooled down according to Newton’s law of cooling, the rate of decrease of temperature from the surrounding. Starting from t = 0, find the time in which the body will lose half the maximum heat it can lose.
or
or
θ1 + θ0 − θ0 = (θ1 − θ0 ) e − kt1 2 1 e − kt1 = 2 ln 2 t1 = . k
) Solution Heat lost by pitcher and water )
R
WO KE O T E
le
Heat
equivalent Q = (9.5 + 0.1) × 1 × 30.2 × 1000 = 20000 cal. Mass of the vaporise water, 20000 m= = 34.48 g 580 Time required = 34.48 min.
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1.36
4. At a temperature of 20ºC, the volume of a certain glass flask, up to a reference mark on the cylindrical stem of the flask is exactly 100 cm3. The flask is filled to this point with a liquid of cubical expansion 120 × 10–5/ºC, when both the flask and liquid is at 20ºC. The cross-section of stem is 1 mm2 and the coefficient of linear expansion of glass is 8 × 10–6/ºC. Find the rise or fall of the liquid level in the stem, when the temperature is raised to 40ºC.
Volume of iron tank after its heating V = V0 (1 + γl ∆T) ...(ii) where, γl = 3α1. The density of kerosene after the heating of the tank is ρ0 ρ= ...(iii) (1 + γ k ∆T)
Substituting expression equations (ii) and (iii) into equation (i), we get
∆m = ρ0 V0 −
Heat
ρ0 V0 (1 + 3α l ∆T) 1 + γ k ∆T
)
) Solution Given that volume = 100 cm
3
m
=
co
fo
a.
≈ ρ0 V0 (γk – 3αl) (T – T0) ∆m = 0.8 × 103 × 5 × 10–2 × (10 × 10–4 – 3 × 1.2 × 10–5) (293 – 273) kg = 0.7712 kg
s_
s
)
e/
un
The required pitch p can be found by dividing the cylinder height h by the rotational speed of drill n, need to drill the cylinder through h p= n In drilling these evolves the heat Q = ms ∆T = Ahρs ∆T πd 2 where, A= 4 Q h= Hence, Aρs ∆T
ht
iv
) Solution
.m
tp s
:// t
ht
rio
ta
er si
e/
tp
s:
//t
.m
ht tp
io
ta r
si
un
s:
)
as a result of heating by 90ºC will be larger than the increase the volume of tank, and a part of kerosene will flow out. We denote by m0 the mass of kerosene in the tank at 273 K and by m the mass of kerosene at room temperature.
U
6. In drilling in a copper cylinder of a bore of diameter 25 mm, the cylinder heats by 43 K. The input torque is 16.2 Nm. 70 per cent of the energy input converts into the internal energy of the cylinder. Compute the pitch of a drill.
er
iv
//w w
w .ja
m
ar
an
) Solution The increase in the volume of kerosene
OLWHUV
ρ0 V0 ( γ k − 3α L ) (T − T0 ) 1 + γ k (T − T0 )
in
5. An iron tank has a capacity of V = 50 liters of kerosene at T0 = 273 K. What amount of kerosene will flow out the full tank if it is brought in a room temperature at T = 293 K. [The density of kerosene ρ = 0.8 × 103, coefficient of volume expansion of kerosene γk = 10 × 10–4 and coefficient of linear expansion of iron αi = 1.2 × 10–5].
Coefficient of cubical expansion of = 235.2 cm.
where, ρ0 is the density at 273 K and ρ and V are the density of kerosene and the volume of tank at room temperature.
The quantity n is to be found from the relationship for the work W done in the drilling of the cylinder. ∴ W = τ 2πn W n= or 2π τ Substituting the values of h and n into the expression for P and considering
Then mass of kerosene flowing out of the tank is ∆m = m0 – m or ∆m = ρ0 V0 – ρV ...(i)
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1.37
Heat
or
60 – θB = 2 (θB – θC) + (θB – θD) 4 θB – 2 θC – θD = 60
For junction C: 2k (θB – θC) = 2k (θC – 10) + 2k (θC – θD)
nf o
rio
si ta
s
rio
ita
...(v)
rs un
4 θB – 8 θC = –40
θB – 2 θC = –10
e/
θC = 20ºC θB = 2 θC – 10 = 30ºC
∴
Thus, the junction temperature of B, C and D are 30º, 20º and 90º respectively. Alternate Method This problem can also be solved by applying Kirchhoff’s Law. Let kx and ky be the thermal conductivities of X and Y and kx and ky their resistance.
(
or
'
Let θB, θC and θD be the temperatures of the junction B, C and D at steady thermal conduction state.
Rx =
1 l kx A
And, R y =
1 l ky A
N
N
...(vi)
5 θC = 100
.m //t
N N
%
er
iv
e/ un
ht tp s:
)
$& N
4 θB – 3 θC = 60
Substracting equation (vi) from (v), we get
& N
θB – 3 θC – θD = θB + 2 θC – 4 θD ...(iv)
From equation (ii), we get or
<
...(iii)
Therefore, from equation (i), we get
(
'
) Solution
ar
m
s: tp
ht
<
θB + 2 θC – 4 θD = –10
From equations (ii) and (iii), we get
.m ;
%
//t
< $
k (θB – θD) + 2k (θC – θD) = k (θD – 10)
∴
.ja
w w
w
s: // tp ht
;
...(ii)
For junction D:
an a
&
θB – 3θC + θD = –10
∴
.c om
7. Three rods of materials X and three rods of material Y are connected as shown in the figure. All the rods are of identical length and cross-sectional area. If the end of A maintained at 60ºC and the junction E at 10ºC, calculate the temperature of the junction B, C, D. The thermal conductivity of X is 0.92 CGS units and that of Y is 0.46 CGS units. Assume all rods are well lagged.
s_ i
≅ .0010 m ≈ 1.0 mm.
;
...(i)
∴
8 × 16.2 × 0.70 380 × 625 × 10−6 × .8900 × 43
iv e
P=
2πτ K 8τ K = 2 2 ρd s ∆T πd ρ s ∆T 4
=
2π τ Q 2πτK ⋅ = ρAs ∆T W ρAs ∆T
P=
∴
If the thermal conductivity of rod Y is taken as k, then the rod X is 2 k. At steady conduction state, heat entering a junction must be equal to that leaving the junction. Therefore, for junction B: k (60 – θB) = 2k (θB – θC) + k (θB – θD)
Q = K, (given), we have W
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1.38
8. A rod CD of thermal resistance 5.0 k/W is joined at the middle of an identical rod AB as shown in the figure. The ends A, B and D are maintained at 100ºC, 0ºC and 25ºC respectively. Find the heat current in CD.
From the figure, no heat current is passing through C and D rod because loop is balanced. So, the figure is reduced to
Heat
&5[ $5\%
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(
% &
&
&
'5\
=
or
m
fo
in
s_
ta
si
s
rio
ita
rs
∆Q3 45°C − 25°C = ∆t 5.0 k/W
50 (k x + k y )k y k x + k y + 2k xy
=
50 × (0.92 + 0.46) 50 × 1.38 = = 30 2.3 3 × 0.46 + 0.92
60 – θB = 30 θB = 30ºC.
50 (k x + k y ) kx + 3 k y
ni
ve
e/ u
.m //t tp s:
100 − θ θ − 0°C θ − 25°C = + 2.5 2.5 5 225°C = 5θ θ = 45°C
iv un .m e/
Thus,
∆Q1 ∆Q 2 ∆Q3 = + ∆t ∆t ∆t
20 K = 4.0 W 5.0 k /W
9. As a result of a temperature rise of 32ºC, a bar with a crack at its centre buckles upward shown in the figure. It the fixed distance L0 = 3.77 m and the coefficient of linear expansion is 25 × 10–6/ºC. Find x, the distance to which the centre rises.
60 − θB =
1 2 + k y kx + k y
or
or ∴
or
=
50
ht
k y (60 − θB ) =
∆Q3 θ − 25°C = ∆t 5.0 k/W
We also have
ht
or
∆Q 2 θ − 0°C = ∆t 2.5 k/W
and
//t
60 − θB 50 = 1 1 1 2⋅ ⋅ ky kx k y 1 + 1 1 ky + kx k y
tp s:
or
∆Q1 100° − θ = ∆t 2.5 k/W
er
w
//w
ht tp
s:
50 2R x R y Ry + Rx + Ry
)
co a. …(ii)
From equations (i) and (ii), we have 60 − θB = Ry
The thermal resistance of AC is equal to that of CB and is equal to 2.5 k/W. Suppose, the temperature at C is θ. The heat current through AC, CB and CD are
rio
60 − 10 2R x R y Ry + Rx + Ry
'
) Solution
ar an
θA − θE = R equivalent
2(R x + R y )
m
i=
2R x 2R y
w .ja
R equivalent = R y +
∴
&
…(i)
θA − θB =i Ry
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Heat
l12 4
…(i)
= l22 −
1.39
3 /
O
) Solution )
$
Length increases when rises the temperature. Let us consider the half length. Then the increased length
When the temperature increased l′1 = l1 (1 + α1 θ) and l2′ = l2 (1 + α2 θ) According to the problem, length of the PQ is fixed, therefore
m
o
co
s_ i
s
rio
l1 α2 =2 . l2 α1
un i
ve r
si ta
l12 ⋅ 2α1θ 4
11. An aluminium plate is fixed in a horizontal position has a hole of diameter da = 2 cm. A steel sphere of diameter ds = 2.005 cm rests on this hole. The temperature of the entire system is slowly increased. At what temperature, the steel plate covers the hole? Given: αa = 23 × 10–6/ºC, and αs = 11 × 10–6/ºC
e/
)
) Solution
When the temperature is increased, diameter of the plate is increased and the diameter of steel ball is also increased. At that point where both the diameters are equal, the steel sphere covers the hole. Given: da = 2 cm, and ds = 2.005 Let at the temperature θ, both diameters are equal. ∴ da (1 + αa ∆θ) = ds (1 + αs∆θ) or da + da αa ∆θ = ds + ds αs ∆θ
PQ = AP 2 − AQ 2
From the figure
ht
)
) Solution
l22 ⋅ 2α 2 θ =
10. An isosceles triangle is formed with a thin rod of length l1 and coefficient of linear expansion α1 as the base and two thin rods each of length l2 and coefficient of the linear expansion α2 as the two sides. If the distance between the apex and the mid-point of the base remain unchanged as the temperature varied, show that l1 α2 =2 . l2 α2
l22 −
t.m s: // tp
= 7.5 cm
l2 l12 = l22 (1 + 2α 2 θ) − 1 (1 + 2α1 θ) 4 4 (because α1 and α2 is very small.)
rio
si ta er iv un
ht
= (188.6508) 2 − (188.5) 2
or
:// t
2
tp s
ht
L x = l ′2 − 0 2
or
.m e/
tp s:
∴
ar a .ja m
//w w
3.77 × 100 (1 + 25 × 10−6 × 32) 2 = 188.5 (1 + 800 × 10–6) = 188.6508
l′ =
l22 −
or
w
/
l12 l2 = [l2 (1 + α 2 θ]2 − 1 (1 + α1θ) 2 4 4
nf
na .
l′ =
[
%
4 O
L0 (1 + αθ) 2 where, α = coefficient of linear expansion
O
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1.40
Heat
mg mgl Young’s modulus = A = dl A dl l Y A dl or m = gl
(da αa – ds αs) ∆θ = ds – da
θ − 10 =
2.005 − 2.000 2 × 23 × 10−6 − 2.005 × 11× 10−6 .005 = 208°C 23.945 × 10−6
θ = (208 + 10) = 218ºC.
Substituting the values we get
1011 × 4 × 10−6 × 0.5 × 10−3 = 400 kg 10 × 0.5 (ii) Since, the wire does not undergo extension or contraction, the energy stored in the rod under the influence of constant weight is E = (mg) dl = (400) × (10) × 5 × 10–4 = 2.0 J m=
12. A thin rod of negligible mass and area of cross-section 4 × 10–6m2, suspended vertically from one end, has a length of 0.5 m at 100ºC. The rod is cooled to 0ºC, but prevented from contracting by attaching a mass at the lower end. Find: (i) This mass. (ii) The energy stored in the rod. Given for the rod, Young’s modules = 1011N/m2, coefficient of linear expansion = 10–5K and g = 10 m/s2.
fo
in
ta
si
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un
iv
er
si
ta
rio
s
er iv
un
ht
.m
tp
s:
//t
ht
.m
tp
e/
s:
(i) Contraction of length of the rod due to the cooling the temperature from 100ºC to 0ºC. dl = l αθ = 0.5 × 10–5 × 100 = .05 × 10–3m
s:
)
//t
) Solution
s_
rio
m
w
//w w
)
) Solution
13. A double-pane window used for insulating a room thermally from outside, consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steadystate, the room-glass interface and the glass-outdoor interface are at constant temperatures at 27ºC and 0ºC respectively. Calculate the rate of heat flow through the window pane. Also find the temperature of other interface. Given that thermal conductivities of glass and air as 0.8 and 0.08 Wm–1K–1, respectively.
an
ar
.ja
a.
co
m
∴
=
ds − da da αa − ds αs
∆θ =
∴
or
ht
G4 GW
7
7
*ODVV
tp
*ODVV
$LU
7
7
G4 GW
P
$LU G4 GW
P P
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Heat
1.41
Here, A = 1 m2, ka = 0.08 Wm–1K–1, A = 1 m2 kg = 0.8 Wm–1K–1 By conduction equation,
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)
m
in
s_
ta
rio
iv
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ta
ρ 1 γ = 5°C − 1 ρ 95°C 90°C = 2 × 10–4/ºC ∴ Coefficient of linear expansion γ = = 6.7 × 10−5/°C. 3 15. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300 K. At time t = 0 the temperature of X is T0 = 400 K. It cools according to Newton’s law of cooling. At time t1, its temperature is found to be 350 K. At this time (t1), the body X is connected to a large box Y at atmospheric temperature TA, through a conducting rod of length L, cross-sectional area A and thermal conductivity k. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1.
un
e/ .m
//t
ht tp
s:
ht
dQ1 = 80 (300 – 299.48) = 41.6 W dt
ρ5°C − ρ95°C ρ95°C (90°C)
s
rio
=
iv
un
e/
.m
//t
s:
tp
Coefficient of volume expansion, V95°C − V5°C γ= V5°C (95°C − 5°C)
si
s:
tp
ht
and T3 = 273.52 K ⇒ Q3 = 0.52ºC Rate of heat flow from window-pane
fo
co a. ar an
m
.ja
w
w
//w
...(ii)
From equation (ii) T2 + T3 = 573 K From equation (i) 51 T2 – T3 = 15000 K ∴ T2 = 299.48 K ⇒ Q2 = 26.48ºC
ρ5°C 52.8 + 49 = = 1.08 ρ95°C 49 + 51
or
er
0.8 × 1 (T3 − 273K) 0.01
and 80 (300 – T2) = 80 (T3 – 273)
&
ρ95ºCg (52.8 cm) – ρ5ºCg (49 cm) + ρ95ºCg (49) – ρ5ºDg (51) = 0
or
dQ1 dQ 2 dQ3 = = dt dt dt This gives 80 (300 – T2) = 1.6 (T2 – T3) ...(i)
&
P0 = P0 + ρAghA – ρBghB + ρCghC – ρDghD
0.08 × 1 (T2 − T3 ) 0.05
At steady-state,
14. The apparatus shown in the figure consists of four glass columns connected by horizontal sections. The height of two central columns B and C are 49 cm each. The two outer columns A and D are open to the atmosphere. A and C are maintained at a temperature of 95ºC while the columns B and D are maintained at 5ºC. The height of the liquid in A and D measured from the base lines are 52.8 cm and 51 cm respectively. Determine the coefficient of thermal expansion of the liquid.
&
Applying Pascal’s law, we have
dQ3 k g A (T3 − T4 ) = dt d
=
'
) Solution
dQ 2 ka A (T2 − T3 ) = dt d =
&
dQ1 k g A (T1 − T2 ) = dt d 0.8 × 1 (300 − T2 ) = 0.01
%
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1.42
Heat
Integrating this expression, we get
)
) Solution
3t
T2
/
%RG\; 7 . DWW
%RG\< .
or
ln
$WPRVSKHUH7 .
dT kA 1 ∫T T − TA = − k + CL ∫t dt 1 1
$
FRQGXFWLQJURG
ln 2 kA = − + (2t1 ) CL t1 2k A t1 = −2 ln 2 − CL
7KHUPDOFRQGXFWLYLW\ N
fo
rio s_ in
ita
rio s
rs
ita
rs
ve
un i
e/
) Solution
dT kA = − k + (T − TA ) dt CL
= (300 K) + (12.5 K) e −2 k A t1 / CL .
)
d T −kA = (T − TA ) dt CL where, C = heat capacity of the body. Taking them together, we get
or
1 = (300 K) + (350 − 300) e −2 k A t1 /CL 4
16. A thermocole vessel contain 0.5 kg of distilled water at 30ºC. A metal coil of area 5 × 10–3m2, number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform time varying magnetic field set up to pass vertically through the coil at time t = 0. The field is first increased from 0 to 0.8 T at a constant rate between 0 and 0.2 second then decreased to zero at the same rate between 0.2 and 0.4 sec. The cycle repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as functions of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axis. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium. Specific heat capacity of the metal = 500 J kg–1K–1 and the specific heat capacity of water = 4200 J kg–1K–1. Neglect the inductance of coil.
s: //t
ht tp
dQ −k A = (T − TA ) dt L
or
T2 = TA + (T1 − TA ) e −2ln 2 e −2 k A t1 /CL
.m
ht tp
s: //t
ht
.m
For the time t > t1, the temperature of the body falls due to Newton’s law of cooling as well as due to conduction of heat through the conducting rod (length L, area of cross-section A, and conductivity k) For cooling via Newton’s Law dT = −k (T − TA ) dt For cooling via conduction
dT kA = − k + dt T − TA CL
or
ve
(350 − 300) = ln 2 (400 − 300)
e/ un i
://
k t1 = −ln
tp s
or
T2 − TA 2k A t1 = exp −2ln 2 − T1 − TA CL
na .c om
T1 − TA = −k t1 T0 − TA
w
ln
0
w
∴
= −k ∫ dt
A
T0
ar a
dT
∫ T−T
w
Integrating
t1
.ja m
dT = −k (T − TA ) dt T1
or
Upto time t1, the body cools under Newton’s law of cooling, for which we write
+HDWFDSDFLW\ &
T2 − TA kA = − k + (2t1 ) T1 − TA CL
Area of the coil, A = 5 × 10–3m2 Number of turns of coil, n = 100
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1.43
Heat
or
−d φ dB = nA dt dt
θ=
Power of the coil = Ei = 1.25 × 2 = 2.5 W
L
m
er s
er si ta rio s
un iv
e/
un
//t
If L1 – L2 is to be independent of ∆T, then
l1α1 – l2α2 = 0 l1 α1 = l2 α 2
2. 70 calories of heat is required to raise the temperature of 2 mole of an ideal gas from 30ºC to 35ºC at constant pressure. The heat required (in calories) to raise the temperature of the same gas through the same range (30ºC to 35ºC) at constant volume will be (a) 30 (b) 50 (c) 70 (d) 90
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⇒
L1 – L2 = (l1 – l2) + (∆T) (l1α1 – l2α2)
e
t
ht tp
D
)
L2 = l2[1 + α2∆T]
n
s
s: //
t.m e/
ht tp s:
α12 α 22 = l1 l2
) Explanation L1 = l1[1 + α1∆T] ∴
W
iv
.m
ht
1. Two rods of length l1 and l2 are made of materials whose coefficients of linear expansion are α1 and α2. If the difference between two lengths is independent of temperature, then l α l α (a) 1 = 1 (b) 1 = 2 l2 α 2 l2 α1 (d)
W
ita
.ja
w
w
//w
s:
tp
SOLVE OBJECTIVE yp qu
(c) l22 α1 = l12 α 2
3 P
0.5 × 4200 (θ – 303 ºK) + 0.06 × 500 (θ – 303 ºK) = 12000
rio s_ in fo
ar a
na .c om
4T = (100) (5.0 × 10–3 m2) =2V s Resistance of the coil = 1.6 Ω E Current of the coil, i = R 2 = = 1.25 A 1.6 Heat of the coil, H = i2 Rt = (1.25)2 × 1.6 × 0.2 = 0.5 J Again decreased the same rate, so the heat produces by decreasing the magnetic field = 0.5 J. Total heat produced in 12000 cycle = (0.5 + 0.5) × 12000 = 12000 J This heat increased both the temperature of water and coil. mwsw∆θ + mcsc∆θ = 12000 J ∴
657390 = 308.63 K = 35.63°C 2130
estio
E=
2100 θ – 636300 + 30 θ – 9090 = 12000 2130 θ = 657390
or or
Rate of increasing magnetic field, dB 0.8T 4T = = dt 0.2s s Electro-magnetic field induced in the coil,
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1.44
) Explanation P = εσAT4
)
)
) Explanation 2CP(35 – 30) = 70
3. A wall is made of two equally thick layers A and B of different materials. The thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36ºC. The temperature difference across the layer A will be (a) 6ºC (b) 12ºC (c) 18ºC (d) 24ºC
(
)
2 −1
(d)
s_
1
(
2 +1
(
2 −1
1
)
)
ta r
io
3
in
fo
)
2 −1
1
si
the given problems is shown in the figure. Since TB > TA, the heat will flow form B to A. To maintain steady state, heat will flow from C to B and A to C Applying the conduction formula,
s
io
si
ta r
er
iv
∆Q KA = (∆T) ∆t l to the sides AC and CB, we get
un e/
t.m
ht
dT 2 1 = = ∆θdt 30 × 4 4
(
(b)
)
://
tp s
dT = K∆θ dt
tp s: //
⇒
K=
∴
dT ∝ ∆θ dt
co m iv
un
t.m
)
) Explanation Rate of cooling ∝ difference in i.e.,
1
) Explanation The diagramatic representation of
e/
://
ht
tp s
w
w
w
4. In a room where the temperature is 30ºC, a body cools from 61ºC to 59ºC is 4 minutes. The time taken by the body to cool from 51ºC to 49ºC will be (a) 4 minutes (b) 6 minutes (c) 5 minutes (d) 8 minutes temperature
.ja m
∆TA = 12ºC
∴
2
(c)
ar
∆TA + ∆TB = 36ºC
(a)
er
and
( )
an a.
)
⇒
4
P1 A1 T1 4π(1) 2 4000 = = =1 P2 A 2 T2 4π(4) 4 2000
6. Three rods of identical cross-sectional area and made from the same metal from the sides of an isosceles triangle ABC, right-angled at B. The points A and B are maintained at temperatures T and 2 T respectively. In the steady state, the temperature of the point C is Tc. Assuming that only heat conduction takes place, Tc/T is
2KA (∆T) A KA(∆T) B = d d 2(∆T)A = (∆T)B
) Explanation
4
∴
Heat required at constant volume = 2Cv(35 – 30) = 2(CP – R) (35 – 30) = 70 – 2 × 2 × 5 = 50 cal
7$ 7
D
D
7%
7& 7
7
⇒
T − TC TC − 2T = a 2a
⇒
T − TC = 2TC − 2T
&
%
5. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The power radiated by the two spheres are in the ratio (a) 16 : 1 (b) 4 : 1 (c) 4 : 1 (d) 1 : 1
ht
In first case, dT = 61 – 59 = 2, ∆θ = 60 – 30 = 30, dt = 4 minutes For second case, dT = 2 ∆θ = 50 – 30 = 20 2 dT ∴ dt = = = 6 min 1 K∆θ × 20 60
∆T ∆T = a CB 2a AC
Heat
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Heat
(
3
This result comes from the fact that Weight = Upthrust
)
2 +1
∴ Vρsolid g = Vsubmerged ρliquid g
∴
(ii) Since ρ ∝
) Explanation A black body is a perfect absorber
=
Vsubmerged
V
ρθ°C
∴
co
V0 °C V0 °C + V0 °C γθ
1 1 + γθ
o
_i
io s
s
rio
er si
ta
iv
&
&
e/ un //t .m
&
In this problem, two concepts are used
)
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(i) When a solid floats in a liquid then Fraction of volume submerged ρ (k ) = solid ρliquid
90 − θ 90 − θ θ − 0 + = R R R Here R = Thermal Resistance ∴ 3θ = 180 or θ = 60ºC
∴
k1 (ρFe )0°C (ρHg )60°C (1 + 60γ Fe ) = = k2 (ρFe )60°C (ρHg )0°C (1 + 60γ Hg )
∴
junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow throught AB + Rate of heat flow though CB = Rate of heat flow through BD
ρ = Density
=
nf
er si ta r
iv
e/ un
Here
1 (mass remaining Volume constant) Vθ°C Vθ°C = = V0 °C V0 °C + ∆V
) Explanation Let θ be the temperature of the
s:
60°C
ρ solid ρliquid
9. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0ºC and 90ºC respectively. The temperature of the junction of the three rods will be (a) 45ºC (b) 60ºC (c) 30ºC (d) 20ºC
t.m
://
; 0°C
tp
and
1 + 60γ Fe 1 − 60γ Hg
ht
ρ k2 = Fe ρ Hg
(d)
ht
)
ρ ) Explanation k1 = Fe ρHg
1 − 60γ Fe 1 + 60γ Hg
tp s
1 + 60γ Fe 1 − 60γ Hg
(b)
s: //w
1 + 60γ Fe 1 + 60γ Hg
ht tp
(c)
w
w
.ja
m
ar an
a.
8. When a block of iron floats in mercury at 0ºC, a fraction k1 of its volume is submerged, while at the temperature 60ºC, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe and that of mercury is γHg, then the ratio k1/k2 can be expressed as (a)
ρ0 ° C
m
)
and a perfect radiator
=
7. An ideal black-body at room temperature is thrown into a furnace. It is observed that (a) initially it is the darkest body and at later times the brightest (b) it is the darkest body at all times (c) it cannot be distinguished at all times (d) initially it is the darkest body and at later times it cannot be distinguished
TC = T1
⇒
)
2 +1
(
3T = TC
⇒
1.45
' &
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1.46
1. Thermal coefficient of volume expansion (V) can be written as (symbols have their usual meanings) dρ dρ (b) −ρ (a) −T dT dT
s
n
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estio
yp qu e
t
objective
d
un
solve
Heat
(d) ρ
1 dρ ρ dT
dρ dT
2. A rod of length 40 cm has the coefficient of linear expansion α1 = 6 × 10–6/ºC. Another rod of length l has the coefficient of linear expansion α2 = 4 × 10–6/ºC. If the difference in length of the two rods always remain same at all temperatures, then the value of l is (a) 26 cm (b) 60 cm (c) 80 cm (d) 32 cm
(b)
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w
w
//w
s:
tp
(c)
4. If E denotes the energy of a particle of mass m and λ be its de-Broglie wavelength then the correct curve between 1 y = 2mE and x = is λ
ht tp
s: //
\
[
5. Two holes of unequal diameters d1 and d2 (d1 > d2) are cut in a metal sheet. If the sheet is heated, (a) both d1 and d2 will decrease (b) both d1 and d2 will increase (c) d1 will increase, d2 will decrease (d) d1 will decrease, d2 will increase
(d)
[
(a)
ita
.ja
3. Three stars A, B and C appears as green, red and blue respectively. The star having minimum temperature is (a) A (b) B (c) D (d) A and B have the same temperature
[ \
ar a
na .c om
(c) −
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Heat
1.47
λB corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 µm. If the temperature of A is 5802 K, then (a) the temperature of B is 1934 K (b) λB = 1.5 µm (c) the temperature of B is 11604 K (d) the temperature of B is 20 K
G
6. In the previous question, the distance between the holes will (a) increase (b) decrease (c) remain constant (d) may either increase or decrease depending on the positions on the sheet and on the ratio d1/d2
9. Left and right ends of a metal conductors of length 2m, cross section area 5 × 10–4m2 and thermal coefficient 80 w/m-K. is maintained at temperature 80ºC and 20ºC respectively. If an electrical resistance delivering a constant power 2 watt is attached to the conductor at a distance 1.2 m from left end, then temperature at the section of resistance will be (a) 90ºC (b) 45ºC
fo
in
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10. A metal wire of length L and radius r is made by copper and aluminium contributing equal length. This wire is now coated by nickel till the radius of wire becomes R. If specific resistances of these materials are ρCu, ρAl and ρni the equivalent conductance of the system across length will be
e/ .m
πr 2 1 1 R2 + + 2 L ρCu ρAl r ρ Ni
(b)
π r2 r 2 2(R 2 − r 2 ) + + L ρCu ρAl r 2ρ Ni
(c)
π 2(R 2 − r 2 ) r 2 − L ρCu + ρAl ρ Ni
(d)
π 2r 2 R2 − r2 − L ρCu + ρAl ρ Ni
//t s: tp
(a)
ht & $
8. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of two bodies are same. The two bodies emit total radiant power at same rate. The wavelength
&
(d) 22.5ºC
un
//t s: tp ht
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ta
1 (c) 66 °C 3
si er
.m
iv
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s:
tp
ht
//w
w
w
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m
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s_
an
7. The two ends of a uniformly tapered rod are maintained at temperatures 100ºC and 0ºC respectively. Under steady state, the temperature of the section AA exactly at the mid-point (a) 50ºC (b) more than 50ºC (c) less than 50 ºC (d) none of these
a.
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13. A black body is at a temperature of 2880 K. The energy of radiation emitted
&
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16. Twelve identical conducting rods form the sides of a uniform cube of side l. In the steady state, B and H ends of the cube area at 100ºC and 0ºC. Find the temperature of the junction A. (a) 80ºC (b) 60ºC (c) 40ºC (d) 70ºC
G
G
ta
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s:
tp
ht
er
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iv
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s:
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ht
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er
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15. Two indentical conducting rods AB and CD are connects to a circular conducting ring at two diametrically opposite points B and C. The radius of the ring is equal to the length of rods AB and CD. The area of cross-section and thermal conductivity of the rod and ring are equal. Points A and D are maintained at temperatures of 100ºC and 0ºC temperature of point C will be (a) 62ºC (b) 37ºC (c) 28ºC (d) 45ºC
an
11. A spherical black body radiating power P and rate of cooling of the body is R. If the radius of the body is r, then (a) P ∝ r (b) P ∝ r2 1 (c) R ∝ r2 (d) R ∝ r 12. Three rods of equal length are joined to form an equilateral triangle ABC. D is the mid-point of AB. The coefficient of linear expansion is α1 for AB and α2 for AC and BC. If the distance DC remains constant for small changes in temperature, then (a) α1 = α2 (b) α1 = 2 α2 1 (c) α1 = 4 α2 (d) α1 = α 2 2
a.
co
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14. 1 kg of ice at 0ºC is mixed with 1.5 kg of water 45ºC [latent heat of fusion = 80 cal g]. Then (a) the temperature of the mixture is 0ºC (b) mixture contains 156.25 gm of ice (c) mixture contains 843.75 gm of ice (d) the temperature of the mixture 15ºC
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17. A solid cube of side ‘a’ and another solid cube with the same material and same side but having spherical cavity of radius b where (b < a) are heated up to the same temperature and allowed to cool in the same surroundings, which one will start cooling faster? (a) both will cool at same rate (b) solid cube (c) hollow cube (d) can’t say
1.49
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18. A solid with coefficient of linear expansion α just floats in a liquid whose coefficient of volume expansion is γ. If the system is heated, the solid will (a) sink in all the cases (b) continue to float in all cases (c) sink if γ > 3α (d) sink if γ < 3α
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21. A mixture of ideal gases filled inside a balloon expands according to the relation PV2/3 = constant. The temperature inside the balloon is (a) increasing (b) decreasing (c) constant (d) can not be defined
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22. A cylindrical vessel of length 10 m and area of cross-section 1 m2 is filled with water and is rotated with angular speed
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20. A tungsten wire with a diameter of 0.20 mm is stretched until the tension becomes 50 N. The wire is then clamped to a stout aluminium rod when both are at a temperature of 150ºC. The tension in the wire when both are brought to a temperature of 20ºC is (Ytungsten = 34 × 1010 N/m2, αT = 4.5 × 10–6/ºC) (Neglect the expansion of aluminium rod) (a) 85 N (b) 57.5 N (c) 100 N (d) 56.2 N
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19. A conducting sphere cools in atmosphere according to Newton’s law of cooling. The correct graph between its temperature θ and time t is
(d)
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1.50
26. To vaporize 4 gm of water by boiling it at 373ºK, 2160 cal of heat is needed. What is specific heat of water in the conduction? (a) 0.36 cal/gm-ºK (b) 5.4 cal/gm-ºK (c) zero (d) infinite
ω = 20 rad/s. Find the rise in temperature of water when it is stopped all of a sudden. (a) no change (b) 2.5ºC (c) 1.58ºC (d) 5ºC
Heat
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27. The two ends of a uniform rod of thermal conductivity K are maintained at different, but constant temperatures. The temperature gradient at any point on the dθ (equal to the difference in temrod is dt perature per unit length). The heat flow per unit time per unit cross-section of the rod is I.
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25. A thermally insulated vessel contains 100 g of water at 0ºC. When air above the water is pumped out, some of the water freezes and some evaporates at 0ºC itself. Calculate the mass of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at 0.C is 2.1 × 106 J/kg and Latent heat of fusion of ice is 3.36 × 105 J/kg. (a) 62 g (b) 86 g (c) 54 g (d) 78 g
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28. Three rods of the same dimensions have thermal conductivities 3 K, 2 K and K. They are arranged as shown, with their ends at 100ºC, 50ºC and 0ºC. The temperature of their junction is 200 (a) 75°C (b) °C 3 100 °C (c) 40°C (d) 3
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29. A and B are two points on a uniform metal ring whose centre is C. The angle ACB = θ. A and B are maintained at two different constant temperatures when θ = 180º, the rate of total heat flow from A to B is 1.2 W. When θ = 90º, this rate will be (a) 0.6 W (b) 0.9 W (c) 1.6 W (d) 1.8 W
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24. A square metallic plate is having a circular hole in it. If the temperature of the plate is increased, the area of the hole (a) decreases (b) increases (c) remains same (d) can not be decided
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23. The power radiated by a black body is Pr and it radiates maximum energy around the wavelength α0. If the temperature of the black body is now changed so that it radiates maximum energy around a 3λ 0 wavelength , the power radiated by 4 it will increase by a factor of 16 4 (a) (b) 9 3
P
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4πK 2 T KP
(b) (d) (a, b) (c)
6. 14. 22. 30.
(d)
(c)
(b)
4πKR 2 T P
(a)
30. A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed T, the thickness of the shell should not be less than
4πKP 2 TP 4πR 2 P KT
ANSWERS 5. 13. 21. 29.
(b) (c) (d) (b)
7. (b) 15. (b) 23. (d)
8. (c) 16. (b) 24. (b)
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1.51
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c h a p t e r
2
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Kinetic Theory of Gases and Thermodynamics
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2.1 FUNDAMENTAL ASSUMPTIONS OF KINETIC THEORY OF GASES
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(i) All gases consists of molecules which are considered as rigid perfectly elastic and solid spheres. These molecules are extremely small i.e., the volume occupied by the molecules is taken negligible as compared to the volume of the containing vessel. (ii) The molecules are in a continuous random motion. They are moving in all directions with all possible velocities. There is no particular direction of motion of the particles. (iii) The molecules during their motion collide with one another and also with the walls of the vessel. At each collision, their velocities change both in magnitude and direction; however in the steady-state, molecular density remains uniform throughout the gas and does not change with time. (iv) No force of attraction or repulsion are exerted on a gas molecule by other molecules or by the container except during collisions. This means, the energy of the gas is purely kinetic. (v) The collisions among the molecules or with the walls of container are purely elastic. A collision is said to perfectly elastic if the energy of the molecules remains same before and after the collision. (vi) Molecules travel in a straight line with a uniform velocity in between the collisions. The average distance travelled by the molecules between the successive collisions is called as mean-free path of a molecule. The time spent during a collision is assumed to be negligibly small as compared with the time taken by the molecule in traversing a mean-free path.
Note: A gas which satisfies all of above set of assumptions of kinetic theory of gases under all conditions of temperature and pressure is called as an ideal gas or perfect gas. A real gas shows appreciable deviations from the perfect gas. At high temperature and low pressure, gases show deviation from the ideal behaviours.
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2.2
Kinetic Theory of Gases and Thermodynamics
2.2 EVIDENCE IN SUPPORT OF MOLECULAR MOTION Diffusion The phenomenon of diffusion presents a remarkable evidance of molecular motion of matter. The particles of one substance diffuse into another against gravity due to the molecular motion. H2 diffuses into CO2 when a jar filled with H2 is held mouth to mouth over a jar filled with CO2. Though H2 is lighter than CO2.
Evaporation
Brownian Motion
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The phenomenon of evaporation is related with the motion of molecules. The molecules of liquid are quite free to move about. At any temperature, all the molecules don’t move with the same velocity, some move faster than the others. Evaporation involves the escape of fast moving molecules (having speed greater than average speed). This results in lowering of average speed, which is turn results in lowering of temperature and thus producing a cooling effect.
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Botanist Robert Brown observed the irregular motion of plane spores floating in water. Such motions are called as Brownian motion. Here the motion of each particle is due to inequality of the impact of surrounding molecules on it. For example: Motion of smoke particles in air is a case of Brownian motion. This supports the view that the molecules of a substance are in constant motion.
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2.3 CALCULATION OF THE PRESSURE OF AN IDEAL GAS
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vy
A1 V vx
O
x
y
Consider an ideal gas enclosed in a cubical vessel of edge L. Take a corner of the vessel as the origin O and the x, y and z-axis along the edges. Let A1 and A2 be the parallel faces perpendicular to the x- axis. Consider a molecule moving with velocity v. The components of the velocity along the axis are vx, vy and vz. When the molecule collides with the face A1, the x-component of the velocity https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.3
is reversed whereas the y and the z-components remain unchanged. This follows from our assumption that the collisions of the molecules with walls are perfectely elastic. The change in momentum of the molecules is ∆p = (–mvx) – (mvx) = 2 mvx
As the momentum remains conserved in a collision, the change in momentum of the wall ...(i)
∆p′ = 2mvx
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After rebound, this molecules travels towards A2 with the x-component of velocity equal to – vx, any collision of the molecule with any other face (except for A2) does not change the value of vx. So, it travels between A1 and A2 with a constant x-component of velocity which is equal to – vx. Note that, we can neglect any collision with the other molecules in view of the last assumption discussed in the previous section. Distance travelled parallel to the x-direction between A1 and A2 = L.
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Thus, the time taken by the molecules to go from A1 to A 2 =
L . vx
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The molecule rebounds from A2, travels towards A1 and collides with it after another L time interval . vx
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Thus, the time between two consecutive collisions of this molecules with A1 is 2L ∆t = . vx
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The number of collisions of this molecule with A1 in unit time is v 1 n= = x ∆ t 2L
…(ii)
tp s
The momentum imparted per unit time to the wall by this molecule is from equations (i) and (ii).
=
ht
ΔF = n Δp’
vx m × 2m vx = vx 2 2L L
This is also the force exerted on the wall A1 due to this molecule. The total force on the wall A1 due to all the molecules is m m F = ∑ vx 2 = ∑ vx 2 …(iii) L L As all directions are equivalent, we have
Σ vx2 = Σ vy2 = Σ vz2 1 = ∑ (v x 2 + v y 2 + v z 2 ) 3 1 = ∑ v2 3
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Kinetic Theory of Gases and Thermodynamics
Thus, from equation (iii) F =
1m ∑ v2. 3L
If N is the total number of molecules in the sample, we can write
F=
1 mN ∑ v 2 3 L N
=
1 mN ∑ v 2 3 L3 N
=
1 M ∑ v2 1 ∑ v2 = ñ 3 L3 N 3 N
co m
Pressure is the force per unit area. Therefore F p= 2 L
a.
fo
v2 is the averN 2 age of the speeds squared. It is written as v and is called mean square speed.
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1 p = ρ v2 3 1 pV = M v 2 3 1 pV = Nm v 2 3
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Thus, the pressure is
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where, M is the total mass of the gas taken and ρ is its density, and ∑
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Root-Mean-Square Velocity (RMS)
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The square root of mean-square speed is called root-mean-square speed or rms speed. It is denoted by the symbol vrms. Thus,
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vrms =
In terms of pressure
1 2 p = ρ vrms 3
∴
2.4
vrms =
3p 3 pV 3RT = = ρ M M
For the same gas at two different temperatures T1 and T2, vrms will be different As vrms ∝ T ∴
vrms (1) T1 = vrms (2) T2
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Kinetic Theory of Gases and Thermodynamics
2.5
For two gases at different temperature T1 and T2:
vrms (1) T M = 1 2 vrms (2) T2 M1 Average speed: (vav)
vav =
v1 + v2 + v3 + ...vN N 8 RT πM
vav =
ar a
2.4 DEDUCTIONS FROM KINETIC THEORY
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Boyle’s Law
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Most probable speed: (vmp)
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At a given temperature, the pressure of a given mass of a gas is inversely proportional to its volume. This is known as Boyle’s law. 1 We know …(i) pV = mN v 2 rms 3
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As for a given gas v2rms ∝ T, the value of v2rms is constant at a given temperature. Also for a given mass of the gas M and N are constant. Thus, from equation (i)
Charles’ Law
As
At a given pressure, the volume of a given mass of a gas is proportional to its absolute temperature. This is known as Charles’ Law. From equation (i), if p is constant, then V ∝ v2rms v2rms ∝ T
or
We get V ∝ T V1 V2 = T1 T2
which is Charles’ law. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
Charles’ Law for Pressure At a given volume, the pressure of a given mass of a gas is proportional to its absolute temperature. This is known as Charles’ law for pressure. In fact, this is the definition of the absolute temperature T. If one starts from the 2 fact that Vrms ∝ T and uses the fact that V is constant, one gets from equation (i)
2 p ∝ vrms
p∝V
or
p1 p2 = V1 V2
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Avogadro’s Law
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At the same temperature and pressure, equal volumes of all gases contain equal number of molecules. This is known as Avogadro’s law. Consider equal volume of two gases kept at the same pressure and temperature.
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1 Now, pV = N1 m1 v12 3 And
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Let m1 = mass of the molecule of the first gas. m2 = mass of the molecule of the second gas. N1 = number of molecules of the first gas. N2 = number of molecules of the second gas. p = common pressure of the two gases. V = common volume of the two gases.
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where, v1 and v2 are root-mean-square speeds of the molecules of the first and second gas respectively. Thus,
As the temperature of the gases are the same, the average kinetic energy of the molecules is same for the two gases, i.e., 1 1 m1v12 = m2 v2 2 2 2
…(ii)
2.6
From equations (i) and (ii), we get N1 = N2 which proves Avogadro’s law.
Graham’s Law of Diffusion When two gases at the same pressure and temperature are allowed to diffuse into each other, the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas. This is known as Graham’s law of diffusion. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.7
It is reasonable to assume that the rate of diffusion is proportional to the rms speed of the molecules of the gas. Then, if r1 and r2 be the rates of diffusion of the two gases, …(i)
r1 v1 , rms = r2 v2 , rms
3p ρ If the pressure of the two gases are the same, then
We know, vrms =
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Dalton’s Law of Partial Pressure
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Dalton’s law of partial pressure says that the pressure exerted by a mixture of several gases equals the sum of the pressures exerted by each gas occupying the same volume as that of the mixture. In kinetic theory, we assume that the pressure exerted by a gas on the walls of a container is due to the collisions of the molecules with the walls. The total force on the wall is the sum of the forces exerted by individual molecules. Suppose, there are N1 molecules of gas 1, N2 molecules of gas 2 etc., in the mixture. Thus, the force on a wall of surface area A is
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F = Force by N1 molecules of gas 1 + force by N2 molecules of gas 2 + ........... = F1 + F2 + .......... Thus, the pressure is F F p = 1 + 2 + ....... A A If the first gas alone is kept in the container, its N1 molecules will exert a force F1 on the wall. If the pressure in this case is p1, then F p1 = 1 A
Similarly, for other gases. p = p1 + p2 + p3 + .......
Ideal Gas Equation
For n moles of an ideal gas at pressure p, absolute temperature T occupying a volume V, the following equation holds: pV = nRT where R = Ideal gas constant
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2.8
Kinetic Theory of Gases and Thermodynamics
The gas equation can also be written in the following form: pM = d RT where, M is the molecular weight and d is the density of the gas at pressure p and absolute temperature T. The molecular weight M of a gas can be given as:
g M = densityin at S.T.P. × 22.4 l
Boltzmann Constant (k)
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pV = NkT
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Gas equation in terms of k:
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In the gas equation pV = nRT, the value of R be taken as: R = 8.314 J/mol/K R = 0.0821 -atm/mol/K R = 2 cal/mol/K R is known as Boltzmann constant. If N0 is the Avogadro number then, k = N0 k = 1.38 × 10–23 J per molecule per degree Kelvin.
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2.5 KINETIC ENERGY OF MOLECULES
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Degrees of Freedom
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The degrees of freedom of a dynamical system are defined as the total number of independent quantities required in order to describe the position or motion of the system completely.
Mono-Atomic Gas Molecule
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The molecules of a mono-atomic gas (He, Ne, Ar) consists of one atom only. It is capable of only translatory motion in free space which has three components namely. x, y and z-directions. Hence, it will have three degrees of freedom per molecule.
Di-Atomic Gas Molecule Each molecule of a di-atomic gas (H2, N2, O2, d2, etc.) consists of two atoms. In the given diagram, a di-atomic molecule A2 (A – A) is shown. A2 is capable of translatory motion in space and, thus possesses three degrees of freedom. Beside this, the molecule may rotate about yy′-axis or zz′-axis (perpendicular to the molecular axis and the plane of paper). Rotation about molecular axis (xx′) is neglected (because the size of atom is negligible). Thus, due to rotational motion, the molecules have two degree of freedom. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.9
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Hence, a di-atomic molecule has five degrees of freedom. Note: A tri-atomic molecule (CO2, H2O, SO2 etc.) has six degrees of freedom (three of them for translatory and three for rotational motion).
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2.6 KINETIC ENERGY OF TRANSLATION PER MOLECULE
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Kinetic energymean
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The average kinetic energy or kinetic energy per molecule due to translatory motion is also known as mean kinetic energy of translation.
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……(as pV = NkT)
Kinetic energy of translation per mole = (KE)mean N0 or
3 3 k N 0 T = RT 2 2
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2.10
Kinetic Theory of Gases and Thermodynamics
2.7 EQUIPARTITION OF ENERGY
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Total energy per molecule = 3 kT Total energy per mole = 3 RT
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The law of equipartition of energy states that for a dynamical system in thermal equilibrium, the total energy of the system is equally divided amongst the various degrees of freedom. 1 The share of each degree of freedom is kT. 2 For Mono-atomic gas: 3 Total energy per molecule = kT 2 3 Total energy per mole = RT 2 For di-atomic gas: 5 Total energy of molecule = kT 2 5 Total energy of mole = RT 2 For tri-atomic gas:
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2.8 THERMAL EXPANSION OF AN ENCLOSED GAS
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When the enclosed gas is heated, its volume and pressure tend to increase. Keeping one of them constant, the thermal expansion of the other can be studied. Since, thermal expansions of gases are appreciable, the coefficient of thermal expension are referred to a reference initial temperature of 0ºC. The coefficient of volume expansion of an enclosed gas at constant pressure is defined as the increase in its volume per unit rise in temperature, per unit volume at 0ºC. The coefficient of pressure increase of an enclosed gas at constant volume is defined as the increase in its pressure, per unit rise in temperature per unit pressure at 0ºC. For an ideal gas, i.e., a gas which obeys the general gas law pV = nRT; Coefficient of pressure increase at constant volume = Coefficient of volume expansion at constant pressure 1 1 = (C°) −1 or K −1 273.15 273.15
=
1 K −1 273
Note: If the mass of the enclosed gas does not remain constant, i.e., the gas leaks out of or leaks into the enclosure, the partial laws (Boyle’s, Charles’ and Gay-Lussac’s laws) are not applicable. But the general gas law holds good and the value of n changes. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.11
2.9 STANDARD CONDITIONS
One mole of any gas occupies a volume of 22.4 litres (2.24 × 10–2 m3) at standard (or normal) temperature and pressure (STP or NTP). Standard temperature = 0ºC = 273.15 K Standard pressure = 1 atm = 1.013 × 105 Nm–2 In practice, standard temperature is taken as 273 K only. Example
)
) Solution MAr = 0.040 kg
ar a
3 RT 3 × 8.3 × (273 + 25) = = 430.7 m/sec. M .040
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T1 1 = 2 298
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vrms (1) T1 = vrms (2) T2
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m
R = 8.3 J/mol/K vrms =
and
rio s_ in fo
na .c om
Calculate the root-mean-square speed of argon atoms at room temperature (25ºC). At what temperature will the root-mean-square speed will be half of the value?
ht tp
Calculate the pressure of Hydrogen (H) in a cylinder of capacity 10 litre, given that the total kinetic energy of translation is 7.5 × 103 J. What is the total kinetic energy of molecules in the cylinder? (1 atm pressure = 105 N/m2). )
) Solution The total kinetic energy of translation in n moles for a di-atomic gas is
3 3 nRT = pV 2 2
or
∴ 7.5 × 103 J =
3 p (10 × 10−3 m3 ) 2
p = 5 × 105 N/m2
Total kinetic energy for di-atomic gas 5 5 = pV = × 5 × 105 × 10 × 10−3 J = 12500 J. 2 2 https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
Example A vertical cylinder of total length 100 cm is closed at the lower end and is fitted with a movable, frictionless gas tight disc at the other end. An ideal gas is trapped under the disc. Initially, the height of the gas column is 90 cm when the disc is in equilibrium between the gas and the atmosphere. Mercury is then slowly pured on top of the disc and it just starts overflowing when the disc has descended through 32 cm. Find the atmospheric pressure. Assume the temperature of the gas to remain constant the neglect the thickness and weight of the disc.
) Solution
Here, V1 = Initial volume of air = 90 A
un iv e/
p1
er si ta rio s
er s
w w //w s:
42 cm
p2
.m
tp
p
ita
.ja
p
rio s_ in fo
m
ar a
na .c om
A = Area of cross-section of the disc V2 = Final volume of air = 58 A p1 = Initial pressure of the gas = Atmospheric pressure (p) (This is because the disc is light and in equilibrium under pressure P and P1.)
58 cm
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
100 90 cm cm
ht
2.12
In the second case, p2 balances the atmospheric pressure and 42 cm of mercury column. ∴ p2 = p + 42 p1V1 = p2V2 ∴ p × 90 A = (p + 42) × 58 A ∴ p = 76.125 cm of Hg. Example An oxygen storage tank has a capacity of 0.05 m3. The gas pressure is 100 atm. at 27ºC. Determine: (i) The total kinetic energy of translation of molecules. (ii) The average kinetic energy of translation of molecules. (iii) The total kinetic energy. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.13
)
) Solution Total kinetic energy of translation =
(ii)
3 (100) × 105 × 0.05 2 = 7.5 × 105 J 3 Average kinetic energy = kT 2 3 = (1.38 × 10−23 ) × 300 2 = 6.21 × 10–21 J.
(i)
3 pV 2
=
The total kinetic energy (for di-atomic gas) 5 5 = nRT = pV 2 2 5 = × 100 × 105 × 0.05 2 = 1.25 × 106 J.
ita rio
.ja m
w
w
rio s
Example
ve rs
ar an
s_ in
fo
a. c
om
(iii)
=
.m //t
n RT1 p1
s:
(i) V1 =
ht tp
)
ht
) Solution
e/
s:
un
//t
iv e
rs
.m e/
The volume of the gas before expansion. The temperature of gas after expansion. The density of gas before expansion. The density of gas after expansion.
tp
ht
(i) (ii) (iii) (iv)
10 × 0.0821× 283 = 2.42 litre 32 × 3
(ii) As pressure is constant
⇒ T2 =
V1 V2 = T1 T2
V2 T1 = 1169.4 K V1
G
PDVV 9
(iv) 'HQVLW\
G
PDVV 9
(iii) 'HQVLW\
ita
tp s:
un i
//w
10 gm of oxygen are subjected to a pressure of 3 atm, at a temperature of 10ºC. Heating at a constant pressure, the gas is expanded to 10 litre. Find:
JP OLWUH
J OLWUH
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Kinetic Theory of Gases and Thermodynamics
Example An open glass tube is immerged in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and raised further by 44 cm. What will be the length of air column above mercury in the tube? (Atmospheric pressure = 76 cm of Hg).
) Solution x (52–x)
52 cm
rio s_ in fo
m
ar a
na .c om
8 cm
er si ta rio s
e/
.m
iv
un
//t
t.m e/
s: //
∴
x = length of the air column finally p2 = New pressure = 76 – (52 – x) = (24 + x) cm of Hg V2 = New Volume = A · x
ht tp s:
tp
s:
un iv
//w
w
er s
w
ita
.ja
p1 = Initial pressure of air in the tube outside the mercury surface = 76 cm of Hg V1 = Initial volume = A × 8 (A = Area of cross-section) After closing the open end, the total length of the glass tube outside is 52 cm. This increases the volume and reduces the pressure of air.
ht
Since T is constant, therefore p1V1 = p2V2 or 76 × A × 8 = (24 + x) Ax or x2 + 24x – 608 = 0
ht tp
2.14
Solving, we get x = 15.4 cm or 39.4 cm ∴ Length of the air column is 15.4 cm. Example A cylindrical diving bell, whose length is 150 cm is lowered to the bottom of a tank. The water is found to rise 50 cm in the bell. Find the depth of the tank. Assume the atmospheric pressure at the surface as equivalent to 1000 cm of water and the temperature as constant. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.15
) Solution When the cylindrical diving bell is in air, it is filled with air at atmospheric pressure p0 = 1000 cm of water ∴ V0 = Initial volume of air = A × 150 (A = Area of cross-section) The hydrostatic pressure at the level of water in the bell is (h – 50) cm of water. p0
p
m
ar a
50 cm
p = Total pressure of air (when the bell is in water) = p0 + (h – 50) = 1000 + h – 50 = 950 + h V = New volume of air = A × 100 Since, the temperature remains constant, therefore p0V0 = pV or 1000 × A × 150 = (950 + h) × A × 100 or h = 550 cm.
s: //
t.m e/
ht tp s:
un
//t
iv
er si ta rio s
e/
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
∴
ht
150 cm
rio s_ in fo
na .c om
h
ht tp
Example A mercury barometer is defective. (It contains some air in the space above the mercury). When an accurate barometer reads 770 mm, the defective one reads 760 mm. When the accurate one reads 750 mm, the defective one reads 742 mm. What is the length of the air column, when the accurate barometer reads 760 mm? What is the reading of the accurate barometer when the defective one reads 752 mm? Assume that the temperature remains constant.
) Solution
The difference between the readings of the accurate and defective instruments represents the pressure exerted by the air above the mercury in the defective barometer. (i) p1 = Pressure of air in the first case = 770 – 760 = 10 mm p2 = Pressure of air in the second case = 750 – 742 = 8 mm
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Kinetic Theory of Gases and Thermodynamics
rio s_ in fo
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
ar a
na .c om
Let, x cm be the length of the air column above 760 mm mark in the first case. Therefore, when it reads 742, the length of the air column would be (x + 1.8) cm. ∴ V1 = Initial volume of air = Ax where A = Area of cross-section V2 = Final volume of air = A (x + 1.8) From Boyle’s law, p1V1 = p2V2 ∴ (10) (Ax) = (8) (A) (x + 1.8) or x = 7.2 cm (ii) V = Volume of air when the defective one reads 752 mm = (x + 0.8) A p = Pressure of air = (p0 – 752) where, p0 = Reading of the accurate barometer. From Boyle’s law p1V1 = p2V2 ∴ 10 × Ax = (p0 – 752) A (x + 0.8) or 10 × 7.2 = (p0 – 752) (7.2 + 0.8) or p0 = 761 mm.
un
t.m e/
p
s: //
ht tp s:
//t
iv
.m
tp
e/
s:
Example A bulb of 500 cm3 is joined by a narrow tube of another bulb B of 250 cm3 and the whole system is initially filled with air at S.T.P. and sealed. If the temperature of the bulb A is now raised to 100ºC and that of B is kept constant, find:
ht
VA
ht tp
2.16
A
p VB
B
(i) The new pressure in the system. (ii) Mass of the air which is transferred from one bulb to another during heating. (Density of air at S.T.P. = 1.29 g/l)
) Solution (i)
Let, p be the final pressure of air in the bulbs. As the bulbs are connected, pressure will be equal in both bulbs. Let, TA and TB be the final temperatures of the bulbs, therefore TA = 373 K TB = 273 K
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Kinetic Theory of Gases and Thermodynamics
2.17
If p0, T0 are the S.T.P. conditions, then Initial moles in the system = Final moles in the system p0 VA p0 VB pVA pVB + = + R T0 R T0 RTA RTB
Using, p0 = 1 atm, T0 = 273 K, we have
1 500 250 p + (500 + 250) = 373 273 273 or p = 1.22 atm
(ii)
M = Molecular weight of air, M = (dSTP) (22.4)
fo in
an
a. co
m
= 1.29 × 22.4 = 28.896 gm Mass transferred = moles × M p VB p0 VB = − × M R TB R T0
rio
ar
w
si
ta
.ja
m
s
er
w
iv
:// w
rio
2.10 THERMODYNAMICS
s_
250 × 10−3 (1.22 − 1) × 28.896 .082 × 273 = 0.0709 gm. =
si ta
iv
ht
//t .m
er
e/
tp s
un
Temperature It may be defined as that thermal property which attains the same value for all bodies placed in thermal contact with each other.
.m
ht tp
e/
s:
un
Heat Energy The transfer or flow of internal energy caused by the difference of temperatures between two bodies is called as Heat Energy.
ht tp
s:
//t
Zeroth Law of Thermodynamics If two bodies A and B are in thermal equilibrium with another body C, then they are also in thermal equilibrium with each other. Thermodynamical system It is a collection of an extremely large numbers of atoms or molecules so that they together have the same pressure (P), Volume (V) and temperature (T). e.g., A gas, vapour, steam, vapour in contact with the liquid.
2.11 WORK DONE AND INTERNAL ENERGY Work Done by a Gas During Expansion Consider an ideal gas enclosed in a perfectly insulated cylinder fitted with a nonconducting and frictionless piston. Let p be the pressure exerted by the gas and V be the volume of the gas at any particular instant. Let A be the area of cross-section of the piston. The force exerted by the gas on the piston = pA https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
dx
gas
m
dW = pdW
er s
w
ita
.ja
Work Done From P-V Diagram
rio s_ in fo
ar a
na .c om
If the piston moves through an infinitesimal distance dx, this force can be assumed constant and the workdone (dW) is given as: dW = (pA) But A dx = dV = infinitesimal change in volume
er si ta rio s
e/
p p1
ht tp
p2
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
s:
un iv
//w
w
An ideal gas contained in a cylinder fitted with a massless and frictionless piston can be considered as a thermodynamical system. Its state can be represented by variables (p, V, T). If p and V are known, T can be calculated by
ht
2.18
O
v1
v2
V
pV V = nRT RT The system’s state at any instant of time can be specified by two variable (p, V). The relation between pressure p and volume V can be studied on a pressure – volume (p – V) graph known as indicator diagram or a p – V diagram. On such a graph, each equilibrium state of a thermodynamical system can be represented by a point whose X co-ordinate represents volume (V) and Y co-ordinate represents pressure (p). https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.19
A process can be represented by a curve on the p – V diagram. If the gas goes from initial state A (p1, V1) to the final state B (p2, V2), the work done W is given by W = Area under the curve AB above the x-axis (shaded portion).
Internal Energy Internal energy (U) of a system is the total of all kinds of energy possessed by the atoms and other particles that comprise the system. It depends only on the temperature difference of initial and final states and not on the process between the states.
2.12 FIRST LAW OF THERMODYNAMICS
rio s_ in fo
∆Q = ∆U + ∆W
m
ar a
na .c om
If some heat is supplied to a system capable of doing some work, the quantity of heat absorbed by the system is equal to the sum of external work done by the system and increase in its internal energy. If ∆Q is the heat gained by the system, ∆W is the work done against external pressure and ∆U is the change in internal energy, then ...(i)
er si ta rio s
w
er s
w
ita
.ja
Heat energy given Increase in Workdone by = + to the system internal energy the system
un iv
//w
This equation is the statement of the first law of thermodynamics.
e/
iv
.m
ht
tp
s:
Note: First law of thermodynamics is a direct consequence of law of conservation of energy.
un t.m e/
ht tp s:
//t
In Thermodynamics
s: //
ht tp
(i) Workdone by a system is taken is positive (+). Workdone on the system is taken as negative (–). (ii) Heat gained by a system is taken as positive (+) and the heat lost by the system is taken as negative (–). (iii) Increase in internal energy is taken positive (+) and decrease in internal energy is taken as negative (–).
2.13 SPECIFIC HEAT OF A GAS The specific heat of a gas is defined as the amount of heat required to raise the temperature of unit mass of substance through 1ºK. Its unit in S.I. system is J/kg/K. When the unit mass is 1 mole, the specific heat is known as molar specific heat and its units are J/mol/K or cal/mol//K. Molar specific heat = molecular weight (specific heat in J/gm/K) In gases, on giving heat to the gas, besides temperature, its pressure and the volume may also change. Thus, specific heat of a gas depends upon the conditions under which the heat is given to the gas. Hence, we will define two principle types of specific heats of a gas. https://t.me/universitarios_infohttps://www.jamarana.com
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2.20
Kinetic Theory of Gases and Thermodynamics
Specific Heat at Constant Volume (CV) Cv of a gas is defined as the amount of heat required to raise the temperature of 1 mole of that gas through 1ºK provided that the volume of the gas remains constant. It is also known as molar specific heat at constant volume. ∆Q CV = n ∆T Constant volume
In general, if n moles of a gas are heated at constant volume to increase the temperature by ∆T.
Heat required = ∆Q = n Cv ∆T
co
m
Specific Heat at Constant Pressure (CP)
_i
io s
∆Q CP = n ∆T Constant pressure
.ja
ta r
m
ar
an
nf
a.
o
CP of a gas is defined as the amount of heat required to raise the temperature of 1 mole of that gas through 1ºK, provided that the pressure of the gas remains constant. It is also known as molar specific heat at constant pressure.
s
rio
iv
si ta
CP is greater than CV (CP > CV)
er
e/
un
s: //
Heat required = ∆Q = n CP ∆T
tp
w
w
er
w
si
In general, if n moles of a gas are heated at constant pressure resulting in a temperature raise by ∆T, then
iv
t.m
ht
s: //
(i) To raise the temperature of the gas. (ii) To do the mechanical work for expansion.
tp
e/
un
s: //t
ht
.m
When we heat the gas at constant volume, the heat is supplied to raise the temperature of gas only. When we heat the gas at constant pressure, volume of the gas apart from temperature of the gas raises. When the gas expands, it does some external work for increasing the volume. So in this case, the heat is supplied:
ht
tp
As a result more amount of heat is required to increase the temperature of the gas through the same amount when the pressure is kept constant. Hence, CP is always greater than the value of CV. C The ratio of P is denoted by γ CV
i.e.,
γ=
CP ( γ > 1) CV
2.14 SPECIFIC HEAT RELATION
Consider n moles of an ideal gas. If its temperature is raised by ∆T at constant volume, then the heat transferred is n CV ∆T, where CV is the specific heat at constant volume. W = 0 (as no volume change is there.) https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.21
By First Law of Thermodynamics
n CV ∆T = ∆U
...(i)
∆Q = ∆U + 0
Next, let us have the same gas at constant pressure. The heat ∆Q required to change the temperature by ∆T will be nCp ∆T, where Cp is the specific heat at constant pressure. W = P ∆V where, ∆V is the change in volume. By First Law of Thermodynamics ...(ii)
om
∆Q = ∆U + p∆V nCP ∆T = ∆U + p∆V From equations (i) and (ii), we get
rio s_ ta
.ja m
ar
an a
in fo
.c
nCP ∆T = n CV ∆T + p∆V From gas equation p∆V = nR ∆T ⇒ nCP ∆T = n CV ∆T + nR ∆T CP = CV + R CP – CV = R
e/
t.m
s: //
…(as Cp Cv = R)
ht tp
ht
tp s:
un
//t
iv
Internal energy per mole =
⇒
rio s
si ta
er
3 RT 2 3 3 3 Cv = ∆U = R (T + 1) − RT = R 2 2 2 3 Cv = R 2 C 5 γ= p = Cv 3
.m
ht
tp
For mono-atomic gas:
e/ un i
s:
//w
ve
w
w
rs i
Cp and CV for Gases Consider 1 mole of a gas heated to raise the temperature by 1ºC. When the gas is heated at constant volume, all heat is converted to increase in internal energy. The workdone is zero, i.e., Cv = ∆U for 1ºC (or 1 K) rise in temperature for 1 mole of a gas.
For Di-atomic gas:
Internal energy per mole =
Cv = ΔU 5 5 5 = R (T + 1) − RT = R 2 2 2 5 Cv = R 2 7 Cp = R 5 CP 7 = Cv 3
…(as Cp – Cv = R)
∴
5 RT, 2
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2.22
Kinetic Theory of Gases and Thermodynamics
2.15 THERMODYNAMIC PROCESSES In a thermodynamic process, the initial and final states are important, but the manner in which the system is taken from initial to final state must also be taken into account. Hence, it is important of distinguish between the following processes:
Isothermal Process
rio s_ in fo er si ta rio s
e/
Heat taken in
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
na .c om
A gas isothermal process is one in which the temperature remains constant.
iv
t.m e/
P
ht tp
s: //
ht tp s:
un
//t
ht
A gas undergoing an isothermal process should remain in perfect thermal contact with a constant temperature heat reservoir. This obeys BOYLE’S LAW, during an isothermal process i.e., pV = constant
V
As the temperature remains constant, internal energy of the system remains constant i.e., ∆U = 0 (For isothermal process).
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Kinetic Theory of Gases and Thermodynamics
2.23
ar a
From First Law: ∆Q = W
rio s_ in fo
na .c om
Heat given out
iv
er si ta rio s
e/ .m p
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
(i) Isothermal expansion: If the gas expands, it takes heat from the heat reservoir and does positive work. In equation ∆Q = W, both ∆Q and W are positive. The process takes place slowly. As the pressure of the piston is reduced, the gas tries to expand. In expansion, it tends to cool down, but the heat reservoir gives heat input to keep temperature constant.
V
(ii) Isothermal compression: If the gas is compressed isothermally, it gives heat to the reservior and the work done is negative, i.e., in the equation ∆Q = W, both ∆Q and W are negative.
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Kinetic Theory of Gases and Thermodynamics
To do a perfectly isothermal process, pressure is increased in very small steps. All the gas is compressed slowly, its temperature tends to rise, but the reservoir extracts the heat from the system to keep the temperature constant.
Work Done by Isothermal Process W = ∫ dW =
V2
∫ pdV
V1
V2 V1
na .c om
= nRT ln
ar a
m
∆Q = Infinity n∆T
er si ta rio s
un iv
//w
w
er s
Adiabatic Process
ita
w
.ja
Cisothermal =
rio s_ in fo
This is also the amount of the heat given, as the volume of the gas is changed from Vinitial to Vfinal. As the change in temperature is zero, the molar heat capacity of such a process is
ht tp s:
un
//t
p
iv
e/ .m
tp
s:
In an adiabatic process, no heat enters or leaves the system. Such a situation is achieved by heavily insulating the system.
ht
s: //
t.m e/
p1
A
Adiabatic (Steeper) B
p2
ht tp
2.24
o
V1
V2
V
p
Adiabatic A From the first law of thermodynamics: p (Steeper) ∆Q = ∆U + ∆W B 1
p2
In adiabatic process ∆Q = 0 ⇒ or
∆U + ∆W = 0 ∆U + pd V = 0
...(i)
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V1
V2
V
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Kinetic Theory of Gases and Thermodynamics
2.25
1dU n dT
We have, CV =
∆U = n CV d T
or
Thus, from equation (i) ...(ii)
nCV d T + Pd V = 0
Since the gas is ideal, therefore pV = nRT or pdV + VdP = nRdT pdV + Vdp nR C where, γ = P CV Let the initial pressure and volume be pi and Vi respectively and the final pressure and volume be pf and Vf respectively. Then dT =
fo
s_ rio ta si
γ
s ta
un
er
e/
iv
pi pf
un
=
tp
e/
s:
Viγ
//t .m
V fγ
rio
pi
p V ln f = ln i p Vi f
s:
tp ht
or
pf
si
w //w
or
Vi
= −ln
er
Vf
w
γln
or
iv
.ja
Vi
dV dp =−∫ V p pi
ar
γ
m
∫
an
pf
Vf
in
a.
co
m
or
.m
Relation between p and T:
//t
ht
Thus, pVγ remains constant in a reversible adiabatice process.
ht
tp
s:
nRT p γ Putting pV = Constant, we have
We have, pV = nRT V = γ
nRT p = Constant p
p1–γ T γ = Constant
Tγ = Constant p γ−1
Relation between V and T: We have, pV = nRT
or
p=
nRT V
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Kinetic Theory of Gases and Thermodynamics
nRT γ Putting in equation pVγ = Constant, we have V = Constant V γ–1 or TV = Constant The adiabatic curve in p – V diagram is much steeper than the isothermal curve.
ar a
p
m
+ve work
er s un iv e/
er si ta rio s
ita
.ja w w //w s:
V
un
//t
iv
.m
tp
rio s_ in fo
na .c om
(i) Adiabatic expansion: As the pressure on the piston is reduced, the gas pusshed the piston upwards due to the expansion, the temperature of the system falls down. The molecules of the gas lose speed after colliding with a receding (moving away piston). The loss in the speed cause a fall in the temperature. In the equation of the Ist Law: ∆U + ∆W = 0 ∆U is negative and W is positive (Internal energy of the gas is being converted into work done).
ht
(ii) Adiabatic compression: As the presence on the piston is increased, it moves downwards compressing the gas. Due to the compression, the temperature of the gas rised. The molecules of the gas gain speed after colliding with an incoming piston. The gain in speed of gas causes a rise in temperature. Work is done on the gas and its internal energy increased.
ht tp
s: //
t.m e/
ht tp s:
2.26
p
–ve work
V
In the equation, ∆U + ∆W = 0, ∆U is positive and ∆W is negative. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.27
Work Done in an Adiabatic Process Let, a sample of gas has initial pressure p1 and initial volume V1. In an adiabatic process, the pressure and volume change to p2 and V2, then pVγ = p1 V1γ = p2 V2γ = k p=
Thus,
...(i)
k Vγ
The work done by the gas in the process is V2
∫
pdV =
V1
V2
1 k
k
∫ V dV = 1 − γ V
γ−1 2
γ
V1
na .c om
From equation (i),
k k = p2 and γ = p1 V2γ V1
k V1γ−1
p V − p2 V2 1 ( p2 V2 − p2 V1 ) = 1 1 γ −1 γ −1
ar a
W=−
m
Thus,
−
rio s_ in fo
W=
er s
w
ita
.ja
Isochoric Process
er si ta rio s
e/
P ∝ T.
.m
ht
tp
s:
un iv
//w
w
A process in which the volume remains constant is known as isochoric process. For an isochoric process, an ideal gas is taken in a container with rigid walls so that neither expansion or compression occurs. The gas obeys the relation
un
//t
iv
In isochoric process, ∆W = 0, so, for Ist law of thermodynamics
t.m e/
ht tp s:
∆Q = ∆U.
ht tp
s: //
(i) If heat is given to the gas, its temperature and pressure rises. Its internal energy also rises. ∆U and ∆Q are both positive. (ii) If heat is taken out from the gas, its temperature and pressure falls. Its internal energy also decreased. ∆Q and ∆U are both negative. p
O
A
B
B
A
w=0
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Kinetic Theory of Gases and Thermodynamics
Isobaric Process A process in which the pressure is kept constant is called as isobaric process. For isobaric process, the external pressure on the piston is kept constant. The gas obeys CHARLES’ LAW:
un iv
//w
w
V ∝ T. W = p(V2 – V1) = nR (T2 – T1)
e/
s:
er si ta rio s
er s
w
ita
.ja
Heat in by the source
Work done,
tp
rio s_ in fo
m
ar a
na .c om
p
ht tp
p
s: //
t.m e/
ht tp s:
un
//t
iv
.m
(i) Isobaric expansion: If heat is given to the gas, isobaric expansion occurs. The volume and the temperature both rise. The gas expands doing positive work. In equation: ∆Q = ∆U + ∆W, all the three terms ∆Q, ∆U and ∆W are positive in the isobaric expansion.
ht
2.28
+ve work A
B
V1
V2
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V
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Kinetic Theory of Gases and Thermodynamics
2.29
(ii) Isobaric compression: If the heat is taken out the gas, isobaric compression occurs. The temperature falls and the gas contracts causing negative work. In the equation: ∆Q = ∆U + W, all the three terms ∆Q, ∆U and W are negative in the isobaric compression. P –ve work B
m
V1
V2
V
er si ta rio s
w
er s
w
ita
.ja
A
rio s_ in fo
ar a
na .c om
P
e/
s:
un iv
//w
2.16 EXPRESSIONS FOR ∆U, W AND ∆Q FOR DIFFERENT PROCESSES INTERNAL ENERGY CHANGE: ∆U
iv
//t
ht
.m
tp
∆U = nCV∆T for every process. It is the state function and depends only the difference of initial and final temperature.
un t.m e/
ht tp s:
∆U = nCV(T1 – T2)
Work Done: (W)
Isothermal process
W = 2303 nRT log10 V2
ht tp
s: //
Process
Heat Exchanged (∆Q)
V1
∆Q = 2.30 nRT log
V2 V1
p1V1 − p2 V2 nR(T2 − T1 ) = ∆Q = 0 γ −1 γ −1 p V − p2 V2 nR(T2 − T1 ) W= 1 1 = γ −1 γ −1
Adiabatic process
W=
Isochoric process
W=0
∆Q = nCV∆T (use definition of CV)
Isobaric process
W = p∆V = p(V2 – V1) W = nR(T2 – T1)
∆Q = nCP∆T (use definition of CP)
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2.30
Kinetic Theory of Gases and Thermodynamics
2.17 RELATION FOR THERMODYNAMIC PROCESS (STATE VARIABLE) Isochoric process: Volume remains constant p∝T p1 p2 = T1 T2
⇒
⇒
Isothermal process: Temperature remains constant 1 p∝ ⇒ V ⇒ p1V1 = p2V2
V1 V2 = T1 T2
⇒
rio s_ in fo
na .c om
Isobaric process: Pressure remains constant ⇒ V∝T
ita un
xample
s: //
∴
)
ht tp
) Solution Let N be the number of molecules.
Given: V = 250 cm3, p = 10–3 mm of Hg, T = 300 K
or
Here, p = 760 mm of Hg, and T0 = 273 K
10−23 × 250 × 6.02 × 1023 × 273 760 × 22400 × 300 = 8.045 × 1015 molecules N=
2. An excellent laboratory vacuum is 10–10 mm of mercury. How many molecules of a gas remain per cubic centimeter at 20ºC in this vacuum?
10−3 250 N 300 × = × 760 22400 6.02 × 1023 273
) Solution 76 cm of Hg pressure = 76 × 13.6 ×
...(ii)
p V N T × = × p0 V0 N 0 T0
)
p0 V0 = N0 k T0
∴
∴ pV = NkT where k is constant ...(i) Now at NTP, one mole of air occupies V0 = 22400 cm3 and contains N0 = 6.02 × 1023 molecules.
S
Dividing equation (i) by equation (ii) we get
t.m e/
ht tp s:
1. An electric bulb of volume 250 cm3 was sealed off during manufacture at a pressure of 10–3 mm of Hg at 27ºC. Find the number of air molecules in the bulb.
//t
ht
WORKED OUT E
iv
e/ .m
tp
s:
un iv
//w
w
T1p11 – γ = T1p21 – γ,
er si ta rio s
er s
w
.ja
T1V1γ = T2V2γ–1,
m
ar a
Adiabatic process: Now heat transfer takes place. p1V1γ = p2V2γ,
980 dyne cm–2
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Kinetic Theory of Gases and Thermodynamics
= 101.3 × 104 dyne cm–2 = 101.3 × 10 Nm 3
The amount of gas is given by
–2
6 2 2 pV (1.6 × 10 N/m ) × ( 0.0083 m ) = = 377 K RT (8.3 J/mol − K )( 300 K )
(∵ 10 dyne cm /µm )
n=
1 v2 Pressure, p = Nm 0 3 V
= 377 K As the gas is kept in a closed vessel, its volume is constant, therefore
2
2 N1 2 m0 v 3 V2 N = kT V
or
m
=
fo
co
in
p1V1 p2 V2 = T1 T2
s_
an
io
ar
w
ta r
e/ un
=
.m
ht tp
677 × 1.6 × 106 N/m 2 300 = 3.6 × 106 N/m2
iv
s: //w
er si
.ja
w
R 8.3 = = 1.38 × 10−23 JK −1 N 0 6.02 × 1023 T = 20° C = 293 K.
er si ta
m
Here, V1 = V2 T ∴ p2 = 2 p1 T1
s
p = 10–10 mm = 10–11 cm of Hg
a.
Final temperature = 300 K + 377 K = 677 K We have,
101.3 × 103 = × 10−11 Nm 2 76 R 8.3 k= = = 1.38 × 10−23 JK −1 N 0 6.02 × 1023
2.49 × 104 J = 377 K ( 5.3 mole ) (1.5 × 8.3 J / mol − K )
3 1 ..... m 0 v 2 = kT 2 2 N p ∴ = V kT where,
∆Q = nCV∆T ∆Q ∆T = nCV
=
rio
–2
k=
2.31
iv
)
)
ht
tp
s:
//t .m
ht
e/ un
s: //t
tp
N 101.3 × 103 × 10−11 1 1 6 4. ×A10sample of ideal gas (γ = 1.4) is heated at = × × = 3.296 3 −23 m 76 1.38 × 10 293 constant pressure. If an amount of 140 J 3 of heat is supplied to the gas, find: × 10−11 1 1 × × = 3.296 × 106 −23 (i) The change in internal energy of the 6 1.38 × 10 293 N 6 gas. or = 3.296 × 10 cm3 (ii) The work done by the gas. Note: Since, this is a statistical calculation, the Solution Suppose, the sample contains n accuracy of the computed result is limited. How- ) moles. Also, suppose the volume changes ever, the order = 106 is reliable. from V1 to V2 and the temperature changes from T1 to T2. 3. An ideal gas has a molar heat capacity pressure CP = 2.5 R. The gas is kept in The heat supplied is given by a closed vessel of volume 0.0083 m3, at a ∆Q = nCP(T2 – T1) temperature of 300 K and a pressure of (i) Change in internal energy 1.6 × 106 N/m2. An amount of 2.49 × 104 J of heat energy is supplied to the gas. Cal∆U = nCV(T2 – T1) culate the final temperature and pressure C = V nCP (T2 − T1 ) of the gas. CP C 140 J ) Solution We have, CV = CP – R = V ∆Q = = 100 J CP 1.4 = 2.5 R – R = 1.5 R
∴
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2.32
)
) Solution
(i) The process is shown in the figure. During the part ab, since the pressure is constant, we have pa Va pb Vb = Ta Tb or
Tb =
Vb Ta Va
= 2aTa = 600 K
(i) What is the final pressure? (ii) How much work is done by the gas in the process?
5. A sample of gas (γ = 1.5) is taken through an adiabatic process in which the volume is compressed from 1600 cm3 to 400 cm3. If the initial pressure is 150 kPa,
(ii) Work done by gas ΔW = ΔQ – ΔU = 140 J – 100 J = 40 J
Kinetic Theory of Gases and Thermodynamics
p
)
) Solution
in
Isothermal
s_ ta
V
s
40l
ita rio
rs
ve ni
paVa = pbVb
pbVbγ = pcVcγ
or
pa(2Va)γ = pcVcγ
Dividing equation (ii) by equation (i), we get 2γ(Va)γ – 1 = (Vc)γ – 1 γ
or
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...(i)
And from the adiabatic curve bc,
ht
(i) Sketch the process in a p–V diagram. (ii) What is the final volume and pressure of the gas? (iii) What is the work done by the gas?
(ii) From the isothermal ac,
20l
si
.m 3
tially at 27º C and occupy a volume of 20 litres. The gas is first expanded at constant pressure untill the volume is doubled. Then it undergoes an adiabatic change untill the temperature returns to its initial value.
tp s:
6. Two moles of Helium gas ( γ = ) are ini
e/ u
s:
240 J − 480 J = − 480 J 0.5
//t
=
ht tp
1.5 − 1
//t .
(150 kPa ) (1600 cm3 ) − (1200 kPa ) ( 400 cm3 )
5
m
=
c
During the part bc, the gas is adiabatically returned to the temperature Ta. The point a and point c are on the same isothermal. Thus, we draw an adiabatic curve bc and an isothermal from a and look for the point of intersection c. That is the final state.
e/ un
s:
p1V1 − p2 V2 γ −1
ht tp
W=
iv
//w
w
er
w
.ja m
1600 = (150 kPa) = 1200 kPa 3 400 1600 2 = (150 kPa) = 1200 kPa 400 (ii) Work done by the gas in an adiabatic process
rio
ar
3 2
Adiabatic
fo
a. c
γ
an
V Thus, p2 = p1 1 V2
b
a
om
For an adiabatic process p1V1γ = p2V2γ
(i)
Vc = 2 γ−1 Va 4 2 Va
= 113 litres
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Kinetic Theory of Gases and Thermodynamics
p′V1 RT′ p′V2 and n′2 = RT′ We have, n1 + n2 = n′1 + n′2 n′1 =
From equation (i),
p1V1 p2 V2 p′V1 P′ V2 + = + RT1 RT2 RT′ RT′
o
rio
si ta
si ta rio
er
e/ u
ni v
m
And Internal energy of the gas after the connection C p′ (V1 + V2 ) = V R
s: //t
.m
e/ u
//t .
ht tp s:
s
er
ni v
//w
ht tp s:
7. Two vessels of volume V1 and V2 contain the same ideal gas. The pressure in the vessels are p1 and p2 and the temperatures are T1 and T2 respectively. The two vessels are now connected to each other through a narrow tube. Assuming that no heat is exchanged between the surroundings and the vessels, find the common pressure and temperature attained after the connection.
Internal energy of the gases before the connection C pV C p V = V 1 1+ V 2 2 R R
w w
Net work done by the gas = 4980 J + 7470 J = 12450 J.
As the vessels have fixed volume, no work done by the gas plus the vessels system. Also, no heat is exchanged with the surroundings. Thus, the internal energy of the total system remains constant. The internal energy of an ideal gas is pV U = nC V T = C V R
in f
an a. co
ar
.ja m
nR(T2 − T1 ) 4980 = = 7470 J 5 γ −1 −1 3
ht tp
Neglecting the change in internal energy of the vessels (the heat capacity of the vessels is assumed negligible). C V p1V1 C V p2 V2 CV p′( V1 + V2 ) + = R R R
)
) Solution
T T (V + V2 ) T′ = 1 2 1 P′ p1V1T2 + p2 V2 T1
m
or
s_
= 2 mole × (8.3 J/mol – K) × (600 K – 300 K) = 4980 J Work done in the adiabatic part bc p V − pc Vc = b b γ −1
p′ 1 p1V1 p2 V2 = + T′ V1 + V2 T1 T2
or
= pa Vb – pa Va = nRT2 – nRT1
or
T
T1T2 p1V1 + p2 V2 p1V1T2 + p2 V2 T1
8. A thermodynamic system is taken through the cycle a b c d a. (i) Calculate the work done by the gas during the parts ab, bc, cd and da.
If p′ and T′ are the common pressure and temperature after the connection is made, the amount are
p1V1 RT1
p1V1 + p2 V2 V1 + V2
From equation (i), T′ =
The amount of gas in vessel 1 is n1 =
p′ =
p2, V2.T2
p1, V1.T1
or
= 4.4 × 104 Pa (iii) Work done by the gas in the part ab = pa (Vb – Va)
pV nRT pc = a a = Vc Vc 2 mol × (8.3 J / mol − k )(300 k ) = 113 × 10−3 m3
=
2.33
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2.34
Kinetic Theory of Gases and Thermodynamics
of 42 J heat is given to the gas. If the temperature rised through 2º C, find the distance moved by piston. Atmospheric pressure = 100 kPa.
(ii) Find the total heat rejected by the gas during the process.
d
c
100 kPa
a
b
) Solution Change in internal energy of the gas )
200 kPa
300 cm3
100 cm3
∆U = 1.5 nRT = 1.5 (1 mole) (8.3 J/mol – K) (2K) = 24.9 J Heat given to the gas = 43 J Work done by the gas is ∆W = ∆Q – ∆U = 42 J – 24.9 J = 17.1 J If the distance moved by the pistion is x, then the work done ∆W = (100 kPa) (8.5 cm2)x = 17.1 J 5 Thus, (10 N/m2) (8.5 × 10–4 m2) x = 17.1 J or x = 0.2 m = 20 cm
ar a
ita
un
s: //
= 2.303 × 60 log 8 (101) J = 1.26 × 104 J
11. The U-tube of uniform cross-section contains mercury to the depth shown. the left side of the tube is closed at the top and the right side is connected to a good vacuum pump (i.e., pressure is made zero). The temperature at air in the closed tube is raised from 27º C to 111º C. How far the level of mercury falls on the left side? Atmospheric pressure is 75 cm of Hg.
or
ht tp
9. The internal energy of a mono-atomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 cm2. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total
V2 V1 V W = 2.303( p1V1 ) log 2 V1 (using p1V1 = nRT) p = 2.303 (20 × 3) log 1 lt. atm p2 W = 2.303nRT log
t.m e/
ht tp s:
perature T,
iv
e/
) Solution In isothermal process at tem-
.m
3
//t
ht
tp
3
er si ta rio s
un iv
s:
= (200 kPa) (100 cm – 300 cm ) = – 40 J The work done during da is zero as the volume does not change. (ii) Total work done by the system during the cycle a b c d a. ∆W = 20 J – 40 J = – 20 J Change in the internal energy, ∆U = 0, as the initial state is the same as the final state. Thus, ∆Q = ∆U + ∆W = – 20 J So, the system rejects 20 J of heat during the cycle.
er s
w
w
//w
∫ pdV d
.ja
m
c
=
10. How much work is done by an ideal gas in expanding isothermally from an initial volume of 3 litres of 20 atm to a final volume of 24 litres?
a
= (100 kPa) (300 cm3 – 100 cm3) = 20 J The work done during bc is zero as the volume does not change. The work done during cd,
∫
= (100 × Pa) dV
)
a
b
na .c om
)
b
(i) Work done during the part ab = ∫ pdV
rio s_ in fo
) Solution
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Kinetic Theory of Gases and Thermodynamics A
2.35
pressure of 1 × 105 N/m2. The total mass of the mixture is 28 gm. If molar masses of neon and argon are 20 and 40 gm/mol respectively, find the masses of the individual gases in the container, assuming them to be ideal (universal gas constant = 8.314 J/mol/L).
B
50 cm
2x
)
) Solution Given that m1 + m2 = 28 gm
)
therefore pA = pB = 0 + 2x ρg Initial pressure, patm = 75 cm Initial volume = 50 Initial temperature = 27º C = 300 K Final pressure = pA = 2xρg Final volume = (50 + x) And, final temperature = 111º C
From gas law pV = nRT
rio s_ in fo 3
er s
…(iii)
er si ta rio s
13. A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air at the same pressure p. When the tube is held at an angle of 60º with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure p in centimetres of mercury. (The temperature of the system is kept at 30º C).
iv
m1 m2 + × 2494.2 20 40
40m1 + 20m2 = 641.48
un
) Solution
or x + 50x – 32 × 75 = 0 Solving, we get x = 30 cm. 2
–5 cm
l
l
p
p
12. A closed container of volume 0.02 m3 contains a mixture of neon and argon gases, at a temperature of 27º C and
2 × 10 =
t.m e/ s: //
2 x (50 + x) = 75 64
or
2
Solving equations (i) and (ii), we get m1 = mass of neon = 4.08 gm and, m2 = mass of argon = 23.92 gm
un iv e/ .m //t
2 xρg (50 + x) 75ρg × 50 = 384 300
1
)
(75 ρg ) × 50 300
ht tp
384
=
3
ita
.ja w ht tp s:
pA ( 50 + x )
or
2
3
ar a
or
p1V1 p2 V2 = T1 T2
or
2
or
w //w s: tp
ht From the gas law
5
m1 m2 + (1 × 10 N/m )(0.02 m )= 20 8.314 × 300 40 5
m
2x
(1×10 N/m )( 0.02 m ) = m20 + m40 8.314 × 300
na .c om
or
pB = 0
50 cm pA
...(i)
) Solution Since pressure of both sides is equal,
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(a)
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2.36
Kinetic Theory of Gases and Thermodynamics
From given condition
)
) Solution p Q
p2 –5
46
cm
p1
cm pincident
co m
p = 5 cos 60° = 2.5 cm
s
rio
ta
si
er un
= 4.695 × 10–24 kg ms–1
Number of collision per sec = 1023 ∴ Change in momentum of the molecules per sec = Rate of gain in momentum by the wall normal to it = Normal force on the wall = 4.695 × 10–24 × 1023 kg ms–2 = 0.4695 N Area of the wall = 2 cm2 = 2 × 10–4 m2 0.4695 ∴ Pressure on the wall = Nm −2 2 × 10−4 = 2.47 × 103 Pascal
14. The mass of a H2 molecule is 3.32 × 10–24 gm. If 1023 hydrogen molecules per second strike a fixed wall of area 3 cm2 at an angle 45º to the normal and rebound elastically with a speed of 105 cms–1, calculate the pressure exerted on the wall.
= 75.4 of Hg.
2 p = 2 × 3.32 × 10−24
e/ .m //t s:
tp ht
p2 × 46 45.25 2.5 × 44.5 × 46 = 1.5 × 45.25
fo
in
s_
s:
tp
ht
ta
si er
//t
ht
( p1 − p2 ) 44.5 2.5 × 44.5 = 1.5 1.5
p=
iv
p2 =
Now, p2 × 46 = p × 45.25
iv
un
or
∆p =
.m
p1 − p2 46 − 44.5 1.5 = = p2 44.5 44.5
∴
= 3.32 × 10–27 × 103 kg ms–1 = 3.32 × 10–24 kg ms–1 Change in momentum normal to the wall per molecule
e/
tp
s: //
w
w
w
∴ (p1) (44.5) = plLeft = plRight = p2 × 46 [from Fig. (b)] p1 46 ∴ = p2 44.5 ∴
R
rio
m
.ja
...(i)
From Fig. (a) pleft = pright p1 = p2 + 2.5
∆p
PQ → Incident momentum direction QR → Rebound momentum direction. Since, the rebounding is elastic and the wall is fixed. pIncident = pRebound
ar an
2l + 5 = 46 + 44.5 + 5 = 95.5 cm 90.5 = 45.25 cm l= 2
a.
(b)
15. An ideal gas is taken through a cycle thermodynamic process through four steps. The amount of heat involved in these steps are Q1 = 5960 J, Q2 = – 5585 J,
.5 44
or
prebound
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Kinetic Theory of Gases and Thermodynamics
2.37
Q3 = – 2980 J and Q4 = 3645 J respectively. The corresponding worked involved are W1 = 2200 J, W2 = – 825 J, W3 = – 1100 J and W4 respectively. (i) Find the value of W4. (ii) What is the efficiency of the cycle? )
) Solution For a cyclic process
∆U = 0
(i) Cyclic ∫ dQ = ∫ dW
)
ta
rio
s_
in
fo
s
rio
si
ta
si er
iv un
er
e/
.m
e/
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or
= P0 Al +
kl nRT1 AP0 p0 Al + kl 2 klT1 + T2 = T1 + nR AP0 = nRT1 + p0 Al + kl 2 +
ht
tp
s:
//t
ht
0
1 2 kl 2 = (100 × 103 Pa) (20 × 10–4 m2) × (10 × 10–2 m) 1 + (200 N/m) (100 × 10−4 m 2 ) 2 = 20 + 1 J = 21 J (ii) Initial temperature, T1 = 300 K. Let the final temperature by T2, then nRT1 = p0V0 kl nRT = pV2 = p0 + (V0 + Al ) A
∫
W = Fdx
un iv
.m
//t
s:
tp
16. 2.00 mole of a mono-atomic ideal gas (U = 1.5 nRT) is enclosed in an adiabatic, vertical cylinder fitted with a smooth light adiabatic piston. The piston is connected to a vertical spring of spring constant 200 N/m as shown in the figure. The area of cross-section of the cylinder is 20.0 cm2. Initially, the spring is at its natural length and the temperature of the gas is 300 K. The atmospheric pressure is 100 kPa. The gas is heated slowly for some time by means of an electric heater so as to move the piston up through 10 cm. Find: (i) The work done by the gas. (ii) The final temperature of the gas. (iii) The heat supplied by the heater.
(i) Force by the gas on the piston is F = p0A + kx where, P0 = 100 kPa is the atmospheric pressure. A = 20 cm2 is the area of the cross-section, k = 200 N/m is the spring constant, and x = the compression of spring. Work done by the gas if the piston moves through l = 10 cm is l
ht
tp s:
//w w
Net work output = 5960 – 5585 – 2980 + 3645 = 1040 J Total heat input = Q1 + Q4 = 5960 + 3645 = 9605 J 1040 ∴ η= × 100 = 10.83% 9605
w .ja
m
(ii) Efficiency of the cycle, Net work output η= Total heat input
) Solution
a. c
or
5960 – 5585 – 2980 + 3645 = 2200 – 825 – 1100 + W4 W4 = 765 J
ar an
or
om
i.e., Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4
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2.38
Kinetic Theory of Gases and Thermodynamics
300 +
B
C
D
300 K
er si ta rio s
ita
e/
T
)
) Solution Number of moles, n = 2
(ii) (∆W)Net = (∆Q)Net = 1153.4 J (iii) (∆U)Net = 0
The gas undergoes cyclic process. Since, internal energy is property of the system, the net change in internal energy during the cyclic process is zero.
18. One mole of a gas is put under a weightless piston of a vertical cylinder at temperature T. The space over the piston is atmosphere. How much work should be performed to increase isothermally the
Helium is a mono-atomic gas. 3 ∴ CV = R 2 5 CP = R 2
( ∆Q) DA = nRT ln
400 K
s: // ht tp
300 K
PD PA 2 = 2 × 8.32 × 300 ln = −3460.2 J 1 ∴ (∆W)Net = 4160 + 4613.6 – 4160 – 3460.2 = 1153.4 J
t.m e/
C
2 1
= 2 × 8.32 × 400 ln = 4613.6 J
iv un
//t ht tp s:
D
PB VC = nRT ln VB PC
= nRT ln
.m
tp ht 1 atm
(i) (∆Q)Net = (∆Q)AB + (∆QBC) + (∆QDA) (∆Q)AB = n × CP × (TB – TA) 5 = 2 × × 8.32(400 − 300) = 4160 J 2 Since Process BC is isothermal, therefore ∆U = 0 (∆Q)BC = (∆W)BC
s:
B
rio s_ in fo
er s
un iv
A
2 atm
T
na .c om
.ja
//w
w
w
m
ar a
p
400 K
Hence, according to the first law of thermodynamics, the net change in the heat energy is equal to the net work done.
1 atm
17. Two moles of helium gas undergo a cyclic process as shown in the figure. Assuming the gas to be ideal, calculate the following quantities in this process: (i) The net change in the heat energy. (ii) The net work done. (iii) The net change in internal energy. [R = 8.32 J mol–1]
A
2 atm
= 300 K + 1.325 K + 30 K = 331 K (iii) Internal energy, U = 1.5 nRT ∴ ∆U = 1.5 nR∆T = 1.5 × 2.00 × 8.3 × 31 = 772 J From the first law, ∆Q = ∆U + ∆W = (772 + 21) J = 793 J
p
20 J + 2 J 200 × 10 × 10−2 × 300 + 20 × 8.3 20 × 10−4 × 100 × 103 Pa
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Kinetic Theory of Gases and Thermodynamics
volume under the piston to twice the volume (neglect friction of piston).
)
) Solution According to the first law of
thermodynamics, ∆Q = ∆U + ∆W For the first vessel: ∆U = 0, (Since, no change in temperature) ∆Q = ∆W
)
) Solution Let A be the area of piston,
therefore F + pA = p0A or F = (p0 – p) A Work done by agent is given by
V2
0
− p )Adx
V2
V
=
− p )dV
s
rio
ta
un
iv
er si
e/ tp s
ht
= nRT [(η – 1) loge η] where, η = 2 and n = 1 W = RT [1 – log e 2]
19. Two vessels contain in each of them one mole of mono-atomic gas. The initial volume of each vessel is 8.3 × 10–3 m3. Equal amount of heat is supplied to each vessel. In one vessel, the volume of gas is doubled without change in its internal energy whereas the volume of the gas is held constant in second vessel. The vessels are now connected to allow free mixing. Find the final temperature and pressure of the combined system.
3 nRT log e 2 = n R ∆T 2 2 or ∆T = × 300 × 0.693 = 138.6 K 3 It is the change in temperature of the second vessel. Now, temperature of the gas in second vessel = T + ∆T = 300 + 138.6 = 438.6 K Let after mixing Tf and pf be the final temperature and pressure, therefore T + (T + ∆T) Tf = 2 300 + 438.6 = = 369.3 K 2
:// t.m
tp ht
= p0 (η – 1)V – nRT logeη
∴
in
si er e/
//t .m
P
s:
ht tp
s:
un
iv
//w
w
F
w
.ja
(since pV = nRT)
Po
s_
ta rio
m
V
dV V
∫ nRT
ηV
= p0 (η − 1)V −
a.
V
fo
co
∫ pdV
V
m
ηV
an
∫
p0 dV −
ar
ηV
=
Since V2 = 2 V1, therefore Q = nRT loge 2, ...(i) For the second vessel: ∆W = 0, (volume is constant) 3 Q = nC V ∆T = n R ∆T ...(ii) 2 3R Since, for mono-atomic gas C V = 2 From equations (i) and (ii), we get
0
(since, pV = nRT)
From the gas equation, pfVf = nRTf
∫ (p
V
=
dV V
∫ nRT
V1
pf =
nRT f Vf
2 × 8.3 × 369.3 2 × 8.3 × 103 + 8.3 × 10−3 = 2.46 ×105 N /m 2 =
ηV
∫ (p
∫ pdV
V1
W=
Q=
ηV
2.39
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2.40
Kinetic Theory of Gases and Thermodynamics
= 0.2 × 8.314 × 400 loge2
p
= 461 J For adiabatic expansion b → c Q=0 ∴ W = – ∆U = nCV(TH – TC) = 0.2 × 20.78 × (400 – 300) = 415.7 J For isothermal compression c → d ∆U = 0
m
a
4 = 13.3 × 10−4 × 3
fo
in
s rio
ta
si
er iv
un
e/
= nCV(TC – TH)
= 0.2 × 20.78 × (300 – 400) = – 415.7 J
s:
13.65 × 10−4 27.3 × 10−4
W=–U
.m
∴
tp ht
Vd Vc
= – 345.8 J For adiabatic expansion d → a Q=0
//t
ht
T Vc = Vb H TC
1 γ−1
er iv
un
.m
//t
s:
tp
For adiabatic expansion b → c THVbγ–1 = Tc Vcγ–1
W = Qc = nRTc log e
∴
= 0.2 × 8.314 × 300 log e
e/
s:
tp
ht
For isothermal expansion a → b paVa = pbVb pV or pb = a a = 5 × 105 pa Vb
∴
rio
s_
si ta
w w
//w
m .ja
nRTH pa 0.2 × 8.314 × 400 = = 6.65 × 10−4 m3 10 × 105
(i) Va =
ar
v
)
) Solution
a.
c
an
d
co
b
The results may be tabulated as follows:
2.5
Q
= 27.3 × 10−4 m3 a→b
461 J
461 J
0J
b→c
0J
415.7 J
– 415.7 J
c→d
– 345.8 J
– 345.8 J
0J
d→a
0J
– 415.7 J
415.7 J
Total
115.2 J
115.2 J
0J
nRTc 0.2 × 8.314 × 300 pc = = = 1.83 × 105 Pa Vc 27.3 × 10−4 nRTc 0.2 × 8.314 × 300 pc = = = 1.83 × 105 Pa Vc 27.3 × 10−4 For adiabatic compression d → a
∆U
W
4 = 6.65 10−4 × 3
2.5
= 13.65 × 10−4
T Vd = Va H TC
(ii) For entire cycle, 1 γ−1
∴
Tc Vdγ–1 = THVaγ–1
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1
2.5 20. Suppose 0.2 mole of an ideal di-atomic nRTc TH γ−1 4 −4 −4 pd = V = V = 6.65 10 × tem = 13.65 × 10 gas (γ = 1.4) undergoes dcyclea with Vd 3 TCK. The perature TH = 400 K and TC = 300 initial pressure is pa = 10 × 105 Pa and 0.2 × 8.314 × 300 = = 3.65 × 105 Pa during isothermal expansion at tempera13.65 × 10−4 ture TH the volume doubles. For isothermal expansion a → b (i) Find Q, W and ∆U from each step in ∆U = 0 the cycle. V ∴ W = Q H = nRTH ⋅ log e b (ii) Find the efficiency. Va
Q=W ∆U = 0
Total work done = 115.2 J https://t.me/universitarios
Kinetic Theory of Gases and Thermodynamics
(10 ×10 N/m )(10 m ) (500 mole) ( 8.314 JK mole ) 4
TC =
2
3
−1
−1
= 481.12 K
21. A sample of 2 kg of mono-atomic Helium ( 5 ×104 N / m2 )( 20 m3 ) (assumed ideal) is taken through the proTD = cess ABC and another sample of 2 kg of (500 mole) ( 8.314 JK −1 mole −1 ) the same gas is taken through the process = 240.50 K ADC. Given relative molecular weight of Helium = 4. (ii) No. (i) What is the temperature of Helium (iii) For the process ABC, we have in each of the states A, B, C and D? QAB = nCVΔT (ii) Is there any way of telling afterwards which sample of Helium went 3 = (500 mole) × 8.314 JK −1 mole −1 (240.56 K − 120.28 through the process ABC and which 2 3 ADC? −1 −1 went through the mole) process = (500 × 8.314 JK mole (240.56 K − 120.28 K) Write yes or no. 2 (iii) How much heat is evolved in each of = 7.5 × 105 J the processes ABC and ADC? QBC = nCPΔT
fo
s_
rio
s
io
ta r
si
QAD = nCPΔT
//t
.m
For the process ADC, we have
ht tp s:
5 Amount of helium = (500 mole) × 8.314 JK −1 mole −1 (240.56 K − 120.2 3 2 m 2 × 10 5 −1 −1 = = = mole 500 mole −1 = 500 × 8.314 JK mole (240.56 K − 120.28 K) ( ) M 4 g mol 2 (i) The temperature of gas at the states A, = 1.25 × 106 J B, C and D are QDC = nCVΔT (5 ×104 N / m2 )(10 m3 ) = 120.28 pV TA = = K −1 −1 nR (500 mole) (8.314 JK mole ) 3 = (500 mole) × 8.314 JK −1 mole −1 (481.12 K − 240.5 4 2 3 5 × 10 N/m 10 m ( )( 3 ) 2 = −1 −1 − −1 1 = 500 mole × 8.314 JK mole (481.12 K − 240.56 K) ( ) (500 mole) ( 8.314 JK mole ) 2 = 120.28 K = 1.5 × 106 J
(10 ×10
4
N / m 2 )(10 m3 )
(500 mole) ( 8.314 JK
= 240.56 K
−1
mole
QADC = QAD + QDC
TB =
−1
)
ve r
= (7.5 × 105 J + 2.5 × 106 J) = 3.25 × 106 J
e/ un i
20V(m3)
ht tp
) Solution
un i
//t .m s:
10 )
e/
:// tp s ht
D
A
w w
5 = (500 mole) × 8.314 JK −1 2
5
5 = (500 mole) × 8.314 JK −1 mole −1 (481.12 K − 240.56 2 mole −1 (481.12 K − 240.56 K) = 2.5 × 106 J QABC = QAB + QBC
ve rs
C
w
B
10
ita
.ja
p(10N/m2)
in
ar an a
m
.c o
m
QH = 461 J W 115.2 η= = = 0.25 ∴ QH 461
2.41
= (1.25 × 106 J + 1.5 × 106 J) = 2.75 × 106 J
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Kinetic Theory of Gases and Thermodynamics
22. One mole of a di-atomic ideal gas (γ = 1.4) is taken through a cyclic process starting from point A. The process A → B is an adiabatic compression, B → C isobaric expansion, C → D is an adiabatic expansion and D → A isochoric expansion. The V VA volume ratios are = 16 and C = 2 VB VB and the temperature at A is TA = 300 K. Calculate the temperature of gas at the points B and D and find the efficiency of the cycle.
Hence,
V TD = C VD
om na .c
ita
s ta rio
er si
un iv
e/
= RTB + CV(TC – TD) – CV(TB – TA)
ht
.m
W = WB → C + WC → D + WB → A
For a di-atomic gas, 5 7 C V = and C P = R. 2 2 Hence,
TD VDγ – 1 = TCVCγ – 1
Work obtained in one cycle Heat absorbed in the process B → C
= (8.314 JK–1 mol–1)
V TD = C TC VD Since, D → A is isochoric process, therefore VD = VA
or
5 W = R TB + (TC − TD − TB − TA ) 2
5 (909 K) + 2 (1818 − 791.4 − 909 + 300)K
V
Now, the work obtained in one cycle is equal to the area within the cycle ABCDA. This work is given as
γ−1
A
rs
ve
un i //t
tp s:
V TC = C TB = 2(909) = 1818 K VB
(iii) Adiabatic expansion from C to D
or
nf s_ i rio
ar a
e/
.m
//t
s:
tp
ht
VC VB = TC TB
(ii) Isobaric expansion from B to C
D
Efficiency of the cycle is defined as η=
TA = (16)1.4−1 (300)
= (3.03) (300 K) = 909 K
o
p
m
.ja
w w
//w
tp s:
ht
γ−1
(1818 K)
V TB = A VB
TC
C
B
)
or
(i) Adiabatic compression from A to B TBVBγ – 1 = TA VAγ – 1
γ−1
= (0.4353) (1818 K) = 791.4 K The given cyclic process is show in the figure.
For the expansion at constant pressure, we have
V = Constant T With this information, temperature of the gas at different stages of the cyclic process may be determined as follows:
V TC = C 16 VB
1.4 −1
2 = 16
) Solution For an ideal gas undergoing adiabatic expansion or compression, we have TVγ – 1 = Constant
γ−1
2.42
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= 16237.2 K mol–1 https://t.me/universitarios
2.43
Kinetic Theory of Gases and Thermodynamics
∆P = −
Force on the piston = (A ∆p)
and
Acceleration =
1/2
γ p A2 Angular frequency, ω = V0 M Oscillation frequency of the piston 1 = ω 2π
∴
fo
io
ta
iv
e/ un
s:
tp
ht
.m
(i) The work done by the gas. (ii) The heat rejected by the gas in the path CA and the absorbed by the gas in the path BC. (iii) The net heat absorbed by the gas in the path BC. (iv) The maximum temperature attained by the gas during the cycle.
//t
ht
x = a small displacement from the equilibrium. p = pressure of the enclosed gas at equilibrium position of the piston. Mg = p0 + A
er si
24. One mole of an ideal gas is taken round the cyclic process ABCA as shown in the figure. Calculate:
.m
tp s: //t
surroundings” must be interpreted as “thermally isolated from the surroundings”. Then the adiabatic gas law to be applied. pVγ = constant
Mg 2 γ p0 + A A V0 M
s
ta r
si
er
un
e/
s:
ht tp
)
) Solution The words “isolated from the
1 = 2π
iv
//w
w w .ja m
ar
s_
in
∴
rio
an a
.c
23. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal crosssectional area A, atmospheric pressure is P0 and when the piston is in equilibrium position. Assuming that the system is completely isolated from its surroundings. Show that the piston executes harmonic motion and find the frequency of oscillation.
om
W 16237.2 = = 0.614 Q 26451.0
γ p A2 = − x V0 M = ω2x ∴ Acceleration ∝ – x Hence motion of the piston is in simple harmonic motion.
= 26451.0 J mole–1 Hence, the efficiency of the cycle is η=
A (∆p ) M
7 = R (TC − TB ) 2 7 = × (8.314 JK–1 mole–1) 2 (1818 K – 909 K)
γp γp ∆V = − ∆x V0 V0
∴
Heat absorbed in the process B → C is given as Q = CP (TC – TB)
3po
po
B
A
C
∆V = small change in volume of the enclosed gas due to x = Ax ∆p = small change in pressure of the gas corresponding to ∆V. Then, pV0γ = (p + ∆p) (V0 + ∆V)γ = (p + ∆p) (V0γ + γ V0γ–1 ∆V + ....) As a first order approximation, i.e., neglecting the terms containing powers of small quantities, higher than 1st order.
p
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Vo
2Vo
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V
Kinetic Theory of Gases and Thermodynamics
) Solution
p=
RT , we get V p 5V0 V T = − 2 0 V2 + V0 R R
s_ in fo
si ta rio
s
ita
er s =
p0 V0 R
=
25 p0 V0 . 8 R
iv un
1 p0 V0 2
2
2 p0 5 5 p0 5 V0 + V0 V0 R 4 R 4
rio
Tmax = −
e/
5 p0 V0 + 3 p0 V0 + Q BC = p0 V0 2
(iv) The path BC is a straight line path. It is represented by the expression
2 p0 5p (2V) + 0 V0 R R
With this Tmax is given by
.m
QCA + QAB + QBC = W
25 25 − 8 + 4
25. The rectangular box shown in the figure has partition which can slide without friction along the length of the box. Initially, each of two chambers of the box has one mole of a mono-atomic ideal gas (γ = 5/3) at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of box and partition are thermally insulated. Heat loss through lead wire of heater is negligible. The gas in left chamber expands, pushing the partition until the final pressure in both chambers 243 p0 . Determine becomes 32
//t
s:
ht
tp
(iii) In a cyclic process, the change in internal energy is zero. Hence
i.e., 0 = −
//t
tp s:
ht
3 = V0 (3 p0 − p0 ) = 3 p0 V0 2
This gives Q BC =
.m
ht tp
5 V ∆p = R 0 2 R
∂T =0 ∂V
5 which gives V = V0 . 4
e/
s:
The path AB is an isochoric expression of one mole of an ideal gas from pressure p0 to 3p0. The heat released in this process is
−
To determine Tmax, we set
un iv er
//w w
w
5 5 = p0 (V0 − 2V0 ) = − p0 V0 2 2
Q2 = n CV ∆T
− 2 p0 V + 5 p0 V0
Replacing p =
a. co m
ar an m
.ja
3 p ∆V = R 0 2 R
or
Q1 = n CP ∆T
(ii) The path CA is an isobaric compression of one mole of an ideal gas from volume 2 V0 to V0. The heat released in this path is
−2 p0 (V − 2 V0 ) V0
=
= p0V0
)
3 p − p0 p − p0 = 0 (V − 2 V0 ) V0 − 2V0
(i) Work done by the gas during a cyclic process is equal to the area enclosed by its P–V diagram. In the present case, W = area of ∆ABC 1 = (AC) (AB) 2 1 = (2 V0 − V0 ) ( 3 p0 − p0 ) 2
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Substituting in equation (i), we get
⇒
x = 0.7037 V0
243 p0 (V0 − x) TR 2.25 T0 × = = 32 p0 V0 T0 T0
Then,
pL VL TL × = p0 V0 T0
If pL, VL and TL are the final pressure, volume and temperature of left chamber respectively.
(i) The final temperature of gas in each chamber.
But, pL = pR =
2.45
VR = V0 – x
rio s_ in fo
ita
243 P0 VRγ 32 32 γ VRγ = V0 243
p0 V0γ =
or
er si ta rio s ∴
iv
er s
TL = 12.937 T0
(ii) Since p0 V0γ = pR VRγ
2 3
3
= × V0 =
8 V0 27
Work done by the gas in right chamber
1
8 32 γ V0 VR = × V0 = 27 243
un
or
=
p0 V0 − pR pR γ −1
=
nR (T0 − TR ) γ −1
1× 8.3(T0 − 2.25T0 ) 5 −1 3 = – 15.562 T0 J mol–1 =
Let x be change in volume in each chamber due to displacement of the partition.
⇒
For one mole of a gas: pV = nRT, and n=1 pR VR TR ∴ × = …(i) p0 V0 T0
∴
243 V0 (1 + 0.7037) TL × = 32 V0 T0
t.m e/
ht tp
= 2.25T0
243 TR = T0 32
5 −1 3 5/3
γ−1 γ
s: //
ht tp s:
//t
γ−1 γ
1 243 × p0 TR = T0 p0 32
TR pR = T0 p0
.m
tp
ht
Then
or
e/
s:
un iv
//w
w
w
.ja
m
ar a
(i) As the gas in the left chamber is heated, it expands and the partition moves to the right. There by expansion takes place in the left chamber and compression in the right chamber. Let pR, VR and TR be the final pressure, volume and temperature in right chamber. No heat transfer takes place in right chamber. Hence there is change is adiabatic.
)
) Solution
243 p0 (V0 + x) TL × = 32 p0 V0 T0
∴
na .c om
(ii) The work done by the gas in the right chamber.
243 p0 32
(negative sign shows that the work done on the gas.)
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p2 V2 p1v1 − nR nR
=
where γ =
in fo
s_
V γ−1 p1V1 1 − 1 R V2
s
γ=
C p, m =
5 R 2
5 3
un iv
e/
.m
//t
s:
Since the volume is held constant, work done by the gas on heating at constant volume, therefore W2 = 0
Total work don by the gas, W = W1 = W 2 2 3 p1V1 V1 3 − =− 1 2 V2
(b) Total change in internal Energy Change in internal energy in adiabatic compression, as derived above,
3 R, and 2
2 3 p1V1 V1 3 − 1 Hence, W1 = − 2 V2
.
Cv , m
rio
ita
er s
∴
ht
tp
Cv , m
Cv , m p1V1γ γ V2 − p1V1 R V2
rio ita
ni e/ u .m //t s:
The curve A to B represents the adiabatic compression of the gas from the volume V1 to V2. In this process the pressure of the gas increases p1 to p2. The line B to C represents increase in pressure of the gas as a result of giving here Q to the gas at constant volume. In this process, the pressure of the gas increases from p2 to p3. (ii) (a) Total work done by the gas Work done by the gas in adiabatic compression. In an adiabatic process, since Q = 0, therefore from the first law of thermodynamics, ∆U1 = – W1 or W1 = ∆V1 = – Cv∆T = – Cv (T2 – T1)
=
Cv , m =
ve
w ht tp
V1
Cp,m
For a mono-atomic gas,
A(p1, V1)
//w tp s: ht V2 V
W1 =
rs
w
p
…(i)
From equation (i),
om
.ja
B(p2, V2)
( p2 V2 − p1V1 )
P1 V1γ = P2 V2γ
an a. c
m ar
C(p3, V3)
R
For a gas underdoes adiabatic process
(i) Figure displays the p-V diagram of the gas undergone the given two processes.
)
) Solution
Cv , m
= − Cv
26. Two moles an ideal mono-atomic gas initially at pressure p1 and volume V1 undergo an adiabatic compression until its volume is V2. Then, the gas is given heat Q at constant volume V2. (i) Sketch the complete process on a p–V diagram. (ii) Find the total work done by the gas, total change in its internal energy and the final temperature of the gas. [Give your answer in terms of p1, V1, V2, Q and R].
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2 3 p1V1 V1 3 − 1 ∆U1 = 2 V2
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Kinetic Theory of Gases and Thermodynamics
at constant pressure until the volume is doubled. Then, it undergoes an adiabatic change until the temperature returns to its initial value.
2 3 p1V1 V1 3 − + Q. = 1 2 V2
(i) Sketch the process on a p-V diagram. (ii) What are the final volume and pressure of the gas? (iii) What is the work done by the gas?
Change in internal energy on heating the gas at constant volume ΔU2 = Q Total change in the internal energy of the gas ΔU = ΔU1 + ΔU2
) Solution
o
nf
s_ i ta si
s io ta r si er iv
un e/
ht
tp
.m s: //t
ht
Change in temperature on heating the gas
3
iv
e/ un
2/3
Q = CV ΔT = CV (T3 – T2)
V2
V1
∴ p1 = =
2 × 8.3 × 300 = 2.49 × 105 Nm −2 20 × 10−3
or
∴
∴
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n RT1 V1
Now, V ∝ T
27. Two moles of helium gas (γ = 5/3) are initially at temperature 27ºC and occupy a volume of 20 litres. The gas is expanded
V
p1V1 = nRT1
2
p1V1 V1 3 Q T3 = + . (3 mole)R (2 mole)R V2
V3
Process 1 → 2 is isobaric expansion
2
p V V 3 Q Q + T2 = + 1 1 1 or T3 = CV nR V2 3 nR 2 Since n = 2, therefore
2
1
er
w
tp s: //t .m
ht
rio
p1
w
tp s
://
− 1
2/3 p1V1 p1V1 V1 − 1 + nR nR V2
pV V = 1 1 1 nR V2
m
a. an ar
2/3
2/3 3 p1V1 V1 − 1 T2 − T1 + 3 V 2 nR 2 2
=
p
m
3 p1V1 V1 2CV V2
.ja
T2 − T1 =
w
or
∆U1 CV
co
therefore, ∆T =
(i) V1 = 20 × 10–3 m3 T1 = 300 K n = 2 moles 5 γ= 3
(c) Final temperature of the gas Change in temperature in adiabatic compression. Since, ΔU = CVΔT
)
2.47
V1 V2 = T1 T2 T2 = T1 ×
V2 2V = 300 × 1 V1 V1
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2.48
Kinetic Theory of Gases and Thermodynamics
s
…(iii)
...(iv)
where, K is constant and is given as K = pA TA = (p1) (2T1) = 2 p1 T1
...(v)
pV = nRT and p T = K
Eliminating T, we get
K …(vi) pV = nR or p 2 V = nRK P (i) The expression of work done by the gas is W = ∫ pdV In the process, we get WAB =
VB
∫ pdV
VA
(ii) The heat absorbed or released by the gas in each of the process.
rio
ita
rs
pT = K
(i) The work done on the gas in the process AB.
nRTC nR(2T1 ) n RT1 = = PC 2 P1 P1
In the process AB, we will have
ht
nRTB nR(2T1 ) 1 n RT1 = = …(ii) PB P1 2 P1
iv e
e/ un
:// t.m
tp s
VB =
It is given that during the process AB,
28. Two mole of an ideal mono-atomic gas is taken through a cycle ABCA as shown in the p – T diagram. During this process AB, pressure and temperature of the gas vary such that pT = constant. If T1 = 300 K, calculate:
fo rio
si ta
−3
(ii) Final volume, V3 = 113.14 × 10–3 m3 Final pressure, p3 = 0.44 × 105 Nm–2 (iii) Total work done by the gas = W = (W)1–2 + (W)2–3 = 4980 + 7472.8 = 12452.8 J.
nRTA nR(2T1 ) 2n RT1 = = …(i) PA P1 P1
VC =
tp
ht
in
)
iv
un
e/
5
( 5 / 3) − 1
= 7472.8 J.
s_
an ar
m .ja
( 2.49 ×10 ) ( 40 ×10 ) − ( 0.44 ×10 ) (113.14 ×10 ) −3
VA =
ht
p2 V2 − p3 V3 γ −1
s: //t .m
tp s
://
w
= 0.44 × 105 N/m 2
(W) 2−3 =
T
2T1
) Solution The volumes of the gas at three states
er
5/3
T1
A, B and C are as follows:
w w
40 113.14
= 2.48 × 105
a.
p3 = p2
∴
V3 V2
γ
A
co m
Now, p2 V2γ = P3 V3γ
C
P1
−1
V3 3 T2 600 = =2 = V T1 300 2
V3 = V2 × (2)3/2 = 40 × 10–3 (2)3/2 = 113.14 × 10–3 m3
∴
5
B
2P1
∴
=
P
Given, T2 V2γ−1 = T3 V3γ−1−1 ,T3 = T1 5
Give answers in terms of the gas constant R.
Work done during process 1 → 2, (W)1–2 = p × ∆V = 2.49 × 105 × (40 – 20) × 10–3 = 4980 J Process 2 → 3 is adiabatic expansion T2 = 600 K p2 = p1 = 2.48 × 105 N/m2 V2 = 40 × 10–3 m3
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=
VB
∫
VA
nRK dV V
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Kinetic Theory of Gases and Thermodynamics
= (3 mole) (R) (2 T1 – T1) = (3 mole) (R) (300 K) = (900 mole K) R QBC = ∆UBC + WBC = (900 mole K) R + (600 mole K) R = (1500 mole K) R The positive sign implies that the heat is absorbed in the process BC. The process CA takes place at constant temperature. Hence,
VB
nRT dV V VA
∫
ita
= (2 mole) (R) (2 × 300 K) In 2 = (1200 mole K) R ln 2 ∆UCA = 0 QCA = ∆UCA + WCA = 0 + (1200 mole K) R ln 2 The positive sign implies that the heat is absorbed in the process CA.
un
29. An ideal mono-atomic is confined in a cylinder by a spring-loaded piston of cross-section 8 × 10–3 m2. Initially, the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is on its relaxed (unstretched, uncompressed) state as shown the figure. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is 8000 Nm–1 and atmospheric pressure is 1 × 105 Nm2. The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. [Assume the spring to be massless].
ht tp
s: //
t.m e/
iv
er si ta rio s
er s
VA …. (where, T = 2 T1) VC
=
rio s_ in fo
∫ pdV
VA
un iv
e/
.m
ht tp s:
//t
ht
tp
s:
VB
= nRT ln
ar a
.ja
w
w
= – (3 mole) T1R = – (3 mole) (300 K) R = – (900 mole K) R Now, from the first law of thermodynamics, QAB = ∆UAB + WAB = – (1200 mole K) R – (900 mole K) R = – (2100 mole K) R The negative sign implies that the heat is released in the process AB. The process BC takes place at constant pressure. Hence, WBC = p V = (2 p1) (VC – VB) nRT1 nRT1 = (2 p1 ) − 2 P1 P1
= nRT1 = (2 mole) (300 K) R = (600 mole K) R Now, ∆UBC = nCv∆T
WCA =
m
3 = (2 mole) R (T1 − 2 T1 ) 2
//w
)
na .c om
(
Using equations (i), (ii) and (v), we get nRT1 2n RT1 WAB = nR(2 p1T1 ) 2 −2 2 p p1 1 1 = 2 nRT1 (2) − 2 2 = – 2n T1R = – 2 (2 mole) (200 K) R = – (1200 mole K) R The negative sign implies that the work is done on the gas. Hence, work done on the gas = (1200 mole K) R (ii) Change in energy of the gas in the process AB is ∆UAB = nCv∆T
= nRK 2 VB − 2 VA
3 = (2 mole) R (TC − TB ) 2
2.49
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2.50
Kinetic Theory of Gases and Thermodynamics
1 ∆W = k ⋅ x 2 + p0 ⋅ ∆V 2 1 = × 800 × (0.1) 2 + 1× 105 × 8 × 10−4 = 120 J 2 1 2 = × 800 × (0.1)Rigid + 1× 105 × 8 × 10−4 = 120 J 2 support ΔQ = ΔU + ΔW = 600 + 120 = 720 J.
Open atmosphere
rio s_ in fo
er s
ita
.ja
un iv
e/
.m
b
iv
i
v
) Solution Path “iaf”∆Q = 50 cal )
s: //
t.m e/
un
//t
ht tp s:
f
er si ta rio s
w
s:
Total pressure acting on the right hand side p′R = p0 + pS = 2 × 105 N/m2 Under equilibrium p′L = p′R
ht
a
F 800 N = = 1× 105 Nm −2 A 8 × 10−3 m 2
tp
pS =
//w
w
∴
∴
p
m
F = kx = 8000 × 0.1 = 800 N Pressure exerted on the piston by the spring
ar a
na .c om
Initallty for the equilibrium of the piston, pL = pR = p0 where pL and pR are the pressures on the left hand and right hand side of the piston. Force exerted by the spring on the piston when it moves
(i) ∆W along the path “ibf”. (ii) If ∆W = – 13 cal for the curved path “fi”, what is the ∆Q for this path? (iii) Taking Ui = 10 cal, what is Uf? (iv) If Ub = 22 cal, what is ∆Q for the process “ib” and the process “bf”?
)
) Solution Let p0 be the atmospheric pressure.
30. A system is taken from state i to the state f (refer to the figure). Along path “iaf”, it is found that ∆Q = 50 cal. ∆W = 20 cal. Along the path “ibf”, ∆Q = 36 cal. Calculate:
Heater
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pL VL p′L V′L = TL T′L
ht tp
∆W = 20 cal ⇒ ∆U = ∆Q – ∆W = 50 – 20 = 30 cal 1× 105 × 2.4 × 10−3 ⇒ U – Ui = 30 cal f = 300 As internal energy change is a state function. 5 −3 ∆U will be same for any path from i to f. 2 × 10 × 3.2 × 10 = (i) Path “ibf” ∆W = ∆Q – ∆U T′L = 36 – (Uf – Ui) T′L = 800 K = 36 – 30 = 6 cal. ∆U = nCv∆T (ii) Path “fi” ∆Q = ∆U + ∆W pL VL 1× 105 × 2.4 × 10−3 = (Uf – Ui) + ∆W where, n= = = 0.09638 mole R T 8.3 × 300 = (– 30) + (– 13) L pL VL 1× 105 × 2.4 × 10−3 = – 43 cal n= = = 0.09638 mole R TL 8.3 × 300 (iii) Uf – Ui = 30 cal 3 ∴ ∆U = 0.09638 × × 8.3 × (800 − 300) = 600 J Uf = Ui + 30 ∴ = 40 cal. 2 or
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Kinetic Theory of Gases and Thermodynamics
2.51
QA → B = n CP (TB – TA)
fo
s
ta rio
si
er
iv
e/ un
ht
B
tp s: //t .m
VB
= (1500 mole K) R. The process B → C occurs at constant temperature. From first law of thermodynamics dU = dQ – dW Since, the internal energy of an ideal gas depends only on temperature, therefore dU = 0 and dQ = dW QB → C = WB → C
rio s_ in
e/ un
.m
:// t C
ht
tp s
D
VD
iv er
w
s: //w
ht tp
V
si ta
m
.ja
w
ar a
na
.c om
5 = (2 mole) R (600 K − 300 K) 2
T
TA = 27°C
VB V = 2 and D = 4 VA VA
Given:
)
) Solution
dV V
= nR TB ln
VC VB
= nR TB ln
VD . . . . .(as VC = VD) VB
V V = nR TB ln D A VA VB
TB
TA
A
VA
= ∫ pdV = nRTB ∫
31. A mono-atomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The V V volume rations are B = 2 and D = 4. VA VA If the temperature TA at A is 27ºC, calculate: (i) The temperature of the gas at point B. (ii) Heat absorbed or released by the gas in each process. (iii) The total work done by the gas during complete cycle. Express your answer in terms of the gas constant R.
(i) The process A → B in which the plot (iv) Process “ib” ∆Q = ∆U + ∆W of V verse T is linear occurs at constant = (Ub – Ui) + (∆W)ibf pressure condition. (∆W)ib = (∆W)ibf V V Hence A = B Because (∆W)bf = 0 TA TA ∆Q = (22 – 10) + 6 V = 18 cal. or TB = B TA = (2) (300 K) = 600 K Process “bf” ∆Q = ∆U + ∆W TA = (Uf – Ub) + 0 T = VB T = (2) (300 K) = 600 K A B TA = (40 – 22) (ii) The process A → B occurs at constant = 18 cal. pressure. Hence,
4 = (2 mole) (R) (600 K) ln 2 = (1200 mole K) R ln 2 The process C → D occurs at constant volume. Hence, QC → D = nCv (TA – TB) 3 = (2 mole) R (300 K − 600 K) 2 = – (900 mole K) R
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2.52
∴
_i nf o
co
ta r
io s
ta r si er
un
iv
B
e/ t.m ://
45°
A
2p0
p0
ht tp s
45°
po
io s
si er
iv
un
.m e/
p
//t
s: ht tp B
a.
an
m ar
.ja
w
w
:// w
tp s
ht
p
P2 – 4p = – KT (Parabola)
= 2 p0 V 0 = 2 × 1 × 105 × 1 × 10–3 = 200 J. (iii) For path ABC (∆Q)ABC = (∆U)AC + (∆W)ABC ⇒ (∆U)AC = 600 – 300 = 300 J. Heat transfer in the process along path AC, (Q)AC = (∆U)AC + (∆W)AC = 300 + 200 = 500 J.
m
(i) Sketch the process on p – T diagram. (ii) Find the work done by the gas along the paths ABC and AC. (iii) Find the heat transfer in the process along the path AC.
KT +4 P
(ii) Workdone along path AC = (∆W)AC = p0 (3 V0 – V0)
32. An ideal gas expands from a volume V0 = 1 litre and pressure p0 = 1 bar to volume 3 litre along two different paths ABC and AC as shown in figure. The heat added to the gas along the path ABC is 600 J.
p=−
∴
(iii) Since, the process ABCDA is a cyclic process, therefore U = 0, W = Q where, Q = QA → B + QB → C + QC → D + QD → A = (1500 mole K) R + (1200 mole K) R ln 2 – (900 mole K) R – (1200 mole K) R ln 2 = (600 mole K) R.
p=–V+4
⇒
1 = (2 mole) (R) (300 K) ln 4 = – (1200 mole K) R ln 2.
At B VB = 2 V0 and pB = 2 p0 Equation of line BC, p – 2 p0 = – tan 45º (V – 2 V0)
The process D → A occurs at constant temperature. Hence, V Q D → A = WD → A = nRTA ln A VD
Kinetic Theory of Gases and Thermodynamics
A
V0
C
45°
45°
2V0
C
3V0
V
p Vo
3Vo
V
B
2p0
A
C
(i) Equation of line AB, p – p0 = tan 45º (V – V0) Hence for ideal gas, p=V Now pV = KT ⇒ p2 = KT (parabola) .....(where K is constant.)
)
) Solution
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T0
3T0
4T0
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T0
33. A 1.00 mole sample of an ideal monoatomic gas originally at a pressure of 1.00 atmosphere undergoes a three-step process:
fo
s_ in
rio
Q2 = n CP(T3 – T2)
5 = × 8.314 JK −1 (231.4 K – 389 K) 2
rio
s
iv er si
w
w
//w
V 45.1 T3 = 1 T2 = (389 K) = 231.4 K 75.8 V2
ta
ar an a.
.ja m
)
co
m
) Solution First step Adiabatic Expansion Q1 = 0
Workdone in the process W2 = – p2 (V1 – V2) = – (0.421 × 101.325 kPa) (45.1 dm3 – 75.8 dm3) = 1309.6 J
(a) Plot these processes on a p-V diagram. (b) Determine T3. (c) Calculate the change in integral energy the workdone by the gas, and heat added to gas for each process. (d) For the complete cycle.
2.53
Second step compression at constant pressure: The final volume in this process will be V1 as in the third step, the system returns to the original state by constant volume process. Hence, in the second step, T2 = (389 K) changes to T3 V2 = (75.8 dm3) changes to V1 = 45.1 dm3 p2 = remains constant.
(i) It is expanded adiabatically from T1 = 550 K and T2 = 389 K. (ii) It is compressed at constant pressure until its temperature reaches T3. (iii) It then returns to its original pressure and temperature by a constantvolume process.
Kinetic Theory of Gases and Thermodynamics
un i
ve rs
t.m
.m e/
s: //
tp
un
e/
tp s:
ht
2
1
T2
550
∆U = 3973.3 J 389 = (1.00 atm) = 0.421 atm. Since, the system return to its original state, 550 −1 −1 K) will have 3 nRT2 (1.0 mole) (8.314 JK mole )(550 we V2 = = = 45.1 ∆U dm =Q+W=0 p2 (1.0 ×101.325 KPa ) Now, W = W1 + W2 + W3 (1.0 mole) ( 8.314 JK −1 mole −1 ) (550 K ) = – 2007.8 J + 1309.6 J + 0 = = – 698.2 J (1.0 ×101.325 KPa ) ∴ Q = – W = 698.6 J = 45.1 dm3 The p-V plot of the given process is shown in ∆U1 = W1 = – 2007.8 J the figure: 5 2
T = p1 1 T2
CP R
tp s:
P
ht
//t
ht
ita
= – 3275.7 J ∆U2 = Q2 + W2 3 −1 −1 – 3275.7 J + 1309.6 J = – 1966.1 J = (1.00 mol) × 8.314 JK mole × (389 K − 550=K) 2 Third step compression at constant volume in 3 mol) × 8.314 JK −1 mole −1 × (389 K − 550 K) this process: 2 W3 = 0 = – 2007.8 J V1 = (45.1 dm3) remains constant For adiabatic expansion of an ideal gas Q3 = n Cv (T1 – T3) p2 T2 (– CP / R )g = p1T1(– CP / R )g 3 C 5 = × 8.314 JK −1 (550 K – 231.4 K) 2 T1 R 389 2 p =p = (1.00 atm) = 0.421 atm. Hence, W1 = n1Cv (T2 – T1)
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2.54
Kinetic Theory of Gases and Thermodynamics
In the complete cycle ∆U = 0
p
(1 atm, 45.1 dm3, 550K)
Q = Q1 + Q2 + Q3 = 0 – 3275.7 J + 3973.3 J = 697.6 J
(0.421 atm, 75.8 dm3, 389K)
W = – Q = 697.6 J (= W1 + W2 + W3)
(0.421 atm, 45.1 dm3, 231.4K)
questions
ype
m
ita
iv
un
3. At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s the gas is (a) H2 (b) F2 (c) O2 (d) Cl2
t.m e/
) Explanation Vrms = )
∴
s: //
ht tp
Vrms, 2 = 2 × Vrms, 1 = 684 m/s
M=
3RT M
3 RT 3 × 8.31× 300 = v 2rms (1.930) 2
)
Vrms =
3 RT 3 × 8.31× 300 = v 2rms (1.930) 2
3RT M
= 2.0078 × 10–3 kg
3 2
) Explanation KE = kT
M=
T2 = 2 T1
er si ta rio s
un iv
e/
2. The average translational energy and the rms speed of molecules in a sample of oxygen at 300 K are 6.21 × 10–21 J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour). (a) 12.42 × 10–21 J, 968 m/s (b) 6.21 × 10–21 J, 968 m/s (c) 8.78 × 10–21 J, 684 m/s (d) 12.42 × 10–21 J, 684 m/s
=
Hence (d) is correct.
//t
ht tp s:
Hence (d) is correct.
Vrms, 2 Vrms,1
∴
.m
ht
er s
s:
tp
)
V p2 = p1 1 V2 = p(8)γ = p(8)5/3 = 32 p
KE2 = 2KE1 = 2 × 6.21 × 10–21 = 12.42 × 10–21 J
.ja
//w
w
w
) Explanation p1 V1γ = p2 V2γ ;
(CV + R)30 VV
∴
dT2 =
rio s_ in fo
i.e.,
ar a
1. A monomatomic gas (γ = 5/3) is suddenly th 1 compressed to its volume adia8 batically. The pressure of the gas will changes to 24 (a) (b) 8 5 40 (c) (d) 32 3
na .c om
SOLVED OBJECTIVE T
= – 2007.8 + 1309.6 J + 0 = 698.2 J
v
= 2.00 gm Hence (a) is correct.
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4. An ideal gas is taken through the cycle A → B → C → A, as shown in the gas in the cycle is 5J, the work done by the gas in the process C → A is (a) – 5J (b) – 10J (c) – 15J (d) – 20J
β
(c)
p
C
2
B
β
in
)
.ja m
s_
s
rio
6. Two insulating cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at temperature 300 K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K. Then the rise in temperature of the gas in B is (a) 30 K (b) 18 K (c) 50 K (d) 42 K
ht
tp s:
//t
ht
.m e/
un
iv
://
tp s
ta
un iv
t.m e/
tp s
ht
⇒
er si
w
w w
://
(dV / dp ) 1 = V p 1 β= p −
er si
⇒
5. Which of the following graph correctly represents the variation of dV â = − V with p for an ideal gas dP at constant temperature?
β
pV = constant pdV + Vdp = 0
ta rio
⇒
)
= (10 + 0 + WCA) J Given: ∆Q = 5J From first law of thermodynamics 5 = 10 + 0 + WCA ⇒ WCA = – 5J
fo
) Explanation As temperature is constant,
) Explanation For the cyclic process ∆U = 0 ∆W = WAB + WBC + WCA
p
ar an
10
(d)
p(N/m2)
a.
A
1
co
m
V(m3)
)
) Explanation For cylinder A For cylinder B
dQ = n CPdT1 dQ = nCVdT2 = n (CV + R) dT1
∴
p
(a)
dT2 =
β
(b)
nCVdT2 = n(CV + R) 30
(CV + R) 30 CV 5 For diatomic gas C V = R 2 ∴ dT2 = 42 K ∴
2.55
Kinetic Theory of Gases and Thermodynamics
p
Hence (d) is correct.
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2.56
)
For monatomic gas g =
Since volume is proportional to length, therefore,
s_
rio
s
rio
iv
si ta
un
P
er
e/
ht
.m
tp
e/
s:
un
//t
iv
.m
tp
ht
2
)
) Explanation In adiabatic process, slope of p-V
curve ∝ γ From the figure slope of curve 2 > slope of curve 1 ∴ γ2 > γ1
s:
tp
ht
L (b) 1 L2
2/3
L (a) 1 L2
1
V
//t
8. A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by
er si
ta
w .ja
w
w
s: //
fo
in
an
ar
m
9. p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to (a) He and O2 (b) O2 and He (c) He and Ar (d) O2 and N2
RT mB Similarly for Gas in B, 1.5 ∆p = M 2V From equation (i) and (ii) we get 2mB = 3mA Hence (c) is the correct.
2/3
m a.
co
)
RT mA ∆p = M 2V
T1 L 2 = T2 L1
Hence (d) is correct.
1 RT 1 ∆p = p2 − p1 = mA − M V1 V2 Putting V1 = V and V2 = 2V, we get ∴
5 3
TV2/3 = constant
⇒
) Explanation TVγ – 1 = constant
) Explanation For gas in A, p1 = m RT p2 = A M V2
2/3
mA RT M V1
L2 (d) L1
L (c) 2 L1
7. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of gas contained in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The change in the pressure in A and B are found to be ∆p and 1.5 ∆p respectively. Then (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB
Kinetic Theory of Gases and Thermodynamics
Now γ for monatomic gas is greater than γ for a diatomic gas. Hence (b) is correct.
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Kinetic Theory of Gases and Thermodynamics
2.57
er si ta rio s
3
T0 (1/V0)
(1/V)
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(4/V0)
7. A thermally insulated vessel containing one mole oxygen at 27ºC, 1 atm is accelerated with an acceleration a0 and is suddenly stopped at the end of 5th second of its motion, there by the temperature of oxygen is increased by 0.5ºC, the value of a0 is nearly (a) 10 ms–2 (b) 6.3 ms–2 –2 (c) 7.2 ms (d) 5 ms–2
(d)
2T0
ht tp
4. Nitrogen gas and ethylene gas are kept in containers 1 and 2 respectively which are at the same temperature. The ratio of velocity of sound in the first cylinder to that in the second is
un t.m e/
2 3
3. How much heat energy should be added to a mixture of 10 gm of hydrogen and 40 gm of He to change the temperature by 50ºC kept in a closed vessel? (a) 2500 Cal (b) 2750 Cal (c) 2000 Cal (d) can’t be defined
A
C1 V1 = C2 V1
T
s: //
3
iv
.m
//t
(c)
ht tp s:
2
(d)
6. Given figure shows a parabolic graph 1 between T and for a mixture of a gas V undergoing an adiabatic process. Then the ratio of vrms and speed of sound in the mixture will be 3 (a) (b) 2 2
er s
un iv
e/
s:
tp
ht
1
C1 V = γ 1 C2 V1
ita
.ja
w
//w
w
2. A gas undergoes A to B through three different processes 1, 2 and 3 as shown in the figure. The heat supplied to the gas is Q1, Q2 and Q3 respectively, then (a) Q1 = Q2 = Q3 (b) Q1 < Q2 < Q3 (c) Q1 > Q2 > Q3 (d) Q1 = Q2 > Q3
B
T
3T
rio s_ in fo
ar a
2T
m
1.5T
T
(c)
D
A
C
B
1 >1 <1 depended on the amounts of the two gases
5. The intensity of sound wave in an ideal gas at temperature T1 K and T2 K are respectively C1 and C2. The rms velocity of gas molecule at these two temperatures are V1 and V2 respectively, then C1 V1 C V = (a) (b) 1 = γ 1 C2 V1 C2 V1
na .c om
P
O
(a) (b) (c) (d)
1. p-T diagram of one mole of an ideal monatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is (a) – 2 RT (b) 3 RT (c) RT (d) – 1.25 RT
UNSOLVED OBJECTIVE TYPE QUESTIONS
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2.58
8. If a monoatomic gas undergoes a thermodynamic process for which its molar heat capacity is equal to the universal gas constant, what is the process in terms of V and T? (a) VT–1/2 = constant (b) VT1/2 = constant (c) V2T2 = constant (d) VT = constant
T
D
C
B A
P
(c)
C
V
9. A gas in found to obey the law p V = constant. The initial temperature and volume are T0 and V0. If the gas expands to a volume 3 V0, then
D B
fo in
s_
s
rio
si ta
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A B
14. A cyclic process is shown in the figure. For the process AB which of the following statement is correct? (a) Heat given to the system will be always –ve throughout the process. (b) Heat given to the system will be always +ve throughout the process.
V P
iv
C
D
13. The specific heat of an ideal gas is proportional to (a) Tº (b) T–1 1/2 (c) T (d) T3
B
(a)
s: //
ht tp
A
e/
t.m
ht
P
12. A vessel X contains 1 mole of O2 gas (molar mass 32) at a temperature T and pressure p. Another identical vessel Y contains one mole of He gas (molar mass 4) at temperature 2T, then (a) pressure in the container Y is p/8 (b) kinetic energy of O2 molecule = kinetic energy of He molecule (c) pressure in the container Y is 2p 6 (d) kinetic energy of He molecule = 5 (kinetic energy of O2 molecule)
un
s: // tp
10. A cyclic process ABCD is shown in the V–T diagram. Which of the following curves represent the same process?
C D
(b)
iv un e/ t.m
ht
T
er
w w
tp
B
A
11. The internal energy of a gas in given by U = 2pV. It expands from V0 to 2V0 against a constant pressure p0. The heat absorbed by the gas in the process is (a) 2p0 V0 (b) 4p0 V0 (c) 3p0 V0 (d) p0 V0
si ta
w C
s: //
V
T
rio
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D
A
(d)
ar
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(a) final temperature becomes 3 T0 (b) internal energy of the gas increases T (c) final temperature becomes 0 3 (d) internal energy of the gas decreases
a. co m
2
Kinetic Theory of Gases and Thermodynamics
V
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Kinetic Theory of Gases and Thermodynamics
(c) First the heat given is +ve then –ve throughout the process. (d) First the heat given is –ve then +ve throughout the process.
B
A
2.59
P C
A
2P0
V
P
C B
C
2V0
A
(a)
T
.c ta rio
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ht
rio
un
//t
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//t
C
ht
tp
s:
V
T
18. n moles of ideal monatomic gas undergoes a process in which the temperature changes with volume as T = KV2. If the temperature of the gas changes from T0 to 4T0, then (a) work done by the gas is 3nRT0 (b) heat supplied to the gas is 4nRT0 3 (c) work done by the gas is nRT0 2 3 (d) heat supplied to the gas is nRT0 2
B A
(d)
T
e/
s:
ht tp
B
A
(c)
(d) None of these
17. For a given thermodynamic process, the P – V diagram is as shown below: Which of the following is the V – T diagram for the process?
(c) 1
ita
iv er s
w
//w
tp s:
V
16. A cylinder of volume V = 10 ℓ contains 64 gm of gas in molecular state. Due to the faulty value, the gas pressure dropped to (1/4)th of the initial pressure such that the temperature remains unchanged. The factor by which velocity of sound has changed in the cylinder is 1 3 (a) (b) 2 2
T
si
p0 V0 4R
er
(d)
B
A
C
ni v
p0 V0 R
(b)
e/ u
2 p0 V0 R
V
.ja m
(c)
(b)
w
p0 V0 2R
(a)
s_
ar a
in f
na
o
C
s
15. One mole of an ideal gas undergoes a p0 , where p0 and v0 process p = 2 V 1+ v0 are constants. The temperature of the gas when v = V0 is
V0
B
V
om
P0
B → C: adiabatic C → A: isothermal
A → B: isothermal
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19. One mole of an ideal monatomic gas (initial temperature T0) is made to go through the cycle abca shown in the figure. If U denotes the internal energy, then (a) U0 – Ua = 10.5 RT0 (b) Ub – Ua = 4.5 RT0 (c) U0 > Ub > Ua (d) all of these
(d) average kinetic energy per mole of hydrogen 5 = × average kinetic energy per mole 3 of helium
fo
in
s
ta rio
er
iv .m
100
50
//t
T
ht
tp
s:
21. When two identical vessel contain helium and hydrogen at same temperature, then (a) average kinetic energy per mole of hydrogen. = average kinetic energy per mole of helium (b) average translational kinetic energy per mole of hydrogen = average translational kinetic energy per mole of helium (c) average kinetic energy per mole of hydrogen 3 = × average kinetic energy per mole 5 of helium
0.1
0.4 x (meters)
24. PV versus T graph of equal masses of H2, He and CO2 is shown in the figure. Choose the correct alternative. (a) 3 corresponds to H2, 2 to He and 1 to CO2 (b) 1 corresponds to He, 2 to H2 and 3 to CO2 (c) 1 corresponds to He, 3 to H2 and 2 to CO2 (d) 1 corresponds to CO2, 2 to H2 and 3 to He
0.3
0.2
si
un e/
s: tp ht
O
.m
F (N)
//t
2
si
er
iv
e/ un
tp s:
ht
1
s_
m
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w
//w w
P
20. An ideal gas undergoes through two processes 1 and 2. If V1 and V2 are the volumes in the process 1 and 2, then (a) V1 = V2 (b) V2 > V1 (c) V1 > V2 (d) none of these
23. The given figure shows the variation of force on an ideal gas (enclosed in a piston-cylinder arrangement) which undergoes a process during which piston position changes from 0.1 to 0.4 m. If the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed during the process is (a) 15 Joules (b) 17.5 Joules (c) 20 Joules (d) 22.5 Joules
ta rio
ar
V
co m
4Vo
Vo
an a.
b
To a
Po
c
2Po
P
22. When the state of a system changes from A to B adiabatically the work done on the system is 322 joule. If the state of the same system is changed from A to B by another process, and 100 calories of heat is required then work done on the system in this process will be (a) 98 Joules (b) 38.2 Joules (c) 15.9 Joules (d) 15.9 Joules
Kinetic Theory of Gases and Thermodynamics
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2.61
(c) equal to work done on the gas 3RT (d) 2
PV
2
T
29. In a thermodynamic process helium, gas obeys the law TP–2/5 = constant. If temperature of 2 moles of the gas is raised from T to 3T, then (a) heat given to the gas is zero (b) increase in internal energy is 6RT (c) work done by the gas is – 6RT (d) all of these
ta rio
er si
iv
s: //
tp
ht
ρ
un t.m
ht
(a) the product of P and V at A is equal to the product of P and V at B (b) pressure at B is greater than the pressure at A (c) work done by the gas during the process AB is negative (d) all of these
e/
tp
26. A vessel contains 1 mole of O2 at a temperature T. The pressure of gas is p. An identical vessel containing 1 mole of the gas at a temperature 2T has pressure of (a) p (b) 2p (c) 3p (d) 4p
un
s: //
ht
Temperature
30. The density (ρ) of an ideal gas varies with temperature T as shown in figure. Then
t.m
tp
e/
s: //
C
s
er iv
w w
B
27. An insulated container is divided into two equal portions. One portion contains an ideal monoatomic gas at pressure P and temperature T, while the other portion is a perfect vacuum. If a hole is opened between the two portions, the change in internal energy of the gas is (a) zero (b) equal to work done by the gas
B
A
o
si ta
.ja w
A
s_ in f
rio
ar
m
Volume
an
a. co m
(c) γ : γ – 1 : 1 (d) γ : 1 : γ – 1
25. The expansion of an ideal gas of mass m at a constant pressure P is given by the straight line B. Then the expansion of the same ideal gas of mass 2m at a pressure 2P is given by the straight line (a) C (b) A (c) B (d) none of these
1
28. A gas of adiabatic exponent γ is supplied at a constant pressure. The ratio between dQ, dW and dU in the process 1 (a) γ : γ − 1: γ (b) 1 : 1 : γ – 1
3
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T
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2.62
Kinetic Theory of Gases and Thermodynamics
ANSWERS 4. 12. 20. 28.
(b) (a) (d) (a)
4. When a piece of metal weighing 48.3 g at 10.7ºC is immersed in a current of steam at 100ºC, 0.762 g of steam is condensed on it. Calculate the specific heat of the metal. [IIT, 1973] [Ans. 0.095 cal/gmºC]
ita
5. A metal piece, which weighs 46 g in air, weighs only 30 g when immersed in a liquid of density 1.24 g/cc at 30ºC. In the same liquid at 42ºC, when its density changes to 1.20 g/cc, it weighs 30.5 g. Find the coefficient of linear expansion of the metal. [Ans. 3.445 × 10–5/ºC]
un iv
e/
iv
.m
un
//t
6. A mixture of 250 gm water and 200 gm ice at 0ºC is kept in a calorimeter of water equivalent 50 gm. If 200 gm of steam at 100ºC is passed through this mixture, calculate the final temperature and the weight of the contents of the calorimeter. Latent heat of fusion of ice = 80 cal/gm Latent heat of vapourization = 540 cal/gm [IIT, 1974] [Ans. Weight of the contents of calorimeter = 572.2 g and final temperature = 100ºC] 7. A vessel of volume V = 20 litre contains a mixture of Argon and Helium at a temperature t = 20ºC and pressure P = 2.0 atm. The total mass of the mixture in the vessel is m = 5.0 g. Find the ratio of the mass of Argon to that of Helium in the mixture. [Ans. 0.61]
3. In an industrial process 10 kg of water per hour is to be heated from 20ºC to 80ºC. To do this, steam at 150ºC is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90ºC. How many kg of steam are required per hour? Specific heat of steam = 1 cal/gmºC = 1 kilo cal/kgºC
ht tp
s: //
t.m e/
ht tp s:
Latent heat of steam = 540 cal/gm = 540 kilo cal/kg. [IIT, 1972] [Ans. 1kg]
er s
w
//w
s:
tp
ht
8. (b) 16. (c) 24. (a)
na .c om
ar a
m
.ja
w
2. A vertical cylinder of total length 100 cm is closed at the lower end and is fitted with a movable, frictionless, gas tight disc at the other end. An ideal gas is trapped under the disc. Initially, the height of the gas column is 90 cm when the disc is in equilibrium between the gas and the atmosphere. Mercury is then slowly poured on the top of the disc and it just starts overflowing when the disc has descended through 32 cm. Find the atmospheric pressure. Assume the temperature of the gas to remain constant and neglect the thickness and weight of the disc. [IIT, 1971] [Ans. 76.125 cm of mercury]
7. (d) 15. (a) 23. (c)
(b) (c) (a) (d)
1. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and raised further by 44 cm. What will be the length of air column above mercury in the tube? Atmospheric pressure = 76 cm of mercury. [IIT, 1967] [Ans. 15.42 cm]
6. 14. 22. 30.
(a) (a) (b, d) (d)
EXERCISE
P
5. 13. 21. 29.
(b) (c, d) (b) (c)
rio s_ in fo
3. 11. 19. 27.
(c) (b) (b, c) (b)
er si ta rio s
2. 10. 18. 26.
(d) (a, b) (b) (c)
ractice
1. 9. 17. 25.
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14. 5 gm of water at 30ºC and 5 gm of ice at –20ºC are mixed together in a calorimeter. Find the final temperature of the mixture. Water equivalent of the calorimeter is negligible, specific heat of ice = 0.5 cal/gm ºC and latent heat of ice = 80 cal/gm. [IIT, 1977] [Ans. 0ºC]
ht tp
(a) pressure of the mixture
nM +n M +n M RT (n1 + n2 + n3 ),(b) 1 1 2 2 3 3 Ans. (a) n1 + n2 + n3 V
(b) mean molar mass M of the mixture which enters into equation of state m pV = RT, where m is the mass M of the mixture.
15. A vertical cylinder closed at both the ends has a freely moving piston that divides its volume into two parts, each containing one mole of air. In equilibrium, at T = 300 K the volume of the upper part is η = 4.0 times greater than that of the lower part. Find the temperature T′ at which the ratio of these two volumes will be equal to η′ = 3.0. [Ans. 421.9 K] 16. A vessel is completely filled with 500 gm of water and 1000 gm of mercury. When 21200 calories of heat are given to it, 3.52 gm of water overflows. Calculate the coefficient of volume expansion of mercury. Expansion of vessel may be neglected.
s: //
t.m e/
ht tp s:
un
//t
iv
.m
er si ta rio s
e/
un iv
er s
ita
.ja
w
w
//w
s:
tp
ht
11. A vessel of volume V contains a mixture of ideal gases at temperature T. The mixture contains n1 mole of oxygen, n2 mole for nitrogen and n3 mole of carbon dioxide. Assuming the gases to be ideal. Find:
na .c om
m
ar a
10. A mixture of nitrogen (m1 gm) and carbon dioxide (m2 gm) is kept in a vessel at temperature T and at a pressure p0. Find the density of the mixture, assuming the gases to be ideal. Ans. p0 (m1 + m2 ) m m RT 1 + 2 M1 M 2
13. A glass capillary tube, sealed at both ends is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube (which are equal in length) contain air at 27ºC and at a pressure of 76 cm of mercury. The tube is kept in a horizontal position such that the air column at one end is at 0ºC and at the other end is maintained at 127ºC. Calculate the length of the air which is at 0ºC and its pressure. Neglect the change in volume of mercury and glass. [IIT, 1975] [Ans. Length of air column at 0ºC, x = 36.5 cm, P2 = 85.3 cm]
rio s_ in fo
9. What should be the lengths of steel and copper rod so that the length of steel rod is 5 cm longer than the copper rod at all temperature? (α for copper = 1.7 × 10–5 per ºC and α for steel = 1.1 × 10–5 per ºC) [IIT, 1975] [Ans. Length of copper rod = 9.17 cm, Length of steel rod = 14.17 cm]
12. The temperatures of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC and when B and C are mixed, it is 23ºC. What should be the temperature when A and C are mixed? [IIT, 1976] [Ans. 20.26ºC]
Latent heat of vapourization of water = 540 cal/g. Latent heat of fusion ice = 80 cal/g Specific heat of copper = 0.092 cal/gºC and, specific heat of ice = 0.5 [Roorkee, 1975] [Ans. 4.6 g]
2.63
8. 20 g of ice in 10 g copper calorimetry is originally at –30ºC. How much steam at 100ºC must be condensed in the calorimeter if the ice is to be changed to water and heated to 40ºC?
Kinetic Theory of Gases and Thermodynamics
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ideal gas is enclosed between the pistons, which are tied together with an inextensible thread. The cross-sectional area of the upper piston is ∆S = 10 cm2 larger than that of the lower one. The combined mass of the two pistons is m = 5.0 kg and the outside air pressure is p0 = 1.0 atm. Find by how many kelvin should the gas between the pistons be heated so as to shift the pistons through ∆l = 5.0 cm. [Ans. 0.9 K]
er s
un iv
e/
un
iv
where T0 and α are constants and V is the volume of one mole of gas?
t.m e/
23. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal crosssectional area A. Atmospheric pressure is P0 and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surrounding, show that the piston executes simple harmonic motion and find the frequency of oscillation. [IIT, 1981]
s: //
ht tp
20. A smooth vertical tube having two different areas of cross sections is open from both the ends and is provided with two freely moving pistons of different cross sections. Each piston slides within its respective cross section. One mole of an
T = T0 + α V2,
.m //t
ht tp s:
(W2 − W1 ) + β(W0 − W1 )(T2 − T1 ) Ans. λ = (W0 − W2 )(T2 − T1 )
22. What is the minimum attainable pressure of an ideal gas in the process
[ Ans. Pm = 2R α T0 ]
19. A sinker of weight W0 has an apparent weight W1 when weighed in a liquid at a temperature T1 and W2 when weighed in the same liquid at a temperature T2. The coefficient of cubical expansion of the material of sinker is β. What is the coefficient of volume expansion of the liquid?
ita
.ja
w
w
//w
s:
tp
ht
log e η Ans. N = ∆V log e 1 + V
rio s_ in fo
m
18. A vessel of volume V is required to be evacuated by means of a piston air pump that captures volume ∆V in each stroke. How many strokes will reduce the pressure in the vessel η times? Assume the process to be isothermal and the gas, ideal.
ar a
na .c om
Given: coefficient of linear expansion of glass = 9 × 10– 6 per degree, coefficient of volume expansion of mercury = 1.8 × 10– 4 per degree. [IIT, 1973] 3 [Ans. 150 cm ]
21. Ice at 0ºC is added to 200 g of water initially at 70ºC in a vacuum flask. When 50 g of ice has been added and has all melted, the temperature of the flask and contents is 40ºC. When a further 80 g of ice has been added and has all melted, the temperature of the whole becomes 10ºC. Neglecting heat loss to surroundings calculate latent heat of fusion of ice. [MNR, 1979] [Ans. L = 90 cal/g]
17. A one litre flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in flask?
Given: Coefficient of volume expansion of water = 1.5 × 10–4 per degree centigrade Density of mercury = 13.6 gm/cm3 Specific heat of mercury = 0.03 cal/gm ºC [IIT, 1976] –4 [Ans. 1.768 × 10 /ºC]
er si ta rio s
Kinetic Theory of Gases and Thermodynamics
1 γ P0 A 2 Ans. n = 2π MV0
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Latent heat of ice = 80 cal/g. [IIT, 1972] [Ans. 2.96 × 10–4 k cal/m. sec ºC]
iv
31. A 2 m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm. When a current of 5 amp flows through the wire, find the temperature difference across the insulation in steady state. Thermal conductivity of plastic = 0.16 × 10–2 cal/s cm ºC. [IIT, 1974] [Ans. 2.05ºC]
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If the final temperature of the container is 27ºC, find the mass of the container. (Latent heat of fusion for water = 8 × 104 cal/kg, specific heat of water = 103 cal/kg K). [IIT, 2001] [Ans. 0.495 kg]
t.m e/
s: //
ht tp
27. An ice cube of mass 0.1 kg is placed in an isolated container which is at 227ºC. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where, A = 100 cal/kg K and B = 2 × 10–2 cal/kg K2.
er si ta rio s
ita
un iv
e/
.m
//t
ht tp s:
2 + (α A + α B )∆T ⋅ d Ans. R = 2 (α A − α B )⋅ ∆T
30. Water is being boiled in a flat bottom kettle on a stove. The area of the bottom is 300 cm2 and the thickness is 2 mm. If the amount of steam produced is 1 g/min, calculate the difference in temperature between the inner and the outer surfaces of the bottom. The thermal conductivity of material of kettle = 0.5 cal (ºC)–1 sec–1 cm–1 and latent heat of steam 540 cal/g. [IIT, 1972] [Ans. 0.012ºC]
un
w
//w
s:
tp
ht
er s
w
26. Two metal strips, A and B each of length l and thickness d at temperature T, are rivetted together. The coefficient of linear expansion of the strips are αA and αB (αA > αB) respectively. When this bimetallic strip is heated to a temperature (T + ∆T), one strip becomes longer than the other and the bimetallic strip bends into an arc of a circle. Find the mean radius of curvature of the strip.
.ja
m
ar a
na .c om
29. A closed cubical box made of perfectly insulating material has walls of thickness 8 cm and the only way for heat to enter or leave the box is through two solid, cylindrical, metallic plugs, each of crosssectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box as shown in the figure. Outer surface A is kept at 100ºC while the outer surface B of the other plug is maintained at 4ºC. The thermal conductivity of material of the plug is 0.5 cal/s/cm/ºC. A source of energy generating 36 cal/s is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. [Ans. θ = 76ºC]
rio s_ in fo
25. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0ºC and at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62ºC. What is the new value of pressure inside the bulbs? The volume of the connection tube is negligible. [IIT, 1985; MNR, 1990] [Ans. 83.75 cm of mercury column]
2.65
28. A slab of stone of area 0.36 m2 and thickness 0.10 m is exposed on the lower surface to steam at 100ºC. A block of ice at 0ºC rests on the upper surface of the slab. If in one hour, 4.8 kg of ice is melted, calculate the thermal conductivity of the stone.
24. An ideal gas of molar mass M is contained in a uniform gravitational field having free-fall acceleration due to g. Find the pressure of the gas as a function of height h if P = P0 at h = 0 and the temperature varies with height h as: T = T0 (1 – ah) where a is a positive constant. [Ans. P = P0 (1 – ah)n]
Kinetic Theory of Gases and Thermodynamics
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32. Two rods A and B are of equal length. Each rod has its ends at temperature T1 and T2. What is the condition that will ensure equal rates of flow of heat through the rods A and B? [IIT, 1976] [Ans. Hence, areas of cross section of the two rods should be inversely proportional to the square root of their thermal conductivity]
C x A
y
rio s_ in fo
e/
er si ta rio s
un iv
er s
37. A block of steel heated to 100ºC is left in a room to cool. Find the curve shown in the figure, which represents the correct behaviour. [IIT, 1980]
ita
m
ar a
na .c om
36. A wall has two layers A and B, each made of different materials. Both the layers have the same thickness. The thermal conductivity of the material A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is 36ºC. Find the temperature difference across the layer A. [IIT, 1980] [Ans. 12ºC]
iv
.m
u Temperature n
C B A
[Ans. Curve A]
38. Radiant energy from the sun strikes the earth at a rate of 1.4 × 103 watt/m2. Calculate the temperature at the surface of sun. Radius of sun = 7 × 108 m Radius of earth orbit = 1.5 × 1011 m Stefan’s constant = 5.7 × 10–8 Wm–2 K–4 [MNR, 1985] 5 [Ans. 1.833 × 10 K]
35. Three rods of material x and three rods of material y are connected as shown in the figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at 60ºC and the junction E at 10ºC, calculate the temperatures of the junctions B, C and D. The thermal conductivity of x is 0.92 CGS unit and that of y is 0.46 CGS unit. [IIT, 1978]
Time
ht tp
s: //
t.m e/
ht tp s:
//t
ht
tp
34. A room is maintained at 20ºC by a heater of resistance 20 ohm connected to 200 volt mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area 1 m2 and thickness 0.2 cm. Calculate the temperature outside (Thermal conductivity of glass is 0.2 cal m–1 ºC sec–1 and mechanical equivalent of heat J = 4.2 joule/cal). [IIT, 1978] [Ans. 15.24ºC]
E
X
[Ans. 30ºC]
.ja
w
w
//w
s:
B
y
D
33. A bar of copper of length 75 cm and a bar of steel of length 125 cm are joined together end to end. Both are of circular cross section with diameter 2 cm. The free ends of copper and steel are maintained at 100ºC and 0ºC respectively. The surface of the bars are thermally insulated. What is the temperature of copper steel junction? Thermal conductivity of copper is 9.2 × 10–2 kilo cal/m (ºC) sec and that of steel is 1.1 × 10–2 kilo cal/m (ºC). [IIT, 1977] [Ans. 93.28ºC]
x
y
2.66
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45. A body cools in 7 minutes from 60ºC to 40ºC. What will be its temperature after the next 7 minutes? The temperature of the surroundings is 10ºC. Assume that the Newton’s law of cooling holds good throughout the process. [MNR, 1992] [Ans. 28ºC] 46. The peak emission from a black body at a certain temperature occurs at a wavelength of 9000 Å. On increasing its temperature the total radiation emitted is increased 81 times. At the initial temperature when the peak radiation from the black body is incident on a metal surface, it does not cause any photo emission from the surface. After the increase of temperature, the peak radiation from the black body caused photo emission.
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43. A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed
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42. A tapered rod of length l and thermal conductivity K has its end radii r1 and r2 maintained at temperature T1 and T2 respectively. Calculate the rate of flow of heat through the rod.
K1 + 3K 2 Ans. 4
44. One end of a copper rod of uniform cross-sectional area and of length 1.5 m is in contact with ice and the other end with water at 100ºC. At what point along its length should a temperature of 200ºC be maintained so that in steady state the mass of ice melting is equal to that of steam produced in the same interval of time? Assume that whole system is insulated from surroundings. [MNR, 1992] [Ans. 1.396 m]
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41. A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. Calculate the effective thermal conductivity of the system. [IIT, 1988]
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40. An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of 20ºC inside it when the outside temperature is –10ºC. The walls have three layers of different materials. The innermost layer is of wood of thickness 2.5 cm, the middle is of cement of thickness 1.0 cm and the outermost layer is brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1.0 watt m–1 (ºC)–1 respectively. [IIT, 1986] [Ans. 9 k W]
coaxially on a thin metal disc of mass 0.4 kg and of the same cross section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If thermal conductivity of the material of the cylinder is 10 watt/m-K and the specific heat of the material of the disc is 600 J/kg-K, how long will it take for the temperature of the disc to increase to 350 K? Assume, for the purpose of calculation, the thermal conductivity of the disc to be very high and system to be thermally insulated except for the upper face of the cylinder. [IIT, 1992] [Ans. 2 min. 46.35 sec]
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Boltzmann constant = 5.67 × 10–8 watt/ m2 (K)4. [MNR, 1986] [Ans. 0.177]
2.67
39. A body which has a surface area 5.00 cm2 and a temperature of 727ºC radiates 300 joules of energy each minute. What is the emissivity?
Kinetic Theory of Gases and Thermodynamics
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(b) temperature of the coil. (Given σ = 17 × 10–8) 3 [IIT, 2003] [Ans. (a) 595 watt/m2, (b) 147ºC]
49. The air density of Mount Everest is less than that at the sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000 cc at sea level (pressure 1 atmosphere, temperature 27ºC). Assuming that the temperature around Mount Everest is –130ºC and that the oxygen cylinder has capacity of 5.2 litre, find the pressure at which oxygen be filled (at site) in the cylinder. [MNR, 1978] [Ans. 2.75ºC]
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52. One gram mole of oxygen at 27ºC and one atmospheric pressure is enclosed in a vessel. (a) Assuming the molecules to be moving with vrms, find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (b) The vessel is next thermally insulated and moved with a constant speed v0. It is then suddenly stopped. The process results in a rise of the tempera-
To
Ta = 27°C Hot Oil
T = 127°C
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51. An electric bulb of volume 250 cm3 was sealed off during manufacture at a pressure of 10–3 mm of Hg at 27ºC. Find the number of molecules in the bulb. [IIT, 1974] [Ans. 8.03 × 1015]
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48. Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity k = 0.149 J/(m ºC-sec), thickness t = 5, emissivity = 0.6. Temperature of the top of the lid is maintained at Tf = 127º. If the ambient temperature Ta = 27ºC, calculate
50. Modern vacuum pumps can bring down the pressure to P = 4.0 × 10–15 atm. at room temperature T = 300 K. Assuming that the gas exhausted is nitrogen, find (a) number of molecules per cc, and (b) average distance between the molecules at this temperature and pressure. [Ans. (a) 105 molecules/cc, (b) 0.2 mm]
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47. A thermocole vessel contains 0.5 kg of distilled water at 30ºC. A metal coil of area 5 × 10–3 m2, number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform, time varying magnetic field is set up to pass vertically through the coil at time t = 0. The field is first increased from zero to 0.8 T at a constant rate between 0 and 0.2 s and then decreased to zero at the same rate between 0.2 and 0.4 s. This cycle is repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as functions of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axes. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium. Specific heat of the metal = 500 J kg–1 K–1 and the specific heat of water = 4200 J kg–1 K–1. Neglect the inductance of the coil. [IIT, 2000] [Ans. 5.63ºC]
(a) rate of heat loss per unit area due to radiation from the laid.
To bring these photo electrons to rest, a potential equivalent to the excitation energy between the n = 2 to n = 3 Bohr levels of hydrogen atom is required. Find the work function of the metal. [Roorkee, 1993] [Ans. 2.25 eV]
Kinetic Theory of Gases and Thermodynamics
2.68
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Kinetic Theory of Gases and Thermodynamics
59. A gas consists of rigid diatomic molecules and has a molar heat capacity C = R. If the volume of this gas is increased η = 3 times in a polytropic process, find how many times will the rate of collisions of molecules against the vessel’s wall be reduced as a result of this process. [Ans. 4.32]
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63. Two perfect gases at absolute temperature T1 and T2 are mixed. There is no loss of energy. Find the temperature of mixture if masses of molecules are m1 and m2 and the number of molecules in the gases are n1 and n2 respectively. [Roorkee, 1989]
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57. A vessel contains m = 15 g of oxygen at T = 300 K. Find the amount of heat that must be transferred to the gas to double the root mean square velocity of its molecules. [Ans. 8.764 kJ]
62. Find the temperature at which the most probable velocities of oxygen and nitrogen molecules differ by ∆v = 40 m/s. [Ans. 0.5 kK]
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56. An astronaut takes a cylinder of volume 10 litres containing nitrogen gas at a temperature of 27ºC and pressure 50 atmosphere. He makes a hole of area 1 sq. cm in this cylinder and places it in open space. Estimate the time it would take for the cylinder to become empty. (Boltzmann constant = 1.38 × 10–23 J/K) [Roorkee, 1985] [Ans. 0.387 sec]
61. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0ºC and a pressure of 76 cm of Hg. One of the bulbs is then placed in water bath maintained at 62ºC. What is the new value of pressure inside the bulbs? [IIT, 1985; MNR, 1990] [Ans. 83.75 cm of Hg]
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55. By how many times has a gas consisting of a rigid diatomic molecules to be expanded adiabatically so as to reduce its root mean square velocity by η = 1.5 times? [Ans. 7.59]
60. A gas consisting of rigid diatomic molecules undergoes expansion in a polytropic process. Find the molar heat capacity of the gas in this process, if the rate of collision of the molecules with the walls of the vessel does not change. [Ans. C = 3R]
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54. A thermally insulated vessel containing gaseous nitrogen at temperature T = 300 K and moving with a velocity v0 = 100 m/s is suddenly stopped. By how much per cent and in what way will the gas pressure change? [Ans. 2.25%]
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58. A gas consists of rigid diatomic molecules and is initially at normal temperature and pressure. If the gas is compressed adiabatically to η = 5.0 times, what will be the mean kinetic energy of a rotating molecule in the final state? [Ans. 7.88 × 10–21 joule]
53. A polyatomic gas having N atoms per molecule is heated isobarically. If all the degrees of freedom of the molecules, viz., translational rotational and vibrational, are excited, find what fraction of heat transferred to the gas in this process is spent to perform the work of expansion. What is its value in case of a monatomic gas? 1 2 Ans. , 3 5 3N − 2
ture of the gas by 1ºC. Calculate the speed v0. [Ans. (a) 1.97 × 1025 (b) 34.35 m/s]
2.69
n1T1 + n2 T2 Ans. n1 + n2
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69. A 0.1 kg steel ball falls from a height of 10 m and bounces to a height of 7 m. (a) Why does it not bounce back to its original height? (b) If all the dissipated energy were absorbed by the ball as heat, how much will its temperature rise? (Specific heat of steel = 0.11 kilo cal/kg ºC, 1 cal = 4.2 joule). [Roorkee, 1978] [Ans. (b) 0.064ºC]
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72. Two different adiabatic paths for the same gas intersect two isothermals at T1 and T2 as shown in P – V diagram. How does Va /Vd compare with Vb / Vc ? [Roorkee, 1983] Va / Vd = 1 Ans. Vb / Vc
68. An insulated container containing monoatomic gas molar mass m is moving with a velocity v0. If the container is suddenly stopped, find the change in temperature.
mv02 Ans. 3R
73. n moles of an ideal gas is contained under a weightless, freely moving piston of a vertical cylinder at a temperature T0. The space over the piston opens into the atmosphere. Find the work that has to be done to increase isothermally the gas volume under the piston η times by slowly raising the piston. [Ans. nRT (η – 1 – loge η)]
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71. Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas is 105 N/m2 and 6 litre respectively. The final volume of the gas is 2 litre. Molar specific heat of the gas at 3 constant volume is R . 2 [IIT, 1982] [Ans. 972 joules]
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67. A cubical box of side 1 meter contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take R = 25/3 J./mol-K and k = 1.38 × 10–23 J/K. Find: (a) temperature of the gas. (b) average kinetic energy per atom. [Ans. (a) 160 K, (b) 3.31 × 10–21 J]
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66. The temperature of a gas is –68ºC. To what temperature should it be heated so that (a) the average kinetic energy of the molecules be doubled? (b) the root mean square velocity of the molecules be doubled? [MNR, 1995] [Ans. (a) 137ºC, (b) 547ºC]
70. Two moles of an ideal gas at a temperature T0 = 300 K were cooled isochorically so that the gas pressure reduced η = 2 times. Thereafter, as a result of isobaric process, the gas expanded till its temperature became T0 again. What is the total amount of heat absorbed by the gas in this process? [Ans. 2.49 kJ]
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65. 0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27ºC. How much heat has to be transferred to the gas to double the rms velocity of its molecules? [Roorkee, 1991] [Ans. 337.5 K]
64. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have? (b) Obtain the work done by the gas during the expansion as a function of initial pressure P and volume V. [IIT, 1990] [Ans. (a) γ = 1.4, (b) W = 1.25 PV]
Kinetic Theory of Gases and Thermodynamics
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78. An ideal gas has a specific heat at con5 stant pressure C p = R. The gas is kept 2 in a closed vessel of volume 0.0083 m3, at a temperature of the 300 K and a pressure of 1.6 × 106 N/m2. An amount of 2.49 × 104 joules of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas. [IIT, 1987] [Ans. 674.78 K; 3.6 × 106 N/m2]
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where a and α are constants. Find the work performed by the gas and the amount of heat transferred to it when its internal energy increases by ∆U. What is the molar heat capacity of the gas in this process? R R γ −1 Ans. 1 + α ∆U; γ − 1 + α
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80. An ideal monatomic gas is confined in a cylinder by a spring loaded piston of cross section 8 × 10–3 m2. Initially, the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is 8000 N/m, atmospheric pressure is 1.0 × 105 N/m2. The cylinder and the piston are thermally
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76. A heat conducting piston divides a thermally insulated cylinder containing an ideal gas at temperature T0 into two equal parts, each of volume V0. The piston is slowly displaced. Find the gas temperature as a function of the ratio η of the volumes of the greater and smaller sections. The ratio of the specific heats of the gas is γ. ( η + 1)2 Ans. T0 4 η
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79. The internal energy U of an ideal gas having adiabatic exponent γ varies with its volume V in a certain thermodynamic process as U = aVα,
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9 Ans. (a ) 4 T0 , (b)12.9T0 75. 70 calories of heat are required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30ºC to 35ºC. Find the amount of heat required (in calories) to raise the temperature of the same gas through the same range (30ºC to 35ºC) at constant volume [IIT, 1985] [Ans. 50 calorie]
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77. A container is divided into two equal portions. One portion contains an ideal gas at pressure P and temperature T while the other portion is a perfect vacuum. If a hole is opened between two portions, find the change in energy and the change in temperature of the gas. [Roorkee, 1987] [Ans. No change]
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(a) final temperature of gas in each chamber and (b) work done by the gas in the right chamber.
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74. Two rectangular boxes shown in the figure has a partition that can slide without friction along the length of the box. Initially, each of the two chambers of the box has one mole of a monatomic ideal gas (γ = 5/3) at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands, pushing the partition until the final pressure in both chambers becomes 243 P0/32. Determine:
Kinetic Theory of Gases and Thermodynamics
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ar a
γR αR αR R +R+ = + , (b) (a ) C = γ − 1 P V γ − 1 P0 V 0 Increase in internal energy for one mole P = 0 ( V f − Vi ) Total work done by the gas γ −1 Vf , Heat transferred = P0 ( V f − Vi ) + α log e Vi Vf λ P0 to the gas = V f − Vi ) + α log e ( λ −1 Vi
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= RK = a constant 83. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume
86. A cycle consists of two isobaric and two adiabatic lines. Assuming the working substance to be an ideal gas with adiabatic exponent γ, find the efficiency of the cycle if the pressure changes n times within the cycle. [Ans. η = 1 – n(γ – 1)/γ] 87. A cycle consists of two isochoric and two isobaric lines. Assuming the working substance to be an ideal gas, having adiabatic exponent γ, find the efficiency
becomes 5.66 V while its temperature T falls to . 2 (a) How many degrees of freedom do the gas molecules have?
85. A diatomic gas is used as a working substance in a Carnot cycle. What is the efficiency of the cycle if as a result of adiabatic expansion, the gas volume increases n times? [Ans. η = 1 – n1 – γ]
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α ( γ − 1) R ⋅ T0 ( η − 1) ; PV γ ⋅ e PV Ans. W = α log e η − γ −1
α(γ −1) PV
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82. The molar heat capacity of an ideal gas, having an adiabatic exponent γ, varies α with temperature as C = where α is T a constant. Find the work performed by one mole of this gas during its heating from temperature T0 to a temperature η times higher, and the equation of the process in the variables P, V.
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81. A piston divides a closed gas cylinder into two parts. Initially, the piston is kept pressed such that one part has a pressure P and volume 5 V and the other part has pressure 8 P and volume V. The piston is now left free. Find the new pressures and volumes for the adiabatic and isothermal processes. For this gas, γ = 1.5. [Roorkee, 1990] [Ans. Under adiabatic change: P1 = 1.837 P, Under isothermal change: P′ = 2.17 P]
[Ans. 800 K, 717.01 J]
Find: (a) heat capacity of the gas as a function of its volume; (b) increase in internal energy of the gas, the work performed by it, and the amount of heat transferred to the gas in the process, if its volume increased from Vi to Vf.
Open atmosphere Rigid support
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84. The pressure P of one mole of an ideal gas having adiabatic exponent γ varies with its volume V as á P = P0 + , where P0 and α are constants. V
(b) Obtain the work done by the gas during the expansion as a function of initial pressure P and volume V. [Ans. (a) 5, (b) 1.25 PV]
Kinetic Theory of Gases and Thermodynamics
insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heat coil is negligible. Assume the spring to be massless. [IIT, 1989]
η − 1);PV γ ⋅ e
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Kinetic Theory of Gases and Thermodynamics
in
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D → A: heating at constant volume
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B → C: cooling at constant volume
A → B: adiabatic expansion C → D: adiabatic compression
tively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation PV19/13 = constant, in adiabatic processes. (a) Find the number of gram moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at T = 300 K. (c) If T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adia1 batically to of its initial volume 5 V. Find the change in its adiabatic compressibility in terms of the given quantities. [IIT, 1995] [Ans. (a) 2, (b) Speed of sound in a gaseous mixture = 400.93 m/s (c) 0.17% (d) 8.278 × 10–5 V]
88. One mole of a monatomic gas is taken through the cycle shown in the given figure:
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The pressure and temperature at A, 90. The pressure in a monatomic gas increases linearly from 4 × 105 Nm–2 to 8 × 105 Nm–2 B etc. are denoted by PA, TA, PB, TB, when its volume increases from 0.2 m3 to etc., respectively. Given that 0.5 m3. Calculate the following: 2 1 TA = 1000 K, PB = PA and PC = PA , (a) work done by the gas, 3 3 (b) increase in the internal energy, calculate the following quantities: (c) amount of heat supplied, (a) the work done by the gas in the pro(d) molar heat capacity of the gas. cess A → B [Roorkee, 1995] (b) heat lost by the gas in the process 5 [Ans. (a) 1.8 × 10 joules, (b) 4.8 × 105 B → C. joules, (c) 6.6 × 105 joules, (d) 17.14 J/ (c) temperature TD. mol-K] 2/5 2 [IIT, 1993] 91. One mole of a diatomic ideal gas (γ = 1.4) Given : = 0.85 3 is taken through a cyclic process starting from point A. The process A → B is an [Ans. (a) 1860.48 joule, (b) 1571.27 joule, adiabatic compression, B → is isobaric (c) 499.72 K] expansion, C → D is an adiabatic expan89. A gaseous mixture enclosed in a vessel of sion and D → A is isochoric. The volume volume V consists of one gram mole of a ratios are VA/VB = 16 and VC/VB = 2 and 5 the temperature at A is TA = 300 K. Calgas A with γ(= C p / C v ) = and another 3 culate the temperature of the gas at the 7 gas B with γ = at a certain temperapoint B and D and find the efficiency of 5 the cycle. [IIT, 1997] ture T. The gram molecular weights of [Ans. TB = 909 K, TD = 791.33 K, 61.4%] the gases A and B are 4 and 32 respec
of the cycle if the absolute temperature of the gas rises α times both in the isochoric heating and in isobaric expansion. α+γ Ans. n = 1 − 1 + αγ
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volume ratios are VB = 2 and VD = 4. If VA VA the temperature TA at A is 27ºC, calculate:
V VB A
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TA
TB T
(a) the temperature of the gas at point B, (b) heat absorbed or released by the gas in each process, (c) the total work done by the gas during the complete cycle. Express your answer in terms of the gas constant R. [IIT, 2001] [Ans. (a) 327ºC, (b) 900 R, (c) 600 R]
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93. A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the given figure. The
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92. A weightless piston divides a thermally insulated cylinder into two parts of volumes V and 3V. 2 moles of an ideal gas at pressure P = 2 atmosphere are confined to the part with volume V = 1 litre. The remainder of the cylinder is evacuated. Initially the gas is at room temperature. The piston is now released and the gas expands to fill the entire space of the cylinder. The piston is then pressed back to the initial position. Find the increase of internal energy in the process and final temperature of the gas. The ratio of the specific heat of the gas γ = 1.5. [Roorkee, 1999] [Ans. No change in the internal energy of the gas and its temperature. Final temperature of the gas is, equal to the room temperature]
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
2. An ideal gas expands according to the law pV2 = const. The molar heat capacity C is (a) CV + R (b) CV – R (c) CV + 2R (d) CV – 3R
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6. An empty pressure cooker of volume 10 liters contains air at atmospheric pressure 105 Pa and temperature of 27ºC. It contains a whistle which has area of 0.1 cm2 and mass of 100 gm. What should be the temp. of air inside so that the whistle is just lifted up? :KLVWOH
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5. The average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is (a) 75 J (b) 100 J (c) 150 J (d) 125 J
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1. Which of the following will have maximum total kinetic energy at temperature 300 K? (a) 1 kg, H2 1 1 (b) kg H 2 + kg He 2 2 1 3 (c) kg H 2 + kg He 2 4 (d) 1 kg, He
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ONLY ONE OPTION IS CORRECT
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6. Explain how it is possible that more than half of the molecules in an ideal gas have kinetic energies less than the average kinetic energy. Shouldn’t half have less and half have more?
7. Explain how an automobile airbag protects the passenger from injury. Why would the airbag be ineffective if the gas pressure inside is too low when the passenger comes into contact with it? What about if it is too high?
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5. Why does a helium weather balloon expand as it rises into the air? Assume the temperature remains constant.
4. Four particles have velocities 1, 0, 2, 3 m/s. The root mean square of the particles is (in m/s) (a) 3.5 (b) 3.5 (c) 1.5 (d) 14/3
4. Suppose we have two tanks, one containing helium gas and the other nitrogen gas. The two gases are at the same temperature and pressure. Which has the higher number density (or are they equal)? Which has the higher mass density (or are they equal)?
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3. Why must we use absolute temperature (temperature in kelvins) in the ideal gas (PV = NkT)? Explain how using the Celsius scale would give nonsensial results.
3. Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa) (a) 0.70 (b) 0.98 (c) 1.40 (d) 210
2. One way to loosen the lid on a glass jar is to run it under hot water. How does that work?
1. Will the temperature of gas in a container increase when we put the container on a moving train? Explain.
CONCEPTUAL QUESTIONS
1
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17. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is (a) 2/5 (b) 3/5 (c) 3/7 (d) 5/7
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13. The pressure and temperature of an ideal gas is a closed vessel are 720 kPa and 40ºC respectively. If 1/4h of the gas is released from the vessel and the temperature of the remaining gas is raised to 353ºC, the final pressure of the gas is
(b) T1 < T2 (d) T1 = T2
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16. An air bubble of volume v0 is released by a fish at a depth h in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure above the lake. The volume of the bubble just before touching the surface will be (density) of water is ρ (a) v0 (b) v0(ρgh/p)
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12. The root mean square speed of the molecules of a diatomic gas is v; then what would be the new root mean square speed when the temperature is doubled and the molecules dissociate into atoms? (a) v (b) 2 (c) 2v (d) 4v
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11. What is the average velocities of particles of a gas? 3RT (a) (b) zero M
15. Maxwell’s velocity distribution curve is given for two different temperature. For the given curves 1
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9. A mixture of ideal gases N2 and He are taken in the mass ratio of 14 : 1 respectively. Molar heat capacity of the mixture at constant pressure is (a) 6R/19 (b) 13R/6 (c) 6R/13 (d) 19R/6
(c)
14. Let V denote the root mean square speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is ‘m’. Neglecting vibrational energy terms, which of these is false? (a) A molecule can have a speed greater than 2 v (b) V is proportional to T (c) The average rotational kinetic energy of a molecule is mv2/4 (d) The average kinetic energy of a molecule is 5mv2/6
8. A gas mixture consists of 2 moles of oxygen and 4 moles of a argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is (a) 4RT (b) 5RT (c) 15RT (d) 11RT
(b) 1080 kPa (d) 540 kPa
7. The specific heat of Ar at constant volume is 0.075 kg–1 K–1. Calculate the atomic weight (R = 2 cal mol–1 K–1) (a) 40 (b) 40.4 (c) 40.2 (d) 40.80
(a) 1440 kPa (c) 720 kPa
(b) 227ºC (d) 600ºC
(a) 400ºC (c) 327ºC
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18. The internal energy of monatomic and diatomic gases are respectively due to (a) Linear motion and rolling motion (b) rolling motion and linear motion (c) Linear motion and rotatory motion (d) rotatory motion and linear motion
1 p1 p2 + 2 T1 T2 p T − p2T1 (d) 1 2 T1 − T2
(b)
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28. The root-mean-square (rms) speed of oxygen molecules (O2) at a certain absolute temperature is v. If the temperature is doubled and the oxygen gas dissociates into atomic oxygen, the rms speed would be
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2v
(d) 2 2v
(c) 2v
(b)
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27. A gas at absolute temperature 300 K has pressure = 4 × 10–10N/m2. Boltzmann constant k = 1.38 × 10–23 J/K. The number of molecules per cm3 is of the order of (a) 100 (b) 105 (c) 108 (d) 1011
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26. Determine the absolute gas temperature at which the root mean square speed of helium molecules exceeds their most probable speed by 200 m/s. (a) 110.2 K (b) 90.2 K (c) 190.2 K (d) 100.2 K
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25. Two containers of equal volume contain the same gas at pressures p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to
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24. The velocity of sound in air is 332 m s–1 at NTP. Find the rms speed of air molecules at NTP. (γ = 1.41) (a) 484 ms–1 (b) 418 ms–1 –1 (c) 248 ms (d) 382 ms–1
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23. Two ideal gases at temperature T1 and T2 are mixed. There is no loss of energy. If the
22. Two identical containers joined by a small pipe initially contain the same gas at P0 pressure and absolute temperature T0. One container is now maintained at the same temperature while the other is heated to 2T0. The common pressure of the gases will be 4 3 (a) p0 (b) p0− 3 2 5 1 (c) p0 (d) p0 3 3
(d)
21. An ideal gas has volume V and pressure P. The total translational kinetic energy of the gas 3 (a) must be PV 2 3 (b) may be PV 2 3 (c) must be ≥ PV 2 3 (d) must be ≤ PV 2
n2T1 + n1T2 n1 + n2
20. The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled if the pressure on the surface is equal to h meter of mercury column and relative density of mercury is ρ then the depth of lake is (a) 2ρhm (b) 4ρhm (c) 8ρhm (d) 7ρhm
(c)
19. The root-mean-square (rms) speed of oxygen (O2) molecules at a certain absolute temperature is v. If the temperature is doubled and the oxygen gas dissociates into atomic oxygen, the rms speed would be (a) v (b) 2v (c) 2 v (d) 2 2v
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masses of molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively, the temperature of the mixture will be T T T +T (a) 1 2 (b) 1 + 2 n1 n2 n1 + n2
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37. The average translational energy and the rms speed of the molecules in a simple of oxygen gas at 300 K are 6.21 × 10–21 J and 484 m/s respectively. Assuming ideal gas behaviors, the corresponding values at 600 K are nearly (a) 12.42 × 10–21 J, 968 m/s (b) 8.78 × 10–21 J, 684 m/s (c) 6.21 × 10–21 J, 968 m/s (d) 12.42 × 10–21 J, 684 m/s
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36. A mixture of n1 moles of monatomic gas and C n2 moles of diatomic gas has p = γ = 1.5 Cv (a) n1 = n2 (b) 2n1 = n2 (c) n1 = 2n2 (d) 2n1 = 3n2
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(2 N + 1)( N + 1) 6 (d) ratio of rms speed of average speed of molecule is 2
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34. Each molecule of a gas has f degrees of C freedom. The ratio p = γ for the gas is Cv
38. Two identical containers A and B have frictionless pistons. They contain the same volume of a ideal gas at the same temperature. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The changes in the pressure in A and B are found to be ∆p and 1.5 ∆p respectively. (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB
33. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ration Cp = γ for the gas is Cv (a) 2 (b) 3/2 (c) 5/3 (d) 4/3
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(2 N + 1)( N + 1) 6N (c) ratio of rms speed of average speed is
32. When an air bubble rises from the bottom to the surface of a lake, its radius becomes double. Find the depth of the lake, given that the atmospheric pressure is equal to the pressure due to a column of water 10 m high. Assume constant temperature and disregard surface tension. (a) 30 m (b) 40 m (c) 70 m (d) 80 m
( f − 1) 3
35. N (< 100) molecules of a gas have velocities 1, 2, 3, ........N km/s respectively. Then (a) rms speed and average speed of molecules is same (b) ratio of rms speed of average speed is
31. A closed vessel is maintained at a constant temperature. It is first evacuated and then vapor is injected into it continuously. The pressure of the vapour in the vessel (a) increases continuously (b) first increases and then remains constant (c) first increases and then decreases (d) none of the above
1 f
30. A gas has volume V and pressure p. The total translational kinetic energy of all the molecules of the gas is (a) (3/2) pV (b) (3/2) pV only if the gas is diatomic (c) > (3/2) pV if the gas is diatomic (d) (3/2) pV in all cases
(d) 1 +
2 f
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29. The average translational kinetic energy of O2 (molar mass 32) at a particular temperature is 0.048 eV. The average translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is (a) 0.0015 (b) 0.003 (c) 0.048 (d) 0.768
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
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4. Hydrogen gas and oxygen gas have volume 1 cm3 each at NTP (a) Number of molecules is same in both the gases (b) The rms velocity of molecules of both the gases is the same (c) The internal energy of each gas is the same (d) The average velocity of molecules of each gas is the same.
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5. Pick the correct statement(s). (a) The rms translational speed for all ideal gas molecules at the same temperature is not the same but it depends on the mass. (b) Each particle in a gas has average translational kinetic energy and the equation 1 2 3 mvrms = kT establishes the relation2 2 ship between the average translational kinetic energy per particle and temperature of an ideal gas. It can be concluded that single particle has a temperature. (c) Temperature of an ideal gas is doubled from 100ºC to 200ºC. The average kinetic energy of each particle is also doubled. (d) It is possible for both the pressure and volume of monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change.
(a) The value of a V0 is 2N (b) The ratio Vavg/V0 is equal to 2/3 (c) The ratio Vrms/V0 is equal to 1/ 2 speed V (d) Three fourth of the total particle has a speed between 0.5V0 and V0.
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3. Let v , Vrms and Vp respectively denote the mean speed, the root-mean-square speed, and the most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. (a) No molecule can have speed greater than Vrms. (b) No molecule can have speed less than Vp . 2 (c) Vp < V < Vrms (d) The average kinetic energy of a mol3 ecule is mv 2p 4
1. Pick the correct statement(s). (a) The rms translational speed for all ideal-gas molecules at the same temperature is not the same but it depends on the mass. (b) Each particle in gas has average translational kinetic energy and the equation 1 3 mvrms = kT establishes the relation2 2 ship between the average translational kinetic energy per particle and temperature of an ideal gas. It can be concluded that single particle has a temperature. (c) Temperature of an ideal gas is doubled from 100ºC to 200ºC. The average kinetic energy of each particle is also doubled. (d) It is possible for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change.
ONE OR MORE THAN ONE CHOICE MAY BE CORRECT
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True. 1. Statement 1: Internal energy of a gas must V increases when its temperature is increased. Statement 2: Internal energy of a gas is proportional to the velocity of the vessel in which gas is contained.
7. The total kinetic energy of translatory motion of all the molecules of 5 litres of nitrogen exerting a pressure P is 3000J. (a) the total KE of 10 litres of N2 at pressure of 2 P is 3000 J (b) the total KE of 10 litres of He at pressure of 2 P is 3000 J (c) the total KE of 10 litres of O2 at pressure of 2 P is 20000 J (d) the total KE of 10 litres of Ne at pressure of 2 P is 12000 J
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6. Statement 1: The maxwell speed distribution graph is symmetric about the most probable speed. Statement 2: The root mean square speed of an ideal gas depends on the type of gas (monatomic, diatomic or polyatomic).
5. Statement 1: State variables (P, V and T) of any gas at low densities obey the equation PV = nRT. Statement 2: Real gases are good approximation of an ideal gas at low density.
7. Statement 1: The r.m.s. speed of an ideal gas depends on type of gas.
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3. Statement 1: Absolute zero temperature is not the temperature of zero energy. Statement 2: Only the translational kinetic energy of the molecules is represented by temperature. C 4. Statement 1: The ratio v for a monaCp tomic gas is more than for a diatomic gas. Statement 2: The molecules of a monatomic gas have more degrees of freedom than those of a diatomic gas.
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ASSERTION AND REASON QUESTIONS Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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10. If the atoms in a diatomic molecule can vibrate, the molecule has (a) 2 rotational degrees of freedom (b) 3 rotational degrees of freedom (c) 1 vibrational degrees of freedom (d) 2 vibrational degrees of freedom
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9. An actual gas deviates from perfect gas behaviour as its molecules (a) Have definite size (b) Attract each other (c) Show Brownian motion (d) Are not spherical
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8. According to the kinetic theory, 0K is that temperature at which (a) pressure of ideal gas is zero (b) volume of ideal gas is zero (c) internal energy of ideal gas is zero (d) ideal gas liquefies
2. Statement 1: Systems are in thermal equilibrium, when their temperature are same or average kinetic energy per molecule is same. Statement 2: Total kinetic energy or internal energy or total energy does not determine the direction of flow of heat.
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6. Two vessels of the same volume contain the same gas at same temperature. If the pressure in the vessels be in the ratio of 1 : 2, then (a) the ratio of the average energy is 1 : 2 (b) the ratio of the root mean square velocity is 1 : 1 (c) the ratio of the average velocity is 1 : 2 (d) the ratio of number of molecules is 1 : 2
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
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Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II.
MATCH THE COLUMN TYPE QUESTIONS
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3. Column II give γ for gas mention in column II, match them correctly. Column I Column II 5 (a) monoatomic (p) = 1.67 3 7 (b) diatomic (q) = 1.4 5 4 = 1.33 (c) linear polyatomic (r) 3 3 (d) non-linear polyatomic (s) 7
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14. Statement 1: Two different gas molecules having same temperature will have the same KE. Statement 2: KE of gas molecules follows the Boltzmann’s law.
2. Column II gives value of Cp for gas mention in column I match them correctly. Column I Column II (a) Helium (p) 7R/2 (b) Oxygen (q) 5R/2 (c) Methane (r) 3R/2 (d) Air (s) 4R
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13. Statement 1: Degree of freedom of an ideal gas is three. Statement 2: We can boil water by convection inside earth’s satellite.
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12. Statement 1: The internal energy of a given sample of an ideal gas depends only its temperature according to kinetic theory of gases. Statement 2: The ideal gas molecules do not exert intermolecular forces.
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11. Statement 1: If temp. of gas increase than most probable speed will be increase. Statement 2: Mean free path of gas 1 λm = 2πd 2 n
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10. Statement 1: The pressure exerted by an enclosed ideal gas does not depends on the shape of the container. Statement 2: The pressure of an ideal gas depends on the number of moles, temperature and volume of the enclosure.
9. Statement 1: In Maxwell’s speed distribution graph, for a given amount of gas, the area under the graph increases as the temperature of the gas increases. Statement 2: Increase in temperature broadening the curve.
Column I Column II (a) In the con(p) Pressure of the gas tainer X is more. (b) In the con(q) RMS speed of gas tainer Y molecules is more (c) Since the (r) Average thermal number of energy of a molmolecules is ecule is the same as more that of a molecule in the other container (d) Since the (s) Internal energy of gas has a the system is more smaller (s) than that of the molecular system in the other mass container.
1. X and Y are two equal size containers. X contains 5 mol of H2 and Y has 10 mol of O2 at the same temperature. Assuming that the gases are ideal, match Columns-I and II:
8. Statement 1: All the poly-atomic gas molecules have six degrees of freedom. Statement 2: The number of degree of freedom depends only on the atomicity of gas.
Statement 2: Different degree of freedom is associated with gases of same nature of gas.
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PASSAGE BASED QUESTIONS PASSAGE-1 The rms speed of an oxygen molecule in a sample at 300 K is about 480 m/s. This does not mean
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m dN = 4πN 2πkT
3/2
v 2 e − mv
2
/ 2 kT
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7. It can be concluded that (a) average translational kinetic energy of a molecule in container X is more than that of a molecule in container Y (b) average translational kinetic energy of a molecule in container Y is more than that of a molecule in container X (c) average translational kinetic energy of a molecule in container X could be more or could also be less than that of a molecule in container Y (d) average translational kinetic energy of a molecule in container X is equal to that of a molecule in container Y
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2. The relation between T2 and T1 is given by (a) T2 > T1 (b) T1 < T2 (c) T1 = T2 (d) None
4. The order of magnitude between vp, vrms and vmean is (a) vp < vrms < vmean (b) vrms < vp < vmean (c) vp < vmean < vrms (d) None
8. It can be concluded that (a) internal energy of gas in container X is more than that of the gas in container Y (b) internal energy of gas in container X is less than that of the gas in container Y
3. Area under graph at T1 is A1 and under T2 is A2 then (a) A2 > A1 (b) A1 < A2 (c) A1 = A2 (d) None
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dN is dv (b) L–1T (d) None
1. The dimensional formula for
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2kT speed. Its value is given by v p = . m
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6. From the given information, we can conclude that (a) the molecules in container X are moving as fast as those in container Y (b) the molecules in container X are moving faster than those in container Y (c) the molecules in container Y are moving faster than those in container X (d) no possibility can be ruled out, i.e., molecules in X could be moving faster than these in Y or the molecules in Y could be moving faster than those in X
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We see that there are some molecules which have speeds many times greater has the mean speed. This fact helps in making nuclear fusion reactions in a laboratory. The speed vp at which dN/dv is maximum is called the most probable
X and Y are two equal size containers. X contains a gas ‘A’ at a temperature 75ºC and Y contains a gas ‘B’ at a temperature 40ºC. Each gas behaves as an ideal gas and specific heat at constant pressure for both gases has the same value. There is no further information about the systems.
dv
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PASSAGE-2
Figure shown plots of dN/dv against v at two different temperatures.
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5. On increasing temperature number of molecules corresponding to most probable speed will (a) Increase (b) Decrease (c) Remain same (d) None
that the speed of each molecule is 480 m/s. Many of the molecules have speed less than 480 m/s and many have speed more than 480 m/s. Maxwell derived an equation giving the distribution of molecules in different speeds. If dN represents the number of molecules with speeds between v and v + dv then
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Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
(c) internal energy of gas in container X could be more or could also be less than that of the gas in container Y (d) internal energy of gas in container X has the same value as that of the gas in container Y
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2. A thermally insulated vessel is divided into two parts by a heat-insulating piston which can move in the vessel without friction.
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5. A smooth vertical tube having two different sections is open from both ends and equipped with two pistons of different areas as shown in figure.
7. A vessel of volume V = 7.5 ℓ contains a mixture of ideal gases at a temperature T = 300 K. η1 = 0.10 mole of oxygen, η2 = 0.20 mole of nitrogen, and η3 = 0.30 mole of carbon dioxide. Assuming the gases to be ideal, find (a) the pressure of the mixture; (b) the mean molar mass M of the given mixture which enters its equation of state pV = (m/M)RT, where m is the mass of the mixture. 8. For silver, the molar specific heat at constant pressure in the range 50 K to 100 K is given by, Cp = (0.076 T – 0.00026 T2 – 0.15) cal/K where T is the temperature. If 2 mole of silver is heated from 50 to 100 K, calculate the heat required in calories?
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4. A vessel of volume V = 30 ℓ contains ideal gas at the temperature 0ºC. After a portion of the gas has been let out, the pressure in the vessel is decreased by ∆p = 0.78 atm (the temperature remaining constant). Find the mass of the released gas. The gas density under the normal conditions ρ = 1.3 g/ℓ.
6. Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into the other if the pressure difference ∆p ≥ 1.10 atm. Initially there was a vacuum in one vessel while the other contained ideal gas at a temperature t1 = 27ºC and pressure p1 = 1 atm. Then both vessels were heated to a temperature t2 = 107ºC. Up to what value will the pressure in the first vessel (which had vacuum initially) increase?
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3. Molar volume is the volume occupied by 1 mole of any ideal gas at STP. Show that it is 22.4 litres.
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The left part of the vessel contains one mole of an ideal monatomic gas, and the right part is empty. The piston is connected to the right wall of the vessel through a spring whose length in free state is equal to the length of the vessel (figure). Determine the heat capacity C of the system, neglecting the heat capacities of the vessel, piston and spring.
Each piston slides within a respective tube section. One mole of an ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional area of the upper piston is ∆S = 10 cm2 greater than that of the lower one. The combined mass of the two pistons is equal to m = 5 kg. The outside air pressure is p0 = 1 atm. By how many kelvins must the gas between the pistons be heated to shift the pistons through l = 5 cm?
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1. Two perfect diatomic gases at absolute temperature T1 and T2 are mixed. There is no loss of energy. Find the temperature of the mixture if masses of the molecules are m1 and m2 and the number of molecules in the gases are n1 and n2.
SUBJECTIVE QUESTIONS
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3. A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 s, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall without any collision with other atoms. Take R = 25/3 J/mol-K and k = 1.38 × 10–23 J/K] [2002]
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12. One mole of an ideal monoatomic gas undergoes a process defined by U = a V where U is internal energy and V is its volume. The molar specific heat of the gas for * R. The numthis process is found to be 12 ber in the numerator is not readable. What may be this number?
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11. For the case of an ideal gas find the equation of the process (in the variables T, V) in which the molar heat capacity varies as (a) C = CV + αT; (b) C = CV + βV; (c) C = CV + aP, and α, β and a are constant
2. A gaseous mixture enclosed in a vessel of volume V consists of one gram mole of gas A with γ = (Cp/CV = 5/3) and another gas B with γ = 7/5 at a certain temperature T. The gram molecular weights of the gases A and B are 4 and 32 respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation, PV19/13 = constant in adiabatic process. (a) Find the number of gram moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at 300 K. (c) If T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 of its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. [1995]
10. One mole of an ideal gas whose adiabatic exponent equals γ undergoes a process p = p0 + α/V, where p0 and α are positive constants. Find (a) heat capacity of the gas as a function of its volume; (b) the internal energy increment of the gas, the work performed by it, and the amount of heat transferred to the gas, if its volume increased from V1 to V2.
9. A vertical cylinder of cross-sectional area S contains one mole of an ideal monatomic gas under a piston of mass M. At a certain instant, a heater which transmits to a gas an amount of heat q per unit time is switched on under the piston. Determine the established velocity v of the piston under the condition that the gas pressure under the piston is constant and equal to p0, and the gas under piston is thermally insulated.
Kinetic Theory of Gases and Thermodynamics
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(a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the box.
4. An insulated box containing a monoatomic gas of molar mass M moving with a speed v0 is suddenly stopped. Find the increment in gas temperature as a result of stopping the box. [2003] 5. Statement 1: The total translational kinetic energy of all the molecules of a given mass
1. A closed container of volume 0.2 m3 contains a mixture of neon and argon gases at a temperature of 27ºC and pressure of 1 × 105 Nm–2.The total mass of the mixture is 28 g If the molar masses of neon and argon are 20 and 40 g mol–1 respectively, find the masses of the individual gases in the container assuming them to be ideal. [Universal gas constant R = 8.314 J/mol-K] [1994]
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PREVIOUS YEAR ’ IIT-JEE QUESTIONS
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Kinetic Theory of Gases and Thermodynamics
(b) Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1 (c) Statement 1 is True, Statement 2 is False (d) Statement 1 is False, Statement 2 is True
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Kinetic Theory of Gases and Thermodynamics
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QUESTION BANK 2 (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
1. Why are heat pumps more often used in mild climates than in areas with severely cold winders? 2. Suppose a body spends its mechanical energy to overcome friction. Will it be a reversible process?
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1. Which of the following process is possible according to the first law of thermodynamics? (a) W > 0, Q < 0 and dU = 0 (b) W > 0, Q < 0 and dU > 0 (c) W > 0, Q < 0 and dU < 0 (d) W < 0, Q > 0 and dU < 0
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8. The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. Explain.
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12. If you leave the refrigerator door open and the refrigerator runs continuously, does the kitchen get colder or warmer? Explain.
4. Is it possible to make a heat pump with a coefficient of performance equal to 1? Explain.
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11. The entropy of a system increases by 10 J/K. Does this mean that the process is necessarily irreversible? Explain.
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CONCEPTUAL QUESTIONS
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2.88
Kinetic Theory of Gases and Thermodynamics
done by the gas is same as increase in internal energy of the gas. The molar heat capacity of gas for the process is (a) C = 4R (b) C = 0 (c) C = 2R (d) C = R
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5. In a process, the pressure of a gas is inversely proportional to the square of the volume. If temperature of the gas is increases, then work done by the gas (a) is positive (b) is negative (c) is zero (d) may be positive
8. Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as energy is added to its heat Q. Graph shows the final volume Vf versus Q. The temperature of the gas is 25 (use ln 9 = 2 and R = J/mol-K) 3
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12. If W1 is the work done in compressing an ideal gas from a given initial state through a certain volume isothermally and W2 is the work done in compressing the same gas from the same initial state through the same volume adiabatically, then
14. The equation of state of a gas is
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15. 5 moles of nitrogen gas are enclosed in an adiabatic cylindrical vessel. The piston itself is a rigid light cylindrical container
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(b) 293 K (d) 412 K
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Kinetic Theory of Gases and Thermodynamics
containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal to the nitrogen gas. A power of 30 cal is transferred to Helium through the bottom surface of the piston. The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly
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18. 0.5 mole of an ideal gas at constant temperature 27ºC kept inside a cylinder of length L and cross-section area A closed by a massless piston.
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Kinetic Theory of Gases and Thermodynamics
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20. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of gas is (a) 2/5 (b) 3/5 (c) 3/7 (d) 5/7
23. A gas is expanded from volume V0 to 2V0 under three different processes. Process 1 is isobaric process, process 2 is isothermal and process 3 is adiabatic. Let ∆U1, ∆U2 and ∆U3 be the change in internal energy of the gas is these three processes. Then
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ONE OR MORE THAN ONE CHOICE MAY BE CORRECT
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Kinetic Theory of Gases and Thermodynamics
(d) Total heat given to the gas in (ii) case is greater than in (i)
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2.98
Kinetic Theory of Gases and Thermodynamics
Statement 2: Initial and final temperature is equal, therefore change in internal energy is zero.
2. Statement 1: If an ideal gas expands in vacuum in an insulated chamber, ∆Q, ∆U and ∆W all are zero. Statement 2: Temperature of the gas remains constant.
MATCH THE COLUMN TYPE QUESTIONS Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II. 1. Match the column.
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Column II (p) –P/V
Column I (a) Adiabatic bulk modulus
9. Statement 1: In cyclic process, initial and final state are same. Therefore net work done is zero.
7. Statement 1: Internal energy change is zero if the temp is constant, irrespective of the process being cyclic or noncyclic. Statement 2: dU = n CvdT for all process and is independent of path. 8. Statement 1: Adiabatic expansion is always accompanied by fall in temperature. Statement 2: In adiabatic process, volume is inversely proportional to temperature.
Column II (Symbols have usual meaning) (p) ∆U = 0 (q) Q = 0 (r) W = 0 (s) ∆U = nCV∆T
2. The P-V diagram of 0.2 mol of a diatomic ideal gas is shown in figure. Process BC is adiabatic γ = 1.4. Column I Column II (a) ∆QAB (J) (p) 602 (b) ∆WBC (J) (q) –644 (c) ∆UCA (J) (r) 1246 (d) ∆UBC (J) (s) –602
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Column I (Processes for ideal gases) (a) Isothermal (b) Isobaric (c) Isochoric (d) Adiabatic free expansions
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compared to that at constant pressure for doing work in expansion.
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Kinetic Theory of Gases and Thermodynamics
(b) Slope of P – V graph in isothermal process (c) Degree of freedom (d) The ratio of molar heat capacity at constant pressure to universal gas constant R
(q) 2/(γ – 1)
PASSAGE BASED QUESTIONS PASSAGE-1 One mole of an ideal gas is taken from an initial state P0, V0 and temperature T0 through the following activities. (γ = CP/CV, CP = 7 R/2, CV = 5 R/2) (i) Heating at constant volume to a temperature three times. (ii) Adiabatic expansion to a volume 2V0. (iii) Cooling at constant volume to a temperature one third. (iv) Adiabatic compression so that it is returned to its initial state.
(r) γP γ γ −1
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Column I Column II (a) Heat given to sys- (p) Positive tem (i.e., ideal gas sample) during isothermal expansion. γ−1 P1 V1 V2 1 − (q) (b) Work done by gas ( γ − 1) V1 during adiabatic γ−1 P1 V1 V2 compression − 1 ( γ − 1) V1 γ−1 P1 V1 V2 (c) Change in inter(r) (1 − γ ) V − 1 1 nal energy of gas γ−1 P1 V1 V2 sample during − 1 adiabatic process. (1 − γ ) V1 (d) Change in inter(s) Negative. nal energy of gas sample from most initial state to the final state.
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4. The temperature at D is (a) T0 21.4 (b) T0 20.4 (c) T0 2–1.4 (d) T0 2–0.4
7. Identify the gas filled in the container A and B. (a) N2, Ne (b) He, H2 (c) O2, Ar (d) Ar, O2
3. The total heat given to the system is (a) 1.25 RT0 (b) 3.75 RT0 (c) 2.25 RT0 (d) 2.275 RT0
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∆PA = 2.5 atm, ∆PB = 1.5 atm Required data for unknown gas: Mono (molar gas) Kr 84g
Xe Rd 131g 222g
O2 32g
Cl2 71g
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He Ne Ar 4g 20g 40g Dia (molar gas) H2 F2 N2 2g 19g 28g
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PASSAGE-2
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8. Total number of molecules in ‘A’ (Here NA = Avagadro number) 125 (a) NA (b) 3.125 NA 64 125 (c) NA (d) 31.25 NA 28
6. Identify the type of gas filled in container A and B respectively. (a) Mono, mono (b) Dia, dia (c) Mono, dia (d) Dia, mono
2. The pressure at point C is given by 3P 3P (a) 0.40 (b) 1.40 2 2 (c) 3P0 × 20.4 (d) 3P0 × 21.4
Kinetic Theory of Gases and Thermodynamics
PASSAGE-3 The PV diagram for a cyclic process performed on an ideal monoatomic gas is shown in figure. The curve AC is one fourth of circle. 3 1P
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(a) 6 RT0 (1 − η−1/3 )
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1. One mole of a certain ideal gas is contained under the weightless piston of a vertical cylinder at a temperature T. The space over the piston opens into the atmosphere of pressure P0. What work has to be done in lifting the piston slowly so that the volume of the gas under the piston increases isothermally n times?
12. Heat absorbed by the gas in process B to C is (a) 150 J (b) –150 J (c) 300 J (d) Can’t be calculated
SUBJECTIVE QUESTIONS
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
Determine the ratio of the energy given out by the gas to its surroundings during the isochoric section of the cycle to the expansion work done during the isobaric section of the circle.
5. A horizontal cylindrical vessel of length 2l is separated by a thin heat-insulating piston into two equal parts each of which contains n moles of an ideal monatomic gas at a temperature T.
7. One mole of an ideal monoatomic gas undergoes the cyclic process as shown. Find out efficiency (in percent) of the cycle. [Take ln 2 = 0.7]
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The piston is connected to the end faces of the vessel by undeformed springs of rigidity k each (figure). When an amount of heat Q is supplied to the gas in the right part, the piston is displaced to the left by a distance x = l/2. Determine the amount of heat Q, given away at the temperature T to a thermostat with which the gas in the left part is in thermal contact all the time.
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P = (3/2) T1/2. It is found that its temperature increases by 100 K in this process. Now it is taken from state B to C through a process for which internal energy is related to volume as U = (1/2) V1/2. Find the total work performed by the gas (in Joule), if it is given that volume at B is 100 m3 and at C it is 1600 m3. [Use R = 8.3 J/mol-K]
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Kinetic Theory of Gases and Thermodynamics
the bubble rises to the surface its diameter increases n = 1.1 times. Find how deep is the pond at that spot. The atmospheric pressure is standard, the gas expansion is assumed to be isothermal.
9. A heat conducting piston can move freely inside a closed, thermally insulated cylinder with an ideal gas (γ = 5/3). At equilibrium the piston divides the cylinder into two equal parts, the temperature of gas being equal to 300 K.
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(b) 3/5 (d) 5/7
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4. Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, calculate the following quantities in this process. [1992]
If p1 = 105 Pa, p0 = 3 × 105 Pa, p2 = 4 × 105 Pa, V2 – V1 = 10 ℓ, and segments 4-3 and 2-1 of the cycle are parallel to the V-axis.
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7. An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5960 J, Q2 = –5585 J, Q3 = –2980 J and Q4 = 3645 J, respectively. The corresponding quantities of work involved are W1 = 2200 J, W2 = –825 J, W3 = –1100 and W4 respectively. [1994]
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5. An ideal gas is taken from the state A (pressure P, volume V) to the state B (pressure P/2, volume 2V) along a straight line path in the P-V diagram. Select the correct statements from the following. [1993]
(a) The net change in the heat energy (b) The net work done (c) The net change in internal energy
The pressure and temperature at A, B, etc are denoted by PA, TA, PB, TB etc respectively. Given that TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA, calculate the following quantities: (a) The work done by the gas in the process A→B (b) The heat lost by the gas in the process B →C (c) The temperature TD. [Given: (2/3)2/5 = 0.85]
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(b) 2 mA = 3 mB (d) 9 mA = 4 mB
11. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in
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10. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be ∆P and 1.5 ∆P respectively. Then [1998]
13. One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate [1998]
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Kinetic Theory of Gases and Thermodynamics
an adiabatic compression until its volume is V2. Then the gas is given heat Q at constant volume is V2. [1999]
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19. In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas [2001]
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(a) the temperature of the gas at point B (b) heat absorbed or released by the gas in each process (c) the total work done by the gas during the complete cycle. [Express your answer in terms of the gas constant R]
(a) the temperature will decrease (b) the volume will increase (c) the pressure will remain constant (d) the temperature will increase 20. P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to [2001]
23. Which of the following graphs correctly dV /dP represent the variation of β = − V with P for an ideal gas at constant temperature? [2002] E
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Column II ∆W > 0 ∆W < 0 ∆Q > 0 ∆Q < 0
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27. P-V graph of cyclic process is shown for an ideal gas. Match the column correctly. [2006]
(a) change in temperature (b) work done and (c) change in internal energy [Given: specific heat 400 J/kg/ºC, coefficient of cubical expansion, γ = 9 × 10–5/ ºC, density ρ = 9000 kg/m3, atmospheric pressure = 105N/m2]
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Kinetic Theory of Gases and Thermodynamics
P0 Mg − 2 πR 2
(d)
P0 Mg + 2 πR 2
(c) An ideal monoatomic gas expands to twice its. (d) An ideal monoatomic gas expands such that its. Column II (p) The temperature of the gas decreases separated by a valve. Chamber I contains an ideal gas and the Chamber II has vacuum. The valve is opened.
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Kinetic Theory of Gases and Thermodynamics
ONLY ONE OPTION IS CORRECT 1. 9. 17. 25. 33.
(c) (c) (b) (a) (a)
2. 10. 18. 26. 34.
3. 11. 19. 27. 35.
(d) (a) (c) (c) (d)
4. 12. 20. 28. 36.
(a) (d) (b) (c) (a)
5. 13. 21. 29. 37.
(b) (b) (d) (d) (a)
6. 14. 22. 30. 38.
(b) (d) (c) (a) (a)
(d) (b) (d) (b) (b)
7. 15. 23. 31. 39.
(b) (a) (a) (c) (c)
8. 16. 24. 32.
(a) (d) (c) (c)
ONE OR MORE THAN ONE CHOICE MAY BE CORRECT 3. (a, c) 4. (a, b, d) 5. (a, c, d) 6. (b, d) 11. (b, c, d) 12. (a, c) 13. (a, d) 14. (c, d)
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3. In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperature becomes equal. The final temperature can be the mean temperature (T1 + T2)/2 only when thermal capacities of the two bodies are equal.
2. When a body is taken from one position to another, it spends mechanical energy to overcome friction. This energy is dissipated as heat. Now, if the body is allowed to go round to its path in reverse direction, its energy spent is not recovered but instead is further spent in the reverse path. Hence the process is not reversible.
CONCEPTUAL QUESTIONS
HINTS AND SOLUTIONS
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Kinetic Theory of Gases and Thermodynamics
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∴ ∆U = 0 ⇒ ∆Q = ∆W = –RT0 ln 2 Process 3-4: Since V ∝ T, thus P = constant 5 ∴ ∆Q = RT0 ; ∆W = RT0 2 Process 4-1: Since temperature is constant
3 WAB = ∫ P d v = ∫ T 1/ 2 dv 2
4. Process A → B
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10. (a) The gas would pass from A to B. The change in pressure and volume will take place under adiabatic conditions. The final pressure in the two cylinders would be 0.5 atm. (b) The change in internal energy of the gas is zero. (c) The change in temperature will be zero. (d) The process is rapid, the intermediate states are non-equilibrium states and hence do not satisfy the gas equation. The intermediate states do not lie on the P-V-T surface.
Process B → C 1 U = V 1/2 2 1 3 1 RT = V 1/2 ⇒ 3 PV1/2 = 1 ∴ p = 3 V 2 2 1600 1 2 dv = V Now WBC = ∫ Pdv = ∫ 3 3 V 100
8. The climate of a Harbour town is more temperate because of the large amount of water vapours present in the air due to its proximity with sea.
SUBJECTIVE QUESTIONS
3 1 = ∫ T 1/2 × RT −1/ 2 dT 2 3 On solving, WAB = 50 R = 50 × 8.3 = 415 J
6. During driving the temperature of the air, inside tyres increases due to friction force between road and tyres. As volume of air in the tyre is constant according to Charle’s law. (P ∝ T)V pressure increases with increase in temperature.
Larger the value of specific heat of the coolent more is the heat absorbed by it.
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Thus, PV = constant ∴ ∆Q = ∆W = 2RT0 ln 2
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= –RT0 – RT0 ln 2 + RT0 + 2RT ln 2 = RT0 ln 2 5 Heat supplied = RT0 + 2 RT0 ln 2 2 ∴ Efficiency RT0 ln 2 = (5 − 2) RT0 + 2 RT0 ln 2
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PREVIOUS YEARS’ IIT-JEE QUESTIONS
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Thus efficiency = (Net work done)/(Heat supplied)
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Kinetic Theory of Gases and Thermodynamics
27. (A) → (s), (B) → (p, r), (C) → (r), (D) → (q, s)
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Kinetic Theory of Gases and Thermodynamics
Q = ∆U + W
∆U = +ve (∆T = +ve)
W = +ve (As volume increases)
PV (d): T = as produced of P and V nR increases, so temperature increases
So, Q = +ve
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In simple terms, we can say that wave motion involves transfer of disturbance (energy) from one point to the other with the particle of medium oscillating about their mean positions i.e., the particles of the medium do not themselves travel along with the wave. Instead, they oscillate back about some equilibrium position. As the wave passes by, only the disturbance is propagated. In this chapter, we will limit discussion to mechanical waves (elastic waves) which require a medium to travel. There are also electromagnetic waves which do not require any medium and can travel in vacuum. The idea of wave motion can be understood by a very simple example. If a stone is dropped in a pool of water, a disturbance is created when it enters the surface. The disturbance is not confined to that place alone but spreads out and eventually it reaches all the parts of water, and brings the particle with which it comes in contact, into motion. These particles then set into motion neighbouring particles. They in turn produce similar motion in other particles and in this fashion disturbance is transmitted from particle to particle. The particles oscillate over short paths about their initial positions and as a result a wave moves through the medium. The medium as a whole does not go in the directions of the motion of the wave.
3.2 TRANSVERSE WAVE In transverse waves, the particle of the medium oscillate perpendicular to the direction in which the wave travels. Travelling waves on tight rope are transverse waves. If one of the rope is rigidly fixed and the other end is given periodic up and down jerks, the disturbance propagates along the length of the rope but the particles of the rope oscillate up and down. Disturbance travels along the rope in form of crests (upward peaks) and troughs (downward peaks). :DYHPRWLRQ
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Wave Mechanics
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In longitudinal waves, the oscillation of the particles is parallel to the direction in which the wave travels. Disturbance travelling in a spring parallel to its length, a pressure variation propagating in a liquid, sound waves travelling in a medium are examples of longitudinal waves. Let us consider a stretched spring. If one end of the spring is suddenly given in and out oscillation parallel to length of the spring, the coils starts exerting force on each other and the compression and the expansion points travel along the length of the spring. The coils oscillate right and left parallel to the spring.
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Compression (crowding together of molecules) and rarefactions (spreading out of molecules away from each other) travel along the tube. The pressure at the compression point is higher and the pressure at a rarefaction point is lower. The molecules of air oscillate right and left i.e., parallel to the wave propagation. Longitudinal waves do not require shearing stress and, hence can travel in any elastic medium: solid, liquid and gases.
Wave Speed (c)
The speed of wave is the distance it covers in one second. It depends on medium. It should be carefully noted that wave speed is completely different from particle speed. Particle speed the speed of the vibrating particles in the medium. On the other hand wave speed is the speed with which (the disturbance/or wave) propagates in the medium.
Wave Frequency (ν) The frequency with which the particle of the medium (through which the wave passes oscillates is known as wave frequency. In transverse waves, frequency is the number of crests (or troughs) that pass through a point in one seconds. In longitudinal waves, frequency is the number of compressions (or rarefactions) that pass through a point in one second. https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
Time Period (T) The time period of the oscillation of the particle in the medium is the time period of the wave.
Amplitude (A) Amplitude of the wave is same as the amplitude of the oscillating particles.
Wave Length (λ) Wave length is the distance between two consecutive crests (or compression) in a wave. Wave length, wave speed and frequency are related as follows:
Phase
rio s_ in fo
na .c om
c = υλ
ita er si ta rio s
er s
//w
$
4
.m
e/
s:
:DYH
ht tp s:
un
//t
5
iv
tp
O
3
un iv
w
w
.ja
m
ar a
When a wave passes through a medium, all particles oscillate with same frequency, but they reach the corresponding positions in their path at different time instants.
ht
s: //
t.m e/
For example, in the above figure, the particle at P is at its top extreme, the particle at Q, is passing through its mean positions, the particle at R is at its bottom extreme. These relative positions represents the phase of motion. (i) If two particles have same position and same velocity at all time instants, they are said to be in same phase (or in phase). (ii) Two particles are said to be in opposite phase (or exactly out of phase) if their displacement from mean position and their velocities are equal in magnitude but opposite in direction.
ht tp
3.4
3
6
7
4 5 P, S and T are in same phase
P and R are is opposite phase
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Wave Mechanics
3.5
■ The distance between particles in same phase = 0, λ, 2 λ ... ■ The distance between particles in opposite phase = λ/2, 3λ/2,
5λ ... 2
3.5 INTENSITy OF A WAVE (I) In a travelling wave, energy is transferred through the medium in the direction in which the wave travels. The transfer of energy per unit time per unit area perpendicular to the direction of motion of the wave is called as the intensity of the wave. If A is the amplitude, c is the speed, ν is the frequency, ρ is the density of the medium, I is the intensity, then I = 2π2 c ρ ν2 A2
...(i)
rio s_ in fo
ita
P 4π r 2
er si ta rio s
er s
//w
w
I=
un iv
w
.ja
m
ar a
na .c om
Note: Intensity is proportional to the square of the amplitude (I ∝ A2). As the wave propagates through the medium, its energy is shared by a larger number of particles and the energy through unit area per second decreases. Hence, the intensity and amplitude go on decreasing as we move from the source of the waves. For a point source giving out wave in all directions (three-dimensional wave), the variations of intensity with distance (r) from the source is given as ...(ii)
A∝
iv
1 r
t.m e/
⇒
un
//t
1 r2
ht tp s:
I∝
.m
ht
tp
e/
s:
where, P is the power (total energy given out per sec.) of the source. Combining equations (i) and (ii), we can deduce that amplitude at a point is inversely proportional to the distance of the point from the source of the waves.
s: //
3.6 WAVE EqUATION
ht tp
If a mathematical equation describes a wave, it must be able to give the position of any particle of the medium at any given time instant. Consider a transverse wave travelling towards right in a tight string lying along x-axis. If we take one point on the string as origin O, the displacement (y) at any particle P located at some 2π x 2π t X-co-ordinate (x) at an instant t is given by y (x, t) = A sin − + φ0 T λ
3 [\ [D[LV 2
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Wave Mechanics
■ φ0 depends upon the initial condition (at φ = 0) and is called initial phase. 2π x 2 xt − + φ0 is called the phase angle and represents the phase λ T of a particle located at x at time instant t. ■ The wave equation can also be written in the following equivalent forms:
■ The angle φ =
y (x, t) = A sin (kx – ω t + φ0) 2π y ( x, t ) = A sin ( x − ct ) + φ0 λ
ta
s_
rio
2π x1 2π x2 2π − = (∆ x) λ λ λ
.ja m
=
ar
an a.
in fo
co
m
2π where, k = angular wave number = , and λ ω = angular frequency. Phase difference between two particles located at x1 and x2 at the same time instant is 2π x1 2π t 2π x2 2π t ∆ φ = φ1 − φ2 = − + φ0 − − + φ0 T T λ λ
un iv
rio
∆ φ = 0, 2π, 4π .... for ∆x = 0, λ, 2λ ....
//w
⇒
s
er
w
w
si
where, ∆ x is the separation between the particles.
ta
si
er
un iv
∆ φ = π, 3 π,5 π, ....for ∆ x =
s:
//t
ht
tp
.m e/
s:
Hence, particle are in same phase if the phase difference is an even multiple of π. Similarly, we can see that λ 3λ , ,... 2 2
dy vp = dt x = constant
ht tp
s:
//t
ht
tp
.m e/
Hence particles are in opposite phase if the phase difference is an odd multiple of π. The velocity of the particle at any instant isgiven by
= – ωA cos (kx – ωt + φ0)
The equation of wave travelling towards left (negative x-axis) y (x, t) = A sin (kx + ωt + φ0).
(i) Find the frequency, wave length and the speed of sound wave in air. (ii) If the equilibrium pressure of air is 1.0 × 105 N/m2, what are the maximum and minimum pressure at a point as the wave passes through that point?
Example The equation of a sound wave in air is given by P = (0.01N/m2) sin [(1000s–1) t – (3.0 m–1) x]
3.6
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Wave Mechanics
3.7
(i) Comparing with the standard form of a travelling wave x P = P0 sin ω t − υ we see that, ω = 1000 s–1 ω 1000 Frequency, υ = = Hz = 160 Hz 2π 2π Also from the same comparison
ω = 3.0 m–1 c ω or c = 3.0 m −1
1000 s −1 = 333 m/sec. 3.0 m −1 c Wavelength, λ = ν
si ta rio
m .ja w
=
s_
an
ar
333m/sec. = 2.1 m 160 Hz
s
ta
s:
un
iv
//w
rio
w
er
(ii) The pressure amplitude is P0 = 0.01 N/m2. Hence, the maximum and minimum pressure at a point in the wave motion will be (1.01 × 105 ± 0.01) N/m2.
in
a.
=
fo
co m
)
) Solution
iv
un
//t
ht
tp
.m e/
er si
Example The density of air at NTP is 1.293 gm/litre and γ = 1.36. Calculate the frequency of a turning fork which emits sound of wavelength 66.5 cm at 17º C.
ht
∴
e/
γP0 1.36 × (1.013 × 106 ) = ρ0 0.001293
//t
s:
c1 =
.m
)
tp s:
) Solution Pressure at NTP = 1.013 × 106 dyne cm–2
tp
= 32640 cm s–1
ht
Let, ct is velocity of sound at 17º C (290 K).
ct T = c1 T0
But
∴
ct = 33640 cm s–1
or
ct 290 = 32640 273
or
ct = νλ
v=
33640 = 506 Hz 66.5
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Wave Mechanics
Example A transverse wave at wavelength 50 cm is travelling towards positive x-axis along a string whose linear density is 0.05 gm/cm. The tension in the string is 450 N. At = 0, the particle at x = 0 is passing through its mean position with upward velocity from an equation describing the wave. The amplitude of the wave is 2.5 cm. )
) Solution Let the wave be described by y (x, t) = A sin (kx – ωt + φ0)
2π 2π = = 4π and A = 2.5 cm λ 0.5 y (0, 0) = 0 A sin φ0 = 0 φ0 = 0, π
in f s_ rio
iv
iv er
.m e/
2πc 2π (300) = = 1200 π rad/sec. λ 0.5 Using all the quantities, the equation is ω = 2πν =
e/
tp
s:
un
//t
⇒
T 450 = = 300 m/sec. m 0.05 × 10−1
un
=
s:
ht
tp
//w
w
Velocity of transverse wave in the string
s
er si
φ0 = π
ta
w .ja
m
– Aω cos φ0 > 0 From equations (i) and (ii), we get
o
dy >0 dt (0,0)
ar a
We also have
si ta rio
∴
na .c om
∴ ⇒
where, k =
//t
ht
.m
y = 2.5 sin (4π x – 1200 πt + π) cm.
ht
tp s:
Example The sound level at a point is increased by 30 dB. By what factor is the pressure amplitude increased? )
) Solution The sound level in dB is I B = 10 log10 I0
If B1 and B2 are the sound levels and I1 and I2 are the intensities in the two cases, then
I I B2 – B1 = 10 log10 2 − log10 1 I0 I0
or or
3.8
I 30 = 10 log10 2 I1 I2 = 103 I1
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Wave Mechanics
3.9
As the intensity is proportional to the square of the pressure amplitude, we have
P2 I = 2 P2 I1 ≅ 32.
= 1000
3.7 SUPERPOSITION OF WAVES Two or more waves can propagate through the same medium simultaneously. At the point where the two waves overlap, the resultant displacement of the particle is the sum of displacement of the individual waves.
m
⇒ y = y1 + y2 + ....
in
s_
rio
m
ta
si
s
er
w w
si ta
un
iv
//w
Stationary Waves
rio
ar an
a.
fo
(i) Stationary waves: These are formed when two waves of same frequency travelling in opposite direction overlap. (ii) Beats: These are formed when two waves of slightly different frequency travelling in same direction overlap.
.ja
co
The two most important situations involving superposition are:
iv
//t .
y = y1 + y 2
un
m
ht tp
The resultant wave is described by
e/
er
e/
//t .m
y1 = A sin (kx – ωt) travelling along positive x-axis. y2 = A sin (kx + ωt) travelling towards negative x-axis.
s:
ht tp
s:
Let us consider equation of two transversewaves travelling along a tight string in opposite directions.
tp
s:
y = A sin (kx – ωt) + A sin (kx + ωt) y (x, t) = (2 A sin kx) cos ωt
ht
Hence, the particle at a location x is oscillating in simple harmonic motion. With angular frequency ω and amplitude 2 A sin kx. As the amplitude depends on location (x). Particles are oscillating with different amplitudes. The points where amplitude is minimum are called nodes and the points where amplitude is maximum are called antinodes.
Nodes: Amplitude = 0
⇒
⇒
⇒
2 A sin kx = 0 π 2π x = 0, , k k
x = 0, λ/2, λ, 3λ/2, 2λ
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Wave Mechanics
Antinodes Amplitude is maximum ⇒ sin kx = ± 1 π 3π ⇒ x= , ,.... 2k 2k λ 3λ 5λ , , ,.... 4 4 4 O $
1
[
$
O
O $
1
$
[ O O
m
ar a
[
1
na .c om
1
O
rio s_ in fo
⇒ x=
[
[ O
ht tp
s: //
er si ta rio s
t.m e/
7 W 7
W
un
W
//t ht tp s: 1
iv
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
■ Nodes are completely at rest. Antinodes are oscillating with maximum amplitude (2A). The points between a node and an antinode have amplitude between 0 and 2 A. ■ Separation between two consecutive nodes (co-ordinates) = λ/2 ■ Separation between a node and the next antinode = λ/4 ■ Nodes and antinodes are alternately placed.
ht
3.10
1
1
1
W 7 W 7
■ N1, N2 are nodes and, hence remain at rest. The segment of the string between N1 and N2 moves up and down. The figure shows the position of this segment at t = T T 3T T 0, , , , , where T is the time period of the wave. 8 4 4 2 ■ The portion between two consecutive nodes is often called as a loop. ■ It is clear from the figure that since nodes are at rest they do not transfer energy. In a stationary wave, energy is not transferred from one point to the other. https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
3.11
Vibrations of a Stretched String Consider a string of length l stretched between two fixed supports. When a wave is set up in the string, the disturbance travels in both directions while being reflected at each end. For a stationary wave to exist, the fixed ends must be nodes. 1
$
1
O
na .c om
As the distance between two consecutive nodes is
ar a
m
⇒ ⇒
rio s_ in fo
nλ if there are (n + 1 nodes in all) 2 2l c nc n T λ = ⇒ ⇒ ν = = ν= n λ 2l 2l m l=
λ , therefore 2
un iv
1 2l
er si ta rio s
tp
s:
where, ν0
T m
e/
//w
w
er s
w
ita
.ja
where, T = tension, and m = mass per unit length. Hence, the string can vibrate with those frequencies which are integral multiples of the fundamental frequency υ0
iv un
//t
ht
.m
The possible frequencies are: ν0, 2ν0, 3ν0 .....
ht tp
s: //
t.m e/
ht tp s:
The fundamental frequency is simply the lowest possible frequency for vibration in the string. The nth multiple of the fundamental is known as the nth harmonics. ν0 is the first harmonic, 2ν0 is the second harmonic, 3ν0 is the third harmonic and so on. The shape of the loops in the strings for these harmonics is shown below: Fundamental first harmonic = υ0 $
1
1
$ 2QHORRS
Second harmonic = 2υ0 (first overtone) https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics $
$
1
1
1
7ZRORRSV
$
$
$ 1
1
rio s_ in fo
1
ar a
1
na .c om
Third harmonic = 3υ0 (second harmonic)
ita
.ja
m
7KUHHORRS
er si ta rio s
w
er s
w
Vibration in an Air Column in a Tube
$
O
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
e/
s:
un iv
//w
1. Open at both ends: The open ends of the tube become antinodes because the particle at the open end can oscillate freely.
ht
$
ht tp
3.12
Let l = length of the tube (or air column). If there are (n + 1) antinodes in all, then l=
nλ 2
because distance between two consecutive antinodes is ⇒
ν=
λ . 2
c nc = λ 2l
where, c is speed of sound wave.
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Wave Mechanics
3.13
The possible vibration frequencies are integral multiples of the fundamental or the lowest frequency ν0 ν0 =
c 2l
Possible frequencies are: ν0, 2ν0, 3ν0 ......
$
$
$ 1
1
$
$
ar a
1 $
1
$
Second harmonic = 2 ν0
er s
un iv
Third harmonic = 3 ν0
er si ta rio s
Fundamental mode first harmonic = ν0
w w //w
$
ita
.ja
m
1
$ 1
rio s_ in fo
na .c om
1
e/
iv
.m
un
//t
ht
tp
s:
2. Closed at one end: The open end becomes antinode and the closed end beλ comes a node. The distance between a node and antinodes is . 4
t.m e/
ht tp s:
⇒ If there are n nodes and n antinodes, then
ht tp
s: //
$
O
1
l = (2n – 1)
λ 4
c (2n −1) c = λ 4l ⇒ The possible vibration frequencies are odd multiples of the fundamental or the lowest frequency v0. ν=
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Wave Mechanics
v0 =
c 4l
Possible frequencies are: v0, 3v0, 5v0 ..... $
$
$ 1 $
1
1 $
na .c om
$ $
Third harmonic = 3 ν0 (first overtone)
m
ar a
Fundamental mode first harmonic = ν0
rio s_ in fo
1
1
Fifth harmonic = 5 ν0 (second overtone)
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
er si ta rio s
e/
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
Note: There are only odd harmonics in a tube closed at one end. Beats: Beats are formed by the superposition of two waves of slightly different frequencies moving in the same direction. If two sources of sound having slightly different frequencies are placed in the same medium, the resultant effect heard at any fixed position will consists of alternate loud and weak sounds. Let us consider two waves of frequencies ν1 and ν2 and amplitude A. We will analyse the net effect of a convenient point (say x = 0) y1 = A sin 2π ν1 t y2 = A sin 2π ν2 t ⇒ y = y1 + y2 ⇒ y = A (sin A 2π ν1 t + sin 2π ν2t) y = [2A cos π (ν1 – ν2) t] sin π (ν1 + ν2) t
ht
3.14
Thus, the resultant wave can be represented as a travelling wave whose frequency ν + ν2 is 1 and amplitude is 2A cos π (ν1 – ν2) t. As the amplitude term contain t, 2 the amplitude (and, hence intensity or loudness in case of sound) varies periodically with time. For loud sounds: net amplitude = ± 2 A ⇒
cos π (ν1 – ν2) t = ± 1
⇒
π (ν1 – ν2) t = 0, π, 2π, 3π ....
⇒
t = 0,
1 2 , .... ν1 − ν 2 ν1 − ν 2
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Wave Mechanics
3.15
For weak sounds, proceeding similarly, we get t= ∴
1 3 , .... 2(ν1 − ν 2 ) 2(ν1 − ν 2 )
Interval between two loud (or weak) sounds =
1 ν1 − ν 2
the number of loud sound per second = ν1 – ν2 or beat per second ν1 – ν2 Example The vibrations of a string of length 60 cm fixed at both ends are represented by the equation
na .c om
πx y = 4 sin cos (96 π t) 15
ita
.ja
m
ar a
What is the maximum displacement of point at x = 5 cm? Where are the nodes located along the string? What is the velocity of the particle at x = 7.5 at t = 0.25 sec? Write equation of the component waves whose superposition gives the above wave.
er s
w
(i) (ii) (iii) (iv)
rio s_ in fo
where, x and y are in cm and t in second.
er si ta rio s
w
) Solution The given equation is represented by the stationary wave.
un iv
π 2π = 15 λ
⇒ λ = 30 cm
iv
k=
e/
A = 2 cm,
un
//t
ht
tp
s:
We get
.m
//w
Comparing this equation by the general equation y (x, t) = (2 A sin kx) cos ωt
ht tp s:
2π = 96 π, T
t.m e/
⇒ T =
s: //
(i) At x = 5 cm
1 sec. 48
ht tp
π ymax = 4 sin × 5 15 = 4 sin
π 4 3 = = 3.464 cm 3 2 λ 30 = = 15 cm 2 2 = distance of two consecutive nodes
(ii) Loop length =
1 $
1
1
FP FP
1
1 %
FP
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Wave Mechanics
πx (iii) y (x, t) = 4 sin cos (96 π t) 15 Velocity =
∂y πx ( x, t ) = 4sin (–1) sin (96 πt) 96 π dt 15
At x = 7.5 cm and t = 0.25 sec. π× 7.5 c(x, t) = – 4 sin sin (96 π × 0.25) 96 π 15 π = − 4sin . sin (24 π) 96 π 2 =0
na .c om
(The particle is at its maximum displacement.)
ita er si ta rio s
er s
π y2 = 2 sin x − 96 t 15
and
//t
iv
.m
e/
s: tp
π y1 = 2 sin x + 96 t 15
un iv
//w
w
w
.ja
Comparing the equation, we get
rio s_ in fo
m
ar a
(iv) Let two waves are y1 = A sin (kx + ω t) and y2 = A sin (kx – ω t) ∴ y = y1 + y2 = 2 A sin kx cos ω t
ht
un
Example A rifle shot is fired in a valley formed between two parallel mountains. The echo from one mountain is heard after 2 seconds and from the other is heard after 2 seconds.
s: //
t.m e/
ht tp s:
(i) What is the width of the valley? (ii) Is it possible to hear the subsequent echoes from the two mountains simultaneously at the same point? If so, after what time? Given c = 360 ms–1.
ht tp
3.16
) Solution
(i) Let the width of valley is x $
[±\
\ 3
%
[
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Wave Mechanics
3.17
Let the distance of point P where the man hears the shot be ‘y’ from A where the man hears the echo. Thus, total distance travelled by the first case is 2 y and the second case is 2 (x – y). So, y = c × 1 = 360 × 1 = 360 m (x – y) = c × 2 = 360 × 2 = 720 m ∴ x = (720 + 360) m = 1080 m (ii) The sound reflected from A reaches P after 2 seconds (first echo). This sound travels toward B and after reflection at B reaches 4 seconds later (third echo). That is after a total time of 6 seconds. The sound reflected from B reaches P after 4 seconds (second echo). This travels towards A and after reflection reaches at A 2 second later (third echo). This shows that the echo is heard after 6 seconds simultaneously.
na .c om
e/
rio s_ in fo
7 %
ht tp s:
un
//t
ht
G G[ O
.m
tp
$
7
[
iv
un iv
s:
7
er si ta rio s
ita
er s
w
//w
w
) Solution
.ja
m
ar a
Example How long will it take the sound waves to travel the distance l between the points A and B if the air temperature between them varies linearly from T1 and T2? The velocity of sound propagation in air is equal to v = α T , where, α is constant.
dx = v ⋅ dt or
l
t
0
0
ht tp
According to kinematics
...(i)
s: //
T −T T = T1 + 2 1 x l
t.m e/
Let the temperature is T at a distance x from the point A. Then,
∫ dx = ∫ v. dt
T −T where, v = α T = α T1 + 2 1 x l
∴
t
t
0
0
∫ dt = ∫
dx T −T α T1 + 2 1 x l
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Wave Mechanics
Put,
T −T T1 + 2 1 x = z , l
T −T ⇒ 2 1 dx = dz l
The limit of z → T1 to T2 T2
t
−
T
1
l dz l 2 z 2 dz t = ∫ dt = ∫ = α . z1/2 (T2 − T1 ) α T∫1 (T2 − T1 ) 0 T1
=
2l T2 − T 1 = α T2 − T1 α
2l
(
T2 + T1
)
.
Example
fo
in
an
a. co
m
The displacement of the medium in a sound wave is given by the equation: y1 = A cos (ax + bt), where A, a, b are positive constant. The wave is reflected by an obstacle situated at x = 0. The intensity of reflected wave is 0.64 times that of the incident wave.
si
ta
.ja
ta
si er un
e/
s:
tp
ht
2π m λ
b Hz 2π (ii) Equation of the reflected wave is y2 = A2 cos (–ax + bt)
⇒λ =
ω = b or 2π f = b
and
∴
2π = a, λ
//t .m
tp s ht
we get,
iv
2π y1 = A cos x + ωt λ
://
(i) Comparing the wave equation
rio
iv
un
)
ht tp
t.m e/
) Solution
s
er
w
w
s: //w
s_
rio
ar
m
(i) What are the wavelength and frequency of the incident wave? (ii) Write the equation of reflected wave. (iii) In the resultant wave formed after reflection, find the maximum and minimum values of the particles speed in the medium. (iv) Express the resultant wave as a super position of a standing wave and a travelling wave. What are the position of the antinodes of the standing wave? What is the direction of the propagation of the travelling wave?
3.18
f =
Here, x is negative because if incident wave has +ive x direction than reflected wave has –ve x direction y = A2 cos (–ax + bt) It is given in the problem that intensity of reflected wave is 0.64 times of incident wave I1 A12 1 = 2 = I 2 A 2 0.64
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Wave Mechanics
(a) When there is zero phase change: Y = yi + yr = A cos (ax + bt) + 0.8 A cos (–ax + bt) = 0.8 A [cos (ax + bt) + cos (–ax + bt)] + 0.2 A cos (ax + bt) = 2 × 0.8 A cos ax. cos bt + 0.2 A cos (ax + bt) This is the mixture of stationary wave (1st term) and a progressive wave (2nd term) progressive in (–x) direction. ∴ At ax = 0, π, 2π Y = 2 × 0.8 A cos bt + 0.2 A cos bt = 1.8 A cos bt with amplitude 1.8 A (maximum) π 3π 5π At ax = , , 2 2 2 y = 0.2 A cos bt, with amplitude 0.2 A (minimum) (b) y = resultant wave when π phase change = 0.8 A cos (ax + bt) – 0.8 cos (–ax + bt) + 0.2 A cos (ax + bt) = – 2 × 0.8 A sin ax. sin bt + 0.2 A cos (ax + bt) This is again a mixture of stationary and progressive wave. At ax = 0, π, 2π, 3π.....
or A2 = 0.8 A Thus y2 = ± 0.8 A cos (– ax + bt) Here, ± depends on whether there is zero or π phase change on reflection.
m
in fo
rio
iv
rs ita
un
e/
y = ± 0.2 A cos bt with ampli tude = 0.2 A (minimum) π 3π 5π ax = , , , .... 2 2 2 y = 2 (0.8 A) sin bt + 0.2 sin bt
un
e/
t.m
ht
= 1.8 = sin bt with amplitude = 1.8 A (maximum)
tp
s:
//t
iv e
.m
tp ht
At
s
er si
ta
.ja
s:
//w
w
w
m
rio
s_
ar an
a. co
3.19
s: //
tp
ht
π 3π 5π , , ..... 2a 2a 2a (c) Progressive wave travels towards the reflecting surface i.e., negative x-axis
(iii) vm = Amax ω = 1.8 Ab and vmin = Amin ω = 0.2 Ab (iv) (a) For resultant wave, y = 1.6 A sin (ax) cos (bt) + 0.2 A cos (ax + bt) or y = – 1.6 A sin (ax) sin (bt) + 0.2 A cos (ax + bt) (b) For zero phase changing, position of antinode are π 2π x = 0, , ,..... a a For π phase change, position of antinode are
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Wave Mechanics
)
Example A wire of density 9 cm–3 is stretched between two clamps 100 cm apart while subjected to an extension of 0.05 cm. What is the lowest frequency of transverse vibrations in the wire, assuming Y = 9 × 1011 dyne cm–2? stress ) Solution Young’s modulus, Y = strain T/A T l Y= = × ∆l / l A ∆l T Y∆l = A l
∴
9 × 1011 × 0.5 100 = 45 × 107 dyne cm–2 Mass per unit length of wire, m =A ρ
m in fo rio s_ si ta
s io
iv
iv
t.m
er
si
un
ta r
1 45 × 107 = 35.3 Hz. 2 × 100 9
e/
s: //w
=
45 × 107 9
er
1 2l
ar
=
T Aρ
.ja m
1 2l
ht tp
v=
w w
an
a. co
=
tp s
:// t.m
ht tp
e/
un
s: //
Example A string 25 cm long and having mass of 2.5 gm is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first over tone and the air in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in is air 320 ms–1. Find the tension in the string. )
ht
) Solution Since by decreasing the tension of the string, the number of beats
decreased, therefore vstring > vpipe
v 320 = = 200 Hz 4 l 4 × 40
vpipe =
1 T l m where, m = mass per unit length
Also,
Fundamental Frequency νstring – νpipe = number of beats = 8 ∴ ν string (1st ovetone) = 200 + 8 = 208Hz
3.20
ν string =
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Wave Mechanics
ν string =
∴
2.5 × 10−3 = 10−2 kg m−1 25 × 10−2
1 T = (208) 2 l2 m
or
3.21
∴
T = (208)2 × (25 × 10–2)2 × 10–2 = 27.04 N
m
Example A sonometer wire of length 100 cm carries a metal block hanging at its free end. When the block is in air, its resonance is in unison with a tuning fork. When the block is in water, the length of vibrating wire in unison with the same fork is 90 cm. Find the specific gravity of the block. )
a. co
) Solution Let w1 and w2 be the weight of the block in air and water respectively.
io s
ta r si
e/ u
tp s:
ni
//w
w
Squaring and dividing, we get w1 l12 = w2 l22
in fo
rs ita rio s_
w2 m
ve
1 2 l2
an
v=
w1 m
ar
1 2 l1
.ja m
v=
w
and
∴
In both the case, the frequency of vibration of the wire remain the same.
But
w1 weight in air = specific gravity = w1 − w2 loss of weight in water
e/ un
iv
//t .m //t
l12 l − l22 2 1
ht tp
s:
ht
.m
tp
s:
Specific gravity =
=
∴
er
w1 l2 = 21 2 w1 − w2 l1 − l2
ht
∴
1002 = 5.26 1002 − 102
Example Two wires of radii r and 2r respectively are welded together end to end. This combination is used as a sonometer wire kept under tension T. The welded point is midway between the two bridges. What would be the ratio of the number of loops formed in the wire such that the joint is a node when stationary vibrations are set up in the wires?
)
) Solution Let m1 and m2 be the masses at per unit length. ∴ m1 = A1 × ρ = πr2 × ρ and m2 = A2 × ρ = π (2r)2 × ρ https://t.me/universitarios_infohttps://www.jamarana.com
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3.22
∴
m1 1 = m2 4
Wave Mechanics
p 2l
T m1
and
q 2l
T m
Now ν1 = ν2 =
fo ita iv
t.m
er si ta
e/ un i
s
ve rs
w
w
w
s: //
tp
ht
un
e/
ht tp s:
//t
.m
tp
s: //
Consider a source of sound (S) producing sound wave of frequency ν. If an observer (O) hears the sound waves, he will record a frequency same as ν, if both the sources and the observers are stationary. If the source or observers or both are moving the apparent frequency as heard by the observer will be different from ν. Let us call this apparent frequency as ν′. Let Vs, V0 be the velocity of the source and the observer (or the components of their velocity along the line joining them). If c is the speed of sound the apparent frequency is given by c V0 ν' =ν c Vs
ht
3.8 DOPPLER’S EFFECT
rio
p 1 = . q 2
rio s_ in
or
.ja m
m1 p = q m2
or
.c o
p p m1 m1 T T m m
ar an a
∴
p p 2l 2l q = q = 2l 2l
m
where, ν1 and ν2 are the frequencies of the vibrations, p and q are the number of loops. Since, it is a continuous wire, therefore ν1 = ν2 (as it vibrates with the same frequency)
V0, Vs are positive, if they are directed from source to the observer. They are negative if they are directed from observer to source. In other words, we will assume that the direction from source to the observer is the positive direction. If the medium is moving relative to ground, the speed of sound will be taken as c + Vm (Vm is the speed of the medium). Again note that Vm will be positive if the medium is moving from source of the observer. So, in general the apparent frequency is
c Vm V0 ν' =ν c Vm Vs https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
3.23
If the sound reaches the observer after being reflected from a stationary surface and the medium is also stationary, the image of the source in the reflecting surface will become the source of the reflected sound. Example Given v = 1000 Hz, c = 330 ms–1.
(i) What are the wave length ahead of and behind the moving source (source moving away from the stationary observer) if Vs = 30 ms–1? (ii) What is the frequency as heard by the observer? (iii) If the source is at rest and the observer is moving away from the source at 30 ms–1, what is the frequency as heard by the observer?
fo
co
m
(i) For the waves in front of the source c − Vs λ= ν 330 − 30 = = 0.30 m 1000 For the waves behind the source
in
rio
ta si
s
er
iv
si
ta
un
er
e/
.m
//t
iv
c c + Vs
s:
un
tp
v′ = ν
rio
330 + 30 = 0.36 m 1000 (ii) Frequency as heard by the observer
ht
s_
an
m
c + Vs ν
//w w
λ=
s:
w
.ja
ar
a.
)
) Solution
.m
//t
(iii) Here
V0 = 30 ms–1, Vs = 0
s:
ht tp
e/
330 = 1000 × = 916.7 Hz 330 + 30
ht
tp
c − V0 v′ = ν c
330 − 30 = 1000 = 909.1 Hz 330
Example An ambulance travels down a highway at a speed of 33 ms–1. Its siren emits sound at a frequency of 400 Hz. Find the change in frequency measured by a passenger in a car travelling at 23 ms–1 in the opposite direction as the car approaches the ambulance and as the car moves away from the ambulance. c = 343 ms–1. )
) Solution Apparent frequency measured by the passenger, when ambulance and car approach each other.
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3.24
c + V0 ν1 = ν c − Vs
Wave Mechanics
343 + 23 = (400) 343 − 23
366 = 400 = 472.2 Hz 310
Apparent frequency measured by the passenger in the car when ambulance and car move away from each other, 343 − 23 = 400 343 + 33
na .c om
c − V0 ν2 = ν = 400 c + Vs
320 = 340.4 hz 376
ar a
rio s_ in fo
Change in frequency measured by the passenger ∆ν = ν1 – ν2
tp
e/
s:
WORKED OUT EXAMPLES
t.m e/
s: //
λ=
or
2 × 1011 8000
= 0.5 × 104 = 5000 m/sec. Equation of the stationary wave is y (x, t) = 2 Asin kx cos ωt Given, 2A = 2 × 10–6
1
$
1
$
1
$
1
$
∴
Since both ends are open, therefore start from antinodes at each side. Since distance between the consecutive λ antinode is , therefore 2
and
P
$
=
Y ρ
)
1
2 = 0.4 m 5
Velocity of sound in rod, c =
ht tp
) Solution $
λ 5 =1 2
iv un
//t
ht tp s:
ht
.m
1. A metallic rod of length 1 m is rigidly clamped at its mid-point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the mid-point. The amplitude of an antinode is 2 × 10–6 m. Write the equation of motion at a point 2 cm from the mid-point and those of the constituent waves in the rod. (Young’s modulus = 2 × 1011 Nm–2, density = 8000 kg m–3)
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
= 472.2 – 340.4 = 131.8 Hz
2π 2π = = 5π λ 0.4 c ω = 2πv = 2π λ 5000 =2π× = 25000 π 0.4 k=
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3.25
Wave Mechanics
λ=
∴
Y = y1 + y 2
5π 2π = L λ
It gives, k =
2L 5
... (ii)
At x = 2cm y (x, t) = 2 × 10–6 sin (0.1 π) cos (25000 π t) The constituent wave is
Velocity of the particle in the string, c =
T Tension = µ Mass per unit length
The two wave travels with same velocity and frequency in opposite directions. y1 = A sin (kx – ω t) and y2 = A sin (kx – ω t) ∴ y1 = 10–6 sin (5 πx – 25000 πt) ∴ y2 = 10–6 sin (5 πx + 25000 πt)
m
= 2π
fo
na .
2 π 5π T = t L µ
ta
Time period =
2L µ . 5 T
er si
rio s
iv
∴
5π T t L µ
3. The following equations represent transverse waves: Z1 = A cos (kx – ω t) Z2 = A cos (kx + ω t)
un
e/ .m
Z3 = A cos (ky – ω t)
//t
y (x, t) = 2 A1 sin kx cos ωt
s:
Identify the combinations of the waves which will produce
tp
Z = Z1 + Z2
5π y0 (x) = A sin x L
(i) Standing waves
)
) Solution
But in the problem 2A1 = A y (x, t) = A sin (kx) cos (ωt) ∴ y0 (x) = A sin (kx) ...(i) On comparing this equation with the given equation, we get
[ /
(i) standing waves (ii) a wave travelling in the direction making an angle if 45º with the positive x-axis and positive y-axis. In each case, find the position at which the resultant intensity is always zero.
ht [
s_ in
s:
ht tp
)
) Solution The equation of stationary wave is
un
e/
//t
.m
ht tp
(ii) What is the period of motion?
[
(ii) Period of motion =
iv
//w
s:
w
5ð x y (x, 0) = y 0 (x) = A sin , L ∂y at (x, 0) = 0, where A is con∂x stant.
5π y (x, t) = A sin x cos L
er si ta rio
ar a
w
.ja
m
(i) Write down the simplest possible expression for the stationary wave y (x, t) describing the motion at all time t > 0 if the initial condition is
c λ
Hence, equation of the stationary wave is
co
2. Consider a vibrating string of length L under tension T and with mass per unit length µ. The string is fixed at both ends (x = 0 and x = L)
= 2π
T 5 5π T . = µ 2L L µ
ω = 2 πν
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= A cos (kx – ωt) + A cos (kx + ωt) = 2 A cos kx · cos ω t
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3.26
Wave Mechanics
No power is dissipated during the propagation of the wave-pulse. Calculate:
Amplitude of the resultant standing wave = 2 A cos kx Since the intensity is directly proportional to the square of the amplitude, therefore I ∝ A2 = 0 So, I=0 ∴ 2 A cos kx = 0 or cos kx = 0 2π or cos x=0 λ 2π Thus, x = (2π − 1) π / 2 λ λ ∴ x = (2n – 1) , where n = 1, 2, 3 .... 4 (ii) Wave travelling in the direction making an angle of 45º degrees with the positive x and positive y. So, resultant equation is the sum of the first and third equation. Z = Z1 + Z3
) Solution
)
o
in f
s_
T m
rio ta
cPQ =
80 × 4.8 = 80 m/sec. 0.06
cQR =
80 × 2.56 = 32 m/sec. 0.2
rio
s
si ta
Time, t = t1 + t2
4.8 2.56 + = (0.06 + 0.08) sec 80 32
er
∴
un iv
=
= 0.14 sec
e/
E 1 2 l 2 1 2 = ωm A = ω t 2 2 t mv A2 The conservation of power at the junction Q gives Pi = P r + Pt P=
m1 v1 (Ai2 – Ar2) = m2 v2 At2 ...(i)
or
4. A long wire PQR is made by joining two wires PQ and QR of equal radii, PQ has a length 4.8 m and mass 0.06 kg. QR has a length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave-pulse of amplitude 3.5 is sent along the wire PQ from the end P.
2π x − y π = (2n − 1) λ 2 2 λ x – y = ± (2n – 1) , where 2 n = 1, 2, 3, ...
(ii) The energy transferred by the wave pulse through the wire 1 = (ω2m) A2 l . 2 Then, average power,
t.m tp
ht
π x− y k = (2n − 1) 2 2
s: //
ht
∴
.m
tp s:
//t
ht
or
er si
iv
un
e/
s:
tp
Hence, if the intensity is zero, then
or
5
where, T = tension, and m = mass per unit length
k ( x − y) k ( x + y) .cos − ωt 2 2
x− y cos k =0 2
4 P NJ
(i) Velocity of the wave in the rod =
ar
m
.ja
w
//w w
= A cos (kx – ωt) + A cos (ky – ωt)
= 2 A cos
P
3P NJ
a.
an
P
co m
(i) The time taken by the wave pulse to reach the other end R of the wire. (ii) The amplitude of reflected and transmitted wave pulses as for the incident wave pulse crosses the joint Q.
The division of amplitude of Q gives Ai = Ar + At Substituting this in equation (i), we get m1 v1 (Ai2 – Ar2) = m2v2 (Ai – Ar)2
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/
Wave Mechanics
a.
s_ i
5λ 5 × 1.6 = (3.6 m) − m = 1.6m 4 4
L−
7λ 7 × 1.6 = (3.6 m) − m = 0.8m 4 4
ta
s
io
si
er
9λ 9 × 1.6 = (3.6 m) − m = 0m 4 4
un
iv
L−
e/
According to the Bernoulli’s equation, the flow of the fluid at any point is given by 1 2 p+ ρv + ρgh = constant 2 For the two points near the top and the bottom of the pipe filled with water, we get 1 1 p + ρv 2 + ρgH 0 = p0 + ρv02 2 2 Taking pressure to be identical at two points, we get
s: //t .m
ht tp
L−
iv
e/ un
:// t.m
tp s
ht
3λ 3 × 1.6 = (3.6 m) − m = 2.4m 4 4
si er
w
w
tp s: //w
ht
1 2 1 ρv + ρgH 0 = v02 2 2 1 2 1 2 or v + gH 0 = v0 …(i) 2 2 Also from the principle of continuity, we get vA = v0 a ...(ii) where, A and a are the areas of cross-section of the pipe and small hole, respectively. Eliminating v0 from equations (i) and (ii), we get
)
L−
rio
m
.ja
5. A 3.6 m long vertical pipe resonates with a source of frequency 212.5 Hz, when water level is at certain height in the pipe. Find the height of water level (from the bottom of pipe) at which resonance occur. Neglect end correction. Now, the pipe is filled to a H of 3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as the function of H. If the radii of the pipe and the hole are 2 × 10–2 m and 1 × 10–3 m respectively, calculate the time interval between the occurrance of first two resonance. Speed of sound in air is 340 m/s and g = 10 m/s2.
) Solution
λ 1.6 m = (3.6 m) − = 3.2m 4 4
Given: Ai = 3.5 cm, ∴ Ar = 0.6 × 3.5 = 2.1 cm Hence, At = Ai – Ar = (3.5 – 2.1) = 1.4 cm
L−
an
A r 1.5 = = 0.6 A i 3.5
ar
co m
1 + (A r / A i ) 0.2 / 2.56 = = 2.5 1 − (A r / A i ) 0.06 / 4.8
o
Substituting the value of m2 and m1, we get
nf
The condition of resonance is shown in the figure. v 340 m/sec. Wave length, λ = = =1.6 m ν 212.5/sec. The height of water level in the vertical pipe at which resonance occur are
ta r
Y2
1 + (A r / A i ) m2 = m1 1 − (A r / A i )
or
Y
O + 2
F , m
Ai + A r m2 = m1 Ai − A r
we get
or
$
O
O
/
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O + 2
Y
Since v =
A i + A r m2 v2 = A i − A r m1 v1
or
3.27
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3.28
1 2 1A v + gH 0 = v 2 2 a
2
t2 =
Wave Mechanics
1/2
2g H a2 v= 2 0 2 A −r In terms of radii, we get
=
or
2gH r v= 4 0 4 R −r Since, r < R, we ignore r4 in comparison to R4, therefore
0.00559 m /s 2 3.6 − 2.4 = 62.3 sec 0.00559
∆ t = t2 – t1 = (62.3 – 19.4) sec
Finally
1/2
= 42.9 sec
r2 2 gH 0 R2 dH v= , we get dt
4
H0 − H2
o
in f
s_
rio
si ta
rio
s
)
er
iv
(i) When the observer receives the sound, the source is closest to him. It means that frequency is emitted by the source sometimes before figure shows and the source approaches the stationary observer with velocity Vs cos θ. Hence, frequency noted by the observer
un e/
s: //
c v = v0 c − Vs cos θ
)
H = H 0 − 0.00559 m /s 2 t
ht
tp
t.m
tp
) Solution
...(iii)
ht
r2 1.0 × 10−3 2 g = 2 × 10 R2 2.0 × 10−2
(
(i) The frequency of sound received by the observer at the moment when the sound gets closest to him. (ii) The distance between the source and the observer at the moment when the observer receives a frequency υ = υ0.
si ta
er iv un e/ .m
)
= .01118 m /s 2 Hence, equation (iii) becomes
m co
ar m
w
//w w
ht
(
2g t
s:
r2 i.e., 2 H 0 − H = 2 R For the given data, we get
H0
t
dH r 2 = 2 2 g ∫ dt H R 0
s:
H
−∫
tp
−
.ja
dH r 2 = 2 2 g dt H R Integrating, we get or
//t
−dH r 2 = 2 2 gH dt R
a. an
v=
Setting
6. A source of sound with natural frequency υ0 = 1.8 kHz moves uniformly along a straight line separated from a stationary observer by a distance l = 250 m. The velocity of the source is equal to η = 0.80 fraction of velocity of sound. Calculate:
6 T
Since for the first resonance H1 = 3.2 m, therefore H 0 − H1 t1 = 0.00559 m /s 2
[
=
3.6 − 3.2 0.00559 m /s 2
= 19.4 sec
Since for the second resonance, H2 = 2.4 m, therefore https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
v″ = v
x l 2 + x2 = Vs c
But
c+u c c+u . =v c c−u c−u c −u λ″ = λ or c+u λ″ c − u or = λ c+u λ″ c−u 2u ∴ 1− = 1− = λ c+u c+u Hence, percentage change in the wavelength
...(i)
V0 c = v0 = c − η c cos θ 1 − η cos θ
3.29
x
Vs =η c
om
s_
an
s
rio
ita
ve rs
ni
∂y = −am ω cos kx sin ωt …(ii) ∂t So, the kinetic energy confined in the string element of length dx is given by or
1 m ∂ y dT = dx 2 l ∂ t
2
Wall will reflect the sound with same frequency v′. Thus, frequency noticed by a stationary observer after reflection from wall. c v″ = v′ c−u
e/ u
c+u c
v′ = v
.m
//t
tp s:
ht
)
) Solution Frequency of sound when it reaches the wall
(i) Let two waves y1 = A cos (ωt – kx) and y2 = A cos (ωt – kx), superimpose and as a result, we have standing wave (the resultant wave) in the string of the form y = 2 A cos kx cos ωt. According to the problem, 2A = am Hence, the standing wave excited in the string is y = am cos kx cos ω t ...(i)
un iv
s: //t
ht tp
7. A stationary source sends forth monochromatic sound. A wall approaches it with velocity u = 33 cm/s. The propagation velocity of sound in the medium is c = 330 m/sec. In what way and how much in per cent, does the wavelength of sound change on reflection from the wall?
) Solution
.m e/
ht
tp s:
The distance between the source and the observer at this time
(i) The maximum kinetic energy of the string. (ii) The mean kinetic energy of the string averaged over one oscillation period.
si ta
er
rio
m ar
w .ja
//w w
Vs t = lη.
in fo
8. A string of mass m is fixed at both ends. The fundamental tone oscillation are excited with circular frequency ω and maximum displacement amplitude am. Calculate:
a. c
(ii) When the source is right in front of O, the sound emitted by it will not be Doppler shifted because θ = 90º. This sound will be received at O at time l t= after the source has passed it. c Then the source will have moved ahead by a distance
= l 1 + η2 = 0.32 km
λ − λ″ λ 2u = × 100% = 0.2 % decrease c+u =
v frequency, v = 0 2 = 5 kHz 1− η
)
we get
or cos θ = η ...(ii) Hence, from the equation (i) and (ii),
l +x
=
2
2
∴
1m = dx am 2 ω2 cos 2 kx sin 2 ωt 2 l
Since, wall behaves as a source of frequency v′, therefore
=
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mam2 ω 2 2π x dx sin ωt cos 2 λ 2l
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3.30
Wave Mechanics
For the interference path difference ∆ = λ, 2λ, 3λ ... c 360 λ= = = 2m v 180 For the minimum distance, from equation (i)
Hence, the kinetic energy contained in the string corresponding to the fundamental tone
T = ∫ dT =
mam2 ω2 sin 2 ωt 2l
λ /2
∫ cos
2
2π x dx λ
2 x 2 22 + − x = λ = 2 string, l2 = λ Because, for the fundamental tone length of the 2 x2 2 λ or 4 4 + = x + 4x + 4 Because, for the fundamental tone length of the string, l = 4 2 ' Integrating, we get
0
o in f rio
[
ta
s rio ta
er si
un
iv
e/
10. A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a horn that has dominant frequency 800 Hz. What will be the apparent frequency heard by driver in the second car? Speed of sound in air = 330 m/sec.
ht
)
) Solution From the figure, path difference between two waves reaching the detector,
First car in source, Vs =
108 × 1000 = 30 m/sec 3600
Second car is observer, 72 × 1000 = 20 m/sec 3600 According to Doppler equation, when the source and observer are moving in the same direction, the apparent frequency V0 =
…(i)
) Solution
tp
9. A source emitting sound of frequency 180 Hz is placed in front of a wall at a distance 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and detector for which the detector detects the maximum sound. Speed of sound in air = 360 m/sec.
2 x = 2 22 + − x 2
16 + x2 = x2 + 4x + 4 4x = 12 x = 3 cm.
.m s: //t
1 < T > = mam 2 ω2 8
or or or
dt
ht tp
0
sin ωt dt
)
∫
un
iv
0 2 π/ ω
6
.m e/
∫
2
//t
=
2 π/ ω
s:
ht tp
1 2 2 mam ω 4
er si
.ja
//w
∫ T dt ∫ dt
s:
=
w w
m
1 ∴ Tmax = mam 2 ω2 4 (ii) Mean kinetic energy averaged over one oscillation period
s_
ar an
a. co
m
1 T = mam 2 ω2 sin 2 ωt 4 Since for maximum kinetic energy, Tm, sin2ωt = 1
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3.31
Wave Mechanics
c − V0 c − Vs
(c + Vm )(c + Vb ) − (c − Vm )(c − Vb ) = ν c 2 − Vb2
c+V = 2 ν Vb 2 m2 . c − Vb
11. A band playing music at a frequency v is moving towards a wall at a speed Vb. A motorist is following the band with a speed Vm. If c is the speed at sound, find an expression for the beat frequency heard by the motorist.
m
) Solution First overtone of the open organ pipe
s
rio
ta
si
iv er
un
.m
e/
lc =
Hence from equations (i) and (ii), we have 330 3 330 − × = ± 2.2 lo 4 3 4 330 − 3 = ± 2.2 lo
ht
or
330 = (300 ± 2.2) Hz lo lo =
330 330 or 332.2 327.8
= 0.993 m
or
or
or 1.0067 m
c + Vm c + Vm = ν − ν c − Vb c + Vb
s_
rio
si ta
c 330 3 = = m 4 × 110 4 × 110 4 Given in the problem | υo – υc | = 2.2 Hz
c + Vm ν″ = ν c − Vb The beat frequency heard by the motorist is ∆υ = ν″ – ν
…(ii)
where, l = length of the closed organ pipe. Fundamental frequency at closed organ pipe. c υ′c = = 110 Hz 4 lc
tp
s:
//t
ht
Thus, the frequency heard by the motorist is
c 3c = 4 lc 4 lc
e/ .m //t
tp
s:
ν . c − Vb
vc = 3 ⋅
er
iv
un
s:
tp ht
where, c = velocity of sound (ii) Frequency indirectly received after reflection from the wall. In this case, the motorist is moving towards reflected frequency which is
…(i)
in
m
//w
w
w
.ja
c + Vm ν′ = ν c + Vb
c c = 2lo lo
where, lo = length of the open organ pipe. First overtone of the closed organ pipe
ar
an
(i) Frequency directly received by the band. In this case, the motorist moves towards the band with velocity Vm and the band moves away from motorist with velocity Vb. The frequency heard by the motorist is
v0 = 2
frequencies:
)
a. co
)
) Solution The motorist receives following two
12. The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the length of the pipes. Velocity of sound c = 330 m/sec.
(330 − 20) 310 = 800 × = 827 Hz (330 − 30) 300
= 800
(330 − 20) 310 = 800 × = 827 Hz (330 − 30) 300
= 800
fo
ν′ = ν
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3.32
14. A column of air at 51º C and a tuningfork produce 4 beats per second when sounded together. As the temperature of the air column is decreased, the number of beats per second tends to decrease and when the temperature is 27º C, the two produce one beat per second. Find the frequency of the tuning-fork.
13. A source of sound moving along a circular orbit of radius 3 metres with an angular velocity of 10 rad s–1. A sound detector located away from the source is executing linear simple harmonic motion along the line BD (shown in the figure) with an amplitude BC = CD = 6 m. The frequency of oscillation of the detector is 5 per second. The source is at the point π A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and the minimum frequencies recorded by the detector (velocity of sound in air = 330 ms–1).
)
) Solution According to the given condition,
o
rio
in f
s_
ta
s
rio
ta
si er
iv
15. A boat is travelling in a river with a speed 10 m/sec along the stream flowing with a speed 2 m/sec. From this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible.
e/
.m
(i) What will be the frequency detected by receiver kept inside the river downstream? (ii) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/sec. in the opposite direction of the river stream. Determine the frequency of the sound detected by the receiver. Given: Temperature of the air and water = 90º C Density of the river water = 103 kg/m3
//t
s:
tp
ht
270 = 255 Hz 360
v = 75 Hz
un
//t
s:
tp
ht
un
∴
.m e/
s:
tp
ht
c − V0 330 − 60 = 340 × c + Vs 330 + 30
= 340 ×
Given that: v1 – v = 1.039 or 1.039 v1 – v = 4 × 1 or .039 v = 2.961
iv
//w
)
both approach each other and the minimum frequency occurs when both move away from each other. Linear velocity of the source, Vs = ωr = 10 × 3 = 30 m/sec. Maximum velocity of the detector at mean position Vo = ωA = (2 π υ) A where, A = amplitude of simple harmonic motion of detector 5 Vo = 2π× × 6 = 60m /sec ∴ π Since the time period of the source and detector are equal, therefore c + V0 330 + 60 vmax = v = 340 × c − Vs 330 − 30 390 = 340 × = 442 Hz 300 vmin = v
…(ii)
where, v1 is the frequency of air column at temperature 21º C
&
) Solution Maximum frequency occurs when
or
m
'
w
P
P
v2 T2 324 = = v1 T1 300
or
w .ja
$
frequency of the air column is greater than the frequency of tuning-fork. Let frequency of the tuning fork is v ∴ v2 – v = 4 ...(i) where, v2 is the frequency of air column at the temperature 51º C.
er si
P
m
%
ar an
a.
co
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Wave Mechanics
Cp,m Cv , m
9D
9V
for air = 1.4
The frequency heard by the receiver is c − V2 ν′ = V0 c − V1
(i) The velocity of sound in water is
)
) Solution
1/ 2
B 2.088 × 109 Pa = ρ 103 kg /m3
344 − 5 5 = 105 Hz = 1.030 × 10 Hz 344 − 15
= 1445 m /sec
c=
5
6
Bulk modulus of the water = 2.088 × 109 Pa Gas constant R = 8.31 J/mol – K Mean molar mass of air = 28.8 × 10–3 kg/mol
3.33
s_ in
ta rio
s
rio
ta er si
iv
un
Dr = Ar sin (k1 x + k2 V1 t)
γ RT M
Dt = At sin (k2 x – k2 V2 t) Ai = Ar + At
V − V2 (ii) Ar = 1 V1 + V2
k1 − k1 Ai = Ai k 2 + k1
(iii) Under what conditions is Ar negative?
The relative velocity of source with respect to the medium (i.e., air) is V1 = Vs – Va = (10 + 5) m/sec. = 15 m/sec
(i)
= 344 m/sec.
Show that:
16. A cord stretched to a tension FT consists of two sections whose linear densities are µ1 and µ2. Taking x = 0 to be the point (a knot) where they are joined, with µ1 referring to the section of cord to the left and µ2 that to the right. A sinusoidal wave Di = A1 sin (k1 x – k1 V1 t), starts at the left end of the cord. When it reaches the knot, part of it is reflected and part is transmitted. Let the equation of the reflected and transmitted waves be
(ii) The velocity of sound in air is
1/2
fo
om
er iv
= 1.007 × 105 Hz
(1.4)(8.314Jk −1mol−1 )(293K) = 28.8 × 10−3 kg mol−1
= 1.037 × 105 Hz
s: //t .m e/
ht tp
1445 + 2 = 105 × 1445 − 8
ht tp
s: //t
ht
.m
The relative velocity of receiver with respect to medium i.e., water = 2 m/sec. The frequency heard by the receiver is c + V0 v′ = v0 c − V1
c=
344 − 5 ν′ = (105 Hz) 344 − 17
∴
9Z
un
9Z
e/
s: //w
5
w
9V
tp
6
Note: In the above solution, speed of 5 m/sec. of air is relative to that of the water. If the value is taken with respect to earth, then V1 = Vs + V2 – Va = 10 m/sec. + 2 m/sec. + 5 m/sec. = 17 m/sec. V2a = V2 – Va = 5 m/sec.
si
w
.ja
m
ar a
na
.c
c 1445 v= = = 105 Hz −3 λ 14.45 × 10 The relative velocity of source with respect to the medium (i.e., water) is V1 = Vs – Vw = (10 – 2) m/sec. = 8 m/sec
The frequency of sound in water is
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3.34 )
At x = 0, we have d dx (Di + D r ) = k1 Ai cos (– k1 V1 t) x =0 + k1 Ar cos (k1 V1 t) = (k1 Ai + k1 Ar) cos (k2 V1 t)
) Solution
k1 V1 = k2 V2
d = k2 At cos [– k2 V2 t] dx Dt x =0 = k2 At cos (k2 V2 t) Equating these two expressions and using the fact that k1 V1 = k2 V2, we get k1 (Ai + Ar) = k2 At Since Ai = Ar + At, we get k1 (Ai + Ar) = k2 (Ai – Ar)
rio s_ in fo
er si ta rio s
ita
k1 V1 , we get V2
V1 − V2 V1 + V2
⇒
FT FT > µ2 µ2
µ2 < µ1
⇒
(b) 180º (d) 270º
(a) 90º (c) 45º
uestions
t – 1.57x) and another wave is y´ = 0.10 sin (314 t – 1.57 x + 1.57) The phase difference between the two wave is ∆φ = (314 t – 1.57x) ~ (314 t – 1.57x + 1.57) = 1.57 radians =
1.57 × 180 = 90° π
Another wave has the equation
where t, x and are in seconds, meters and centimeters respectively.
Then the phase difference between the two waves is
pe
t
) Explanation One wave is, y0 = 0.30 sin (413
y = 0.30 sin (314 t – 1.57x)
y´ = 0.10 sin (314 t – 1.57x + 1.57)
q
)
1. A simple harmonic wave has the equation
SOLVED OBJECTIVE y
x 2. The equation y = a cos2 2 πnt − 2 π α represents a wave with
ht tp
s: //
t.m e/
V2 > V1
un
iv
(iii) For Ar to be negative, we must have
.m
//t
d (Dt) = k2 At cos (k2x – k2 V2 t) dx
ht tp s:
k − k A r = Ai 2 1 k2 + k1
Putting k2 =
or
e/
s:
tp ht
Ai (k1 – k2) = – Ar (k1 + k2)
A r = Ai
d (Di + Dr) = k1 Ai cos (k1 x – k1 dx V1 t) + k1 Ar cos (k1 x + k1 V1 t)
or
er s
un iv
//w
w
w
(ii) Since, the cord is continuous, slope of the cord at the boundary (x = 0) will be identical in either side of the knot.
na .c om
ar a
.ja
Since, ∴ or
sin (k1 V1 t) [– Ai + Ar] = sin (k2 V2 t) [– At] k1 V1 = k2 V2, we have – Ai + Ar = – At A i = Ar + A t
m
or
Since, x = 0 at the knot, we will have Ai (– k1 V1 t) + Ar sin (k1 V1 t) = At sin (– k2 V2 t)
Di + Dr = Dt
Since, the rope is continuous, the displacement at infinitesimal distance on either of the knot will have the same value. Hence,
(i) Since frequency of the wave will be same in both sections, therefore
Wave Mechanics
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Wave Mechanics
wave-
nf
o
5. When two simple harmonic motions of same periods, same amplitude, having 3π , and at right angles to each phase of 2 other super imposed, the resultant wave form is a (a) circle (b) parabola (c) ellipse (d) figure of eight
s_ i
ta rio
si ta rio
s
rs i
ve
)
iv er
motions of different amplitudes and same period are superimposed, the resulting motion is in general elliptical. When the phase difference is 0 or 180º, the motion will be a straight line. When the phase difference is 90 or 270º, the motion will be an ellipse whose axes coincide with coordinate axes. If the amplitudes are the same and the phase difference is 90º or 270º, the motion will be circular.
un e/
(c) 2
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1 2 19 (d) 18 (b)
18 19
(a)
t.m
s: //
tp
ht
6. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered is f2. If the speed of sound is 340 m/s, then the ratio f1/f2 is
un i
e/
T π (2r ) 2 d
) Explanation When two simple harmonic
t.m
s: //
tp
ht
T m
p2 = 2p1 p 1 : p2 = 1 : 2
∴ or
om
ar
m
w .ja
//w w
tp s:
ht
of a SHM of amplitude y0 and frequency f is 2πfy0. The wave velocity is fλ. For 2π fy0 to be πy0 equal to 4fλ, λ has to be (since λ = a) 2 4. Two wires of radii r and 2r respectively are welded together end to end. The combination is used as a sonometer wire and is kept under a tension. The welded point is midway between the bridges. What should be the ratio of number of loops formed in the wires such that the joint is a node when stationary vibrations are set up in the wires?
p T = 2 πr 2 d 2l
an a
p1 2l
i.e.,
The maximum particle velocity is equal to four times the wave velocity if a is equal to (a wavelength) πyo πyo (a) (b) 4 2 (c) πy0 (d) 2πy0
p 2l
When the string is vibrating with p loops as the frequency of wave in both they will be the same,
wave-
3. A transverse wave is described by the equation
)
f =
) Explanation The maximum particle velocity
)
string is given by
wave-
1 2πx = a 1 + cos α 2 (2n)t − 2 λ /2 a Hence Amplitude is , frequency is 2n and 2 λ wavelength is 2
y = y0 sin 2π (ft – x/a).
(b) 1 : 4 (d) 1 : 1
) Explanation Frequency of transverse wave in a
.c
2πx y = a cos 2 2πnt − λ
∴
(a) 2 : 3 (c) 1 : 2
wave-
)
(a) amplitude a, frequency n and length λ (b) amplitude a, frequency 2n and length 2λ a (c) amplitude , frequency 2n and 2 length λ a (d) amplitude , frequency 2n and 2 λ length 2 1 2 2 ) Explanation From cos θ = (1 + cos θ) 2
3.35
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3.36
(a) (b) (c) (d)
340 ) Explanation f1 = f ;
340 f2 = f 340 − 17
) Explanation After 2 seconds, the pulses will be
)
7. A flat horizontal platform moves up and down in S.H.M. with an amplitude of 1 cm. A small object is placed on the platform. What is the maximum frequency the platform can have, if the object is not to separate from it during any part of the motion?
at the same location, giving zero displacement at each point. Thus, the energy will not be in the form of potential energy.
om
s_ i
nf o
an a. c
rio
s
ta
rs i
ta rio
ve rs i
un i e/
v0′ =2 v0
10. The amplitude of a wave disturbance propagating in the positive x-direction is 1 given by y = at time t = 0 and by y 1 + x2 1 = at t = 2 seconds, 1 + ( x − 1) 2
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where x and y are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is (a) 1 ms–1 (b) 0.5 ms–1 –1 (c) 1.5 ms (d) 2 ms–1
v + v′0 6 = 5 v
1.2v = v + v0′
.m
tp s:
8. Two pulses in a stretched string whose centers are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be
⇒ ∴
ht
1 g 1 f = = 980 2π a 2π
and
//t
ht
tp
s:
//t
simple harmonic motion is the weight of the object in the platform. If A is the amplitude, we have mω2A = mg, where, ω = 2π f.
)
ve
⇒
.m
)
) Explanation Maximum restoring force of
v + v0 5.5 = 5 v 1.1 v = v + v0
ni e/ u
ht tp s:
(d) 2π × 980 per second
FP
(b) 2 (d) 11/6
) Explanation From Doppler effect,
//w
980 per second 2π
(a) 242/252 (c) 5/6
w
(c)
ar
980 per second 2πn
9. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B, he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is
m
(b)
.ja
980 per second 2π
w
(a)
f1 19 = f 2 18
)
340 − 34
zero purely kinetic purely potential partly kinetic and partly potential
Wave Mechanics
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3.37
Wave Mechanics
) Explanation Writing the general expression for
er si ta rio s
iv
un
t.m e/
(L
(b) 80 m/s (d) 120 m/s
s: //
ht tp
7. What will be the ratio of amplitude of the reflected wave to the incident wave in case a sinusoidal wave produced in a stretched string gets reflected and losses 36% of its energy in getting reflected from the fixed end of the string? (a) 6 : 10 (b) 8 : 10 (c) 1 : 1 (d) None of these
rio s_ in fo
ita
//t
ht tp s:
4. A 1 metre long sonometer wire is fixed at both ends and a bridge is placed at onetenth of length of wire from one of its ends. It is set into transverse vibrations. The velocity of the wave in the wire if the frequency is 400 Hz, is
6. The 4th overtone of an open organ pipe resonates with the first overtone of semi-open organ pipe. The ratio of the length of the open pipe to the semi-open pipe is (a) 5 : 3 (b) 4 : 1 (c) 10 : 3 (d) none of these
e/
.m
un iv
er s
//w
s:
tp
ht
E 100
(d)
Q
Y
ar a
m
.ja
w
w
(b) 4π (d) none of these
3. If the speed of longitudinal waves equals 10 times the speed of the transverse waves in a stretched wire of material which has modulus of elasticity E, then the stress in the wire is (a) 10 E (b) 100 E
(a) 40 m/s (c) 100 m/s
2
5. A plane progressive wave of frequency 25 Hz, amplitude 2.5 × 10–5 m and initial phase zero moves along the negative x-direction with a velocity of 300 m/s. A and B two points 6 m apart on the line of propagation of the wave. At any instant the phase difference between A and B is φ. The maximum difference in the displacement at A and B is ∆. (a) φ = π (b) φ = 0 (c) ∆ = 0 (d) ∆ = 5 × 10–5 m
na .c om
unsolved
2. A particle oscillates according to equation x = 10 sin (0. 5πt). The speed of the particle when it is at a distance 4 m from the mean position
E 10
1[ x − v(2) ]
OBJECTIVE T PE UESTIONS
1. A closed organ pipe resonates in its fundamental mode at a frequency of 200 Hz in O2 at a certain temperature. If the pipe contains 2 moles of O2 and, 3 moles of Ozone is now added to it, then what will be the fundamental frequency of same pipe at same temperature? (a) 268.23 (b) 172.7 (c) 149.45 (d) None of these
(c)
1
Comparing with the given equation we get 2v = 1 or v = 0.5 m/s
At t = 0
(a) 2π 3 (c) 2π
y=
At t = 2 s,
1 1 + ( x − vt ) 2
y=
1 1 + x2
y=
)
y in terms of x as
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(U
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3.38
11. Three coherent sources S1, S2, S3 are placed at the three vertices of a right angled, triangle as shown in the figure. A detector is placed at P (mid-point of hypotenuse). The intensity of light as detected by detector is (a) maximum (b) zero (c) neither maximum nor minimum (d) insufficient data
9. A transverse wave travels along a string of material of specific gravity 10. If maximum particle velocity is one-tenth of wave velocity v, then intensity of the wave in the string (in SI units) is (a) 0.05v3 (b) 5v3 (c) 50v3 (d) 5 × 105 v3
Q
(d)
W
W
er si ta rio s
iv
15. A string of length L is stretched along the x-axis and is rigidly clamped at its two ends. It undergoes transverse vibration. If n is an integer, which of the following relation may represent the shape of the spring at any time t?
(c)
Q
un
W
14. Sound waves of frequency 600 Hz fall normally on a perfectly reflecting wall. The distance from the wall at which the air particles have the maximum amplitude of vibration is (speed of sound in air = 330 m/s) (a) 13.75 cm (b) 41.25 cm (c) 68.75 cm (d) all of the these
s: // ht tp
(b)
t.m e/
W
Q
13. The frequency of a sonometer wire is f. The frequency becomes f/2 when the mass producing the tension is completely immersed in water and on immersing the mass in a certain liquid, frequency becomes f/3. The relative density of the liquid is (a) 1.32 (b) 1.03 (c) 1.41 (d) 1.18
er s
un iv
e/ .m //t
ht tp s:
(a)
(b) 3 I (d) 5 I
ita
.ja
w
w
//w
s: tp ht
Q
(a) 2 I (c) 4 I
m
ar a
10. A railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside a railway track. The frequency (n) of the sound heard by the observer is plotted against time (t). Which of the following best represents the resulting curve?
rio s_ in fo
12. Two source of sound of the same frequency produce sound intensities I and 4I at a point P when used individually. If they are used together such that the sounds from them reach P with a phase 2π difference of , the intensity at P will be 3
na .c om
8. A source of sound of frequency 1020 Hz is moving towards a wall with a velocity 20 m/s. The position of a doctor D is such that the detector is in between the source and the wall, as shown in figure. If the speed of sound is 320 m/s, the number of beats recorded by the detector per second is (a) 128 (b) 68 (c) 60 (d) zero
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(a) (b) (c) (d)
nπx (a) y = A sin cos ωt L nπx (b) y = A sin sin ωt L
16. A wire having a mass per unit length m and tension T resonates at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. The length of the wire is
in f
s_
rio
ita
ita
e/ u
iv
un
.m
e/
π (c) y ( x, t ) = A sin ωt − kx − 2
//t
π (d) y ( x, t ) = A sin ωt + kx − 2
22. A man wants to produce sound of frequency 680 Hz using a closed organ pipe of length 34 cm. Suggest him. (a) break the close part of pipe. (b) apply more and more pressure to produce then open end. (c) make a hole at a distance 25 cm from open end. (d) closed the other part of pipe also because it is not possible.
s:
ht tp
er s
π (b) y ( x, t ) = A sin ωt + kx + 2
//t
ht
(b) 0.5 m/s (d) 50 m/s
23. A tuning fork of frequency 340 Hz is sounded above an organ pipe of length 120 cm. Water is now slowly poured in
19. A standing wave is produced in a vapour of a chemical of atomic mass (127 g mol–1) at 400 K. Nodes are found to be 9.57 cm apart when frequency of source was 1000 Hz. The vapour used for the experiment is (R = 8.31 JK–1 mol–1)
(a) 5 m/s (c) 15 m/s
rio
π (a) y ( x, t ) = A sin ωt − kx + 2
.m
s:
tp
18. A stationary observer receives sound waves from two tuning forks, one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears beats with frequency 2 Hz. What is velocity of each tuning fork if their oscillation frequency v = 680 Hz and the velocity of sound in σy air is vs = 340 m/s?
s
ni ve
//w w
s:
tp
ht
rs
w
17. An air column in a pipe which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 330 Hz if the length of the column is (a) 50 cm (b) 75 cm (c) 150 cm (d) 100 cm
21. A sinusoidal wave (longitudinal or transverse) is propagating through a medium in the direction of negative x-axis. The parameters of the waves are A, ω and k. The particle at x = λ/4 executes the motion y (t) = A sin ωt. Possible equation of the wave is
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24. An open pipe of sufficient length is dipping in water with a speed v vertically. If at any instant l is length of tube above water, then the rate at which fundamental frequency of pipe changes is [c is the speed of sound in air]
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25. A railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside the railway track. The frequency (n) of the sound heard by the observer is plotted against time (t). Which of the following best represents the resulting curve?
27. Two sound waves of same amplitude and same frequency of 100 Hz travel in opposite direction in nitrogen gas at 200 K. What is the approximate distance between two consecutive nodes of the standing wave so formed? (R = 8 J mol–1 K–1) (a) 2.8 m (b) 1.4 m (c) 0.7 m (d) 5.6 m
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26. A plane wave of sound travelling in air is incident upon a plane surface of a liquid. The angle of incidence is 60º. The speed of sound in air is 300 m/s and in the liquid it is 600 m/s. Assume Snell’s law to be valid for sound waves, the wave will (a) refract into liquid away from normal (b) refract into liquid towards the normal (c) reflect back into air (d) none of these
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(c) b 2 = 4ac,
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Wave Mechanics
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Wave Mechanics
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5. A wave is expressed by the equation y = 0.5 sin π (0.01 x – 3 t)
where y and x are in metres and t in seconds. Find the speed of propagation. [Roorkee, 1981] [Ans. 300 m/s]
3. The velocity of sound in a tube containing air at 27ºC and a pressure of 76 cm of mercury is 330 m/s. What will be its velocity when the pressure is increased to
2. An engine approaches a hill with a constant speed. When it is at a distance of 0.9 km, it blows a whistle whose echo is heard by the driver after 5 sec. If the speed of sound in air is 330 m/s, calculate the speed of the engine. [IIT, 1966] [Ans. 30 m/s]
4. A tuning fork whose frequency as given by the manufacturer is 512 Hz is being tested with an accurate oscillator. It is found that the fork produces a beat of 2 Hz when oscillator reads 514 Hz but produces a beat of 6 Hz when oscillator reads 510 Hz. Find the actual frequency of the fork. [Ans. 516 Hz]
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EXERCISE 1
1. The temperature between two points A and B, separated by a distance l, varies linearly from T1 to T2. If the velocity of sound propagation in air is equal to v = α T , where α is a constant. Find how long will it take the sound waves to travel from the point A to point B? 2l Ans. t = α T2 + T1
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30. When an observer is moving towards a stationary sound source with uniform velocity vo, the fractional change in frequency is found to be x. Now, if the same observer moves away from the stationary sound source with the same uniform speed v0, then the fractional change in frequency will be numerically (a) equal to x (b) greater than x (c) less than x (d) nothing can be predicted
A static siren at Y is emitting sound of frequency n. How many times (approximately) in an hour will the motor cyclist hear the sound of actual frequency Y? (a) 24 (b) 287 (c) 600 (d) 573
3.41
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Molar heat capacity at constant volume = Cv = 2.5 R
12. Calculate the speed of sound in oxygen from the following data:
6. A stone is dropped into a well and its splash is heard at the mouth of the well after an interval of 1.45 seconds. Find the depth of the well. Given that velocity of sound in air at room temperature is equal to 332 m/s. [MNR, 1981] [Ans. 9.9 m]
Molar heat capacity at constant pressure = Cp = 3.5 R
Pressure = P = 1.013 × 105 N/m2
7. The equation of a wave travelling in a string can be written as
Temperature = T = 273 K
Mass of 22.4 litre of oxygen at NTP = 32 g. Ans. 315.08 m/s] 13. The intensity of sound from a point source is 1.0 × 10–8 W/m2 at a distance of 10 m from the source. Find the intensity at a distance of 100 m from the source. [Ans. 1.0 × 10–10 W/m2]
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15. A string tied between x = 0 and x = l vibrates in fundamental mode. The amplitude A, tension T and mass per unit length µ are given. Find the total energy of the string. [IIT, 2003] A 2 π2 T ] [ Ans. E = 4l 16. A wire having a linear density of 0.05 gm/cc is stretched between two rigid supports with a tension of 4.5 × 107 dynes. It is observed that the wire resonates at a frequency of 420 cycles/sec. The next highest frequency at which the same wire resonates is 490 cycles/sec. Find the length of the wire. [IIT, 1971] [Ans. 214.28 cm]
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11. Two speakers connected to the same source of fixed frequency are placed 2.0 m apart in a box. A sensitive microphone placed at a distance of 4.0 m from their mid-point along the perpendicular bisector shows maximum response. The box is slowly rotated till the speakers are in a line with the microphone. The distance between the mid-point of the speakers and the microphone remains unchanged. Exactly 5 maximum responses are observed in the microphone in doing this. Calculate the wavelength of sound wave. [Roorkee, 1989] [Ans. λ = 0.4 m]
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10. The speed of sound in hydrogen is 1270 m/s. Calculate the speed of sound in a mixture of oxygen and hydrogen in which they are mixed in the ratio 1 : 3. [Ans. 582.72 m/s]
14. Two audio speakers, separated from one another by a certain distance, are driven by the same amplifier system. A listener, in line with the speakers and in between them is at a distance of 6 m from one speaker and 6.4 m from the other. If the frequency of the source is varied continuously from 1000 Hz to 4000 Hz, find the frequencies for which there is destructive interference at the place of the listener. The speed of sound in air is 320 m/s. [Ans. υ = 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz.]
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9. A progressive wave of frequency 550 Hz is travelling with a velocity of 360 m/s. How far apart are two points 60º out of phase? [MNR, 1993] [Ans. 10.9 cm]
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where x and y are in cm and t is in seconds. Find its frequency. [MNR, 1986] [Ans. 1 sec–1]
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8. The equation of a transverse wave is given by
y = 3 cos π (100 t – x). Find its wavelength. [MNR, 1985] [Ans. λ = 2 cm]
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24. A copper wire is held at the two ends by rigid supports. At 30ºC, the wire is just taut, with negligible tension. Find the speed of transverse waves in this wire at 10ºC (α = 1.7 × 10–5/ºC, Y = 1.3 × 1011 N/ m2, d = 9 × 103 kg/m3). [IIT, 1979] [Ans. 70.08 m/s.]
25. A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of vibration is found to be 320 Hz. The velocity of sound in air is 320 m/s. Estimate the diameter of the tube. One end of the tube is now closed. Calculate the lowest frequency of resonance for the tube. [IIT, 1980] [Ans. 163.27 Hz.]
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20. A tuning fork having frequency of 340 vib/sec is vibrated just above a cylindrical tube. The height of the tube is 120 cm. Water is slowly poured in it. What is the minimum height of water required for resonance? Velocity of sound in air = 340 m/s. [IIT, 1975] [Ans. 0.45 m.]
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23. AB is a cylinder of length 1.0 m fitted with a thin flexible diaphragm C at the middle and two other thin flexible diaphragms A and B at the ends. The portions AC and BC contain hydrogen and oxygen gases respectively. The diaphragms A and B are set into vibrations of same frequency. What is the minimum frequency of these vibrations for which the diaphragm C is a node. Under the conditions of the experiment, the velocity of sound in hydrogen is 1100 m/s and in oxygen is 300 m/s. [IIT, 1978] [Ans. 11 × 150 = 1650 Hz.]
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22. A column of air at 51ºC and a tuning fork produces 4 beats per second, when sounded together. As the temperature of air column is decreased, the number of beats per second tends to decrease and when the temperature is 16ºC, the two produce 1 beat per second. Find the frequency of the tuning fork. [IIT, 1977] [Ans. n = 50 Hz]
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19. A pipe of length 1.5 m closed at one end is filled with a gas and it resonates in its fundamental mode with a tuning fork. Another pipe of the same length but open at both ends is filled with air and it resonates in its fundamental with the same tuning fork. Calculate the velocity of sound at 0ºC in the gas, given that the velocity of sound in air is 360 m/s at 30ºC where the experiment is performed. [Ans. 683.42 m/s]
21. A wire of density 9 gm/cm3 is stretched between two clamps 1.00 m apart while subjected to an extension of 0.05 m. What is lowest the frequency of transverse vibrations in the wire? Young’s modulus Y = 9 × 1010 N/m2. [IIT, 1975] [Ans. 36.27 Hz]
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18. An underwater swimmer sends a sound signal to the surface. If it produces 5 beats/sec when compared with the fundamental tone of a pipe 20 cm length closed at one end, what is the wavelength of sound in water? (Velocity of sound in air = 360 m/s, velocity of sound in water = 1450 m/s) [IIT, 1973] [Ans. λ´ = 3.258 m or 3.187 m]
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17. A sonometer wire fixed at one end has a solid mass M hanging from its other end to produce tension in it. It is found that a 70 cm length of the wire produces a certain fundamental frequency when plucked. When the same mass M is hanging in water completely submerged, it is found that the length of the wire has to be changed by 5 cm in order that it will produce the same fundamental frequency. Calculate the density of the material of mass M hanging from the wire. [IIT, 1972] 3 3 [Ans. 7.26 × 10 kg/m ]
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26. A string 25 cm long and having a mass of 2.5 gm is under tension. A pipe closed at one end is 40 cm long. When the string is vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats/sec are heard.
It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string. [IIT, 1982] [Ans. T = 27.04 N]
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12. Water waves can travel in any direction when far offshore, but they always arrive at the beach nearly head on. Explain.
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1. A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s. (a) 1 m/s (b) 2 m/s (c) 0.5 m/s (d) 1.5 m/s
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8. Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, Explain.
3. If the equation of a progressive wave is t x π given by y = 4sin π − + then which 5 9 6 of the following is correct? (a) v = 5 cm/s (b) λ = 18 m (c) A = 0.04 cm (d) f = 50 Hz
7. A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, Explain.
6. Why is it possible to understand the words spoken by two people at the same time?
2. A note has a frequency of 128 Hz. The frequency of a note which is two octave higher than this is (a) 256 Hz (b) 320 Hz (c) 400 Hz (d) none of these
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11. The shape of a pulse gets distorted during propagation in a dispersive medium.
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3. The piano strings that vibrate with the lowest frequencies consist of a steel wire around which a thick coil of copper wire is wrapped. Only the inner steel wire is under tension. What is the purpose of the copper coil?
10. The wavelength of the fundamental standing wave on a cello string depends on which of these quantities: length of the string, mass per unit length of the string, or tension? The wavelength of the sound wave resulting from the string’s vibration depends on which of the same three quantities?
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2. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all (a) y = 2 cos (3x) sin (10t) (b) y = 2 x − vt (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2t
9. Why is transverse wave sometimes called a shear wave?
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1. When connecting speakers to a stereo, it is important to connect them with the correct polarity so that, if the same electrical signal is sent, they both move in the same direction. If the wires going to one speaker are reversed, the listener hears a noticeably weaker bass (low frequencies). Explain what causes this and why low frequencies are affected more than high frequencies.
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3.46
(b)
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13. Equation of a stationary and travelling waves are as follows y1 = a sin kx cos ωt and y2 = a sin (ωt – kx). The phase difference π 3π between two points x1 = and x2 = 3k 2k is φ1 in the standing wave (y1) and is φ2 in
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14. In a sonometer wire, the tension is maintained suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m3. The fundamental frequency of the wire is 260 Hz. If the suspended mass is completely submerged
9. The extension in a string, obeying Hooke’s law, is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be (a) 1.22 v (b) 0.61 v (c) 1.50 v (d) 0.75 v
(a) 1 (c) 3/4
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8. A string fixed at both ends has consecutive standing wave modes for which the distances between adjacent nodes are 18 cm and 16 cm respectively. The minimum possible length of the string is (a) 144 cm (b) 152 cm (c) 176 cm (d) 200 cm
12. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is (a) 25 Kg (b) 5 Kg (c) 12.5 Kg (d) 1/25 Kg
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7. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is (a) 242/252 (b) 2 (c) 5/6 (d) 11/6
11. The intensity of a plane progressive wave of frequency 1 kHz is 10–10 watt/m2. If the density of air is 1.3 kg/m3 and the speed of sound is equal to 330 m/s, then pressure amplitude of the wave is (a) 3 × 10–5 N/m2 (b) 3 × 10–4 N/m2 (c) 3 × 10–6 N/m2 (d) 3 × 10–3 N/m2
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6. A sound wave of frequency 440 Hz is passing through air. An O2 molecule (mass = 5.3 × 10–26 kg) is set in oscillation with an amplitude of 10–6 m. Its speed at the centre of its oscillation is (a) 1.70 × 10–5 m/s (b) 17.0 × 10–5 m/s (c) 2.76 × 10–3 m/s (d) 2.77 × 10–5 m/s
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5. The frequency of a man’s voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child’s voice is 1.5 m, then the frequency of the child’s voice is (a) 200 Hz (b) 150 Hz (c) 400 Hz (d) 350 Hz
10. If the source is moving towards right, wave front of sound waves get modified to
4. Two coherent sources of different intensities send waves which interfere. The ratio of the maximum intensity to the minimum intensity is 25. The intensities are in the ratio (a) 25 : 1 (b) 5 : 1 (c) 9 : 4 (d) 625 : 1
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in water, the fundamental frequency will become (a) 240 Hz (b) 220 Hz (c) 230 Hz (d) 280 Hz
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24. Two pulses is a stretched string whose centers are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 second, the total energy of the pulses will be
23. A man standing between two cliffs hears the first echo of a sound after 2 sec and the second echo 3 sec after the initial sound. If the speed of sound be 330 m/s, the distance between the two cliffs should be (a) 1650 m (b) 990 m (c) 825 m (d) 660 m
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18. A taut string at both ends vibrates in its nth overtone. The distance between adjacent Node and Antinode is found to be ‘d’. If the length of the string is L, then (a) L = 2 d (n + 1) (b) L = d (n + 1) (c) L = 2 dn (d) L = 2 d (n – 1)
21. The average density of earth’s crust 10 km beneath the surface is 2.7 gm/cm3. The speed of longitudinal seismic waves at that depth is 5.4 km/s. The bulk modulus of earth’s crust considering its behaviour as fluid at that depth is (a) 7.9 × 1010 Pa (b) 5.6 × 1010 Pa (c) 7.9 × 107 Pa (d) 1.46 × 107 Pa
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16. In the resonance tube experiment, the first resonance is heard when length of air column is ℓ1 and second resonance is heard when length of air column is ℓ2. What should be the minimum length of the tube so that third resonance can also be heard? (a) 2ℓ2 – ℓ1 (b) 2ℓ1 (c) 5ℓ1 (d) 7ℓ1
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20. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is y1 = A sin (πx/L) sin ωt and energy is E1 and in another experiment its displacement is y2 = A sin(2πx/L) sin 2ωt and energy is E2. Then (a) E2 = E1 (b) E2 = 2E1 (c) E2 = 4E1 (d) E2 = 16E1
15. The beat frequency produced by two tuning forks when sounded together is observed to be 4 Hz. One of the forks makes 384 vibrations per second. When the other fork is loaded with a small piece of wax, the beats disappear first. The frequency of the second tuning fork is (a) 388 Hz (b) 380 Hz (c) more than 388 Hz (d) less than 380 Hz
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28. Organ pipe P1 closed at one end vibrating in its first harmonic and another pipe P2 open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is (a) 8/3 (b) 1/6 (c) 1/2 (d) 1/3
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25. A source S emits electromagnetic waves which are detected by the detector D kept at a distance of d/4 from the source. The waves directly received from S and those reflected by a layer at a distance d above the ground are found to be in phase. If the layer moves up by a distance h = 0.01d, the interfering waves are once again in phase. The wavelength of the waves from the source is about
(a) (b) (c) (d)
27. A source of frequency f is stationary and an observer starts moving towards it at t = 0 with constant small acceleration. Then the variation of observed frequency f registered by the observer with time is best represented as
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(b) 1/2 (d) 19/18
35. The amplitude of a wave disturbance propagating in the positive x-direction is given by 1 1 y= at t = 0 and y = at t 2 1+ x 2 + x2 − 2x = 2s, where x and y are in metre. Assuming that the shape of the wave disturbance does not change during the propagation, the speed of the wave is (a) 0.5 m/s (b) 1 m/s (c) 1.5 m/s (d) 2 m/s
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37. A racing car moving towards a cliff sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If V = the velocity of sound, the velocity of the car is (b) V/2 (a) V / 2 (c) V/3 (d) V/4
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36. Two vibrating tuning forks produce progressive waves given by y1 = 4 sin (500πt) and y2 = 2 sin (506πt). These tuning forks are held near the ear of a person. The person will hear (a) 3 beats/s with intensity ratio between maxima and minima equal to 2 (b) 3 beats/s with intensity ratio between maxima and minima equal to 9 (c) 6 beats/s with intensity ratio between maxima and minima equal to 2 (d) 6 beats/s with intensity ratio between maxima and minima equal to 9
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31. In resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air columns is 19 cm. If the 400 Hz tuning fork is replaced by 1600 Hz tuning fork then to get resonance, the water level in the tube should be further lowered by (take end correct = 1 cm.) (a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm
3.49
(a) 18/19 (c) 2
30. A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/sec. If sound travels at a speed of 330 m/sec, then number of beats per second heard by an observer between the wall and the source is (a) 7.7 Hz (b) 9 Hz (c) 4 Hz (d) None of these
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Wave Mechanics FP
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46. A source of frequency f gives 5 beat/s when sounded with source of frequency 200 Hz. The second harmonic of the source gives 10 beat/s when sounded with a source of frequency 420 Hz. The value of f is (a) 200 Hz (b) 210 Hz (c) 205 Hz (d) 195 Hz
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54. When a sound wave is reflected from a wall, the phase difference between the reflected and incident pressure wave is (a) 0 (b) π (c) π/2 (d) π/4
55. A wave travels uniformly in all directions from a point source in an isotropic medium. The displacement of the medium at any point distance r from the source may be
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50. A composite string is made up by joining two strings of different masses per unit length → µ and 4µ. The composite string is under the same tension. A transverse wave pulse Y = (6 mm) sin (5 t + 40 x), where ‘t’ is in seconds and ‘x’ is in metres, is sent along the lighter string towards the joint.
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53. A closed organ pipe has length ℓ. The air it is vibrating in 3rd overtone with maximum amplitude ‘a’. The amplitude at a distance of ℓ/7 from closed end of the pipe is equal to (a) a (b) a/2 a 3 (c) (d) zero 2
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49. An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is (a) 200 Hz (b) 300 Hz (c) 240 Hz (d) 480 Hz
48. When beats are produced by two progressive waves of nearly the same frequency, which one of the following is correct? (a) The particles vibrate simple harmonically, with the frequency equal to the difference in the component frequencies. (b) The amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two waves (c) The frequency of beats depends upon the position, where the observer is (d) The frequency of beats changes as the time progresses
Thus the periodic motion consists of three components.
51. A pipe open at the top end is held vertically with some of its lower portion dipped in water. At a certain depth of immersion, the air column of length (3/8) m in the pipe resonates with a tuning fork of frequency 680 Hz. The speed of sound in air is 340 m/s. The pipe is now raised up by a distance ‘a’ until it resonates in the “next overtone” with the same tuning fork. The value of ‘x’ is (a) 20 cm (b) 40 cm (c) 50 cm (d) 25 cm
47. The displacement y of a particle executing periodic motion is given by y = 4 cos2 (t) sin (1000t). This expression may be considered to be a result of the superposition of waves (a) two (b) three (c) four (d) five
The joint is at x = 0. The equation of the wave pulse reflected from the joint is (a) (2 mm) sin (5 t – 40 x) (b) (4 mm) sin (40 x – 5 t) (c) –(2 mm) sin (5 t – 40 x) (d) (2 mm) sin (5 t – 10x)
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Wave Mechanics
beat frequency of 4 Hz is heard. For forks A and Y, the beat frequency is 7 Hz. For forks X and Y, the beat frequency is 3 Hz. (a) The frequency of X is 436 Hz (b) The frequency of Y is 447 Hz (c) The frequency of X is 444 Hz (d) The frequency of Y is 433 Hz
represented by (A is a constant representing strength of source) A (a) sin (kr − ωt ) r (b) A r sin (kr − ωt )
(c) Ar sin(kr – ωt) A /r sin (kr − ωt )
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5. Standing waves are produced on a stretched string of length L with fixed ends. When there is a node at a distance L/3 from one end, then (a) minimum and next higher number of nodes excluding the ends are 2, 5 respectively (b) minimum and next higher number of nodes excluding the ends are 2, 4 respectively (c) frequency produced may be V/3 L (d) frequency produced may be 3V/2 L [V = Velocity of waves in the string]
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1. In a resonance tube experiment, a close organ pipe of length 120 cm resonates when tune with a turning fork of frequency 340 Hz. If water is poured in the pipe then (given vair = 340 m/sec) (a) minimum length of water column to have the resonance is 45 cm. (b) the distance between two successive nodes is 50 cm. (c) the maximum length of water column to create the resonance is 95 cm. (d) None of these
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4. A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is traveling on a stretched string. The maximum speed of any point on the string is v/10 where v is the speed of propagation of the wave. If a = 10–3 m and v = 10 ms–1, then λ and f are given by (a) λ = 2π × 10–2 m (b) λ = 10–3 m 3 (c) f = 10 /(2π) Hz (d) f = 104 Hz
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11. In a wave motion y = a sin(kx – ωt), y can represent (a) electric field (b) magnetic field (c) displacement (d) pressure
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10. y-x curve at an instant for a wave travelling along x-axis on a string is shown. Slope at the point A on the curve, as shown, is 53º.
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3.53
9. The vibration of a string fixed at both ends are described by Y = 2 sin (πx) sin (100πt) where Y is in mm, x is in cm, t in sec, then (a) Maximum displacement of the particle at x = 1/6 cm would be 1 mm. (b) velocity of the particle at x = 1/6 cm at time t = 1/600 sec will be 157 3 mm/s (c) If the length of the string be 10 cm, number of loop in it would be 5 (d) None of these
7. A string of length L is stretched along the x-axis and is rigidly clamped at its two ends. It undergoes transverse vibration. If n is an integer, which of the following relations may represents the shape of the string at any time t?
Wave Mechanics
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3.54
Q
ASSERTION AND REASON UESTIONS
Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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a is reflected from a rigid wall b is reflected from a yielding surface c is reflected from a yielding surface d is reflected from a rigid wall
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2. Statement 1: An 80 dB sound has twice the intensity of a 40 dB sound. Statement 2: Loudness of a sound of a certain intensity I is defined as: L (in dB) = I 10 log10 I0 3. Statement 1: An open tube having fundamental frequency f generates the same frequency when half immersed in a dense liquid. Statement 2: On immersing the tube in a liquid, the open tube behaves as a closed pipe.
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1. Statement 1: All points on a wavefront vibrate in same phase with same frequency. Statement 2: Two sources are said to be coherent if they produce waves of same frequency with a constant phase difference.
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(a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True.
13. In the figure, A is an incident pulse and a, b, c, d are the possible forms of the reflected pulse. Tick the correct answer.
(b) 240 Hz (d) 400 Hz
(a) 80 Hz (c) 320 Hz
5. Statement 1: A plane wave of sound travelling in air is incident upon a plane water surface. Angle of incidence is θ. If Snell’s law is valid for sound waves, it follows that
Wave Mechanics
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Wave Mechanics
Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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10. Statement 1: A person is standing near a railway track A train is moving on the track. As train is approaching the person, apparent frequency keeps on increasing and when train has passed the person then apparent frequency keeps on decreasing. Statement 2: When train is approach
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9. Statement 1: A transverse waves are produced in a very long string fixed at one end. Only progressive wave is observed near the free end. Statement 2: Energy of reflected wave does not reach the free end.
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1. For four sine waves, moving on a string along positive x-direction, displacementdistance curves (y-x curves) are shown at time t = 0. In the right column, expressions for y as function of distance x and time t for sinusoidal waves are given. All terms in the equations have general meaning. Correctly match y-x curves with corresponding equations.
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7. Statement 1: Like light; sound waves can be polarized. Statement 2: Polarization is a phenomenon related to all waves.
8. Statement 1: The intensity of a plane progressive wave does not change with change in distance from the source. Statement 2: The wavefronts associated with a plane progressive wave are planar.
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6. Statement 1: A tuning fork vibrated near the open end of the tube vibrate the air column above the water in it. At the surface of water in the tube, with reference to pressure, the status will be antinode. Statement 2: Displacement minima corresponds to pressure maxima.
sound will be refracted into water away from normal. Statement 2: From Snell’s law, angle of refraction is more than angle of incidence when wave travel from denser to rarer medium.
3.55
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3.56
Column II y = A cos (ωt – kx) y = – A cos (kx – ωt) y = A sin (ωt – kx) y = A sin (kx – ωt)
(c) y = 10 sin (2πx – 120t) + 10 cos (120t + 2πx) (d) y = 10 sin (2πx – 120t) + 8 cos (118t – 59/30πx)
2. A string of linear mass density 5 × 10–3 kg/m is stretched under a tension of 72 N between two rigid supports 60 cm apart. The string is vibrating in second overtone so that amplitude at one of its antinode is 0.25 cm (values in Column II are in CGS unit) Column II Column I 150 π (a) maximum (p) 2 velocity of a particle at antinode 2 (b) maximum (q) 90000π 2 velocity of a particle at 5 cm from any node (c) maximum (r) (150 π) acceleration of a particle at antinode (d) maximum (s) 90000 π2 acceleration of particle at 5 cm from any node
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5. A closed organ pipe P1 vibrating in its first overtone and another open pipe P2 vibrating in its third overtone are in resonance with the tuning fork of frequency 12 Hz. In column II, the beats frequency in Hz is given when produced in conditions provided in column I. Match column I and II.
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Column II Particles at every position are performing SHM Equation of travelling wave Equation of standing wave Equation of Beats
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Column I (a) P1 resonate in 5th harmonic and P2 resonate in 3rd overtone (b) P1 resonate in 1st overtone and P2 resonate in 3rd harmonic (c) P1 resonate in 9th harmonic and P2 resonate in 7th overtone (d) P1 resonate in 2nd overtone and P2 resonate in 4th overtone
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Wave Mechanics
PASSAGE–1 A plane pressure pulse triangular in shape approaches a rigid wall along normal at a speed of 400 m/s. At time t = 0, situation is shown in the figure. The peak pressure is 100 P. By the wall pulse gets reflected and pressure near the wall
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Wave Mechanics
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11. The transverse of the particle at x = 13 cm and t = 0.015 s will be (a) –250 cm/s (b) –500 cm/s (c) 500 cm/s (d) –1000 cm/s
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10. The vertical displacement of the particle of the string at x = 7 cm and t = 0.01 s will be (a) 0.75 cm (b) 0.5 cm (c) 0.25 cm (d) zero
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9. The shape of the string is drawn at t = 0 and the area of the pulse enclosed by the string and the x-axis is measured. It will be equal to (a) 2 cm2 (b) 2.5 cm2 2 (c) 4 cm (d) 5 cm2
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Here f and x are in centimeters. The linear mass density of the string is 50 g/m and it is under a tension of 5 N.
The figure represents the instantaneous picture of a transverse harmonic wave travelling along the negative x-axis.
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4. Total impulse imparted by the pulse on the wall will be (a) 300 Ns (b) 150 Ns (c) 750 Ns (d) none of these
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16. If m2 < m1/2, then (a) vD > vB > vA > vC (b) vB > vA > vD = vC (c) vD > vA = vB > vC (d) vD > (vA = vB) > vC
13. The phenomena of beats can take place for (a) Only transverse waves (b) Only longitudinal waves (c) Both longitudinal and transverse waves (d) For sound waves only
17. If m1 > m2 > m1/2, then (a) vD > vC > vB > vA (b) vA = vB > vD > vC (c) vB > vA > vC > vD (d) vA = vB > vC > vD
14. The frequency of beats produced air when two sources of sound are activated, one emitting wavelength 32 cm, other 32.2 cm is (Take Vsound = 350 m/s)
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12. If two sound sources of frequency difference 25 Hz are sounded together. Then which of the following is correct? (a) A normal human ear will hear 25 Hz beat frequency (b) A normal human ear will hear only 10 Hz beat frequency (c) A normal human ear-cannot hear this frequency (d) A normal human ear can hear maximum of the two frequency sounded together
PASSAGE–5
ω − ω2 ω1 + ω2 x = 0, y = 2 A cos 1 .t . t sin 2 2
Thus amplitude frequency is small and fluctuates slowly. A beat i.e., a maximum of intensity occurs, also intensity depends on square of amplitude. The beat frequency is given by ωbeat = |ω1 – ω2| Number of beats per second is called as beat frequency. A normal ear can detect only upto 15 Hz of frequency because of persistence of ear.
15. A tuning fork of unknown frequency makes 3 beats per second with a standard fork of frequency 384 Hz. The beat frequency decreases when wax is put on prongs of first fork and the frequency of this fork is (a) 381 Hz (b) 387 Hz (c) 384 Hz (d) None of above
Superposition of waves results in maximum and minimum of intensities such as in case of standing waves. This phenomenon is called as interference. Another type of super-position result in interference in time which is called as beats. In this case waves are analysed at a fixed point as a function of time. If the two waves are of nearby same frequency are superimposed, at a particular point, intensity of combined waves gives a periodic peak and fall. This phenomenon is beats. If ω1 and ω2 are the frequencies of two waves then by superimposed y = y1 + y2, we get at
(b) 18 (d) 10
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1. In the figure shown a source of sound of y frequency 510 Hz moves with constant velocity vs = 20 m/s in the direction shown. The wind is blowing at a constant velocity vw = 20 m/s towards an observer who is at rest at point B. Find the frequency (in Hz) detected by the observer corresponding to the sound emitted by the source at initial position A. [Speed of sound relative to air = 330 m/s]
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SUBJECTIVE UESTIONS
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8. How long will it take sound waves to travel the distance ℓ between the points A and B if the air temperature between them varies linearly from T1 to T2? The velocity of sound propagation in air is equal to v = α T , where α is a constant.
11. A sound source of frequency f0 = 1300 Hz is dropped from a height H = 505 m above the ground. At the same time a detector is thrown upwards with a velocity v = 50 m/s along the same line. If the velocity of sound is v = 300 m/s, find the frequency (in Hz) detected by the detector at t = 5 s.
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9. In the given figure, a string of linear mass density 3 × 10–2 kg/m and length L = 1 m, is stretched by a force F = (3 – kt) N, where ‘k’ is a constant and ‘t’ is time in sec. At the time t = 0, a pulse is generated at the end P of the string. Find the value of k (in N/s) if the value of force becomes zero as the pulse reaches point Q.
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1. A band playing music at a frequency f is moving towards a wall at a speed vb. A motorist is following the band with a speed vm. If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist.
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3. The air column is a pipe closed at one end is made to vibrate in its second overtone by tuning fork of frequency 440 Hz. The speed of sound in air is 330 m/s. End corrections may be neglected. Let P0 denote the mean pressure at any point in the pipe and ∆P0 the maximum amplitude of pressure variation. [1999]
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11. A boat is travelling in a river with a speed 10 m/s along the stream flowing with a speed 2 m/s. From this boat a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible. [2001]
(a) What will be the frequency detected by a receiver kept inside the river downstream? (b) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/s in the direction opposite the river stream. Determine
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8. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered is f2. If the speed of sound is 340 m/s then the ratio f1/f2 is [2000]
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(b) magnetic field (d) pressure
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10. A 3.6 m long pipe resonates with a source of frequency 212.5 Hz when water level is at certain heights in the pipe. Find the heights of waver level (from the bottom of the pipe) at which resonances occur. Neglect end correction. Now the pipe is filled to a height H (≈ 3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of H. If the radii of the pipe and the hole are 2 × 10–2 m and 1 × 10–3 m respectively. Calculate the time interval between the occurrence of first two resonances. Speed of sound in air is 340 m/s and g = 10 m/s2. [2000]
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9. Two vibrating strings of the same material but of length L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency v1 and the other with frequency v2. The ratio v1/v2 is given by [2000] (a) 2 (c) 8
0.8 represents a mov[(4 x + 5t ) 2 + 5] ing pulse where x and y are in metres and t is in second. Then [1999]
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(a) The time taken by the wave pulse to reach the other end R and (b) the amplitude of the reflected and transmitted wave pulse after the incident wave pulse crosses the joint Q.
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length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of the wave pulse. Calculate [1999]
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Wave Mechanics
12. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displacement of the wire
15. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by mass M. The wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is [2002]
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16. Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both end and is filled with a monoatomic gas of molar mass MA. Pipe B is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass MB. Both gases are at the same temperature. [2002]
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13. Two pulses in a stretched string, whose centres are initially 8 cm apart, are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds the total energy of the pulses will be [2001]
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a frequency of 5.5 kHz, while the train approaches the siren. During his return journey in a different train B, he records a frequency of 6 kHz while approaching the same siren. The ratio of the velocity of train B to that train A is [2002]
the frequency of the sound detected by the receiver. (Temperature of the air and water = 20ºC; Density of river water = 103 kg/m3; Bulk modulus of the water = 2.088 × 109 Pa; Gas constant R = 8.31 J/mol-K; Mean molecular mass of air = 28.8 × 10–3 kg/mol; Cp/Cv for air = 1.4)
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17. In a resonance tube experiment to determine the speed of sound (in air, a pipe of diameter 5 cm is used. The air column in pipe resonates with a tuning fork of frequency 480 Hz when the minimum) length
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14. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records
(a) (b) (c) (d)
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The frequency of sound recorded by an observer who is standing in air is [2004]
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24. An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct option. [2005] 5 3 (a) n = 3, f 2 = f1 (b) n = 3, f 2 = f1 4 4 5 3 (c) n = 5, f 2 = f1 (d) n = 5, f 2 = f1 4 4
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21. A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s.
25. A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance is 30.7 cm and at second resonance is 63.2 cm. The error in calculating velocity of sound is [2005]
(b) 0.025 m (d) 0.024 m
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20. In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction. [2003]
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22. A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is [2004]
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19. A police car moving at 22 m/s chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle. If it is given that the motorcyclist does not observe any beats. [2003]
(a) 200 Hz (c) 120 Hz
18. A string of mass per unit length µ is clamped at both ends such that one end of the string is at x = 0 and the other is at x = ℓ. When string vibrates in fundamental mode amplitude of the mid-point of the string is a, and tension in the string is T. Find the total oscillation energy stored in the string. [2003]
of the air column is 16 cm. Find the speed of sound in air at room temperature. [2003]
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31. At x = 0, how many times does the net amplitude between 0 and 1 s (a) 46 (b) 42 (c) 50 (d) 100
27. A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m/s and 90 m/s2 respectively. Velocity of the wave is 20 m/s. Find the waveform. [2005]
32. In the experiment to determine the speed of sound using a resonance column. [2007]
26. An observer standing on a railway crossing receives frequency of 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. Find the velocity of the train. [The speed of the sound in air is 300 m/s] [2005]
(a) Prongs of the tuning fork are kept in a vertical plane. (b) Prongs of the tuning fork are kept in a horizontal plane. (c) In one of the two resonances observed, the length of the resonating air columns is close to the wavelength of sound in air. (d) In one of the two resonances observed, the length of resonating air column is close to half of the wavelength of sound in air.
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(a) 340 m/s for passengers in A and 310 m/s for passengers in B.
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Two trains A and B are moving with speed 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engines of train A blows a long whistle. Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz as shown in the figure. The speed in the frequency (highest frequency lowest frequency) is thus 320 Hz. Speed of sound in still air is 340 m/s. [2007]
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(b) 360 m/s for passengers in A and 310 m/s for passengers in B. (c) 310 m/s for passengers in A and 360 m/s for passengers in B. (d) 340 m/s for passengers in both the trains. 34. The distribution of sound intensity of the whistle as observed by the passengers in trains A is best separated by
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36. A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is [2008]
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8. For the propagation of transverse waves a medium should have young modulus and coefficient of rigidity. But gases do not have both factor so transverse waves can not propagate in gases. On the other hand propagation of longitudinal waves a medium should have bulk modulus. Hence longitudinal waves can propagate in all the three mediums.
11. In a dispersive medium the wavelength of a pulse gets changed and hence the shape of the pulse gets distoned.
5. Bats produce ultrasonic waves and receive them after reflection. During the time interval, between the production of ultrasonic and its reception back bats estimate the size direction etc. of the obstacle.
4. At the modes pressure is maximum. Therefore a displacement node is a pressure antinode.
7. Two notes of the same frequency differ in overtones and their relative strength. As a result, timber of the two notes are different and two notes can be distinguished.
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2. Imagine a cyclinder of radius 7 m and length 10 m. Intensity of sound at the surface of cylinder is same everywhere.
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Lets check at any point, say at x = 0, y = (10 cos t) sin (100t) at any point amplitude is changing sinusoidally. so this is equation of beats. (c) y = 10 sin (2πx – 120t) + 10 cos (120t + 2πx) = superposition of two coherent waves travelling in opposite direction. ⇒ equation of standing waves (d) y = 10 sin (2πx – 120t) + 8 cos (118t – 59/30πx) = superposition of two waves whose frequency are slightly different (ω1 = 120, ω2 = 118) ⇒ equation of beats
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300 + 20 350 = × 510 = 525 Hz 330 + 20 − 10 340
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1. f =
Q
SUBJECTIVE UESTIONS
3.67
10−3 = 2 × 10−2 0.05
1 1 dK = dm VP2 = µ dx 2 2
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∂y −Vω ∂x
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3.68
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5. (b, c, d) 6. (a, b, c, d)
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PREVIOUS EARS’ IIT-JEE UESTIONS
Vair 330 = = 0.33 m f 1000 9
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d 1 |Path difference| = − 1 d = 2 2
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4. (b) the even harmonics are missing, (c) 25 Hz, (d) 3rd, 5th, 7th harmonic, (e) 4 m
d V =λ= ⇒ f (2n + 1) f
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1 K = T 10−4 (0.1) + 4 × 10−4 (0.5) 2
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F= f =
1 K = T m12 (0.1) + m22 (0.5) 2
0.15 2 0.1 2 m dx + m 1∫ 2 ∫ dx 0.1 0
Wave Mechanics
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8. (c)
27. y = (0.1 m) sin [(30 rad/s)t + (1.5 m–1)x + φ]
9. (d)
13. (b)
14. (b)
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11. (a) 1.0069 × 105 Hz, (b) 1.0304 × 105 Hz
28. (a)
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26. VT = 30 m/s
7. (a, b, c)
Wave Mechanics
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C H A P T E R
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4.1 REFLECTION
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Laws of Reflection
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Light (Geometrical Optics)
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(i) The angle of incident equals the angle of reflection, i.e.,
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In going from one point to another, a light ray travels a path that requires equal or less time than the time required for neighbouring paths. In other words, the light ray takes the path that requires the minimum time.
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Concave mirror
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Light (Geometrical Optics)
Real/Virtual Object (Image)
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When a group of incident rays strike a surface, the point from which they diverge or the point at which they try to converge is known as object. After reflection we get a group of reflected rays. The point where reflected rays converge or the point from which the reflected rays appear to diverge is known as the image of the object. Objects and images may be real or virtual. If the light, after reflection, actually passes through the point where the image is located, the image is called as a real image, if the light only appears to come from the point where the image is called a virtual image. A real image will appear on a screen placed at its position while a virtual image will not.
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Similarly, if the object is a point where the light energy is actually present, it is a real object. If the light rays only tend to go towards the point where the object is located, it is a virtual object.
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Sign convention (i) All the distances will be measured from the pole (or the centre of the surface) as origin. Distance of the object from the surface is called the object distance (u) and that of the image will be called image distance (v). (ii) All real distances will be taken positive. Distances of all virtual points will be taken negative.
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4A.2
Concave Mirror and Convex Mirror $ 5 3ULQFLSDO D[LV &
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Concave Mirror C: Centre of Curvature angles AOB: Aperture https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.3
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Convex mirror O: Pole of mirror OC = AC = BC = R = Radius of curvature
Focus If a group of rays parallel (and close) to the principal axis strike a concave mirror, the rays converges after reflection to a point which is known as focus (F). In case of a convex mirror, these incident rays will diverge away from focus (after reflection). The distance of focus from pole is called as focal length (f). (i) In a concave mirror, focus is a real point and, hence focal length is positive. (ii) In convex mirror, focus is a virtual point and, hence focal length is negative. (iii) The magnitude of focal length in spherical mirrors is half the radius of curvature, i.e., R f = . 2 https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
Mirror Equation (Magnification) Object distance, image distance and focal length in mirrors are related by the equation 1 1 1 + = u v f
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Ray Diagrams
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For a linear object, the ratio of image size (I) to the object size (O) is called magnification (M). It can also be proved that this ratio is equal to the ratio of v and u.
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The reflection of light rays and formation of images are shown with the help of ray diagram. Some typical incident rays and the corresponding reflected rays are shown below:
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■ A light ray parallel to principal axis passes through focus after reflection.
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■ A light ray passing through focus becomes parallel to principal axis after reflection.
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Light (Geometrical Optics)
4A.5
■ A light ray striking the Mirror normally i.e., collinear with C will be reflected along the same path.
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■ A light ray going towards focus becomes parallel to principal axis after reflection from a convex mirror.
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4.2 IMAGE FORMATION FOR A REAL LINEAR OBJECT
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Light (Geometrical Optics)
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Object: at F Image: at infinity
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Object: beyond C Image: between F and C, real inverted, diminished.
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Object: between F and pole Image: behind the mirror, virtual erect and enlarged.
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Light (Geometrical Optics)
4A.7
Convex Mirror
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(i) Reversibility of path: The path of a ray is completely reversible. If a light ray is turned back, it will take its original path. (ii) Whenever, a light ray strikes a reflecting surface normally, it comes back along the same path. (iii) In the case of plane mirror, focal length can be taken as infinite v = – u. We get a real image for virtual object and virtual image for a real object. The object and the image are equidistant from the mirror. (iv) If the incident ray on a plane mirror is kept fixed and the mirror is rotated through angle θ, the reflected ray gets rotated through an angle 2θ.
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(v) When a light ray is reflected, it deviates through an angle 180º – 22i,, where i = angle of incidence. Angle of deviation = D = angle I’ OR = 180 – (i + r) = 180 – 2i
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Light (Geometrical Optics)
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(vi) If two plane mirrors are inclined at an angle θ with each other, the total number of images of a point object placed between them is equal to the smallest integer 360° − 1 . greater than or the equal to θ (vii) The concave mirror is also called as converging mirror and the convex mirror is also called a diverging mirror.
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Example A 2.0 cm high object is placed on the principal axis of a concave mirror at a distance of 12 cm from the pole. If the image is inverted, real and 5.0 cm high, find the location of the image and the focal length of the mirror.
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Since the image is formed at 30 cm from the pole on the side of the object, we have 1 1 1 1 1 7 = + = + =− f v u −30cm −12cm 60cm https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
f =−
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4A.9
60cm = –8.6 cm. 7
Hence, focal length of the mirror is 8.6 cm. Example Rays of light are incident on a plane mirror at 45º. At what angle with the first should a second plane mirror be placed such that the rays emerge from the second mirror parallel to the first mirror?
) Solution The incident ray AB at an angle of incidence 45º is reflected along BC.
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This ray BC now suffers reflection at the second mirror at C and emerges along CD, where, ∠BCN′ = ∠N′CD (law of reflection). Since, ray CD is parallel to the first mirror. ∴ ∠M1CD = ∠BCO = θ Now in triangle BOC 2θ + 45º = 180º 135° or θ= = 67.5º 2
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Example An object is placed 25 cm from the surface of a convex mirror and a plane mirror is set so that the image formed in the two mirrors lie adjacent to each other. The plane mirror is then 20 cm from the object. What is the radius of curvature of the convex mirror?
) Solution Since, the image formed in plane mirror is as far behind the mirror as the object is in front of it. Image distance = |v| = (2d – u) = 40 – 25 = 15 cm (virtual)
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Light (Geometrical Optics)
| v | = –15 cm
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Light (Geometrical Optics)
4A.11
Case II: For virtual image: m = +3, since the image is erect. −v m= 2 =3 ∴ v2 = –3u2 u2 1 1 1 Now, = + f v2 u2 1 1 1 = + 30 3u2 u2 u2 = 20 cm
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Example Find the angle between two plane mirrors such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror, parallel to the first mirror.
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The ray AB parallel to OM2 is reflected along BC, where ∠ABM1 = ∠CBO = θ At C, the ray is reflected along CD, where CD is parallel to OM1. ∴ ∠M2CD = ∠BCO = θ Now in triangle BOC, 3θ = 180º ∴ θ = 60º Example A ray of light is incident on a plane reflecting surface at an angle of 30º to the normal. Find the deviation in the incident ray. What will be the deviation if the ray suffers a reflection again at a surface inclined at 60º to the first surface?
) Solution By the geometry
∠B′CM2 = 60º
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Light (Geometrical Optics)
∠M2CD = 60º
∠B′CD = 60º + 60º = 120º
∠DCN2 = ∠N2CB = 30º
∠ABN1 = ∠N1BC = 30º
∠CBA′ = 120º = D1
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The incident ray of light AB on the first mirror M1 at an angle of incidence of 30º is reflected along BC at an angle of reflection 30º. ∴ Deviation, D1 in the incident ray = 60º + 60º = 120º The reflected ray BC is incident on the second mirror M2.
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Example A rod of length 10 cm lies along the principal axis at a concave mirror of focal length 10 cm in such a way that the end closer to the pole is 20 cm away from it. Find the length of the image.
) Solution The situation is shown in the figure.
Radius of curvature of the mirror is R = 2f = 20 cm. Thus, the nearer end B of the rod AB is at the centre of the curvature and hence, its image will be formed at B itself. We shall now locate the image of A.
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Light (Geometrical Optics)
4A.13
Here, u = –30 cm and f = –10 cm. We have, =
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1 1 1 = − v f u
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v = –15 cm
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Light (Geometrical Optics)
Example A plane mirror is placed at a distance of 50 cm from a concave mirror of focal length 16 cm. Where should a short object be placed between the mirrors and facing both the mirrors so that its virtual image in the plane mirror coincides with the real image in concave mirror? What is the ratio of the sizes of the two images?
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Let the object be at a distance x from plane mirror. ∴ Distance of object from concave mirror u = (50 – x) For the plane mirror, since object and image distances are equal, therefore
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1 1 1 = + 16 50 − x 50 + x
1 100 = 16 502 − x 2 or x2 = 502 – 16 × 100 = (2500 – 1600) cm2. = 900 cm2. ∴ x = 30 cm. Hence the object must be placed at a distance of 30 cm. from the plane mirror. or
AB A′B′ u 50 − x 50 − 30 20 1 = = = = = v 50 + x 50 + 30 80 4
Ratio of image sizes =
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Light (Geometrical Optics)
4A.15
The real image formed by the concave mirror is 4 times the size of virtual image formed by the plane mirror.
4.3 REFRACTION AT A PLANE SURFACE Refractive Index of a Medium (i) Refractive index of a medium 1 with respect to air µ1 = =
Speed of light in air Speed of light in medium 1 cair c1
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Light (Geometrical Optics)
Rarer Medium and Denser Medium When two mediums are in contact, the one with the lesser refractive index is known as rarer medium and other with greater refractive index is known as denser medium.
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If the angle of incidence is increased, the angle of refraction may eventually becomes 90º. The angle of incidence for which the angle of refraction is 90º is called the optical angle for that interface. Let µr be the refractive index of the rarer medium and µd be the refractive index of the denser medium. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.17
Obviously, µr < µd From Snell’s Law: µ sin C = r sin 90° µ d
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If a ray of light travelling in a denser medium strikes a rarer medium at an angle of incidence i which is greater than the critical angle C, it gets totally reflected back into the same medium. This phenomenon is called a total internal reflection.
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4.4 REFRACTION THROUGH A GLASS SLAB When a light ray passes through a glass slab having parallel faces, it gets refracted twice and finally emerges out parallel to its initial direction. i.e., the angle of emergence (e) is equal to the angle of incidence (i). https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
Incident ray
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⇒ e=i Lateral displacement of the ray = d = distance between incident ray and emergent ray. It can be shown that
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When there is a layer of denser medium between a real object and an observer (in air), the object appears to be shifted towards the observer. If t is the thickness of the layer of denser medium, apparent shift is given as
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1 ∆t = 1 − t µ (i) A coin lying at the bottom of a vessel containing water upto depth t appears uplifted. Let, t = real depth of coin O. t ∴ Apparent depth = µ
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1 and Apparent Shift = OI = 1 − t µ (ii) (a) A glass slab of thickness t is placed between a real object O and an observer. The object appears to be shifted towards the slab by the distance
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4A.18
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Light (Geometrical Optics)
4A.19
(b) Whenever, a glass slab is placed in the path of converging rays, the point of convergence is shifted away from the slab by the distance 1 t 1 − µ W
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t t1 t2 + + 3 + .... µ1 µ 2 µ3
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(iii) Multiple slabs: (Near normal rays) These relations hold also for images formed by successive refraction through a number of parallel sided slabs. It t1, t2, t3, ............ are the thicknesses of each of the slabs and µ1, µ2, µ3 ........... are the absolute refractive indexes of material of each slab, then object O viewed from rarer air medium is seen at I. Then, as seen from air
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AO = real depth, AI = apparent depth, OI = apparent shift. Example Consider the situation shown in the figure. Find the maximum angle θ for which the light suffers total internal reflection at the vertical surface.
) Solution For this case, optical angle θ '' = sin −1
1 4 = sin −1 1.25 5
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Light (Geometrical Optics)
4 5 π Since, θ '' = − θ ', we have 2 3 sin θ′ = cos θ′′ = 5
or sin θ '' =
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Maximum value of θ, for which the total reflection takes place at the vertical 3 surface, is sin −1 . 4 Example When a ray of light in air enters a liquid, it is found that the angle between the incident and the refracted rays is 150º and the angle between the reflected and refracted ray is 60º. Calculate the refractive index of the liquid.
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4A.20
) Solution If the liquid is denser than air, the refracted ray bends towards the normal. Given: ∠AOB = 150º ∴ ∠AOQ + 90º + r = 150º or 90º – i + 90º + r = 150º or i – r = 30º Also ∠BOC = 60º (Given) ∴ ∠BOP + ∠POC = 60º or 90º – r + 90º – i = 60º ∴ i + r = 120º
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Light (Geometrical Optics)
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4A.21
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Show that the lateral displacement of a ray of light passing obliquely through a parallel glass slab of thickness t and absolute refractive index µ is given by
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cos i sin i. t 1 − µ 2 − sin 2 i
) Solution
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Light (Geometrical Optics)
…(i)
In triangle BCE, x sin(i − r ) = BC ∴ x = BC sin (i – r) In triangle BCD,
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t [ BD = t ] cos r Substituting in equation (i), we get t x= sin(i − r ) cos r BC =
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4A.22
cos i x = t 1 − sin i 2 2 m − sin i
Example A layer of ice (µ = 1.33) lies on a glass plate (µ = 1.5). A ray of light makes an angle of incidence 60º on the surface of ice. Find the angle between this ray and the normal in glass.
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Light (Geometrical Optics)
4A.23
) Solution
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sin r1 µ g = sin r2 µice
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…(ii)
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sin 60° = µg sin r2
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sin 60° µice = sin r1 1
m
At A:
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In the figure, the ray gets deflected twice from air to ice at A and from ice to glass at B. Find the angle r2.
A glass cube of refractive index 1.5 is 4 cm. on a side. A ray of light is incident on one face 1 cm from the edge at an angle of incidence 60º and directed towards the near side. Where and at what angle will the ray emerge from the cube?
) Solution Let the ray strike the cube at A. After refraction into the cube, it again strikes the other surface at B as shown. If r is the angle of refraction at A, the angle of incident at B is (90º – r). At point A, or
sin 60° 1.5 = sin r 1 sin r =
1 3
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Light (Geometrical Optics)
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∴ r = 35º Angle of incident at B, i = 90 – r = 55º
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At point C,
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1 2 Critical angle for glass air interface = sin −−11 = sin −−11 ≈ 42° 1.5 3 As i > critical angle, the ray will be total internally reflected at B. Finally, the ray strike at C as shown.
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∴ θ = 60º Now CQ = BQ tan r = (PQ – BP) tan r = (PQ – AP cot r) tan r = PQ tan r – AP 4 = − 1 = 1.83 cm 2 The light ray emerges out from the opposite face at a distance of 1.83 cm from the edge at an angle of 60º.
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4A.24
Example A beaker contains water upto a height h1 and kerosene oil above water upto another height h2. Find the apparent shift in the position of the bottom of the beaker when viewed from above. Refractive index of water is µ1 and that of kerosene oil is µ2.
) Solution The apparent shift of the bottom due to the water is 1 ∆t1 = 1 − h1 µ 1 https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.25
And apparent shift of bottom due to kerosene oil is 1 ∆t2 = 1 − h2 µ2 1 1 Total shift = ∆t1 + ∆t2 = 1 − h1 + 1 − µ µ 1 2
∴
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Example A concave mirror of radius 40 cm lies on a horizontal table and water is filled in it up to a height of 5.0 cm as shown in the figure. A small dust particle flows on the water surface at a point P vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The refractive index of water is 1.33.
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Let us first locate the image formed by the concave mirror. Let us take vertically upwards as the negative axis. Then, R = 40 cm. The object distance, u = –5 cm. Using the mirror equation,
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) Solution The ray diagram is shown in the figure.
3
3 3
or
1 1 2 + = u v R 1 2 1 = −2 − 1 = 6 = − v R u 40 cm −5 cm 40 cm
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or
v = 6.67 cm
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Light (Geometrical Optics)
The positive sign shows that the image P1 is formed below the mirror and hence, it is virtual. These reflected rays are refracted at the water surface and go to the observer. Depth of the point P1 from the surface = 6.67 cm + 5.00 cm = 11.67. Due to refraction at the water surface, the image P1 will be shifted above by a distance 1 = (11.67 cm) 1 − = 2.92 cm 1.33 Thus, the final image is formed at a point (11.67 – 2.92) cm = 8.75 cm below the water surface.
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Example The base of a tank is horizontal plane of glass 8 cm thick (µ = 1.6). Above this is a layer of liquid of thickness 4.5 cm (µ = 1.5) and upon this floats a layer of water 12 cm thick (µ = 4/3). Looking from above, what is the apparent position of the spot on the bottom tank and where should the eye be held to see this spot?
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) Solution O is the object on the bottom and its image appears at I.
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4A.26
8 4.5 12 + + 1.6 1.5 4 3 = 5 + 3 + 9 = 17 cm Apparent position of the spot = 24.5 – 17 = 7.5 cm above O. Position of eye: Since, the least distance of distinct vision is 25 cm, therefore the eye must be held at a distance = 25 cm from I or at a distance (25 – 17) = 8 cm above A. =
Example A slab of glass 20 cm thick and refractive index 1.5 is kept in front of a plane mirror and a pin is kept in front of it in air at a distance of 40 cm from the mirror. Find the position of the image as formed by the mirror. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.27
) Solution 0LUURU FP
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The virtual image of pin at O is formed by the slab I1, where OI1 is the shift due to slab 1 2 OI1 = t 1 − = 20 1 − µ 3
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and
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This image I1 now acts as an object for the plane mirror at a distance = 40 – 6.67 = 33.33 cm. ∴ Distance of image from the plane mirror = 33.33 cm.
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4.5 REFRACTION THORUGH A PRISM
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A light ray striking at one face of a triangular glass prism gets refracted twice and emerges out from the other face as shown above. The angle between the emergent and the incident rays is called the angle of deviation (D). The angle between the two refracting faces involved is called the refracting angle (A) of the prism. From AXY, we have A + (90 – r1) + (90 – r2) = 180º ⇒ r1 + r2 = A ...(i) Deviation, D = (i – r1) + (e – r2) = (i + e) – (r1 + r2) = i + e – A ⇒ A+D=i+e ...(ii) https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
We also have two equations from Snell’s Law at X and Y sin r2 1 sin i = µ and = sin r1 sin e µ
Angle of Deviation It can be easily seen that if we reverse the emergent ray, it goes back along the same path. The angle of incidence and emergence get interchanged but the angle of deviation remains same.
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angle of incidence = θ1 angle of emergence = θ2 θ1 + θ2 = A + D
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4A.28
angle of incidence = θ2 angle of emergence = θ1 θ2 + θ1 = A + D
Hence, the same angle of deviation D is possible for two different angles of incidence: θ1 and θ2 where, θ1 + θ2 = A + D
Minimum Deviation The angle of deviation is minimum when the path of light ray through the prism is symmetrical. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.29
i.e., angle of incidence = angle of emergence i=e ⇒ r1 = r2 Hence, A + Dmin = 2 i Also, A = r1 + r2 = 2 r sin i =µ sin r1 A + D min ) 2 =µ A sin 2 $
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Grazing Incidence
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When i = 90º the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass-air. This is known as grazing angle. When e = 90º, the emerged ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for the glass-air. This is known as grazing emergence. $
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Maximum Deviation The angle of deviation is same for both the above cases (grazing incidence and grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
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Dispersion Through a Prism
µred < µviolet because λred > λviolet
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The refractive index of a medium is different for light rays of different wavelengths. More is the wavelength,
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A ray of white light passing through a prism gets splits into two different colours because the deviation is more for violet and less for red. This phenomenon is known as dispersion.
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For a prism with very small refracting angle A, the deviation D is given by: D ≈ (µ – 1) A
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4A.30
Hence, deviation of violet = Dv = (µv – 1) A and deviation of red = (µr – 1) A The angle between the red and violet rays is known as the angle of dispersion. Angle of dispersion, φ = Dv – Dr = (µv – µr) A Angle of mean deviation, D m =
Dv + D r 2
µ + µr = v − 1 A 2 Dispersion power of glass, w =
µv − µr φ = Dm µv + µr − 1 2
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Light (Geometrical Optics)
4A.31
Combination of Prisms
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Condition for achromatism is φ1 = φ2 ⇒ (µ1v – µ1r) A1 = (µ2v – µ2r) A2
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µ + µ1r µ + µ2r = 1v − 1 A1 − 2 v − 1 A 2 2 2
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Net deviation produced, Dnet = Dm1 – Dm2
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(i) If two prisms are placed opposite to each other (as shown) and their angles of dispersion are equal, the combination is known as achromatic combination. The dispersive effect of one prism is cancelled by the other.
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(ii) If we want to have only dispersion and no deviation, we combine two prisms having same mean deviation. Dm1 = Dm2
µ + µ1r µ + µ2r ⇒ 1v − 1 A1 = 2 v − 1 A 2 2 2 Such a combination is known as non-deviation or direct-vision combination. The net angle of dispersion produced by the combination is Qnet = φ1 – φ2 = (µ1v – µ1r) A1 – (µ2v – µ2r) A2. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
Example A glass prism has a reflecting angle 60º and refractive index 1.5. A ray of light is incident at 48.6º. Find the angle of emergence and angle of deviation. )
) Solution Given: A = 60º, i = 48.6º, µ = 1.5 Now,
µ=
or
1.5 =
sin i sin r1 sin 48.6° sin r1
0.7501 = 0.5 1.5 ∴ r1 = 30º Also, r1 + r2 = A ∴ r2 = (60º – r1) = 60º – 30º = 30º Which shows that r = r1, i.e., the ray passes symmetrically through the prism. ∴ e = i = 48.6º D=i+e–A = 48.6º + 48.6º – 60º = 37.2º This deviation is minimum at i = e sin r1 =
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Example The refractive indexes of silicate flint glass for wavelength 400 nm and 700 nm are 1.66 and 1.61 respectively. Find the minimum angles of deviation of an equilateral prism made of this glass for light of wavelengths 400 nm and 700 nm.
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s:
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ht
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A + Dm sin 2 µ= A sin 2
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)
s: //
) Solution The minimum angle of deviation Dm is given by
ht
D sin 30° + m D 2 = = 2sin 30° + m sin30° 2
For 400 nm light,
or or
or
D 1.66 = 2sin 30° + m 2
4A.32
D sin 30° + m 2 30° +
= 0.83
Dm = 56° 2 Dm = 52°
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Light (Geometrical Optics)
4A.33
For 700 nm light, D 1.61 = 2sin 30° + m 2 Solving, we get Dm = 48°. Example A prism of angle 60º deviate a ray of light through 40º for two angles of incidence which differ by 11º. What is refractive index of the glass of the prism?
) Solution The incident ray PQ is deviated through 40º and the principle of
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Hence other angle of incidence causing same deviation of 40º is e = i2 where, i1 – i2 = 11º (Given) But i + e = A + D = 60º + 40º = 100º ∴ i1 + i2 = 100 ∴ i1 = 55º . 30′ and i2 = 44º . 30′
s: //
sin i1 sin i2 = sin r1 sin r2
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Now, µ =
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reversibility of ray through optical system. Shows that the ray SR would also suffer the same deviation of 40º.
where, r2 = 60° –r1 ∴
µ=
sin 55° ⋅ 30′ sin 44° ⋅ 30′ = sin r1 sin(60° − r1 )
…(i)
From equation (i), we first calculate r1, therefore 0.8241 = sin r1
0.7009 3 1 cos r1 − sin r1 2 2 Dividing both sides in denominator by sin r1, we get 0.8241 0.7009 = sin r1 3 1 sin r1 cot r1 − sin r1 2 2 sin r1 https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
1 0.7009 ( 3 cot r1 − r2 ) = 2 0.8241
or
3 cot r1 = 2.701
cot r1 = 1.559 r1 = 32.40′
or or
or
∴ µ=
sin 55° ⋅ 30′ 0.8241 = = 1.526 sin 32° ⋅ 40′ 0.5400
Example
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A ray of white light incident at an angle of 35º on one face a crown glass of refracting angle 60º. Find the angle of dispersion and mean deviation produced by the prism. Given for the crown glass, µv = 1.538 and µr = 1.52. )
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) Solution The angle of prism is 60º and hence it is not a small angle prism,
⇒ rv 1 = 21.9º
∴ rv 2 = (60 – 21.9) = 38.1º
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i = 35º, A = 60º. (i) For the violet colour at the first face, sin i sin 35° µv = = 1.538 = sin rv1 sin rv1
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un
//t
si
sin ev sin ev = 1.538 = sin rv2 sin 38.1°
er
e/
µv =
.m
tp s:
Again at the second face,
ht
e/
s:
tp s:
sin i sin 35° = 1.52 = sin γ r1 sin γ r1
µr =
ht
//t
.m
ht tp
⇒ ev = 71.6º ∴ Deviation for the violet ray, DV = (i + ev) – A = (35 + 71.6) – 60 = 46.6º. (ii) For the red colour at the first face,
γ r1 = 22.2°
Then, γ r2 = (60 − 22.2) = 37.8° Again at the second face, sin er sin er µr = = 1.52 = sin γ r1 sin 37.8°
⇒
⇒
4A.34
er = 68.7º
Deviation for the red ray, Dr = (i + er) – A = (35 + 68.7) – 6.0 = 43.7º
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Light (Geometrical Optics)
4A.35
Again of dispersion, φ = (Dv – Dr)
= (46.6 – 43.7) = 2.9º 46.6 + 43.7 Angle of the mean deviation = 2 = 45.15°.
om
Example Crown glass has µv = 1.538, µr = 1.52; quartz glass has µv = 1.557; µr = 1.542. Given a quratz prism of refracting angle 4º. Calculate (i) The angle of the crown prism to form a direct vision combination for white light. (ii) The angular width of the spectrum product by this combination.
an a. c
)
) Solution The prisms are small angled. For direct vision combination
ta
0.5495 × 4 = 4.16° 0.5290
si
s ita rio
w
A crown =
w
∴
un iv er
.ja
rio
m ar
s_
in fo
1.557 + 1.542 − 1 µ − ( 1) A crown 2 m quartz m quartz = 0.5495 crown = = A quartz (µ mm crown 1.538 + 1.520 0.5290 quartz crown − 1) − 1 2
ve rs
ni
//t .m
ht
tp
e/
s: //
w
Then, angular width of the spectrum produced by the combination = (µv – µr)quariz – (µv – µr)crown = |(0.015) 4 – (0.018) × 4.16 | = 0.01488º
s: //
ht
t.m
e/ u
tp s:
Note: The net angle of dispersion is the angular width of the spectrum and is the modulus of the difference between the angle of dispersion produced by the individual prisms.
ht
tp
Example A 60º glass prism has a refractive index of 1.5. Calculate: (i) The angle of incidence for minimum deviation. (ii) The minimum deviation. (iii) The angle of emergence of the light at maximum deviation. )
) Solution
(i) At minimum deviation, the ray passes symmetrically through the prism. A 60° r= = = 30° ∴ 2 2 sin i Now µ = sin r ∴ sin i = µ sin r = 1.5 sin 30° = 0.75 ∴ i = 48°.35′ https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
A+D sin 2 µ = (ii) A sin 2 60° + D sin 2 or 1.5 = sin 30° or sin
∴
na .c om
60 + D = 1.5sin 30° = 0.75 2 60 + D = 48.35° 2 D = 37.10°
or
m
ar a
sin 90° = µ = 1.5 sin r1 sin r1 =
1 = 0.6667 1.5
∴ r1 = 41.49′ and r2 = (A – r1) = 60º – 41.49º = 18.11º sin e Now, = µ = 1.5 sin r2
un iv
er si ta rio s
er s
w
w
//w
un
//t
sin e = 1.5 sin r2 = 1.5 sin 18.55º = 0.4681 or e = angle of emergence = 27.55º
ht tp
s: //
t.m e/
ht tp s:
∴
iv
.m
e/
s: tp
ita
∴
.ja
rio s_ in fo
(iii) Maximum deviation occur when the light is incident at 90º on one face.
ht
4A.36
U
U
H
Example Light falls at grazing incidence on one of the surfaces of a prism of refractive index µ and refracting angle A. Show that the angle of emergence is given by sin e =
(
µ 2 − 1 sin A − cos A
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)
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Light (Geometrical Optics)
4A.37
) Solution The ray PQ is incident at grazing incidence (Since, i = 90º) and emerges
along RS with angle of emergence e. Now because of the reversibility of the ray, through the optical system, the ray RQ would emerge along QP with angle of refraction = 90º. ∴ r1 = C (critical angle) But r1 + r2 = A ∴ r2 = (A – r1) = (A – C) Now for refraction at R,
sin e =µ sin r2
iv
…(iii)
t.m e/
un
//t
cos C =
er si ta rio s
un iv e/
µ2 − 1 µ Substituting in equation (i), we get
∴
ht tp s:
ht
.m
µ2 − 1 µ2
=
tp
s:
//w
w
er s
w
ita
.ja
Also, cos2C = 1 – sin2C 1 = 1− 2 µ
rio s_ in fo
m
ar a
na .c om
∴ sin e = µ sin (A – C) or sin e = µ (sin A cos C – cos A sin C) ...(i) Note that in the expression asked in the problem, since the terms sin C and cos C are absent, therefore, we must find the values of sin C and cos C in terms of µ and A. Since, C is the critical angle, therefore 1 sinCC= or sin = …(ii) µ
ht tp
s: //
µ2 − 1 1 sin e = µ sin A − cos A µ µ = µ 2 − 1 sin A − cosA.
$
4
5 U
U
H 6
3
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4A.38
Light (Geometrical Optics)
4.6 REFRACTION AT A SPHERICAL SURFACE Consider a spherical surface of radius R separating two media with refractive indexes µ1 and µ2. The radius of curvature R will be taken positive when incident rays strike the convex side of the surface. Let u is the object distance and v is the image distance. $
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m
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,
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er si ta rio s
er s
w //w
J
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rio s_ in fo
3
D
ar a
2
na .c om
U
e/
tp
s:
For light rays going from medium 1 (µ1) to medium 2 (µ2)
iv
.m
…(i)
t.m e/
ht tp s:
un
//t
ht
µ1 µ 2 µ 2 − µ1 + = u v R The corresponding equation for refraction at a plane surface will be µ1 µ 2 + = 0 (taking R = ∞) u v
ht tp
s: //
Derivation of equation Let ∠DOP = α, ∠DCP = β and ∠DIC = γ. For paraxial rays, D is close to P and α, i, r, β, and γ are all small. From triangle ODC α + β = i And from ∆DCI, r+γ=β Also from Shell’s Law,
...(i) ...(ii)
sin i µ 2 = sin r µ1
We can write sin e ≈ i and sin r ≈ r, so that the above equation becomes µ1i = µ2r Putting values of i and r from equation (i) and (ii) into equation (iii), we get µ1 (α + β) = µ2 (β – γ) or µ1 α + µ2 γ = (µ2 – µ1) β https://t.me/universitarios_infohttps://www.jamarana.com
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...(iii)
...(iv)
Light (Geometrical Optics)
4A.39
As α, β, and γ are small, hence from figure DP DP DP α≈ ,β= and γ = PO PC PI Putting these values in equation (iv), we get DP DP DP µ1 + µ 2 PI = (µ 2 − µ1 ) PC PO µ1 µ 2 µ 2 − µ1 + = PO PI PC
or where,
rio s_ in fo
na .c om
R = radius of curvature. µ1 µ 2 µ 2 − µ1 + = u v R
ar a
∴
PO = Object distance, PI = Image distance, and
ita
4
&
2
iv
.m
$
er si ta rio s
e/
3
2
un
//t
ht
tp
s:
un iv
//w
w
er s
w
.ja
m
4.7 EXTENDED OBJECTS: LATERAL MAGNIFICATION
4
P %
s: //
t.m e/
ht tp s:
P
ht tp
The lateral or transverse magnification is defined as h m= 2 h1 where, h2 = height of the image and h1 = height of the object In the figure, OQ = + h1 and O′Q′ = – h2. h O′Q′ m= 2 =− h1 OQ Since the triangles OCQ and O′CQ′ are similar, therefore −O′Q′ −O′C −PO′ − PC m= =− = OQ OC PO + PC In the figure, PO = – u, PC = + R and PO′ = + v From equation (i), we have R −v v−R m=− = −u + R R − u https://t.me/universitarios_infohttps://www.jamarana.com
…(i)
…(ii) https://t.me/universitarios
Light (Geometrical Optics)
µ 2 µ1 µ 2 − µ1 − = v u R
Also or
µ 2u − µ1v µ 2 − µ1 = uv R
or
R=
…(Here, u = negative)
(µ 2 − µ1 )uv µ 2u − µ1v
This gives, R − v =
µ1v(v − u ) µ µ (v − µ ) and R − u = 2 µ 2u − µ1v µ 2u − µ1v µ1v µ 2u
na .c om
By equation (ii), we get m =
e/
&
.m
3
, 2
P
s: //
t.m e/
ht tp s:
un
//t
iv
tp
s:
un iv
//w
w
$
er si ta rio s
er s
w
ita
.ja
m
ar a
rio s_ in fo
Example There is a small air bubble inside a glass sphere (µ = 1.5) of radius 10 cm. The bubble is 4.0 cm below the surface and is viewed normally from the outside (in figure). Find the apparent depth of the bubble.
ht
) Solution The observer sees the image formed due to refraction at the spherical
ht tp
4A.40
surface when the light from the bubble goes from the glass to the air. Here, u = –4.0 cm, R = –10 cm, µ1 = 1.5 and µ2 = 1 We have µ 2 µ1 µ 2 − µ1 + = v u R or
1 1.5 1 − 1.5 + = v 4.0 cm −10 cm
or
1 0.5 1.5 = − v 10 cm 4.0 cm
or v = –3.0 cm Thus, the bubble will appear 3.0 cm below the surface. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.41
Example A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 9 mm situated in water. Assuming the light rays to be paraxial: (i) Find the position of the image due to refraction at the first surface and the position of the final image. (ii) Draw a ray diagram showing the positions of both the images.
rio s_ in fo
m
ar a
na .c om
) Solution
er si ta rio s
…(i)
un
1 1 − 1.33 = v 2 mm
t.m e/
ht tp s:
Thus,
//t
.m
e/
For the first refraction (water to air). µ1 = 1.33, µ2 = 1, u = ∞, R = + 2 mm
tp ht
un iv
s:
//w
µ 2 µ1 µ 2 − µ1 + = v u R
iv
w
er s
w
ita
.ja
The ray diagram is shown in the figure. The equation for refraction at a spherical surface is
ht tp
s: //
or v = – 6 mm The negative sign shows that the image I, is virtual and forms at 6 mm from the surface of the bubble on the water side. The refracted rays (which seen to come from I1) are incident on the farther surface of the bubble. For this refraction, µ1 = 1, The object distance,
From equation (i), or
µ2 = 1.33, R = – 2
u = (6 mm + 4 mm) = 10 mm
µ 2 µ1 µ 2 − µ1 + = v u R
1.33 1 1.33 − 1 + = v 10 −2
or
1.33 0.33 1 =− − v 2 10
or
v = –5 mm
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4A.42
Light (Geometrical Optics)
This minus sign shows that the image is formed on the air side at 5 mm from the refracting surface. Measuring from the centre of the bubble, the first image is formed at 8.0 mm from the centre and the second image is formed at 3.0 mm from the centre. Both images are formed on the side from which the incident rays are coming.
4.8 LENSES
na .c om
A lens is one of the most familiar optical device for human being, the lenses are made of a transparent material bounded by two spherical surfaces. The surface may be both convex, both concave or one convex and one concave. When the thickness of the lens is small as compared to the other dimension like object distance etc., we call it a thin lens.
Convex Lens
rio s_ in fo
er si ta rio s
3
e/
)
iv
.m t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
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m
ar a
It a beam of light parallel to principal axis strikes a convex lens, it converges after refraction to a point called as focus (F). Focus in a convex lens is a real point and, hence focal length (f) is positive. Due to the converging effect, convex lens is also called as a converging lens.
ht tp
s: //
Lenses have two foci, because light can strike a lens from both the sides. As there are two spherical surfaces, there are two centres of curvatures C1 and C2 and correspondingly two radii of curvature R1 and R2. The line joining C1 and C2 is called the principal axis of the lens. The centre P of the thin lens which lies on the principal axis, is called the optical centre.
&
3
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,
&
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Light (Geometrical Optics)
4A.43
Ray Diagrams for Convex Lens Some important rays and their paths after refraction are shown below.
er s
w
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.ja
m
ar a
rio s_ in fo
na .c om
)
er si ta rio s
e/ )
&
t.m e/
)
%
)
)
$
ht tp
s: //
ht tp s:
%
un
iv
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$
//t
ht
tp
s:
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//w
w
Consider a real linear object AB placed at different locations in front of a convex lens.
Object beyond 2F: Image between F and 2F, real inverted and diminished.
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Object at 2F: Image at 2F, real, inverted and of same size as object.
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Light (Geometrical Optics)
$ & )
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)
%
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$
Object between 2F and F: Image beyond 2F, real, inverted and enlarged.
na .c om
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Object at F: Image at infinity.
tp
)
)
)
ht
s: //
Object between F and lens: Image of the same size as object, virtual, erect and enlarged.
Concave Lens
ht tp
4A.44
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)
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Light (Geometrical Optics)
4A.45
When a beam of light parallel to principal axis strikes a concave lens, it diverges away from a point known as focus (F). As the focus is a virtual point, the focal length (f) of a concave lens is negative. A concave lens is also called as a diverging lens.
Ray Diagrams for Concave Lens
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s:
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s: //
ht tp s:
//t
ht
tp
Image formation for a real object AB in a concave lens is shown below:
)
&
)
ht tp
%
Object anywhere in front of lens: Image between F and lens on the same side, virtual, erect diminished.
Sign Conventions The optical centre is taken as the origin and the principal axis as the x-axis. The positive direction of the axis is generally taken along the incident rays. The quantities u, v, f, R1 and R2 represent the x-coordinate of the object, the image, the focus, first centre of curvature and second centre of curvature respectively. The table below shows the signs of u, v, f, R1 and R2 in certain cases shown in the figure. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
na .c om
)
2
& )
&
un iv
e/
, ) & 2
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
&
er si ta rio s
er s
w w //w s:
(c)
tp
,
& ,
ita
.ja
m
ar a
(b)
&
rio s_ in fo
(a)
ht
4A.46
(d) 2
Figure a b c d
u – – + –
v + – + –
)
f + – + –
,
&
R1 + – + +
&
R2 – + – +
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Light (Geometrical Optics)
4A.47
Generally, the incident rays and hence the positive direction of the axis is taken from left to right. Heights measured upwards are taken to be positive and the heights measured downward are taken to be negative. With the usual choice of axis, f of a lens is positive for converging lens and is negative for a diverging lens.
4.9 LENS MAKER’S FORMULA AND LENS FORMULA $
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'
&
2 2
er si ta rio s
un iv
//w
w
µ 2 µ1 µ 2 − µ1 + = v u R
er s
w
The general equation for refraction at a spherical surface is …(i)
e/
un
…(ii)
t.m e/
ht tp s:
iv
.m
µ1 µ 2 µ 2 − µ1 + = u v1 R2
//t
ht
tp
s:
For the first refraction, the object is at O, the image is at O1 and the centre of curvature is at C1. If u, v and R denote their x-co-ordinates, then
s: //
For the second refraction, the light goes from the medium µ2 to medium µ1. From equation (ii), we get µ1 µ 2 µ1 − µ 2 + = v v1 R2
ht tp
…(iii)
From equations (ii) and (iii),we get 1 1 1 1 µ1 + = (µ 2 − µ1 ) − v u R1 R 2 or
1 1 1 µ2 1 + = − 1 − v u µ1 R1 R 2
…(iv)
If the object O is taken far away from the lens, the image is formed close to the focus. Thus, for u = ∞, v = f. ∴
1 1 µ2 1 = − 1 − f µ1 R R 2 1
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Light (Geometrical Optics)
If the refractive index of the material is µ and its placed in air, µ2 = µ and µ1 = 1, then 1 1 1 = ( µ − 1) − f R R 2 1
…(vi)
This is called Lens Maker’s formula because it tells what curvature will be needed to make a lens of desired focal length. Combining equations (iv) and (v), we get 1 1 1 + = u v f which is known as Lens formula.
rio s_ in fo
I v f v− f + = O u u− f f
na .c om
Magnifi Magnification, cation, M M ==
un iv
//w
) Solution
er si ta rio s
w
er s
w
ita
.ja
m
ar a
Example (i) Find the focal length of a biconvex lens in air if the radius of its surfaces are 60 cm and 15 cm. Refractive index of glass is 1.5. (ii) Find the focal length of a concave lens in water (µ = 4/3) if the surface have radii equal to 40 cm and 30 cm. (µg = 1.5).
e/
FP
ht tp
s: //
t.m e/
un
//t ht tp s:
FP
iv
.m
tp
s:
(i) Consider a light ray going through the lens as shown. If strikes the convex side of 60 cm radius and concave side of 15 cm radius while coming out.
ht
4A.48
Here, R1 = + 60, R2 = –15 cm Now
1 1 µ2 1 = − 1 − f µ1 R1 R 2 1 1.5 1 = − 1 − 1 60 −15 1 5 1 = 0.5 + = 0.5 × 60 15 60
f = 24 cm.
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Light (Geometrical Optics)
4A.49
(ii) If we image a light ray going through the lens as shown.
FP
FP
na .c om
Here, R1 = –30, R2 = + 40 cm and µ1 = 4/3, µ2 = 1.5
ar a
1 1 µ2 1 = − 1 − f µ1 R1 R 2
960 = −137.1 cm.. 7
t.m e/
un
iv
er si ta rio s
er s
un iv
e/
f =
.m
∴
//t
9 −2 = − 1 8 120
ht tp s:
ht
tp
s:
//w
w
w
ita
.ja
m
1.5 1 1 = − 1 − 4 −30 +40 3
rio s_ in fo
Using the formula,
ht tp
s: //
Example A point source S is placed at a distance of 15 cm from a converging lens of focal length 10 cm on its principal axis. Where should a diverging mirror of focal length 12 cm be placed, so that a real image is formed on the source itself?
) Solution
O
6
FP
/
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0
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Light (Geometrical Optics)
Equation for the lens is 1 1 1 + = u v f
Here, u = 15 cm and f = + 10 cm 1 1 1 ∴ + = 15 v 10
in
rio s_
ita
.ja m
ar
an a.
fo
co
m
1 1 1 2 = − = v 10 15 60 or v = 30 cm The positive sign of v shows that the image I1 is formed to the right of the lens in the figure. The diverging mirror is to be placed to the right in such a way that the light rays fall on the mirror perpendicularly. Then only the rays will retrace their path and form the final image on the object. Thus, the image I1 formed by the lens should be at the centre of curvature of the mirror. We have, LI1 = 30 cm. MI1 = R = 2f = 24 cm Hence, LM = LI – MI1 = 6 cm Thus, the mirror should be placed 6 cm to the right of the lens. or
s
ita rio
ni ve rs
e/ u
tp s:
//w
w
ni ve rs
w
Example A convex lens in air produces a real image having the same size as object. When the object and the lens are immersed in a liquid, the real image formed is enlarged two times the object size. Find the refractive index of the liquid (Take µg = 1.5). )
m
) Solution Combining the lens equation and Lens Maker’s formula, we have
e/ u
tp s:
//t .
1 1 1 µ2 1 + = − 1 − u v µ1 R1 R 2
t.m
v =1 u v=u 1 1 1.52 1 1 + = − 1 − u u 1 R1 R 2
ht
ht
∴
tp s
⇒
m=
://
In air,
v =2 u v = 2u 1 1 1.5 1 1 + = − 1 − u 2u µ1 R1 R 2
...(i)
ht
∴
...(ii)
⇒
In liquid, m =
From equations (i) and (ii), we get 1 1 + u u = 1.5 − 1 1 1 1.5 + − 1 u 2u µ 1
4A.50
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Light (Geometrical Optics)
2×
or or
4A.51
0.5µ1 2 = 3 1.5 − µ1
µ1 =
12 11
Hence, refractive index of liquid is
12 11
Example
rio s_ in fo O
tp
e/
s:
un iv
//w
6
3
0
er si ta rio s
w
w
.ja
6
ita
O
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A converging lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axis coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens as shown in the figure. It is found that the final beam comes out parallel to the principal axis. Find the separation of the mirror and the lens.
t.m e/ s: //
1 1 1 = − v f u
ht tp
or
1 1 1 + = u v f
un
ht tp s:
We have,
iv
.m
Here, u = 12 cm and f = 15 cm.
//t
ht
) Solution Let us first locate the image of S formed by the lens.
1 1 −1 − = 15 12 60 or v = –60 cm The negative sign shows that the image is formed to the left of the lens as suggested in the figure. The image I1 acts as the source for the mirror. The mirror forms an image I2 of the source I1. This image I2 then acts as the source for the lens and the final beam comes out parallel to the principal axis. Clearly, I2 must be at the focus of the lens. We have I1I2 = I1P + PI2 =
= 60 cm + 15 cm = 75 cm. Let, the distance of mirror from I2 is x cm. Then for the reflection from the mirror, u = MI1 = – (75 + x) cm, v = – x and f = –20 cm. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
1 1 1 + = u v f
From the relation,
1 1 1 + = x 75 + x 20 75 + 2 x 1 = (75 + x) x 20
or
x2 + 35x –1500 = 0
or
−35 ± (35) 2 + 4 × 1500 2 ∴ x = 25 or –60 As the negative sign has no physical meaning, only positive sign should be taken. Taking x = 25, the separation between the lens and the mirror = (15 + 25) = 40 cm x=
an
in
a.
fo
co
m
or
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rio
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er si
ta
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m
rio
ar
s_
Example A convex lens mounted on an optical bench forms an image of an object on a screen, the magnification being 2.5. The object and the screen are kept fixed and the lens is moved through a distance of 10 cm, when a sharp image is again formed on the screen, the magnification now being 0.4. What is the focal length of the lens?
ta er iv
un
s:
//t
f u1 − f + 10
tp
0.4 =
e/
tp
ht
and
…(i)
.m
u −f 1 = 1 2.5 f
∴
s:
//t
.m
tp
e/
According to the given condition, f m= = 2.5 u1 − f
si
)
s:
un
) Solution Given: m = 2.5 and m′ = 0.4
ht
5 u1 − f + 10 = 2 f
ht
5 2 10 − = 2 5 f
or
12 10 = 10 f
or
∴
…(ii)
or
4A.52
f = 4.76 cm.
Example Photographs of the ground are taken from an aircraft flying at an altitude of 2000 metres, by a camera with a lens of focal length 50 cm. The size of the film https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.53
in the camera is 18 cm × 18 cm. What area of the ground can be photographed by this camera at any one time?
) Solution Since, the object (ground) is at a very large distance from the lens,
therefore, the image of the ground can be assumed to be formed at the focal length distance ‘f ’. Hence for u = 2000 m v = 0.5 m Let the length of ground photographed be x.
x=
2000 × 0.18 = 720 m. 0.5
m
ar a
or
na .c om
0.18 0.5 = x 2000
or
w
er s
w
ita
.ja
Y I I P
un iv
//w
X P
ht tp s:
un
//t
iv
.m
e/
s: tp ht
rio s_ in fo
I v = O u
er si ta rio s
m=
Now,
t.m e/
∴ Area of ground photographed = 720 m × 720 m
s: //
4.10 CONJUGATE POINTS AND PLANES
ht tp
A consequence of the reversibility of light rays is that the positions of a real object and its real image are interchangeable. Any pair of such real object point and its real image point on principal axis are called conjugate points. The planes passing through these points and perpendicular to the axis are called conjugate planes.
Power of a Lens The reciprocal of focal length expressed in metres is known as the power of lens. It is expressed in Dioptres. I P = (m −1 ) f
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4A.54
Light (Geometrical Optics)
4.11 THIN LENSES IN CONTACT
2
2
/
/
3
3
,
u = object distance for the first lens v = final image distance for the second lens. v1 = image distance of the first image O1 for the first lens. As the lenses are assumed to be thin, v1 is also object distance to the second lens. 1 1 1 Then, − = …(i) u v1 f1
//w
1 1 1 1 − = + v u f1 f 2
e/
s:
un iv
From equations (i) and (ii), we get
…(ii)
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ita
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1 1 1 − = v v1 f 2
w
And,
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Let
iv
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s: //
t.m e/
ht tp s:
un
//t
ht
.m
tp
If the combination is replaced by the single lens, then 1 1 1 − = v u F 1 1 1 From equations (i) and (ii), we get = + F f1 f 2 1 1 1 1 In general, = + + + ... F f1 f 2 f 3
Two Thin Lenses Separated by a Distance G $
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1 1 1 d = + − F f1 f 2 f1 f 2 https://t.me/universitarios_infohttps://www.jamarana.com
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…(iii)
Light (Geometrical Optics)
4A.55
If we place a mirror at the end of a group of lenses, the combination behaves as a mirror. Any light ray entering the combination passes through each lens twice before reflection and the after reflection from mirror. The equivalent focal length of the combination behaving as a mirror is given by: 1 1 1 1 1 = 2 + + + ... + F f f f 2 3 1 fm where, f1, f2, f3, ... are the focal lengths of the lenses and fm is the focal length of mirror.
Displacement Method
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Consider a real object and a screen fixed at a distance D apart. A convex lens is placed between them. Real images are obtained on the screen for two positions of the lens. One at a distance x1 and the other at a distance x2 from the object. 6FUHHQ
6
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)
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ht
/ /
.m
tp
e/
s:
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'±[
ht tp
s: //
)
[
'±[D
'±[
7
Two object distances ((x1 and x2) are roots of the equation 1 1 1 + = x D−x f or
x2 – Dx + Df = 0
or
x=
∴
x1 =
D ± D 2 − 4Df 2 D − D 2 − 4Df D + D 2 − 4Df and x2 = 2 2
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Light (Geometrical Optics)
Under the condition D > 4f the displacement of lens is L = x2 – x1 D 2 − L2 4D The above formula is frequently used to calculate the focal length of lens by measuring D and L. L = D 2 − 4Df
Also i.e.,
f =
D−L D+L + =D 2 2 x1 = D – x2 and x2 = D – x1. x1 + x2 =
Hence, object and image distances are interchanged in the two positions of the lens. Let the magnifications be m1 and m2, then
⇒
m1m2 =
⇒
I1 I 2 ⋅ =1 O O
na .c om
m1 =
ar a
(D − x1 ) (D − x2 ) =1 x1 x2
rio s_ in fo
D − x1 D − x2 and m2 = x1 x2
⇒
ita
.ja
m
⇒ Object size = I1I 2
//t
iv
er si ta rio s
e/
.m
tp
s:
un iv
//w
w
er s
w
Example A thin equiconvex lens is placed on a horizontal plane mirror and a pin held 20 cm above the lens coincides in position with its one image. The space between the lens and mirror is filled with water (µ = 4/3) and then to coincide with its own image as before, the pin has to be raised until its distance from the lens is 27.5. Find the radius of curvature of lens.
ht
un
) Solution In the first case, we get the focal length of the convex lens
ht tp s:
s: //
t.m e/
∴ f1 = 20 cm In the second case, we get the focal length F of the combination of the convex lens and water plano-convex lens.
ht tp
4A.56
)
I FP
∴ F = 27.5 cm Let f2 = focal length of water lens ∴
1 1 1 = + F f1 f 2
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Light (Geometrical Optics)
4A.57
1 1 1 = + 27.5 20 f 2 −220 ∴ f2 = cm 3 For the liquid lens, 1 1 µ2 1 = − 1 − f 2 µ1 R1 R 2 −3 4 1 1 = −1 − (µ1 = 1) or 220 3 R1 ∞ − 220 ∴ R = cm (R1 = R) 9 ∴ Radius of curvature of the convex lens, 220 R= cm. 9 [When the image and object coincide it is known as auto-collimation].
rio s_ in fo
ar a
na .c om
or
ht tp
s: //
t.m e/
ht tp s:
un
iv
.m
e/
1 1 1 = ( a µ g − 1) − f R1 R 2
//t
ht
tp
s:
) Solution We know,
er si ta rio s
un iv
//w
w
er s
w
ita
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m
Example An equiconvex lens of refractive index 1.5, focal length 10 cm is held with its axis vertical and its lower surface immersed in water. The upper surface being in air. At what distance from the lens will a vertical beam of parallel light incident on the lens be focused? Refractive index of water is 4/3.
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:DWHU 7
O
Here, R2 = –R1 ∴
1 2 = 0.5 × 10 R
or R = 10 cm.
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Light (Geometrical Optics)
Now for air-glass interface µ 2 µ1 µ 2 − µ1 + = v u R 1.5 1 1.5 − 1 + = v ∞ 10 ∴ v = 30 cm. For glass-water interface u′ = – v = –30 cm, and R = –10 cm. µ1 µ 2 µ 2 − µ1 Now, + = u v' R 2 4 2 4 − 3+ 3 = 3 3 or v ' −30 −10 ∴ v′ = 20 cm
rio s_ in fo
na .c om
∴
er si ta rio s
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
Example The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in the figure. (i) Where should a pin be placed on the optic axis such that its image is formed at the same place? (ii) The concave part is filled with water of refractive index 4/3. Find the distance through which the pin should be moved, so, that the image of the pin coincides with the pin.
ht
4A.58
) Solution (i)
1 1 µg 1 = − 1 − f g µa R1 R 2 1 1 1 = (1.5 − 1) − = 20 60 60 Now
1 1 1 1 = + + F f g fm f g =
=
1 1 1 + + 60 R 2/2 60
2 2 1 1 1+ 3 4 + + + = = 60 20 30 10 30 30
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Light (Geometrical Optics)
∴
F = 7.5 cm
But F =
4A.59
u , since, at u object and image coincide. 2
1 4 1 1 − − 1 × = f w 3 60 180
fw = 180 cm. = Focal length of combination, F′ is given by
co
∴
m
or
1 1 1 µw = − 1 − fw µa 60 ∞
Now
∴ u = 2F = 2 × 7.5 = 1.5 cm (ii) Here, fg = 60 cm, fm = 10 cm
s_
an
rio
ar
2 1 2 + + fw fm f g
1 1 1 + + 90 10 30
=
1 + 9 + 3 13 = 90 90
s si er iv
un
//t s:
io ta r
iv
un
.m
tp ht
90 13
But,
F′=
u′ , since, at u′ object and image coincide. 2
∴
u′ = 2 F ′ =
.m
//t
tp
s:
tp
e/
F′=
∴
ht
ta
=
si
2 1 2 + + 180 10 60
er
=
e/
s:
//w
w w
.ja
m
=
in
a.
fo
1 1 1 1 1 1 = + + + + F′ f w f g f m f g f w
180 cm = 13.85 cm 13
ht
Therefore, the pin has to be moved towards the lens by (15 – 13.85) cm = 1.15 cm
Example A thin equiconvex lens of glass of refractive index µ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water (µ = 4/3). On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in the figure. The separation between lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics) P
P
na .c om
0LUURU
) Solution The rays coming from the object will suffer refraction at air-glass
un iv
//w
w
µa µ g µ g − µa + = u v′ R
.m
iv
P:
u
u
ht tp
2
u u u u u u
Y
t.m e/
PD
u u u u u u PuJ u u u u u u
s: //
ht tp s:
//t
Y P
…(i)
un
tp
e/
s:
For refraction of glass-water surface,
er si ta rio s
er s
w
ita
.ja
m
ar a
rio s_ in fo
surface. It may be supposed to from a virtual image in the direction of refracted light. This image acts as an object for the subsequent refraction at glass-water surface to form the image. If this image is formed beyond the mirror, the actual image will lie in front of mirror at a distance identical to that of image behind the mirror. For refraction at air-glass surface,
ht
4A.60
O
u u u u u
O
O
µ g µw µw − µ g + = −v′ v −R Adding equations (i) and (ii), we get
…(ii)
µa µw µ g − µa µw − µ g + = − u v R R
…(iii)
The radius of curvature R of the surface of equiconvex lenses may be determined by the formula. 1 1 1 = µg −1 − f R1 R 2
(
)
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Light (Geometrical Optics)
4A.61
Here, Rg = –R1 = R 1 2 = (1.5 − 1) 0.3 R
∴
or R = 0.5 × 2 × 0.3 = 0.3 m Substituting this value of R along with other data in equation (iii), we get 3 4 3 −1 − 1 4/3 2 + = − 3 2 0.9 v 0.3 0.3 m 1 1 4 − = 0.6 0.8 1.8
or
4/3 4 1 2 = − = v 1.8 0.9 1.8
or
v=
rio s_ in fo
na .c om
=
m
ar a
4 1.8 × = 1.2 m 3 2
e/
tp
s:
Example
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
The distance v = 1.2 m lies 0.4 m behind the mirror. Hence, the image is formed 0.4 m in front of the mirror. This point is also 0.4 m on the right of the convex lens.
iv
t.m e/
ht tp s:
un
//t
ht
.m
A converging lens of 20 cm focal length is arranged coaxially with a diverging lens of focal length 8 cm. A point object lies on the same side as the converging lens and very far away on the axis.
) Solution
ht tp
s: //
(i) What is the smallest possible distance between the lenses if the combination is to form a real image? (ii) If the lenses are placed 6 cm apart, what is the position and nature of the final image of the object? (i)
I ±FP
I FP
) [ I±[ I
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Light (Geometrical Optics)
For convex lens: Since, u1 = ∞, the rays would converge toward its focus F1, but the concave lens diverges and produces the final image. For concave lens: The point F1 would act as the virtual object for concave lens u2 = – (f1 – x) = – (20 – x)
1 1 1 1 1 −(20 − x) + 8 =− + = = − v2 f 2 u2 8 (20 − x) 8(20 − x)
∴
∴ v2 =
8(20 − x) x − 12
ta rio
s
er s
w
w
Example
s_ i
56 cm (virtual) (from concave lens) 3
.ja m
=−
nf
8(20 − x) 8(20 − 6) = x − 12 6 − 12
ita rio
∴ v2 =
ar an a
o
.c om
For the image to be real, v2 ≥ 0 Hence the minimum value of x should be 12 cm. (ii) Given: x = 6 cm
er si
ht
tp s:
//t
ht
.m e/
tp s:
//t
un iv
tp
.m e/
s:
un iv
//w
A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and image again coincide at a distance 25 cm from the lens. Calculate the refractive index of liquid.
ht
4A.62
) Solution Using Lens Maker’s equation 1 1 1 − = ( µ − 1) R R f 2 1 For the bi-convex lens µ = µg, R1 = R and R2 = –R
(
)
1 1 1 2 µg −1 = µg −1 − = R f1 R ( −R)
(
)
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Light (Geometrical Optics)
4A.63
For the plano concave liquid lens µ = µw, R = –R and R2 = ∞ 1 1 1 − ( µ w − 1) = ( µ w − 1) − = f2 R ( − R) ∞
∴
The focal length of the combination is
=
(
)
2 µ g − 1 − ( µ w − 1) R
...(i)
1 1 1 = + f f1 f 2
s_
ita rio
f = 15 cm
s ita rio ve rs
e/ un i
1 1 1 = + f 15 ∞
t.m
or
ht tp
s:
we get
//w
w
w
1 1 1 = + f u v
ve rs
.ja
m
ar an
in f
a.
o
co
m
With the above combination of lenses, it is given that the object and its image coincide with each other. This is possible when the light starting from the object point and after passing through the combination of lenses emerges perpendicular to the mirror so that the light retraces its path and reach the same starting material. Hence, in such a case we will have u = 15 cm and v = ∞ Thus, from the lens equation
e/ u
s: //
ni
Using f = 15 cm and the values of µg (= 1.5) and µw = 4/3 in equation (i), we get
.m
ht tp
R = f [2 (µg – 1) – (µw – 1)]
tp s:
//t
4 = 15 [2 (1.5 − 1) − ( − 1)] 3
∴
or
ht
2 = (15 × ) cm = 10 cm. 3 For the second case R = f[2 (µg – 1) – (µ1 – 1)] 10 = 25 [2 (1.5 – 1) – (µ1 – 1)]
µ1 = 2 −
10 40 = = 1.6 25 25
Example An object is kept at a distance of 100 cm from a screen. A convex lens is place between them produces a real magnified image on the screen. If the lens is shifted 30 cm towards the screen, real image is again obtained. Find the focal length of the lens. Calculate the size of the object, the image size are 16 mm and 9 mm respectively. https://t.me/universitarios_infohttps://www.jamarana.com
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4A.64
Light (Geometrical Optics)
) Solution The situation is based on the displacement method. Here, D = 100 cm, L = 30 cm Now, f =
D 2 − L2 1002 − 302 91 = = cm. 4D 4 × 100 4
We must verify that, D = 100 cm > 4f Image sizes are I1 = 16 mm and I2 = 9 mm. Object sizes = I1I 2 = 16 × 9 = 12mm
er si ta rio s
er s
iv un
//t
ht
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s:
//w
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m
Simple Magnifier
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4.12 OPTICAL INSTRUMENTS
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Object sizes = I1I 2 = 16 × 9 = 12mm
)
2EVHUYHU
t.m e/
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)
s: //
FP
ht tp
A simple magnifier (or magnifying glass) consists of a convex lens to be placed before an object that is to be examined. The object is adjusted between the focus and lens, so, that a virtual magnified image is seen by eye. For most distinct vision, the image is usually adjusted to be at a distance of 25 cm. Let, Magnification obtained = M ∴
M=
25 v− f v 25 = − 1 = − + 1 ≈ − f f f f
Thus, the image size is almost
25 times the object size. f
Microscope A compound microscope basically consisting of two converging lenses, the objective and the eye-piece, mounted on a common axis, in two coaxial tubes, one telescoping into another. The distance between the lenses can be altered at will. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.65
( %
2
Y
$ IR
Y
IH
$
IH
$
X %
X
%
rio s_ in fo
m
Angle subtended by the final image(at eye) Angle subtenede by the object at least distance of distinct vision from the eye
ita
.ja
=
ar a
na .c om
AB is the object held slightly beyond the principal focus of the objective. A′B′ is the real image produced by the objective, which serves as the real object for the eye-piece. A ′B′ lies within the focal length of the eye-piece. A “B” is the final image (virtual with respect to A ′B′) produced by the eye-piece. The final image is inverted with respect to the original object AB. The eye-piece works as a simple microscope. Magnifying power of a compound Microscope
er si ta rio s
e/
s:
un iv
//w
w
er s
w
= Magnification by the objective × magnifying power of the eye-piece v v = 1 1 + 2 u1 f2
iv
s: //
Special Cases
t.m e/
ht tp s:
un
//t
ht
.m
tp
where, u1 = Distance of the object from the objective. v1 = Image distance of the image produced by the objective. v2 = Distance of the final image from the eye-piece. f2 = Focal length of the eye-piece.
ht tp
(i) The final image at the distance of distinct vision from the eye (eye-piece) v2 = D, the distance of distinct vision from the eye.
∴ Magnifying Power =
v1 D 1 + u1 f2
(ii) Final distance at infinite distance from the eye-piece, D >>> f2 Magnifying Power =
v1 D × u1 f 2
(iii) To get maximum magnification, the object is very nearly at the focal point of the objective.
∴ u1 = f1 Magnification Power =
v1 D × for D >>> f 2 u1 f 2
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Light (Geometrical Optics)
Telescope It is similar to microscope with the difference that in microscope the objective has smaller focal length and aperture, while in the telescope the eye-piece has smaller focal length and aperture. X
Y $
D
D
$
E
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%
Y
rio s_ in fo
%
m
ar a
The object is at infinite distance for an astronomical telescope. The image A′ B′ produced by the objective is in the focal plane. The distance of eye-piece is so adjusted that the final image is
er si ta rio s
w
er s
w
ita
.ja
(i) at the distance of distinct vision, or (ii) at infinity.
e/
iv
.m
un
t.m e/
ht tp s:
∴
A′B′ u v β Magnifying Power = = 2 = 1 A ′ B ′ α u2 v1
//t
tp
s:
un iv
//w
Magnifying power of the telescope is defined as the ratio of the visual subtended by the image at the eye to that subtended by the object, directly at the eye.
ht ∴
s: //
For astronomical telescope, v1 = f1, (the focal length of the objective). If the final image is also at infinity, u2 = f2 (the focal length of the eye-piece). Magnifying Power =
ht tp
4A.66
f1 f2
If the telescope is used to view nearer objects as in the case of field glasses, A′B′ A′B′ α= ,β= v1 u2 ∴
Magnifying Power =
Now, v1 = ∴
v1 u2
u1 f1 v f and u2 = 2 2 u1 − f1 v2 − f 2
u f v − f2 Magnifying power = 1 1 2 u1 − f1 v2 f 2
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Light (Geometrical Optics)
4A.67
Special Cases (i) Final image of at distance of distinct vision v2 = D u f D − f2 ∴ Magnifying power = 1 1 u1 − f1 Df 2 (ii) Final image at ∞, v2 = ∞ u f Magnifying power = 1 1 u1 − f1
f2 v2 u1 f1 1 × f 2 u1 − f1 f 2
1−
f1 D − f 2 f 2 D
ar a
Magnifying power =
ita
.ja
m
4.13 CHROMATIC ABERRATION
rio s_ in fo
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(iii) Object at infinity, final image at distinct vision. u1 = ∞, v2 = D
er si ta rio s
iv
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
un iv
//w
w
er s
w
The refractive index of glass is different for different wavelengths (colours). Thus, when a white light ray strikes a lens, smaller wavelengths get refracted more than the larger wavelengths.
Voilet V oilet
fv
Red Fv
Fr
fr
In the above figure, violet rays are focused closer to a converging lens than are red rays. This property of a lens to converge of different wavelengths to different foci is known as chromatic aberration. As
µred < µviolet
∴
fred > fviolet
The colour spread on the axis of lens = fr – fv ≈ wfm where, fm is the mean focal length (usually for yellow colour).
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Light (Geometrical Optics)
Achromatic Combination of Lenses Two lenses-one converging and one diverging can be combined in such a way that the combined lens will not have chromatic aberration. If w1, w2 are dispersive powers of lenses and f1, f2 are their focal lengths, chromatic aberration will be eliminated if w1 w2 + =0 f1 f2 Such a combination is known as an achromatic combination of lenses. The equivalent focal length (F) can be calculated by
Example
rio s_ in fo
na .c om
1 1 1 = + F f1 f 2
er s
w
ita
.ja
m
ar a
A microscope of an objective lens of focal length 0.9 cm placed 13 cm from an eye-piece of 5 cm focal length. An insect .05 cm long is placed 1 cm from the objective. Find the nature, position and the size of the final image.
er si ta rio s
un iv
u1 f1 1× 0.9 = 9 cm.. = u1 − f1 1 − 0.9
.m
Objective
O
t.m e/
ht tp s:
V1
u2
ht tp
s: //
1 cm l2
Eye-piece
un
//t
iv
v1 =
tp
s:
∴
e/
//w
w
) Solution For the objective: u1 = 1 cm, f1 = 0.9 cm
ht
4A.68
l1 13 cm
For eye-piece: Distance between image of object and eye-piece, u2 = 13 – 9 = 4 cm, f2 = 5 cm v2 =
u2 f 2 4×5 = = − 20 cm u2 − f 2 4−5
Size of image =
v1 v 9 9 × size of object =1 × size × 0.05 of =object 0.45 cm. = × 0.05 = 0.45 cm. u1 u1 1 1
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Light (Geometrical Optics)
4A.69
v2 −20 × (size of first image) = × 0.45 = −2.25 cm u2 4
Size of final image =
v2 −20 × (size of first image) = × 0.45 = −2.25 cm u2 4 Hence, the final image is virtual at a distance of 20 cm from eye-piece and 2.25 cm long.
na .c om
1. A man stands 2 m in front of a concave spherical mirror. He finds that his image is one and a half times greater than that in a flat mirror placed at the same distance from the face. Find the radius of the concave mirror.
ita
When h < < 2a, α and β are small.
ht tp
H 3 h = a + b 2 2a
∴
v=b v=b
⇒
u=a
And for small angles, tan α = α, tan β = β
H 3 ( a + b) = h 4a
I I
u=a
er si ta rio s
un
t.m e/
β
O
s: //
β
β 3 = α 2 (i.e., the image of the man on the concave 3 mirror is times greater than that on the flat 2 mirror.) h H tan α = and tan β = ∴ 2a a+b Given that,
iv
.m //t ht tp s:
hO
…(ii)
er s
un iv
e/
s:
tp
ht h
H b = h a
m
.ja
w
//w
)
w
) Solution When the man (object) sees his own image by two different mirrors, the ratio of the sizes equals the ratio of the angles subtended at the position of the object.
rio s_ in fo
1 1 1 = + …(i) f a b Now for concave mirror, the linear magnification of the mirror is For concave mirror
ar a
Solving equations (i), (ii) and (iii), we get f =
h
R = 2f = 6 m
∴
h I
α a a
a a
I
2. A prism of flint glass with an angle of refraction 30º is placed inside water. (i) At what angle should a ray of light fall on the face of the prism so that
O
α
h O
h
3 3 a = ×2 = 3 m 2 2
WORKED OUT EXAMPLES
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4A.70
Light (Geometrical Optics)
For the extreme rays r
θ
inside the prism, the ray is perpendicular to the bisector of the angle of the prism? (ii) Through what angle with the ray turn after passing through both faces of the prism?
P 2r
α
) Solution The ray inside the prism QR is
)
m
3 2
8 3
ta
rio
ta
cos α =
s
s_
sin θ 1 2 1 = × = µ 2 3 3
rio
si
er
iv
un .m e/
2r r = 0 sin(90 + θ − α) sin α
//t
C
r 1 = 2r 2
sin α =
r0 =
s:
sin i = µ sin 15º i = 18.7º
or ∴
⇒
iv un S
s: ht tp
B
Q
In ∆ OPQ, angle ∠OQP = 90 + θ – α Applying sin e rule
//t
ht
i2
.m e/
r2
Q r1
si er
w R
3 2
in
a. m ar
⇒
w //w s: tp
D
r0
sin θ sin α
µ=
.ja
A
P
cos θ =
an
sin i = sin r
i1
⇒
µ
fo
co
Also A = 30º ∴ r = 15º. µ
sin θ =
O
Deviation, D = i1 + i2 – A = 2i – A A = r1 + r2 = 2r,
α
θ
evidently parallel to the base of the prism BC. Hence, we can conclude that the ray is undergoing minimum deviation. ∴ i1 = i2 and r1 = r2 But
θ
sin α sin α ⋅ 2r = 2r cos (θ − α) cos θ cos α + sin θ sin α
ht tp
sin α sin α ⋅ 2r = 2r cos (θ − α) cos cos sin sin θ α + θ α 3. A vertical beam of cross-sectional radius 1 r is incident symmetrically on the curved × 2r 2 surface of a glass hemisphere (µ = 3/2) 3 = = 2r of radius 2r placed with its base on a 3 8 1 1 2 6 +1 × + × horizontal table. Find the radius of the 3 2 3 2 luminous spot formed on the table. = 0.678r r0 =
)
The path of beam within the hemisphere is as shown. Let, θ = angle of incidence α = angle of refraction
4. A ray of light parallel to the axis of a concave mirror is incident at an angle of 30º. This ray after reflection meets the axis at I. Find PI, where P is the pole of the mirror. Focal length of the mirror is 10 cm.
) Solution
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Light (Geometrical Optics)
) Solution
1
3 2 η (y) ky 2 + 1 , where k = 1.0 (metre)–3/2 The refractive index of air is 1.0 (i) Obtain a relation between the slop of the trajectory of the ray at a point B (x, y) in the medium and the incident angle at that point.
)
For the mirror R = 2f = 20 cm ∴ AC = 20 cm
A 30° 30°
P
30°
60°
I
F
Y
C
P(r1, y1)
AIR
B(x, y)
m
K
3
s_ rio
s
rio
ta
si
er
e/ s:
= 10(2 − 3 +
1 3
)=
)
.m
//t
Y
10 3
B 1m
(2 3 − 2)
i
(r, y)
= 10(2 − 3) +
10
ht tp
From equations (i) and (ii), we get PI = PK + KI
i + θ = 90º
Slop, m = tan θ = tan (90 – i) = cot i
tp
ht
…(ii)
= 20 − 10 3 = 10(2 − 3)
un iv
s:
(i) From the figure,
PK = PC – KC
Now,
ta un
.m e/
) Solution
//t
ht
KC = 10 3
∴
…(i)
3 KC KC cos 30° = = AC 20
X
(ii) Obtain an equation for the trajectory y(x) of the ray in the medium. (iii) Determine the co-ordinate (x1, y1) of the point P, where the ray intersects the upper surface of the slab-air boundary. (iv) Indicate the path of the ray subsequently.
si
er
10
KI =
In ∆AKC,
iv
tp s: //w
w
w .ja
AK 10 = KI KI
tan 60° =
In ∆AKI
AIR
A(0, 0)
m
∴
Medium
in
a.
fo
co
t = 1.0m
ar an
AK AK = AC 20 AK = 10 cm.
In ∆ AKC, sin 30º =
∴
4A.71
3
)=
10
(2 3 − 2) 3 = 8.45 cm
5. A ray of light travelling in air is incident at grazing angle (incident angle = 90º) on a long rectangular slabs of a transparent medium of thickness t = 1.0 cm. (seen the figure). The point if incidence is the origin A (0, 0). The medium has a variable index of refraction η (y) given by
O
(ii) As the light passes through different medium, hence by Snell’s law sin α1 η2 = sin α 2 η1
1
∴ η1 sin α1 = η2 sin α2 = η3 sin α3 =...
3
= 10(2 − 3 +
10
3) +
i.e., at any point in the medium with variable refractive index η (y) sin λy = C = Constant
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4A.72
Light (Geometrical Optics)
where, η(y) = refractive index at any point y. αy = angle of incidence at any point y. At y = 0, Refractive index, η1 = [k ⋅ (0) + 1]1/ 2 = 1
P
mR
R
)
) Solution 3
α3
α2
2 1
α1
(iii) At y = 1 m x = 4 m
or
1.5 0.5 = R (1.5m + 1) R (1.5 m + 1) 0.5 = 1.5 2 4 m= = 1.5 3
7. The x – y plane is the boundary between two transparent media. Medium 1 with z ≥ 0 has a refractive index 2 and medium 2 with z ≤ 0 has a refractive index 3 . A ray of light in medium 1 given by
6. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance of mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure.
or
or
(iv) Let αe be the angle of emergence in the air. Then by Snell’s Law ηe sin αe = η1 sin α1 = 1 where, ηe = refractive index of air = 1 ∴ sin αe = 1 or αe = 90º Thus, it emerges parallel to the incident ray.
x 256
ht
y=
4
(∵k = 1)
∴
ht
∴ 4y1/4 = x
tp s
1 1 1.5 − 1.0 + = ∞ mR v or v = – 1.5 mR Since for the refraction at the curved surface, the image I acts as an object for this refraction, we will have µ1 = 1.5, µ2 = 1.0, u = 1.5 mR + R v = ∞, R = – R 1.5 1.0 1.0 − 1.5 + = ∴ −R 1.5 mR + R ∞ ∴
At x = 0, y = 0, Therefore c = 0
y +1/ 4 = kx+c (+1/4)
For the refraction at the plane surface, we have µ1 = 1.0, µ2 = 1.5, u = mR, R = ∞
dy = k y 3/ 4 dx
:// w
µ1 µ 2 µ 2 − µ1 + = u v R
2
∴ y −3/ 4 dy = k dx ⋅
The refraction formula is
∴ m=
R
V
or ky + 1 = 1 + m 3/2
u
P
I
w tp w s: .ja //t m .m ht ar tp e an /u s: a. ni //t co v .m e m rs e/ i un ta rio iv er s_ si in ta fo rio s
since, α1 = 90º, therefore, C = 1 × 1 = 1 ∴ η (y) = sin αy = 1 Substituting the values, we get 1 [ky 3/ 2 + 1]1/ 2 ⋅ =1 1 + m2
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4A.73
Light (Geometrical Optics)
Since, the refracted ray lies in the plane of incident ray (the plane containing the vector A) and the normal to the incidence plane, the refracted vector may be express as A r = ai + b j + ck
the vector A = 6 3i + 8 3j − 10 k is incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium 2.
) Solution )
with the value of a and b are equal to the corresponding values in the incident vector, i.e., A r = 6 3 i + 8 3 j + ck The value of c can be determined from the fact that A r ⋅ (−k ) cos r = | Ar |
z
si
(6 3) 2 + (8 3) 2 + c 2
1 2
−c
o
2
=
ta rio s_
ar a
w
2
−c
=
.ja m
1
1
(6 3) 2 + (8 3) 2 + c 2
in f
Thus,
na .
r
x
θ
co m
i
y
1 2
=
−c 1
2 2 300 + c
−c
=
1
2 2 300 + c
ve r
tp
)
ht
s:
//t
ht
.m
tp
e/
s:
un i
//t
ht
.m
tp
e/
si
un i
ta rio s
ve r
w
//w
s:
The given diagram in the figure. c = ± 10 3 The incident ray represented by the vector A and the refracted wave represented by Ar shown on the plane (x – y) of incidence (z = 0). Selecting, c = − 10 3 (as the refracted rays The cosine of the angle between the vector proceeds downward like the incident ray) A and the positive z direction is We get A r = 6 3i + 8 3 j − 10 3k A⋅k cos θ = |A| The unit vector of Ar is −10 (6 3 i + 8 3 j − 10k ) ⋅ k 1 =− = =− 1 A 6 3 i + 8 3 j − 10 3 k 3 i + 4 j − 5k r 20 2 = = 1 (6 3) 2 + (8 3) 2 + (−10) 2 2 | Ar | 5 2 (6 3) 2 + (8 3) 2 + (10 3) 2 2 −10 (6 3 i + 8 3 j − 10k ) ⋅ k 1 =− = =− 1 A 6 3 i + 8 3 j − 10 3 k 3 i + 4 j − 5k r 20 2 = = 1 (6 3) 2 + (8 3) 2 + (−10) 2 2 | Ar | 5 2 (6 3) 2 + (8 3) 2 + (10 3) 2 2 ∴ θ = 120º 8. When a telescope is in normal adjustAngle of incidence, ∠i = 180º – 120º ment, the magnification produced is 30. = 60º. The eye-piece has a focal length of 5 cm. From Snell’s Law of refraction, Find the magnification if the final image is to be formed at the distance of distinct sin i µ 2 3 = = vision. What will be the length of the telesin r µ1 2 scope in that case? 2 2 1 = (sin 60°) = Hence, sin r = (sin i ) ) Solution In normal adjustment, final image is 3 3 2 f formed at infinity and magnification is o 1 2 fe Also, cos r = 1 − sin r = 2 https://t.me/universitarios_infohttps://www.jamarana.com
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4A.74
fo = 150 cm.
or
M=
) Solution
fo = 30 fe
∴
)
Light (Geometrical Optics)
v1
A'
For the final image at distinct vision (25 cm), the magnification is
A v1=D
25 + f o 25 + 5 M = + fo = 150 = +36 25 × 5 25 f e
B'
Length of telescope = fo + object distance for eye-piece ve f e ve − f e
B
u1
The object is within the focal distance
m
β=
s_
Hence, the magnification will be 36 and the length of telescope = 154.17 cm.
rio
Magnifying power =
β D 25 = 1+ = 1+ =6 α f 5
er
//t
11. A telescope has an objective of focal length 50 cm and an eye-piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 200 cm away from the objective. Calculate: (i) The separation between the objective and the eye-piece. (ii) The magnification produced.
10. A convex lens of focal length f is used in simple microscope. Find its magnifying power if the final image is at the distance of distinct vision. Find also the magnification and the distance of the object f = 5 cm, D = 25 cm.
25 × 6 =6 25
12 25 31 31 = 10.33 1+ = 2× = 6 6 6 3
ht
M=
125 25 = cm. 30 6
Magnification =
∴
s:
=
v1 f −25 × 5 = v1 − f −25 − 5
.m
e/ un
iv
Also, u1 =
tp
where, u = 6 cm, fo = 4 cm, fe = 6 cm, D = 256 uf o 6× 4 = = 12 cm ∴ v= 2 u − fo
si ta
un
e/
m
//t .
s:
tp
v D 1 + u fe
ht
M=
rio
iv
w
://
tp s
ht
)
produced by the objective × angular magnification produced by the eye-piece. ∴
s
er
w w
si ta
.ja
9. The focal lengths of the objective and the eye-piece of a compound microscope are power = β = 1 + D = 1 + 25 = 6 Magnifying 4 cm and 6 cm respectively. If an object α f 5 is placed at a distance of 6 cm in front of A′B′ v1 25 the objective, what is the magnification = = Magnification = AB u1 41 produced by the microscope?
m
ar
∴
) Solution Magnification = magnification
A′B′ AB AB = ; α= v′ u1 D
in
an
a.
co
(−25)(5) 25 = 150 + = 150 + = 154.17 cm. −25 − 5 6
fo
= fo +
β α
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Light (Geometrical Optics)
) Solution
)
)
) Solution I
E
fe
B'
Object and image will coincide if the rays strike the mirror normally and come back along the same path. Hence, the ray must diverges away from the centre of the curvature C of the mirror. The concave lens must produce a virtual image at C for the lens:
B''
rio
f =
rs
ita
uv 12 × (−7.5) = = − 20 cm. u+v 12 − 7.5 = –20 cm.
un
iv e
uv 12 × (−7.5) = = − 20 cm. u+v 12 − 7.5
13. An equiconvex lens of focal length 30 cm in air is kept stationary on the surface of water (µ = 4/3). A point object is placed at a height 90 cm above the centre of the lens in air. If the refractive index of glass is 1.5, find the position of the final image.
.m
e/
//t
s:
tp
v v′ 200 1 6 = × = × × 25 × =2 u u′ 3 200 25
12. An object placed at 12 cm from a concave lens coincide with its image if a concave mirror of focal length 15 cm is placed at a distance of 22.5 from the lens on the other side. Find the focal length of the lens.
)
) Solution As the medium is not same on both sides, we cannot use
1 1 1 + = u v f
ht tp s:
//t
ht
−25 × 5 25 v′ f ′ = = cm v′ − f ′ −25 − 5 6
Total magnification produced = mo × me
er e/ un
f =
.m
tp
ht
where l is the separation between the lenses. The image (final) distance, v′ = 25 cm from the eye-piece and f′ = 5 cm.
u = 12 cm v = – (30 – 22.5) = –7.5 cm.
iv
w
s: //
si ta
.ja
w w
∴
This serves as an object for the eye-piece. The distance between A′B′ and the eye-piece is 200 u′ = l − , 3
s_ i
an a
m
ar
∴ and
200 × 50 1000 200 = = cm 200 − 50 15 3
200 × 50 1000 200 = = = cm 200 − 50 15 3
u′ =
nf o
.c om
Let AB be the position of the object. Here, u = 200 cm, f = 50 cm Let the image A′ B′ formed by the objective is at a distance v from it. uf v= ∴ u− f =
C
O.I.
s
A'
fo A''
B
rio
Q
Let us first calculate the ratio of the curvature of lens. Let x = magnitude of radius of curvature of each side.
A
∴
4A.75
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4A.76
Light (Geometrical Optics)
Refraction at first surface of the lens. µ µ µ − µ1 Using the relation, 1 + 2 = 2 u v R 1 1.5 1.5 − 1 + = 90 v1 +30
O
or
1.5 1 1 3− 2 1 = − = = v1 60 90 180 180
∴
v1 = 180 × 1.5 = 270 cm.
Air
Glass
Refraction at second surface: Glass to water µ µ µ − µ1 Using the relation, 1 + 2 = 2 u v R 4 − 1.5 1.5 4/3 3 + = or −270 v2 −30
fo
in
ta
rio s ta si er
uestions
un
iv
pe
)
2h µ
2. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 or L2 having refractive indices n1 and n2
1 (d) h 1 + µ
2h µ
= x2 − x1 =
2h µ −1
rio s_
si er iv un
ht tp
ht tp
s: //t .m e/
(either plane or spherical) does not depend on the medium The image of P will be formed at a distance h below the mirror If d = depth of the liquid in the tank, then d −h apparent depth of P = x1 = µ apparent depth of the image of P d +h = x2 = µ apparent distance between P and its image
h
(b)
(c)
q
) Explanation Image formation by a mirror
O
(a) 2µh
a.
w s: //t .m e/
tp s
:// w
w
ht
SOLVED OBJECTIVE y
1. A plane mirror is placed at the bottom of a tank containing a liquid of refractive indeed µ. P is a small object at a height h above the mirror. An observer O, vertically above P, outside the liquid-sees P and its image in the mirror. The apparent distance between these two will be
P
4 −0.5 1.5 = + 3v2 −90 270
or v2 = 120 cm. Hence, the final image is formed 120 cm below the water surface.
t
x = 30 cm.
or
or
.ja
1 1 1 = (1.5 − 1) − 30 x −x
or
m ar an
1 µ 1 1 = − 1 − f 1 R1 R 2
co
m
Water
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Light (Geometrical Optics)
respectively (n2 > n1 > 1). The lens will diverge a parallel beam of light if it is filled with (a) air and placed in air (b) air and immersed in L1 (c) L1 and immersed in L2 (d) L2 and immersed in L1
B
A
θ
θ p
θ
4A.77
C
s
rio
ta
si
v=
4/3 8 = 3/2 9
( f0 + fe ) fe f0
f v image size l = e = = u f 0 object size L
f0 L = = magnification of telescope in normal adjustment l fe
= magnification of telescope in nomal adjusment 5. A ray of light passes through four transparent media with refractive indices
−18 9
iv
un or
∴
f0 1 1 1 = − = v fe f0 + fe ( f0 + fe ) fe
θ ≤ sin
or
Magnification =
n sin θ = w sin 90° ng
⇒
e/
)
internally at the point P (say). Therefore critical angle ≅ θ. Applying Snell’s Law of refraction at the critical angle of incidence (θ) we get
sin θ =
1 1 1 − = v −( f 0 + f e ) f e
ht
(d) None of these
) Explanation The light ray is just reflected
⇒
of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) = f0 + fe. Treating the line on objective as the object, and the eyepiece as the lens, µ = – (f0 + fe) and f = fe.
:// t.m
2 3
tp s
2 8 < sin θ > 3 9
L+1 L−l
er
e/
(b) sin θ <
(c)
(d)
) Explanation Let f0 and fe be the focal lengths
//t s: ht tp
8 9
(a) sin θ ≥
rio
ita rs un
iv e
A
.m
ht tp C
s_ in
an
ar
m
.ja w w θ
L −1 l
(c)
s:
//w
B
fo
a. co
m
4. In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is L L +1 (b) (a) l l
)
)
3. A glass prism of refractive index 1.56 is immersed in water (R.I. = 4/3). The beam of light incident normally on the face AB is totally reflected to reach the face BC, if
Hence (d) is the correct answer.
2 1 µ2 = − 1 − f µ1 R µ Here f is negative if 2 > 1, i.e., µ2 = n2 and µ1 µ1 = n1 Hence (d) is the correct answer.
) Explanation
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4A.78
µ1
µ4
µ3
30º at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is (a) 28 (b) 30 (c) 32 (d) 34
µ1, µ2, µ3 and µ4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have
Light (Geometrical Optics)
2√3 m
C
B
B 30°
in s_
ta
.ja
= 0.2 tan 30° =
s
rio
er
(0.2/ 3)
iv
8. A screen beaming a real image of magnification m1 formed by a convex lens is moved a distance x. The object is the moved until a new image of magnification m2 is formed on the screen. The focal length of the lens is x x (a) (b) m2 − m1 m2 − m1
e/
un
x
(d) None of these
) Explanation In first case,
⇒
1 + m1 =
q f
1 1 1 q + = and = m1 p q f p
7. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle
m1m2
)
…(ii)
f0 =m=5 fe where, m = magnification = 5 Solving equations (i) and (ii) we get 5 fe + fe = 36 ⇒ fe = 6 cm and f0 = 5fe = 30 cm
l = length of the tube = 36 cm
(c)
...(i)
.m
//t
s:
tp
ht
ta
un
e/
.m
//t
s:
tp
ht
)
where
= 30
si
iv
//w w
s:
tp
ht
2 3
=
) Explanation We know f0 + fe = 36
0.2
3 Hence required number of reflections
er
w
si
)
6. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye piece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the length fe of the eye piece respectively are (a) f0 = 45 cm and fe = –9 cm (b) f0 = 50 cm and fe = 10 cm (c) f0 = 7.2 cm and fe = 5 cm (d) f0 = 30 cm and fe = 6 cm
rio
)
m
due to a single reflection
) Explanation From Snell’s law, µ sin θ = constant ⇒ µ1 = µ2
A
) Explanation Forward shift of point of incidence
ar
(b) µ2 = µ3 (d) µ4 = µ1
(a) µ1 = µ2 (c) µ3 = µ4
an
a.
A
fo
co
m
0.2 m
In the second case,
1 1 1 q+x + = and = m2 q + x p′ f p′
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…(i)
Light (Geometrical Optics)
q+x f Solving equations (i) and (ii), we get
…(ii)
m2 =
⇒
x f
O 3h
m2 − m1 =
x f = m2 − m1
r
i
r
2h
h
9. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is
i
10. A beam of light, consisting of red, green and blue colours, is incident on a right angled prison, as shown. The refractive indices of the material of the prison for the above red, green and blue wavelength are 1.39, 1.44 and 1.47 respectively. The prison will
2h
s_
si ta .m
tp
1
) Explanation The angle of incidence of all the
5
=
rays is 45º at the hypotencese. For a critical angle of 45º, the refractive index must be,
5h
2
)
h
h + (2h) 2
)
2
=
h 2
=
2
1
(sin 45°) −1 = 2 = 1.414
2
h +h
s
s: tp ht h
) Explanation sin i =
45°
(a) separate part of the red colour from the green and blue colours. (b) separate part of the blue colour from the red and green colours. (c) separate all the three colours from one another. (d) not separate even partially any colour from the other two colour.
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2h
90°
e/
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s: //
iv
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er
e/
si ta
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sin r =
45°
er
w
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(d)
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a.
m
3 2
rio
3 2
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5 2
(b)
(c)
5 2
(a)
co m
⇒
4A.79
2
or µ =
5
5 2
1 sin i = = µ sin r
For red light, µ = 1.39 < 1.414. Hence, its critical angle > 45º Therefore, red light will pass through the surface into air.
According to Shell’s law,
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4A.80
Light (Geometrical Optics) P
60°
r1
11. A ray incident at a point at an angle of incidence of 60º enters a glass spheres of R.I. n = 3 and is reflected and refracted at the further surface of the spheres. The angle between the reflected and refracted rays at this surface is (a) 50º (b) 60º (c) 90º (d) 40º
r2
α
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s:
µ2
tp
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ht
µ1
ht
uestions
er si
pe
t
tion of a ray in the two media are along the vectors r1 = aiˆ + bjˆ and r2 = ciˆ + djˆ , then
//t
s:
tp
q
ta
un
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ht
tp
s:
//w
w
iv er
si
ta
w .ja
m
UNSOLVED OBJECTIVE y
1. A point source emits sound equally in all directions in a no-absorbing medium. The ratio of amplitudes of the waves by the source at P to that of Q is 2 : 3. Ratio of distance of P and Q from the source is (a) 2 : 3 (b) 3 : 2 (d) 3 : 2 (c) 2 : 3
fo
in
a.
Putting r2 = 30º we get i2 = 60º Reflection at Q r2 = r1 = 30º ∴ α = 180º – (r2 + i2) = 180º – (30º + 60º) = 90º
ar an
sin r2 1 = sin i2 3
Refraction at Q
co m
)
sin 60° = 3 sin r1 1 sin r1 = ⇒ 2 ⇒ r1 = 30º Since r2 = r1, therefore r2 = 30º
(a) µ1a = µ2b
µ1a
(b)
=
µ2 d
a +b c2 + d 2 2 2 (c) µ1(a + b ) = µ2 (c2 + d2) (d) None of these
2
2
4. A liquid of refractive index 1.33 is placed between two identical palno-convex lenses, with refractive index 1.50. Two possible arrangement P and Q are shown. The system is
3. Two thin slabs of refractive indices µ1 and µ2 are placed parallel to each other in the x-z plane. If the direction of propaga-
x
2. A point object is placed at a distance of 12 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that image formed by the combination coincides with the object itself. The focal length of the convex mirror is (a) 20 cm (b) 25 cm (c) 15 cm (d) 30 cm
i1
) Explanation Refraction at P.
Q
r'2
For green and blue lights, µ > 1.414. Hence their critical angles are < 45º. They will be reflected iternally and emerge from the surface at the bottom.
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Light (Geometrical Optics)
4A.81
Q
ar a
vA + v
2 B
vA − v
D2
7. A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other side of the lens, the required distance from the lens
D1
C
(a)
(d)
2
D
2 B
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(b) |vA – vB|
(c)
2
10. A ray of light travels from a medium of refractive index µ into air. If the angle of incidence at the plane surface of separation is θ and the corresponding angle of deviation is D, the variation D with θ is shown correctly by
B
(a) vA + vB
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t.m e/
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VS
move to the left move to the right remain at P move either to the left or right, depending upon focal lengths of the lenses
9. Half the surface of a transparent sphere of refractive index 2 is silvered. A narrow, parallel beam of light is incident on the unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from the sphere will be a (a) parallel beam (b) converging beam (c) slightly divergent beam (d) widely divergent beam
.m
//t
ht tp s:
Normal
45°
m
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45°
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w
w
//w
VA
m
P
(a) (b) (c) (d)
6. Two bodies A and B are moving towards a plane mirror with speeds vA and vB respectively as shown in the figure. The speed of image of A respect to the body B is
B
rio s_ in fo
5. A ray of light on a slab of transparent material is party reflected from the surface and partly refracted into the slab. The reflected and refracted rays are mutually perpendicular. The incident ray makes on angle i with the normal to the slab. The refractive index of the slab is (a) tan–1(i) (b) cot–1(i) –1 (c) sin (i) (d) cos–1(i)
A
A
divergent in P, convergent in Q convergent in P, convergent in Q convergent in both divergent in both
(a) (b) (c) (d)
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P
8. Two convex lenses placed in contact form the image of a distant object at P. If the lens B is moved to the right, the image will
of a convex mirror of radius of curvature 10 cm, in order to have an image of the object coincidet with it, is (a) 70 cm (b) 30 cm (c) 50 cm (d) 60 cm
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(0, 0)
θ
π/2
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4A.82
D D1 D2 C
iv
m (c) 2v k
2 5 *
(a) (b) (c) (d)
m (d) 4v k
12. When the object is at distances u1 and u2 from the optical center of a convex lens, a real and a virtual image of the same magnification is obtained. The focal length of the lens is
s
ta
un iv e/
14. A person is looking at the flat surface of a transparent hemisphere of refractive index µ = 2 . Half of the flat surface is coloured black and half of curved surface is coloured in six equal strips as shown in the figure. The person can see
.m
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s: tp ht
(a) separate part of the red colour from the green and blue colours. (b) separate part of the blue colour from the red and green colours. (c) separate all the three colours from one another. (d) not separtate even partially any colour from the other two colours.
m k
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s:
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v (b) 2
m (a) v k
er si
w
w
//w
s:
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wall
L
rio
π/2
11. A plane mirror having a mass m is tied to the free end of a massless spring of spring constant k. The other end of the spring is attached to a wall. The spring with the mirror held vertically to the floor can slide along it smoothly. When the spring is at its natural length, the mirror is found to be moving at a speed of v cm/s. The separation between the images of a man standing before the mirror, when the mirror is in its extreme positions, will be
k
fo ta rio
θ
in s_
ar an m
C
(0, 0)
.ja
D2
si
D1
a. c
D
(d)
π/2
θ
om
(0, 0)
13. A beam of light consisting of red, green and blue colours is incident on a rightangled prism. The refractive indices of the material of prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will.
er
(c)
u1 + u2 2
D
θ
(d)
UL
(0, 0)
C
(b)
(c) 2u2
(b) 2u1
D2
D1
u1 − u2 2
(a)
D
Light (Geometrical Optics)
%
,
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all the colours only yellow and indigo only yellow, indigo and blue only orange, yellow and green
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4A.83
close very close to each other in vacuum. A point source is placed at the center point C1 of first sphere. Image formed by paraxial rays when viewed through second sphere is
15. A graph is plotted between angle of deviation (δ) and angle of incidence (i) for a prism, where prism is being used to study dispersion. The plot will be like
Light (Geometrical Optics)
G
P 5
(a)
,
P
F
5 F
G
o
in f
s_
s
rio
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ZDWHU
2
t.m
tp ht
2 (a) A < sin −1 3
µ (a) 2 − 1 t µ1
µ (b) 1 − 1 t µ2
2 (b) A = sin −1 3
(c) (µ 2 − µ1 )t
s: //
3
ht
tp
W
19. A lens of any transparent material in air will behave as (a) converging lens in water (b) diverging lens in water (c) neither converging nor diverging in water (d) may be none of these
(d) (µ 2 − 1)t
2 (d) A = 2sin −1 3 17. Two transparent spheres of refractive index 2 and radius R are placed very
(\H
2 (c) sin −1 (2/3) < A < 2sin −1 3
P
6
%
ZDWHU
6
e/
s:
//t
ht
.m
tp
s: //
16. A black spot B is marked very near to base of the prism (µ = 3/2) as shown in the figure. If the image of black spot appears on the surface of the prism, the angle of the prism A will be $
rio
si
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w
w w
G
(d)
18. A Young’s double slit experiment is conducted in water (µ1) as shown in the figure, and a glass plate of thickness t and refractive index µ2 is placed in the path of S2. The magnitude of the optical path difference at O, will be
ta
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m
,
m
(c)
ar an a.
co
, G
at point c1 at point c2 at point m outside of the spheres
(a) (b) (c) (d)
(b)
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4A.84
FP
(b) 7.5 cm (d) none of these
23. If final image formed after two refractions through the lens and one reflection
s
ita rio
rs
(d) can’t be said
in fo
s_
rio
&RLQ
26. A man stands on a glass slab of height 1 and inside an elevator accelerated upwards with ‘a’. The bottom of the slab appears to have shifted with respect to the man by a distance (if the R.I. of the glass is µg) 1 (a) less than µg 1 (b) greater then µg 1 (c) equal to µg
FP
(a) 1.6 cm (c) 8 cm
ve
ni
e/ u
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//t
s:
tp ht FP
25. A beam of light coming from infinity is passing through a convex lens having radius of curvature R = 20 cm each; is focused at a certain distance from lens. Find the radius of convex of emergent wave front from lens: (µlens = 1.5, µ = 1) (a) 10 m (b) 5 m (c) 2 m (d) 1 m
s:
tp
ht
er iv un
.m e/
//t
22. A man observes a coin placed at the bottom of a beaker which contain two immiscible liquids of refractive index 1.2 and 1.4 as shown in the figure. The depth of the coin below the surface, as observed from above, is
si
ta
.ja
w w
(b) 1.47 (d) 1.23
ht
(a) 1.3 (c) 1.02
&
//w
tp s:
%
24. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiec is 36 cm. The final image is formed at infinity. The focal length f0 of the objective and te of the eyepiece respectively are (a) 45 cm and –9 cm (b) 50 cm and 10 cm (c) 7.2 cm and 5 cm (d) 30 cm and 6 cm
om
.c
an a
m
ar
/LTXLG
5 FP
(b) 110 cm (d) 80 cm
(a) 100 cm (c) 90 cm
FP
FP
21. On the hypotenuse of a right angled prism of refractive index 1.5, a drop of liquid is placed as shown. A ray of light falls normally on the AB. What should be the maximum value of refractive index of liquid, so that it gets totally reflected at face AC? $
G
(µ − 1)R (µ − 1)
(d)
µR (µ + 1)
(c)
from the mirror forming at same point O, then will be
20. A sphere of radius R made of transparent material of refractive index µ works like a thin lens having its optical center coinciding with the center of the sphere. The focal length of this thin lens is µR µR (a) (b) (µ1 − 1) 2(µ − 1)
Light (Geometrical Optics)
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29. A ray of light incident on a transparent glass slab of refractive index 3 . If the reflected and refracted rays are mutually perpendicular, then angle of incidence is
28. A lens of refractive index µ is put in a liquid of refractive index µ′. If the focal length of the lens in air is f, its focal length in liquid will be
er si ta rio s
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(a) (c) (b) (b)
7. (c) 15. (a) 23. (b)
8. (b) 16. (d) 24. (d)
un t.m e/
2. A person looking through a telescope just sees the point A on the rim of bottom of a cylindrical vessel when it is empty shown in the figure. When the vessel is filled with a liquid of refractive index 1.5, he observes a mark B at the centre of the vessel, without moving the telescope or the vessel. What is the height of the vessel if the diameter of its cross section is 10 cm? [IIT, 1977] [Ans. 8.45 cm] 3. In a river 2 m deep, a water level measuring post embedded into the river stands vertically with 1 m of it above the water
s: //
ractice
ht tp
6. 14. 22. 30.
EXERCISE 1
1. A point source of light is placed at the bottom of a vessel containing a liquid of 5 refractive index . A person is viewing 3 the source from above the surface. There is an opaque disc of radius 1 cm floating on the surface. The centre of the disc lies vertically above the source S. The liquid from the vessel is gradually drained out through a tap. What is the maximum height of the liquid for which the source cannot at all be seen from above? [IIT, 1970] [Ans. 1.33 cm]
(a) (a) (a) (a)
iv
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ht tp s:
P
er s 5. 13. 21. 29.
(c) (d) (b) (a)
e/
4. 12. 20. 28.
//t
(b) (d) (d) (c)
ita
.ja w w 3. 11. 19. 27.
//w
(b) (a) (c) (c)
s:
2. 10. 18. 26.
tp
(b) (a) (d) (b)
ht
1. 9. 17. 25.
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ar a
f µ′µ (µ − µ′)
(b) 45º (d) None of these
30. Images formed by an object when placed between two plane mirrors making an angle of 90º with one another lie on a (a) straight line (b) circle (c) ellipse (d) parabola
na .c om
(d)
f (µ′ − µ) µ′(µ − 1)
(a) 60º (c) 30º
m
(b)
µ′(µ − 1) f (µ′ − µ)
(c)
− f µ′(µ − 1) µ′ − µ
(a)
ANSWERS
4A.85
27. A thin prism P1 of angle 4º, and made from a glass of refractive index 1.54, is combined with another thin prism P2 made from a glass of refractive index 1.72, to produce dispersion without deviation. The angle of P2 is (a) 5.33º (b) 4º (c) 3º (d) 2.6º
Light (Geometrical Optics)
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4A.86
Light (Geometrical Optics)
reflected ray and refracted ray are mutually perpendicular, what is the angle of incidence? Give calculations. [MNR, 1987] [Ans. 58.19′]
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iv
t.m e/ s: //
ht tp
7. A ray of light falls on a transparent glass slab of refractive index 1.62. If the
rio s_ in fo
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ht tp s:
11. A man standing on the edge of a water tank looks at a stone lying at its bottom. Find the position where he sees the image of the stone if the line of vision makes an angle θ with the normal to the water surface. Take the depth of water in the tank to be H. (µ 2 − 1) H sin 3 θ Ans. l = − (µ 2 − sin 2 θ)3/ 2
12. A convex lens of focal length 15 cm is placed in front of a convex mirror. Both are coaxial and the lens is 5 cm from the apex of the mirror. When an object is placed on the axis at a distance of 20 cm from the lens, it is found that the image coincides with the object. Calculate the radius of curvature of mirror. [Ans. 55 cm]
6. A ray of light is incident from a denser medium on to a rarer medium. If θ1C is the critical angle and θ1B is the Brewster’s angle of incidence, find the relative refractive index of the two mediums. Given sin θ1c that, = η = 1.28 [IIT, 1983] sin θ1B [Ans. 0.798]
[Ans. 2 ]
4
8 Ans. sin θ ≥ 9
10. Light is incident at an angle α on one planar end of a transparent cylindrical rod of refractive index n. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of α. [IIT, 1992]
er s
un iv
//t
T&
&
e/
.m
tp
5
$
T
9. A ray of light falls on a glass slab making an angle of 45º with the vertical shown in the figure. Find the refractive index of the glass if total internal reflection takes place at the vertical face AB. [Ans. µ = 1.22]
ita
ar a
m
.ja
w
w
//w s:
3
ht
%
8. Glycerine (refractive index 1.4) is poured into a large jar of radius 0.2 m. to a depth of 0.1 m. There is a small light source at the centre of the bottom of the jar. Find the area of the surface of glycerine through which the light passes. [Roorkee, 1998] [Ans. 0.0327 m2]
na .c om
5. A glass prism of refractive index 1.5 is 4 immersed in water (refractive index ). A 3 light beam incident normally on the face AB shown in the figure is totally reflected to reach the face BC if
4. Explain how a fish inside a pond full of water can see everything above the water near the pond. Show that the field of vision of the fish is concentrated within a cone of half angle 49º inside the water 4 (refractive index of water = and (sin 3 49º = 0.75). [Roorkee, 1980) [Ans. 49º]
surface. If the angle of inclination of sun above the horizon is 30º, calculate the length of the post on the bottom 4 surface of the river. µ for water = 3 [Roorkee, 1978] [Ans. 3.44 m]
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4A.87
18. A pin is placed 10 cm in front of convex lens of focal length 20 cm made of material of refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature 22 cm. Determine the position of the final image. Is the image real or virtual? [IIT, 1978] [Ans. 11 cm; real]
13. An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm at a distance of 15 cm from the lens. A plane mirror is placed inclined at 45º to the lens axis, 10 cm to the right of the lens. Find the position and size of the image formed by the lens and mirror combination. Trace the path of the rays forming the image. [IIT, 1972] [Ans. Both the points A and B are at a distance of 20 cm from the lens axis. Size of image = 8 cm]
19. The radius of curvature of the convex face of a plano convex lens is 12 cm and its µ = 1.5. (a) Find the focal length of the lens. The plane surface of the lens is now silvered. (b) At what distance from the lens will parallel rays incident on the convex surface converge? (c) When a point object is placed on the axis 20 cm from the lens, Calculate the image distance when the object is placed. [IIT, 1979] [Ans. (a) 24 cm, (b) 12 cm, (c) 30 cm]
ar a
iv
t.m e/
21. The convex surface of a thin concavo convex lens of glass of refractive index 1.5 has a radius of curvature 60 cm. The convex side is silvered and placed at horizontal surface. Find: (a) Where should a pin be placed on the optic axis such that its image is formed at the same place? (b) If the concave part is filled with water 4 of refractive index , find the distance 3 through which the pin should be moved
s: //
ht tp
20. An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. You may take the distance of the near surface of the slab from the mirror to be 1 cm. [IIT, 1980] [Ans. 7.56 cm]
un
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17. A concave mirror of radius of curvature 1 m is placed at the bottom of a tank of water. The mirror forms an image of the sun when it is directly overhead. Calculate the distance of the images from the mirror for different depths, 80 cm and 40 cm of the water in the tank. [IIT, 1968] [Ans. When the depth of water 80 cm, distance = 50 cm, When the depth of water 40 cm, distance = 47.5 cm]
er si ta rio s
er s
un iv
e/
.m
tp
ht
16. Photographs of the ground are taken from an airplane flying at an altitude of 2000 m by a camera with a lens of focal length 50 cm. The size of the film of the camera is 18 cm × 18 cm. What area of the ground can be photographed by this camera at any instant? [IIT, 1976] [Ans. 720 m × 720 m]
ita
m
.ja
w
w
//w
s:
15. An object is placed at 20 cm left of the convex lens of focal length 10 cm. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens, find the magnification of the final image. [IIT, 1974] [Ans. 1]
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14. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm it is found that there is no parallex formed by the two mirrors. What is the radius of curvature of the convex mirror? [IIT, 1973] [Ans. 25 cm]
Light (Geometrical Optics)
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4A.88
22. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. Find the power of combination. [Ans. –1.5 diopter]
sin
23. A convex lens of focal length 20 cm and a concave mirror with a radius of curvature 16 cm are placed at a distance of 25 cm from each other. A luminous object is placed at a distance of 25 cm from the lens. Where will be the image formed and what will be its magnification? [Roorkee, 1984] [Ans. In-between the lens and mirror and at a distance of 7.23 cm from the mirror. Magnification, –0.39]
θ b = 2 µ R2
2 2 2 1/2 2 2 1/2 (µ R − b ) − ( R − b )
in
fo
28. A parallel beam of light travelling in 4 water (refractive index = ) is refracted 3 by a spherical air bubble of radius 2 mm, situated in water. Assuming the light ray to be parallel, find the position of the image due to refraction at the first surface and position of final image? [Ans. Image is formed at a distance of 6 mm on the object side. Final image will be formed at a distance of 5 mm to the left of the second surface.]
s_
rio
ta
si
si
ta
rio
s
er
iv
un
er
e/
un
29. The distance between two point source of light is 24 cm. Find out where would you place a converging lens of focal length 9 cm so that the images of both sources are formed at the same point. [Roorkee, 1988] [Ans. Lens, should be placed at a distance of 18 cm or 6 cm from the source.]
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e/
iv
.m
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s:
tp
30. In a compound microscope the objective and eyepiece have focal lengths of 0.95 cm and 5 cm respectively and are kept at a distance of 20 cm. The last image is formed at a distance of 25 cm from the eyepiece. Calculate the position of the object and the total magnification. [Roorkee, 1988]
FP
ht
tp
s:
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27. Two lenses of power +12 and –2 diopters are placed in contact. What will be the focal length of combination? [MNR, 1987] [Ans. 10 cm]
m
ar
m
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w
w
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s:
tp
ht
25. A convex lens of focal length 20 cm and a concave mirror with a radius of curvature of 16 cm are placed as shown in the figure, at a distance of 25 cm from each other. A luminous object is placed at a distance of 25 cm from the lens. Where will be the image formed and what would be its magnification? [Roorkee, 1986] FP
[Roorkee, 1987]
co
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24. A plano-convex lens has a thickness of 4 cm. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottommost point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face of 25 cm. Find the the lens is found to be 8 focal length of the lens. [IIT, 1984] [Ans. 75 cm]
26. A ray of light passes through a transparent sphere of refractive index µ and radius R. If b is the distance between the incident ray and a parallel diameter of sphere, show that the angle of deviation θ is given by expression
so that the image of the pin again coincides with the pin. [IIT, 1981] [Ans. (a) 15 cm, (b) 1.06 cm]
Light (Geometrical Optics)
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31. A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate:
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15. Why does a diamond sparkle?
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18. Explain why the planets do not twinkle. 19. Why does the Sun appear reddish early in the morning?
20. Why does the sky appear dark instead of blue to an astronaut?
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21. Does short-sightedness (myopia) or longsightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
22. The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense does then a magnifying glass provide angular magnification?
23. In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back? 24. Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power.
10. A concave spherical mirror has a radius of curvature of 40 cm. Draw ray diagrams to locate the image (if one is formed) for an object at a distance of (a) 100 cm (b) 40 cm (c) 20 cm and (d) 10 cm from the mirror. For each case, state whether the image is real or virtual, erect or inverted, and enlarged, reduced, or the same size as the object.
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14. Why do different coloured rays deviate differently in the prism?
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13. What are fibre cables?
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12. When a monochromatic light passes through a prism, will it show dispersion?
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11. In previous question instead of concave if convex mirror (same radius of curvature) is used what will be the answer.
1. A concave mirror forms a real image of a object. What change in the image do you expect if a fly (a) is sitting on the object. (b) is between object and mirror and (c) is sitting on the mirror?
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2. A ray of light is passing from air into glass. If the angle of incidence, with respect to the normal to the interface, is increased (a) Total internal reflection will occur when the angle of incidence equals the critical angle (b) Total internal reflection will occur when the angle of incidence is less than the critical angle (c) Total internal reflection will occur when the angle of incidence is greater than the critical angle (d) Angle of refraction angle will increase but there will be no total internal reflection
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12. A mango tree is at the bank of river and one of the branch of tree extends over the river. A tortoise lives in river. A mango falls just above the tortoise. The acceleration of the mango falling from tree appearing to the tortoise is (Refractive index of water is 4/3 and the tortoise is stationary) (a) g (b) 3 g/4 (c) 4 g/3 (d) None of these
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19. Liquid is filled in a vessel of height 2H/3. At the bottom of the vessel there is a spot P and a hole from which liquid is coming out. Let d be the distance of image of P from an eye at height H from bottom at an instant when level of liquid in the vessel is x. If we plot a graph between d and x it will be like
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35. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of
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38. An air bubble is inside water. The refractive index of water is 4/3. At what distance from the air bubble should a point be placed so as form a real image at the same distance from the bubble? (a) 2R (b) 3R (c) 4R (d) the air bubble cannot form a real image of real object
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44. A pendulum of length ℓ is free to oscillate in vertical plane above point O. An observer is viewing the bob of the pendulum directly from above. The pendulum is performing small oscillations in water (refractive index is µ) about its equilibrium position. The equation of trajectory of bob as seen by observer is
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52. A concave mirror has a focal length 20 cm. The distance between the two positions of the object for which the image size is double of the object size is (a) 20 cm (b) 40 cm (c) 30 cm (d) 60 cm
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Light (Geometrical Optics)
index 1.5. The wavelength of refracted will be (a) dependent on intensity of refracted light (b) same (c) smaller (d) larger
60. Prism angle of a prism is 10º. Their refractive index for red and violet colour is 1.51 and 1.52 respectively. Then dispersive power will be (a) 0.5 (b) 0.15 (c) 0.019 (d) 0.032
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73. Power of a lens is –4D and for second lens, power is +2D, the total power for the couple is (a) –2D (b) 6D (c) –6D (d) –8D
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72. Light travels through a glass plate of thickness t and having a refractive index µ. If c is the velocity of light in vacuum, the time taken by light to travel this thickness of glass is (a) tµc (b) tc/µ (c) t/µc (d) µt/c
66. A beam of monochromatic light is refracted from vacuum into a medium of refractive
71. The refractive index of water is 1.33. What will be the speed of light in water? (a) 3 × 108 m/s (b) 2.25 × 108 m/s 8 (c) 4 × 10 m/s (d) 1.33 × 108 m/s
70. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together, what will be their resulting power (a) +6.5D (b) –6.55D (c) +7.5D (d) –0.75D
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65. An object is situated at a distance of f/2 from a convex lens of focal length f. Distance of image will be (a) + (f/2) (b) + (f/3) (c) + (f/4) (d) – f
69. Focal length of a convex lens will be maximum for (a) blue light (b) yellow light (c) green light (d) red light
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64. One light wave is incident upon a plate of refracting index µ. Incident angle i, for which refractive and reflective waves are mutually perpendicular will be (a) i = 45º (b) i = sin–1(µ) –1 (c) i = cosec (µ) (d) i = tan–1(µ)
68. The focal length of the convex lens depends upon (a) frequency of the light ray (b) wavelength of the light ray (c) both (a) and (b) (d) None of these
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63. A telescope consisting of an objective of focal length 60 cm and a single–lens eye piece of focal length 5 cm is focussed at a distant object in such a way that parallel rays emerge from the eye piece. If the object subtends an angle of 2º at the objective, then the angular width of the image will be (a) 10º (b) 24º (c) 50º (d) 1/6º
62. Diameter of the moon is 3.5 × 103 km and its distance from earth is 3.8 × 105 km. It is seen by a telescope whose objective and eyepiece have focal lengths 4 m and 10 cm respectively. The angular diameter of the image of the moon will be nearly (a) 5º (b) 10º (c) 20º (d) 25º
67. Time taken by the sunlight to pass through a window of thickness 4 mm whose refractive index is 1.5 is (a) 2 × 10–8 second (b) 2 × 108 second (c) 2 × 10–11 second (d) 2 × 1011 second
61. Prism angle and refractive index for a prism for a 60º and 1.414. Angle of minimum deviation will be (a) 15º (b) 30º (c) 45º (d) 60º
(b) 15 cm (d) None of these
(a) 10 cm (c) 50 cm
4A.100
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74. A ray of light from air is incident in water then which property of light will not change in water (a) velocity (b) frequency (c) amplitude (d) colour
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(a) The rays are focussed at 12 cm from the point P to the right, in the situation as shown in figure (i) (b) The rays are focussed at 16 cm from the point P to the right, in the situation as shown in figure (i) (c) If the rays are incident at the curved surface (figure (ii)) then these are focused at distance 18 cm from point P to the right (d) If the rays are incident at the curved surface (figure (ii)) then these are focused at distance 14 cm from point P to the right
1. The quantity quoted first in the options is plotted on y-axis against the quantity quoted latter in the option on x-axis. Then for which of the options will the graph have the form as shown?
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3. In vacuum, the speed of light does not depend on (a) Wavelength (b) Frequency (c) Intensity (d) Speed of observer
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2. A parallel beam of light is incident normally on the flat surface of a hemisphere of radius 6 cm and refractive index 1.5, placed in air as shown in figure (i). Assume paraxial ray approximation.
76. Radius of curvature of convex mirror is 40 cm and the size of object is twice as that of image, then the image distance is (a) 10 cm (b) 20 cm (c) 40 cm (d) 30 cm
75. The focal length of convex lens is 30 cm and the size of image is quarter of the object, then the object distance is (a) 150 cm (b) 60 cm (c) 30 cm (d) 40 cm
77. The combined power of two lenses is contact is +10D. When they are separated by 20 cm, their power becomes +6.25D. The powers of these lenses are (a) –3.5D, +6.5D (b) –7.5D, +2.5D (c) +7.5D, +2.5D (d) +9.0D, +1.0D
4A.101
(a) The distance of object from focus of a convex mirror plotted against the distance of the image from the same point (b) pressure plotted against the temperature for an ideal gas undergoing process P V = constant. (c) electric field due to charged conducting sphere plotted against distance from the centre of the sphere. (d) the tension in a string plotted against its fundamental frequency
Light (Geometrical Optics)
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9. In case of three plane-mirrors meeting at a point to form a corner of a cube, if incident light suffers one reflection on each mirror (a) The emergent ray is antiparallel to incident one (b) The emergent ray is perpendicular to incident one (c) The emergent ray is in phase with incident one (d) The emergent ray is in opposite phase with incident one
4. A man of height 170 cm wants to see his complete image in a plane mirror (while standing). His eyes are at a height of 160 cm from the ground. (a) Minimum length of the mirror = 80 cm (b) Minimum length of the mirror = 85 cm (c) Bottom of the mirror should be at a height 80 cm (d) Bottom of the mirror should be at a height 85 cm
Light (Geometrical Optics)
5. When light passes from air to water which of the following changes? (a) Wavelength (b) Velocity (c) Frequency (d) Colour
10. A ray of light is incident normally on one face of 30º – 60º – 90º prism of refractive index 5/3 immersed in water of refractive index 4/3 as shown in figure.
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6. In case of reflection by a plane-mirror, which of the following statements are not correct? (a) It can never give real image (b) It can never give inverted image (c) It changes left into right (d) It changes front into back
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11. A plane mirror, reflecting a ray of incident light, is rotated through an angle θ about an axis through the point of incidence in the plane of the mirror perpendicular to the plane of incidence, then
8. If two mirrors are inclined to each other at 90º, the image seen may be (a) One (b) Two (c) Three (d) Four
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7. A biconvex thin lens of refractive index µ2 = 1.4 has a radii of curvature R1 = 20 cm and R2 = 60 cm, µ1 = 1.5 and µ3 = 1.2. Focal length are f1 and f2 in both the media
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4A.102
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the reflected ray does not rotate the reflected ray rotates an angle θ the reflected ray rotates an angle 2θ the incident ray is fixed
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12. An object is placed in front of a converging lens at a distance equal to twice the focal length f1 of the lens. On the other side of the lens is a concave mirror of focal length f2 separated from the lens by a distance 2(f1 + f2). Light from the object passes rightward through the lens, reflects from the mirror, passes leftwards through the lens, and forms a final image of the object.
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13. Which of the following can form diminished, virtual and erect image of your face? (a) Converging mirror (b) Diverging mirror (c) Converging lens (d) Diverging lens
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16. In displacement method, the distance between object and screen is 96 cm. The
14. The radius of curvature of the left and right surface of the concave lens are 10 cm and 15 cm respectively. The radius of curvature of the mirror is 15 cm.
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(a) The distance between the lens and the final image is equal to 2f1. (b) The distance between the lens and the final image is equal to 2(f1 + f2). (c) The final image is real, inverted and of same size as that of the object. (d) The final image is real, erect and of same size as that of the object.
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15. A fish, F in the pond is at a depth of 0.8 m from the water surface and is moving vertically upwards with velocity 2 m/s. At the same instant a bird B is at a height of 6 m from the water surface and is moving downwards with velocity 3 m/s. At this instant both are on the same vertical line as shown in the figure. Which of the following statements are correct?
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(a) Equivalent focal length of the combination is –18 cm (b) Equivalent focal length of the combination is +36 cm (c) The system behaves like a concave mirror (d) The system behaves like a convex mirror
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4A.103
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(c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True. 1. Statement 1: Maximum possible angle of refraction in a medium is critical angle. Statement 2: The critical angle θc (or µ) depends upon pair of medium, colour of light (or λ) and temperature.
ratio of length of two images formed by a convex lens placed between them is 4.84. (a) Ratio of the length of object to the length of shorter image is 11/5 (b) Distance between the two positions of the lens is 36 cm (c) Focal length of the lens is 22.5 cm (d) Distance of the lens from the shorter image is 30 cm
4A.104
2. Statement 1: A point object is placed at a distance of 26 cm from a concave mirror of focal length 26 cm. The image will form at infinity. Statement 2: For above given system the 1 1 1 gives v = ∞. equation + = u v f
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1.
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4. Statement 1: A convex mirror is used as a driver’s mirror. Statement 2: Because convex mirror’s field of view is large and images formed are virtual, erect and diminished.
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6. Statement 1: In visible light µr < µv Statement 2: This follows from Cauchy’s B C formula µ = A + 2 + 4 λ λ
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5. Statement 1: Focal length of a lens for red colour is greater than its focal length for violet colour. 1 1 1 Statement 2: Because = (µ − 1) − f R1 R2 and µ1 < µv.
7. Statement 1: Refractive index of material of a prism depends on angle of prism A and angle of minimum deviation δm. Statement 2: Refractive index of a prism,
Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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ASSERTION AND REASON UESTIONS
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(a) As ‘x’ is increased from zero, intensity continuously decreases (b) As ‘x’ is increased from zero, intensity first increases then decreases (c) Intensity at centre of screen for x = 90 cm and x = 110 cm is same (d) Radius of bright circle obtained on screen is equal to 1 cm for x = 200 cm
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3. Statement 1: When a concave mirror is held under water, its focal length will increase. Statement 2: The focal length of a concave mirror is independent of the medium in which it is placed.
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17. A point source of light S is placed on the axis of a lens of focal length 20 cm as shown. A screen is placed normal to the axis of lens at a distance ‘x’ from it. Treat all rays as paraxial.
A + δm sin 2 µ= sin( A/2)
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16. Statement 1: A virtual image can be photographed. Statement 2: Only real objects are photographed.
15. Statement 1: If a spherical mirror is dipped in water, its focal length remains unchanged. Statement 2: A laser light is focused by a converging lens. There will be a significant chromatic aberration.
17. Statement 1: The small object, to be seen in a microscope, is kept between F and 2F of its objective. Statement 2: In this case, the image formed by the objective is nearer to the eyepiece.
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19. Statement 1: Light rays retrace their length when their direction is reversed (Law of reversibility of light rays). Statement 2: For the refraction light, water is denser than air, but for the refraction of sound, water is rarer than air.
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20. Statement 1: If the angle between the two plane mirror is 72º and the object is asymmetrically placed between the two mirrors, then 5 images of the object will be formed.
13. Statement 1: As the temperature of a medium increases, the refractive index decreases. Statement 2: When a ray travels from vacuum to a medium, then µ is known as absolute refractive index of the medium. (µvacuum = 1)
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12. Statement 1: If both plane mirror and object are moved through a distance x, then the image moves through a distance 3x. Statement 2: When the object is fixed and plane mirror is moved through a distance x. Then the image is also moved through the distance x.
14. Statement 1: Critical angle is minimum for violet colour. Statement 2: Because critical angle
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11. Statement 1: When the object moves with a velocity v , its image in the plane mirror moves with a velocity of −2v . Statement 2: The minimum height of the mirror to be required to see the full image of man of height h is h/2.
18. Statement 1: A point source of light is placed at a distance of 2f from a converging lens of focal length f. The intensity of the other side of the lens is maximum at distance 2f. Statement 2: In chromatic aberration, the rays of different colours fail to converge at a point after going through a converging lens.
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10. Statement 1: Keeping a point object fixed, if a plane mirror is moved, the image will also move. Statement 2: In case of a plane mirror, distance of object and its image is equal from any point on the mirror.
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9. Statement 1: Two convex lenses joined together cannot produce an achromatic combination. Statement 2: The condition for achromaω ω tism is 1 + 2 = 0 where symbols have f1 f 2 their usual meaning.
4A.105
8. Statement 1: Spherical aberration of a lens can be reduced by blocking the central portion or peripheral portion of the lens. Statement 2: Spherical aberration arises on account of inability of the lens to focus central (paraxial) and peripheral (marginal) rays at the same point.
Light (Geometrical Optics)
1 1 θc = sin −1 and µ ∝ µ λ
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Statement 2: For given system of mirror, the total number of images formed due
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Light (Geometrical Optics)
to successive reflection is equal to either
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MATCH THE COLUMN T PE UESTIONS Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II
1. An executed object is moving in front of concave mirror as shown in figure. On L.H.S. various velocity of object and position is given. On R.H.S. some properties of image and its velocity is given. Column I Object
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2. Column I shows four diagrams of real object point O, image point I and principal axis have been sketched. Select the proper optical system from column II which can produce the required image. (Image may be real or virtual).
26. Statement 1: If P1 and P2 be the powers of two thin lenses located coaxially in a medium of refractive index µ at a distance d, then the power P of the combination is P = P1 + P2 – P1P2d/µ Statement 2: Because for above given system equivalent focal length is given by f1 f 2 1 and P = F= F f1 + f 2 − d /µ
(p) +ve velocity
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(a) +ve velocity and object is between focus and centre of curvature (b) –ve velocity and object is between focus and pole (c) –ve velocity and object is beyond centre of curvature (d) –ve velocity and object is virtual
25. Statement 1: Rainbow is an example of the dispersion of sunlight by the water droplets. Statement 2: Light of shorter wavelength is scattered much more than light of larger wavelength.
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24. Statement 1: A man wearing glasses of focal length +1 m cannot see beyond 1 m. Statement 2: A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black, the intensity of the image will decrease but the image will not be shifted upward or downward.
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23. Statement 1: When we seen an object, the image formed on the retina is real and inverted. Statement 2: If the magnification of a system is less than one, then the image formed in inverted.
Column II Image
22. Statement 1: All large telescopes use parabolic mirrors rather than spherical mirrors. Statement 2: A parabolic mirror brings parallel rays (rays parallel to its axis) to an exact focus at an point.
21. Statement 1: Optical fibres are used to transmit light without any appreciable loss in its intensity over distance of several kilometers. Statement 2: Optical fibres are very thick and all the light is passed through it without any loss.
360° 360° 360° is or − 1 accordingly as θ θ θ odd or even respectively.
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4A.106
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4. Match the following:
(b)
(s) −10iˆ
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(a) VA′, A (b) VB′, B (c) VC ′, C (d) VD′, D
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6. Match the following: Column I (a) Inverted crown-flint glass prism (b) Achromatism (c) Hollow prism (d) Glass slab Column II 1 (p) Deviation ∝ dispersive power
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5. A convex lens (f) forms an image on a screen considering the object to be at the zero mark in a scale, match the following: Column I (a) Image (b) Additional lens in contact (c) Reduction in refractive index (d) Slicing the lens to have one plane and another convex surface Column II (p) Moves the image of infinite object further away (q) Not unique as lens is moved between object and source. (r) Virtual for screen position at a distance <4f from the object. (s) Object at d forms real image further nearer plano-convex lens.
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3. Four particles are moving with different velocities in front of stationary plane mirror (lying in y-z plane). At t = 0, velocity of A is v A = iˆ, velocity of B is vB = −iˆ + 3 ˆj , velocity of C is vC = 5iˆ + 6 ˆj , velocity of D is vD = 3iˆ − ˆj. Acceleration of particle A is a A = 2iˆ + ˆj and acceleration of par ticle C is aC = 2t ˆj. The particle B and D move with uniform velocity (Assume no collision to take place till t = 2 seconds). All quantities are in S.I. units. Relative velocity of image of object A with respect to object A is denoted by VA′, A . Velocity of images relative to corresponding objects are given to column I and their values are given in column II at t = 2 second. Match column I with corresponding values in column II. Column I Column II
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Column II Diverging lens Converging lens Concave mirror Convex mirror
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(a) Power of convex mirror (b) Power of concave mirror (c) Power of plane mirror (d) Power of convex lens
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4A.107
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4A.108
(r) Absence of chromatic aberration (s) Dispersion without deviation 7. Match the features of a telescope listed in Column I with the quantities that they depend upon listed in Column II.
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Column I Intensity of light received by lens Angular magnification Length of telescope Sharpness of image
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PASSAGE BASED UESTIONS Q
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8. The graphs given apply to convex lens of focal length f, producing a real image at a distance v from the optical centre when self luminuous object is at distance u from the optical centre. The magnitude of magnification is m. Identify the following graphs with the first named quantity being plotted along y-axis. Column I (a) v against u 1 1 (b) against v u (c) m against v v (d) (m + 1) against f
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Column II (p) Focal lengths of objective and eye-piece (q) Diameter (aperture) of the objective (r) Dispersion of light by objective and eyepiece (s) Spherical aberration
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The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature
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PASSAGE–2
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10 14 22 40 200
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Age, (years) 10
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1. Minimum focal length of eye lens of a normal person is (a) 25 cm (b) 2.5 cm (c) 25/9 cm (d) 25/11 cm
2. Maximum focal length of eye lens of normal person is (a) 25 cm (b) 2.5 cm (c) 25/9 cm (d) 25/11 cm
3. A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be (a) +1 (b) –1 (c) +3 (d) –3
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In the normal human eye, light from an object is refracted by the cornea-lens system at the front of the eye and produces a real image on the retina at the rear of the eye. For a given eye, its lens-toretina distance is fixed at about 2.5 cm. Most of the focusing of an image is done by the cornea, which has a fixed curvature that is convex with respect to incoming light. The importance of the lens is that its radius of curvature can be changed, allowing the lens to fine-tune the focus. The lens is surrounded by the ciliary muscle. Contraction of the muscle decreases tension on the lens. This allows the natural elasticity of the lens to produce an increase in the radius of curvature. When the muscle relaxes, the lens flattens out, decreasing its radius of curvature. Unfortunately, the lens loses elasticity with age and the ability to alter curvature decreases. The range over which clear vision is possible is bounded by the far point and the near point. In normal vision, the far point is infinity and the near point depends on the radius of curvature of the lens. For normal eyes the average near point for reading is 25 cm.
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decreases) and focal length decreases. For a clear vision, the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person. A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grownup person, minimum distance of object should be around 25 cm. A person suffering for eye defects uses spectacles (Eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina. The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example, power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3. For all the calculations required you can use the lens formula and lens maker’s formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.
4A.109
In the myopic (nearsighted) eye, the lens-toretina length, is too long and/or the radius of curvature of the cornea is too great. This causes rays from an object at infinity to focus at a point in front of the retina. The far point is closer than normal. A corrective, lens will put a virtual image of a distant object at the position of the actual far point of the eye. In the hyperopic (farsighted) eye, the lens-to-retina length is too short and/or the radius of the curvature of the cornea is not great enough. This causes rays from an object at infinity to focus at a point behind the retina. The
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Light (Geometrical Optics)
(d) relaxing the ciliary muscle and decreasing the radius of curvature
near point is farther away than normal. A corrective lens will put a virtual image of the close object at the position of the actual near point. The relation among the object (o) and image (i) distances from the eye and the focal length (f) of the lens is given by the lens-distance rule
PASSAGE–3 Figure shows a convex lens of focal length 15 cm. A point object is placed on the principle axis of lens at a distance 20 cm from it as shown. On the other side of lens two observer eyes O1 and O2 are situated at a distance 100 cm from the lens at some distance above and below the principal axis.
1 1 1 + + o i f When using this equation, all distances are given in centimeters. The power of corrective lenses is usually given in units called diopters. Power, in diopters, is the reciprocal of the focal length in meters 1 Pdropter = f meter
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8. After activity–I, which of the following observer will not be able to see the image of object, if before this activity both were seeing the image? (a) O1 (b) O2 (c) both O1 and O2 (d) neither O1 nor O2
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7. In initial setup before the activities, which of the following statement is correct? (a) Observer O1 will see a real image at 60 cm from the lens but observer O2 will not be able to see it. (b) Observer O2 will see a real image at 60 cm from the lens but observer O1 will not be able to see it. (c) Both the observers will see a real image at 60 cm from lens irrespective the positions of O1 and O2. (d) Both the observers may or may not be able to see the image at 60 cm from lens depending on the positions of O1 and O2.
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6. In a mildly hyperopic eye, the focal length of the eye’s natural lens can be corrected by (a) contracting the ciliary muscle and increasing the radius of curvature (b) contracting the ciliary muscle and decreasing the radius of curvature (c) relaxing the ciliary muscle and increasing the radius of curvature
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I. Converging lenses have positive focal lengths, and diverging lenses have negative focal lengths. II. Real images have positive distances from the lens, and virtual images have negative distances from the lens.
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4A.110
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4A.111
Light (Geometrical Optics)
PASSAGE–4 The laws governing the behaviour of the rays namely rectilinear propagation, laws of reflection and refraction can be summarized in one fundamental law known as Fermat’s principle. According to this principle, a ray of light travels from one point to another such that the time taken is at a stationary value (maximum or minimum). If c is the velocity of light in a vacuum, the velocity c in a medium of refractive index n is , hence n n time taken to travel a distance l is . If the light c passes through a number of media, the total time
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1. A 4 cm thick layer of water covers a 6 cm thick glass slab. A coin is placed at the bottom of the slab and is being observed from the air side along the normal to the surface. Find the apparent position of the coin from the surface.
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11. The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is
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SUBJECTIVE UESTIONS
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Light (Geometrical Optics)
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4. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
10. In the figure shown, the speed of image in cm/sec with respect to mirror is *, then find the value of ‘*’
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7. Two persons A and B wear glasses of optical powers (in air) P1 = + 2 D and P2 = + 1 D respectively. The glasses have refractive index 1.5. Now they jump into a swimming pool and look at each other. B appears to be present at distance 2 m (from A) to A. ‘A’ appears to be present at distance 1 m (from B) to B. Find out the refractive index of water in the swimming pool, in the from X/10 and fill value of X.
13. A concave mirror and a convex mirror are placed co-axially, their reflecting surfaces facing each other. Their focal lengths are 15 and 12 cm respectively. An object placed between them is 20 cm from the concave mirror. The image formed by it is at the object itself. Calculate the distance of the concave mirror from the object. 14. The powers of the objective and eye piece of a compound microscope is 100D and 20D respectively. The microscope is adjusted for normal observation (eye muscles least strained). The magnifying power of the microscope is 50. What will be the magnifying power if the distance between the objective and eye piece is increased by 2 cm and under normal observation adjustment?
11. The angle of crown glass (µ = 1.52) prism is 5º. What should be angle of flint glass (µ = 1.63) prism, so that the two prism together may be used in direct vision spectroscope?
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6. You hold a plane mirror 1.0 m in front of your eyes and are able to see a 15 m high tree behind your. If the mirror is 20.0 cm high, and the tree image completely fills the mirror, how far are you standing from the tree?
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4A.112
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16. An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.
2. An eye specialist prescribes spectacles having combination of convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm. The power of this lens combination in diopters is [1997]
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3. A thin equiconvex lens of glass of refractive index µ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water µ = 4/3. On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. [1997]
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18. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
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17. A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.
4A.113
15. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.
Light (Geometrical Optics)
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1. A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half lens is 2. Find the focal length of the lens and separation between the halves. Draw the ray diagram for image formation. [1996]
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Light (Geometrical Optics)
4A.114
(a) is absent (b) is smaller if the curved surface of the lens faces the object (c) is smaller if the plane surface of the lens faces the object (d) is the same whichever side of the lens faces the object
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9. The x–y plane is the boundary between two transparent media. Medium-1 with z ≥ 0 has a refractive index 2 and medium-2 with z ≤ 0 has a refractive index 3. A ray of light in medium-1 given by vector A = 6 3 iˆ + 8 3 ˆj − 10kˆ is incident on the plane of separation. Find the unit vector in the direction of the separation. Find the unit vector in the direction of the refracted ray in medium-2. [1999]
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7. A prism of refractive index n1 and another prism of refractive index n2 are stuck together with a gap as shown in the figure. The angles of the prism are as shown. n1 and n2 depend on λ, the wavelength of 10.8 × 104 light according to n1 = 1.20 + and λ2 1.80 × 104 n2 = 1.45 + where λ is in nm. λ2 [1998]
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8. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a
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6. A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to [1998]
(a) real and will remain at C (b) real and located at a point between C and ∞ (c) virtual and located at a point between C and O (d) real and located at a point between C and O
5. A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be [1998]
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4A.115
Light (Geometrical Optics)
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15. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A’ B’ is the image after refraction from the lens and the reflection from the mirror, find the distance of A’ B’ from the pole of the mirror and obtain its magnification. Also locate positions of A’ and B’ with respect to the optic axis RS. [2000]
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13. A diverging beam of light from a point source S having divergence angle α falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and its refractive index is n, then the divergence angle of the emergent beam is [2000]
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16. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 (n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax, such that the ray comes out only from the surface CD, is given by [2000] $
18. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will suffer [2001]
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(i) Determine the angle of the flint glass prism. (ii) Calculate the net dispersion of the combined system.
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17. A ray of light passes through four transparent media with refractive indices µ1, µ2, µ3 and µ4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have
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20. A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image
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Light (Geometrical Optics)
4A.116
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Light (Geometrical Optics)
again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid [2001]
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23. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30º at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
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24. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image [2003] (a) (b) (c) (d)
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25. A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of µg would be [2003]
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4A.117
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Light (Geometrical Optics)
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A light of wavelength 6600Å is incident on face AB such that angle of incidence is 60º. Find (a) the angle of emergence and (b) the minimum value of thickness of the coated film on the face AC for which the light emerging from the face has maximum intensity. [2003]
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4A.118
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Light (Geometrical Optics)
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37. Focal length of the plano-convex lens is 15 cm. The object is at A as shown in the figure. The plane side is silvered. The image is [2006]
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34. A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths? [2005]
35. AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence of Q, so that the ray is totally reflected by both the slabs. [2005]
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Light (Geometrical Optics)
38. Match the following: Column I Column II (a) Angular magnifi- (p) Dispersion of cation lenses (q) f0 and fe
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(a) STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is a correct explanation for STATEMENT 1 (b) STATEMENT 1 is True, STATEMENT 2 is True; STATEMENT 2 is NOT a correct explanation for STATEMENT 1 (c) STATEMENT 1 is True, STATEMENT 2 is False (d) STATEMENT 1 is False, STATEMENT 2 is True
43. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60º). In the position of minimum deviation, the angle of refraction will be [2008]
(a) 30º for both the colours (b) greater for the violet colour
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40. A ray of light traveling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be [2007] (a) only a reflected ray and no refracted ray (b) only a refracted ray and not reflected ray (c) a reflected ray and a refracted ray and the angle between them would be less than 180º – 2θ (d) a reflected ray and a refracted ray and the angle between them would be greater than 180º – 2θ
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39. Parallel rays of light from Sun falls on a biconvex lens of focal length f and the circular image of radius r is formed on the focal plane of the lens. Then which of the following statement is correct? [2006] 2 (a) Area of image is πr and area is directly proportional to f (b) Area of image is πr2 and area is directly proportional to f2 (c) Intensity of image increases if f is increased. (d) If lower half of the lens is covered with black paper, area will become half.
(a) (b) (c) (d)
(r) Aperture of lenses (s) Spherical aberration
(b) Sharpness of the image (c) Light gathering power (d) Length of telescope
41. In an experiment to determine the focal length f of a concave mirror by the u – v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in the line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then
60 cm to the left of lens 12 cm to the left of lens 60 cm to the right of lens 30 cm to the left of lens
(a) (b) (c) (d)
4A.120
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4A.121
Light (Geometrical Optics) 6
(c) greater for the red colour (d) equal but not 30º for both the colours
44. An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. Column I
45. A light beam is travelling from Region IV (figure). The refractive index in regionals I, n n n II, III and IV are n0 , 0 , 0 and 0 respec2 6 8 tively. The angle of incidence θ for which the beam just misses entering region IV is [2008]
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(b) (b) (a) (a) (b) (d) (c) (b) (d) (a)
2. 10. 18. 26. 34. 42. 50. 58. 66. 74.
(d) (c) (d) (a) (a) (d) (c) (c) (c) (b)
3. 11. 19. 27. 35. 43. 51. 59. 67. 75.
(b) (a) (c) (a) (b) (c) (a) (a) (c) (a)
4. 12. 20. 28. 36. 44. 52. 60. 68. 76.
(d) (c) (c) (a) (c) (d) (a) (c) (b) (a)
5. 13. 21. 29. 37. 45. 53. 61. 69. 77.
(c) (a) (b) (b) (c) (b) (d) (b) (d) (c)
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6. 14. 22. 30. 38. 46. 54. 62. 70.
(b) (b) (c) (c) (d) (d) (c) (c) (d)
7. 15. 23. 31. 39. 47. 55. 63. 71.
(a) (a) (a) (b) (b) (a) (a) (b) (b)
8. 16. 24. 32. 40. 48. 56. 64. 72.
(b) (a) (c) (b) (a) (c) (c) (d) (d)
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Light (Geometrical Optics)
4A.122
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ONE OR MORE THAN ONE CHOICE MA BE CORRECT 1. (a, b)
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7. (a) → q (b) → p (c) → r, s (d) → p
8. (a) → r, (b) → s, (c) → q, (d) → p
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PREVIOUS EARS’ IIT-JEE UESTIONS
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43. (a) At minimum deviation for any wavelength r1 = r2 = A/2, Because r1 + r2 = A
19. (i) 4º, (ii) –0.04º
4A.123
18. (c)
21. (b)
22. (c)
23. (b)
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44. (a) → (p, q, r, s), (b) → (q), (c) → (p, q, r, s), (d) → (p, q, r, s) uf f u (a) f < 0, v = and = = u − f 1 − f /4 u /f − 1 v m = − values of v may be positive, negau tive or infinity, also it can have values less than or greater than u. (a) → (p, q, r, s) (b) In this case f is positive.
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Light (Geometrical Optics)
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4.1 Wave optics: n introduction
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Light (Wave Optics)
4.2 Huygens’ principle
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In geometrical optics, we have represented light as rays which travel in straight lines in a homogeneous medium. By doing this, we have studied a variety of phenomenon involving mirrors and lenses. The phenomenon like interference and diffraction can not be explained on the bases of particle nature of light. These phenomenon can only be explained on the basis of wave nature of light. This part of optics is called physical optics. The wave theory of light was presented by Christian Huygens in 1678. During that period, Newton’s corpuscular theory had satisfactorily explained the phenomenon of reflection, refraction and rectilinear propagation of light. So the scientists believed in the corpuscular theory; no one really believed in Huygen’s wave theory. The wave characteristics of light was not really accepted until the interference experiments of Young in 1801. It should be pointed out that Huygens did not know whether the light waves were longitudinal or transverse and also how they propagate through vacuum. It was then explained by Maxwell by introducing electromagnetic wave theory in nineteenth century.
(i) Each point of a given wavefront is a source of new disturbance which is called secondary disturbance. The wavelets originated from these points spread out in all directions with the speed of light. (ii) The envelope of these wavelets in the forward direction gives the shape and position of the new wavefront at any subsequent time. To understand this, consider a spherical wave front AB as shown in figure. Every point such as 1, 2, etc., on AB becomes the source of secondary spherical wavelets. After time t the radius of each wavelet will be ct, where c is the speed
Huygens principle provides a geometrical method which allows us to determine the shape of the wavefront at any time, if the shape of the wavefront at an earlier time is known. A wave front is the locus of the points which are in the same phase. Huygens’ principle can be stated as follows:
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Light (Wave Optics)
of the light. Thus from the points 1, 2, 3,.... etc., draw spheres of radii equal to ct. These spheres represent the secondary wavelets. According to Huygens, the common envelope A1B1 in forward direction gives the position of new wavefront (see fi g.). $ $
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In the time, the disturbance at A reaches C, the secondary waves from B will travel a distance BD such that BD = AC. With the point B as centre and radius equal to AC draw an arc. From the point C, draw the tangent CD. In triangles BAC and BDC; BC is common and BD = AC https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.3
∠BAC = ∠BDC = 90º The two triangles are congruent, and so ∠ABC = ∠BCD i = r.
∴ or
Thus angle of incidence is equal to angle of reflection. This proves the law of reflection.
proof of law of refraction
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Let xy is the interface between two media 1 and 2 of refractive indexes µ1 and µ2 respectively. Suppose v1 and v2 are the velocities of light in two media. The second medium is optically denser than first and so v2 < v1. AMB is the plane wavefront incident at an angle i. In the time disturbance at B reaches C, the secondary waves from A will travel a distance AD = v2t, where t is the time taken by the waves to travel the distance BC. Thus
With A as the centre and radius AD draw an arc. Then draw a tangent CD to the arc. CD represents the refracted wavefront r be the angle of refraction. We have BC v1t v1 = = . AD v2t v2
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In triangles ABC and ACD, we have sin i BC /AC = sin r AD /AC BC . AD From equation (i) and (ii), we have sin i v1 = sin r v2
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4B.4
Light (Wave Optics)
v1 µ 2 = , v2 µ1
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or µ1 sin i = µ2 sin r. This proves the law of refraction, which is called Snell’s law.
4.3 interference
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When two or more coherent waves superimpose, the resultant intensity in the region of superposition is different from the intensity of individual waves. This modification in the distribution of intensity in the region of superposition is called interference.
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Thomas Young in 1801 devised an ingenious method of producing coherent sources. In this method, a single wavefront is divided into two; these two split wavefronts act as if they originated from two sources having a constant phase relationship and therefore, when they were allowed to interfere, a stationary interference pattern was obtained. In the experiment, light from a source S fell on a cardboard which contained two pinholes (or slits) S1 and S2 which were very close to one another. The spherical waves originating from S1 and S2 were coherent and so beautiful interference fringes or bands were obtained on the screen.
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Light (Wave Optics)
4B.5
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1. Two independent laser sources of equal wavelengths can be coherent. Because they can maintained the constant phase difference for long time.
2. Two ordinary sources can not maintain the constant phase difference so they can not be coherent and hence will not interfere.
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Consider a monochromatic source of light S emitting light waves of wavelength λ and two narrow slits S1 and S2. S1 and S2 are separated a distance d and equidistance from S. S1 and S2 then becomes two virtual coherent sources of light waves. Let φ is the phase difference between the two waves reaching at point P. The equation of wave for any fixed position (say screen at x = 0) can be written as: y = a sin (ωt – kx), where x = 0 and so, we
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I = I1 + I2 + 2 I1 I 2 cos φ.
...(3)
As intensity I of wave is proportional to square of the amplitude, and so Also dividing equation (ii) by (i), we get a2 sin φ . a1 + a2 cos φ
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In Young’s interference experiment, incident monochromatic light is diffracted by slit S, which then acts as a point source of light that emits semicircular wavefronts. As that light reaches screen B, it is diffracted by slits S1 and S2, which then act as two point sources of light. The light waves traveling from slits S1 and S2 overlap and undergo interference, forming an interference pattern of maxima and minima on viewing screen C. Depending on the phase difference φ between the two waves, the intensity of resulting wave may be minimum or maximum. Accordingly, there are two types of interference. These are:
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(ii) Destructive interference (dark point) The intensity I will be minimum, when cos or
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Light (Wave Optics)
∆x = (2n − 1)
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Interference between two waves with φ = π.
Special cases: When two identical waves interfere,
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Light (Wave Optics)
intensity distribution It has been obtained that intensity at bright points is 4a2 and at dark points is zero. According to law of conservation of energy, the energy of the interfering waves as a whole remains constant. Thus the energy from points of minimum intensity transfers to the points of maximum intensity. The intensity variation with phase difference is shown in fi gure. , D
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Consider two sources S1 and S2 emitting monochromatic light of wavelength λ. The separation between them is d. The interference fringes are obtained on a screen placed at a distance D from the sources. The fringes are of equal width and alternatively bright and dark. The centre to centre distance between two consecutive bright or dark fringes is called fringe width β. Consider a point P on the screen at a distance yn from the centre of the screen O. The angular position of the point P is θ from the centre of the sources (see fi g. 4.8). The path difference between the waves on arriving at point P, is S2P – S1P, which is equal to ∆x. From the figure, ∆x = d sin θ. For small θ, we can write sin θ ≃ tan θ. Thus
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(i) Bright fringes There will be bright fringe at P, when ∆x = nλ. Thus path difference dyn = nλ D n Dλ or yn = ; n = 0, 1, 2, d
.........(9)
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4B.9
Light (Wave Optics)
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Diagram showing fringe width.
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//t ht tp s:
β = yn – yn – 1
s: //
∴ Fringe width
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iv
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tp
Q 6HFRQGGDUN
er si ta rio s
er s
w w //w s:
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ht
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ita
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m
6
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rio s_ in fo
na .c om
G 6
ht tp
=
or
β=
n Dλ Dλ − (n − 1) d d Dλ . d
...(10)
(ii) Dark fringes There will be dark fringe at P, when ∆x = (2n – 1)
λ . Thus 2
dyn λ = (2n − 1) D 2 or yn =
(2n − 1) Dλ ; n = 1, 2, ... 2 d
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...(11)
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Light (Wave Optics)
Equation (11) represents the position of nth dark fringe. The (n – 1)th fringe will be at a distance 2(n − 1) − 1 Dλ yn−1 = d 2 ∴ Fringe width β = yn – yn – 1
2n − 1 Dλ 2(n − 1) − 1 Dλ = − d 2 2 d or
β=
Dλ . d
na .c om
It shows that the fringe width is equal for bright and dark fringe.
w
rio s_ in fo
er s
w
angular fringe Width
er si ta rio s
ita
.ja
m
ar a
note The maximum path difference ∆xmax = d, when sin θ = 1. If n are the number of d brights fringes on one side of the central bright, then d = nλ or n = . Thus total λ number of fringes that can be on the screen are = 2 n + 1, including central fringe.
un iv
e/
iv
ht tp
s: //
t.m e/
ht tp s:
un
Dλ /d D λ α = radian. d =
or
β D
.m
α=
//t
tp
s:
//w
Sometimes, it is required to represent fringe width in terms of angle subtended at the centre of the sources. If α is the angular fringe width, then
ht
4B.10
6
D
E
6
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Light (Wave Optics)
4B.11
Special case: If YDSE is performed in water, and observer is in air, then fringe width
Dλ water . d λ = air , µω
βwater =
As λwater
∴ βwater =
1 Dλ air βair = . µ ω d µ ω
I
mportant points
nf
a.
s_ i
an
rio
ar
m
ta
iv
s:
un
ondition of observable interference
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s
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si ta
w .ja
//w
w
o
co m
1. In YDSE, the central fringe is bright, and all the bright fringes are of same intensity. Colour of bright fringes are of the colour of incident light. 2. If slits are of equal size, the intensity of all the dark frings are zero. 3. If slits are of unequal size, then the intensity of dark fringe is not zero. 4. All the fringes are of equal width. 5. If sources have random phase difference, then there will be no interference. The intensity at any point will be I = a2 + a2 = 2a2. 6. If white light is used in the experiment, then the central fringe will be white, and other fringes are overlapped colour fringes.
er
e/
iv
un
e/
s:
//t
ht
.m
tp
//t
.m
tp
ht
si
1. The sources must be coherent. 2. The separation between the slits should be small (order of mm), so that size of fringe is large enough to observe. 3. The amplitudes of interfering waves are equal or nearly equal, otherwise the intensities of bright and dark fringes are not differentiable.
ht
tp
s:
E
xample 1 Consider interference between two sources of intensities I and 4I. Obtain intensities at points where the phase difference is (i) π/2 and (ii) π.
)
) Solution
We know that resultant intensity
For φ = π/2;
(i)
I R = I + 4 I + 2 I × 4 I cos
π 2
= 5I
(ii)
I R = I1 + I 2 + 2 I1 I 2 cos φ
For φ = π;
I R = I + 4 I + 2 I × 4 I cos π = 5I – 4I = I
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Light (Wave Optics)
example 2 A parallel wavefront of monochromatic light is incident on double slit arrangement at an angle α as shown in fi gure. Find the position of zero order maxima from the centre of the screen. The distance of screen from the slits is D (d << D).
) Solution
3 6
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G D
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ar a
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6
m
%
rio s_ in fo
D
na .c om
\ T
2
6FUHHQ
er s
w
ita
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∴
yo = D sin α
iv
t.m e/
yo D
s: //
0 = d sin α – d
yo and for zero order maxima, ∆x = 0. D
ht tp
or
ht tp s:
For small and sin θ ≃ tan θ =
un
//t
= d sin α – d sin θ
er si ta rio s
un iv
e/
∆x = (BS2 + S2P) – (AS1 + S1P) = (S2P – S1P) – (AS1 – BS2)
.m
s:
tp
//w
w
Suppose the position of zero order maxima is at P at a distance yo from O. The path difference between two waves at P is
ht
4B.12
example 3 The two coherent sources of monochromatic light of wavelength λ are located at a separation λ. The two sources are placed on a horizontal line and screen is placed perpendicular to the line joining the sources (see figure). Find position of the farthest minima from the centre of the sources.
) Solution Suppose at P the farthest minima will occur. Let it subtends an angle θ at the centre of the sources.
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Light (Wave Optics)
4B.13
3
'[ T
6
6FUHHQ 6
G O
un iv
//w
Thus in between these two positions there is only one minima for which ∆x = λ . Thus 2 λ = λ cos θ 2 1 or cos θ = 2 ∴ θ = 60º
e/
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp ht
er si ta rio s
w
∆xm = 0; when cos θ = 0 or θ = 90º
s:
er s
w
and minimum path difference
ita
∆xmax = λ; when cos θ = 1 or θ = 0º
.ja
m
The maximum path difference can be
rio s_ in fo
∆x = S2P – S1P ≃ d cos θ = λ cos θ
ar a
na .c om
The path difference
example 4 Two slits in Young’s interference experiment have width in the ratio 1 : 4. Find the ratio of intensity at the maxima and minima in their interference.
) Solution The intensity of the wave is proportional to the area of the slit. Thus I1 b1 b1 1 = = = . If a1 and a2 are the amplitudes of the waves, then I 2 b2 b2 4 I1 a12 1 = = I 2 a2 2 4
∴
a1 1 = . a2 2
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Light (Wave Optics)
The ratio
I max (a1 + a2 ) 2 = I min (a1 − a2 ) 2 (1 + 2) 2 9 = . (1 − 2) 2 1
=
example 5 In double slit arrangement, the source S is not symmetrically placed from the slits. It is located as shown in the figure. Find the position of the zero order maxima from the centre of the screen. The separation between slits and screen is D (d << D).
w
un iv
//w s:
e/
//t
iv
.m
tp
GVLQT
6
un
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2
6FUHHQ
t.m e/
ht tp s:
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3
ita
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er s
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T
er si ta rio s
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m
6
rio s_ in fo
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) Solution
ht
s: //
Suppose the position of zero order maxima is at P at a distance yo from O. The path difference between two waves at P is
ht tp
4B.14
∆x = (SS2 + S2P) – (SS1 + S1P) = –(SS1 – SS2) + (S2P – S1P) =−
(
)
2 d − d + d sin θ
For small θ, sin θ ≃ tan θ =
∴
∆x = −
(
)
2d −d +
yo . D
dyo D
For zero order maxima, ∆x = 0 or
∴
0=− yo =
(
(
)
2d −d +
)
dyo D
2 −1 D
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Light (Wave Optics)
4B.15
example 6 In an interference arrangement similar to Young’s double slit experiment, slits S1 and S2 are illuminated with coherent microwave sources each of frequency 1 MHz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity Iθ is measured as a function of θ where θ is defined as shown in figure. If Io is the maximum intensity, calculate Iθ for (a) θ = 0º (b) θ = 30º and (c) = 90º.
) Solution
T
T
m
ar a
G
un iv e/
un
t.m e/
s: //
∆x = d sin θ.
ht tp
iv
.m
//t
ht tp s:
c 3 × 108 = = 300 m f 106
The path difference
er si ta rio s
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w //w s: tp ht
The wavelength of microwave λ=
2
ita
'[
w
.ja
6
rio s_ in fo
6
na .c om
3
The corresponding phase difference φ= =
2π 2π ∆x = (d sin θ) λ 300 2π (150sin θ) 300
= π sin θ. The resultant intensity in interference is given by
IR = I1 + I 2 + 2 I1 I 2 cos φ I1 + I 2 + 2 I1 I 2 cos(π sin θ).
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Light (Wave Optics)
For
I1 = I2 = I.
IR = 2I + 2I cos (π sin θ).
(a)
(b)
For θ = 0º, IR = 2I + 2I cos 0º = 4I I Given I0 = 4I; ∴ I = 0 . Ans. 4 For θ = 30º
IR = 2I + 2I cos (π sin 30º) = 2I + 2I cos
(c)
For θ = 90º
IR = 2I + 2I cos (π sin 90º)
ar a
I0 I0 = Ans. 4 2
na .c om
=2
π = 2I 2
rio s_ in fo
4B.16
= 2I – 2I = 0.
er s
w
ita
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4.4 displacement of fringes (fringes sHift and numBer of sHifted fringes)
er si ta rio s
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t.m e/
ht tp s:
un
∆x = S2P – [S1P + (µ – 1) t] = (S2P – S1P) – (µ – 1) t
//t
ht
.m
tp
e/
s:
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//w
w
Suppose a transparent sheet of thickness t and refractive index µ is introduced in front of one of the slits of YDSE. The optical path of the light waves emerging from slit will increase by an amount (µ – 1) t. In the arrangement shown the optical path of S1P + (µ – 1) t. Thus path difference between waves at P
S2P – S1P = d sin θ
s: //
From the geometry of the figure
ht tp
For small angle θ, sin θ ≃ tan θ =
yn . D 3
W P
6
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6FUHHQ
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Light (Wave Optics)
∴ ∆x =
4B.17
dyn − (µ − 1)t. D
For bright fringes the path difference ∆x = nλ. Thus dyn − (µ − 1)t = nλ; D or yn =
n = 0, 1, 2, ....
n Dλ D(µ −1)t + . d d
...(1)
In the absence of the sheet, the position of nth bright, yn = n
rio s_ in fo er si ta rio s '
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2
iv un
//t s: //
t.m e/
ht tp s:
6
.m
tp ht
ita er s
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//w
6
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of fringes
Dλ . Thus displacement d
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Displacement of the fringes.
∆=
D(µ − 1)t d
...(2)
The position of (n – 1)th order bright fringe yn −1 =
(n − 1) Dλ D(µ − 1)t + d d
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Light (Wave Optics)
The fringe width β = yn – yn – 1 or
β=
Dλ . d
This shows that when a transparent sheet is introduced in the path of the slit, the entire fringe pattern will shift towards that side but fringe width remains same. The number of fringe shifted N=
∆ D(µ − 1)t /d = β Dλ d
N=
or
(µ − 1)t . λ
...(3)
rio s_ in fo
ita un iv
e/
tp
2
un
P
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t.m e/
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T
W
'
ht tp
s: //
6
3
er si ta rio s
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6
s:
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example 7 Two transparent sheets of thickness t1 and t2 and refractive indexes µ1 and µ2 are placed infront of the slits as shown in the fi gure. If D is the distance of the screen from the slits, then find the distance of zero order maxima from the centre of the screen. What is the condition that zero order maxima is formed at the centre O?
ht
4B.18
) Solution Suppose P is the position of zero order maxima. The distance of P from the centre O of the screen is y0. the optical path of light waves from source S1
x1 = S1P + (µ1 – 1) t1
The optical path of light waves from source S2
x2 = S2P + (µ2 – 1) t2
The path difference
∆x = x2 – x1 = (S2P – S1P) + (µ2 – 1) t2 – (µ1 – 1) t1
From the geometry,
S2P – S1P = d sin θ ≃d tan θ =
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dy0 . D
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Light (Wave Optics)
dy0 + (µ2 – 1) t2 – (µ1 – 1) t1 D
∆x =
∴
4B.19
For zero order maxima, ∆x = 0.
dy0 + (µ2 – 1) t2 – (µ1 – 1) t1 D
∴
0=
or
y0 =
D[(µ1 − 1)t1 − (µ 2 − 1)t2 ] . d
For zero order maxima at the centre O, y0 = 0
or
(µ1 – 1) t1 = (µ2 – 1) t2.
ita
.ja
4.5 fresnel’s Biprism
ar a
na .c om
0=
rio s_ in fo
D[(µ1 − 1)t1 − (µ 2 − 1)t2 ] d
∴
m
iv
er si ta rio s
e/
un
Dλ . d
6 G
ht tp
s: //
t.m e/
ht tp s:
β=
//t
ht
.m
tp
s:
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//w
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er s
w
Fresnel developed another arrangement to get interference phenomenon. He used two small angled prisms (prism angle about 0.5º) placed base to base. With this arrangement he produced two coherent sources by using a single source. The given fi gure shows a source S of monochromatic light and biprism arrangement. S1 and S2 are the virtual images of S as formed by two prisms. These sources are originated from the same source, and so they are coherent. If λ is the wavelength of light used, then fringe width is
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In the arrangement D = (a + b). The value of d can be determined by two methods. These are: https://t.me/universitarios_infohttps://www.jamarana.com
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4B.20
Light (Wave Optics)
(i) By displacement method In this method, a convex lens is used to form real images of the sources. If d1 and d2 are the separations between images in two positions of the lenses, then d = d1d 2 . (ii) If A is the angle of prism, then angle of deviation produced by any prism
δ = (µ – 1) A.
From the geometry of the figure, for small angle δ=
d /2 a
d 2a ∴ d = 2a (µ – 1) A
ar a
4.6 lloyd’s mirror arrangement
rio s_ in fo
na .c om
or (µ – 1) A =
er si ta rio s
iv
t.m e/
∴
ht tp s:
0 = I + I + 2 II cos φ φ = π rad.
un
//t
ht
.m
tp
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s:
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//w
w
er s
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ita
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In Lloyd’s mirror arrangement, a plane mirror is used to produce another source, which together with real source constitutes two coherent sources. This set-up was developed by Lloyd in 1834. In this arrangement the interference pattern similar to Young’s double slit experiment is obtained on the screen. But the central fringe is dark instead of being bright. It means there is destructive interference at the centre O. If we assume that mirror is perfectly reflecting, then the intensity of light waves from S1 and S2 is equal, let it is I. Then for destructive interference (zero resultant intensity)
ht tp
s: //
It shows that the wave after reflecting from mirror undergone a phase change of π rad. This experiment proves, that a light wave after reflection from an optically denser medium undergoes a phase change of π rad.
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Light (Wave Optics)
4B.21
example 8 A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment: (i) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å. (ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm. nDλ . d
For third bright, n = 3
3Dλ 3 × (120 × 10−2 ) × 6500 × 10−10 = d 2 × 10−3 = 1.17 × 10–3 m.
y3 =
(ii)
Let n1th bright of 6500 Å concides with the n2th bright of 5200 Å, then
ar a
∴
4 5
un
=
t.m e/
5200 6500
ht tp
s: //
ht tp s:
=
iv
.m
e/
n1 λ 2 = n2 λ1
er si ta rio s
un iv
Dλ1 Dλ 2 = n2 d d
//t
or
n1
er s
w
w //w s: tp ht
ita
.ja
m
rio s_ in fo
yn =
na .c om
) Solution (i) For bright fringe,
O c
O c
\
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Light (Wave Optics)
The minimum of value of n1 and n2 are 4 and 5 respectively. Therefore y4 =
4Dλ1 4 × (120 × 10−2 ) × (6500 × 10−10 ) = d 2 × 10−3
= 1.56 × 10–3 m.
m
ar a
) Solution
un iv
3
6FUHHQ
s: //
t.m e/
ht tp s:
un
//t
iv
6
e/
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er si ta rio s
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er s
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6 '[
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example 9 A parallel beam of monochromatic light is used in a Young’s double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. The light is incident an angle θ with the normal to the plane of slits. Find the value of θ so that, there will be dark fringe at the centre P of the pattern.
ht
The path difference between two wavefronts at P
ht tp
4B.22
∆x = d sin θ
For dark fringe at P, ∆x =
λ 2
d sin θ =
λ 2
Therefore or
λ θ = sin −1 . 2d
example 10 A narrow slit S transmitting light of wavelength λ is placed a distance d above a large plane mirror as shown in the fi gure. The light coming directly from the slit and that coming after reflection interfere at a screen Σ placed at a https://t.me/universitarios_infohttps://www.jamarana.com
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4B.23
Light (Wave Optics)
distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e., just above O? (b) At what distance from O does the first maximum will occur?
) Solution
6
6 G
na .c om
G
ar a
6
E
rio s_ in fo
2
er si ta rio s
er s
un iv
//w
w
ita
β Dλ /2d Dλ = . = 2 2 4d
w
y=
.ja
m
(a) There will be dark fringe at O, therefore intensity of light at O will be zero. (b) It is clear from the figure, that the distance of first maximum from O
6
iv
ht tp
s: //
) Solution
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
example 11 Two plane mirrors M1 and M2 are inclined to each other at an angle θ and an illuminated slit S is placed infront of them at a distance of 12 cm away from and parallel to the line of intersection of the mirror. An eye piece is mounted at a distance of 60 cm from the line of intersection of the mirrors. If the bandwidth obtained is 0.16 mm and wavelength of light used is 5460 Å. Find the value of θ.
0
G
T
T 6
FP
0
6 FP
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Light (Wave Optics)
From the figure D = 60 + 12 = 72 cm Let separation between the sources formed by mirrors is d. We have β= or
0.16 × 10–3 =
Dλ d (72 × 10−2 ) × (5460 × 10−10 ) d
d = 2.46 × 10–3 m.
∴
d 0.24 2.46 × 10−3 θ= 0.24 = 10.25 × 10–3 rad.
For small angle tan θ ≃ θ =
na .c om
or
rio s_ in fo
er si ta rio s
un iv
//w
w
er s
w
ita
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m
ar a
example 12 S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis 50 with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO.
un
//t s: //
ht tp
/
t.m e/
ht tp s:
) Solution
iv
.m
tp
e/
s:
(i) If the third intensity maximum occurs at the point A on the screen, find the distance OA. (ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase decrease, or remains the same?
ht
4B.24
3
6
PP 3
6 G
2
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PP
P
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6FUHHQ
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Light (Wave Optics)
4B.25
For any half u = –0.15 m f = 0.10 m We have,
1 1 1 − = v u f
1 1 1 − = v −0.15 0.10
or
v = 0.30 m.
∴
Now from similar triangles S S1 S2 and S P1 P2 we have
d = 1.5 × 10–3 m
ar a
3 × 1.0 × 500 × 10−9 1.5 × 10−3
un iv
er s
w //w
w
=
= 1 × 10–3 m
e/
s: tp
iv
.m
(ii) As we have seen d is proportional to P1 P2, that is gap between L1 and L2, therefore with decrease in gap, d decreases and hence OA increases, because 1 y∝ . d
t.m e/
ht tp s:
un
//t
ht
Dλ d
ita
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m
(i) For third maximum OA = y3 = 3
er si ta rio s
or
rio s_ in fo
na .c om
d (0.30 + 0.15) = −3 0.5 × 10 0.15
ht tp
s: //
example 13 Consider the arrangement shown in the fi gure. The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe width.
) Solution 3 [ G 2 '
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Light (Wave Optics)
(a) The path difference at O, ∆x = 2 D 2 + d 2 − 2 D λ 3λ For the dark fringe at O, ∆x = , , 2 2 For minimum value of d, λ 2 λ or ( D 2 + d 2 )1/2 − D = 4 2 D2 + d 2 − 2D =
1/2
m
Dλ . 2
er si ta rio s iv
.m
[
un '
s: //
'
3
2
6
t.m e/
ht tp s:
6
6 G
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tp
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w
er s
w
(b)
ita
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or d =
ar a
na .c om
d2 λ or D 1 + −D= 2 2 D 4 2 d λ or D + −D= 2D 4
rio s_ in fo
d2 λ or D 1 + 2 − D = 4 D
ht
Here the path difference ∆x = (SS1 + S1P) – (SS2 + S2P)
ht tp
4B.26
= ( D 2 + d 2 ) + ( x − d )2 + D 2 − D + D 2 + x 2 For the next bright fringe after first dark fringe, ∆x = λ
∴ D 2 + d 2 + ( x − d )2 + D 2 − D + D 2 + x 2 = λ 1/2
1/2
( x − d )2 d2 or D 1 + 2 + D 1 + D2 D
1/2 x2 − D + D 1 + 2 = λ D
d2 ( x − d )2 x2 + D + or D + − D + D + =λ 2D 2D 2D https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.27
d 2 + ( x − d )2 =λ 2D or d2 + x2 + d2 – 2xd – x2 = 2λ D
or
or 2d2 – 2xd = 2λD Dλ , we get 2
For d =
2
in fo rio s_
ar an a
Dλ = − Dλ 2
2x
or
2
ve rs ita rio
iv
ni
.m
:// t
Dλ . d
e/ u
tp s
(c) Fringe width β =
ht
Dλ =d 2
e/ un
://
w
which gives x =
er si ta
w .ja
w
m
Dλ 2 or 2 x 2 = (− Dλ)
s
Dλ Dλ − 2x = 2λ D 2 2
.c o
or 2
m
Dλ Dλ 2 = 2λD − 2 x 2 2
t.m
ht tp
s: //
ht
E
tp s
xample 14 Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength λ. The separation between the sources is 2λ. Consider a line passing through S2 and perpendicular to the line S1S2. What is the smallest distance S2 where a minimum of intensity occurs? )
) Solution Path difference at P, ∆x = d 2 + x 2 − x
∆xmin = 0, when x → ∝, and maxima will occur. ∆xmax = 2λ, when x = 0, again maxima will occur.
3λ 2
d 2 + x2 − x =
3λ 2
or
or
or
For nearest minima ∆x =
3λ d 2 + x2 = + x 2 d 2 + x2 =
2
9λ 2 + x 2 + 3λx 4
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Light (Wave Optics)
6 G [
6
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0LQLPD
0D[LPD
0LQLPD
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s:
ita
un iv
//w
7λ . 12
er si ta rio s
x=
w
or
3λ
9λ 2 4
er s
ar a m w
.ja
=
(2λ) 2 −
rio s_ in fo
'[ O
na .c om
'[ O
or
[
3
'[ O
2 9λ 2 d − 4 x= 3λ
(a)
λD 2d
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
e/
example 15 Consider the situation shown in the fi gure. The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength λ. The separation between the slits is d. The light transmitted by the slits falls on a screen E1 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3. Another screen E2 is placed a further distance D away from E1. Find the ratio of the maximum to minimum intensity observed on E2 if z is equal to
ht
4B.28
(b)
λD d
(c)
λD . 4d
6
G
6
] T
6
6 '
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Light (Wave Optics)
4B.29
)
) Solution Light from sources S1 and S2 get interfered and thereafter S3 and S4
become new sources. At S3 the path difference between the lights coming from S1 and S2 is zero. Therefore they interfere constructively and so a3 = (a + a) = 2a
(a) At S4 the path difference ∆x = d sin θ ≃ d tan θ λD d× z 2d =d = D D λ = 2
s_
an
in
a.
I max (a3 + a4 ) 2 (2a + 0) 2 = =1 = 2a − 0 I min (a3 − a4 ) 2
rio
ar
iv un
tp
Corresponding phase difference =
ht
s:
//t
.m e/
tp
s:
//t
ht
er s
.m e/
tp
I max (2a + 2a ) 2 = =∞ I min (2a − 2a) 2
and
λD d dz 4d λ (c) ∆x = = = D D 4
A42 = a 2 + a 2 + 2aa cos
rio
ita
Now a4 = a + a = 2a
ht
un
iv
//w
Corresponding phase difference = 2π radian
s:
w
w
er si
ta
.ja m
λD d dz d (b) ∆x = = =λ D D
s
The ratio
fo
a4 = 0
∴
co m
Corresponding phase difference = π radian
or
λ radian 2
π = 2a 2 2
A4 = 2a
( (
) )
2 2
2a + 2a I max = I min 2a − 2a
=
(3.414) 2 (0.586) 2
= 34.
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Light (Wave Optics)
example 16 In a modified Young’s double slit experiment, a monochromatic uniform and 10 parallel beam of light of wavelength 6000 Å and intensity W/m2 is incident π normally on two circular apertune A and B of radii 0.001 m and 0.002 m respectively. A perfect transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength of 6000 Å is placed infront of aperture A (see fi gure).
$
)
na .c om
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rio s_ in fo
m
ar a
Calculate the power (in watts) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focus spot.
ita
.ja
) Solution The intensities of light of sources S1 and S2 are
er si ta rio s
un iv
e/
10 I 2 = × π (0.02)2 = 4 × 10–5 W. π
.m
tp
s:
//w
w
er s
w
10 I1 = × π (0.01)2 = 10–5 W π
un
//t
IA = 0.10 × 10–5 W = 10–6 W IB = 0.10 × 4 × 10–5 W = 4 × 10–6 W.
t.m e/
iv
The intensities of sources after emerging from the lenses are
ht tp s:
ht
s: //
The path difference produced due to film
∆x = (µ – 1) t = (1.5 – 1) × 2000 × 10–10 = 10–7 m
and
φ=
ht tp
4B.30
2π 2π π × ∆x = × 10−7 = radian λ 6000 × 10−10 3
Now power received at F I + I A − I B + 2 I A I B cos φ −6 −6 = 10−6 + 4 × 10−6 +2 10 × 4 × 10 cos
π 3
= 7 × 10–6 W
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Light (Wave Optics)
4B.31
4.7 interference in tHin films Thin film may be an oil film spread over water, paint on glass, or air film between two glass plates etc. The phenomenon of interference was first explained by Young. It has been observed that interference in the case of thin film takes place due to (1) reflected light (2) and transmitted light.
na .c om
(1) Interference in reflected light Consider a thin film of transparent material of thickness t and refractive index µ is situated in air. A ray incident at A on the upper surface of the film is partly reflected and partly refracted along AB. At B it is partly reflected along BC and finally emerges out along BG. This process of reflection and refraction goes continuously at several points on the film (theoretically infinity).
L
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un
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W
)
t.m e/ U
W
*
7UDQVPLWWHGOLJKW
ht tp
s: //
ht tp s:
//t
$LU
.m
ht
tp
e/
s:
P
UU
er si ta rio s
un iv
//w
w
er s
w
.ja
UU
iv
$
$LU
ita
m
ar a
'
rio s_ in fo
5HIOHFWHGOLJKWIURPWZRSRLQWV RQWKHILOP
To get the path difference between the light waves from points A and C, drop a perpendicular at D on the reflected ray from A and at E on the ray BC. Thus optical path difference,
∆x = (AB + BC) in medium – AD in air = (AB + BE + EC) in medium – AD in air
We know that AB distance travels in medium of refractive index µ becomes equal to µAB in air, and so for BE and EC. Therefore
∆x = (AB + BE + EC)µ in air – AD in air
...(i)
By Snell’s law; µ=
sin i AD /AC AD = = ∴ AD = µ EC. sin r EC /AC EC
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...(ii)
Light (Wave Optics)
Substituting this value in equation (i), we get ∆x = (AB + BE)µ As AB = A′B, ∴ ∆x = (A’B + BE)µ = (A′E)µ
From the triangle AA′E, A′E = 2t cos r, and so ∆x = 2µt cos r
m
fo
s_
λ = nλ 2
rio
2µt cos r ±
si
er
iv
.m
...(3)
un
//t
tp s:
or
rio s
λ λ = (2n – 1) 2 2 2µt cos r = nλ; n = 1, 2, .....
∴ 2µt cos r ±
...(2)
si ta
un i
e/
tp
s:
//w
w
ve r
w
ta
.ja m
λ 2µt cos r = (2n – 1) ; n = 1, 2, … 2 λ If we take 2µt cos r = (2n + 1) , then n will be 0, 1, 2, 2 (ii) The minima will occur when ∆x = (2n – 1) λ/2
∴ or
e/
//t
ht
Here n = 0 is discarded because path difference 2µt cos r can not be zero. When this condition is fulfilled the film will appear dark in reflected light.
.m
ht
in
a.
ar an
(i) The maxima will occur when ∆x = nλ.
...(1)
co
λ 2
The path difference in equation (iii) is the optical path difference. It has been proved that when light is reflected from optical denser medium (upper surface of the film), it undergoes an abrupt phase change of π rad; an equivalent path difference λ/2. Therefore the effective path difference becomes,
∆xe = 2µt cos r ±
...(iii)
∴ 2µt cos r = nλ; n = 1, 2, (ii) The minima will occur when ∆x = (2n – 1) λ/2. ∴ 2µt cos r = (2n – 1) λ/2; n = 1, 2,
...(4)
ht tp
N
s:
ote For normal incidence, angle of incidence i and hence r become zero and so cos r = 1, then 2µt cos r → 2µt. (2) Interference in transmitted light The optical path difference between the light waves originated from B and F can be calculated as in the case of reflected light. Thus ∆x = 2 µt cos r. In this case, the light rays emerge from B and F has no phase change because light emerges from F has gone two reflections, each with a phase change of π rad. Thus ∆xe = 2µt cos r + 0 = 2 µt cos r. (i) The maxima will occur when ∆x = nλ.
4B.32
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...(5)
Light (Wave Optics)
4B.33
Conclusion: The condition of maxima and minima for interference in reflected and transmitted light are opposite to each other. It means, if any colour is strongly reflected, then it will be absent in transmitted light.
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E
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un
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iv
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D
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er si ta rio s
er s
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tp
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na .c om
example 17 A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in fi gure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25% of the incident on it and transmits the remaining. Find the ratio of the minimum and the maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.
t.m e/
) Solution The intensities of reflected beams at two points on the first plates are shown in figure. From which
s: //
I 9I and I 2 = . 4 64
ht tp
I1 =
The intensities of maxima and minima after their interference are: I max = I1 + I 2 + 2 I1 I 2 and
I min = I1 + I 2 − 2 I1 I 2
∴
I min I1 + I 2 − 2 I1 I 2 = I max I1 + I 2 − 2 I1 I 2
After substituting the values of I1 and I2 and simplifying, we get I min = 1/49. I max https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
colours of thin film When white light is incident on thin film, beautiful colours are observed. The incident light will split up due to reflection at the top and bottom of the film. The splitted rays interfere and gives bright colours. The interfering rays of only certain wavelengths will have a path difference satisfying the conditions of bright fringe. Hence only such wavelengths will have a path difference satisfying the conditions of bright fringe. Hence only such wavelengths will be present there. Other wavelengths will be present with diminished intensity.
2 µt cos r = nλ.
...(i)
L
er si ta rio s
tp
e/
s:
D
un iv
//w
w
er s
w
ita
.ja
m
ar a
For normal incidence r → 0, cos r = 1. 2 µt = nλ
rio s_ in fo
na .c om
interference in wedge shaped film Consider a thin film of refractive index µ is enclosed between the two surfaces inclined at an angle α. For small angle α, the effective path difference between two light waves originated from upper surface can be taken equal to 2 µt cos r. Thus for minima
U
t.m e/
ht tp s:
un
//t
iv
.m
where t is the thickness of the film at the position of nth minima t From the figure tan α = xn or t = xn tan α
ht
s: //
From equation (i), 2µ (xn tan α) = nλ nλ or xn = 2µ tan α
...(ii)
ht tp
4B.34
and
xn – 1
(n − 1)λ 2µ tan α
...(ii) E W D
∴
β = xn – xn – 1 =
[Q± [Q
λ 2µ tan α
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Light (Wave Optics)
4B.35
Newton’s rings: Newton’s rings are formed due to interference between the waves reflected from the top and bottom surfaces of the air film enclosed between the lens and the plate. From the property of circle, we have
r × r = t × (2R – t)
or
r2 = 2Rt – t2 ≃ 2Rt [Since t << R] t=
∴
r2 2R
na .c om
5
U
rio s_ in fo
ar a
$LU
W
W
er si ta rio s iv
'Q
s: //
For bright ring:
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
1HZWRQ¶VULQJE\UHIOHFWHGOLJKW
or 2µ or or or
ht tp
2µt = (2n – 1)λ/2 n = 1, 2, 3,.... r2 λ = (2n − 1) 2R 2 (2 n − 1) λ R r2 = 2µ
D 2 (2n − 1)λR = 4 2µ 2(2n − 1)λR Dn2 = µ
...(i)
Similarly, for dark ring: 2µt = nλ, n = 1, 2, 3
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Light (Wave Optics)
E
or
r2 = nλ 2R nλR r2 = µ Dn2 nλR = 4 µ
Dn2 =
or
or
or 2µ
4nλR . µ
xample 18
si ta
.ja
m
ar
rio s_ i
an
nf
a.
o
co
m
In Young’s double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron is introduced in the path of one of interfering waves. The mica-sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift on the introduction of mica-sheet. Calculate the wavelength of the monochromatic light used in the experiment.
s rio
iv
ta
un
//w
∆=
er
D (µ − 1)t. d
w
)
w
) Solution The shift produced in the fringes is
er si
e/
e/
un
//t
s:
.m
tp
ht
tp
s:
//t
ht
D (2 D)λ (µ − 1)t = c d (µ − 1)t ∴ λ= 2
iv
.m
(2 D)λ . d According to the given condition β=
ht
tp
s:
When distance between slits and screen is doubled, the fringe width
(1.6 − 4) × 1.964 × 10−6 2 = 5892 Å.
=
xample 19 A glass plate of refractive index 1.5 is coated with a thin layer of thickness of t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If the λ = 648 nm. Obtain the least value of t for which the rays interfere constructively. E
4B.36
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Light (Wave Optics)
4B.37
) Solution
W
P P
ar a
2µt = (2n – 1)λ/2, n = 1, 2, 3, λ ∴ t = (2n − 1) 4µ
.ja
er s
w
un iv
w //w
648 × 10−9 = 90 × 10–9 m 4 × 1.8
s:
=
λ λ = 4µ 4µ
er si ta rio s
tmin = (2 × 1 − 1)
∴
ita
m
For least value of t, n = 1
rio s_ in fo
na .c om
Condition of constructive interference in the situation shown in figure is
e/
iv
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
.m
tp
example 20 A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in figure. Calculate the position of the central fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled upto OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid.
) Solution $ 6
3 FP 6
' 2
4
6 P
%
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FP
&
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Light (Wave Optics)
' 3 6
FP
T
T
2
T
4
P
ar a
∆x = (SS1 + S1P) – (SS2 + S2P)
.ja
= (SS1 – SS2) – (S2P – S1P)
un iv
//w
For central bright ∆x = 0
er si ta rio s
w
er s
w
= ∆x1 – ∆x2
0 = ∆x1 – ∆x2
tp
e/
s:
or
∆x1 = ∆x2
or
d sin θ1 = d sin θ2
or
sin θ1 = sin θ2
0.4 + 2
2
=
y 2
un
t.m e/
2
0.1 + y 2
s: //
ht tp s:
//t
iv
.m
or
0.4
FP
ita
m
The path difference,
rio s_ in fo
na .c om
6
6
or
T
'[
T
\
'[
ht
After solving, we get y = 0.02 m = 2 cm.
ht tp
4B.38
When water is filled in the vessel, it causes shift in position of fringe
or
∆ = 2 cm D (µ – 1)t = 0.02 d
0.10 (µ – 1) × 0.1 = 0.02 0.8 × 10−3 which gives
µ = 1.002.
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&
4B.39
Light (Wave Optics)
example 22 In a modified YDSE the region between screen and slits is immersed in a liquid 5 T whose refractive index varies with time as µ l = − until it reaches a steady 2 4 5 3 state value of . A glass plate of thickness 36 µm and refractive index is in4 2 troduced infront of one of the slits. (a) Find the position of central maxima as a function of time and the time when it is at point O, located symmetrically on the x-axis. (b) What is the speed of the central maxima when it is at O?
3 \
er si ta rio s
ita un iv
2
[
iv
' P G PP
t.m e/
ht tp s:
un
//t
ht
.m
tp
W
e/
//w
w
PJ
6
s:
er s
w G
\
P[
.ja
6
m
ar a
rio s_ in fo
na .c om
) Solution (a)
∆x = S 2 liquid P + (µ g − µ )t – (S1P)liquid
ht tp
s: //
The optical path difference between lights start from S1 and S2
= (S2P – S1P)liquid + (µg – µℓ)t = µl(S2P – S1P)air + (µg – µℓ)t or
∆x = µl d
y + (µg – µℓ)t D
For central maxima, ∆x = 0 y 0 = µ d or + (µg – µℓ)t D which gives
y=
− D(µ g − µ )t d µ
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Light (Wave Optics)
3 5 T − D − − t 2 2 4 = 5 T d − 2 4 T D 1 − t D(4 − T )t 4 = = 5 T d (10 − T ) d − 2 4
na .c om
T D 1 − t 4 0= ⇒ T = 4 s. 5 T d − 2 4
dy 6 Dt = dt (10 − T ) 2 d Central maxima is at O at T = 4 s Speed of central maxima v =
ita
6 Dt 6 Dt Dt = = 2 (10 − 4) d 36d 6d
un iv
//w
e/
.m
s:
1× (36 × 10−6 ) = 3 × 10–3 m/s. 6 × (2 × 10−3 )
er si ta rio s
v=
w
w
Thus
er s
.ja
m
ar a
(b)
tp
rio s_ in fo
The time when y becomes zero,
ht
un
//t
iv
example 23 A glass surface is coated by an oil film of uniform thickness 1.00 × 10–4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm – 750 nm) which are completely transmitted by the oil film under normal incidence.
s: //
t.m e/
ht tp s:
) Solution Optical path difference for the light transmitted through oil is = 2µt cos r for normal incidence
ht tp
4B.40
2LO P *ODVV P
∴
r = 0, cos r → 1. ∆x = 2µt
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Light (Wave Optics)
4B.41
But at the interface between oil and glass will produce an additional path difference of λ/2. Therefore effective path difference ∆x = 2µt + λ/2 For constructive interference in transmitted light λ = nλ, n = 1, 2, …. 2 2µt = (2n – 1) λ/2
λ=
4µt (2n − 1)
=
4 × 1.25 × 1× 10−6 2n − 1
=
5 × 10−6 (2n − 1)
o in f
n = 4, λ =
5 × 10−6 m = 714.29 nm 7
n = 5, λ =
n = 6, λ =
ita
5 × 10−6 m = 555.55 nm 9
e/
un
//t
s:
tp
rs
iv e
.m
e/
un
iv
//w
s:
s
rio
5 × 10−6 m = 1000 nm 5
w
n = 3, λ =
5 × 10−6 m = 454.54 nm 11
ht
//t .m
s_
5 × 10−6 m = 1666.67 nm 3
si ta
n = 2, λ =
ht tp
er
w .ja
m
rio
ar
n = 1, λ = 5 × 10–6 m = 5000 nm
For
co
a.
an
m
or
or
2µt +
ht
tp
s:
The wavelength which are strongly transmitted in visible range are : 714.29 nm, 555.55 nm and 454.54 nm
E
xample 24 Two square surface of a transparent plastic block having a shape of as shown in fig. 4.41 have thickness t1 at one edge and t2 at the other. The refractive index of the material is 1.6. When viewed at normal incidence, using a light of wavelength 5000 Å, 20 fringes are observed. Find the difference (t2 – t1). )
) Solution
...(i) ...(ii)
2µt1 = n λ 2µt2 = (n + 20)λ
and
Let, at the thickness t1, nth number of fringe is formed, then at thickness t2, (n + 20)th number of fringe will form. For reflected light
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4B.42
Light (Wave Optics)
W
W
2µ(t2 – t1) = 20λ 10λ (t2 – t1) = µ 10 × (5000 × 10−10 ) = 1.6 = 3.125 × 10–6 m.
ar a
na .c om
∴
er s
w
ita
.ja
m
4.8 diffraction
rio s_ in fo
Subtracting (i) from (ii), we get
iv
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The bending of waves around the edges of an obstacle or aperture is called diffraction. This phenomenon was discovered by Grimaldi. The theorical explanation was first given by Fresnel. According to him, diffraction results from the superposition of secondary wavelets originating from the different parts of the same wavefront. For diffraction to occur, the size of the aperture should be small enough. It should be comparable to the wavelength of wave. In case when size of aperture is very large in comparison to the wavelength, then there is no appreciable diffraction will occur, and so infront of the aperture, there is uniform illumination on the screen (see fi g.).
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Light (Wave Optics)
4B.43
If an obstacle with a small gap is placed in almost like a point source. If the gap is large however, the diffraction is much more limited. Small, in this context, means that the size of the obstacle is comparable to the wavelength of the waves.
difference Between interference and diffraction
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1. Interference takes place between two or more wavefronts originating from coherent sources. While in diffraction superposition takes place between the secondary wavelets originating from the same wavefront.
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2. In the interference pattern the regions of minimum intensity are usually almost perfectly dark. While it is not so in diffraction pattern. 3. The fringe width in interference may or may not be equal, while in diffraction pattern fringe widths are never equal. 4. In interference all the maxima are of same intensity but in diffraction they are of varying intensity.
4.9 fraunHoffer diffraction at single slit In this type of diffraction, a plane wavefront falls on a slit and its response is seen on a plane screen placed at very large distance from the slit. Suppose a plane wavefront falls on a slit of width d. Every point of the exposed part of the wavefront acts as a source of secondary wavelets. These wavelets superpose (interfere) and produce diffraction.
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Light (Wave Optics)
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If the intensity of the principal maxima is I0, then the intensity of the first maxima I I and second order maxima are found to be 0 and 0 respectively. The diffraction 22 61 fringes are of decreasing width and decreasing intensity. In general, for any angular position θ, the intensity I is given by
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4B.44
2
sin α I (θ) = I 0 , α πa where α = sin θ. λ
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Light (Wave Optics)
4B.45
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principal maxima When wavefront strikes the slit, the secondary wavelets from all point in the slit travel about the same distance to reach the centre of the diffraction pattern and thus are in same phase there. So they interfere constructively and produce principal maxima. To understand higher order maximas, we can use a clever technique and the slit can be divided hypothetically into odd number of zones; three zones for first order maxima and five zones for second order maxima and so on.
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Light (Wave Optics)
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positions of first and Higher order minimas To understand first order minima, we can divide the slit hypothetically into two identical zones Z1 and Z2 each of width d/2. Take a wavelet from top of the zone Z1 and other from the top of the zone Z2. The path difference between them is d/2 sin θ. All other similar pairs will have the same path difference. If these wavelets interfere destructively, then they will produce minima. This will happen when d/2 sin θ is equal to λ/2. Thus for first order minima or
d/2 sin θ = λ/2 d sin θ = λ
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Light (Wave Optics)
4B.47
The position of first order minima,
sin θ =
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The principal maxima spread symmetrically about central line in angle 2θ. Thus angular width of principal maxima is ...(2)
λ 2θ = 2sin −1 . d
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λ Linear width of principal maxima; β = (2θD) = 2D sin −1 . d For second order minima, we can divide the slit hypothetically into four zones. The wavelets of nearest two zones interfere destructively and again produce minima. Thus for second order minima, we have
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To understand first order maxima, we can divide the slit hypothetically into three identical zones Z1, Z2, Z3 each of width d/3. The wavelets of two neighbouring zones Z1 and Z2 interfere destructively and cancel their mutual effect. The wavelets of remaining third zone will cause first order maxima. Thus for first order maxima the path difference between two wavelets, one from top of the zone Z1 and other from top of zone the Z2 is d/3 sin θ. For destructive interference between them
Similarly for second order maxima, we can get
a sin θ =
5λ . 2
In general, for higher order maxima, we can write; λ d sin θ = (2n + 1) ; n = 1, 2, 2
N
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The technique used to explain the positions of minima and maxima can not be used to get the intensity of maxima. https://t.me/universitarios_infohttps://www.jamarana.com
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4B.48
Light (Wave Optics)
4.10 diffraction grating A diffraction grating consists of equally spaced, parallel slits. If a is the width of the slit and b is the width of the opaque portion between two adjacent slits, then (a + b) is the distance between the adjacent slits. This is known as grating element. If there are N parallel slits (rulings) in each centrimetre, then grating element is given by
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The grating equation can be written as:
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The above equation gives the position of nth maxima.
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example 25 Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30º.
sin θ = Given,
θ = 30º
∴ sin 30º = or
λ d
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λ 5
1 2 = 2.5 cm.
λ = 5×
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Light (Wave Optics)
4B.49
diffraction by a circular aperture When a plane wavefront falls on a circular hole, it is diffracted by the hole. The response, if received on a screen at a large distance, the pattern is a bright disc surrounded by alternate dark and bright rings of decreasing intensity as shown in fi gure. The position of first dark ring from the hole at an angle θ with the axis 1.22λ , d where λ → wavelength of the light used.
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example 26 Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and falls on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?
R = 1.22 =
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) Solution The radius of central dark can be obtained as λD d
1.22 × ( 560 × 10−9 ) × 2 0.20 × 10−3
= 0.68 cm.
example 27 Angular width of central maximum in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 Å. When the slit is illuminated by light of another wavelength, the angular width is decreases by 30%. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid.
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) Solution Angular width of central maxima = 2λ ′ d According to given condition, we have
2λ . d
For the other light it is = 2λ′ 2λ = (0.70) d d
λ′ = 0.7 λ = 0.7 × 6000 Å = 4200 Å.
⇒
When the apparatus is immersed in liquid, the same decrease is observed in the angular width.
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λ (2λ air )/µ = d d 1 µ = = 1.43 0.7
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example 28 A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?
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4B.50
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1.22 × 620 × 10−9 0.08 = 9.455 × 10–6 rad. =
Since θ is small, so we can take tan θ ≃ sin θ = 9.455 × 10–6 rad By geometry, we have, tan θ = ∴
R = 9.45 × 10–6 D
R D
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Light (Wave Optics)
or
4B.51
R = 9.45 × 10–6 × 0.20 = 1.89 × 10–6 m.
example 29 A wedge of angle 0.5º is illuminated with sodium light whose two lines corresponds to the wavelengths 5890 Å and 5896 Å. Find the distance from the apex at which the maxima due to the two wavelengths first coincide when observed in the reflected light. (The wedge contains air).
) Solution Let the thickness of the wedge at the point where the maximums of both
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coincides be t. For constructive interference in reflected light, we have
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and
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t = (2n + 1)
Let x be the required distance from the apex, then
∴
t = tan θ ≃ θ x t x= θ 0.004 = = 4.58 cm. π 0.5 × 180
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resolution limit and resolving power When we decrease the separation between the two point objects, a limit is reached when our eyes can not differentiate them separately, even they are not in contact. The minimum separation between two closely placed objects, the eyes can distinguish them separately is known as resolution limit (RL). The reciprocal of resolution limit is known as resolving power (RP). Thus RP =
1 . RL
rayleigh Criteria
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According to Rayligh, two images of two objects are said to be just resolved when central maxima of one diffraction pattern falls on first minima of other (see figure).
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4B.52
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(i) The resolution limit of normal human eye is one minute (1´). (ii) The resolution limit of a telescope is given by angle θ, where 1.22λ θ= rad. d Here λ is the wavelength of light used and d is the diameter of objective lens of the telescope. (iii) The resolution limit of a microscope is given by 1.22λ x= metre, 2µ sin θ https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.53
where µ is the refractive index of medium between objects and lens; θ is the angle subtended by the objects at objective lens of the microscope. µ sin θ is called numerical aperture of the lens.
4.11 transverse nature of ligHt
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Interference and diffraction phenomenon prove the wave character of the light. These phenomenon do not give any idea; whether light waves are transverse or longitudinal. Polarisation proves the transverse character of light waves. Transverse character of light waves was known in the early nineteenth century; however the association of electric and magnetic vectors with light wave was known only after Maxwell. The electric field in a light wave propagating in free space is perpendicular to the direction of propagation. Natural light consists of many pulses emitted by different atoms; in general have electric field in different directions. Hence the resultant electric field at a point changes continuously and randomly. Such a light is called unpolarized light. If Eo is the amplitude of the resultant of all the waves, then it can be resolved into two components; Ex and Ey, where Ex = Eo cos θ and Ey = Eo sin θ
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I0 = Ix + Iy = E02 cos 2 θ + E02 sin 2 θ
or
I0 =
E02 E02 + = E02 . 2 2
Thus unpolarized light can be assumed of two components; each of intensity
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I0 . 2
4B.54
Light (Wave Optics)
4.12 polariZation When electric field vector (or magnetic) vibrates only along one direction perpendicular to the direction of propagation, the light is said to be linearly polarized light and the phenomenon associated is called polarization. Thus when unpolarized light is incident on a polarizer, the emerging light becomes linearly polarized of I intensity 0 . 2 ,\
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Linearly polarized light.
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4.13 polariZation By reflection: BreWster’s laW In 1808, Malus found that when natural light is incident of a glass slab, the reflected light becomes partially polarized. Brewter forwarded his study and found that when light is incident on glass slab at certain angle, called polarising angle, the reflected light becomes linearly polarized. This happens when reflected and refracted rays are perpendicular to each other. Consider a beam of natural light is incident on a glass slab at an angle ip as shown in fi gure. A part of it is reflected and a part is refracted. If r is the angle of refraction, then https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.55
r + iP = 90º r = 90º – iP.
∴
By Snell’s law; sin i sin r sin iP = sin(90° − iP )
µ=
=
sin iP cos iP
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polarization by double refraction When unpolarized light (natural light) is incident on a calcite crystal, it splits into two rays; O-ray (ordinary ray) and E-ray (extra-ordinary ray). O-ray has same speed in all directions and hence obeys law of refraction. While E-ray has different speed in different directions in a medium and so does not obey law of refraction. These two rays are plane polarized and their planes of vibrations are mutually perpendicular. https://t.me/universitarios_infohttps://www.jamarana.com
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4B.56
Light (Wave Optics)
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Iθ = (E cos θ)2 = E2 cos2 θ Iθ = I cos2 θ. Malus’ law
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In complete rotation of analyser, we will see two maximum and two zero intensities. For
θ = 0º, Iθ = I cos20º = I θ = 90º, Iθ = I cos290º = 0 θ = 180º, Iθ = I cos2 180º = I θ = 270º, Iθ = I cos2 270º = 0.
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Light (Wave Optics)
4B.57
example 30 Two polaroids are placed at 90º to each other and the transmitted intensity zero. What happens when one more polaroid is placed between these two bisecting the angle between them?
) Solution
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Figure shows the orientation of the polaroids. If I is the intensity of the light emerging from the first polaroid, then intensity of emerging light from second and third are I cos2 45º and (I cos2 45º).sin2 45° These are: I/2 and I/4 respectively.
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example 31 Three nicols prisms are placed such that, first and third are mutually perpendicular. Unpolarized light is incident on first nicol’s prism, the intensity of light emerges from third nicol’s prism is 1/16 the intensity of incident light. Find the angle between first and third nicol’s prisms.
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± T
Thus according to the given condition I0 I cos2 θ sin2 θ = 0 2 16 https://t.me/universitarios_infohttps://www.jamarana.com
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4B.58
1 8 1 (2 sin θ cos θ)2 = 2 1 sin2 (2θ) = 2 (sin θ cos θ)2 =
or
or
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= x2 + d 2 − x
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8 n − n 2 It is easily seen that the path difference decreases from ∆x ≃ 4 m to (∆x → 0 at x → ∞) Hence, n can have value n = 3, 2, 1. Going away from O towards the x-axis, 8 3 7 first maxima occurs at x = − = m (for 3 2 6 n = 3) 8 2 second maxima occurs at x = − = 3 m (for 2 2 n = 2) 8 1 third maxima occurs at x = − = 7.5 m (for 1 2 n = 1)
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Let
= x2 + d 2 − x
Using d = 4 m, λ = 1 m, we have
) Solution Let the detector be at the point P (x, 0) S1 S2 = d = 4 m
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1. Two point sources of radiation: S1 and S2 radiate waves of same frequency and are excited by the same oscillator. They are also in phase with each other. S1 is placed at orgin O, while S2 is placed at the point (0, 4) on y-axis. Find the points of maxima of received intensity if a detector is moved towards positive x-axis starting from origin. The wavelength of radiation is 1 m.
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2. The distance between the coherent sources is 0.3 mm and the screen is 90 cm from the source. The second dark band is 0.3 cm from central bright fringe. Find the wavelength and the distance of the fourth bright fringe from CBF.
6
6 2
)
) Solution Here, d = 0.3 mm = 3 × 10–2 cm [
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1 2
sin 2θ =
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Light (Wave Optics)
D = 90 cm, x2 = 0.3 cm
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)
) Solution
λD 6 × 10−5 × 80 = = 0.096 cm d 0.05 (ii) Distance of the 10th bright fringe from CBF λD x10 = n × d = 10 × 0.096 = 0.96 cm Distance of the 10th dark fringe on the other side of CBF λD ′ = (2n − 1) × x10 d
∴ w=
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4. A biprism is placed at a distance of 5 cm in front of a narrow slit illuminated by monochromatic light of wave length 6 × 10–5 cm. The virtual images formed by the biprism are 0.05 cm apart. Find the fringe width and a screen 75 cm from the biprism. Also
d=
180 × 6000 × 10−10 = 3.44 ×10-4 m 0.1 π
6. A beam of light constituting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment. (i) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å.
s:
10 × 589.3 × 10−9 m × 100 × 10−2 12 × 10−3 = 4.9 × 10–4 m = 0.049 cm
= 1.872 cm.
5. In a Young’s double slit experiment, the angular width of a fringe formed on a distant screen is 0.1º. The wavelength of the light used is 6000 Å. What is the spacing between the slits? w λ ) Solution Angular width = = D d λ 0.1π ⇒ = d 180
=
rs
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nλD d Given xn = 12 mm, n = 10 λ = 589.3 mm,
s
rs
un
.m e/
//t
s:
)
distance between the central line
D = 100 cm nλD d = xn
1 = (2 × 10 − 1) × × 0.096 2 = 0.912 cm ∴ Required distance = x10 + x´10
iv e
w
w
s: //
tp
ht
) Solution For the nth maximum fringe, the xn =
in f
o
ar
w
3. A double slit experiment is performed with sodium (yellow) light of wavelength 589.3 mm and the interference pattern is observed an a screen 100 cm away. The length bright fringe has its centre at a distance of 12 mm from the central maximum. Find the separation between the slits.
.ja
m
an
a.
co
m
or
s_
2 λ = × 10−4 cm 3 2 or λ = ×104 Å 3 th (ii) For n bright fringes, λD xn = n d 2 90 −4 Putting n = 4, x4 = 4 × × 10 × 3 3 × 10−2 = 0.8 cm
3 90λ 0.3 = × 2 3 × 10−2
⇒
(i) Here, d = 0.05 λ = 6 × 10–5 m D = distance between the screen and virtual images = 75 + 5 = 80 cm
Putting n = 2,
λD 2d x2 = (2 × 2 – 1) × λ × 90 2 × 3 × 10−2
xn =(2n – 1) ×
4B.59
find the distance between the 10th bright band on one side and 10th dark band on the other side of the central bright fringe.
(i) For nth dark fringes,
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4B.60
∴
Icenter = Imax (∆ φ = 0)
2π (Path λ difference) 2π λ π = × = λ 8 4
At the given point, ∆ φ =
I = I max cos 2
(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.
Light (Wave Optics)
)
) Solution Given: D = 120 cm. d = 0.2 cm
∆φ π = I max cos 2 2 8
I 1 + (1/ 2 ) 2 +1 = = I max 2 2 2
∴
in
s_
an
ht
Hence, fourth bright fringe due to λ1 coincides with fifth bright fringe due to λ2 at distance
x=
si ta
un
iv er
si ta
rio
s
er
t.m e/
s: //
s:
m 4 = n 5
6500 m = 5200 n
P
$
tp
⇒
//t
tp
⇒ ⇒
Dλ1 Dλ 2 and x = n d d m λ1 = m λ2 x=m
ht
∴
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ht
tp
e/
un
s: //
iv
w
w
w
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m
rio
ar
fo
co
a.
3Dλ1 3 × 120 × 6500 × 10−8 = = 0.117 cm d 0.2 Dλ1 w1 = (ii) d Dλ 2 and w2 = d th Let the m bright fringe due to λ1 and nth bright fringe due to λ2 coincide at a distance x from central maximum.
8. In the figure shown, S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two are placed symmetrically about the central axis so with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO. (i) If the third intensity maximum occurs at the point A on the screen, find the distance OA. (ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease or remain the same?
m
Let λ1 = 6500 Å and λ2 = 5200 Å (i) Distance of nth bright fringe from central maximum nDλ = d ∴ Distance of third bright fringe
P
6
6FUHHQ
4Dλ1 d
P
)
) Solution At central maximum, constructive interference occurs.
)
) Solution
For L1 and L2: u1 = u2 = 0.15 m, and f = 0.1 m
1 1 1 + = u v f
or
1 1 1 1 1 = − = − v f u 0.1 0.15
∴
4 × 120 × 6500 × 10−8 = = 0.156 cm. 0.2 7. The path difference of a certain point on screen in a double-slit experiment is oneeight of the wavelength. Find the ratio of intensity at this point and the intensity at central maximum.
2
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Light (Wave Optics)
4B.61
6 [ / I 6
[
[
3
2 [ /
I
ita er s
er si ta rio s
un iv
G
t.m e/
un
iv
G
D FP
s: //
= 2 × 10 ×
∴
w=
=
λD d
π π (1.5 –1) = cm 90 9
5890 × 108 × 100 = 0.01687cm. π 9
10. In Young’s experiment, the source is red
)
) Solution Here, λ = 5890 × 10–8 cm
d 2a d = 2a × A (µ – 1)
tan δ ≅ δ =
∴
rio s_ in fo
ar a e/
ht tp
∴
(ii) If gap between L1 and L2 is reduced from its original value of 0.5 mm, then d will reduce and hence OA will increase. 9. A Fresnel’s biprism of angle 2º is illuminated by light of wavelength λ = 5890 Å by a slit source 10 cm from it. Calculate the fringe width on a screen 90 cm from biprism, if the R.I. of biprism is 1.5.
G
The minimum deviation of the thin prism is given as δ = A(µ – 1)
3 × 500 × 10−9 × 1 = 1 × 10–3 m. 1.5 × 10−3
D = 90 + 10 = 100 cm
6
$
6
.m
ht tp s:
=
µ = 1.5
π radians 90 The distance d is found as shown:
//t
ht
tp
s:
//w
w
w
Hence distance between two source S1 and S2 = 3 x = 1.5 mm (i) The maximum intensity on the screen is observed, when the distance of maximum from O, λD xn = n d Given: n = 3, D = 1.3 – 0.3 = 1m d = S1S2 = 1.5 mm, λ = 500 nm ∴ OA = x3
6
Angle of biprism, A = 2º =
m
⇒
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v = 0.3 m v1 v2 = =2 u1 u2
na .c om
X X P
or
[
Y Y P
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4B.62
11. In Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index. 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum earlier, lies below the point P while the sixth minimum lies above P. Calculate the thickness of the glass plate. (Absorption of light by glass plate may get neglected.)
6
ta
si
ta rio er si
iv un e/
= (1.7 – 1.4) t = 0.3t
...(i)
Since, P lies in between the original fifth maximum and sixth minimum, this path difference will also be given as Path difference = 5λ + ∆x
...(ii)
Equating equations (i) and (ii), we get
The change in fringe width due to the change in wavelength D w = w2 – w1 = (λ g − λ r ) d = 285.71 × (5 × 10–7 – 7 × 10–7) m
s
er
Path difference = (µ2 – µ1) t
6 × 7 × 10−7 + 1 = 1.6 7 × 10−6
6FUHHQ
µg =
WKPLQLPXP
0.3t = 5λ + ∆x
tp s
ht
or
3
The optical path difference between the two rays reaching P from the two slits S1 and S2 after the insertion of glass plates of equal thickness and having refractive indexes µ1 and µ2 respectively is given as
ht
6 × 7 × 10–7 = (µg – 1) × 7 ×10–7
or
WKPLQLPXP
WP
:// t.m
tp
s:
Now, for green light (λ = 5 × 10–7 m), we have ∆x = (µg – 1) t
= – 5.71 × 10–5 m.
WP
iv
un
e/
ht
5 × 7 × 10−8 = 7 × 10–6 m 1.5 –1 It is also given that the central bright fringe shift by 10–3 m, hence, from the expression λD xn = n d 10−3 D xn = = 285.71 ∴ = d nλ 5 × 7 × 10−7 =
6
//t .m
tp
s:
//w
w
w
.ja
where, µr (=1.5) is the refractive index of glass plate for red light and t is the thickness of the plate. Given: ∆x = 5λ. ∆x ∴ t= µr − 1
s_
)
m
ar
) Solution
in
an
by the insertion of glass plate in the path of one of the interfering beam is given as ∆x = (µr – 1) t
rio
)
a.
) Solution The optical path difference introduced
fo
co m
light of wavelength 7 × 10–7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beam, the central bright fringe shift by 10–3 m to the position previously occupied by the fifth bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10–7 m, the central fringe shift to a position initially occupied by the sixth bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength.
Light (Wave Optics)
...(iii)
The path difference ∆x can be determined from the given intensity at P, which is
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) Solution
)
According to the expression I ∆φ = cos 2 I0 2 2π ∆x Here ∆φ = λ 2
4B.63
(ii) If the incident beam makes an angle of 30º with the x-axis (as in the dotted arrow in the figure), find the y co-ordinates of the first minima on either side of the central maximum.
I 3 = I0 4
Light (Wave Optics)
I 2π ∆x 3 = cos 2 . = I0 λ 2 4
∴
\ T
G
3 π∆x cos = λ 2 π∆x π or = λ 6 λ or ∆x = 6 Now, thickness of the glass plates t is given by
2
rio
ta
si
un iv er e/
Thus, the allowed values of m are + 1, 0, –1 and – 2. Hence, four minima will be observed on the screen; distance on the screen is given by
;
For m = 1, sin θ =
3 3 , y= m 4 7
For m = 0, sin θ =
1 1 , y= m 4 15
For m = –1, sin θ =
1 –1 m , y=– 4 15
For m = –2, sin θ =
3 –3 m , y=– 4 7
6FUHHQ
' P
y = D tan θ = 1.0 sin θ sin θ = 1.0 1 − sin 2 θ cos θ
G
1 λ 1 sin θ = m + = m + 2D 2 0.5 mm 1 1 = m+ 1.0 mm 2 2
ht
tp
s: //
t.m
ht tp s:
e/
.m
//t
ht
12. A coherent parallel beam of microwave of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 mm is as shown in the figure. (i) If the incident beam falls normally on the double slit apparatus, find the y co-ordinates of all the interference minima on the screen.
1 d sin θ = m + λ; m = 0, 1, 2, … 2 Only those values of m are allowed for which sin θ does not exceed ± 1. For the given data, we get
s
si ta
un iv er
w
w
tp
s:
//w
or
(i) The interference minima will occur on the screen provided
λ 6 1 31 t= × λ 0.3 6 1 31 = × × 5400Å 0.3 6 = 9.3 × 104 Å = 9.3 × 10–6 m
0.3t = 5λ +
' P
rio s_ in fo
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ar
an a
.c
om
or
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4B.64
(ii) Find the deviation of the beams transmitted through the face AC. (iii) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and the lower beams, immediately after transmission from the face AC, are 4I and I respectively, find the resultant intensity at the focus.
$
an a.
s_
ta
si
rio s
ta
.m e/
4
3
tp s:
H
G
c
G c
4
&
%
= 0.8 × 10–14 m2
∴ b = (1.25 – 1.20) (4 × 10–7)2 = 0.05 × (4 × 10–7)2
0.8 × 10−14 25 × 10−14 = 1.20 + .032 = 1.232 Let, e = angle of emergence of the 5000 Å ray of light
(ii) µ5000 = 1.20 +
where, λ is in Å and b is a positive constant. The value of b is such that the condition for total reflection at the face AC is just satisfied for one wavelength and is not satisfied for the other. (i) Find the value of b.
$
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ht
13. Two parallel beams of light P and Q (separation d) containing radiation of wavelengths 4000 Å and 5000 Å (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in the figure. The refractive index of the prism as a function of wavelength is given by the relation. b µ(λ) = 1.20 + 2 λ
1 1 = sin θ 0.8 b = 1.20 + (4 × 10−7 ) 2
(i) µ 4000 =
//t
ht tp
)
m
//t .
s:
First minimum below the central maximum occurs when –1 1 m = –2, sin θ = , y = – m 4 15
and smaller the critical angle. For just total internal reflection, θ = critical angle. Critical angle for 4000 Å = θ
er
e/ un
tp s:
1 1 m = –1, sin θ = , y = m 4 15
ht
&
) Solution Greater the µ, smaller the wavelength
iv
//w
w
w
1 1 This gives sin θ = , y = m 2 3 First minimum above the central maximum occurs when
%
rio
ar
For the central maximum, we have d sin θ – d sin α = mλ, where, m = 0
.ja m
1 3 = m+ 2 2
6LQT
G
fo
1 11 +m+ 2 22
m
=
co
1 0.5mm = sin 30° + m + 2 1.0 mm
in
1λ sin θ = sin α + m + 2d
si
1 d sin θ – d sin α = m + λ, 2 m = 0, 1, 2, … From given data, we get
er
(ii) For angle of incidence, α = 30º, the condition for minima is
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= EF sin θ (µ5000) in air
or sin e = 0.8 × 1.232 = 0.9856 ∴ e = 80.27º Alow, θ = sin–1 (0.8) = 53.13º Then δ = deviation for 5000 Å = 80.27º – 53.13º = 27.14º
sin e = EF sin θ in air sin θ = EF sin e (in air)
Hence, when the converging lens foci P and Q at its principal focus, the beams have equal paths, i.e., phase difference = zero, but different intensities IP and IQ respectively. ∴ Intensity at the focus
$ T *
E
IO = I P + IQ + 2 I P IQ cos θ
ar a m
N
&
ita
er si ta rio s
■ The formula I0 = I1 + I2 + 2 I1I 2 cos θ is applicable only if the interfering beams are of the same single frequency.
type questions )
) Explanation Due to introduction of glass
plate, additional path difference = (µ – 1) t. For the intensity to be same as before, (µ – 1) t = nλ For minimum value of t, n = 1 ⇒ t = 2λ
2. Two coherent monochromatic light beams of intensities I and 4I are superimposed. The maximum and minimum possible intensities in the resulting beam are
ECTIVE
O
BJ
1. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is 2λ (a) 2λ (b) 3 λ (c) (d) λ 3
SOLVED
ht tp
s: //
t.m e/
un
//t
ht tp s:
ote: ■ If the data means P is 4000 Å and Q is 5000 Å and the lens refracts the two beams: (i) They will not come to the same focus. (ii) The interference affect is not constant at any position.
er s
.m
ht
un iv
e/
s:
tp
//w
w
w
.ja
(iii) The two beams of 5000 Å (P and Q) emerging from the face AC, have same path FG in air, till they reach FG (a) The beam P passes KE in prism + EG in air up to G. (b) The beam Q passes (LG + GH) in prism to reach F. It can be shown that the two paths of P and Q up to FG are equal as shown below: KE (in glass) = DH (in prism) EG (in air) = EF cos α
=9I
%
= 4I + I + 2 4I × 1
)
iv
+
4
rio s_ in fo
G
na .c om
D
(
3
4B.65
= EF sin e (in air) HF (in glass) = EF sin θ (in prism)
sin e = 1.232 sin θ
∴
Light (Wave Optics)
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4B.66
5I and I 5I and 3I 9I and I 9I and 3I
(b) 1/2 (d) 16
(a) 1 (c) 4
) Explanation Two waves of a single source )
(a) (b) (c) (d)
Light (Wave Optics)
having an amplitude A interfere. The resulting amplitude Ar is given by A 2r = A12 + A 22 + 2A1A cos δ
)
) Explanation Intensity ∝ (Amplitude)2 i.e., I ∝ A2 When two waves (beams) of amplitude A1 and A2 superimpose, at maxima and minima, the amplitude of the resulting wave are (A1 + A2) and (A1 – A2) respectively. If the maximum and minimum possible intensities are Imax and Imin respectively, then
where A1 = A2 = A and δ = phase difference between the waves
When the maxima occurs at the centre, δ = 0 ∴ Ir = 4I …(i) Since the phase difference between two successive fringes is 2π, the phase difference between two points separated by a distance equal to one quarter of the distance between the two successive fringes,
in f
a. an
s_
si ta
s
si
4I =2 2I
er
I r1
I r2
iv
un
=
4. The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringes from the centre is
)
//t
s:
tp
ht
600 n2 = 12 = 18 400
.m
e/
5. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference π between the beams is at point A and π 2 at point B. Then the difference between the resultant intensities at A and B is (a) 2 I (b) 4 I (c) 5 I (d) 7 I π ) Explanation IA = I + 4I + 2 1.41 cos = 5I 2
s: //
tp
ht
…(ii)
ta
un
t.m e/
s:
tp
ht
)
rio
er iv
//w w
3. In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is (a) 12 (b) 18 (c) 24 (d) 30 Dλ and ) Explanation Fringe width, w = d n1w1 = n2w2 ∴
π/ 2 I r2 = 4 I cos 2 = 2I 2 From equations (i) and (ii), we get ⇒
Imax = 9I, and Imin = I
⇒
1 π δ = ( 2π ) = radians 4 2
rio
m ar
I max I min
w .ja
Now,
2
A1 + 1 A1 + A 2 A 2 9 = = =A A1 – A 2 1 − 1 1 A 2 2
A1 I 1 = = A2 4I 2
o
co
m
2
Imax ∝ (A1 + A2) and Imin ∝ (A1 – A2) 2
I r = I1 + I 2 + 2 I1I 2 cos δ
∴
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IB = I + 4I + 2 1.4I cos π = I IA – IB = 4I
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+
er si ta rio s
iv
un
7. In Young’s double slit experiment, the separation between two coherent sources S1 and S2 is d and the distance between the source and screen is D. In the interference pattern, it is found that exactly in front of one slit, there occurs a minimum. Then the possible wavelengths used in experiment are
t.m e/
ita
6. In Young’s double slit experiment, we get 60 fringes in the field of view of monochromatic light of wavelength 4000 Å. If we use monochromatic light of wavelength 600 Å, then the number of fringes obtained in the same field of view is (a) 60 (b) 90 (c) 40 (d) 1.5
un iv
e/
.m
//t
ht tp s:
d2 , d2 , d2 ... D 3D 5D
(b) λ =
d2 , d2 , d2 ... D 5D 9D
(c) λ =
d2 , d2 , d2 ... D 2D 3D
(d) λ =
d2 , d2 , d2 ... 3D 7D 11D
s: //
ht tp
(a) λ =
8. A parallel beam of light is incident on a thin film of air of uniform thickness.
5. A monochromatic light source of wavelength λ is placed at S. Three slits S1, S2 and S3 are equidistant from the source S and the point P on the screen. λ 2λ . S1 P – S 2 P = and S1 P – S 3 P = 6 6
If I be the intensity at P when only one slit is open, the intensity at P when all the three slits are open is (a) 3 I (b) 5 I (c) 8 I (d) zero
er s
tp
ht
) &
rio s_ in fo
na .c om
ar a
.ja
w
w
s:
//w
E
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4. In a double slit experiment, the distance between the slits is d. The screen is at a distance D from the slits. If a bright fringe is formed opposite to a slit, its order is (λ is wavelength of light used) d D2 (a) (b) 2Dλ 2dλ 2 d2 d (c) (d) 2Dλ 2Dλ
QUESTIONS
4
m
3. If white light is used in a Young’s doubleslit experiment, (a) bright white fringe is formed at the centre of the screen (b) fringes of different colours are observed clearly only in the first order. (c) the first-order violet fringes are closer to the centre of the screen than the firstorder red fringes (d) all of these
*
D
G
x (b) I0 cos2 β I 2 πx (d) 0 cos 4 β
(
3
x (a) I0 cos β πx (c) I0 cos2 β
T
2. In a Young’s double-slit experiment, let β be the fringe width, and let I0 be the intensity at the centre bright fringe. At a distance x from the central bright fringe, the intensity will be
4B.67
$
TYPE
1. In Young’s double slit experiment, we get 60 fringes in the field of view of monochromatic light of wavelength 4000 Aº. If we use monochromatic light of wavelength 6000 Aº, then the number of fringes obtained in the same field of view is (a) 60 (b) 90 (c) 40 (d) 1.5
O
unsolved BJ
ECTIVE
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4B.68
fo
in
s_
rio
s
rs ita
ve
15. Microwaves of frequency 3 × 104 MHz and ultrasonic waves of wavelength 1cm are passed through a slit of width 2 cm. Diffraction will occur (a) only in the microwaves. (b) only in the ultrasonic waves. (c) in both, but the diffraction patterns will be different. (d) in both and the diffraction patterns will be identical.
( µ 2 − 1) t
16. Let S1 and S2 be the two slits in Young’s double slit experiment. If central maxima is observed at P and angle ∠S1PS2 = θ, then the fringe width for the light of wavelength λ will be
11. A double slit of separation 0.1 cm is illuminated by white light. A coloured interference pattern is formed on a screen 100 cm away. If a pin hole is located in this screen at a distance of 2 mm from the central fringe, the wavelengths in the visible spectrum which will be absent in the light transmitted through the pinhole are
e/ //t s: tp
ht
(d)
( µ 2 − µ1 ) t
(c)
.m
ht tp µ (b) 1 − 1 t µ 2
µ2 (a) − 1 t µ1
ta rio
si
un i
s:
2
W
14. In the Young’s experiment with sodium light, the slits are 0.589 m apart. What is the angular width of the forth maximum? Given that λ = 589 nm (a) sin–1(3 × 10–6) (b) sin–1(3 × 10–8) (c) sin–1(0.33 × 10–6) (d) sin–1 (0.33 × 10–8)
un
e/ .m //t
ht
w
w
s: //
tp
6
6
13. In a two slit experiment with white light, a white fringe is observed on a screen kept behind the slits. When the screen is moved away by 0.05m, this white fringe (a) does not move at all (b) gets displaced from its earlier position (c) becomes coloured (d) disappears
iv er
w
.ja
m
10. A Young’s double slit experiment is conducted in water (µ1 as shown in the figure, and a glass plate of thickness t and refractive index µ2 is placed in the path of S1. Find the magnitude of the optical path difference at ‘O’. :DWHU
12. A plane monochromatic light wave falls normally on a diaphragm with two narrow slits separated by 2.5 mm. The fringe pattern is formed on a screen 100 cm behind the diaphragm. By what distance will these fringes be displaced, when one of the slits is covered by a glass plate (µ = 1.5) of thickness 10 µm? (a) 2 mm (b) 1 mm (c) 3 mm (d) 4 mm
co m
ar an
3 λ (d) 1 4 λ2
a.
4 λ1 3 λ2
3 λ (c) 2 4 λ1
(b)
4 λ2 3 λ1
5714 Å and 4444 Å 6000 Å and 5000 Å 5500 Å and 4500 Å 5200 Å and 4200 Å
9. In Young’s double slit experiment, the 8th maximum with wavelength µ1 is at a distance d1 from the central maximum and the 6th minimum with a wavelength λ2 is at a distance d2. Then d1/d2 is equal to (a)
(a) (b) (c) (d)
Wavelengths of 7200 Å and 5400 Å are observed to be missing from the spectrum of reflected light viewed normally. The other wavelengths in the visible region missing in the reflected spectrum is (a) 6000 Å (b) 4320 Å (c) 5500 Å (d) 6500 Å
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8. (b)
14. (a)
15. (d)
16. (a)
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7. (a)
t.m e/ 3. In a Young’s double slits experiment two narrow slits 0.8 mm apart are illuminated by the same source of light (λ = 5893 Å). How far apart are adjacent bright bands in the interference pattern observed on a screen 2 m away? [Roorkee, 1982] [Ans. 0.147 cm]
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4. In Young’s double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron is introduced in the path of one of the interfering waves. The mica sheet is then removed
2. Two slits are separated by a distance of 0.03 cm. An interference pattern is produced on a screen 1.5 m away. The fourth bright fringe is at a distance of 1 cm from the central maximum. Determine the wavelength of light used. [Roorkee, 1980] [Ans. 5000 Å]
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1. In Young’s experiments for interference of light, the slits are 0.2 cm apart and illuminated by yellow light (λ = 5896 Å). What will be the fringe width on a screen placed 1 m from the slit? [Roorkee, 1977] [Ans. 0.295 mm]
6. (c)
1
ERCISE
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ANSWERS
(a) 2λ
18. In a double-slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern the intensity of
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X (m1 + m2 )
(d)
X m1 + m2
19. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is
iv
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(d)
17. In the displacement method, a convex lens is placed in between an object and a screen. If the magnification in the two positions be m1 and m2 and the displacement of the lens between the two positions is X, then the focal length of the lens is X X (a) (b) m1 × m2 m1 − m2 (c)
4B.69
(a) both the maxima and the minima increase (b) the maxima increases and the minima has zero intensity (c) the maxima decreases and that of the minima increases (d) the maxima decreases and the minima has zero intensity.
(b) λθ
λ θ 2λ (c) θ (a)
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4B.70
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10. In a modified Young’s double slit experiment, a monochromatic, uniform and parallel beam of light of wavelength 6000 10 Å and intensity W/m2 is incident π normally on two circular apertures A and B of radii 0.001 m and 0.002 m, respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength of 6000 Å is placed in front of aperture A Calculate the power of the lens (in watt) received at the focal spot F. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot. [IIT, 1998] [Ans. 7 × 10–6 W]
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7. A two-slit Young’s experiment is done with monochromatic light of wavelength 6000 Å. The slits are 2 mm apart and the fringes are observed on the screen placed 10 cm away from the slits. Now a transparent plate of thickness 0.5 mm is placed in front of one of the slits and it is found that the interference pattern shifts by 5 mm. What is the refractive index of the transparent plate? [Roorkee, 1985] [Ans. 1.2]
9. A Young’s double-slit arrangement produces interference fringes for sodium light (λ = 5890 Å) that are 0.20º apart. What is the angular fringe separation if the entire fringe separation is immersed in water? (Refractive index of water = 4/3) [Roorkee, 1988] [Ans. 0.15º]
8. Consider interference between two source intensities I and 4 I. Obtain intensities at points where the phase difference is
[Roorkee, 1987]
[Ans. (a) 5 I, (b) I]
5. In a Young’s double slit experiment, the angular width of a fringe formed on a distant screen is 0.1º. The wavelength of the light used is 6000 Å. What is the spacing between the slits? [Roorkee, 1983] [Ans. 3.44 × 10–4 m]
6. A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of slits and the screen is 120 cm. [IIT, 1985] [Ans. (a) 1.17 mm, (b) 1.56 mm]
(b) π
(a) π 2
and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or, minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of monochromatic light used in the experiment. [IIT, 1983] [Ans. 5892 Å]
11. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in the figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25 percent of the light incident on it and transmits the remaining. Find the ratio of the minimum and maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate. [IIT, 1990]
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4B.71
2
1
12. In a two-slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10–2 m towards the slits, the change in fringe width is 3 × 10–5 m. If the distance between the slits is 10–3 m, calculate the wavelength of light used. [Roorkee, 1992] [Ans. 6000 Å]
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16. A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If λ = 648 nm, obtain the least value of t for which the rays interfere constructively. [IIT, 2000] [Ans. 4µt = (2n – 1) λ, 90 nm]
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15. Unpolarized light of intensity 32 Wm– 2 passes through three polarisers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is 3 Wm–2, what is the angle between the transmission axes of the first two polarizers? At what angle will the transmitted intensity be maximum? [Roorkee, 1995] [Ans. 30º, 45º]
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13. In the given figure, S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 cm. The screen at O is normal to SO.
1.30 m
(a) If the third intensity maximum occurs at the point A on screen find the distance OA.
17. A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass pates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and
L2 0.15 m
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14. A monochromatic light of λ = 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. The interference pattern is seen on a screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10–6 m and refractive index µ = 1.5 is placed between one of the slits and the screen. Find the intensity at the centre of the screen if the intensity is I0 in the absence of the plate. Also find the lateral shift of the central maximum. [Roorkee, 1993] [Ans. 0, 1.5 mm]
(b) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease or remain the same? [IIT, 1993] [Ans. (a) 1.0 mm, (b) Increases]
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4B.72
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18. A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB (see figure). The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. (a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure) (c) If the intensity at point P corresponds to a maximum calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum. [IIT, 2002] [Ans. (a) circular, (b) 16, (c) 300 nm]
passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure below. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled up to OQ. The central bight fringe is found to be at Q. Calculate the refractive index of the liquid. [IIT, 2001]
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(a) the speed of light in vacuum, (b) the speed of light in a medium (say, glass or water), depend?
7. In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? 8. When a tiny circular obstacle is placed in the path of light from a distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
On which of these factors, if any, does
4. Let us list some of the factors, which could possibly influence the speed of wave propagation (i) nature of the source. (ii) direction of propagation. (iii) motion of the source and/or observer. (iv) wavelength. (v) intensity of the wave.
9. Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
10. Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
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6. In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
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3. Huygens’ principle leads to the laws of reflection and refraction, use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
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2. Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
5. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
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1. What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations? (a) The screen is moved away from the plane of the slits (b) The (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength (c) The separation between the two slits is increased (d) The source slit is moved closer to the double-slit plane (e) The width of the source slit is increased (f) The monochromatic source is replaced by a source of white light.
QUESTIONS
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4B.73
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4B.74
11. When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
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8. In a YDSE experiment if a slab whose refraction index can be varied is placed in front of one of the slits then the variation of resultant intensity at mid-point of screen with ‘µ’ will be best represented by (µ ≥ 1). [Assume slits of equal width and there is no absorption by slab; mid-point of screen is the point where waves interfere with zero phase difference in absence of slab]
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4. Two coherent light sources each of wavelength λ are separated by a distance 3λ. The maximum number of minima formed on line AB which runs from –∞ to +∞ is 3
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3. In Young’s slit experiment 10th order maximum is obtained at the point of observation in the interference pattern for λ = 7000 Å. If the source is replaced by another one of wavelength 5000 Å then the order of maximum at the same point will be (a) 12 th (b) 14 th (c) 16 th (d) 18 th
7. A beam of light parallel to central line AB is incident on the plane of slits in a YDSE experiment as shown. The number of minima obtained on the large screen is n1. Now if the beam is titled by some angle (≠ 90º) as shown in the figure, then the number of minima obtained is n2. Then
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6. A parallel beam of monochromatic light is incident on a narrow rectangular slit of width 1mm. When the diffraction pattern is seen on a screen placed at a distance of 2 m the width of principal maxima is found to be 2.5 mm. The wave length of light is (a) 6250 Å (b) 6200 Å (c) 5890 Å (d) 6000 Å
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5. When wave of wavelength 0.2 cm is made incident normally on a slit of width 0.004 m, then the semi-angular width of central maximum of diffraction pattern will be (a) 60º (b) 30º (c) 90º (d) 0º
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1. The two coherent sources of intensity that ratio 2 : 8 produce an interference pattern. The values of maximum and minimum intensities will be respectively. (a) Il and 9 Il (b) 9Il and Il (c) 2Il and 8Il (d) 8Il and 2Il 2. In young’s double slit experiment, the distance between two slits is made three times then the fringe width will become (a) 9 times (b) 1/9 times (c) 3 times (d) 1/3 times
12. The principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
(b) 4 (d) 8
(a) 2 (c) 6
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14. In the figure shown, if a parallel beam of white light is incident on the plane of the slits then the distance of the nearest white spot on the screen from O is (assumed d << D, λ << d]
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9. A thin oil film of refractive index 1.2 floats on the surface of water (µ = 4/3). When a light of wavelength λ = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is (a) 10–7 m (b) 2 × 10–7 m (c) 3 × 10–7 m (d) 5 × 10–7 m
13. In Young’s double slit experiment the two slits are illuminated by light of wavelength 5890Å and the distance between the fringes obtained on the screen is 0.2º. If the whole apparatus is immersed in water then the angular fringe width will be, it the refractive index of water is 4/3. (a) 0.30º (b) 0.15º (c) 15º (d) 30º
10. Light of wavelength 6328 Å is incident normally on slit having a width of 0.2 mm. The width of the central maximum measured from minimum to minimum of diffraction pattern on a screen 9.0 metres away will be about
2
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(b) d/2 (d) d/6
(a) 0 (c) d/3
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12. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is (a) 2I (b) 4I (c) 5I (d) 7I
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11. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-length λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is (a) 2λ (b) 2λ/3 (c) λ/3 (d) λ
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4B.75
(b) 0.18º (d) 0.09º
(a) 0.36º (c) 0.72º
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4B.76
15. The intensities of two sources are I and 9I respectively. If the phase difference between the waves emitted by them is π then the resultant intensity at the point of observation will be (a) 3I (b) 4I (c) 10I (d) 82I
19. When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The
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23. The equation of two light waves are y1 = 6 cos ωt, y2 = 8 cos (ωt + φ). The ratio of maximum to minimum intensities produced by the superposition of these waves will be (a) 49 : 1 (b) 1 : 49 (c) 1 : 7 (d) 7 : 1
18. When a mica sheet of thickness 7 microns and µ = 1.6 is placed in the path of one of interfering beams in the biprism experiment then the central fringe gets at the position of seventh bright fringe. The wavelength of light used will be (a) 4000 Å (b) 5000 Å (c) 6000 Å (d) 7000 Å
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22. The intensity ratio of two waves is 9 : 1. These waves produce the event of interference. The ratio of maximum to minimum intensity will be (a) 1 : 9 (b) 9 : 1 (c) 1 : 4 (d) 4 : 1
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17. In Fresnel’s biprism experiment, the width of 10 fringes is 2 cm which are formed at a distance of two 2 metre from the slit. If the wavelength of light is 5100 Å then the distance between two coherent sources will be (a) 5.1 × 10–4 m (b) 5.1 × 104 cm (c) 5.1 × 10–4 mm (d) 10.1 × 10–4 cm
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21. Consider interference between waves from two sources of intensities I and 4I. Find intensities at points where the phase difference is π. (a) I (b) 5I (c) 4I (d) 3I
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24. Interference fringes were produced using white light in a double slit arrangement. When a mica sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some a distance. This distance is equal to the width of 30 interference bands if light of wavelength 4800Å is used. The thickness (in 1 µm) of mica is
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20. From a medium of index of refraction n1, monochromatic light of wavelength λ is incident normally on a thin film of uniform thickness L (where L > 0.1λ) and index of refraction n2. The light transmitted by the film travels into a medium with refractive index n3. The value of minimum film thickness when maximum light is transmitted if (n1 < n2 < n3) is
16. A ray of light is incident on a thin film. As shown in figure M, N are two reflected rays and P, Q are two transmitted rays. Rays N and Q undergo a phase change of π. Correct ordering of the refracting indices is Q
thickness of the film will be, if the wavelength of light is 5890Å. (a) 6.544 × 10–4 cm (b) 6.544 × 10–4 m (c) 6.54 × 10–4 cm (d) 6.5 × 10–4 cm
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25. In a Young’s experiment, the separation between the slits is 0.10 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1.0 m away. Find the separation between the successive bright fringes. (a) 6.6 mm (b) 6.0 mm (c) 6 m (d) 6 cm
29. Figure shows two coherent sources S1 – S2 vibrating in same phase. AB is an irregular wire lying at a far distance from the sources λ S1 and S2. Let = 10−3 ∠BOA = 0.12º. d How many bright spots will be seen on the wire, including points A and B.
(b) 12 (d) 24
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4B.77
Light (Wave Optics)
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32. In the figure shown in a YDSE, a parallel beam of light is incident on the slits from a medium of refractive index n1. The wavelength of light in this medium is λ1. A transparent slab of thickness t and refractive index is put infront of one slit. The medium between the screen and the plane of the slits is n2. The phase difference between the light waves reaching point O (symmetrical, relative to the slit) is
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28. In Young’s experiment, the wavelength of red light is 7.5 × 10–5 cm and that of blue light 5.0 × 10–5 cm. The value of n for which (n + 1)th the blue bright band coincides with nth red band is (a) 8 (b) 4 (c) 2 (d) 1
31. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 × 10–7m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between third dark fringe and the fifth bright fringe is equal to (a) .65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.87 mm
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27. Two waves originating from source S1 and S2 having zero phase difference and common wavelength λ will show completely destructive interference at a point P if (S1 P – S2 P) is (a) 5λ (b) 3λ/4 (c) 2λ (d) 11λ/2
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30. In Young’s slit experiment, carried out with lights of wavelength λ = 5000 Å, the distance between the slit is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum will be at x equal to (a) 1.67 cm (b) 1.5 cm (c) 0.5 cm (d) 5.0 cm
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26. A broad source of light (I = 680 nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.034 mm in diameter at the other end. The total number of bright fringes that appear over the 120 mm distance is
6
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4B.78
33. In an experiment, the two slits are 0.5 mm apart and the fringes are observed to 100 cm from the plane of the slits. The distance of the 11th bright fringe from the 1st bright fringe is 9.72 mm. Calculate the wavelength. (a) 4.85 × 10–5 cm (b) 4.85 × 10–5 m –7 (c) 4.86 × 10 cm (d) 4.86 × 10–5 cm
38. Interference fringes were produced in young’s double slit experiment using light of wave length 5000Å. When a film of material 2.5 × 10–3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe width. The refractive index of the material of the film is (a) 1.25 (b) 1.33 (c) 1.4 (d) 1.5
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42. Find the minimum thickness of a film which will strongly reflect the light of wavelength 589 nm. The refractive index of the material of the film is 1.25 (a) 118 nm (b) 120 nm (c) 118 m (d) 188 mm
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41. If the ratio of the intensity of two coherent I −I sources is 4 then the visibility m ax min I m ax + I min of the fringes is (a) 4 (b) 4/5 (c) 3/5 (d) 9
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40. The path difference between two interfering waves at a point on screen is 171.5 times the wavelength. If the path difference is 0.01029 cm. Find the wavelength. (a) 6000 × 10–10 cm (b) 6000 Å (c) 6000 × 10–8 mm (d) None
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37. A beam of light consisting of two wavelength 6500Å and 5200Å is used to obtain
39. The path difference between two interfering waves at a point on the screen is λ/6. The ratio of intensity at this point and that at the central bright fringe will be (Assume that intensity due to each slit in same) (a) 0.853 (b) 8.53 (c) 0.75 (d) 7.5
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35. Two light waves are given by E1 = 2 sin (100πt – kx + 30º) and E2 = 3 cos (200πt – k′x + 60º). The ratio of intensity of first wave to that of second wave is (a) 2/3 (b) 4/9 (c) 1/9 (d) 1/3
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34. In a Young’s experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one meter away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be (a) 60 × 10–4 cm (b) 10 × 10–4 cm –5 (c) 10 × 10 cm (d) 6 × 10–5 cm
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interference fringes in a young’s double slit experiment. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. What is the least distance from the central maximum where the bright fringes due to both the wave length coincide? (a) 0.156 cm (b) 0.152 cm (c) 0.17 cm (d) 0.16 cm
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43. In Young’s double slit interference experiment, the distance between two sources is 0.1/π mm. The distance of the screen from the source is 25 cm. Wavelength of light used is 5000Å. Then the angular position of the first dark fringe is (a) 0.10º (b) 0.15º (c) 0.30º (d) 0.45º
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52. Select the right option of the following. (a) Christian Hygens a contemporary of Newton established the wave theory of light by assuming that light waves were transverse (b) Maxwell provided the compelling theoretical evidence that light is transverse wave (c) Thomas Young experimentally proved the wave behaviour of light and Huygens assumption (d) All the statements given above, answers the question “what is light”
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53. Two slits separated by a distance of 1 mm, are illuminated with red light of wavelength 6.5 × 10–7m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (a) 0.65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.88 mm
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48. The initial phase difference between the light waves from the two coherent sources S1 and S2 is π. Then the central fringe will be (a) a bright fringe (b) a dark fringe
51. Huygen wave theory allows us to know (a) The wavelength of the wave (b) The velocity of the wave (c) The amplitude of the wave (d) The propagation of wave fronts
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47. Mark the correct statement(s). (a) An interference pattern can be formed on a screen by two independent phase locked laser sources (b) An interference pattern can not be formed by two independent monochromatic sources of same wavelength (c) Interference pattern using white light contains few coloured fringes also (d) All of the above
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46. In Young’s double slit experiment (a) only diffraction takes place (b) only interference takes place (c) both diffraction and interference takes place (d) neither interference nor diffraction takes place
50. Which one of the following phenomena is not explained by Huygen’s construction of wavefront? (a) Refraction (b) Reflection (c) Diffraction (d) Origin of spectra
49. Wavefront means (a) All particles in it have same phase (b) All particles have opposite phase of vibrations (c) Few particles are in same phase, rest are in opposite phase (d) None of these
44. In Young’s double slit experiment, if the width of one of the slits is increased, then (a) contrast of the fringes decreases (b) contrast of the fringes increases (c) contrast of the fringes remains the same (d) interference pattern is destroyed immediately 45. One important similarity between light waves and sound waves is that both (a) can travel through vacuum (b) are transverse wave (c) have same speed in air (d) can interfere
4B.79
(c) bright or dark depending upon the wavelength of light (d) neither bright, nor dark
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4B.80
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60. Two Nicol prisms are first crossed and then one of them is rotated through 60º. The percentage of incident light transmitted is
4. If one of the slits of a standard Young’s double slit experiment is covered by a thin parallel sided glass slab so that it transmits only one half the light intensity of the other, then (a) The fringe pattern will get shifted towards the covered slit (b) The fringe pattern will get shifted away from the covered slit
59. Two points separated by a distance of 0.1 mm can just be inspected in a microscope when light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used, this limit of resolution will become (a) 0.80 mm (b) 0.12 mm (c) 0.10 mm (d) 0.08 mm
3. The intensity of a plane progressive wave in loss free medium is (a) directly proportional to the square of amplitude of the wave (b) directly proportional to the velocity of the wave (c) directly proportional to the square of frequency of the wave (d) inversely proportional to the density of the medium
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58. The first diffraction minima due to a single slit diffraction is at θ = 30º for a light of wavelength 5000 Å. The width of the slit is (a) 5 × 10–5 cm (b) 10 × 10–5 cm –5 (c) 2.5 × 10 cm (d) 1.25 × 105 cm
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2. An interference pattern is formed on the screen, when light from two different monochromatic sources are allowed to interfere. Then, it is true that (a) frequencies of light from the two sources are equal to each other (b) the sources are coherent (c) the sources should be located in the same medium (d) the path difference should either be an even or, an odd multiple of λ/2, where λ is the wavelength of light
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57. In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength λ1 is found to be coincident with the third maximum at λ2. So (a) 3 λ1 = 0.3λ2 (b) 3λ1 = λ2 (c) λ1 = 3.5λ2 (d) 0.3λ1 = 3λ2
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55. A double slit is illuminated by two wavelength 450 nm and 600 nm. What is the lowest order at which the maxima of one wavelength coincides with the minima of the other wavelength? (a) 1 (b) 2 (c) 3 (d) 4 56. x1 and x2 are the separations between the two images of slit obtained on a fixed screen, in a biprism set-up, using a convex lens, the separation in the two interfering virtual sources is (a) x1/x2 (b) x2/x1
(b) 25.0 (d) 50
1. If the first minima in a Young’s slit experiment occurs directly in front of one of the slits, (distance between slit and screen D = 12 cm and distance between slits d = 5 cm) then the wavelength of the radiation used can be (a) 2 cm (b) 4 cm (c) 2/3 cm (d) 4/3 cm
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54. The central fringe of the interference pattern produced by light of wavelength 6000 Å is found to shift to the position of 4th bright fringe after a glass plate of refractive index 1.5 is introduced. The thickness of the glass plate would be (a) 4.8 µm (b) 8.23 µm (c) 14.98 µm (d) 3.78 µm
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12. A person views the interference pattern, produced by two slits illuminated by white light, on placing a green filter in front of his eyes. Then (a) He will see sharply distinguishable dark and bright fringes (b) The fringes will not be sharp but are differentiable (c) On replacing the green filter by a blue filter, the fringes will be again sharp but closer than those by the green filter (d) On using both the filters simultaneously, the central bands will be maximum bright
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10. A Young’s double slit experiment is performed with white light. (a) The central fringe will be white (b) There will not be a completely dark fringe (c) The fringe next to the central will be red (d) The fringe next to the central will be violet.
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7. A parallel beam of light (λ = 5000 Å) is incident at an angle α = 30º with the normal to the slit plane in a Young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is I0. Point O is equidistant from S1 and S2. The distance between slits is 1mm.
(a) the intensity at O is 4I0 (b) the intensity at O is zero (c) the intensity at a point on the screen 1 m below O is I0 (d) the intensity at a point on the screen 1 m below O is zero
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8. White light is used to illuminate the two slits in Young’s double slit experiment. The separation between the slits is b and the screen is at a distance d (>> b) from the slits. At a point on the screen directly in front of any one of slits, certain wavelengths are missing. Some of these missing wavelengths are (a) λ = b2/d (b) λ = 2b2/d (c) λ = b2/3d (d) λ = 2b2/3d
6. Huygen’s principle of secondary wavelets may be used to (a) find the velocity of light in vacuum (b) explain the particle behavior of light (c) find the new position of a wavefront (d) explain Snell’s law.
4B.81
5. If instead of monochromatic light, white light is used in Young’s double slit experiment, then (a) a bright white fringe is formed at centre of screen (b) fringes of different colours are clearly observed only in the first order place (c) the first order red fringes are relatively close to the screen than the first order violet fringes. (d) the first order violet fringes are relatively close to the centre of the screen than the first order red fringes
(c) The bright fringes will become less bright and the dark ones will become more bright (d) The fringe width will remain unchanged
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4B.82
14. Two sources are called coherent if they produce waves (a) of equal wavelength (b) of equal velocity (c) having same shape of wavefront (d) having a constant phase difference
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4. Statement 1: When two coherent waves of intensity I1 and I2 are superimposed with a constant phase difference φ, then the intensity of resultant wave is given by I = I1 + I2 + 2 I1 I 2 cos φ Statement 2: For two incoherent sources, resultant intensity is given by I = I1 + I2.
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QUESTIONS
5. Statement 1: In YDSE central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen. Statement 2: In an interference pattern, whatever energy disappears at the minimum, appears at the maximum. 6. Statement 1: Red light travels faster in glass than green light. Statement 2: The refractive index of glass is less for red light than for green light.
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3. Statement 1: All points on a wavefront vibrate in same phase with same frequency. Statement 2: Two sources are said to be coherent if they produce waves of same frequency with a constant phase difference.
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.
2. Statement 1: In YDSE, number of bright fringe or dark fringe can not be unlimited. Statement 2: In YDSE, path difference between the superposing waves can not be more than the distance between the slits.
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16. If one of the slits of a standard YDSE apparatus is covered by a thin parallel sided glass slab so that it transmit only one half of the light intensity of the other, then (a) the fringe pattern will get shifted towards the covered slit (b) the fringe pattern will get shifted away from the covered slit (c) the fringe pattern will be less bright and the dark ones will be more bright (d) the fringe width will remain unchanged.
ASSERTION
1. Statement 1: In Young’s double slit experiment if wavelength of incident monochromatic light is just doubled, number of bright fringe on the screen will increase. Statement 2: Maximum number of bright lunge on the screen is inversely proportional to the wavelength of light used.
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15. To observe a stationary interference pattern formed by two light waves, it is not necessary that they must have (a) the same frequency (b) same amplitude (c) a constant phase difference (d) the same intensity
Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
(b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True.
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13. In a YDSE apparatus, we use white light. Then (a) the fringe next to the central will be red (b) the central fringe will be white (c) the fringe next to the central will be violet (d) there will not be a completely dark fringe
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and Imin = ( I1 − I 2 ) 2 15. Statement 1: In YDSE, if a thin film is introduced in front of the upper slit, then the fringe pattern shifts in the downward direction. Statement 2: In YDSE if the slit widths are unequal, the minima will be completely dark.
QUESTIONS
TYPE
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1. In the figure shown A, B and C are the three slits each of which individually produces the same intensity ℓ0, at point P0 when they are illuminated by parallel beam of light of wavelength λ.
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12. Statement 1: The fringe visibility will be maximum when amplitude of light waves from two coherent sources is exactly same. Statement 2: Fringe visibility I −I = max min I max + I min 13. Statement 1: In YDSE, the fringe width (β) depends on the medium in which the experiment is carried out. λD Statement 2: Because β = and λ d depends on the medium.
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Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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11. Statement 1: Diffraction takes place for all types of waves mechanical or non-mechanical, transverse or longitudinal. Statement 2: Diffraction’s effect are perceptible only if wavelength of wave is comparable to dimensions of diffracting device.
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9. Statement 1: YDSE, the intensity at the maxima is observed to be I0. If one of the slits is closed, then the intensity at the location of the maxima reduces to I0/4. Statement 2: In YDSE, fringes with blue light are thicker than those for red light. 10. Statement 1: If white light is used in YDSE, then the central bright fringe will be white. Statement 2: Because all the wavelengths produce their zero order maxima at the same position.
Statement 2: In YDSE Imax = ( I1 + I 2 ) 2
8. Statement 1: According to Ray light, intensity of scattered light is inversely proportional to the fourth power of the wavelength of light. Statement 2: One can not see in fog because intensity of light decreases due to scattering.
maximum when both the sources have same intensity. Statement 2: In YDSE, the nth maxima always comes before the nth minima.
14. Statement 1: In YDSE, if I1 = 9I0 and I I2 = 4I0 then max = 25 I min
I max is I min
4B.83
7. Statement 1: In YDSE, the ratio
Light (Wave Optics)
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Match the quantities of column I with their values in column II. Column I Column II (a) Resultant intensity (p) 4 at P0 in (W/m2)
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Light (Wave Optics)
(d) All fringes are equally spaced
(s) Lloyd’s mirror experiment 4. In Young’s double slit experiment, if distance between slits is d, distance between slit and screen is D, wavelength of light used is λ. Then match column I and II. Column I (a) condition for bright fringe
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5. In a YDSE experiment d = 1 mm, D = 2000 mm, λ = 10–5 mm positions are measured from central maxima on screen Column 1 1 (a) Optical path difference at x = 2 (b) Optical path difference at x = 2000 mm (c) Path difference is almost (λ) (d) Maximum path difference
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3. Match the observation given in Column I with the experiments given in Column II where these observation occur.
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Column II (p) 1 mm (q) Just a distance below central maxima (.02 mm).
(r) (1/2 2) × 10−3 mm (s) Far away from central maxima
Column I Column II (a) Central fringes (p) Interference is bright and all in wedgefringes are equally shaped thin spaced film (b) Central fringes (q) Young’s is dark and all double slit fringes are equally experiment spaced (c) Central fringes is (r) Fraunhofer bright and fringes single slit are not equally diffraction spaced experiment
6. The arrangement of Lloyd’s mirror experiment is shown in figure. S is a point source of frequency 6 × 1014 Hz. A and B are the two ends of a mirror placed horizontally, and LOM represents the screen.
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(b) Resultant (q) zero amplitude at P0 (in unit) (c) If slit C is closed (r) 2 then resultant intensity at P0 (d) If slit B is (s) 2 2 closed resulting amplitude at P0 (in SI unit)
4B.84
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Column I The minimum height, above point O, fringes will form is (in mm) The maximum height, above point O, fringes will be formed is (in mm) Total no. of fringes are Fringe width (in µm) is
1. Initially, the distance of third minima from principal maxima will be
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coming out of S1 and S2 are same and is I0. In this situation, the principal maximum is formed at the point P. At the point on screen where principal maximum is formed, phase difference between two interfering waves will be zero.
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Light (Wave Optics)
PASSAGE–2 The YDSE setup shown in figure, is lighted by a bright point source S, placed at a distance D from the plane of slits of equal widths on the perpendicular bisector of the slits.
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4B.86
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1. Central maxima will remain at the centre of screen 2. Fringe pattern will slowly shift in the direction opposite to motion of source 3. Fringe pattern will slowly shift in the direction same as that of motion of source 4. Fringe width changes
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5. During activity – II, which of the following statements are true?
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4. During activity – I, which of the following statement is/are true?
On the basis of the phenomenon of polarization, it was established that light is a form of transverse wave motion.
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Activity–IV: Width of one of the slit is slightly increased.
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Activity– II: The wavelength of light of source is continuously decreased. Activity–III: Source is moved towards the plane of slits slowly.
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Activity– I:
1. Central maxima will remain at the centre of screen 2. Fringe pattern will slowly shift in the direction opposite to motion of source 3. Fringe pattern will slowly shift in the direction same as that of motion of source 4. Fringe width changes (a) only 1 (b) only 2 (c) only 3 (d) 1 and 4
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6. During activity – III, which of the following statements are true?
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(a) only 1 (c) only 3
Light (Wave Optics)
Since light is an electromagnetic wave, it con sists of electric vector and magnetic vector E H varying periodically in phase, at right angles to each other and also perpendicular to the direction of propagation. In ordinary light, the electric vector vibrates in all planes with equal probability at right angles to the direction of propagation. Hence the ordinary light also called unpolarized light can be represented by a star as shown. Hence unpolarized light is symmetrical about the direction of propagation. By the phenomenon of reflection, light can be polarized. The polarized light has vibrations confined only to a single line in a plane perpendicular to the direction of propagation and is called plane polarized. To polarize light by reflection we make use of Brewster’s law which states that the tangent of the angle of
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The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.
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1. An interference pattern is obtained by using a Fresnel’s biprism. If the fringe width is 4 mm when air is the surrounding medium, then find the fringe width if water is the surrounding medium. Keeping the same source. Assume nglass = 1.5, nwater = 4/3, nair = 1.
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11. If the intensity at P be maximum then the minimum distance through which the reflecting surface MN should be displaced so that at P again becomes maximum? (a) 3 × 10–10m (b) 6 × 10–10m –10 (c) 1.5 × 10 m (d) 12 × 10–10 m
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9. An analyser examines two adjacent planepolarized beams A and B whose planes of vibration are mutually perpendicular. In one position of the analyser, beam B shows zero intensity. From this position, a rotation of 30º shows the two beams as matched in intensity. The intensity ration IA/IB of the two beams is (a) 3 (b) 1/3 (d) 5 (c) 1/ 2
10. The shape of the interference fringes on the screen is (a) circle (b) ellipse (c) parabola (d) straight line
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polarization (p) is numerically equal to the refractive index µ of the medium. The device that polarizes unpolarized light is called the polarizer and the device that analyses the polarized light is called the analyser. The amplitude of a plane polarized light incident on the analyser can be resolved into two components, one parallel to the plane of transmission of the analyser and the other perpendicular to it. The parallel component, namely, (a cosθ) is transmitted through the analyser. 7. At the angle of polarization, the angle of inclination between the reflected and refracted rays is (a) π/8 (b) π/6 (c) π/4 (d) π/2 8. If in an unpolarized light E = 2iˆ + 3 ˆj and H = 3iˆ − 2 ˆj , then the direction of propagation is given by (a) −13k (b) −10 ˆj ˆ ˆ (c) −6i + 6k (d) 5 ˆj − 4kˆ
4B.87
Light (Wave Optics)
Slit S1 is covered by a transparent sheet of thickness t1 = 2.5µm and S2 by another sheet of thickness t2 = 1.25µm as shown in figure. Both sheets are made of same material having refractive index µ = 1.40. Water is filled
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4B.88
The widths of S1 and S2 are w and 2w respectively. A thin transparent film of thickness 4 mm and R.I. 3/2 is placed infront of S1. It absorbs 50% light energy and transmits the remaining. The interference is observed on the screen. Point O is equidistant from S1 and S2. If the maximum intensity on the screen is I, then find (i) intensity at O (ii) Minimum intensity (iii) fringe width (iv) Distance of nearest maxima from O (v) Distance of central maxima from O (vi) intensity at 4 mm from O upwards.
3. In a Young’s double slit experiment, a parallel light beam containing wavelength λ1 = 4000Å and λ2 = 5600Å is incident on a diaphragm having two narrow slits. Separation between the slits is d = 2 mm. If difference between diaphragm and screen is D = 40 cm, calculate (i) distance of first black line from central bright fringe (ii) distance between two consecutive black lines.
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7. Two slits S1 and S2 are on a plane inclined at an angle of 45º with horizontal. The distance between the slits is 2 mm. A monochromatic point source S of wavelength, λ = 5000Å is placed at a distance 1/ 2 mm from the mid-point of slits as shown in figure. The screen is placed at a distance of
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4. In a YDSE, the distance between the slits and the screen is 100 cm. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width 0.25 mm. When the distance between the slits is increased by ∆d = 1.2 mm, the fringe width decreases to n = 2/3 of the original value. In the final position, a thin glass plate of refractive index 1.5 is kept in front of one of the slits and the shift of central maximum is observed to be 20 fringe width. Find the thickness of the plate and wavelength of the incident light.
6. (a) Thin glass plate of thickness t and refractive index µ is inserted between screen and one of the slits in a Young’s experiment. If the intensity at the centre of the screen is I, what was the intensity at the same point prior to the introduction of the sheet. (b) One slit of a Young’s experiment is covered by a glass plate (µ1 = 1.4) and the other by another glass plate (µ2 = 1.7) of the same thickness. The point of central maxima on the screen, before the plates were introduced is now occupied by the third bright fringe. Find the thickness of the plate, the wavelength of light used is 4000Å.
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in space between diaphragm and screen. A monochromatic light beam of wavelength λ = 5000 Å is incident normally on the diaphragm. Assuming intensity of interference pattern obtained on the screen, where C is foot of perpendicular bisector of S1S2. (Refractive index of water, µw = 4/3)
Light (Wave Optics)
2 m. Find the fringe width of interference pattern on the screen.
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9. In a Young’s double experiment, the upper slit is covered by a thin glass plate of refractive index µ1 = 1.4 while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index µ2 = 1.7. Interference pattern is observed using light of wavelength λ = 5400 Å. It is found that the point P on the screen where the central maxima fell before the glass plates were inserted, now has 3/4 the original intensity. It is further observed that what used to be the fifth maxima earlier, lies below the point P while the sixth minima lies above the point P. Neglecting absorption of light by glass plates, calculate thickness of the glass plates.
13. A beam of light consisting of two wavelengths, 6500Å and 5200Å is used to obtain interference fringe in a Young’s double slit experiment. (i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500Å. (ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.
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11. A Young’s double slit arrangement produces interference fringes for sodium light (λ = 5890 Å) that are 0.20º apart. What is the angular fringe separation, if the entire arrangement is immersed in water. (R.I. of water = 4/3)
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10. A plane monochromatic light wave falls normally on a single slit. At a distance D = 4 m from it there is a screen showing a certain diffraction pattern. The slit width is decreased n = 2 times. Calculate the new distance D’ at which the screen should be positioned to obtain the diffraction pattern similar to previous one but diminished n times.
14. In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000Å and intensity (10/π) W/m2 is incident normally on two apertures A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000Å and refractive index 1.5 for the wavelength of 6000Å is placed in front of aperture A (see figure). Calculate the power (in W) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.
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The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
12. In Young’s double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves.
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4B.89
8. In a Young’s double slit experiment, when a lens is interposed at two points between slits and screen, the distance between the two images of the slits are 4 mm and 9 mm. When the screen is shifted by 10 cm, the number of fringes is reduced from 100 to 80 in a 0.4 cm. Find the wavelength of the light used in the experiment.
Light (Wave Optics)
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4B.90
apparatus is immersed in a liquid. Find refractive index of the liquid. [1996]
QUESTIONS
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1. Consider Fraunhoffer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radian) between the wavelets from the opposite edges of the slit is [1995]
4. A double-slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of the slits and screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300Å. [1996] (a) Calculate the fringe width. (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minima on the axis.
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6. In a Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400Å. It is found that the point P on the screen where the central maximum (n = 0) fall before the glass plates were inserted, now has ¾ the original intensity. It is further observed that what used to be the 5th maximum earlier, lies below the point P while the 6th minima
I(θ) = I0 / 2 for θ = 30º I(θ) = I0 / 4 for θ = 90º I(θ) = I0 for θ = 0º I(θ) is constant for all values of θ
3. Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000Å. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original
(a) (b) (c) (d)
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The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150 m. The intensity 1(θ) is measured as a function of θ, where θ is defined as shown. If I0 is the maximum intensity, then I(θ) for 0 ≤ θ ≤ 90º is given by [1995]
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5. In Young’s experiment, the source is red light of wavelength 7 × 10–7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10–3m to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10–7m, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of the glass for the green light. Also estimate the change in fringe width due to the change in wavelength. [1997]
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2. In an interference arrangement similar to Young’s double-slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106Hz.
(b) π/2 (d) π
(a) π/4 (c) 2π
PREVIOUS
Light (Wave Optics)
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7. A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on screen placed parallel to the plane of the slits at a distance of 1.0 m from it, as shown in the figure.
9. The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. \ 6
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10. Yellow light is used in a single slit diffraction experiment with slit width of 0.6. mm. If yellow light is replaced by X-rays, then the observed pattern will reveal [1999]
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(a) If the incident beam falls normally on the double slit apparatus, find the y-coordinates of all the interference minima on the screen. (b) If the incident beam makes an angle of 30º with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum. [1998]
8. A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is [1998]
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(a) Find the location of the central maximum (bright fringe with zero path difference) on the y-axis. (b) Find the light intensity at point O relative to the maximum fringe intensity. (c) Now, if 600 nm light is replaced by white light of range 400 to 700 nm, find the wavelengths of the light that form maxima exactly at point O. [All wavelengths in this problem are for the given medium of refractive index 4/3. Ignore dispersion.]
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The lower slit S2 is covered by a thin glass sheet of thickness 10.4 µm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown. [1999]
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4B.91
(a) zero (c) π
lies above P. Calculate the thickness of the glass plate. [1997]
Light (Wave Optics)
that the central maximum is narrower more number of fringes less number of fringes no diffraction pattern
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4B.92
11. A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat plate as shown. [1999]
is changed to 400 nm, number of fringes observed in the same segment of the screen is given by [2001]
Light (Wave Optics)
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17. A point source S emitting light of wavelength 600 nm is placed at a very small height h above the flat reflecting surface AB (see figure).
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13. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is [2001] (a) 2I (c) 5I
16. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is [2002]
12. In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern [2000] (a) the intensities of both the maxima and the minima increases (b) the intensity of the maxima increases and the minima has zero intensity (c) the intensity of maxima decreases and that of minima increases (d) the intensity of maxima decreases and the minima has zero intensity
14. In young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light
15. A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure below. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled up to OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid. [2001]
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(b) 18 (d) 30
(a) 12 (c) 24
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distance between the plane of the slits and the screen is I m. The minimum distance between two successive regions of complete darkness is [2004]
The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. [2002]
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Directions (Questions 22 to 24): The figure shows a surface XY separating two transparent media, medium–1 and medium–2. The lines ab and cd represent wavefronts of a light wave travelling in medium–1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium–2 after refraction. [2007]
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20. In a Young’s double slit experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again? Take D/d = 103. Symbols have their usual meanings. [2004]
2
(b) 5.6 mm (d) 28 mm
(a) 4 mm (c) 14 mm
(a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure) (c) If the intensities at point P corresponds to a maximum, calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum. 18. In the adjacent diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition on θ for constructing interference at P between the ray BP and reflected ray OP [2003] 3λ (a) cos θ = 2d λ (b) cos θ = 4d λ (c) sec θ − cos θ = d 4λ (d) sec θ − cos θ = d
4B.93
Light (Wave Optics)
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19. In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the
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25. In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [2008]
23. The phases of the light wave at c, d, e and f are φc, φd, φe and φf respectively. It is given that φc ≠ φf (a) φc cannot be equal to φd (b) φd can be equal to φe (c) (φd – φf) is equal to (φc – φe) (d) (φd – φc) is not equal to (φf – φe)
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(c) (b) (d) (a) (b) (a) (d)
8. (a, c) 16. (a, c, d)
QUESTIONS
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1. (a, c) 2. (a, b) 3. (a, b, c) 4. (a, c, d) 5. (a, b, d) 6. (c, d) 7. (a, c) 9. (a, c, d) 10. (a, b, d) 11. (b, c, d) 12. (a, c) 13. (b, c, d) 14. (a, b, c, d) 15. (b, d)
1. (d) 9. (d)
8. 16. 24. 32. 40. 48. 56.
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(b) (b) (a) (b) (d) (b) (a) (c)
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24. Speed of the light is (a) the same in medium–1 and medium–2 (b) larger in medium–1 than in medium–2 (c) larger in medium–2 than in medium–1 (d) different at b and d
A
(a) If d = λ, the screen will contain only one maximum (b) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen (c) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase (d) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase
22. Light travels as a (a) parallel beam in each medium (b) convergent beam in each medium (c) divergent beam in each medium (d) divergent beam in one medium and convergent beam in the other medium.
Light (Wave Optics)
3. (b) 11. (b)
4. (b) 12. (a)
5. (b) 13. (a)
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6. (a) 14. (a)
7. (c) 15. (d)
8. (b)
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1. (a) → r, (b) → r, (c) → q, (d) → q
4. (a) → r, (b) → r, (c) → q, (d) → p 6. (a) → q, (b) → p, (c) → s, (d) → r
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25. (a,b). For at least one maxima, sin θ = λ/d If λ = d, sin θ = 1 and y → ∞ So there will be only central maximum on the screen. If λ < d < 2λ, sin θ exists and y is finite so there will be two more maxima on screen. Intensity of dark fringe becomes zero when intensities of two slits are equal. Initial intensity at both the slits are unequal so there will be some brightness at dark fringe. Hence when intensity of both slits is made same the intensity at dark on screen shall decrease to zero. So both (c) and (d) are false.
3. (a) 11. (a)
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5. (a) → r, (b) → r, (c) → q, (d) → p,s
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4B.95
QUESTIONS
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General Physics Particle Kinematics S.C. Pandey Dynamics of Particle Circular Motion Energy and Momentum
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Waves, Optics and Thermodynamics
Chandigarh • Delhi • Chennai https://t.me/universitarios_infohttps://www.jamarana.com
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Production Editor: Nitkiran Bedi Composition: Tantla Composition Services, Chandigarh Printer: Sanat Printers
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The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertant omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book.
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Copyright © 2010 Dorling Kindersley (India) Pvt. Ltd
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This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated, without the publisher’s prior written consent, in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise) without the prior written permission of both the copyright owner and the abovementioned publisher of this book. ISBN 978-81-317-3411-7 10 9 8 7 6 5 4 3 2 1 Published by Dorling Kindersley (India) Pvt. Ltd, licensees of Pearson Education in South Asia.
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Preface
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Contents Chapter 2
Kinetic Theory of Gases and Thermodynamics
Chapter 3
Wave Mechanics
1.1–1.51 2.1–2.114 3.1–3.69 4A.1–4A.123 4B.1–4B.95
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Chapter 4B Light (Wave Optics)
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Heat
Chapter 1
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Salient feature
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For a science student, physics is the most important subject as it requires logical reasoning and high imagination. Without improving the level of physics, it is difficult to achieve a goal with the kind of competition that exists today. This five part volume covers all parts of general physics—Mechanics, Heat, Wave, Light, Electromagnetism and Modern Physics—which is written in accordance with the latest syllabus of the IIT-JEE and AIEEE. There is no single book that is available in the market that contains a large amount of solved examples.
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■ Entire syllabus is covered in five volumes. ■ Content of each chapter is well defined and builds new concepts from the scratch. ■ Each chapter describes the theory in a simple and lucid style. ■ Covers a wide spectrum of questions to enable the student to develop enough expertise to tackle any problem. ■ Helps students in building analytical and quantitative skills, which, in turn, develop confidence in problem solving. ■ Practice exercises are given at the end of each chapter. ■ Numerous diagrams in every chapter. After studying the entire chapter, students will be able to learn different tricks and techniques of problem solving with suitable level of analytical ability. Suggestions for improving the book are always welcome.
All the best!
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S. C. PANDEY
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c h a p t e r
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Heat
1
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Zeroth Law of Thermodynamics
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It is a form of energy which determines the change in thermal state of the body and is defined as the flow of energy from one body to the other body due to difference in the degree of hotness of two bodies (temperature). It flows from the body which is at a higher temperature to the other at low temperature.
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If two bodies A and B are in thermal equilibrium and A and C are also in thermal equilibrium then B and C are also in thermal equilibrium. Calorimetry It is the branch of heat transfer that deals with the measurement of heat. The heat is usually measured in calories or kilocalories. One Calorie 1 calorie is the quantity of heat required to raise the temperature of 1 gram of water by 1ºC. Specific heat The specific heat capacity of a substance is defined as the heat supplied per unit mass of the substance per unit rise in the temperature. s=
∆θ m∆T
Its unit is cal/gm/ºC. The heat lost by the body or gained from the body depends upon the difference in the temperature. Heat lost or gained by a body = ms θ
where, m = Mass of the body.
s = Specific heat. θ = Rise or fall in the temperature of body.
1.1 HEA
a
Heat
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Heat
Heat Capacity The heat capacity of the body is the quantity of heat required by the body to raise its temperature by 1ºC. It is also known as thermal capacity. Heat capacity = ms (mass × specific heat)
Water Equivalent It is the quantity of water whose thermal capacity is same as the heat capacity of the body. It is denoted by W.
rinciple of Calorimetry
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W = ms = Heat capacity of the body.
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m1s1 (θ2 – θ) = m2s2 (θ – θ1)
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If two bodies at different temperatures are made to come in contact with each other in the calorimeter. As a result, heat is exchanged between the objects as well as with the calorimeter. The principle of calorimetry states that the total heat given by the hot bodies equals the total heat received by the cold bodies. Suppose, two bodies are in the contact. The temperature of one body is θ1 and the other body temperature is θ2 (θ2 > θ1). Let the temperature of equilibrium is θ. Let also the mass of bodies is m1 and m2 respectively and specific heat s1 and s2. Then, from principle of calorimetry, Heat lost by hot body = Heat gained by cold body
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Latent heat During this process of melting or vaporization, the temperature remains constant. The amount of heat needed to melt a solid of mass m
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Latent Heat of Fusion
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Q = mL where, L = Latent heat of fusion or vaporization.
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It is the quantity of heat required to change the unit mass of a solid substance to the liquid state at its melting point. For ice, latent heat of fusion is 80 cal/gram.
Latent Heat of Vaporization It is the quantity of heat required to change the unit mass of a liquid to gaseous state at the boiling point of the liquid. For water, latent heat of vaporization is 540 cal/gm. Example The temperature of 100 gm of water is to be raised from 24ºC to 90ºC by adding steam to it. Calculate the mass of steam required for this purpose. )
) Solution
Let the mass required of stream is m gm Heat lost by steam = m × 540 + ms (100 – 90) = (m × 540 + m × 10) calorie
1.2
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Heat
1.3
Specific heat of water is 1 calorie/gm/ºC.
Heat lost = 550 m calorie Heat gained by water = 100 × 1 × (90 – 24) calories = 6600 calories By the principle of calorimetry Heat lost = heat gained 550 m = 6600 m=
∴
6600 = 12 gm. 550
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Example Three liquids A, B and C are given. The temperature of A, B and C are 60ºC, 55ºC and 50ºC respectively. When 4g of A and 3g of C are mixed the temperature is 55ºC, while when 2g of A and 3g of B are mixed, the temperature is 57ºC. What would be the temperature of the mixture when equal masses of B and C are mixed? )
m
) Solution Let the specific heat capacity of the liquid is sA, sB and sC respectively.
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Thus, when the liquids A and C are mixed, Heat lost by A = 4 × sA (60 – 55) = 4 × sA × 5
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sA 3 = sC 4
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4 × sA × 5 = 3 × sC × 5
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Heat gained by C = 3 × sC (55 – 50) = 3 × sC × 5
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When, the liquids A and B are mixed. Heat lost by A = 2 × sA (60 – 57) = 2 × sA × 3
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sA = sB
...(ii)
or
2 × sA × 3 = 3 × sB × 2
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∴
Heat gained by B = 3 × sB (57 – 55) = 3 × sB × 2
...(iii)
When the liquids B and C are mixed Heat lost by B = m × sB (55 – θ) Heat gained by C = m × sC (θ – 50) ∴ sB (55 – θ) = sC × (θ – 50) From equations (i), (ii) and (iii), we get
4 sA (θ − 50) 3 165 – 3θ = 4θ – 200
sA × (55 − θ) =
or
365 = 7θ
∴
θ=
or
365 = 52.14°C 7
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Heat
Example When a block of metal of specific heat 0.1 cal/gm/ºC and weighing 110 gm is heated to 100ºC and then quickly transferred to a calorimeter containing 200 gm of a liquid at 10ºC, the resulting temperature is 18ºC. On repeating the experiement with 400 gm of same liquid in the same calorimeter at some initial temperature, the resulting temperature is 14.5ºC. Find (i) Specific heat of liquid. (ii) The water equivalent at calorimeter. )
) Solution
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Let the specific heat capacity of liquid is s and water equivalent of calorimeter is W. Then, the heat lost by block in first case 110 × 0.1 × (100 – 18) = (110 × 0.1 × 82) calorie Heat gained by (liquid + calorimeter) = 200 × s × (18 – 10) + W (18 – 10) = 200 × s × 8 + 8 W From the principle of calorimetry Heat lost = Heat gained or 110 × 0.1 × 82 = 200 × s × 8 + 8 W ...(i) From second case 110 × 0.1 × 85.5 = 400 × s × 4.5 + 4.5 W ...(ii) Solving equations (i) and (ii), we get s = 0.48 cal/g/ºC and W = 16.6 gm.
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Example A lead ball at 30ºC is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. Calculate the latent heat of fusion of lead. Specific heat capacity of lead = 126 J/kg.ºC and melting point of lead = 330ºC. Assume that any mechanical energy lost is using heat for of ball. Use, g = 10 m/s2. )
) Solution
Initial gravitational potential energy of the ball = mgh = m × 10 × 6.2 × 103 = m × (6.2 × 104 J/kg) All this energy is used to heat the ball as it reaches the ground with a small velocity. Energy required to take the ball from 30ºC to 330ºC is = m × (126 J/kg – ºC) × 300ºC = m × 37800 J/kg And energy required to melt the ball at 330ºC = mL where, L = latent heat of fusion of lead. Thus, m × (6.2 × 104) = m × 37800 + mL or L = 2.4 × 104 J/kg.
1.4
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Heat
1.5
Example An aluminium container of mass 100 g contains 200 g of ice at – 20ºC. Heat is added to the system at a rate of 100 cal/s. What is the temperature of the system after 4 minutes? Draw a rough sketch showing the variation in the temperature of the system as a function of time. Specific heat capacity of ice = 0.5 cal/gm-ºC. Specific heat capacity of aluminium = 0.2 cal/gm-ºC. Specific heat capacity of water = 1 cal/gmºC and latent heat of fusion of ice = 80 cal/gm.
) Solution
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Total heat supplied to the system in 4 minutes Q = 100 cal/s × 240 s = 2.4 × 104 calorie. Heat required to take the system from – 20ºC to 0ºC. = (100 g) × (0.2 cal/gm–ºC) × (20ºC) + (200 g) × (0.5 cal/gm–ºC) × (20ºC)
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Thus, 2.4 × 104 cal = 2400 cal. + 16000 cal. + (220 cal/ºC) θ 5600 or θ= = 25.5°C. 220 Variation in the temperature as a function of time is sketched in the figure below.
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= 400 cal + 2000 cal = 2400 cal. 2400 Time taken in this process = = 24sec. 100 Heat required to melt the ice at 0ºC = 200 × 80 = 16000 cal. 16000 Time taken in this process = sec. = 160 sec. 100 If the final temperature is θ, the heat required to take the system to the final temperature = (100 g) (0.2 cal/gm-ºC) θ + (200 g) (1 cal/gm-ºC) θ
±
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Example 300 gm of ice at a temperature of – 10ºC is dropped into a calorimeter containing 200 gm of water at 30ºC. What will be the temperature and its contents, after thermal equilibrium is established? The specific heat of ice is 0.48 cal gm–1 (ºC)–1 and the latent heat of fusion of ice is 80 cal gm–1. The heat capacity of the calorimeter is negligible. https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
) Solution
Final temperature of the mixture can be greater than 0ºC or less than 0ºC. Let us analyse each of these cases:
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(i) Final temperature greater than 0ºC: All the ice in the calorimeter absorbs heat and rises to 0ºC. Ice at 0ºC absorbs latent heat and becomes water at 0ºC. Now, the temperature of this water is raised to θºC. Heat required for the ice at –10º to become water at θºC = (300 × 0.8 × 10) + (300 × 80) + 300 × 1 × θ = (26880 + 300 θ) calorie Heat lost by water at 30ºC = 200 (30 – θ) = (6000 – 200 θ) calorie Since, the heat required is greater than the heat lost by water for any temperature between 0 and 30ºC, the system cannot attain a temperature greater than 0ºC. (ii) Final temperature less than 0ºC: For the final temperature to be less than 0ºC, the entire water in the calorimeter should condense and the temperature becomes – θºC. Hence, to be released by water attain the temperature – θºC. = (200 × 1 × 30) + (200 × 80) + (200 × 0.4 × θ) = (22000 + 96 θ) calorie Heat absorbed by ice = 300 × 0.4 × (10 – θ) = (1440 – 144 θ) calorie Since, the heat which can be absorbed by ice is less than the heat to be released by water for all temperatures between 0 and – 10ºC, the system cannot attain a temperature less than 0ºC. (iii) Thus, the final temperature of the mixture can be neither greater than 0ºC nor less than 0ºC. Therefore, the calorimeter will contain both ice and water the final temperature of the mixture will be 0ºC.
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Example A solid material is supplied with heat at a constant rate. The temperature of the material is changing with the heat input as shown in the graph. Study the graph carefully and answer the following questions.
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Heat
(i) (ii) (iii) (iv)
1.7
What do the horizontal regions AB and CD represent? If CD = 2 AB, what do you infer? What does the slope DE represent? The slope of OA > slope of BC. What does this indicate?
)
) Solution
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(i) AB represents change of state from solid to liquid (at constant temperature). CD represents the change of state from liquid to gas (at constant temperature). (ii) For the given mass, heat required to change liquid state into gaseous state is double that required to change the solid state into the liquid state. Hence, the latent heat of vaporization of the material is double the latent heat of fusion. dT (for gaseous state) (iii) Slope DE = dθ dθ is the heat required to change the temperature of the given mass Now dT through 1ºC, assuming the gas is enclosed at constant volume or at constant pressure. 1 = Reciprocal of the heat capacity under the given ∴ Slope = heat capacity
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AL EX A SIO P
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Solid have length, breadth and thickness. When heated, in case of a thin rod of larger length, linear expansion is more important. In case of a thin plate having length and breadth with negligible thickness, area expansion (superficial expansion) is important. When there is a change in length breadth and thickness of a solid, when heated, volume expansion is considered. Accordingly, a solid can have linear expansion, superficial expansion, and volume (cubical) expansion. Linear expansion (expansion in length of a solid) Let l1 and l2 be the length of rod at temperature θ1ºC and θ2 ºC respectively. Then
l2 = l1 [1 + α (θ2 – θ1)] = l1 (1 + αθ)
where, α = Coefficient of linear expansion θ = (θ2 – θ1) If l0 is the length of the rod at 0ºC, then the length of the rod at θºC is
lt = l0 (1 + αθ)
or
1.2 THE
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assumption. (iv) Slope OA > slope BC Heat capacity for the solid state < Heat capacity for the liquid state. But since the mass is the same (heat capacity = ms), specific heat of the liquid state is more than that of the solid state.
α=
lt − l0 l0 θ
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Heat
Coefficient of linear expansion is defined as the change in length per unit length per unit degree change in temperature. Superficial expansion (expansion in surface area) If A1 is the area of solid at θ1 ºC and A2 is the area at θ2 ºC, then A2 = A1 (1 + βθ) where, θ = θ2 – θ1 β = coefficient of superficial (areal) expansion. Volume expansion (expansion in volume) If V1 is the volume of solid at θ1ºC and V2 is the volume at θ2ºC, then
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V2 = V1 (1 + γθ) where, γ = Coefficient of cubical (volume expansion)
d1 (1 + γ θ)
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d2 =
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note: For isotropic solids: β = 2α, γ = 3α, [6α = 3β = 2γ] As temperature increases, density of solid decreases. If d1 is the density at θ1ºC, d2 is the density at θ2ºC, then
s:
Heating a metallic Scale
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A metallic scale (linear) expands in length when heated. As a result all the marking are displaced from their usual (correct) positions. A reading of 1 unit on a heated scale is equivalent to an actual length of 1 (1 + αθ), where, α is a coefficient of linear expansion of material of scale, and θ is rise in the temperature of the scale.
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1.8
If the reading is x, actual length = x (1 + αθ) and actual length = reading (1 + αθ) https://t.me/universitarios_infohttps://www.jamarana.com
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1.9
ifference in Length of Two ods R
D
Heat
Consider two rods 1 and 2 of lengths l1 and l2. Let, they be heated through a temperature θ. If l1′ and l2′ are their expanded lengths, then and
l2′ = l2 (1 + α2 θ) l1′ = l1 (1 + α1 θ)
where α2 = Coefficient of linear expansion of rod 2, and α1 = Coefficient of linear expansion of rod 1. Since, the difference of lengths of two rods is constant, therefore l1 α1 = l2 α2
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Time eriod of endulum
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and
l2′ – l1′ = l2 – l1
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Time period (T) of a simple pendulum of length l is given by
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If there is rise in temperature by θ, length of the pendulum increases and, hence time period increases. As the result, the clock supporting such a simple pendulum slows down. If l0 be the length of pendulum and corresponding time period be T0, then
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If the pendulum heated by θ (rise in temperature), the new time period Tt will be given by Tt = 2π
where lt = Expanded length
lt g
⇒
Tt 1 = 1 + αθ = 1 + αθ (as α is very small) T0 2
⇒
⇒
Tt − T0 1 = αθ T0 2
⇒
⇒
Tt l l (1 + αθ) = t = 0 T0 l0 l0
⇒
∆T 1 = αθ T0 2
∆T = fractional increases in the time period = time lost in 1 second. T0 1 Time lost in one day = αθ × 86400 s. 2
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Heat
Example A steel tube of coefficient of linear expansion 18 × 10–6/ºC contains mercury, whose coefficient of absolute volume expansion is 180 × 10–6/ºC. The volume of mercury contained in the tube is 10–5m3 at 0ºC and it is desired that the length of the mercury column should remain constant at all normal temperature. This is achieved by fixing a rod of silica, whose thermal expansion is negligible at the bottom of the tube, to be immersed fully in mercury. Calculate the volume of the silica rod. )
) Solution At 0ºC:
m
At θº C:
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Let, V0 = Volume of mercury. V = Volume of silica rod. l0 = Length of mercury column. A0 = Length of mercury column. A0 = Area of cross-section of the steel tube. A0l0 = V0 + V
...(i)
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V0 ( γ Hg − 2.α) 2α
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180 × 10−16 − 2 × 18 × 10−6 = 10−5 2 × 18 × 10−16 –5 3 = 4 × 10 m
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∴
(V0 + V) (1 + 2 · α · θ) = (V0 + V) + (V0 · γHg · θ)
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...(ii)
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A0 (1 + 2 · α θ) l0 = Vθ + V = V0 (1 + γHg · θ) + V From equation (i), A0 · l0 = V0 + V or
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Aθ = A0 (1 + 2 · α θ) Since, l0 should remain constant, therefore Aθ (l0) = Vθ + V
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Example A glass bottle of volume 50 cm3 at 0ºC, is filled with paraffin at 15ºC. What is the mass of paraffin? Given d0 = 0.82 g cm–3 and γparaffin = 9 × 10–4/ºC. Coefficient of linear expansion of glass = 9 × 10–6/ºC.
) Solution )
1.10
V0 = Volume of glass bottle at 0ºC = 50 cm3 V15 = Volume of the bottle at 15ºC = 50 (1 + γg θ)
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Heat
1.11
= 50 (1 + 3 × 9 × 10–6 × 15) = 50 × 1.000405 cm3 d0 = Density of paraffin at 0ºC = 0.82 g cm–3 d15 = Density of paraffin at 15ºC d0 1 + γ ρθ
=
0.82 1 + 9 × 10−4 × 15
=
0.82 1.0135
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=
ar a
= 40.47 g
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∴ Mass of the paraffin filling the bottle at 15ºC = V15 d15
iv
The length of iron rod at 100ºC is l1 = (50) [1 + (12 × 10–6) (100 – 20)] = 50.048 cm The length of the aluminium rod at 100ºC is l2 = (100) [1 + (24 × 10–6) (100 – 20)] = 100.192 cm The length of the composite system at 100ºC is = (50.048 + 100.192) cm = 150.24 cm Since length of the composite system at 20ºC is 150 cm, hence, the average coefficient of linear expansion of the composite rod is,
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) Solution
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Example A iron rod of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurement refer to 20ºC. Find the length of the composite system at 100ºC and its average coefficient of linear expansion. The coefficient of linear expansion of iron and aluminium are 12 × 10–6/ºC and 24 × 10–6/ºC respectively.
α=
0.24 150 × (100 − 20)
= 20 × 10–6/ºC.
Example A glass vessel of volume 100 cm3 is filled with mercury and is heated from 25ºC to 75ºC. What volume of mercury will over flow? Coefficient of linear expansion of glass = 1.8 × 10–6/ºC and the coefficient of volume expansion of mercury is 1.8 × 10–4/ºC. https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
)
) Solution
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Volume of mercury at 25ºC is V0 = 100 cm3 Coefficient of volume expansion of mercury γL = 1.8 × 10–4/ºC Coefficient of volume expansion of glass γS = 3 × 1.8 × 10–6/ºC = 5.4 × 10–6/ºC Thus, the volume of mercury at 75ºC is VLθ = V0 (1 + γL θ) and the volume of vessel at 75ºC is VSθ = V0 (1 + γS θ) Volume of mercury overflow = VLθ – VSθ = V0 (γL – γS) θ = 100 cm3 (1.8 × 10–4 – 5.4 × 10–6)/ºC × 50ºC = 0.87 cm3
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N
ote: γa = (γL – γS) acts as the effective coefficient of expansion of the liquid with respect to the solid. The expansion of mercury as seen from the glass can be written as Vθ – V0 = V0 γa θ or Vθ = V0 (1 + γa θ) The constant γa is called the apparent coefficient of expansion of the liquid with respect to the solid.
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Example Find the difference in levels of mercury in two identical vertical communicating open glass tubes with horizontal base, if the left hand tube is maintained at 0ºC and the right hand tube is heated to 100ºC. The height of mercury column in the left hand tube is 90 cm. g = 1.82 × 10–4/ºC. Neglect the expansion of glass. )
) Solution
Given, h0 = 90 cm to find (h100 – h0). Let d0 and d100 be the corresponding densities. Therefore P0 = Pressure due to the left hand column = h0d0g P100 = Pressure due to the right hand column = h100d100g Since, the column are balanced, therefore P0 = P100 or h0d0g = h100d100g
or
d0 h = 100 d100 h0
But
1.12
d0 = 1 + γθ d100
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Heat
1.13
h100 = 1 + γθ h0
∴
h100 – h0 = h0 γθ = 90 × 1.82 × 10–4 × 100 = 1.638 cm
∴
Example A pendulum clock loses 12 sec in a day if the temperature is 40ºC and goes fast by 4 sec a day if the temperature is 20ºC. Find the temperature at which the clock will show correct time and the coefficient of linear expansion of the metal of the pendulum shaft.
) Solution
rio s_ in fo
er si ta rio s
un iv
e/
32 86400 × 20
iv
α=
.m
∴
32 = 86400 α (40 – 20)
un
//t
ht
tp
s:
//w
w
er s
w
ita
.ja
m
ar a
na .c om
Let θ be the temperature at which the clock is correct. 1 Time lost per day = α (rise in temperature) × 86400 2 1 ⇒ 12 = α (40 – θ) × 86400 2 1 Time gained by day = α (drop in temperature) × 86400 2 1 ⇒ 4 = α (θ – 20) × 86400 2 Adding equations (i) and (ii), we get
ht tp
s: //
t.m e/
ht tp s:
= 1.85 × 10–5/ºC
G
5
I
Dividing equations (i) by equation (ii), we get 12 (θ – 20) = 4 (40 – θ) or 12θ – 240 = 160 – 4θ or 16θ = 400 ∴ θ = 25ºC Hence, clock shows correct time at 25ºC. https://t.me/universitarios_infohttps://www.jamarana.com
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...(i)
...(ii)
Heat
Example A copper and a tungsten plate having a thickness δ = 2 mm each are riveted together so that at 0ºC they form a flat bimetallic plate. Find the average radius of the curvature of this plate at θ = 200ºC. The coefficient of linear expansion for copper and tungsten are αcu = 1.7 × 10–5/k and αt = 0.4 × 10–5/k, respectively. )
) Solution
fo
…(i)
rio
m
δ φ R + = l0 (1 + α c θ) 2
s_ in
ar an a
.c om
The average length of copper plate at a temperature θ = 200ºC is lc = l0 (1 + αc θ) where, l0 is the length of copper plate at 0ºC. Length of the tungsten plate is lt = l0 (1 + αtθ) From figure we have, lc = φ (R + δ/2) and lt = φ (R – δ/2) Consequently
rio s
ita
rs ve
.m
e/ un i
s: //
δ [Neglecting (αc + αt) in numerator.] (α c − α t )θ
R=
t.m
2 + (α c + α t )θ R = δ [2(α c − αt )θ]
ht tp
ht
tp
s:
or
or
si ta
iv er
un
e/
d 2 = (1 + α c θ) δ (1 + α t θ) R− 2 R
//t
tp s:
//w
w
w .ja
δ φ R − = l0 (1 + α t θ) …(ii) 2 We shall assume that the edges of plates are not displaced during deformation and that an increase in the plate thickness due to heating can be neglected. From equations (i) and (ii), we have
ht
Substituting the values in above relation, we get R = 0.769 m.
1.14
Example A mercury thermometer is to be made with glass tubing of internal bore 0.5 mm diameter and the distance between the fixed points is to be 20 cm. Estimate the internal volume of the bulb and stem below the lower fixed point. Coefficient of cubical expansion of mercury is 18 × 10–5/ ºC and coefficient of linear expansion of glass is 9 × 10–6/ºC.
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Heat
1.15
)
) Solution
Let, V0 = Volume of mercury upto the lower fixed mark (0ºC). When heated to 100ºC, the increase in volume is given as Expansion = πr2 · L 2
=
39.3 × 10−3 V0 × 100
=
39.3 × 10−5 V0
rio s_ in fo
Vθ − V0 V0 θ
ar a
er s
w
.m
iv un
t.m e/
V0 = 2.57 cm3.
s: //
ht tp s:
R
N
39.3 V0
//t
15.3 =
1.3 HEAT T A SFE R
39.3 × 10−5 + 2.7 × 10−5 (when there is no over flow.) V0
e/
tp or
ht
or
un iv
//w
w
18 × 10−5 =
s:
or
Heat can be transferred from one place to another by three different methods, namely, conduction, convection and radiation.
ht tp
ita
.ja
m
Coefficient of cubical expansion of glass γg = 3α = 3 × 9 × 10–6 = 27 × 10–6/ºC ∴ γ = γa + γg
er si ta rio s
γa =
na .c om
∴
0.5 = π × 20 20 = 39.3 × 10–3cm3 Coefficient of apparent expansion
Conduction It is the process by which the heat is transferred in solids. In conduction, molecules vibrate about a fixed location and transfer the heat by collision. Convection It is the process by which heat is transferred in fluids (liquid and gases). In convection, transfer of heat takes place by transporting of matter (in form of motion of particles). When a liquid in container is heated, the molecules at the lower layers are heated up as their densities decreases. As a result the molecules rise up and heavier ones come down, hence a continuous movement of molecules takes place giving rise to convection currents. In this manner the whole of liquid gets heated.
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1.16
Heat
radiation It is a process in which the heat is transferred in the form of electromagnetic waves without the aid of any material medium. Heat is radiated from a body due to its own temperature (greater than) is called as radiant energy.
1.4 THErmAL COnduCTIVITy
ar a er s
w w
;
un iv
//w s:
$
7
ita
.ja
m
$
er si ta rio s
7
rio s_ in fo
na .c om
The ability of a material to conduct heat is measured by thermal conductivity. Consider a slab of uniform cross-section A and length x. Let one face of the slab be maintained at temperature T1 and other at T2. Also suppose the remaining surface is covered with a non-conducting material so that no heat is transferred through the sides. After sufficient time, steady state is reached and the temperature at any point remains unchanged as time passes. In such a case, the amount of heat crossing per unit time through any cross-section of the slab is equal.
e/
.m
ht
tp
If ∆Q amount of heat crosses through any cross-section in time ∆t, then
∆Q is ∆t
…(i)
s: //
A(T1 − T2 ) ∆Q =k ∆t x
ht tp
Thus,
t.m e/
ht tp s:
un
//t
iv
called the heat current. It is found that in steady-state the heat current is proportional to the area of crosssection A, proportional to the temperature difference (T1 – T2) between the ends and inversely proportional to the length x.
where, k is a constant for the material of the slab and is called the thermal conductivity of the material. If area is non-uniform ∆Q dT = − kA …(ii) ∆t dx dT The quantity is called the temperature gradient. The minus sign indicates that dx dT is negative along the direction of heat flow. dx The unit of thermal conductivity can be easily worked using equations (i) and (ii). The SI unit is J/s-m-K or W/m-K.
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Heat
1.17
Heat Conduction Through a Composite Slab (rod) (A) Two rods connected in series. (B) Three rods connected in series. (A) Consider a composite rod made up of two rods of lengths d1 and d2 and each of cross-section A, joined end-to end (as shown). Let, k1 and k2 be the coefficients of thermal conductivities of two rods and T1 and T2 be the temperature of two ends of the composite rod. 7
N
N
G
ita
G
un iv
//w
Let us assume T as the temperature of the junction of two rods and T1 > T2. Heat flows from left to the right. In steady-state, heat flow per second is ∆Q same through each rod. It is given by H = ∆t k For Ist rod: H = 1 A(T1 − T) …(i) d1
un
k2 A (T − T2 ) d2
…(ii)
s: //
For IInd rod: H =
t.m e/
ht tp s:
//t
iv
.m
e/
s: tp ht
$
er si ta rio s
w
er s
w
.ja
m
ar a
$
7
rio s_ in fo
na .c om
7
ht tp
Eliminating T from equations (i) and (ii), we get H=
A (T1 − T2 ) d1 d 2 + k1 k2
Equating the equations (i) and (ii), we get k1 k A (T1 − T1 ) = 2 A (T − T2 ) d1 d2
⇒
k1 k T1 + 2 T2 d1 d2 T= k1 k2 + d1 d 2
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Heat
(B) Consider a composite slab made up of three rods of lengths d1, d2 and d3 joined end to end. Let area of cross-section of each rod be A. The left side of the composite rod is maintained at T1 and right side at T2 (T1 > T2). Let the coefficient of thermal conductivities of three rods be k1, k2 and k3 and junction temperature of rods 1 and 2 be TA and that of 2 and 3 be TB. 7
7$
+
7%
N
N
7
+
N
$
ar a
G
G
rio s_ in fo
na .c om
$
G
iv
un
//t
ht tp s:
k3 A (TB − T2 ) d3
t.m e/
.m
tp
For IIIrd rod: H =
er si ta rio s
ita
k2 A (TA − TB ) d2
e/
s:
For IInd rod: H =
k1 A (T1 − TA ) d1
er s
H=
//w
w
w
For Ist rod:
un iv
.ja
m
As T1 > T2, heat flows from left to the right side. At steady-state let H be the heat flowing per second.
ht
H d1 A k1
s: //
From equation (i) T1 − TA =
…(i) …(ii) …(iii) …(iv)
From equation (ii) TA − TB =
H d2 A k2
…(v)
From equation (iii) TB − T2 =
H d3 A k3
…(vi)
ht tp
1.18
Since, T1 and T2 is known, so TA and TB can be determined. To eliminate TA and TB, adding the equations (iv), (v) and (vi), we get T1 − T2 =
⇒
H=
H d1 d 2 d3 + + A k1 k2 k3 (T1 − T2 ) A d1 d 2 d3 + + k1 k2 k3
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Heat
Conduction of Heat Through a Cylindrical Shell U
U
rio s_ in fo er si ta rio s
er s
w
U
un iv
w +
iv
.m
+
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
+
U
e/
//w s: tp
+
ita
.ja
m
ar a
na .c om
/
Let, T1 = Temperature of the inner shell (T1 > T2). T2 = Temperature of the outer surface. L = Length of the cylinder. k = Thermal conductivity. H = −kA
∴ or
H
dT dT = −k (2π rL) dr dr
dr = −2π kL dT r
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1.19
Heat
integrating
r2
∫H r1
or
H In
T
2 dr = −2π kL ∫ dT r T1
r2 = −2π kL (T2 − T1 ) r1 = 2π kL (T1–T2)
or
H=
2π L (T1 − T2 ) r 1 log 2 k r1
un iv e/
iv
.m t.m e/
un
//t
2π L (T1 − T2 ) r r 1 1 log 2 + log 3 k1 r1 k2 r1
ht tp
s: //
ht tp s:
U
er si ta rio s
er s
w w //w s: tp
H=
U
ita
.ja
m
ar a
U
rio s_ in fo
na .c om
In case of two cylindrical layers:
ht
1.20
Conduction of Heat Through a Spherical Shell H=
4π (T1 − T2 ) 1 1 1 − k r1 r2
In case of multiple shells: H=
4π (T1 − T2 ) 11 1 1 1 1 − + − k1 r1 r2 k2 r2 r3
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Heat
1.21
U 7
U 7
na .c om
1.5 GrOWTH OF ICE On THE SurFACE OF A LAKE
rio s_ in fo
er si ta rio s
un iv
e/
∆Q kA (θ1 − θ2 ) = ∆t d
iv
ht
We know,
.m
tp
s:
//w
w
er s
w
ita
.ja
m
ar a
A layer of ice of thickness x is formed on the surface of a lake. The water below the ice layer is at a temperature of 0ºC. The air above the ice layer is at temperature of – θºC (below 0ºC). Heat flows from water to air through ice. As a result of this heat loss, the layer increase in thickness from below. Let dx be the increase in thickness in time dt. Heat loss from water to air = mass × latent heat = (A dx) ρ L where, ρ = Density of ice. L = Latent heat of ice. A = Area of cross-section of ice slab.
t.m e/
ht tp s:
un
//t
dx kAθ − ⇒ Heat flow per second = ΑρL = x dt
s: //
Rate of growth of thickness =
dx k θ = dt ρLx
⇒
2 2
2 1
ht tp
If the thickness changes from x1 to x2 in t seconds, then
x2
kθ
t
∫ x dx = ρL ∫ dt
x1
0
x −x kθ t = 2 ρL
DLU
Tq& LFHOD\HU
:DWHU&
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Heat
Example Water is boiled in a rectangular steel tank, the bottom of which is 5 mm thick. If the water level falls steady at the rate of 1 cm in 5 minutes, what is the temperature of the lower surface of the tank. Thermal conductivity of steel = 0.12 CGS units. Latent heat of steam = 540 cal g–1. )
) Solution
Let area of the bottom plate of the tank be A cm2 and the temperature of the lower surface be θºC. Since, the upper surface is in contact with the boiling water, its temperature may be assumed to be 100ºC. θ − 100 °C cm −1. 0.5 The heat passing by conduction through the plate is sufficient to vapourise A cm3 of water in 5 minutes or A gram in 5 minutes. This requires 540 A calories in 5 minutes. The rate of flow of heat through the plate is
ar an
in
a.
fo
co
m
Temperature gradient of steel plate =
m
rio
s_
θ 540 A = cal s −1 t 5 × 60
si
ta
.ja
Substituting in the conductivity equation
rio
s
er iv
s:
un
//t
iv
er
si
ta
un
e/
θ = 107.5ºC
⇒
540 A 0.12 × A(θ − 100) = 5 × 60 0.5
.m
tp
s:
we have
//w w
w
θ kA (θ1 − θ2 ) = t d
ht
ht
tp
s:
//t
ht
.m
tp
e/
Example A closed cubical box is made of perfectly insulating material and the only way for heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross-sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface of one plug is kept at a temperature of 100ºC while the outer surface of the other plug is maintained at a temperature of 4ºC. the thermal conductivity of the material of the plug is 2.0 W/m–ºC A source of energy generated 13 W is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is same at all points on the inner surface.
) Solution )
1.22
The situation is shown in the figure. Let the temperature inside the box be θ. The rate at which heat enters the box through the left plug is ∆Q1 kA (θ1 − θ) = ∆t x Rate of heat generation in the box = 13 W.
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Heat
1.23
The rate at which heat flows out of the box through the right plug is ∆Q 2 kA (θ − θ2 ) = ∆t x In the steady-state
T &
6
T &
na .c om
∆Q1 ∆Q 2 + 13 = ∆t ∆t
or
2 kA kA θ= (θ1 + θ2 ) + 13W x x
or
θ=
m
un
//t
iv
.m
= 52º + 216.67ºC ≅ 269ºC
er si ta rio s
er s
un iv
100 + 4 13 × 0.08 + 2 2 × 2.0 × 12 × 10−4
e/
=
ita
.ja
w
w //w s: tp ht
θ1 + θ2 (13 W) x + 2 2 kA
rio s_ in fo
kA (θ1 − θ) kA + 13 W = (θ − θ2 ) x x
ar a
or
ht tp
s: //
t.m e/
ht tp s:
Example Two parallel plates A and B are joined together to form a compound plate. The thickness of the plates are 4.0 cm and 2.5 cm respectively and the area of crosssection is 100 cm2 for each plate. A thermal conductivities are kA = 200 W/mºC for the plate A and kB = 400 W/m-ºC for the plate B. The outer surface of the plate A is maintained at 100ºC and the outer surface of the plate B is maintained at 0ºC. Find
&
$
%
&
(i) The rate of heat flow through any cross-section. (ii) The temperature at the interface. (iii) The equivalent thermal conductivity of the compound plate.
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Heat
)
) Solution
(i) Let the temperature of the interface be θ. The area of cross-section of each plate is A = 100 cm2 = 0.01 m2. The thickness are xA = 0.04 m and xB = 0.025 m. 1 xA The thermal resistance of the plate A is R1 = kA A 1 xB and that of the plate B is R 2 = kB A
The equivalent thermal resistance is
m
fo si ta er iv
un
t.m e/
s: tp
ht
3810 W =
s
rio
iv
un
.m e/
∆Q A (θ − θ2 ) = xB ∆t kB
//t
(0.01m 2 )(θ − 0°C) (0.025m) (400 W/m − °C)
s: //
in
si ta
er
w //w
s:
(0.01 m 2 ) (100°C) (0.04) (0.025m) + (200 W/m − °C) (400 W/m − °C)
=3810 W.
tp
s_
an
.ja
m
ar
w
=
rio
co
∆Q θ1 − θ2 A (θ1 − θ2 ) = = xA xB ∆t R + kA kB
Thus
or
…(i)
a.
1 xA xB + A kA kB
=
R= R1 + R2
(ii) We have,
ht
tp
xA + xB xA xB + kA kB Substituting the value of kA, kB, xA and xB, we get k = 248 W/m-ºC.
or
ht
or θ = 24ºC (iii) If k is the equivalent thermal conductivity of the compound plate, its thermal resistance is 1 xA + xB R= A k Comparing with equation (i), we get xA + xB xA xB = + k kA kB
1.24
k=
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Heat
1.25
Example One end of a uniform brass rod 15 cm long and 20 cm2 cross-sectional and is kept at 100ºC. The other end is in perfect thermal contact with an iron rod of identical cross-section and length 8 cm. The free end of iron rod is kept in melting ice and when the composite rod is well lagged and steady-state has been reached it is found that 684 g of ice melt per hour. Calculate the thermal conductivity of iron, given that the thermal conductivity of brass is 0.25 cal s–1cm–1/ºC. L = 80 cal/gm.
) Solution
G
G
For the iron rod, ∴ or
er si ta rio s
un
t.m e/
0.25 × 20(100 − θ) 15
s: //
or
15.2 =
θ = 54.4ºC Q k2 A(θ − θ2 ) = t d2
ht tp
∴
ht tp s:
//t
iv
.m
e/
s:
un iv
//w
w
er s
w
Here, Q = Heat required to melt 684 g of ice = 684 × 80 cal t = 1 hour = 3600 sec. θ 684 × 80 ∴ = = 15.2 cal s −1 t 3600 Since, the rods are well lagged, there is no loss of the heat from the surface, thereθ fore is the same for both. t Q k1 A (θ1 − θ) For the brass rod, = t d1
tp ht
T &
ita
.ja
m
ar a
T &
rio s_ in fo
na .c om
%UDVVN ,URQN "
15.2 =
k2 × 20(54.4 − 0) 8
k2 = 0.112 cal s–1cm–1/ºC.
Example On a cold winter day, the atmospheric temperature is –θ (on celsius scale) which is below 0ºC. A cylindrical drum of height h made of a bad conductor is completely filled with water at 0ºC and is kept outside without any lid. Calculate the time taken for the whole mass of water to freeze. Thermal conductivity of ice is k and its latent heat of fusion is L. Neglect expansion of water on freezing. https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
) Solution Suppose, the ice starts forming at time t = 0 and a thickness is formed
at time t. The amount of heat flown from the water to the surrounding in the time interval t to t + dt is kAθ dt x The Mass of ice formed due to the loss of this amount of heat is ∆Q=
dm =
∆Q kAθ = dt L xL
The thickness dx of ice formed in time dt is
m
±T
ita
.ja
er s
w
un iv
w //w
K
.m
G[
t.m e/
ht tp s:
un
//t
&
[
iv
tp
e/
s:
rio s_ in fo
ρL x dx kθ
kθ dt ρx L
er si ta rio s
dt =
=
na .c om
dm Aρ
ar a
or
dx =
ht T
∫ dt = 0
or
T=
h
s: //
Thus, the time T taken for the whole mass of the water to freeze is given by ρL x dx k θ ∫0
ht tp
1.26
ρ Lh 2 . 2k θ
Example Figure shows a large tank of water at a constant temperature θ0 and a small vessel containing a mass m of water at an initial temperature θ1 (< θ0). A metal rod of length L, area of cross-section A and thermal conductivity k connects the two vessels. Find the time taken for the temperature of the water in the smaller vessel to become θ2 (θ1 < θ2 < θ0). Specific heat capacity of water is s and all other heat capacities are negligible.
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Heat
1.27
T /
T
) Solution
kA (θ0 − θ) dt L
rio s_ in fo
∆Q=
na .c om
Suppose, temperature of the water in the smaller vessel is θ at time t. In the next time interval dt, a heat ∆Q is transferred to it where …(i)
er si ta rio s
un iv
ht tp s:
0
θ
Lms 2 d θ kA θ∫1 θ0 − θ
iv
∫ dt =
un
T
or
Lms d θ kA θ0 − θ
e/
dt =
.m
ht
tp
s:
or
//t
//w
w
er s
w
ita
.ja
m
ar a
This heat increases the temperature of the water of mass m to θ + dθ, where ∆Q = ms dθ ...(ii) From equations (i) and (ii), we have kA (θ0 − θ) dt = ms d θ L
Lms θ0 − θ1 ln kA θ0 − θ2
s: //
T=
ht tp
Thus,
t.m e/
where, T is the time required for the temperature of the water to become θ2.
1.6 HEAT TrAnSFEr THrOuGH rAdIATIOn In radiation, heat is transferred from one body to other or to the surroundings even in the absence of any medium in the intervening space. Heat energy of the sun is transmitted to earth through radiation. Radiation is only a mode of transfer of energy by transverse electromagnetic waves. While studying heat radiation (Radiant energy), we are concerned with thermal radiations which form the infra-red region of electromagnetic waves. All bodies emit heat to the surrounding at all temperatures and at all times. When the temperature of a body remains constant, it emits as much heat to the surrounding as it gains from them. The body is then in a state of dynamic (thermal) equilibrium.
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Heat
Black Body A black body may be described as one that completely absorbs the radiations of all wave-lengths falling upon it. There is no substance which exactly behaves like a black body as described.
Hollow Enclosure
Emissive power
rio s_ in fo
na .c om
A hollow enclosure, kept in constant temperature behaves like a black body. If a small hole is made such a enclosure, will acts as perfectly black body towards the incident radiations. Any ray getting into the enclosure through the hole will be internally reflected not be able to escape. Any radiation coming out of a narrow hole made in a hollow enclosure is generally returned to as black body radiation or cavity radiation.
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
er si ta rio s
'Z
e/
s:
un iv
//w
¨$
er s
w
w
ita
.ja
m
ar a
It is the total amount of energy radiated by a body (at a particular temperature) per second per unit area of the surface.
ht
Consider a small area ∆A of the body emitting the thermal radiation. Consider a small angle ∆ω about the normal to the radiating surface. Energy radiated per second per unit area of the surface.
ht tp
1.28
E= where,
1 ∆Q A ∆t
A = Surface area ∆Q = Heat radiated per sec. from the surface. ∆t
Absorptive power Absorption power of a body is defined as the fraction of the incident radiation that is absorbed by the body. If we denote the absorption power by a, then a=
energy absorbed energy incident
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Heat
1.29
As all the radiation incident on a black body is absorbed, the absorptive power of a black body is unity. Note that the absorptive power is a dimensionless quantity but the emissive power is not.
Kirchhoff ’s Law The ratio of emissive power to absorptive power is the same for all bodies at a given temperature and is equal to the emissive power of a black body at that temperature. E (body) = E (black body) a (body)
Thus,
Stefan’s Law
er si ta rio s
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
rio s_ in fo
na .c om
The energy emitted per second per unit area of a black body is proportional to the fourth power of the absolute temperature. i.e., E = σ T4 where, σ = Stefan’s constant = 5.67 × 10–8 J/m2/s/K4 For any other body: E = ∈ σ T4 where, ∈ = Emissivity of body (∈ = 1 for a black body)
e/
.m
ht
tp
net Heat Loss From the Surface of the Body
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
The rate at which body radiates energy is determined by the temperature of the body, but the rate at which it absorbs energy by radiation depends on the temperature of surrounding. When a body is hotter than its surroundings, the rate of absorption is greater than the rate of emission and its temperature rises. At thermal equilibrium the two rates are equal. :DOOV 7
7
Hence, for a body at a temperature of T1, surrounded by walls at a temperature T2 (as in the figure), the net rate of loss (or gain) of energy per sec by radiation is Hnet = A ∈ σ T14 – A ∈ σ T24 = A ∈ σ (T14 – T24) https://t.me/universitarios_infohttps://www.jamarana.com
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1.30
Heat
1.7 nEWTOn’S LAW OF COOLInG dQ It states that the rate of heat loss from a hot body to the surroundings is given dt dQ by = kA (θ − θ0 ), if (θ – θ0) << absolute temperature of surroundings dt where,
A = Surface area. k = Constant. θ = Absolute temperature of the body. θ0 = Temperature of surroundings. θ – θ0 = Difference in temperature.
na .c om
dθ dQ = ms dt dt
s:
Wien’s Law
er si ta rio s
er s
un iv
//w
ita
.ja
dθ ∝ A(θ − θ0 ) dt
w
ms
w
⇒
rio s_ in fo
m
ar a
where, m = Mass of the body. s = Specific heat of the body. dθ = Rate of fall of temperature (rate of cooling). dt
e/
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
The distribution of energy among the wavelengths of radiations emitted from a “Black Body” was studied by Lummer and Pringsheion. The energy contained over a wide range of wavelengths of radiations emitted by black body is not equally divided. It was found that the relative intensities of different wavelengths depend upon the absolute temperature of the radiator. The energy distributed (E) in the spectrum for the temperature range 893 K to 1893 K was found and plotted against the wavelengths (λ). The general shape of the curve is same for all temperature. From the graph it was analysed that: 7!7!7!7 7 (
7 7 7 OP
O
■ For a particular temperature, the graph is continuous on spreading from a minimum to a maximum at the other end and has a wavelength where the emission is maximum (λm). https://t.me/universitarios_infohttps://www.jamarana.com
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Heat
1.31
■ As the temperature increases, the minimum cut-off wavelength shifts to a lower value and the wavelength where emission is maximum (λm) also shift to a lower value. Then deduced that, λmT = constant where, λm = Wavelength corresponding to maximum energy at a particular temperature. The value of constant is 2.9 × 10–3 mK = 0.29 cm K. This statement is known as “Wien’s Displacement Law”. If E (λm) is the energy associated with λm at a given temperature, then E (λm) ∝ T5
na .c om
ote: Wien’s Law is fairly true for shorter wavelengths.
rio s_ in fo
N
This is known as Wien’s Fifth Power Law or simply Wien’s Displacement Law (II)
ita
er s
w
)
) Solution
.ja
m
ar a
Example A copper ball cools from 62ºC to 50ºC in 10 minutes and to 42ºC in the next 10 minutes. Calculate its temperature at the end of next 10 minutes.
θ1 + θ2 62 + 50 = = 56°C 2 2
iv
.m
The rate of fall temperature
d θ 62 − 50 = = 1.2°C/min dt 10 dθ ∝ (θ − θ0 ) dt where, θ0 = temperature of surroundings ∴ 1.2 = kA (56 – θ0) ...(i) Since, the ball cools from θ2 = 50ºC to θ3 = 42ºC in time t2 = 10 minutes, therefore
⇒
θ′ =
t.m e/
ht tp
s: //
ht tp s:
−
un
//t
ht
tp
e/
s:
θ=
er si ta rio s
un iv
//w
w
Average temperature of the liquid in the first case
θ2 + θ3 = 46°C 2
d θ′ θ2 − θ3 50 − 42 = = = 0.8 °C/min dt t2 10
d θ′ = 0.8 = kA (46 − θ0 ) dt Dividing the equation (i) by equation (ii), we get
∴
1.2 56 − θ0 = 0.8 46 − θ0 θ0 = 26ºC
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…(ii)
Heat
Since, the ball cools from θ3 = 42º to θ4 in t3 = 10 minutes, therefore θ3 + θ4 42 + θ4 = 2 2
θ″ =
d θ″ θ3 − θ4 42 − θ4 = = dt t3 10 42 + θ4 42 + θ4 = kA − 26 2 2
…(iii)
Now,
⇒
6 θ4 = 220 θ4 = 36.7ºC
in s_
m
rio
ar
an
a.
or ∴
fo
co
m
42 − θ4 42 + θ4 − 26 Dividing the equation (iii) by equation (ii), we have 10 2 = 0.8 46 − 26 42 − θ4 42 + θ4 − 52 = or 8 2 × 20 or 5 (42 – θ4) = θ4 – 10
s
ta
e/
2
//t 2
s:
ht
.m
tp
A1 r1 = A 2 r2
∴
si
er
un
//t
iv
A1 = Surface area of the first sphere = 4πr12 A2 = Surface area of the second sphere = 4πr22
s:
)
) Solution
.m
rio
iv
un
e/
s:
(i) The initial rates of loss of heat. (ii) The initial rates of fall of temperature in the two cases.
tp
//w
w
er
w
si
ta
.ja
Example Two solid copper spheres of radii 6 cm and 12 cm are suspended in two evacuated chambers, each being at a temperature 15ºC above its surroundings. Compare:
ht
ht
tp
6 1 = = 4 12
4 Mass of the first sphere, M1 = volume × density = π r13 d 3
4 M 2 = π r23 d 3 3
(i)
3
M1 r1 6 1 = = = M 2 r2 12 8
∴
1.32
dQ1 = kA1 (θ − θ0 ) dt dQ 2 = kA 2 (θ − θ0 ) dt
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Heat
1.33
dQ1 dt = A1 = 1 . dQ 2 A 2 4 dt
dθ (ii) M1s 1 ∝ A1 (θ − θ0 ) dt
dθ M 2 s 2 ∝ A 2 (θ − θ0 ) dt d θ1 M1 dt A ∴ ⋅ = 1 d θ M2 A2 2 dt
o
co
nf
a.
s_ i
an
ta r
m
io
ar
tp
e/
s er si
ta
un
iv
w
s:
//w
w
or
er si
.ja
d θ1 dt = 2 . d θ2 1 dt
rio
d θ1 1 dt 1 or ⋅ = 8 d θ2 4 dt
m
iv
//t
.m
ht tp
e/
un
s: //
ht
t.m
Example A blackened metal sphere of thermal capacity 1000 cal/ºC and surface area 500 cm2 is kept inside an evacuated enclosure with blackened walls maintained at 27ºC. Find the rate of cooling of the sphere when its temperature is at 127ºC, σ = 1.36 × 10–12 cal. cm–2s–1K–4. )
s:
) Solution
ht
tp
Given: Thermal capacity = ms = 1000 cal/ºC A = 500 cm2 T = 273 + 127 = 400 K T0 = 27 + 273 = 300 K It is required to find d θ = Rate of cooling
dt d θ ∈σA (T4 – T 4) ...(where, ∈ = 1 for black body) ms = 0 dt
= 1.36 × 10–12 × 500 × 108 (256 – 81)
∴
dt
or 1000 d θ = 1 × 1.36 × 10–12 × 500 [(400)4 – (300)4] d θ 1.36 × 5 × 10–5 × 175 = dt
= 0.0119ºC/sec.
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Heat
Example Estimate the temperature of the earth, assuming it to be in radioactive equilibrium with the sun. The radius of the sun = 7 × 108 m, temperature of the solar surface = 6000 K, distance of earth from sun = 1.5 × 1011 m.
) Solution Assuming both the sun and the earth as black bodies.
where,
K=
π rE2 4π R 2
rio s_ in fo
Q2 = K × Q1
∴
na .c om
Energy radiated from the sun in unit time Q1 = σ AT4 where, A is the surface of the sun Q1 = σ × 4π rs2 × Ts4 At a distance R from the sun, where the earth is situated, the entire energy radiated by the sun is spread into an area of 4π R2. Out of this area, the earth blocks an effective area of πrE2 (cross-sectional area of the earth) and the radiation spread in this area is incident upon the earth. Let Q2 be the energy absorbed by the earth in unit time
ar a
The fraction of the total radiated energy absorbed by the earth, π rE2 × σ 4π rs2 × Ts4 4π R 2
6XQ¶VUD\V
t.m e/
ht tp s:
un
//t
iv
.m
e/
s: tp
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
Q2 =
m
∴
ht
s: //
(DUWK
Energy radiated by the earth = σ 4π rE2 × TE4 When, there is a radioactive equilibrium, then Energy radiated by the earth = Energy absorbed by the earth
ht tp
1.34
or or or
σ 4π rE 2 × TE 4 =
π rE2 σ 4π rS2 × TS4 4π R 2
r2 TE 4 = TS4 s 2 4R 1/2
r TE = TS s 2R
1/2
7 × 108 = 6000 × 11 2 × 1.5 × 10
∴
TE = 290 K.
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)
) Solution We have, dθ = − k (θ − θ0 ) dt where, θ0 is the temperature of the surrounding and θ is the temperature of the body at time t. Suppose, θ = θ1 at t = 0, then
ita
∴ 10800 + 2000 – 20 θ = 8000 + 500 + 110 θ
= 33ºC.
un
3. An earthen pitcher loses one gram of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water. Calculate the time required for the water in the pitcher cool to 28ºC from its original temperature of 30ºC. Neglect radiation effects. Average latent heat of vaporization of water in this range of temperature is 580 cal g–1.
iv
er si ta rio s
ht tp
s: //
∆Qms θ1 − θ0 = 2ms 2 If the body loses this heat in time t1, the temperature at t1 will be
θ1 − θ0 θ1 + θ0 = 2 2 Putting these values of time and temperature in equation (i), we get
θ1 −
130 θ = 4300 4300 θ= 130
t.m e/
ht tp s:
//t
.m
e/
or
or
un iv
//w
s:
tp
rio s_ in fo
ar a m
w
...(i)
The body continues to lose heat till its temperature becomes equal to the surrounding. The loss of heat in this entire period is ∆Qm = ms (θ1 – θ0) It is the maximum heat, that a body can lose. If the body loses half this heat, the decrease in its temperature will be
ht
xa
)
Heat lost by steam = mL + ms (100 – θ) where, θ is the equilibrium temperature ∴ Heat lost by steam = 20 × 540 + 20 × 1 (100 – θ) = 10800 + 2000 – 20 θ Heat gained by (ice + calorimeter) = 100 × 80 + 100 × 0.5 × 10 + 110 × θ Now Heat lost = Heat gained
er s
θ – θ0 = (θ1 – θ0) e–kt
w
or
ln
) Solution
.ja
θ − θ0 = −kt θ1 − θ0
or
dθ ∫θ θ − θ0 = −k ∫0 dt 1
S
2. 20 gm steam at 100ºC is let into a closed calorimeter of water equivalent 10 gm containing 100 gm ice at – 10ºC. Find the final temperature of the calorimeter and its contents. Latent heat of steam is 540 cal/gm, latent heat of fusion of ice = 80 cal/gm, specific heat of ice = 0.5 cal/ºC gm.
na .c om
t
θ
mp
1.35
U
D
1. A hot body placed in air is cooled down according to Newton’s law of cooling, the rate of decrease of temperature from the surrounding. Starting from t = 0, find the time in which the body will lose half the maximum heat it can lose.
or
or
θ1 + θ0 − θ0 = (θ1 − θ0 ) e − kt1 2 1 e − kt1 = 2 ln 2 t1 = . k
) Solution Heat lost by pitcher and water )
R
WO KE O T E
le
Heat
equivalent Q = (9.5 + 0.1) × 1 × 30.2 × 1000 = 20000 cal. Mass of the vaporise water, 20000 m= = 34.48 g 580 Time required = 34.48 min.
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1.36
4. At a temperature of 20ºC, the volume of a certain glass flask, up to a reference mark on the cylindrical stem of the flask is exactly 100 cm3. The flask is filled to this point with a liquid of cubical expansion 120 × 10–5/ºC, when both the flask and liquid is at 20ºC. The cross-section of stem is 1 mm2 and the coefficient of linear expansion of glass is 8 × 10–6/ºC. Find the rise or fall of the liquid level in the stem, when the temperature is raised to 40ºC.
Volume of iron tank after its heating V = V0 (1 + γl ∆T) ...(ii) where, γl = 3α1. The density of kerosene after the heating of the tank is ρ0 ρ= ...(iii) (1 + γ k ∆T)
Substituting expression equations (ii) and (iii) into equation (i), we get
∆m = ρ0 V0 −
Heat
ρ0 V0 (1 + 3α l ∆T) 1 + γ k ∆T
)
) Solution Given that volume = 100 cm
3
m
=
co
fo
a.
≈ ρ0 V0 (γk – 3αl) (T – T0) ∆m = 0.8 × 103 × 5 × 10–2 × (10 × 10–4 – 3 × 1.2 × 10–5) (293 – 273) kg = 0.7712 kg
s_
s
)
e/
un
The required pitch p can be found by dividing the cylinder height h by the rotational speed of drill n, need to drill the cylinder through h p= n In drilling these evolves the heat Q = ms ∆T = Ahρs ∆T πd 2 where, A= 4 Q h= Hence, Aρs ∆T
ht
iv
) Solution
.m
tp s
:// t
ht
rio
ta
er si
e/
tp
s:
//t
.m
ht tp
io
ta r
si
un
s:
)
as a result of heating by 90ºC will be larger than the increase the volume of tank, and a part of kerosene will flow out. We denote by m0 the mass of kerosene in the tank at 273 K and by m the mass of kerosene at room temperature.
U
6. In drilling in a copper cylinder of a bore of diameter 25 mm, the cylinder heats by 43 K. The input torque is 16.2 Nm. 70 per cent of the energy input converts into the internal energy of the cylinder. Compute the pitch of a drill.
er
iv
//w w
w .ja
m
ar
an
) Solution The increase in the volume of kerosene
OLWHUV
ρ0 V0 ( γ k − 3α L ) (T − T0 ) 1 + γ k (T − T0 )
in
5. An iron tank has a capacity of V = 50 liters of kerosene at T0 = 273 K. What amount of kerosene will flow out the full tank if it is brought in a room temperature at T = 293 K. [The density of kerosene ρ = 0.8 × 103, coefficient of volume expansion of kerosene γk = 10 × 10–4 and coefficient of linear expansion of iron αi = 1.2 × 10–5].
Coefficient of cubical expansion of = 235.2 cm.
where, ρ0 is the density at 273 K and ρ and V are the density of kerosene and the volume of tank at room temperature.
The quantity n is to be found from the relationship for the work W done in the drilling of the cylinder. ∴ W = τ 2πn W n= or 2π τ Substituting the values of h and n into the expression for P and considering
Then mass of kerosene flowing out of the tank is ∆m = m0 – m or ∆m = ρ0 V0 – ρV ...(i)
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1.37
Heat
or
60 – θB = 2 (θB – θC) + (θB – θD) 4 θB – 2 θC – θD = 60
For junction C: 2k (θB – θC) = 2k (θC – 10) + 2k (θC – θD)
nf o
rio
si ta
s
rio
ita
...(v)
rs un
4 θB – 8 θC = –40
θB – 2 θC = –10
e/
θC = 20ºC θB = 2 θC – 10 = 30ºC
∴
Thus, the junction temperature of B, C and D are 30º, 20º and 90º respectively. Alternate Method This problem can also be solved by applying Kirchhoff’s Law. Let kx and ky be the thermal conductivities of X and Y and kx and ky their resistance.
(
or
'
Let θB, θC and θD be the temperatures of the junction B, C and D at steady thermal conduction state.
Rx =
1 l kx A
And, R y =
1 l ky A
N
N
...(vi)
5 θC = 100
.m //t
N N
%
er
iv
e/ un
ht tp s:
)
$& N
4 θB – 3 θC = 60
Substracting equation (vi) from (v), we get
& N
θB – 3 θC – θD = θB + 2 θC – 4 θD ...(iv)
From equation (ii), we get or
<
...(iii)
Therefore, from equation (i), we get
(
'
) Solution
ar
m
s: tp
ht
<
θB + 2 θC – 4 θD = –10
From equations (ii) and (iii), we get
.m ;
%
//t
< $
k (θB – θD) + 2k (θC – θD) = k (θD – 10)
∴
.ja
w w
w
s: // tp ht
;
...(ii)
For junction D:
an a
&
θB – 3θC + θD = –10
∴
.c om
7. Three rods of materials X and three rods of material Y are connected as shown in the figure. All the rods are of identical length and cross-sectional area. If the end of A maintained at 60ºC and the junction E at 10ºC, calculate the temperature of the junction B, C, D. The thermal conductivity of X is 0.92 CGS units and that of Y is 0.46 CGS units. Assume all rods are well lagged.
s_ i
≅ .0010 m ≈ 1.0 mm.
;
...(i)
∴
8 × 16.2 × 0.70 380 × 625 × 10−6 × .8900 × 43
iv e
P=
2πτ K 8τ K = 2 2 ρd s ∆T πd ρ s ∆T 4
=
2π τ Q 2πτK ⋅ = ρAs ∆T W ρAs ∆T
P=
∴
If the thermal conductivity of rod Y is taken as k, then the rod X is 2 k. At steady conduction state, heat entering a junction must be equal to that leaving the junction. Therefore, for junction B: k (60 – θB) = 2k (θB – θC) + k (θB – θD)
Q = K, (given), we have W
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1.38
8. A rod CD of thermal resistance 5.0 k/W is joined at the middle of an identical rod AB as shown in the figure. The ends A, B and D are maintained at 100ºC, 0ºC and 25ºC respectively. Find the heat current in CD.
From the figure, no heat current is passing through C and D rod because loop is balanced. So, the figure is reduced to
Heat
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(
% &
&
&
'5\
=
or
m
fo
in
s_
ta
si
s
rio
ita
rs
∆Q3 45°C − 25°C = ∆t 5.0 k/W
50 (k x + k y )k y k x + k y + 2k xy
=
50 × (0.92 + 0.46) 50 × 1.38 = = 30 2.3 3 × 0.46 + 0.92
60 – θB = 30 θB = 30ºC.
50 (k x + k y ) kx + 3 k y
ni
ve
e/ u
.m //t tp s:
100 − θ θ − 0°C θ − 25°C = + 2.5 2.5 5 225°C = 5θ θ = 45°C
iv un .m e/
Thus,
∆Q1 ∆Q 2 ∆Q3 = + ∆t ∆t ∆t
20 K = 4.0 W 5.0 k /W
9. As a result of a temperature rise of 32ºC, a bar with a crack at its centre buckles upward shown in the figure. It the fixed distance L0 = 3.77 m and the coefficient of linear expansion is 25 × 10–6/ºC. Find x, the distance to which the centre rises.
60 − θB =
1 2 + k y kx + k y
or
or ∴
or
=
50
ht
k y (60 − θB ) =
∆Q3 θ − 25°C = ∆t 5.0 k/W
We also have
ht
or
∆Q 2 θ − 0°C = ∆t 2.5 k/W
and
//t
60 − θB 50 = 1 1 1 2⋅ ⋅ ky kx k y 1 + 1 1 ky + kx k y
tp s:
or
∆Q1 100° − θ = ∆t 2.5 k/W
er
w
//w
ht tp
s:
50 2R x R y Ry + Rx + Ry
)
co a. …(ii)
From equations (i) and (ii), we have 60 − θB = Ry
The thermal resistance of AC is equal to that of CB and is equal to 2.5 k/W. Suppose, the temperature at C is θ. The heat current through AC, CB and CD are
rio
60 − 10 2R x R y Ry + Rx + Ry
'
) Solution
ar an
θA − θE = R equivalent
2(R x + R y )
m
i=
2R x 2R y
w .ja
R equivalent = R y +
∴
&
…(i)
θA − θB =i Ry
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Heat
l12 4
…(i)
= l22 −
1.39
3 /
O
) Solution )
$
Length increases when rises the temperature. Let us consider the half length. Then the increased length
When the temperature increased l′1 = l1 (1 + α1 θ) and l2′ = l2 (1 + α2 θ) According to the problem, length of the PQ is fixed, therefore
m
o
co
s_ i
s
rio
l1 α2 =2 . l2 α1
un i
ve r
si ta
l12 ⋅ 2α1θ 4
11. An aluminium plate is fixed in a horizontal position has a hole of diameter da = 2 cm. A steel sphere of diameter ds = 2.005 cm rests on this hole. The temperature of the entire system is slowly increased. At what temperature, the steel plate covers the hole? Given: αa = 23 × 10–6/ºC, and αs = 11 × 10–6/ºC
e/
)
) Solution
When the temperature is increased, diameter of the plate is increased and the diameter of steel ball is also increased. At that point where both the diameters are equal, the steel sphere covers the hole. Given: da = 2 cm, and ds = 2.005 Let at the temperature θ, both diameters are equal. ∴ da (1 + αa ∆θ) = ds (1 + αs∆θ) or da + da αa ∆θ = ds + ds αs ∆θ
PQ = AP 2 − AQ 2
From the figure
ht
)
) Solution
l22 ⋅ 2α 2 θ =
10. An isosceles triangle is formed with a thin rod of length l1 and coefficient of linear expansion α1 as the base and two thin rods each of length l2 and coefficient of the linear expansion α2 as the two sides. If the distance between the apex and the mid-point of the base remain unchanged as the temperature varied, show that l1 α2 =2 . l2 α2
l22 −
t.m s: // tp
= 7.5 cm
l2 l12 = l22 (1 + 2α 2 θ) − 1 (1 + 2α1 θ) 4 4 (because α1 and α2 is very small.)
rio
si ta er iv un
ht
= (188.6508) 2 − (188.5) 2
or
:// t
2
tp s
ht
L x = l ′2 − 0 2
or
.m e/
tp s:
∴
ar a .ja m
//w w
3.77 × 100 (1 + 25 × 10−6 × 32) 2 = 188.5 (1 + 800 × 10–6) = 188.6508
l′ =
l22 −
or
w
/
l12 l2 = [l2 (1 + α 2 θ]2 − 1 (1 + α1θ) 2 4 4
nf
na .
l′ =
[
%
4 O
L0 (1 + αθ) 2 where, α = coefficient of linear expansion
O
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1.40
Heat
mg mgl Young’s modulus = A = dl A dl l Y A dl or m = gl
(da αa – ds αs) ∆θ = ds – da
θ − 10 =
2.005 − 2.000 2 × 23 × 10−6 − 2.005 × 11× 10−6 .005 = 208°C 23.945 × 10−6
θ = (208 + 10) = 218ºC.
Substituting the values we get
1011 × 4 × 10−6 × 0.5 × 10−3 = 400 kg 10 × 0.5 (ii) Since, the wire does not undergo extension or contraction, the energy stored in the rod under the influence of constant weight is E = (mg) dl = (400) × (10) × 5 × 10–4 = 2.0 J m=
12. A thin rod of negligible mass and area of cross-section 4 × 10–6m2, suspended vertically from one end, has a length of 0.5 m at 100ºC. The rod is cooled to 0ºC, but prevented from contracting by attaching a mass at the lower end. Find: (i) This mass. (ii) The energy stored in the rod. Given for the rod, Young’s modules = 1011N/m2, coefficient of linear expansion = 10–5K and g = 10 m/s2.
fo
in
ta
si
e/
un
iv
er
si
ta
rio
s
er iv
un
ht
.m
tp
s:
//t
ht
.m
tp
e/
s:
(i) Contraction of length of the rod due to the cooling the temperature from 100ºC to 0ºC. dl = l αθ = 0.5 × 10–5 × 100 = .05 × 10–3m
s:
)
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) Solution
s_
rio
m
w
//w w
)
) Solution
13. A double-pane window used for insulating a room thermally from outside, consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steadystate, the room-glass interface and the glass-outdoor interface are at constant temperatures at 27ºC and 0ºC respectively. Calculate the rate of heat flow through the window pane. Also find the temperature of other interface. Given that thermal conductivities of glass and air as 0.8 and 0.08 Wm–1K–1, respectively.
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m
∴
=
ds − da da αa − ds αs
∆θ =
∴
or
ht
G4 GW
7
7
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tp
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7
7
G4 GW
P
$LU G4 GW
P P
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Heat
1.41
Here, A = 1 m2, ka = 0.08 Wm–1K–1, A = 1 m2 kg = 0.8 Wm–1K–1 By conduction equation,
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ρ 1 γ = 5°C − 1 ρ 95°C 90°C = 2 × 10–4/ºC ∴ Coefficient of linear expansion γ = = 6.7 × 10−5/°C. 3 15. A solid body X of heat capacity C is kept in an atmosphere whose temperature is TA = 300 K. At time t = 0 the temperature of X is T0 = 400 K. It cools according to Newton’s law of cooling. At time t1, its temperature is found to be 350 K. At this time (t1), the body X is connected to a large box Y at atmospheric temperature TA, through a conducting rod of length L, cross-sectional area A and thermal conductivity k. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1.
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s:
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dQ1 = 80 (300 – 299.48) = 41.6 W dt
ρ5°C − ρ95°C ρ95°C (90°C)
s
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=
iv
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//t
s:
tp
Coefficient of volume expansion, V95°C − V5°C γ= V5°C (95°C − 5°C)
si
s:
tp
ht
and T3 = 273.52 K ⇒ Q3 = 0.52ºC Rate of heat flow from window-pane
fo
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m
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w
w
//w
...(ii)
From equation (ii) T2 + T3 = 573 K From equation (i) 51 T2 – T3 = 15000 K ∴ T2 = 299.48 K ⇒ Q2 = 26.48ºC
ρ5°C 52.8 + 49 = = 1.08 ρ95°C 49 + 51
or
er
0.8 × 1 (T3 − 273K) 0.01
and 80 (300 – T2) = 80 (T3 – 273)
&
ρ95ºCg (52.8 cm) – ρ5ºCg (49 cm) + ρ95ºCg (49) – ρ5ºDg (51) = 0
or
dQ1 dQ 2 dQ3 = = dt dt dt This gives 80 (300 – T2) = 1.6 (T2 – T3) ...(i)
&
P0 = P0 + ρAghA – ρBghB + ρCghC – ρDghD
0.08 × 1 (T2 − T3 ) 0.05
At steady-state,
14. The apparatus shown in the figure consists of four glass columns connected by horizontal sections. The height of two central columns B and C are 49 cm each. The two outer columns A and D are open to the atmosphere. A and C are maintained at a temperature of 95ºC while the columns B and D are maintained at 5ºC. The height of the liquid in A and D measured from the base lines are 52.8 cm and 51 cm respectively. Determine the coefficient of thermal expansion of the liquid.
&
Applying Pascal’s law, we have
dQ3 k g A (T3 − T4 ) = dt d
=
'
) Solution
dQ 2 ka A (T2 − T3 ) = dt d =
&
dQ1 k g A (T1 − T2 ) = dt d 0.8 × 1 (300 − T2 ) = 0.01
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1.42
Heat
Integrating this expression, we get
)
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3t
T2
/
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%RG\< .
or
ln
$WPRVSKHUH7 .
dT kA 1 ∫T T − TA = − k + CL ∫t dt 1 1
$
FRQGXFWLQJURG
ln 2 kA = − + (2t1 ) CL t1 2k A t1 = −2 ln 2 − CL
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dT kA = − k + (T − TA ) dt CL
= (300 K) + (12.5 K) e −2 k A t1 / CL .
)
d T −kA = (T − TA ) dt CL where, C = heat capacity of the body. Taking them together, we get
or
1 = (300 K) + (350 − 300) e −2 k A t1 /CL 4
16. A thermocole vessel contain 0.5 kg of distilled water at 30ºC. A metal coil of area 5 × 10–3m2, number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform time varying magnetic field set up to pass vertically through the coil at time t = 0. The field is first increased from 0 to 0.8 T at a constant rate between 0 and 0.2 second then decreased to zero at the same rate between 0.2 and 0.4 sec. The cycle repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as functions of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axis. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium. Specific heat capacity of the metal = 500 J kg–1K–1 and the specific heat capacity of water = 4200 J kg–1K–1. Neglect the inductance of coil.
s: //t
ht tp
dQ −k A = (T − TA ) dt L
or
T2 = TA + (T1 − TA ) e −2ln 2 e −2 k A t1 /CL
.m
ht tp
s: //t
ht
.m
For the time t > t1, the temperature of the body falls due to Newton’s law of cooling as well as due to conduction of heat through the conducting rod (length L, area of cross-section A, and conductivity k) For cooling via Newton’s Law dT = −k (T − TA ) dt For cooling via conduction
dT kA = − k + dt T − TA CL
or
ve
(350 − 300) = ln 2 (400 − 300)
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k t1 = −ln
tp s
or
T2 − TA 2k A t1 = exp −2ln 2 − T1 − TA CL
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T1 − TA = −k t1 T0 − TA
w
ln
0
w
∴
= −k ∫ dt
A
T0
ar a
dT
∫ T−T
w
Integrating
t1
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dT = −k (T − TA ) dt T1
or
Upto time t1, the body cools under Newton’s law of cooling, for which we write
+HDWFDSDFLW\ &
T2 − TA kA = − k + (2t1 ) T1 − TA CL
Area of the coil, A = 5 × 10–3m2 Number of turns of coil, n = 100
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1.43
Heat
or
−d φ dB = nA dt dt
θ=
Power of the coil = Ei = 1.25 × 2 = 2.5 W
L
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//t
If L1 – L2 is to be independent of ∆T, then
l1α1 – l2α2 = 0 l1 α1 = l2 α 2
2. 70 calories of heat is required to raise the temperature of 2 mole of an ideal gas from 30ºC to 35ºC at constant pressure. The heat required (in calories) to raise the temperature of the same gas through the same range (30ºC to 35ºC) at constant volume will be (a) 30 (b) 50 (c) 70 (d) 90
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⇒
L1 – L2 = (l1 – l2) + (∆T) (l1α1 – l2α2)
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)
L2 = l2[1 + α2∆T]
n
s
s: //
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α12 α 22 = l1 l2
) Explanation L1 = l1[1 + α1∆T] ∴
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iv
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ht
1. Two rods of length l1 and l2 are made of materials whose coefficients of linear expansion are α1 and α2. If the difference between two lengths is independent of temperature, then l α l α (a) 1 = 1 (b) 1 = 2 l2 α 2 l2 α1 (d)
W
ita
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w
w
//w
s:
tp
SOLVE OBJECTIVE yp qu
(c) l22 α1 = l12 α 2
3 P
0.5 × 4200 (θ – 303 ºK) + 0.06 × 500 (θ – 303 ºK) = 12000
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4T = (100) (5.0 × 10–3 m2) =2V s Resistance of the coil = 1.6 Ω E Current of the coil, i = R 2 = = 1.25 A 1.6 Heat of the coil, H = i2 Rt = (1.25)2 × 1.6 × 0.2 = 0.5 J Again decreased the same rate, so the heat produces by decreasing the magnetic field = 0.5 J. Total heat produced in 12000 cycle = (0.5 + 0.5) × 12000 = 12000 J This heat increased both the temperature of water and coil. mwsw∆θ + mcsc∆θ = 12000 J ∴
657390 = 308.63 K = 35.63°C 2130
estio
E=
2100 θ – 636300 + 30 θ – 9090 = 12000 2130 θ = 657390
or or
Rate of increasing magnetic field, dB 0.8T 4T = = dt 0.2s s Electro-magnetic field induced in the coil,
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1.44
) Explanation P = εσAT4
)
)
) Explanation 2CP(35 – 30) = 70
3. A wall is made of two equally thick layers A and B of different materials. The thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36ºC. The temperature difference across the layer A will be (a) 6ºC (b) 12ºC (c) 18ºC (d) 24ºC
(
)
2 −1
(d)
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2 −1
1
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the given problems is shown in the figure. Since TB > TA, the heat will flow form B to A. To maintain steady state, heat will flow from C to B and A to C Applying the conduction formula,
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ta r
er
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∆Q KA = (∆T) ∆t l to the sides AC and CB, we get
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dT 2 1 = = ∆θdt 30 × 4 4
(
(b)
)
://
tp s
dT = K∆θ dt
tp s: //
⇒
K=
∴
dT ∝ ∆θ dt
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)
) Explanation Rate of cooling ∝ difference in i.e.,
1
) Explanation The diagramatic representation of
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tp s
w
w
w
4. In a room where the temperature is 30ºC, a body cools from 61ºC to 59ºC is 4 minutes. The time taken by the body to cool from 51ºC to 49ºC will be (a) 4 minutes (b) 6 minutes (c) 5 minutes (d) 8 minutes temperature
.ja m
∆TA = 12ºC
∴
2
(c)
ar
∆TA + ∆TB = 36ºC
(a)
er
and
( )
an a.
)
⇒
4
P1 A1 T1 4π(1) 2 4000 = = =1 P2 A 2 T2 4π(4) 4 2000
6. Three rods of identical cross-sectional area and made from the same metal from the sides of an isosceles triangle ABC, right-angled at B. The points A and B are maintained at temperatures T and 2 T respectively. In the steady state, the temperature of the point C is Tc. Assuming that only heat conduction takes place, Tc/T is
2KA (∆T) A KA(∆T) B = d d 2(∆T)A = (∆T)B
) Explanation
4
∴
Heat required at constant volume = 2Cv(35 – 30) = 2(CP – R) (35 – 30) = 70 – 2 × 2 × 5 = 50 cal
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D
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7& 7
7
⇒
T − TC TC − 2T = a 2a
⇒
T − TC = 2TC − 2T
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5. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The power radiated by the two spheres are in the ratio (a) 16 : 1 (b) 4 : 1 (c) 4 : 1 (d) 1 : 1
ht
In first case, dT = 61 – 59 = 2, ∆θ = 60 – 30 = 30, dt = 4 minutes For second case, dT = 2 ∆θ = 50 – 30 = 20 2 dT ∴ dt = = = 6 min 1 K∆θ × 20 60
∆T ∆T = a CB 2a AC
Heat
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Heat
(
3
This result comes from the fact that Weight = Upthrust
)
2 +1
∴ Vρsolid g = Vsubmerged ρliquid g
∴
(ii) Since ρ ∝
) Explanation A black body is a perfect absorber
=
Vsubmerged
V
ρθ°C
∴
co
V0 °C V0 °C + V0 °C γθ
1 1 + γθ
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90 − θ 90 − θ θ − 0 + = R R R Here R = Thermal Resistance ∴ 3θ = 180 or θ = 60ºC
∴
k1 (ρFe )0°C (ρHg )60°C (1 + 60γ Fe ) = = k2 (ρFe )60°C (ρHg )0°C (1 + 60γ Hg )
∴
junction (say B). Thermal resistance of all the three rods is equal. Rate of heat flow throught AB + Rate of heat flow though CB = Rate of heat flow through BD
ρ = Density
=
nf
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1 (mass remaining Volume constant) Vθ°C Vθ°C = = V0 °C V0 °C + ∆V
) Explanation Let θ be the temperature of the
s:
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ρ solid ρliquid
9. Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0ºC and 90ºC respectively. The temperature of the junction of the three rods will be (a) 45ºC (b) 60ºC (c) 30ºC (d) 20ºC
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tp
and
1 + 60γ Fe 1 − 60γ Hg
ht
ρ k2 = Fe ρ Hg
(d)
ht
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ρ ) Explanation k1 = Fe ρHg
1 − 60γ Fe 1 + 60γ Hg
tp s
1 + 60γ Fe 1 − 60γ Hg
(b)
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8. When a block of iron floats in mercury at 0ºC, a fraction k1 of its volume is submerged, while at the temperature 60ºC, a fraction k2 is seen to be submerged. If the coefficient of volume expansion of iron is γFe and that of mercury is γHg, then the ratio k1/k2 can be expressed as (a)
ρ0 ° C
m
)
and a perfect radiator
=
7. An ideal black-body at room temperature is thrown into a furnace. It is observed that (a) initially it is the darkest body and at later times the brightest (b) it is the darkest body at all times (c) it cannot be distinguished at all times (d) initially it is the darkest body and at later times it cannot be distinguished
TC = T1
⇒
)
2 +1
(
3T = TC
⇒
1.45
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1. Thermal coefficient of volume expansion (V) can be written as (symbols have their usual meanings) dρ dρ (b) −ρ (a) −T dT dT
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1 dρ ρ dT
dρ dT
2. A rod of length 40 cm has the coefficient of linear expansion α1 = 6 × 10–6/ºC. Another rod of length l has the coefficient of linear expansion α2 = 4 × 10–6/ºC. If the difference in length of the two rods always remain same at all temperatures, then the value of l is (a) 26 cm (b) 60 cm (c) 80 cm (d) 32 cm
(b)
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4. If E denotes the energy of a particle of mass m and λ be its de-Broglie wavelength then the correct curve between 1 y = 2mE and x = is λ
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5. Two holes of unequal diameters d1 and d2 (d1 > d2) are cut in a metal sheet. If the sheet is heated, (a) both d1 and d2 will decrease (b) both d1 and d2 will increase (c) d1 will increase, d2 will decrease (d) d1 will decrease, d2 will increase
(d)
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3. Three stars A, B and C appears as green, red and blue respectively. The star having minimum temperature is (a) A (b) B (c) D (d) A and B have the same temperature
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Heat
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λB corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 µm. If the temperature of A is 5802 K, then (a) the temperature of B is 1934 K (b) λB = 1.5 µm (c) the temperature of B is 11604 K (d) the temperature of B is 20 K
G
6. In the previous question, the distance between the holes will (a) increase (b) decrease (c) remain constant (d) may either increase or decrease depending on the positions on the sheet and on the ratio d1/d2
9. Left and right ends of a metal conductors of length 2m, cross section area 5 × 10–4m2 and thermal coefficient 80 w/m-K. is maintained at temperature 80ºC and 20ºC respectively. If an electrical resistance delivering a constant power 2 watt is attached to the conductor at a distance 1.2 m from left end, then temperature at the section of resistance will be (a) 90ºC (b) 45ºC
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10. A metal wire of length L and radius r is made by copper and aluminium contributing equal length. This wire is now coated by nickel till the radius of wire becomes R. If specific resistances of these materials are ρCu, ρAl and ρni the equivalent conductance of the system across length will be
e/ .m
πr 2 1 1 R2 + + 2 L ρCu ρAl r ρ Ni
(b)
π r2 r 2 2(R 2 − r 2 ) + + L ρCu ρAl r 2ρ Ni
(c)
π 2(R 2 − r 2 ) r 2 − L ρCu + ρAl ρ Ni
(d)
π 2r 2 R2 − r2 − L ρCu + ρAl ρ Ni
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8. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of two bodies are same. The two bodies emit total radiant power at same rate. The wavelength
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(d) 22.5ºC
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7. The two ends of a uniformly tapered rod are maintained at temperatures 100ºC and 0ºC respectively. Under steady state, the temperature of the section AA exactly at the mid-point (a) 50ºC (b) more than 50ºC (c) less than 50 ºC (d) none of these
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13. A black body is at a temperature of 2880 K. The energy of radiation emitted
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16. Twelve identical conducting rods form the sides of a uniform cube of side l. In the steady state, B and H ends of the cube area at 100ºC and 0ºC. Find the temperature of the junction A. (a) 80ºC (b) 60ºC (c) 40ºC (d) 70ºC
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15. Two indentical conducting rods AB and CD are connects to a circular conducting ring at two diametrically opposite points B and C. The radius of the ring is equal to the length of rods AB and CD. The area of cross-section and thermal conductivity of the rod and ring are equal. Points A and D are maintained at temperatures of 100ºC and 0ºC temperature of point C will be (a) 62ºC (b) 37ºC (c) 28ºC (d) 45ºC
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11. A spherical black body radiating power P and rate of cooling of the body is R. If the radius of the body is r, then (a) P ∝ r (b) P ∝ r2 1 (c) R ∝ r2 (d) R ∝ r 12. Three rods of equal length are joined to form an equilateral triangle ABC. D is the mid-point of AB. The coefficient of linear expansion is α1 for AB and α2 for AC and BC. If the distance DC remains constant for small changes in temperature, then (a) α1 = α2 (b) α1 = 2 α2 1 (c) α1 = 4 α2 (d) α1 = α 2 2
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14. 1 kg of ice at 0ºC is mixed with 1.5 kg of water 45ºC [latent heat of fusion = 80 cal g]. Then (a) the temperature of the mixture is 0ºC (b) mixture contains 156.25 gm of ice (c) mixture contains 843.75 gm of ice (d) the temperature of the mixture 15ºC
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by this object with wavelength between 499 nm and 500 nm is u1, between 999 nm and 1000 nm is u2 and between 1499 nm and 1500 nm is u3. The Wien constant is b = 2.88 × 106 nmk. Then (a) u1 = 0 (b) u3 = 0 (c) u1 > u2 (d) u2 > u1
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17. A solid cube of side ‘a’ and another solid cube with the same material and same side but having spherical cavity of radius b where (b < a) are heated up to the same temperature and allowed to cool in the same surroundings, which one will start cooling faster? (a) both will cool at same rate (b) solid cube (c) hollow cube (d) can’t say
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18. A solid with coefficient of linear expansion α just floats in a liquid whose coefficient of volume expansion is γ. If the system is heated, the solid will (a) sink in all the cases (b) continue to float in all cases (c) sink if γ > 3α (d) sink if γ < 3α
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21. A mixture of ideal gases filled inside a balloon expands according to the relation PV2/3 = constant. The temperature inside the balloon is (a) increasing (b) decreasing (c) constant (d) can not be defined
(b)
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22. A cylindrical vessel of length 10 m and area of cross-section 1 m2 is filled with water and is rotated with angular speed
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20. A tungsten wire with a diameter of 0.20 mm is stretched until the tension becomes 50 N. The wire is then clamped to a stout aluminium rod when both are at a temperature of 150ºC. The tension in the wire when both are brought to a temperature of 20ºC is (Ytungsten = 34 × 1010 N/m2, αT = 4.5 × 10–6/ºC) (Neglect the expansion of aluminium rod) (a) 85 N (b) 57.5 N (c) 100 N (d) 56.2 N
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19. A conducting sphere cools in atmosphere according to Newton’s law of cooling. The correct graph between its temperature θ and time t is
(d)
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1.50
26. To vaporize 4 gm of water by boiling it at 373ºK, 2160 cal of heat is needed. What is specific heat of water in the conduction? (a) 0.36 cal/gm-ºK (b) 5.4 cal/gm-ºK (c) zero (d) infinite
ω = 20 rad/s. Find the rise in temperature of water when it is stopped all of a sudden. (a) no change (b) 2.5ºC (c) 1.58ºC (d) 5ºC
Heat
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27. The two ends of a uniform rod of thermal conductivity K are maintained at different, but constant temperatures. The temperature gradient at any point on the dθ (equal to the difference in temrod is dt perature per unit length). The heat flow per unit time per unit cross-section of the rod is I.
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25. A thermally insulated vessel contains 100 g of water at 0ºC. When air above the water is pumped out, some of the water freezes and some evaporates at 0ºC itself. Calculate the mass of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at 0.C is 2.1 × 106 J/kg and Latent heat of fusion of ice is 3.36 × 105 J/kg. (a) 62 g (b) 86 g (c) 54 g (d) 78 g
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tp s
ht
28. Three rods of the same dimensions have thermal conductivities 3 K, 2 K and K. They are arranged as shown, with their ends at 100ºC, 50ºC and 0ºC. The temperature of their junction is 200 (a) 75°C (b) °C 3 100 °C (c) 40°C (d) 3
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29. A and B are two points on a uniform metal ring whose centre is C. The angle ACB = θ. A and B are maintained at two different constant temperatures when θ = 180º, the rate of total heat flow from A to B is 1.2 W. When θ = 90º, this rate will be (a) 0.6 W (b) 0.9 W (c) 1.6 W (d) 1.8 W
ht
24. A square metallic plate is having a circular hole in it. If the temperature of the plate is increased, the area of the hole (a) decreases (b) increases (c) remains same (d) can not be decided
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m
256 81
t.m
(d)
dθ is the same for all points on the dt rod (b) I will decrease as we move from higher to lower temperature dθ (c) I = K. dl (d) None of these
(a)
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64 27
(c)
in
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fo
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23. The power radiated by a black body is Pr and it radiates maximum energy around the wavelength α0. If the temperature of the black body is now changed so that it radiates maximum energy around a 3λ 0 wavelength , the power radiated by 4 it will increase by a factor of 16 4 (a) (b) 9 3
P
P
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4πK 2 T KP
(b) (d) (a, b) (c)
6. 14. 22. 30.
(d)
(c)
(b)
4πKR 2 T P
(a)
30. A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed T, the thickness of the shell should not be less than
4πKP 2 TP 4πR 2 P KT
ANSWERS 5. 13. 21. 29.
(b) (c) (d) (b)
7. (b) 15. (b) 23. (d)
8. (c) 16. (b) 24. (b)
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(a) (a, d) (c) (a)
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4. 12. 20. 28.
(b) (d) (d) (a, c)
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3. 11. 19. 27.
(b) (d) (c) (d)
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2. 10. 18. 26.
m
(b) (b) (c) (b)
tp
1. 9. 17. 25.
ht
Heat
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1.51
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iv
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c h a p t e r
2
rio s_ in fo
na .c om
Kinetic Theory of Gases and Thermodynamics
ar a
2.1 FUNDAMENTAL ASSUMPTIONS OF KINETIC THEORY OF GASES
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(i) All gases consists of molecules which are considered as rigid perfectly elastic and solid spheres. These molecules are extremely small i.e., the volume occupied by the molecules is taken negligible as compared to the volume of the containing vessel. (ii) The molecules are in a continuous random motion. They are moving in all directions with all possible velocities. There is no particular direction of motion of the particles. (iii) The molecules during their motion collide with one another and also with the walls of the vessel. At each collision, their velocities change both in magnitude and direction; however in the steady-state, molecular density remains uniform throughout the gas and does not change with time. (iv) No force of attraction or repulsion are exerted on a gas molecule by other molecules or by the container except during collisions. This means, the energy of the gas is purely kinetic. (v) The collisions among the molecules or with the walls of container are purely elastic. A collision is said to perfectly elastic if the energy of the molecules remains same before and after the collision. (vi) Molecules travel in a straight line with a uniform velocity in between the collisions. The average distance travelled by the molecules between the successive collisions is called as mean-free path of a molecule. The time spent during a collision is assumed to be negligibly small as compared with the time taken by the molecule in traversing a mean-free path.
Note: A gas which satisfies all of above set of assumptions of kinetic theory of gases under all conditions of temperature and pressure is called as an ideal gas or perfect gas. A real gas shows appreciable deviations from the perfect gas. At high temperature and low pressure, gases show deviation from the ideal behaviours.
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2.2
Kinetic Theory of Gases and Thermodynamics
2.2 EVIDENCE IN SUPPORT OF MOLECULAR MOTION Diffusion The phenomenon of diffusion presents a remarkable evidance of molecular motion of matter. The particles of one substance diffuse into another against gravity due to the molecular motion. H2 diffuses into CO2 when a jar filled with H2 is held mouth to mouth over a jar filled with CO2. Though H2 is lighter than CO2.
Evaporation
Brownian Motion
rio s_ in fo
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na .c om
The phenomenon of evaporation is related with the motion of molecules. The molecules of liquid are quite free to move about. At any temperature, all the molecules don’t move with the same velocity, some move faster than the others. Evaporation involves the escape of fast moving molecules (having speed greater than average speed). This results in lowering of average speed, which is turn results in lowering of temperature and thus producing a cooling effect.
er si ta rio s
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w
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Botanist Robert Brown observed the irregular motion of plane spores floating in water. Such motions are called as Brownian motion. Here the motion of each particle is due to inequality of the impact of surrounding molecules on it. For example: Motion of smoke particles in air is a case of Brownian motion. This supports the view that the molecules of a substance are in constant motion.
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s: //
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ht tp s:
z
un
iv
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//t
ht
2.3 CALCULATION OF THE PRESSURE OF AN IDEAL GAS
vz
A2
vy
A1 V vx
O
x
y
Consider an ideal gas enclosed in a cubical vessel of edge L. Take a corner of the vessel as the origin O and the x, y and z-axis along the edges. Let A1 and A2 be the parallel faces perpendicular to the x- axis. Consider a molecule moving with velocity v. The components of the velocity along the axis are vx, vy and vz. When the molecule collides with the face A1, the x-component of the velocity https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.3
is reversed whereas the y and the z-components remain unchanged. This follows from our assumption that the collisions of the molecules with walls are perfectely elastic. The change in momentum of the molecules is ∆p = (–mvx) – (mvx) = 2 mvx
As the momentum remains conserved in a collision, the change in momentum of the wall ...(i)
∆p′ = 2mvx
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ar a
in f
na .
o
co m
After rebound, this molecules travels towards A2 with the x-component of velocity equal to – vx, any collision of the molecule with any other face (except for A2) does not change the value of vx. So, it travels between A1 and A2 with a constant x-component of velocity which is equal to – vx. Note that, we can neglect any collision with the other molecules in view of the last assumption discussed in the previous section. Distance travelled parallel to the x-direction between A1 and A2 = L.
.ja m
rio
Thus, the time taken by the molecules to go from A1 to A 2 =
L . vx
s
rio
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w
iv er s
w
ita
The molecule rebounds from A2, travels towards A1 and collides with it after another L time interval . vx
ita
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ht
t.m
tp s
e/ un
Thus, the time between two consecutive collisions of this molecules with A1 is 2L ∆t = . vx
:// t.m
tp s
ht
://
The number of collisions of this molecule with A1 in unit time is v 1 n= = x ∆ t 2L
…(ii)
tp s
The momentum imparted per unit time to the wall by this molecule is from equations (i) and (ii).
=
ht
ΔF = n Δp’
vx m × 2m vx = vx 2 2L L
This is also the force exerted on the wall A1 due to this molecule. The total force on the wall A1 due to all the molecules is m m F = ∑ vx 2 = ∑ vx 2 …(iii) L L As all directions are equivalent, we have
Σ vx2 = Σ vy2 = Σ vz2 1 = ∑ (v x 2 + v y 2 + v z 2 ) 3 1 = ∑ v2 3
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Kinetic Theory of Gases and Thermodynamics
Thus, from equation (iii) F =
1m ∑ v2. 3L
If N is the total number of molecules in the sample, we can write
F=
1 mN ∑ v 2 3 L N
=
1 mN ∑ v 2 3 L3 N
=
1 M ∑ v2 1 ∑ v2 = ñ 3 L3 N 3 N
co m
Pressure is the force per unit area. Therefore F p= 2 L
a.
fo
v2 is the averN 2 age of the speeds squared. It is written as v and is called mean square speed.
rio
ar
si
s rio
iv
…(ii)
er si
ta
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…(iii)
iv
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un
//t
ht
tp
or
…(i)
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s:
or
//w w
w
1 p = ρ v2 3 1 pV = M v 2 3 1 pV = Nm v 2 3
ta
.ja
m
Thus, the pressure is
s_
an
in
where, M is the total mass of the gas taken and ρ is its density, and ∑
tp
e/
s:
Root-Mean-Square Velocity (RMS)
s:
//t
ht
.m
The square root of mean-square speed is called root-mean-square speed or rms speed. It is denoted by the symbol vrms. Thus,
ht
tp
v 2 + v22 + ...vN2 ∑ v2 = 1 N N
vrms =
In terms of pressure
1 2 p = ρ vrms 3
∴
2.4
vrms =
3p 3 pV 3RT = = ρ M M
For the same gas at two different temperatures T1 and T2, vrms will be different As vrms ∝ T ∴
vrms (1) T1 = vrms (2) T2
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Kinetic Theory of Gases and Thermodynamics
2.5
For two gases at different temperature T1 and T2:
vrms (1) T M = 1 2 vrms (2) T2 M1 Average speed: (vav)
vav =
v1 + v2 + v3 + ...vN N 8 RT πM
vav =
ar a
2.4 DEDUCTIONS FROM KINETIC THEORY
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Boyle’s Law
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2 RT M
vmp =
na .c om
Most probable speed: (vmp)
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s:
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At a given temperature, the pressure of a given mass of a gas is inversely proportional to its volume. This is known as Boyle’s law. 1 We know …(i) pV = mN v 2 rms 3
iv
ht tp s:
ht tp
p∝
s: //
p1V1 = p2V2
1 V which is Boyle’s law. or
t.m e/
pV = constant
or
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//t
ht
.m
tp
e/
As for a given gas v2rms ∝ T, the value of v2rms is constant at a given temperature. Also for a given mass of the gas M and N are constant. Thus, from equation (i)
Charles’ Law
As
At a given pressure, the volume of a given mass of a gas is proportional to its absolute temperature. This is known as Charles’ Law. From equation (i), if p is constant, then V ∝ v2rms v2rms ∝ T
or
We get V ∝ T V1 V2 = T1 T2
which is Charles’ law. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
Charles’ Law for Pressure At a given volume, the pressure of a given mass of a gas is proportional to its absolute temperature. This is known as Charles’ law for pressure. In fact, this is the definition of the absolute temperature T. If one starts from the 2 fact that Vrms ∝ T and uses the fact that V is constant, one gets from equation (i)
2 p ∝ vrms
p∝V
or
p1 p2 = V1 V2
or
na .c om
Avogadro’s Law
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At the same temperature and pressure, equal volumes of all gases contain equal number of molecules. This is known as Avogadro’s law. Consider equal volume of two gases kept at the same pressure and temperature.
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t.m e/
1 pV = N 2 m2 v2 2 3
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ht
tp
s:
1 Now, pV = N1 m1 v12 3 And
er si ta rio s
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Let m1 = mass of the molecule of the first gas. m2 = mass of the molecule of the second gas. N1 = number of molecules of the first gas. N2 = number of molecules of the second gas. p = common pressure of the two gases. V = common volume of the two gases.
s: //
N1 m1 v12 = N2 m2 v22
...(i)
ht tp
where, v1 and v2 are root-mean-square speeds of the molecules of the first and second gas respectively. Thus,
As the temperature of the gases are the same, the average kinetic energy of the molecules is same for the two gases, i.e., 1 1 m1v12 = m2 v2 2 2 2
…(ii)
2.6
From equations (i) and (ii), we get N1 = N2 which proves Avogadro’s law.
Graham’s Law of Diffusion When two gases at the same pressure and temperature are allowed to diffuse into each other, the rate of diffusion of each gas is inversely proportional to the square root of the density of the gas. This is known as Graham’s law of diffusion. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.7
It is reasonable to assume that the rate of diffusion is proportional to the rms speed of the molecules of the gas. Then, if r1 and r2 be the rates of diffusion of the two gases, …(i)
r1 v1 , rms = r2 v2 , rms
3p ρ If the pressure of the two gases are the same, then
We know, vrms =
rio s_ in fo
r1 ρ2 = r2 ρ1 which is Graham’s law.
na .c om
v1 , rms ρ2 = v2 , rms ρ1 Hence from equation (i)
ar a
Dalton’s Law of Partial Pressure
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Dalton’s law of partial pressure says that the pressure exerted by a mixture of several gases equals the sum of the pressures exerted by each gas occupying the same volume as that of the mixture. In kinetic theory, we assume that the pressure exerted by a gas on the walls of a container is due to the collisions of the molecules with the walls. The total force on the wall is the sum of the forces exerted by individual molecules. Suppose, there are N1 molecules of gas 1, N2 molecules of gas 2 etc., in the mixture. Thus, the force on a wall of surface area A is
un
//t
ht tp
s: //
t.m e/
ht tp s:
F = Force by N1 molecules of gas 1 + force by N2 molecules of gas 2 + ........... = F1 + F2 + .......... Thus, the pressure is F F p = 1 + 2 + ....... A A If the first gas alone is kept in the container, its N1 molecules will exert a force F1 on the wall. If the pressure in this case is p1, then F p1 = 1 A
Similarly, for other gases. p = p1 + p2 + p3 + .......
Ideal Gas Equation
For n moles of an ideal gas at pressure p, absolute temperature T occupying a volume V, the following equation holds: pV = nRT where R = Ideal gas constant
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2.8
Kinetic Theory of Gases and Thermodynamics
The gas equation can also be written in the following form: pM = d RT where, M is the molecular weight and d is the density of the gas at pressure p and absolute temperature T. The molecular weight M of a gas can be given as:
g M = densityin at S.T.P. × 22.4 l
Boltzmann Constant (k)
in fo
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pV = NkT
R T = n N 0 kT N0
s
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pV = nRT = n N 0
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Gas equation in terms of k:
rio
ar an
a.
co m
In the gas equation pV = nRT, the value of R be taken as: R = 8.314 J/mol/K R = 0.0821 -atm/mol/K R = 2 cal/mol/K R is known as Boltzmann constant. If N0 is the Avogadro number then, k = N0 k = 1.38 × 10–23 J per molecule per degree Kelvin.
iv e un
//t
ht
.m
2.5 KINETIC ENERGY OF MOLECULES
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Degrees of Freedom
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tp
s:
//t .m
The degrees of freedom of a dynamical system are defined as the total number of independent quantities required in order to describe the position or motion of the system completely.
Mono-Atomic Gas Molecule
ht
ta rs i
tp
e/
s:
un i
where, N = n N0 = number of molecules.
The molecules of a mono-atomic gas (He, Ne, Ar) consists of one atom only. It is capable of only translatory motion in free space which has three components namely. x, y and z-directions. Hence, it will have three degrees of freedom per molecule.
Di-Atomic Gas Molecule Each molecule of a di-atomic gas (H2, N2, O2, d2, etc.) consists of two atoms. In the given diagram, a di-atomic molecule A2 (A – A) is shown. A2 is capable of translatory motion in space and, thus possesses three degrees of freedom. Beside this, the molecule may rotate about yy′-axis or zz′-axis (perpendicular to the molecular axis and the plane of paper). Rotation about molecular axis (xx′) is neglected (because the size of atom is negligible). Thus, due to rotational motion, the molecules have two degree of freedom. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.9
Y Z
A
A
X'
X
Y'
rio s_ in fo
na .c om
Z'
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w
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w
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Hence, a di-atomic molecule has five degrees of freedom. Note: A tri-atomic molecule (CO2, H2O, SO2 etc.) has six degrees of freedom (three of them for translatory and three for rotational motion).
un iv
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2.6 KINETIC ENERGY OF TRANSLATION PER MOLECULE
iv
1 1 1 m0 v12 + m0 v22 + ... + m0 vN2 2 2 = 2 N 1 31 2 2 = m0 vrms = m0 vrms 2 23
s: //
t.m e/
ht tp s:
Kinetic energymean
un
//t
ht
.m
tp
e/
s:
The average kinetic energy or kinetic energy per molecule due to translatory motion is also known as mean kinetic energy of translation.
ht tp
1 31 2 2 = m0 vrms = m0 vrms 2 23 3 pV = 2 N 3 = kT 2
……(as pV = NkT)
Kinetic energy of translation per mole = (KE)mean N0 or
3 3 k N 0 T = RT 2 2
Kinetic energy of translation in n moles of a gas at pressure p and volume V 3 3 = n RT = pV 2 2 https://t.me/universitarios_infohttps://www.jamarana.com
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2.10
Kinetic Theory of Gases and Thermodynamics
2.7 EQUIPARTITION OF ENERGY
rio s_ in fo
er si ta rio s
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Total energy per molecule = 3 kT Total energy per mole = 3 RT
ita
m
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na .c om
The law of equipartition of energy states that for a dynamical system in thermal equilibrium, the total energy of the system is equally divided amongst the various degrees of freedom. 1 The share of each degree of freedom is kT. 2 For Mono-atomic gas: 3 Total energy per molecule = kT 2 3 Total energy per mole = RT 2 For di-atomic gas: 5 Total energy of molecule = kT 2 5 Total energy of mole = RT 2 For tri-atomic gas:
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s:
2.8 THERMAL EXPANSION OF AN ENCLOSED GAS
iv
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s: //
t.m e/
ht tp s:
un
//t
ht
.m
tp
When the enclosed gas is heated, its volume and pressure tend to increase. Keeping one of them constant, the thermal expansion of the other can be studied. Since, thermal expansions of gases are appreciable, the coefficient of thermal expension are referred to a reference initial temperature of 0ºC. The coefficient of volume expansion of an enclosed gas at constant pressure is defined as the increase in its volume per unit rise in temperature, per unit volume at 0ºC. The coefficient of pressure increase of an enclosed gas at constant volume is defined as the increase in its pressure, per unit rise in temperature per unit pressure at 0ºC. For an ideal gas, i.e., a gas which obeys the general gas law pV = nRT; Coefficient of pressure increase at constant volume = Coefficient of volume expansion at constant pressure 1 1 = (C°) −1 or K −1 273.15 273.15
=
1 K −1 273
Note: If the mass of the enclosed gas does not remain constant, i.e., the gas leaks out of or leaks into the enclosure, the partial laws (Boyle’s, Charles’ and Gay-Lussac’s laws) are not applicable. But the general gas law holds good and the value of n changes. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.11
2.9 STANDARD CONDITIONS
One mole of any gas occupies a volume of 22.4 litres (2.24 × 10–2 m3) at standard (or normal) temperature and pressure (STP or NTP). Standard temperature = 0ºC = 273.15 K Standard pressure = 1 atm = 1.013 × 105 Nm–2 In practice, standard temperature is taken as 273 K only. Example
)
) Solution MAr = 0.040 kg
ar a
3 RT 3 × 8.3 × (273 + 25) = = 430.7 m/sec. M .040
er s
w
For the same gas
s: //
Example
t.m e/
T1 = 74.5 K = – 198.5 °C
un
//t
iv
.m
e/
un iv
//w
or
T1 1 = 2 298
ht tp s:
ht
or
tp
s:
w
vrms (1) T1 = vrms (2) T2
er si ta rio s
ita
.ja
m
R = 8.3 J/mol/K vrms =
and
rio s_ in fo
na .c om
Calculate the root-mean-square speed of argon atoms at room temperature (25ºC). At what temperature will the root-mean-square speed will be half of the value?
ht tp
Calculate the pressure of Hydrogen (H) in a cylinder of capacity 10 litre, given that the total kinetic energy of translation is 7.5 × 103 J. What is the total kinetic energy of molecules in the cylinder? (1 atm pressure = 105 N/m2). )
) Solution The total kinetic energy of translation in n moles for a di-atomic gas is
3 3 nRT = pV 2 2
or
∴ 7.5 × 103 J =
3 p (10 × 10−3 m3 ) 2
p = 5 × 105 N/m2
Total kinetic energy for di-atomic gas 5 5 = pV = × 5 × 105 × 10 × 10−3 J = 12500 J. 2 2 https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
Example A vertical cylinder of total length 100 cm is closed at the lower end and is fitted with a movable, frictionless gas tight disc at the other end. An ideal gas is trapped under the disc. Initially, the height of the gas column is 90 cm when the disc is in equilibrium between the gas and the atmosphere. Mercury is then slowly pured on top of the disc and it just starts overflowing when the disc has descended through 32 cm. Find the atmospheric pressure. Assume the temperature of the gas to remain constant the neglect the thickness and weight of the disc.
) Solution
Here, V1 = Initial volume of air = 90 A
un iv e/
p1
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er s
w w //w s:
42 cm
p2
.m
tp
p
ita
.ja
p
rio s_ in fo
m
ar a
na .c om
A = Area of cross-section of the disc V2 = Final volume of air = 58 A p1 = Initial pressure of the gas = Atmospheric pressure (p) (This is because the disc is light and in equilibrium under pressure P and P1.)
58 cm
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
100 90 cm cm
ht
2.12
In the second case, p2 balances the atmospheric pressure and 42 cm of mercury column. ∴ p2 = p + 42 p1V1 = p2V2 ∴ p × 90 A = (p + 42) × 58 A ∴ p = 76.125 cm of Hg. Example An oxygen storage tank has a capacity of 0.05 m3. The gas pressure is 100 atm. at 27ºC. Determine: (i) The total kinetic energy of translation of molecules. (ii) The average kinetic energy of translation of molecules. (iii) The total kinetic energy. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.13
)
) Solution Total kinetic energy of translation =
(ii)
3 (100) × 105 × 0.05 2 = 7.5 × 105 J 3 Average kinetic energy = kT 2 3 = (1.38 × 10−23 ) × 300 2 = 6.21 × 10–21 J.
(i)
3 pV 2
=
The total kinetic energy (for di-atomic gas) 5 5 = nRT = pV 2 2 5 = × 100 × 105 × 0.05 2 = 1.25 × 106 J.
ita rio
.ja m
w
w
rio s
Example
ve rs
ar an
s_ in
fo
a. c
om
(iii)
=
.m //t
n RT1 p1
s:
(i) V1 =
ht tp
)
ht
) Solution
e/
s:
un
//t
iv e
rs
.m e/
The volume of the gas before expansion. The temperature of gas after expansion. The density of gas before expansion. The density of gas after expansion.
tp
ht
(i) (ii) (iii) (iv)
10 × 0.0821× 283 = 2.42 litre 32 × 3
(ii) As pressure is constant
⇒ T2 =
V1 V2 = T1 T2
V2 T1 = 1169.4 K V1
G
PDVV 9
(iv) 'HQVLW\
G
PDVV 9
(iii) 'HQVLW\
ita
tp s:
un i
//w
10 gm of oxygen are subjected to a pressure of 3 atm, at a temperature of 10ºC. Heating at a constant pressure, the gas is expanded to 10 litre. Find:
JP OLWUH
J OLWUH
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Kinetic Theory of Gases and Thermodynamics
Example An open glass tube is immerged in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and raised further by 44 cm. What will be the length of air column above mercury in the tube? (Atmospheric pressure = 76 cm of Hg).
) Solution x (52–x)
52 cm
rio s_ in fo
m
ar a
na .c om
8 cm
er si ta rio s
e/
.m
iv
un
//t
t.m e/
s: //
∴
x = length of the air column finally p2 = New pressure = 76 – (52 – x) = (24 + x) cm of Hg V2 = New Volume = A · x
ht tp s:
tp
s:
un iv
//w
w
er s
w
ita
.ja
p1 = Initial pressure of air in the tube outside the mercury surface = 76 cm of Hg V1 = Initial volume = A × 8 (A = Area of cross-section) After closing the open end, the total length of the glass tube outside is 52 cm. This increases the volume and reduces the pressure of air.
ht
Since T is constant, therefore p1V1 = p2V2 or 76 × A × 8 = (24 + x) Ax or x2 + 24x – 608 = 0
ht tp
2.14
Solving, we get x = 15.4 cm or 39.4 cm ∴ Length of the air column is 15.4 cm. Example A cylindrical diving bell, whose length is 150 cm is lowered to the bottom of a tank. The water is found to rise 50 cm in the bell. Find the depth of the tank. Assume the atmospheric pressure at the surface as equivalent to 1000 cm of water and the temperature as constant. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.15
) Solution When the cylindrical diving bell is in air, it is filled with air at atmospheric pressure p0 = 1000 cm of water ∴ V0 = Initial volume of air = A × 150 (A = Area of cross-section) The hydrostatic pressure at the level of water in the bell is (h – 50) cm of water. p0
p
m
ar a
50 cm
p = Total pressure of air (when the bell is in water) = p0 + (h – 50) = 1000 + h – 50 = 950 + h V = New volume of air = A × 100 Since, the temperature remains constant, therefore p0V0 = pV or 1000 × A × 150 = (950 + h) × A × 100 or h = 550 cm.
s: //
t.m e/
ht tp s:
un
//t
iv
er si ta rio s
e/
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
∴
ht
150 cm
rio s_ in fo
na .c om
h
ht tp
Example A mercury barometer is defective. (It contains some air in the space above the mercury). When an accurate barometer reads 770 mm, the defective one reads 760 mm. When the accurate one reads 750 mm, the defective one reads 742 mm. What is the length of the air column, when the accurate barometer reads 760 mm? What is the reading of the accurate barometer when the defective one reads 752 mm? Assume that the temperature remains constant.
) Solution
The difference between the readings of the accurate and defective instruments represents the pressure exerted by the air above the mercury in the defective barometer. (i) p1 = Pressure of air in the first case = 770 – 760 = 10 mm p2 = Pressure of air in the second case = 750 – 742 = 8 mm
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Kinetic Theory of Gases and Thermodynamics
rio s_ in fo
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
ar a
na .c om
Let, x cm be the length of the air column above 760 mm mark in the first case. Therefore, when it reads 742, the length of the air column would be (x + 1.8) cm. ∴ V1 = Initial volume of air = Ax where A = Area of cross-section V2 = Final volume of air = A (x + 1.8) From Boyle’s law, p1V1 = p2V2 ∴ (10) (Ax) = (8) (A) (x + 1.8) or x = 7.2 cm (ii) V = Volume of air when the defective one reads 752 mm = (x + 0.8) A p = Pressure of air = (p0 – 752) where, p0 = Reading of the accurate barometer. From Boyle’s law p1V1 = p2V2 ∴ 10 × Ax = (p0 – 752) A (x + 0.8) or 10 × 7.2 = (p0 – 752) (7.2 + 0.8) or p0 = 761 mm.
un
t.m e/
p
s: //
ht tp s:
//t
iv
.m
tp
e/
s:
Example A bulb of 500 cm3 is joined by a narrow tube of another bulb B of 250 cm3 and the whole system is initially filled with air at S.T.P. and sealed. If the temperature of the bulb A is now raised to 100ºC and that of B is kept constant, find:
ht
VA
ht tp
2.16
A
p VB
B
(i) The new pressure in the system. (ii) Mass of the air which is transferred from one bulb to another during heating. (Density of air at S.T.P. = 1.29 g/l)
) Solution (i)
Let, p be the final pressure of air in the bulbs. As the bulbs are connected, pressure will be equal in both bulbs. Let, TA and TB be the final temperatures of the bulbs, therefore TA = 373 K TB = 273 K
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Kinetic Theory of Gases and Thermodynamics
2.17
If p0, T0 are the S.T.P. conditions, then Initial moles in the system = Final moles in the system p0 VA p0 VB pVA pVB + = + R T0 R T0 RTA RTB
Using, p0 = 1 atm, T0 = 273 K, we have
1 500 250 p + (500 + 250) = 373 273 273 or p = 1.22 atm
(ii)
M = Molecular weight of air, M = (dSTP) (22.4)
fo in
an
a. co
m
= 1.29 × 22.4 = 28.896 gm Mass transferred = moles × M p VB p0 VB = − × M R TB R T0
rio
ar
w
si
ta
.ja
m
s
er
w
iv
:// w
rio
2.10 THERMODYNAMICS
s_
250 × 10−3 (1.22 − 1) × 28.896 .082 × 273 = 0.0709 gm. =
si ta
iv
ht
//t .m
er
e/
tp s
un
Temperature It may be defined as that thermal property which attains the same value for all bodies placed in thermal contact with each other.
.m
ht tp
e/
s:
un
Heat Energy The transfer or flow of internal energy caused by the difference of temperatures between two bodies is called as Heat Energy.
ht tp
s:
//t
Zeroth Law of Thermodynamics If two bodies A and B are in thermal equilibrium with another body C, then they are also in thermal equilibrium with each other. Thermodynamical system It is a collection of an extremely large numbers of atoms or molecules so that they together have the same pressure (P), Volume (V) and temperature (T). e.g., A gas, vapour, steam, vapour in contact with the liquid.
2.11 WORK DONE AND INTERNAL ENERGY Work Done by a Gas During Expansion Consider an ideal gas enclosed in a perfectly insulated cylinder fitted with a nonconducting and frictionless piston. Let p be the pressure exerted by the gas and V be the volume of the gas at any particular instant. Let A be the area of cross-section of the piston. The force exerted by the gas on the piston = pA https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
dx
gas
m
dW = pdW
er s
w
ita
.ja
Work Done From P-V Diagram
rio s_ in fo
ar a
na .c om
If the piston moves through an infinitesimal distance dx, this force can be assumed constant and the workdone (dW) is given as: dW = (pA) But A dx = dV = infinitesimal change in volume
er si ta rio s
e/
p p1
ht tp
p2
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
s:
un iv
//w
w
An ideal gas contained in a cylinder fitted with a massless and frictionless piston can be considered as a thermodynamical system. Its state can be represented by variables (p, V, T). If p and V are known, T can be calculated by
ht
2.18
O
v1
v2
V
pV V = nRT RT The system’s state at any instant of time can be specified by two variable (p, V). The relation between pressure p and volume V can be studied on a pressure – volume (p – V) graph known as indicator diagram or a p – V diagram. On such a graph, each equilibrium state of a thermodynamical system can be represented by a point whose X co-ordinate represents volume (V) and Y co-ordinate represents pressure (p). https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.19
A process can be represented by a curve on the p – V diagram. If the gas goes from initial state A (p1, V1) to the final state B (p2, V2), the work done W is given by W = Area under the curve AB above the x-axis (shaded portion).
Internal Energy Internal energy (U) of a system is the total of all kinds of energy possessed by the atoms and other particles that comprise the system. It depends only on the temperature difference of initial and final states and not on the process between the states.
2.12 FIRST LAW OF THERMODYNAMICS
rio s_ in fo
∆Q = ∆U + ∆W
m
ar a
na .c om
If some heat is supplied to a system capable of doing some work, the quantity of heat absorbed by the system is equal to the sum of external work done by the system and increase in its internal energy. If ∆Q is the heat gained by the system, ∆W is the work done against external pressure and ∆U is the change in internal energy, then ...(i)
er si ta rio s
w
er s
w
ita
.ja
Heat energy given Increase in Workdone by = + to the system internal energy the system
un iv
//w
This equation is the statement of the first law of thermodynamics.
e/
iv
.m
ht
tp
s:
Note: First law of thermodynamics is a direct consequence of law of conservation of energy.
un t.m e/
ht tp s:
//t
In Thermodynamics
s: //
ht tp
(i) Workdone by a system is taken is positive (+). Workdone on the system is taken as negative (–). (ii) Heat gained by a system is taken as positive (+) and the heat lost by the system is taken as negative (–). (iii) Increase in internal energy is taken positive (+) and decrease in internal energy is taken as negative (–).
2.13 SPECIFIC HEAT OF A GAS The specific heat of a gas is defined as the amount of heat required to raise the temperature of unit mass of substance through 1ºK. Its unit in S.I. system is J/kg/K. When the unit mass is 1 mole, the specific heat is known as molar specific heat and its units are J/mol/K or cal/mol//K. Molar specific heat = molecular weight (specific heat in J/gm/K) In gases, on giving heat to the gas, besides temperature, its pressure and the volume may also change. Thus, specific heat of a gas depends upon the conditions under which the heat is given to the gas. Hence, we will define two principle types of specific heats of a gas. https://t.me/universitarios_infohttps://www.jamarana.com
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2.20
Kinetic Theory of Gases and Thermodynamics
Specific Heat at Constant Volume (CV) Cv of a gas is defined as the amount of heat required to raise the temperature of 1 mole of that gas through 1ºK provided that the volume of the gas remains constant. It is also known as molar specific heat at constant volume. ∆Q CV = n ∆T Constant volume
In general, if n moles of a gas are heated at constant volume to increase the temperature by ∆T.
Heat required = ∆Q = n Cv ∆T
co
m
Specific Heat at Constant Pressure (CP)
_i
io s
∆Q CP = n ∆T Constant pressure
.ja
ta r
m
ar
an
nf
a.
o
CP of a gas is defined as the amount of heat required to raise the temperature of 1 mole of that gas through 1ºK, provided that the pressure of the gas remains constant. It is also known as molar specific heat at constant pressure.
s
rio
iv
si ta
CP is greater than CV (CP > CV)
er
e/
un
s: //
Heat required = ∆Q = n CP ∆T
tp
w
w
er
w
si
In general, if n moles of a gas are heated at constant pressure resulting in a temperature raise by ∆T, then
iv
t.m
ht
s: //
(i) To raise the temperature of the gas. (ii) To do the mechanical work for expansion.
tp
e/
un
s: //t
ht
.m
When we heat the gas at constant volume, the heat is supplied to raise the temperature of gas only. When we heat the gas at constant pressure, volume of the gas apart from temperature of the gas raises. When the gas expands, it does some external work for increasing the volume. So in this case, the heat is supplied:
ht
tp
As a result more amount of heat is required to increase the temperature of the gas through the same amount when the pressure is kept constant. Hence, CP is always greater than the value of CV. C The ratio of P is denoted by γ CV
i.e.,
γ=
CP ( γ > 1) CV
2.14 SPECIFIC HEAT RELATION
Consider n moles of an ideal gas. If its temperature is raised by ∆T at constant volume, then the heat transferred is n CV ∆T, where CV is the specific heat at constant volume. W = 0 (as no volume change is there.) https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.21
By First Law of Thermodynamics
n CV ∆T = ∆U
...(i)
∆Q = ∆U + 0
Next, let us have the same gas at constant pressure. The heat ∆Q required to change the temperature by ∆T will be nCp ∆T, where Cp is the specific heat at constant pressure. W = P ∆V where, ∆V is the change in volume. By First Law of Thermodynamics ...(ii)
om
∆Q = ∆U + p∆V nCP ∆T = ∆U + p∆V From equations (i) and (ii), we get
rio s_ ta
.ja m
ar
an a
in fo
.c
nCP ∆T = n CV ∆T + p∆V From gas equation p∆V = nR ∆T ⇒ nCP ∆T = n CV ∆T + nR ∆T CP = CV + R CP – CV = R
e/
t.m
s: //
…(as Cp Cv = R)
ht tp
ht
tp s:
un
//t
iv
Internal energy per mole =
⇒
rio s
si ta
er
3 RT 2 3 3 3 Cv = ∆U = R (T + 1) − RT = R 2 2 2 3 Cv = R 2 C 5 γ= p = Cv 3
.m
ht
tp
For mono-atomic gas:
e/ un i
s:
//w
ve
w
w
rs i
Cp and CV for Gases Consider 1 mole of a gas heated to raise the temperature by 1ºC. When the gas is heated at constant volume, all heat is converted to increase in internal energy. The workdone is zero, i.e., Cv = ∆U for 1ºC (or 1 K) rise in temperature for 1 mole of a gas.
For Di-atomic gas:
Internal energy per mole =
Cv = ΔU 5 5 5 = R (T + 1) − RT = R 2 2 2 5 Cv = R 2 7 Cp = R 5 CP 7 = Cv 3
…(as Cp – Cv = R)
∴
5 RT, 2
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2.22
Kinetic Theory of Gases and Thermodynamics
2.15 THERMODYNAMIC PROCESSES In a thermodynamic process, the initial and final states are important, but the manner in which the system is taken from initial to final state must also be taken into account. Hence, it is important of distinguish between the following processes:
Isothermal Process
rio s_ in fo er si ta rio s
e/
Heat taken in
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
na .c om
A gas isothermal process is one in which the temperature remains constant.
iv
t.m e/
P
ht tp
s: //
ht tp s:
un
//t
ht
A gas undergoing an isothermal process should remain in perfect thermal contact with a constant temperature heat reservoir. This obeys BOYLE’S LAW, during an isothermal process i.e., pV = constant
V
As the temperature remains constant, internal energy of the system remains constant i.e., ∆U = 0 (For isothermal process).
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Kinetic Theory of Gases and Thermodynamics
2.23
ar a
From First Law: ∆Q = W
rio s_ in fo
na .c om
Heat given out
iv
er si ta rio s
e/ .m p
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
tp
s:
un iv
//w
w
er s
w
ita
.ja
m
(i) Isothermal expansion: If the gas expands, it takes heat from the heat reservoir and does positive work. In equation ∆Q = W, both ∆Q and W are positive. The process takes place slowly. As the pressure of the piston is reduced, the gas tries to expand. In expansion, it tends to cool down, but the heat reservoir gives heat input to keep temperature constant.
V
(ii) Isothermal compression: If the gas is compressed isothermally, it gives heat to the reservior and the work done is negative, i.e., in the equation ∆Q = W, both ∆Q and W are negative.
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Kinetic Theory of Gases and Thermodynamics
To do a perfectly isothermal process, pressure is increased in very small steps. All the gas is compressed slowly, its temperature tends to rise, but the reservoir extracts the heat from the system to keep the temperature constant.
Work Done by Isothermal Process W = ∫ dW =
V2
∫ pdV
V1
V2 V1
na .c om
= nRT ln
ar a
m
∆Q = Infinity n∆T
er si ta rio s
un iv
//w
w
er s
Adiabatic Process
ita
w
.ja
Cisothermal =
rio s_ in fo
This is also the amount of the heat given, as the volume of the gas is changed from Vinitial to Vfinal. As the change in temperature is zero, the molar heat capacity of such a process is
ht tp s:
un
//t
p
iv
e/ .m
tp
s:
In an adiabatic process, no heat enters or leaves the system. Such a situation is achieved by heavily insulating the system.
ht
s: //
t.m e/
p1
A
Adiabatic (Steeper) B
p2
ht tp
2.24
o
V1
V2
V
p
Adiabatic A From the first law of thermodynamics: p (Steeper) ∆Q = ∆U + ∆W B 1
p2
In adiabatic process ∆Q = 0 ⇒ or
∆U + ∆W = 0 ∆U + pd V = 0
...(i)
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V1
V2
V
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Kinetic Theory of Gases and Thermodynamics
2.25
1dU n dT
We have, CV =
∆U = n CV d T
or
Thus, from equation (i) ...(ii)
nCV d T + Pd V = 0
Since the gas is ideal, therefore pV = nRT or pdV + VdP = nRdT pdV + Vdp nR C where, γ = P CV Let the initial pressure and volume be pi and Vi respectively and the final pressure and volume be pf and Vf respectively. Then dT =
fo
s_ rio ta si
γ
s ta
un
er
e/
iv
pi pf
un
=
tp
e/
s:
Viγ
//t .m
V fγ
rio
pi
p V ln f = ln i p Vi f
s:
tp ht
or
pf
si
w //w
or
Vi
= −ln
er
Vf
w
γln
or
iv
.ja
Vi
dV dp =−∫ V p pi
ar
γ
m
∫
an
pf
Vf
in
a.
co
m
or
.m
Relation between p and T:
//t
ht
Thus, pVγ remains constant in a reversible adiabatice process.
ht
tp
s:
nRT p γ Putting pV = Constant, we have
We have, pV = nRT V = γ
nRT p = Constant p
p1–γ T γ = Constant
Tγ = Constant p γ−1
Relation between V and T: We have, pV = nRT
or
p=
nRT V
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Kinetic Theory of Gases and Thermodynamics
nRT γ Putting in equation pVγ = Constant, we have V = Constant V γ–1 or TV = Constant The adiabatic curve in p – V diagram is much steeper than the isothermal curve.
ar a
p
m
+ve work
er s un iv e/
er si ta rio s
ita
.ja w w //w s:
V
un
//t
iv
.m
tp
rio s_ in fo
na .c om
(i) Adiabatic expansion: As the pressure on the piston is reduced, the gas pusshed the piston upwards due to the expansion, the temperature of the system falls down. The molecules of the gas lose speed after colliding with a receding (moving away piston). The loss in the speed cause a fall in the temperature. In the equation of the Ist Law: ∆U + ∆W = 0 ∆U is negative and W is positive (Internal energy of the gas is being converted into work done).
ht
(ii) Adiabatic compression: As the presence on the piston is increased, it moves downwards compressing the gas. Due to the compression, the temperature of the gas rised. The molecules of the gas gain speed after colliding with an incoming piston. The gain in speed of gas causes a rise in temperature. Work is done on the gas and its internal energy increased.
ht tp
s: //
t.m e/
ht tp s:
2.26
p
–ve work
V
In the equation, ∆U + ∆W = 0, ∆U is positive and ∆W is negative. https://t.me/universitarios_infohttps://www.jamarana.com
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Kinetic Theory of Gases and Thermodynamics
2.27
Work Done in an Adiabatic Process Let, a sample of gas has initial pressure p1 and initial volume V1. In an adiabatic process, the pressure and volume change to p2 and V2, then pVγ = p1 V1γ = p2 V2γ = k p=
Thus,
...(i)
k Vγ
The work done by the gas in the process is V2
∫
pdV =
V1
V2
1 k
k
∫ V dV = 1 − γ V
γ−1 2
γ
V1
na .c om
From equation (i),
k k = p2 and γ = p1 V2γ V1
k V1γ−1
p V − p2 V2 1 ( p2 V2 − p2 V1 ) = 1 1 γ −1 γ −1
ar a
W=−
m
Thus,
−
rio s_ in fo
W=
er s
w
ita
.ja
Isochoric Process
er si ta rio s
e/
P ∝ T.
.m
ht
tp
s:
un iv
//w
w
A process in which the volume remains constant is known as isochoric process. For an isochoric process, an ideal gas is taken in a container with rigid walls so that neither expansion or compression occurs. The gas obeys the relation
un
//t
iv
In isochoric process, ∆W = 0, so, for Ist law of thermodynamics
t.m e/
ht tp s:
∆Q = ∆U.
ht tp
s: //
(i) If heat is given to the gas, its temperature and pressure rises. Its internal energy also rises. ∆U and ∆Q are both positive. (ii) If heat is taken out from the gas, its temperature and pressure falls. Its internal energy also decreased. ∆Q and ∆U are both negative. p
O
A
B
B
A
w=0
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Kinetic Theory of Gases and Thermodynamics
Isobaric Process A process in which the pressure is kept constant is called as isobaric process. For isobaric process, the external pressure on the piston is kept constant. The gas obeys CHARLES’ LAW:
un iv
//w
w
V ∝ T. W = p(V2 – V1) = nR (T2 – T1)
e/
s:
er si ta rio s
er s
w
ita
.ja
Heat in by the source
Work done,
tp
rio s_ in fo
m
ar a
na .c om
p
ht tp
p
s: //
t.m e/
ht tp s:
un
//t
iv
.m
(i) Isobaric expansion: If heat is given to the gas, isobaric expansion occurs. The volume and the temperature both rise. The gas expands doing positive work. In equation: ∆Q = ∆U + ∆W, all the three terms ∆Q, ∆U and ∆W are positive in the isobaric expansion.
ht
2.28
+ve work A
B
V1
V2
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Kinetic Theory of Gases and Thermodynamics
2.29
(ii) Isobaric compression: If the heat is taken out the gas, isobaric compression occurs. The temperature falls and the gas contracts causing negative work. In the equation: ∆Q = ∆U + W, all the three terms ∆Q, ∆U and W are negative in the isobaric compression. P –ve work B
m
V1
V2
V
er si ta rio s
w
er s
w
ita
.ja
A
rio s_ in fo
ar a
na .c om
P
e/
s:
un iv
//w
2.16 EXPRESSIONS FOR ∆U, W AND ∆Q FOR DIFFERENT PROCESSES INTERNAL ENERGY CHANGE: ∆U
iv
//t
ht
.m
tp
∆U = nCV∆T for every process. It is the state function and depends only the difference of initial and final temperature.
un t.m e/
ht tp s:
∆U = nCV(T1 – T2)
Work Done: (W)
Isothermal process
W = 2303 nRT log10 V2
ht tp
s: //
Process
Heat Exchanged (∆Q)
V1
∆Q = 2.30 nRT log
V2 V1
p1V1 − p2 V2 nR(T2 − T1 ) = ∆Q = 0 γ −1 γ −1 p V − p2 V2 nR(T2 − T1 ) W= 1 1 = γ −1 γ −1
Adiabatic process
W=
Isochoric process
W=0
∆Q = nCV∆T (use definition of CV)
Isobaric process
W = p∆V = p(V2 – V1) W = nR(T2 – T1)
∆Q = nCP∆T (use definition of CP)
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2.30
Kinetic Theory of Gases and Thermodynamics
2.17 RELATION FOR THERMODYNAMIC PROCESS (STATE VARIABLE) Isochoric process: Volume remains constant p∝T p1 p2 = T1 T2
⇒
⇒
Isothermal process: Temperature remains constant 1 p∝ ⇒ V ⇒ p1V1 = p2V2
V1 V2 = T1 T2
⇒
rio s_ in fo
na .c om
Isobaric process: Pressure remains constant ⇒ V∝T
ita un
xample
s: //
∴
)
ht tp
) Solution Let N be the number of molecules.
Given: V = 250 cm3, p = 10–3 mm of Hg, T = 300 K
or
Here, p = 760 mm of Hg, and T0 = 273 K
10−23 × 250 × 6.02 × 1023 × 273 760 × 22400 × 300 = 8.045 × 1015 molecules N=
2. An excellent laboratory vacuum is 10–10 mm of mercury. How many molecules of a gas remain per cubic centimeter at 20ºC in this vacuum?
10−3 250 N 300 × = × 760 22400 6.02 × 1023 273
) Solution 76 cm of Hg pressure = 76 × 13.6 ×
...(ii)
p V N T × = × p0 V0 N 0 T0
)
p0 V0 = N0 k T0
∴
∴ pV = NkT where k is constant ...(i) Now at NTP, one mole of air occupies V0 = 22400 cm3 and contains N0 = 6.02 × 1023 molecules.
S
Dividing equation (i) by equation (ii) we get
t.m e/
ht tp s:
1. An electric bulb of volume 250 cm3 was sealed off during manufacture at a pressure of 10–3 mm of Hg at 27ºC. Find the number of air molecules in the bulb.
//t
ht
WORKED OUT E
iv
e/ .m
tp
s:
un iv
//w
w
T1p11 – γ = T1p21 – γ,
er si ta rio s
er s
w
.ja
T1V1γ = T2V2γ–1,
m
ar a
Adiabatic process: Now heat transfer takes place. p1V1γ = p2V2γ,
980 dyne cm–2
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Kinetic Theory of Gases and Thermodynamics
= 101.3 × 104 dyne cm–2 = 101.3 × 10 Nm 3
The amount of gas is given by
–2
6 2 2 pV (1.6 × 10 N/m ) × ( 0.0083 m ) = = 377 K RT (8.3 J/mol − K )( 300 K )
(∵ 10 dyne cm /µm )
n=
1 v2 Pressure, p = Nm 0 3 V
= 377 K As the gas is kept in a closed vessel, its volume is constant, therefore
2
2 N1 2 m0 v 3 V2 N = kT V
or
m
=
fo
co
in
p1V1 p2 V2 = T1 T2
s_
an
io
ar
w
ta r
e/ un
=
.m
ht tp
677 × 1.6 × 106 N/m 2 300 = 3.6 × 106 N/m2
iv
s: //w
er si
.ja
w
R 8.3 = = 1.38 × 10−23 JK −1 N 0 6.02 × 1023 T = 20° C = 293 K.
er si ta
m
Here, V1 = V2 T ∴ p2 = 2 p1 T1
s
p = 10–10 mm = 10–11 cm of Hg
a.
Final temperature = 300 K + 377 K = 677 K We have,
101.3 × 103 = × 10−11 Nm 2 76 R 8.3 k= = = 1.38 × 10−23 JK −1 N 0 6.02 × 1023
2.49 × 104 J = 377 K ( 5.3 mole ) (1.5 × 8.3 J / mol − K )
3 1 ..... m 0 v 2 = kT 2 2 N p ∴ = V kT where,
∆Q = nCV∆T ∆Q ∆T = nCV
=
rio
–2
k=
2.31
iv
)
)
ht
tp
s:
//t .m
ht
e/ un
s: //t
tp
N 101.3 × 103 × 10−11 1 1 6 4. ×A10sample of ideal gas (γ = 1.4) is heated at = × × = 3.296 3 −23 m 76 1.38 × 10 293 constant pressure. If an amount of 140 J 3 of heat is supplied to the gas, find: × 10−11 1 1 × × = 3.296 × 106 −23 (i) The change in internal energy of the 6 1.38 × 10 293 N 6 gas. or = 3.296 × 10 cm3 (ii) The work done by the gas. Note: Since, this is a statistical calculation, the Solution Suppose, the sample contains n accuracy of the computed result is limited. How- ) moles. Also, suppose the volume changes ever, the order = 106 is reliable. from V1 to V2 and the temperature changes from T1 to T2. 3. An ideal gas has a molar heat capacity pressure CP = 2.5 R. The gas is kept in The heat supplied is given by a closed vessel of volume 0.0083 m3, at a ∆Q = nCP(T2 – T1) temperature of 300 K and a pressure of (i) Change in internal energy 1.6 × 106 N/m2. An amount of 2.49 × 104 J of heat energy is supplied to the gas. Cal∆U = nCV(T2 – T1) culate the final temperature and pressure C = V nCP (T2 − T1 ) of the gas. CP C 140 J ) Solution We have, CV = CP – R = V ∆Q = = 100 J CP 1.4 = 2.5 R – R = 1.5 R
∴
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2.32
)
) Solution
(i) The process is shown in the figure. During the part ab, since the pressure is constant, we have pa Va pb Vb = Ta Tb or
Tb =
Vb Ta Va
= 2aTa = 600 K
(i) What is the final pressure? (ii) How much work is done by the gas in the process?
5. A sample of gas (γ = 1.5) is taken through an adiabatic process in which the volume is compressed from 1600 cm3 to 400 cm3. If the initial pressure is 150 kPa,
(ii) Work done by gas ΔW = ΔQ – ΔU = 140 J – 100 J = 40 J
Kinetic Theory of Gases and Thermodynamics
p
)
) Solution
in
Isothermal
s_ ta
V
s
40l
ita rio
rs
ve ni
paVa = pbVb
pbVbγ = pcVcγ
or
pa(2Va)γ = pcVcγ
Dividing equation (ii) by equation (i), we get 2γ(Va)γ – 1 = (Vc)γ – 1 γ
or
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...(i)
And from the adiabatic curve bc,
ht
(i) Sketch the process in a p–V diagram. (ii) What is the final volume and pressure of the gas? (iii) What is the work done by the gas?
(ii) From the isothermal ac,
20l
si
.m 3
tially at 27º C and occupy a volume of 20 litres. The gas is first expanded at constant pressure untill the volume is doubled. Then it undergoes an adiabatic change untill the temperature returns to its initial value.
tp s:
6. Two moles of Helium gas ( γ = ) are ini
e/ u
s:
240 J − 480 J = − 480 J 0.5
//t
=
ht tp
1.5 − 1
//t .
(150 kPa ) (1600 cm3 ) − (1200 kPa ) ( 400 cm3 )
5
m
=
c
During the part bc, the gas is adiabatically returned to the temperature Ta. The point a and point c are on the same isothermal. Thus, we draw an adiabatic curve bc and an isothermal from a and look for the point of intersection c. That is the final state.
e/ un
s:
p1V1 − p2 V2 γ −1
ht tp
W=
iv
//w
w
er
w
.ja m
1600 = (150 kPa) = 1200 kPa 3 400 1600 2 = (150 kPa) = 1200 kPa 400 (ii) Work done by the gas in an adiabatic process
rio
ar
3 2
Adiabatic
fo
a. c
γ
an
V Thus, p2 = p1 1 V2
b
a
om
For an adiabatic process p1V1γ = p2V2γ
(i)
Vc = 2 γ−1 Va 4 2 Va
= 113 litres
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Kinetic Theory of Gases and Thermodynamics
p′V1 RT′ p′V2 and n′2 = RT′ We have, n1 + n2 = n′1 + n′2 n′1 =
From equation (i),
p1V1 p2 V2 p′V1 P′ V2 + = + RT1 RT2 RT′ RT′
o
rio
si ta
si ta rio
er
e/ u
ni v
m
And Internal energy of the gas after the connection C p′ (V1 + V2 ) = V R
s: //t
.m
e/ u
//t .
ht tp s:
s
er
ni v
//w
ht tp s:
7. Two vessels of volume V1 and V2 contain the same ideal gas. The pressure in the vessels are p1 and p2 and the temperatures are T1 and T2 respectively. The two vessels are now connected to each other through a narrow tube. Assuming that no heat is exchanged between the surroundings and the vessels, find the common pressure and temperature attained after the connection.
Internal energy of the gases before the connection C pV C p V = V 1 1+ V 2 2 R R
w w
Net work done by the gas = 4980 J + 7470 J = 12450 J.
As the vessels have fixed volume, no work done by the gas plus the vessels system. Also, no heat is exchanged with the surroundings. Thus, the internal energy of the total system remains constant. The internal energy of an ideal gas is pV U = nC V T = C V R
in f
an a. co
ar
.ja m
nR(T2 − T1 ) 4980 = = 7470 J 5 γ −1 −1 3
ht tp
Neglecting the change in internal energy of the vessels (the heat capacity of the vessels is assumed negligible). C V p1V1 C V p2 V2 CV p′( V1 + V2 ) + = R R R
)
) Solution
T T (V + V2 ) T′ = 1 2 1 P′ p1V1T2 + p2 V2 T1
m
or
s_
= 2 mole × (8.3 J/mol – K) × (600 K – 300 K) = 4980 J Work done in the adiabatic part bc p V − pc Vc = b b γ −1
p′ 1 p1V1 p2 V2 = + T′ V1 + V2 T1 T2
or
= pa Vb – pa Va = nRT2 – nRT1
or
T
T1T2 p1V1 + p2 V2 p1V1T2 + p2 V2 T1
8. A thermodynamic system is taken through the cycle a b c d a. (i) Calculate the work done by the gas during the parts ab, bc, cd and da.
If p′ and T′ are the common pressure and temperature after the connection is made, the amount are
p1V1 RT1
p1V1 + p2 V2 V1 + V2
From equation (i), T′ =
The amount of gas in vessel 1 is n1 =
p′ =
p2, V2.T2
p1, V1.T1
or
= 4.4 × 104 Pa (iii) Work done by the gas in the part ab = pa (Vb – Va)
pV nRT pc = a a = Vc Vc 2 mol × (8.3 J / mol − k )(300 k ) = 113 × 10−3 m3
=
2.33
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2.34
Kinetic Theory of Gases and Thermodynamics
of 42 J heat is given to the gas. If the temperature rised through 2º C, find the distance moved by piston. Atmospheric pressure = 100 kPa.
(ii) Find the total heat rejected by the gas during the process.
d
c
100 kPa
a
b
) Solution Change in internal energy of the gas )
200 kPa
300 cm3
100 cm3
∆U = 1.5 nRT = 1.5 (1 mole) (8.3 J/mol – K) (2K) = 24.9 J Heat given to the gas = 43 J Work done by the gas is ∆W = ∆Q – ∆U = 42 J – 24.9 J = 17.1 J If the distance moved by the pistion is x, then the work done ∆W = (100 kPa) (8.5 cm2)x = 17.1 J 5 Thus, (10 N/m2) (8.5 × 10–4 m2) x = 17.1 J or x = 0.2 m = 20 cm
ar a
ita
un
s: //
= 2.303 × 60 log 8 (101) J = 1.26 × 104 J
11. The U-tube of uniform cross-section contains mercury to the depth shown. the left side of the tube is closed at the top and the right side is connected to a good vacuum pump (i.e., pressure is made zero). The temperature at air in the closed tube is raised from 27º C to 111º C. How far the level of mercury falls on the left side? Atmospheric pressure is 75 cm of Hg.
or
ht tp
9. The internal energy of a mono-atomic ideal gas is 1.5 nRT. One mole of helium is kept in a cylinder of cross-section 8.5 cm2. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total
V2 V1 V W = 2.303( p1V1 ) log 2 V1 (using p1V1 = nRT) p = 2.303 (20 × 3) log 1 lt. atm p2 W = 2.303nRT log
t.m e/
ht tp s:
perature T,
iv
e/
) Solution In isothermal process at tem-
.m
3
//t
ht
tp
3
er si ta rio s
un iv
s:
= (200 kPa) (100 cm – 300 cm ) = – 40 J The work done during da is zero as the volume does not change. (ii) Total work done by the system during the cycle a b c d a. ∆W = 20 J – 40 J = – 20 J Change in the internal energy, ∆U = 0, as the initial state is the same as the final state. Thus, ∆Q = ∆U + ∆W = – 20 J So, the system rejects 20 J of heat during the cycle.
er s
w
w
//w
∫ pdV d
.ja
m
c
=
10. How much work is done by an ideal gas in expanding isothermally from an initial volume of 3 litres of 20 atm to a final volume of 24 litres?
a
= (100 kPa) (300 cm3 – 100 cm3) = 20 J The work done during bc is zero as the volume does not change. The work done during cd,
∫
= (100 × Pa) dV
)
a
b
na .c om
)
b
(i) Work done during the part ab = ∫ pdV
rio s_ in fo
) Solution
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Kinetic Theory of Gases and Thermodynamics A
2.35
pressure of 1 × 105 N/m2. The total mass of the mixture is 28 gm. If molar masses of neon and argon are 20 and 40 gm/mol respectively, find the masses of the individual gases in the container, assuming them to be ideal (universal gas constant = 8.314 J/mol/L).
B
50 cm
2x
)
) Solution Given that m1 + m2 = 28 gm
)
therefore pA = pB = 0 + 2x ρg Initial pressure, patm = 75 cm Initial volume = 50 Initial temperature = 27º C = 300 K Final pressure = pA = 2xρg Final volume = (50 + x) And, final temperature = 111º C
From gas law pV = nRT
rio s_ in fo 3
er s
…(iii)
er si ta rio s
13. A thin tube of uniform cross-section is sealed at both ends. It lies horizontally, the middle 5 cm containing mercury and the two equal ends containing air at the same pressure p. When the tube is held at an angle of 60º with the vertical direction, the length of the air column above and below the mercury column are 46 cm and 44.5 cm respectively. Calculate the pressure p in centimetres of mercury. (The temperature of the system is kept at 30º C).
iv
m1 m2 + × 2494.2 20 40
40m1 + 20m2 = 641.48
un
) Solution
or x + 50x – 32 × 75 = 0 Solving, we get x = 30 cm. 2
–5 cm
l
l
p
p
12. A closed container of volume 0.02 m3 contains a mixture of neon and argon gases, at a temperature of 27º C and
2 × 10 =
t.m e/ s: //
2 x (50 + x) = 75 64
or
2
Solving equations (i) and (ii), we get m1 = mass of neon = 4.08 gm and, m2 = mass of argon = 23.92 gm
un iv e/ .m //t
2 xρg (50 + x) 75ρg × 50 = 384 300
1
)
(75 ρg ) × 50 300
ht tp
384
=
3
ita
.ja w ht tp s:
pA ( 50 + x )
or
2
3
ar a
or
p1V1 p2 V2 = T1 T2
or
2
or
w //w s: tp
ht From the gas law
5
m1 m2 + (1 × 10 N/m )(0.02 m )= 20 8.314 × 300 40 5
m
2x
(1×10 N/m )( 0.02 m ) = m20 + m40 8.314 × 300
na .c om
or
pB = 0
50 cm pA
...(i)
) Solution Since pressure of both sides is equal,
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(a)
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2.36
Kinetic Theory of Gases and Thermodynamics
From given condition
)
) Solution p Q
p2 –5
46
cm
p1
cm pincident
co m
p = 5 cos 60° = 2.5 cm
s
rio
ta
si
er un
= 4.695 × 10–24 kg ms–1
Number of collision per sec = 1023 ∴ Change in momentum of the molecules per sec = Rate of gain in momentum by the wall normal to it = Normal force on the wall = 4.695 × 10–24 × 1023 kg ms–2 = 0.4695 N Area of the wall = 2 cm2 = 2 × 10–4 m2 0.4695 ∴ Pressure on the wall = Nm −2 2 × 10−4 = 2.47 × 103 Pascal
14. The mass of a H2 molecule is 3.32 × 10–24 gm. If 1023 hydrogen molecules per second strike a fixed wall of area 3 cm2 at an angle 45º to the normal and rebound elastically with a speed of 105 cms–1, calculate the pressure exerted on the wall.
= 75.4 of Hg.
2 p = 2 × 3.32 × 10−24
e/ .m //t s:
tp ht
p2 × 46 45.25 2.5 × 44.5 × 46 = 1.5 × 45.25
fo
in
s_
s:
tp
ht
ta
si er
//t
ht
( p1 − p2 ) 44.5 2.5 × 44.5 = 1.5 1.5
p=
iv
p2 =
Now, p2 × 46 = p × 45.25
iv
un
or
∆p =
.m
p1 − p2 46 − 44.5 1.5 = = p2 44.5 44.5
∴
= 3.32 × 10–27 × 103 kg ms–1 = 3.32 × 10–24 kg ms–1 Change in momentum normal to the wall per molecule
e/
tp
s: //
w
w
w
∴ (p1) (44.5) = plLeft = plRight = p2 × 46 [from Fig. (b)] p1 46 ∴ = p2 44.5 ∴
R
rio
m
.ja
...(i)
From Fig. (a) pleft = pright p1 = p2 + 2.5
∆p
PQ → Incident momentum direction QR → Rebound momentum direction. Since, the rebounding is elastic and the wall is fixed. pIncident = pRebound
ar an
2l + 5 = 46 + 44.5 + 5 = 95.5 cm 90.5 = 45.25 cm l= 2
a.
(b)
15. An ideal gas is taken through a cycle thermodynamic process through four steps. The amount of heat involved in these steps are Q1 = 5960 J, Q2 = – 5585 J,
.5 44
or
prebound
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Kinetic Theory of Gases and Thermodynamics
2.37
Q3 = – 2980 J and Q4 = 3645 J respectively. The corresponding worked involved are W1 = 2200 J, W2 = – 825 J, W3 = – 1100 J and W4 respectively. (i) Find the value of W4. (ii) What is the efficiency of the cycle? )
) Solution For a cyclic process
∆U = 0
(i) Cyclic ∫ dQ = ∫ dW
)
ta
rio
s_
in
fo
s
rio
si
ta
si er
iv un
er
e/
.m
e/
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or
= P0 Al +
kl nRT1 AP0 p0 Al + kl 2 klT1 + T2 = T1 + nR AP0 = nRT1 + p0 Al + kl 2 +
ht
tp
s:
//t
ht
0
1 2 kl 2 = (100 × 103 Pa) (20 × 10–4 m2) × (10 × 10–2 m) 1 + (200 N/m) (100 × 10−4 m 2 ) 2 = 20 + 1 J = 21 J (ii) Initial temperature, T1 = 300 K. Let the final temperature by T2, then nRT1 = p0V0 kl nRT = pV2 = p0 + (V0 + Al ) A
∫
W = Fdx
un iv
.m
//t
s:
tp
16. 2.00 mole of a mono-atomic ideal gas (U = 1.5 nRT) is enclosed in an adiabatic, vertical cylinder fitted with a smooth light adiabatic piston. The piston is connected to a vertical spring of spring constant 200 N/m as shown in the figure. The area of cross-section of the cylinder is 20.0 cm2. Initially, the spring is at its natural length and the temperature of the gas is 300 K. The atmospheric pressure is 100 kPa. The gas is heated slowly for some time by means of an electric heater so as to move the piston up through 10 cm. Find: (i) The work done by the gas. (ii) The final temperature of the gas. (iii) The heat supplied by the heater.
(i) Force by the gas on the piston is F = p0A + kx where, P0 = 100 kPa is the atmospheric pressure. A = 20 cm2 is the area of the cross-section, k = 200 N/m is the spring constant, and x = the compression of spring. Work done by the gas if the piston moves through l = 10 cm is l
ht
tp s:
//w w
Net work output = 5960 – 5585 – 2980 + 3645 = 1040 J Total heat input = Q1 + Q4 = 5960 + 3645 = 9605 J 1040 ∴ η= × 100 = 10.83% 9605
w .ja
m
(ii) Efficiency of the cycle, Net work output η= Total heat input
) Solution
a. c
or
5960 – 5585 – 2980 + 3645 = 2200 – 825 – 1100 + W4 W4 = 765 J
ar an
or
om
i.e., Q1 + Q2 + Q3 + Q4 = W1 + W2 + W3 + W4
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2.38
Kinetic Theory of Gases and Thermodynamics
300 +
B
C
D
300 K
er si ta rio s
ita
e/
T
)
) Solution Number of moles, n = 2
(ii) (∆W)Net = (∆Q)Net = 1153.4 J (iii) (∆U)Net = 0
The gas undergoes cyclic process. Since, internal energy is property of the system, the net change in internal energy during the cyclic process is zero.
18. One mole of a gas is put under a weightless piston of a vertical cylinder at temperature T. The space over the piston is atmosphere. How much work should be performed to increase isothermally the
Helium is a mono-atomic gas. 3 ∴ CV = R 2 5 CP = R 2
( ∆Q) DA = nRT ln
400 K
s: // ht tp
300 K
PD PA 2 = 2 × 8.32 × 300 ln = −3460.2 J 1 ∴ (∆W)Net = 4160 + 4613.6 – 4160 – 3460.2 = 1153.4 J
t.m e/
C
2 1
= 2 × 8.32 × 400 ln = 4613.6 J
iv un
//t ht tp s:
D
PB VC = nRT ln VB PC
= nRT ln
.m
tp ht 1 atm
(i) (∆Q)Net = (∆Q)AB + (∆QBC) + (∆QDA) (∆Q)AB = n × CP × (TB – TA) 5 = 2 × × 8.32(400 − 300) = 4160 J 2 Since Process BC is isothermal, therefore ∆U = 0 (∆Q)BC = (∆W)BC
s:
B
rio s_ in fo
er s
un iv
A
2 atm
T
na .c om
.ja
//w
w
w
m
ar a
p
400 K
Hence, according to the first law of thermodynamics, the net change in the heat energy is equal to the net work done.
1 atm
17. Two moles of helium gas undergo a cyclic process as shown in the figure. Assuming the gas to be ideal, calculate the following quantities in this process: (i) The net change in the heat energy. (ii) The net work done. (iii) The net change in internal energy. [R = 8.32 J mol–1]
A
2 atm
= 300 K + 1.325 K + 30 K = 331 K (iii) Internal energy, U = 1.5 nRT ∴ ∆U = 1.5 nR∆T = 1.5 × 2.00 × 8.3 × 31 = 772 J From the first law, ∆Q = ∆U + ∆W = (772 + 21) J = 793 J
p
20 J + 2 J 200 × 10 × 10−2 × 300 + 20 × 8.3 20 × 10−4 × 100 × 103 Pa
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Kinetic Theory of Gases and Thermodynamics
volume under the piston to twice the volume (neglect friction of piston).
)
) Solution According to the first law of
thermodynamics, ∆Q = ∆U + ∆W For the first vessel: ∆U = 0, (Since, no change in temperature) ∆Q = ∆W
)
) Solution Let A be the area of piston,
therefore F + pA = p0A or F = (p0 – p) A Work done by agent is given by
V2
0
− p )Adx
V2
V
=
− p )dV
s
rio
ta
un
iv
er si
e/ tp s
ht
= nRT [(η – 1) loge η] where, η = 2 and n = 1 W = RT [1 – log e 2]
19. Two vessels contain in each of them one mole of mono-atomic gas. The initial volume of each vessel is 8.3 × 10–3 m3. Equal amount of heat is supplied to each vessel. In one vessel, the volume of gas is doubled without change in its internal energy whereas the volume of the gas is held constant in second vessel. The vessels are now connected to allow free mixing. Find the final temperature and pressure of the combined system.
3 nRT log e 2 = n R ∆T 2 2 or ∆T = × 300 × 0.693 = 138.6 K 3 It is the change in temperature of the second vessel. Now, temperature of the gas in second vessel = T + ∆T = 300 + 138.6 = 438.6 K Let after mixing Tf and pf be the final temperature and pressure, therefore T + (T + ∆T) Tf = 2 300 + 438.6 = = 369.3 K 2
:// t.m
tp ht
= p0 (η – 1)V – nRT logeη
∴
in
si er e/
//t .m
P
s:
ht tp
s:
un
iv
//w
w
F
w
.ja
(since pV = nRT)
Po
s_
ta rio
m
V
dV V
∫ nRT
ηV
= p0 (η − 1)V −
a.
V
fo
co
∫ pdV
V
m
ηV
an
∫
p0 dV −
ar
ηV
=
Since V2 = 2 V1, therefore Q = nRT loge 2, ...(i) For the second vessel: ∆W = 0, (volume is constant) 3 Q = nC V ∆T = n R ∆T ...(ii) 2 3R Since, for mono-atomic gas C V = 2 From equations (i) and (ii), we get
0
(since, pV = nRT)
From the gas equation, pfVf = nRTf
∫ (p
V
=
dV V
∫ nRT
V1
pf =
nRT f Vf
2 × 8.3 × 369.3 2 × 8.3 × 103 + 8.3 × 10−3 = 2.46 ×105 N /m 2 =
ηV
∫ (p
∫ pdV
V1
W=
Q=
ηV
2.39
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2.40
Kinetic Theory of Gases and Thermodynamics
= 0.2 × 8.314 × 400 loge2
p
= 461 J For adiabatic expansion b → c Q=0 ∴ W = – ∆U = nCV(TH – TC) = 0.2 × 20.78 × (400 – 300) = 415.7 J For isothermal compression c → d ∆U = 0
m
a
4 = 13.3 × 10−4 × 3
fo
in
s rio
ta
si
er iv
un
e/
= nCV(TC – TH)
= 0.2 × 20.78 × (300 – 400) = – 415.7 J
s:
13.65 × 10−4 27.3 × 10−4
W=–U
.m
∴
tp ht
Vd Vc
= – 345.8 J For adiabatic expansion d → a Q=0
//t
ht
T Vc = Vb H TC
1 γ−1
er iv
un
.m
//t
s:
tp
For adiabatic expansion b → c THVbγ–1 = Tc Vcγ–1
W = Qc = nRTc log e
∴
= 0.2 × 8.314 × 300 log e
e/
s:
tp
ht
For isothermal expansion a → b paVa = pbVb pV or pb = a a = 5 × 105 pa Vb
∴
rio
s_
si ta
w w
//w
m .ja
nRTH pa 0.2 × 8.314 × 400 = = 6.65 × 10−4 m3 10 × 105
(i) Va =
ar
v
)
) Solution
a.
c
an
d
co
b
The results may be tabulated as follows:
2.5
Q
= 27.3 × 10−4 m3 a→b
461 J
461 J
0J
b→c
0J
415.7 J
– 415.7 J
c→d
– 345.8 J
– 345.8 J
0J
d→a
0J
– 415.7 J
415.7 J
Total
115.2 J
115.2 J
0J
nRTc 0.2 × 8.314 × 300 pc = = = 1.83 × 105 Pa Vc 27.3 × 10−4 nRTc 0.2 × 8.314 × 300 pc = = = 1.83 × 105 Pa Vc 27.3 × 10−4 For adiabatic compression d → a
∆U
W
4 = 6.65 10−4 × 3
2.5
= 13.65 × 10−4
T Vd = Va H TC
(ii) For entire cycle, 1 γ−1
∴
Tc Vdγ–1 = THVaγ–1
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1
2.5 20. Suppose 0.2 mole of an ideal di-atomic nRTc TH γ−1 4 −4 −4 pd = V = V = 6.65 10 × tem = 13.65 × 10 gas (γ = 1.4) undergoes dcyclea with Vd 3 TCK. The perature TH = 400 K and TC = 300 initial pressure is pa = 10 × 105 Pa and 0.2 × 8.314 × 300 = = 3.65 × 105 Pa during isothermal expansion at tempera13.65 × 10−4 ture TH the volume doubles. For isothermal expansion a → b (i) Find Q, W and ∆U from each step in ∆U = 0 the cycle. V ∴ W = Q H = nRTH ⋅ log e b (ii) Find the efficiency. Va
Q=W ∆U = 0
Total work done = 115.2 J https://t.me/universitarios
Kinetic Theory of Gases and Thermodynamics
(10 ×10 N/m )(10 m ) (500 mole) ( 8.314 JK mole ) 4
TC =
2
3
−1
−1
= 481.12 K
21. A sample of 2 kg of mono-atomic Helium ( 5 ×104 N / m2 )( 20 m3 ) (assumed ideal) is taken through the proTD = cess ABC and another sample of 2 kg of (500 mole) ( 8.314 JK −1 mole −1 ) the same gas is taken through the process = 240.50 K ADC. Given relative molecular weight of Helium = 4. (ii) No. (i) What is the temperature of Helium (iii) For the process ABC, we have in each of the states A, B, C and D? QAB = nCVΔT (ii) Is there any way of telling afterwards which sample of Helium went 3 = (500 mole) × 8.314 JK −1 mole −1 (240.56 K − 120.28 through the process ABC and which 2 3 ADC? −1 −1 went through the mole) process = (500 × 8.314 JK mole (240.56 K − 120.28 K) Write yes or no. 2 (iii) How much heat is evolved in each of = 7.5 × 105 J the processes ABC and ADC? QBC = nCPΔT
fo
s_
rio
s
io
ta r
si
QAD = nCPΔT
//t
.m
For the process ADC, we have
ht tp s:
5 Amount of helium = (500 mole) × 8.314 JK −1 mole −1 (240.56 K − 120.2 3 2 m 2 × 10 5 −1 −1 = = = mole 500 mole −1 = 500 × 8.314 JK mole (240.56 K − 120.28 K) ( ) M 4 g mol 2 (i) The temperature of gas at the states A, = 1.25 × 106 J B, C and D are QDC = nCVΔT (5 ×104 N / m2 )(10 m3 ) = 120.28 pV TA = = K −1 −1 nR (500 mole) (8.314 JK mole ) 3 = (500 mole) × 8.314 JK −1 mole −1 (481.12 K − 240.5 4 2 3 5 × 10 N/m 10 m ( )( 3 ) 2 = −1 −1 − −1 1 = 500 mole × 8.314 JK mole (481.12 K − 240.56 K) ( ) (500 mole) ( 8.314 JK mole ) 2 = 120.28 K = 1.5 × 106 J
(10 ×10
4
N / m 2 )(10 m3 )
(500 mole) ( 8.314 JK
= 240.56 K
−1
mole
QADC = QAD + QDC
TB =
−1
)
ve r
= (7.5 × 105 J + 2.5 × 106 J) = 3.25 × 106 J
e/ un i
20V(m3)
ht tp
) Solution
un i
//t .m s:
10 )
e/
:// tp s ht
D
A
w w
5 = (500 mole) × 8.314 JK −1 2
5
5 = (500 mole) × 8.314 JK −1 mole −1 (481.12 K − 240.56 2 mole −1 (481.12 K − 240.56 K) = 2.5 × 106 J QABC = QAB + QBC
ve rs
C
w
B
10
ita
.ja
p(10N/m2)
in
ar an a
m
.c o
m
QH = 461 J W 115.2 η= = = 0.25 ∴ QH 461
2.41
= (1.25 × 106 J + 1.5 × 106 J) = 2.75 × 106 J
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Kinetic Theory of Gases and Thermodynamics
22. One mole of a di-atomic ideal gas (γ = 1.4) is taken through a cyclic process starting from point A. The process A → B is an adiabatic compression, B → C isobaric expansion, C → D is an adiabatic expansion and D → A isochoric expansion. The V VA volume ratios are = 16 and C = 2 VB VB and the temperature at A is TA = 300 K. Calculate the temperature of gas at the points B and D and find the efficiency of the cycle.
Hence,
V TD = C VD
om na .c
ita
s ta rio
er si
un iv
e/
= RTB + CV(TC – TD) – CV(TB – TA)
ht
.m
W = WB → C + WC → D + WB → A
For a di-atomic gas, 5 7 C V = and C P = R. 2 2 Hence,
TD VDγ – 1 = TCVCγ – 1
Work obtained in one cycle Heat absorbed in the process B → C
= (8.314 JK–1 mol–1)
V TD = C TC VD Since, D → A is isochoric process, therefore VD = VA
or
5 W = R TB + (TC − TD − TB − TA ) 2
5 (909 K) + 2 (1818 − 791.4 − 909 + 300)K
V
Now, the work obtained in one cycle is equal to the area within the cycle ABCDA. This work is given as
γ−1
A
rs
ve
un i //t
tp s:
V TC = C TB = 2(909) = 1818 K VB
(iii) Adiabatic expansion from C to D
or
nf s_ i rio
ar a
e/
.m
//t
s:
tp
ht
VC VB = TC TB
(ii) Isobaric expansion from B to C
D
Efficiency of the cycle is defined as η=
TA = (16)1.4−1 (300)
= (3.03) (300 K) = 909 K
o
p
m
.ja
w w
//w
tp s:
ht
γ−1
(1818 K)
V TB = A VB
TC
C
B
)
or
(i) Adiabatic compression from A to B TBVBγ – 1 = TA VAγ – 1
γ−1
= (0.4353) (1818 K) = 791.4 K The given cyclic process is show in the figure.
For the expansion at constant pressure, we have
V = Constant T With this information, temperature of the gas at different stages of the cyclic process may be determined as follows:
V TC = C 16 VB
1.4 −1
2 = 16
) Solution For an ideal gas undergoing adiabatic expansion or compression, we have TVγ – 1 = Constant
γ−1
2.42
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= 16237.2 K mol–1 https://t.me/universitarios
2.43
Kinetic Theory of Gases and Thermodynamics
∆P = −
Force on the piston = (A ∆p)
and
Acceleration =
1/2
γ p A2 Angular frequency, ω = V0 M Oscillation frequency of the piston 1 = ω 2π
∴
fo
io
ta
iv
e/ un
s:
tp
ht
.m
(i) The work done by the gas. (ii) The heat rejected by the gas in the path CA and the absorbed by the gas in the path BC. (iii) The net heat absorbed by the gas in the path BC. (iv) The maximum temperature attained by the gas during the cycle.
//t
ht
x = a small displacement from the equilibrium. p = pressure of the enclosed gas at equilibrium position of the piston. Mg = p0 + A
er si
24. One mole of an ideal gas is taken round the cyclic process ABCA as shown in the figure. Calculate:
.m
tp s: //t
surroundings” must be interpreted as “thermally isolated from the surroundings”. Then the adiabatic gas law to be applied. pVγ = constant
Mg 2 γ p0 + A A V0 M
s
ta r
si
er
un
e/
s:
ht tp
)
) Solution The words “isolated from the
1 = 2π
iv
//w
w w .ja m
ar
s_
in
∴
rio
an a
.c
23. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal crosssectional area A, atmospheric pressure is P0 and when the piston is in equilibrium position. Assuming that the system is completely isolated from its surroundings. Show that the piston executes harmonic motion and find the frequency of oscillation.
om
W 16237.2 = = 0.614 Q 26451.0
γ p A2 = − x V0 M = ω2x ∴ Acceleration ∝ – x Hence motion of the piston is in simple harmonic motion.
= 26451.0 J mole–1 Hence, the efficiency of the cycle is η=
A (∆p ) M
7 = R (TC − TB ) 2 7 = × (8.314 JK–1 mole–1) 2 (1818 K – 909 K)
γp γp ∆V = − ∆x V0 V0
∴
Heat absorbed in the process B → C is given as Q = CP (TC – TB)
3po
po
B
A
C
∆V = small change in volume of the enclosed gas due to x = Ax ∆p = small change in pressure of the gas corresponding to ∆V. Then, pV0γ = (p + ∆p) (V0 + ∆V)γ = (p + ∆p) (V0γ + γ V0γ–1 ∆V + ....) As a first order approximation, i.e., neglecting the terms containing powers of small quantities, higher than 1st order.
p
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Vo
2Vo
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V
Kinetic Theory of Gases and Thermodynamics
) Solution
p=
RT , we get V p 5V0 V T = − 2 0 V2 + V0 R R
s_ in fo
si ta rio
s
ita
er s =
p0 V0 R
=
25 p0 V0 . 8 R
iv un
1 p0 V0 2
2
2 p0 5 5 p0 5 V0 + V0 V0 R 4 R 4
rio
Tmax = −
e/
5 p0 V0 + 3 p0 V0 + Q BC = p0 V0 2
(iv) The path BC is a straight line path. It is represented by the expression
2 p0 5p (2V) + 0 V0 R R
With this Tmax is given by
.m
QCA + QAB + QBC = W
25 25 − 8 + 4
25. The rectangular box shown in the figure has partition which can slide without friction along the length of the box. Initially, each of two chambers of the box has one mole of a mono-atomic ideal gas (γ = 5/3) at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of box and partition are thermally insulated. Heat loss through lead wire of heater is negligible. The gas in left chamber expands, pushing the partition until the final pressure in both chambers 243 p0 . Determine becomes 32
//t
s:
ht
tp
(iii) In a cyclic process, the change in internal energy is zero. Hence
i.e., 0 = −
//t
tp s:
ht
3 = V0 (3 p0 − p0 ) = 3 p0 V0 2
This gives Q BC =
.m
ht tp
5 V ∆p = R 0 2 R
∂T =0 ∂V
5 which gives V = V0 . 4
e/
s:
The path AB is an isochoric expression of one mole of an ideal gas from pressure p0 to 3p0. The heat released in this process is
−
To determine Tmax, we set
un iv er
//w w
w
5 5 = p0 (V0 − 2V0 ) = − p0 V0 2 2
Q2 = n CV ∆T
− 2 p0 V + 5 p0 V0
Replacing p =
a. co m
ar an m
.ja
3 p ∆V = R 0 2 R
or
Q1 = n CP ∆T
(ii) The path CA is an isobaric compression of one mole of an ideal gas from volume 2 V0 to V0. The heat released in this path is
−2 p0 (V − 2 V0 ) V0
=
= p0V0
)
3 p − p0 p − p0 = 0 (V − 2 V0 ) V0 − 2V0
(i) Work done by the gas during a cyclic process is equal to the area enclosed by its P–V diagram. In the present case, W = area of ∆ABC 1 = (AC) (AB) 2 1 = (2 V0 − V0 ) ( 3 p0 − p0 ) 2
2.44
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Kinetic Theory of Gases and Thermodynamics
Substituting in equation (i), we get
⇒
x = 0.7037 V0
243 p0 (V0 − x) TR 2.25 T0 × = = 32 p0 V0 T0 T0
Then,
pL VL TL × = p0 V0 T0
If pL, VL and TL are the final pressure, volume and temperature of left chamber respectively.
(i) The final temperature of gas in each chamber.
But, pL = pR =
2.45
VR = V0 – x
rio s_ in fo
ita
243 P0 VRγ 32 32 γ VRγ = V0 243
p0 V0γ =
or
er si ta rio s ∴
iv
er s
TL = 12.937 T0
(ii) Since p0 V0γ = pR VRγ
2 3
3
= × V0 =
8 V0 27
Work done by the gas in right chamber
1
8 32 γ V0 VR = × V0 = 27 243
un
or
=
p0 V0 − pR pR γ −1
=
nR (T0 − TR ) γ −1
1× 8.3(T0 − 2.25T0 ) 5 −1 3 = – 15.562 T0 J mol–1 =
Let x be change in volume in each chamber due to displacement of the partition.
⇒
For one mole of a gas: pV = nRT, and n=1 pR VR TR ∴ × = …(i) p0 V0 T0
∴
243 V0 (1 + 0.7037) TL × = 32 V0 T0
t.m e/
ht tp
= 2.25T0
243 TR = T0 32
5 −1 3 5/3
γ−1 γ
s: //
ht tp s:
//t
γ−1 γ
1 243 × p0 TR = T0 p0 32
TR pR = T0 p0
.m
tp
ht
Then
or
e/
s:
un iv
//w
w
w
.ja
m
ar a
(i) As the gas in the left chamber is heated, it expands and the partition moves to the right. There by expansion takes place in the left chamber and compression in the right chamber. Let pR, VR and TR be the final pressure, volume and temperature in right chamber. No heat transfer takes place in right chamber. Hence there is change is adiabatic.
)
) Solution
243 p0 (V0 + x) TL × = 32 p0 V0 T0
∴
na .c om
(ii) The work done by the gas in the right chamber.
243 p0 32
(negative sign shows that the work done on the gas.)
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Kinetic Theory of Gases and Thermodynamics
p2 V2 p1v1 − nR nR
=
where γ =
in fo
s_
V γ−1 p1V1 1 − 1 R V2
s
γ=
C p, m =
5 R 2
5 3
un iv
e/
.m
//t
s:
Since the volume is held constant, work done by the gas on heating at constant volume, therefore W2 = 0
Total work don by the gas, W = W1 = W 2 2 3 p1V1 V1 3 − =− 1 2 V2
(b) Total change in internal Energy Change in internal energy in adiabatic compression, as derived above,
3 R, and 2
2 3 p1V1 V1 3 − 1 Hence, W1 = − 2 V2
.
Cv , m
rio
ita
er s
∴
ht
tp
Cv , m
Cv , m p1V1γ γ V2 − p1V1 R V2
rio ita
ni e/ u .m //t s:
The curve A to B represents the adiabatic compression of the gas from the volume V1 to V2. In this process the pressure of the gas increases p1 to p2. The line B to C represents increase in pressure of the gas as a result of giving here Q to the gas at constant volume. In this process, the pressure of the gas increases from p2 to p3. (ii) (a) Total work done by the gas Work done by the gas in adiabatic compression. In an adiabatic process, since Q = 0, therefore from the first law of thermodynamics, ∆U1 = – W1 or W1 = ∆V1 = – Cv∆T = – Cv (T2 – T1)
=
Cv , m =
ve
w ht tp
V1
Cp,m
For a mono-atomic gas,
A(p1, V1)
//w tp s: ht V2 V
W1 =
rs
w
p
…(i)
From equation (i),
om
.ja
B(p2, V2)
( p2 V2 − p1V1 )
P1 V1γ = P2 V2γ
an a. c
m ar
C(p3, V3)
R
For a gas underdoes adiabatic process
(i) Figure displays the p-V diagram of the gas undergone the given two processes.
)
) Solution
Cv , m
= − Cv
26. Two moles an ideal mono-atomic gas initially at pressure p1 and volume V1 undergo an adiabatic compression until its volume is V2. Then, the gas is given heat Q at constant volume V2. (i) Sketch the complete process on a p–V diagram. (ii) Find the total work done by the gas, total change in its internal energy and the final temperature of the gas. [Give your answer in terms of p1, V1, V2, Q and R].
2.46
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2 3 p1V1 V1 3 − 1 ∆U1 = 2 V2
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Kinetic Theory of Gases and Thermodynamics
at constant pressure until the volume is doubled. Then, it undergoes an adiabatic change until the temperature returns to its initial value.
2 3 p1V1 V1 3 − + Q. = 1 2 V2
(i) Sketch the process on a p-V diagram. (ii) What are the final volume and pressure of the gas? (iii) What is the work done by the gas?
Change in internal energy on heating the gas at constant volume ΔU2 = Q Total change in the internal energy of the gas ΔU = ΔU1 + ΔU2
) Solution
o
nf
s_ i ta si
s io ta r si er iv
un e/
ht
tp
.m s: //t
ht
Change in temperature on heating the gas
3
iv
e/ un
2/3
Q = CV ΔT = CV (T3 – T2)
V2
V1
∴ p1 = =
2 × 8.3 × 300 = 2.49 × 105 Nm −2 20 × 10−3
or
∴
∴
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n RT1 V1
Now, V ∝ T
27. Two moles of helium gas (γ = 5/3) are initially at temperature 27ºC and occupy a volume of 20 litres. The gas is expanded
V
p1V1 = nRT1
2
p1V1 V1 3 Q T3 = + . (3 mole)R (2 mole)R V2
V3
Process 1 → 2 is isobaric expansion
2
p V V 3 Q Q + T2 = + 1 1 1 or T3 = CV nR V2 3 nR 2 Since n = 2, therefore
2
1
er
w
tp s: //t .m
ht
rio
p1
w
tp s
://
− 1
2/3 p1V1 p1V1 V1 − 1 + nR nR V2
pV V = 1 1 1 nR V2
m
a. an ar
2/3
2/3 3 p1V1 V1 − 1 T2 − T1 + 3 V 2 nR 2 2
=
p
m
3 p1V1 V1 2CV V2
.ja
T2 − T1 =
w
or
∆U1 CV
co
therefore, ∆T =
(i) V1 = 20 × 10–3 m3 T1 = 300 K n = 2 moles 5 γ= 3
(c) Final temperature of the gas Change in temperature in adiabatic compression. Since, ΔU = CVΔT
)
2.47
V1 V2 = T1 T2 T2 = T1 ×
V2 2V = 300 × 1 V1 V1
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2.48
Kinetic Theory of Gases and Thermodynamics
s
…(iii)
...(iv)
where, K is constant and is given as K = pA TA = (p1) (2T1) = 2 p1 T1
...(v)
pV = nRT and p T = K
Eliminating T, we get
K …(vi) pV = nR or p 2 V = nRK P (i) The expression of work done by the gas is W = ∫ pdV In the process, we get WAB =
VB
∫ pdV
VA
(ii) The heat absorbed or released by the gas in each of the process.
rio
ita
rs
pT = K
(i) The work done on the gas in the process AB.
nRTC nR(2T1 ) n RT1 = = PC 2 P1 P1
In the process AB, we will have
ht
nRTB nR(2T1 ) 1 n RT1 = = …(ii) PB P1 2 P1
iv e
e/ un
:// t.m
tp s
VB =
It is given that during the process AB,
28. Two mole of an ideal mono-atomic gas is taken through a cycle ABCA as shown in the p – T diagram. During this process AB, pressure and temperature of the gas vary such that pT = constant. If T1 = 300 K, calculate:
fo rio
si ta
−3
(ii) Final volume, V3 = 113.14 × 10–3 m3 Final pressure, p3 = 0.44 × 105 Nm–2 (iii) Total work done by the gas = W = (W)1–2 + (W)2–3 = 4980 + 7472.8 = 12452.8 J.
nRTA nR(2T1 ) 2n RT1 = = …(i) PA P1 P1
VC =
tp
ht
in
)
iv
un
e/
5
( 5 / 3) − 1
= 7472.8 J.
s_
an ar
m .ja
( 2.49 ×10 ) ( 40 ×10 ) − ( 0.44 ×10 ) (113.14 ×10 ) −3
VA =
ht
p2 V2 − p3 V3 γ −1
s: //t .m
tp s
://
w
= 0.44 × 105 N/m 2
(W) 2−3 =
T
2T1
) Solution The volumes of the gas at three states
er
5/3
T1
A, B and C are as follows:
w w
40 113.14
= 2.48 × 105
a.
p3 = p2
∴
V3 V2
γ
A
co m
Now, p2 V2γ = P3 V3γ
C
P1
−1
V3 3 T2 600 = =2 = V T1 300 2
V3 = V2 × (2)3/2 = 40 × 10–3 (2)3/2 = 113.14 × 10–3 m3
∴
5
B
2P1
∴
=
P
Given, T2 V2γ−1 = T3 V3γ−1−1 ,T3 = T1 5
Give answers in terms of the gas constant R.
Work done during process 1 → 2, (W)1–2 = p × ∆V = 2.49 × 105 × (40 – 20) × 10–3 = 4980 J Process 2 → 3 is adiabatic expansion T2 = 600 K p2 = p1 = 2.48 × 105 N/m2 V2 = 40 × 10–3 m3
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=
VB
∫
VA
nRK dV V
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Kinetic Theory of Gases and Thermodynamics
= (3 mole) (R) (2 T1 – T1) = (3 mole) (R) (300 K) = (900 mole K) R QBC = ∆UBC + WBC = (900 mole K) R + (600 mole K) R = (1500 mole K) R The positive sign implies that the heat is absorbed in the process BC. The process CA takes place at constant temperature. Hence,
VB
nRT dV V VA
∫
ita
= (2 mole) (R) (2 × 300 K) In 2 = (1200 mole K) R ln 2 ∆UCA = 0 QCA = ∆UCA + WCA = 0 + (1200 mole K) R ln 2 The positive sign implies that the heat is absorbed in the process CA.
un
29. An ideal mono-atomic is confined in a cylinder by a spring-loaded piston of cross-section 8 × 10–3 m2. Initially, the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is on its relaxed (unstretched, uncompressed) state as shown the figure. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is 8000 Nm–1 and atmospheric pressure is 1 × 105 Nm2. The cylinder and the piston are thermally insulated. The piston is massless and there is no friction between the piston and cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heater coil is negligible. [Assume the spring to be massless].
ht tp
s: //
t.m e/
iv
er si ta rio s
er s
VA …. (where, T = 2 T1) VC
=
rio s_ in fo
∫ pdV
VA
un iv
e/
.m
ht tp s:
//t
ht
tp
s:
VB
= nRT ln
ar a
.ja
w
w
= – (3 mole) T1R = – (3 mole) (300 K) R = – (900 mole K) R Now, from the first law of thermodynamics, QAB = ∆UAB + WAB = – (1200 mole K) R – (900 mole K) R = – (2100 mole K) R The negative sign implies that the heat is released in the process AB. The process BC takes place at constant pressure. Hence, WBC = p V = (2 p1) (VC – VB) nRT1 nRT1 = (2 p1 ) − 2 P1 P1
= nRT1 = (2 mole) (300 K) R = (600 mole K) R Now, ∆UBC = nCv∆T
WCA =
m
3 = (2 mole) R (T1 − 2 T1 ) 2
//w
)
na .c om
(
Using equations (i), (ii) and (v), we get nRT1 2n RT1 WAB = nR(2 p1T1 ) 2 −2 2 p p1 1 1 = 2 nRT1 (2) − 2 2 = – 2n T1R = – 2 (2 mole) (200 K) R = – (1200 mole K) R The negative sign implies that the work is done on the gas. Hence, work done on the gas = (1200 mole K) R (ii) Change in energy of the gas in the process AB is ∆UAB = nCv∆T
= nRK 2 VB − 2 VA
3 = (2 mole) R (TC − TB ) 2
2.49
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2.50
Kinetic Theory of Gases and Thermodynamics
1 ∆W = k ⋅ x 2 + p0 ⋅ ∆V 2 1 = × 800 × (0.1) 2 + 1× 105 × 8 × 10−4 = 120 J 2 1 2 = × 800 × (0.1)Rigid + 1× 105 × 8 × 10−4 = 120 J 2 support ΔQ = ΔU + ΔW = 600 + 120 = 720 J.
Open atmosphere
rio s_ in fo
er s
ita
.ja
un iv
e/
.m
b
iv
i
v
) Solution Path “iaf”∆Q = 50 cal )
s: //
t.m e/
un
//t
ht tp s:
f
er si ta rio s
w
s:
Total pressure acting on the right hand side p′R = p0 + pS = 2 × 105 N/m2 Under equilibrium p′L = p′R
ht
a
F 800 N = = 1× 105 Nm −2 A 8 × 10−3 m 2
tp
pS =
//w
w
∴
∴
p
m
F = kx = 8000 × 0.1 = 800 N Pressure exerted on the piston by the spring
ar a
na .c om
Initallty for the equilibrium of the piston, pL = pR = p0 where pL and pR are the pressures on the left hand and right hand side of the piston. Force exerted by the spring on the piston when it moves
(i) ∆W along the path “ibf”. (ii) If ∆W = – 13 cal for the curved path “fi”, what is the ∆Q for this path? (iii) Taking Ui = 10 cal, what is Uf? (iv) If Ub = 22 cal, what is ∆Q for the process “ib” and the process “bf”?
)
) Solution Let p0 be the atmospheric pressure.
30. A system is taken from state i to the state f (refer to the figure). Along path “iaf”, it is found that ∆Q = 50 cal. ∆W = 20 cal. Along the path “ibf”, ∆Q = 36 cal. Calculate:
Heater
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pL VL p′L V′L = TL T′L
ht tp
∆W = 20 cal ⇒ ∆U = ∆Q – ∆W = 50 – 20 = 30 cal 1× 105 × 2.4 × 10−3 ⇒ U – Ui = 30 cal f = 300 As internal energy change is a state function. 5 −3 ∆U will be same for any path from i to f. 2 × 10 × 3.2 × 10 = (i) Path “ibf” ∆W = ∆Q – ∆U T′L = 36 – (Uf – Ui) T′L = 800 K = 36 – 30 = 6 cal. ∆U = nCv∆T (ii) Path “fi” ∆Q = ∆U + ∆W pL VL 1× 105 × 2.4 × 10−3 = (Uf – Ui) + ∆W where, n= = = 0.09638 mole R T 8.3 × 300 = (– 30) + (– 13) L pL VL 1× 105 × 2.4 × 10−3 = – 43 cal n= = = 0.09638 mole R TL 8.3 × 300 (iii) Uf – Ui = 30 cal 3 ∴ ∆U = 0.09638 × × 8.3 × (800 − 300) = 600 J Uf = Ui + 30 ∴ = 40 cal. 2 or
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Kinetic Theory of Gases and Thermodynamics
2.51
QA → B = n CP (TB – TA)
fo
s
ta rio
si
er
iv
e/ un
ht
B
tp s: //t .m
VB
= (1500 mole K) R. The process B → C occurs at constant temperature. From first law of thermodynamics dU = dQ – dW Since, the internal energy of an ideal gas depends only on temperature, therefore dU = 0 and dQ = dW QB → C = WB → C
rio s_ in
e/ un
.m
:// t C
ht
tp s
D
VD
iv er
w
s: //w
ht tp
V
si ta
m
.ja
w
ar a
na
.c om
5 = (2 mole) R (600 K − 300 K) 2
T
TA = 27°C
VB V = 2 and D = 4 VA VA
Given:
)
) Solution
dV V
= nR TB ln
VC VB
= nR TB ln
VD . . . . .(as VC = VD) VB
V V = nR TB ln D A VA VB
TB
TA
A
VA
= ∫ pdV = nRTB ∫
31. A mono-atomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the figure. The V V volume rations are B = 2 and D = 4. VA VA If the temperature TA at A is 27ºC, calculate: (i) The temperature of the gas at point B. (ii) Heat absorbed or released by the gas in each process. (iii) The total work done by the gas during complete cycle. Express your answer in terms of the gas constant R.
(i) The process A → B in which the plot (iv) Process “ib” ∆Q = ∆U + ∆W of V verse T is linear occurs at constant = (Ub – Ui) + (∆W)ibf pressure condition. (∆W)ib = (∆W)ibf V V Hence A = B Because (∆W)bf = 0 TA TA ∆Q = (22 – 10) + 6 V = 18 cal. or TB = B TA = (2) (300 K) = 600 K Process “bf” ∆Q = ∆U + ∆W TA = (Uf – Ub) + 0 T = VB T = (2) (300 K) = 600 K A B TA = (40 – 22) (ii) The process A → B occurs at constant = 18 cal. pressure. Hence,
4 = (2 mole) (R) (600 K) ln 2 = (1200 mole K) R ln 2 The process C → D occurs at constant volume. Hence, QC → D = nCv (TA – TB) 3 = (2 mole) R (300 K − 600 K) 2 = – (900 mole K) R
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2.52
∴
_i nf o
co
ta r
io s
ta r si er
un
iv
B
e/ t.m ://
45°
A
2p0
p0
ht tp s
45°
po
io s
si er
iv
un
.m e/
p
//t
s: ht tp B
a.
an
m ar
.ja
w
w
:// w
tp s
ht
p
P2 – 4p = – KT (Parabola)
= 2 p0 V 0 = 2 × 1 × 105 × 1 × 10–3 = 200 J. (iii) For path ABC (∆Q)ABC = (∆U)AC + (∆W)ABC ⇒ (∆U)AC = 600 – 300 = 300 J. Heat transfer in the process along path AC, (Q)AC = (∆U)AC + (∆W)AC = 300 + 200 = 500 J.
m
(i) Sketch the process on p – T diagram. (ii) Find the work done by the gas along the paths ABC and AC. (iii) Find the heat transfer in the process along the path AC.
KT +4 P
(ii) Workdone along path AC = (∆W)AC = p0 (3 V0 – V0)
32. An ideal gas expands from a volume V0 = 1 litre and pressure p0 = 1 bar to volume 3 litre along two different paths ABC and AC as shown in figure. The heat added to the gas along the path ABC is 600 J.
p=−
∴
(iii) Since, the process ABCDA is a cyclic process, therefore U = 0, W = Q where, Q = QA → B + QB → C + QC → D + QD → A = (1500 mole K) R + (1200 mole K) R ln 2 – (900 mole K) R – (1200 mole K) R ln 2 = (600 mole K) R.
p=–V+4
⇒
1 = (2 mole) (R) (300 K) ln 4 = – (1200 mole K) R ln 2.
At B VB = 2 V0 and pB = 2 p0 Equation of line BC, p – 2 p0 = – tan 45º (V – 2 V0)
The process D → A occurs at constant temperature. Hence, V Q D → A = WD → A = nRTA ln A VD
Kinetic Theory of Gases and Thermodynamics
A
V0
C
45°
45°
2V0
C
3V0
V
p Vo
3Vo
V
B
2p0
A
C
(i) Equation of line AB, p – p0 = tan 45º (V – V0) Hence for ideal gas, p=V Now pV = KT ⇒ p2 = KT (parabola) .....(where K is constant.)
)
) Solution
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T0
3T0
4T0
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T0
33. A 1.00 mole sample of an ideal monoatomic gas originally at a pressure of 1.00 atmosphere undergoes a three-step process:
fo
s_ in
rio
Q2 = n CP(T3 – T2)
5 = × 8.314 JK −1 (231.4 K – 389 K) 2
rio
s
iv er si
w
w
//w
V 45.1 T3 = 1 T2 = (389 K) = 231.4 K 75.8 V2
ta
ar an a.
.ja m
)
co
m
) Solution First step Adiabatic Expansion Q1 = 0
Workdone in the process W2 = – p2 (V1 – V2) = – (0.421 × 101.325 kPa) (45.1 dm3 – 75.8 dm3) = 1309.6 J
(a) Plot these processes on a p-V diagram. (b) Determine T3. (c) Calculate the change in integral energy the workdone by the gas, and heat added to gas for each process. (d) For the complete cycle.
2.53
Second step compression at constant pressure: The final volume in this process will be V1 as in the third step, the system returns to the original state by constant volume process. Hence, in the second step, T2 = (389 K) changes to T3 V2 = (75.8 dm3) changes to V1 = 45.1 dm3 p2 = remains constant.
(i) It is expanded adiabatically from T1 = 550 K and T2 = 389 K. (ii) It is compressed at constant pressure until its temperature reaches T3. (iii) It then returns to its original pressure and temperature by a constantvolume process.
Kinetic Theory of Gases and Thermodynamics
un i
ve rs
t.m
.m e/
s: //
tp
un
e/
tp s:
ht
2
1
T2
550
∆U = 3973.3 J 389 = (1.00 atm) = 0.421 atm. Since, the system return to its original state, 550 −1 −1 K) will have 3 nRT2 (1.0 mole) (8.314 JK mole )(550 we V2 = = = 45.1 ∆U dm =Q+W=0 p2 (1.0 ×101.325 KPa ) Now, W = W1 + W2 + W3 (1.0 mole) ( 8.314 JK −1 mole −1 ) (550 K ) = – 2007.8 J + 1309.6 J + 0 = = – 698.2 J (1.0 ×101.325 KPa ) ∴ Q = – W = 698.6 J = 45.1 dm3 The p-V plot of the given process is shown in ∆U1 = W1 = – 2007.8 J the figure: 5 2
T = p1 1 T2
CP R
tp s:
P
ht
//t
ht
ita
= – 3275.7 J ∆U2 = Q2 + W2 3 −1 −1 – 3275.7 J + 1309.6 J = – 1966.1 J = (1.00 mol) × 8.314 JK mole × (389 K − 550=K) 2 Third step compression at constant volume in 3 mol) × 8.314 JK −1 mole −1 × (389 K − 550 K) this process: 2 W3 = 0 = – 2007.8 J V1 = (45.1 dm3) remains constant For adiabatic expansion of an ideal gas Q3 = n Cv (T1 – T3) p2 T2 (– CP / R )g = p1T1(– CP / R )g 3 C 5 = × 8.314 JK −1 (550 K – 231.4 K) 2 T1 R 389 2 p =p = (1.00 atm) = 0.421 atm. Hence, W1 = n1Cv (T2 – T1)
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2.54
Kinetic Theory of Gases and Thermodynamics
In the complete cycle ∆U = 0
p
(1 atm, 45.1 dm3, 550K)
Q = Q1 + Q2 + Q3 = 0 – 3275.7 J + 3973.3 J = 697.6 J
(0.421 atm, 75.8 dm3, 389K)
W = – Q = 697.6 J (= W1 + W2 + W3)
(0.421 atm, 45.1 dm3, 231.4K)
questions
ype
m
ita
iv
un
3. At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s the gas is (a) H2 (b) F2 (c) O2 (d) Cl2
t.m e/
) Explanation Vrms = )
∴
s: //
ht tp
Vrms, 2 = 2 × Vrms, 1 = 684 m/s
M=
3RT M
3 RT 3 × 8.31× 300 = v 2rms (1.930) 2
)
Vrms =
3 RT 3 × 8.31× 300 = v 2rms (1.930) 2
3RT M
= 2.0078 × 10–3 kg
3 2
) Explanation KE = kT
M=
T2 = 2 T1
er si ta rio s
un iv
e/
2. The average translational energy and the rms speed of molecules in a sample of oxygen at 300 K are 6.21 × 10–21 J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour). (a) 12.42 × 10–21 J, 968 m/s (b) 6.21 × 10–21 J, 968 m/s (c) 8.78 × 10–21 J, 684 m/s (d) 12.42 × 10–21 J, 684 m/s
=
Hence (d) is correct.
//t
ht tp s:
Hence (d) is correct.
Vrms, 2 Vrms,1
∴
.m
ht
er s
s:
tp
)
V p2 = p1 1 V2 = p(8)γ = p(8)5/3 = 32 p
KE2 = 2KE1 = 2 × 6.21 × 10–21 = 12.42 × 10–21 J
.ja
//w
w
w
) Explanation p1 V1γ = p2 V2γ ;
(CV + R)30 VV
∴
dT2 =
rio s_ in fo
i.e.,
ar a
1. A monomatomic gas (γ = 5/3) is suddenly th 1 compressed to its volume adia8 batically. The pressure of the gas will changes to 24 (a) (b) 8 5 40 (c) (d) 32 3
na .c om
SOLVED OBJECTIVE T
= – 2007.8 + 1309.6 J + 0 = 698.2 J
v
= 2.00 gm Hence (a) is correct.
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4. An ideal gas is taken through the cycle A → B → C → A, as shown in the gas in the cycle is 5J, the work done by the gas in the process C → A is (a) – 5J (b) – 10J (c) – 15J (d) – 20J
β
(c)
p
C
2
B
β
in
)
.ja m
s_
s
rio
6. Two insulating cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at temperature 300 K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K. Then the rise in temperature of the gas in B is (a) 30 K (b) 18 K (c) 50 K (d) 42 K
ht
tp s:
//t
ht
.m e/
un
iv
://
tp s
ta
un iv
t.m e/
tp s
ht
⇒
er si
w
w w
://
(dV / dp ) 1 = V p 1 β= p −
er si
⇒
5. Which of the following graph correctly represents the variation of dV â = − V with p for an ideal gas dP at constant temperature?
β
pV = constant pdV + Vdp = 0
ta rio
⇒
)
= (10 + 0 + WCA) J Given: ∆Q = 5J From first law of thermodynamics 5 = 10 + 0 + WCA ⇒ WCA = – 5J
fo
) Explanation As temperature is constant,
) Explanation For the cyclic process ∆U = 0 ∆W = WAB + WBC + WCA
p
ar an
10
(d)
p(N/m2)
a.
A
1
co
m
V(m3)
)
) Explanation For cylinder A For cylinder B
dQ = n CPdT1 dQ = nCVdT2 = n (CV + R) dT1
∴
p
(a)
dT2 =
β
(b)
nCVdT2 = n(CV + R) 30
(CV + R) 30 CV 5 For diatomic gas C V = R 2 ∴ dT2 = 42 K ∴
2.55
Kinetic Theory of Gases and Thermodynamics
p
Hence (d) is correct.
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2.56
)
For monatomic gas g =
Since volume is proportional to length, therefore,
s_
rio
s
rio
iv
si ta
un
P
er
e/
ht
.m
tp
e/
s:
un
//t
iv
.m
tp
ht
2
)
) Explanation In adiabatic process, slope of p-V
curve ∝ γ From the figure slope of curve 2 > slope of curve 1 ∴ γ2 > γ1
s:
tp
ht
L (b) 1 L2
2/3
L (a) 1 L2
1
V
//t
8. A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by
er si
ta
w .ja
w
w
s: //
fo
in
an
ar
m
9. p-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to (a) He and O2 (b) O2 and He (c) He and Ar (d) O2 and N2
RT mB Similarly for Gas in B, 1.5 ∆p = M 2V From equation (i) and (ii) we get 2mB = 3mA Hence (c) is the correct.
2/3
m a.
co
)
RT mA ∆p = M 2V
T1 L 2 = T2 L1
Hence (d) is correct.
1 RT 1 ∆p = p2 − p1 = mA − M V1 V2 Putting V1 = V and V2 = 2V, we get ∴
5 3
TV2/3 = constant
⇒
) Explanation TVγ – 1 = constant
) Explanation For gas in A, p1 = m RT p2 = A M V2
2/3
mA RT M V1
L2 (d) L1
L (c) 2 L1
7. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of gas contained in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The change in the pressure in A and B are found to be ∆p and 1.5 ∆p respectively. Then (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB
Kinetic Theory of Gases and Thermodynamics
Now γ for monatomic gas is greater than γ for a diatomic gas. Hence (b) is correct.
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Kinetic Theory of Gases and Thermodynamics
2.57
er si ta rio s
3
T0 (1/V0)
(1/V)
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(4/V0)
7. A thermally insulated vessel containing one mole oxygen at 27ºC, 1 atm is accelerated with an acceleration a0 and is suddenly stopped at the end of 5th second of its motion, there by the temperature of oxygen is increased by 0.5ºC, the value of a0 is nearly (a) 10 ms–2 (b) 6.3 ms–2 –2 (c) 7.2 ms (d) 5 ms–2
(d)
2T0
ht tp
4. Nitrogen gas and ethylene gas are kept in containers 1 and 2 respectively which are at the same temperature. The ratio of velocity of sound in the first cylinder to that in the second is
un t.m e/
2 3
3. How much heat energy should be added to a mixture of 10 gm of hydrogen and 40 gm of He to change the temperature by 50ºC kept in a closed vessel? (a) 2500 Cal (b) 2750 Cal (c) 2000 Cal (d) can’t be defined
A
C1 V1 = C2 V1
T
s: //
3
iv
.m
//t
(c)
ht tp s:
2
(d)
6. Given figure shows a parabolic graph 1 between T and for a mixture of a gas V undergoing an adiabatic process. Then the ratio of vrms and speed of sound in the mixture will be 3 (a) (b) 2 2
er s
un iv
e/
s:
tp
ht
1
C1 V = γ 1 C2 V1
ita
.ja
w
//w
w
2. A gas undergoes A to B through three different processes 1, 2 and 3 as shown in the figure. The heat supplied to the gas is Q1, Q2 and Q3 respectively, then (a) Q1 = Q2 = Q3 (b) Q1 < Q2 < Q3 (c) Q1 > Q2 > Q3 (d) Q1 = Q2 > Q3
B
T
3T
rio s_ in fo
ar a
2T
m
1.5T
T
(c)
D
A
C
B
1 >1 <1 depended on the amounts of the two gases
5. The intensity of sound wave in an ideal gas at temperature T1 K and T2 K are respectively C1 and C2. The rms velocity of gas molecule at these two temperatures are V1 and V2 respectively, then C1 V1 C V = (a) (b) 1 = γ 1 C2 V1 C2 V1
na .c om
P
O
(a) (b) (c) (d)
1. p-T diagram of one mole of an ideal monatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is (a) – 2 RT (b) 3 RT (c) RT (d) – 1.25 RT
UNSOLVED OBJECTIVE TYPE QUESTIONS
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2.58
8. If a monoatomic gas undergoes a thermodynamic process for which its molar heat capacity is equal to the universal gas constant, what is the process in terms of V and T? (a) VT–1/2 = constant (b) VT1/2 = constant (c) V2T2 = constant (d) VT = constant
T
D
C
B A
P
(c)
C
V
9. A gas in found to obey the law p V = constant. The initial temperature and volume are T0 and V0. If the gas expands to a volume 3 V0, then
D B
fo in
s_
s
rio
si ta
er
A B
14. A cyclic process is shown in the figure. For the process AB which of the following statement is correct? (a) Heat given to the system will be always –ve throughout the process. (b) Heat given to the system will be always +ve throughout the process.
V P
iv
C
D
13. The specific heat of an ideal gas is proportional to (a) Tº (b) T–1 1/2 (c) T (d) T3
B
(a)
s: //
ht tp
A
e/
t.m
ht
P
12. A vessel X contains 1 mole of O2 gas (molar mass 32) at a temperature T and pressure p. Another identical vessel Y contains one mole of He gas (molar mass 4) at temperature 2T, then (a) pressure in the container Y is p/8 (b) kinetic energy of O2 molecule = kinetic energy of He molecule (c) pressure in the container Y is 2p 6 (d) kinetic energy of He molecule = 5 (kinetic energy of O2 molecule)
un
s: // tp
10. A cyclic process ABCD is shown in the V–T diagram. Which of the following curves represent the same process?
C D
(b)
iv un e/ t.m
ht
T
er
w w
tp
B
A
11. The internal energy of a gas in given by U = 2pV. It expands from V0 to 2V0 against a constant pressure p0. The heat absorbed by the gas in the process is (a) 2p0 V0 (b) 4p0 V0 (c) 3p0 V0 (d) p0 V0
si ta
w C
s: //
V
T
rio
m
.ja
D
A
(d)
ar
an
(a) final temperature becomes 3 T0 (b) internal energy of the gas increases T (c) final temperature becomes 0 3 (d) internal energy of the gas decreases
a. co m
2
Kinetic Theory of Gases and Thermodynamics
V
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Kinetic Theory of Gases and Thermodynamics
(c) First the heat given is +ve then –ve throughout the process. (d) First the heat given is –ve then +ve throughout the process.
B
A
2.59
P C
A
2P0
V
P
C B
C
2V0
A
(a)
T
.c ta rio
.m
ht
rio
un
//t
.m
//t
C
ht
tp
s:
V
T
18. n moles of ideal monatomic gas undergoes a process in which the temperature changes with volume as T = KV2. If the temperature of the gas changes from T0 to 4T0, then (a) work done by the gas is 3nRT0 (b) heat supplied to the gas is 4nRT0 3 (c) work done by the gas is nRT0 2 3 (d) heat supplied to the gas is nRT0 2
B A
(d)
T
e/
s:
ht tp
B
A
(c)
(d) None of these
17. For a given thermodynamic process, the P – V diagram is as shown below: Which of the following is the V – T diagram for the process?
(c) 1
ita
iv er s
w
//w
tp s:
V
16. A cylinder of volume V = 10 ℓ contains 64 gm of gas in molecular state. Due to the faulty value, the gas pressure dropped to (1/4)th of the initial pressure such that the temperature remains unchanged. The factor by which velocity of sound has changed in the cylinder is 1 3 (a) (b) 2 2
T
si
p0 V0 4R
er
(d)
B
A
C
ni v
p0 V0 R
(b)
e/ u
2 p0 V0 R
V
.ja m
(c)
(b)
w
p0 V0 2R
(a)
s_
ar a
in f
na
o
C
s
15. One mole of an ideal gas undergoes a p0 , where p0 and v0 process p = 2 V 1+ v0 are constants. The temperature of the gas when v = V0 is
V0
B
V
om
P0
B → C: adiabatic C → A: isothermal
A → B: isothermal
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2.60
19. One mole of an ideal monatomic gas (initial temperature T0) is made to go through the cycle abca shown in the figure. If U denotes the internal energy, then (a) U0 – Ua = 10.5 RT0 (b) Ub – Ua = 4.5 RT0 (c) U0 > Ub > Ua (d) all of these
(d) average kinetic energy per mole of hydrogen 5 = × average kinetic energy per mole 3 of helium
fo
in
s
ta rio
er
iv .m
100
50
//t
T
ht
tp
s:
21. When two identical vessel contain helium and hydrogen at same temperature, then (a) average kinetic energy per mole of hydrogen. = average kinetic energy per mole of helium (b) average translational kinetic energy per mole of hydrogen = average translational kinetic energy per mole of helium (c) average kinetic energy per mole of hydrogen 3 = × average kinetic energy per mole 5 of helium
0.1
0.4 x (meters)
24. PV versus T graph of equal masses of H2, He and CO2 is shown in the figure. Choose the correct alternative. (a) 3 corresponds to H2, 2 to He and 1 to CO2 (b) 1 corresponds to He, 2 to H2 and 3 to CO2 (c) 1 corresponds to He, 3 to H2 and 2 to CO2 (d) 1 corresponds to CO2, 2 to H2 and 3 to He
0.3
0.2
si
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s: tp ht
O
.m
F (N)
//t
2
si
er
iv
e/ un
tp s:
ht
1
s_
m
.ja
w
//w w
P
20. An ideal gas undergoes through two processes 1 and 2. If V1 and V2 are the volumes in the process 1 and 2, then (a) V1 = V2 (b) V2 > V1 (c) V1 > V2 (d) none of these
23. The given figure shows the variation of force on an ideal gas (enclosed in a piston-cylinder arrangement) which undergoes a process during which piston position changes from 0.1 to 0.4 m. If the internal energy of the system at the end of the process is 2.5 J higher, then the heat absorbed during the process is (a) 15 Joules (b) 17.5 Joules (c) 20 Joules (d) 22.5 Joules
ta rio
ar
V
co m
4Vo
Vo
an a.
b
To a
Po
c
2Po
P
22. When the state of a system changes from A to B adiabatically the work done on the system is 322 joule. If the state of the same system is changed from A to B by another process, and 100 calories of heat is required then work done on the system in this process will be (a) 98 Joules (b) 38.2 Joules (c) 15.9 Joules (d) 15.9 Joules
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
2.61
(c) equal to work done on the gas 3RT (d) 2
PV
2
T
29. In a thermodynamic process helium, gas obeys the law TP–2/5 = constant. If temperature of 2 moles of the gas is raised from T to 3T, then (a) heat given to the gas is zero (b) increase in internal energy is 6RT (c) work done by the gas is – 6RT (d) all of these
ta rio
er si
iv
s: //
tp
ht
ρ
un t.m
ht
(a) the product of P and V at A is equal to the product of P and V at B (b) pressure at B is greater than the pressure at A (c) work done by the gas during the process AB is negative (d) all of these
e/
tp
26. A vessel contains 1 mole of O2 at a temperature T. The pressure of gas is p. An identical vessel containing 1 mole of the gas at a temperature 2T has pressure of (a) p (b) 2p (c) 3p (d) 4p
un
s: //
ht
Temperature
30. The density (ρ) of an ideal gas varies with temperature T as shown in figure. Then
t.m
tp
e/
s: //
C
s
er iv
w w
B
27. An insulated container is divided into two equal portions. One portion contains an ideal monoatomic gas at pressure P and temperature T, while the other portion is a perfect vacuum. If a hole is opened between the two portions, the change in internal energy of the gas is (a) zero (b) equal to work done by the gas
B
A
o
si ta
.ja w
A
s_ in f
rio
ar
m
Volume
an
a. co m
(c) γ : γ – 1 : 1 (d) γ : 1 : γ – 1
25. The expansion of an ideal gas of mass m at a constant pressure P is given by the straight line B. Then the expansion of the same ideal gas of mass 2m at a pressure 2P is given by the straight line (a) C (b) A (c) B (d) none of these
1
28. A gas of adiabatic exponent γ is supplied at a constant pressure. The ratio between dQ, dW and dU in the process 1 (a) γ : γ − 1: γ (b) 1 : 1 : γ – 1
3
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T
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2.62
Kinetic Theory of Gases and Thermodynamics
ANSWERS 4. 12. 20. 28.
(b) (a) (d) (a)
4. When a piece of metal weighing 48.3 g at 10.7ºC is immersed in a current of steam at 100ºC, 0.762 g of steam is condensed on it. Calculate the specific heat of the metal. [IIT, 1973] [Ans. 0.095 cal/gmºC]
ita
5. A metal piece, which weighs 46 g in air, weighs only 30 g when immersed in a liquid of density 1.24 g/cc at 30ºC. In the same liquid at 42ºC, when its density changes to 1.20 g/cc, it weighs 30.5 g. Find the coefficient of linear expansion of the metal. [Ans. 3.445 × 10–5/ºC]
un iv
e/
iv
.m
un
//t
6. A mixture of 250 gm water and 200 gm ice at 0ºC is kept in a calorimeter of water equivalent 50 gm. If 200 gm of steam at 100ºC is passed through this mixture, calculate the final temperature and the weight of the contents of the calorimeter. Latent heat of fusion of ice = 80 cal/gm Latent heat of vapourization = 540 cal/gm [IIT, 1974] [Ans. Weight of the contents of calorimeter = 572.2 g and final temperature = 100ºC] 7. A vessel of volume V = 20 litre contains a mixture of Argon and Helium at a temperature t = 20ºC and pressure P = 2.0 atm. The total mass of the mixture in the vessel is m = 5.0 g. Find the ratio of the mass of Argon to that of Helium in the mixture. [Ans. 0.61]
3. In an industrial process 10 kg of water per hour is to be heated from 20ºC to 80ºC. To do this, steam at 150ºC is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90ºC. How many kg of steam are required per hour? Specific heat of steam = 1 cal/gmºC = 1 kilo cal/kgºC
ht tp
s: //
t.m e/
ht tp s:
Latent heat of steam = 540 cal/gm = 540 kilo cal/kg. [IIT, 1972] [Ans. 1kg]
er s
w
//w
s:
tp
ht
8. (b) 16. (c) 24. (a)
na .c om
ar a
m
.ja
w
2. A vertical cylinder of total length 100 cm is closed at the lower end and is fitted with a movable, frictionless, gas tight disc at the other end. An ideal gas is trapped under the disc. Initially, the height of the gas column is 90 cm when the disc is in equilibrium between the gas and the atmosphere. Mercury is then slowly poured on the top of the disc and it just starts overflowing when the disc has descended through 32 cm. Find the atmospheric pressure. Assume the temperature of the gas to remain constant and neglect the thickness and weight of the disc. [IIT, 1971] [Ans. 76.125 cm of mercury]
7. (d) 15. (a) 23. (c)
(b) (c) (a) (d)
1. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and raised further by 44 cm. What will be the length of air column above mercury in the tube? Atmospheric pressure = 76 cm of mercury. [IIT, 1967] [Ans. 15.42 cm]
6. 14. 22. 30.
(a) (a) (b, d) (d)
EXERCISE
P
5. 13. 21. 29.
(b) (c, d) (b) (c)
rio s_ in fo
3. 11. 19. 27.
(c) (b) (b, c) (b)
er si ta rio s
2. 10. 18. 26.
(d) (a, b) (b) (c)
ractice
1. 9. 17. 25.
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14. 5 gm of water at 30ºC and 5 gm of ice at –20ºC are mixed together in a calorimeter. Find the final temperature of the mixture. Water equivalent of the calorimeter is negligible, specific heat of ice = 0.5 cal/gm ºC and latent heat of ice = 80 cal/gm. [IIT, 1977] [Ans. 0ºC]
ht tp
(a) pressure of the mixture
nM +n M +n M RT (n1 + n2 + n3 ),(b) 1 1 2 2 3 3 Ans. (a) n1 + n2 + n3 V
(b) mean molar mass M of the mixture which enters into equation of state m pV = RT, where m is the mass M of the mixture.
15. A vertical cylinder closed at both the ends has a freely moving piston that divides its volume into two parts, each containing one mole of air. In equilibrium, at T = 300 K the volume of the upper part is η = 4.0 times greater than that of the lower part. Find the temperature T′ at which the ratio of these two volumes will be equal to η′ = 3.0. [Ans. 421.9 K] 16. A vessel is completely filled with 500 gm of water and 1000 gm of mercury. When 21200 calories of heat are given to it, 3.52 gm of water overflows. Calculate the coefficient of volume expansion of mercury. Expansion of vessel may be neglected.
s: //
t.m e/
ht tp s:
un
//t
iv
.m
er si ta rio s
e/
un iv
er s
ita
.ja
w
w
//w
s:
tp
ht
11. A vessel of volume V contains a mixture of ideal gases at temperature T. The mixture contains n1 mole of oxygen, n2 mole for nitrogen and n3 mole of carbon dioxide. Assuming the gases to be ideal. Find:
na .c om
m
ar a
10. A mixture of nitrogen (m1 gm) and carbon dioxide (m2 gm) is kept in a vessel at temperature T and at a pressure p0. Find the density of the mixture, assuming the gases to be ideal. Ans. p0 (m1 + m2 ) m m RT 1 + 2 M1 M 2
13. A glass capillary tube, sealed at both ends is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube (which are equal in length) contain air at 27ºC and at a pressure of 76 cm of mercury. The tube is kept in a horizontal position such that the air column at one end is at 0ºC and at the other end is maintained at 127ºC. Calculate the length of the air which is at 0ºC and its pressure. Neglect the change in volume of mercury and glass. [IIT, 1975] [Ans. Length of air column at 0ºC, x = 36.5 cm, P2 = 85.3 cm]
rio s_ in fo
9. What should be the lengths of steel and copper rod so that the length of steel rod is 5 cm longer than the copper rod at all temperature? (α for copper = 1.7 × 10–5 per ºC and α for steel = 1.1 × 10–5 per ºC) [IIT, 1975] [Ans. Length of copper rod = 9.17 cm, Length of steel rod = 14.17 cm]
12. The temperatures of equal masses of three different liquids A, B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC and when B and C are mixed, it is 23ºC. What should be the temperature when A and C are mixed? [IIT, 1976] [Ans. 20.26ºC]
Latent heat of vapourization of water = 540 cal/g. Latent heat of fusion ice = 80 cal/g Specific heat of copper = 0.092 cal/gºC and, specific heat of ice = 0.5 [Roorkee, 1975] [Ans. 4.6 g]
2.63
8. 20 g of ice in 10 g copper calorimetry is originally at –30ºC. How much steam at 100ºC must be condensed in the calorimeter if the ice is to be changed to water and heated to 40ºC?
Kinetic Theory of Gases and Thermodynamics
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ideal gas is enclosed between the pistons, which are tied together with an inextensible thread. The cross-sectional area of the upper piston is ∆S = 10 cm2 larger than that of the lower one. The combined mass of the two pistons is m = 5.0 kg and the outside air pressure is p0 = 1.0 atm. Find by how many kelvin should the gas between the pistons be heated so as to shift the pistons through ∆l = 5.0 cm. [Ans. 0.9 K]
er s
un iv
e/
un
iv
where T0 and α are constants and V is the volume of one mole of gas?
t.m e/
23. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal crosssectional area A. Atmospheric pressure is P0 and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surrounding, show that the piston executes simple harmonic motion and find the frequency of oscillation. [IIT, 1981]
s: //
ht tp
20. A smooth vertical tube having two different areas of cross sections is open from both the ends and is provided with two freely moving pistons of different cross sections. Each piston slides within its respective cross section. One mole of an
T = T0 + α V2,
.m //t
ht tp s:
(W2 − W1 ) + β(W0 − W1 )(T2 − T1 ) Ans. λ = (W0 − W2 )(T2 − T1 )
22. What is the minimum attainable pressure of an ideal gas in the process
[ Ans. Pm = 2R α T0 ]
19. A sinker of weight W0 has an apparent weight W1 when weighed in a liquid at a temperature T1 and W2 when weighed in the same liquid at a temperature T2. The coefficient of cubical expansion of the material of sinker is β. What is the coefficient of volume expansion of the liquid?
ita
.ja
w
w
//w
s:
tp
ht
log e η Ans. N = ∆V log e 1 + V
rio s_ in fo
m
18. A vessel of volume V is required to be evacuated by means of a piston air pump that captures volume ∆V in each stroke. How many strokes will reduce the pressure in the vessel η times? Assume the process to be isothermal and the gas, ideal.
ar a
na .c om
Given: coefficient of linear expansion of glass = 9 × 10– 6 per degree, coefficient of volume expansion of mercury = 1.8 × 10– 4 per degree. [IIT, 1973] 3 [Ans. 150 cm ]
21. Ice at 0ºC is added to 200 g of water initially at 70ºC in a vacuum flask. When 50 g of ice has been added and has all melted, the temperature of the flask and contents is 40ºC. When a further 80 g of ice has been added and has all melted, the temperature of the whole becomes 10ºC. Neglecting heat loss to surroundings calculate latent heat of fusion of ice. [MNR, 1979] [Ans. L = 90 cal/g]
17. A one litre flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in flask?
Given: Coefficient of volume expansion of water = 1.5 × 10–4 per degree centigrade Density of mercury = 13.6 gm/cm3 Specific heat of mercury = 0.03 cal/gm ºC [IIT, 1976] –4 [Ans. 1.768 × 10 /ºC]
er si ta rio s
Kinetic Theory of Gases and Thermodynamics
1 γ P0 A 2 Ans. n = 2π MV0
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Latent heat of ice = 80 cal/g. [IIT, 1972] [Ans. 2.96 × 10–4 k cal/m. sec ºC]
iv
31. A 2 m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm. When a current of 5 amp flows through the wire, find the temperature difference across the insulation in steady state. Thermal conductivity of plastic = 0.16 × 10–2 cal/s cm ºC. [IIT, 1974] [Ans. 2.05ºC]
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If the final temperature of the container is 27ºC, find the mass of the container. (Latent heat of fusion for water = 8 × 104 cal/kg, specific heat of water = 103 cal/kg K). [IIT, 2001] [Ans. 0.495 kg]
t.m e/
s: //
ht tp
27. An ice cube of mass 0.1 kg is placed in an isolated container which is at 227ºC. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where, A = 100 cal/kg K and B = 2 × 10–2 cal/kg K2.
er si ta rio s
ita
un iv
e/
.m
//t
ht tp s:
2 + (α A + α B )∆T ⋅ d Ans. R = 2 (α A − α B )⋅ ∆T
30. Water is being boiled in a flat bottom kettle on a stove. The area of the bottom is 300 cm2 and the thickness is 2 mm. If the amount of steam produced is 1 g/min, calculate the difference in temperature between the inner and the outer surfaces of the bottom. The thermal conductivity of material of kettle = 0.5 cal (ºC)–1 sec–1 cm–1 and latent heat of steam 540 cal/g. [IIT, 1972] [Ans. 0.012ºC]
un
w
//w
s:
tp
ht
er s
w
26. Two metal strips, A and B each of length l and thickness d at temperature T, are rivetted together. The coefficient of linear expansion of the strips are αA and αB (αA > αB) respectively. When this bimetallic strip is heated to a temperature (T + ∆T), one strip becomes longer than the other and the bimetallic strip bends into an arc of a circle. Find the mean radius of curvature of the strip.
.ja
m
ar a
na .c om
29. A closed cubical box made of perfectly insulating material has walls of thickness 8 cm and the only way for heat to enter or leave the box is through two solid, cylindrical, metallic plugs, each of crosssectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box as shown in the figure. Outer surface A is kept at 100ºC while the outer surface B of the other plug is maintained at 4ºC. The thermal conductivity of material of the plug is 0.5 cal/s/cm/ºC. A source of energy generating 36 cal/s is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. [Ans. θ = 76ºC]
rio s_ in fo
25. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0ºC and at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62ºC. What is the new value of pressure inside the bulbs? The volume of the connection tube is negligible. [IIT, 1985; MNR, 1990] [Ans. 83.75 cm of mercury column]
2.65
28. A slab of stone of area 0.36 m2 and thickness 0.10 m is exposed on the lower surface to steam at 100ºC. A block of ice at 0ºC rests on the upper surface of the slab. If in one hour, 4.8 kg of ice is melted, calculate the thermal conductivity of the stone.
24. An ideal gas of molar mass M is contained in a uniform gravitational field having free-fall acceleration due to g. Find the pressure of the gas as a function of height h if P = P0 at h = 0 and the temperature varies with height h as: T = T0 (1 – ah) where a is a positive constant. [Ans. P = P0 (1 – ah)n]
Kinetic Theory of Gases and Thermodynamics
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32. Two rods A and B are of equal length. Each rod has its ends at temperature T1 and T2. What is the condition that will ensure equal rates of flow of heat through the rods A and B? [IIT, 1976] [Ans. Hence, areas of cross section of the two rods should be inversely proportional to the square root of their thermal conductivity]
C x A
y
rio s_ in fo
e/
er si ta rio s
un iv
er s
37. A block of steel heated to 100ºC is left in a room to cool. Find the curve shown in the figure, which represents the correct behaviour. [IIT, 1980]
ita
m
ar a
na .c om
36. A wall has two layers A and B, each made of different materials. Both the layers have the same thickness. The thermal conductivity of the material A is twice that of B. Under thermal equilibrium, the temperature difference across the wall is 36ºC. Find the temperature difference across the layer A. [IIT, 1980] [Ans. 12ºC]
iv
.m
u Temperature n
C B A
[Ans. Curve A]
38. Radiant energy from the sun strikes the earth at a rate of 1.4 × 103 watt/m2. Calculate the temperature at the surface of sun. Radius of sun = 7 × 108 m Radius of earth orbit = 1.5 × 1011 m Stefan’s constant = 5.7 × 10–8 Wm–2 K–4 [MNR, 1985] 5 [Ans. 1.833 × 10 K]
35. Three rods of material x and three rods of material y are connected as shown in the figure. All the rods are of identical length and cross-sectional area. If the end A is maintained at 60ºC and the junction E at 10ºC, calculate the temperatures of the junctions B, C and D. The thermal conductivity of x is 0.92 CGS unit and that of y is 0.46 CGS unit. [IIT, 1978]
Time
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s: //
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ht tp s:
//t
ht
tp
34. A room is maintained at 20ºC by a heater of resistance 20 ohm connected to 200 volt mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area 1 m2 and thickness 0.2 cm. Calculate the temperature outside (Thermal conductivity of glass is 0.2 cal m–1 ºC sec–1 and mechanical equivalent of heat J = 4.2 joule/cal). [IIT, 1978] [Ans. 15.24ºC]
E
X
[Ans. 30ºC]
.ja
w
w
//w
s:
B
y
D
33. A bar of copper of length 75 cm and a bar of steel of length 125 cm are joined together end to end. Both are of circular cross section with diameter 2 cm. The free ends of copper and steel are maintained at 100ºC and 0ºC respectively. The surface of the bars are thermally insulated. What is the temperature of copper steel junction? Thermal conductivity of copper is 9.2 × 10–2 kilo cal/m (ºC) sec and that of steel is 1.1 × 10–2 kilo cal/m (ºC). [IIT, 1977] [Ans. 93.28ºC]
x
y
2.66
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un
45. A body cools in 7 minutes from 60ºC to 40ºC. What will be its temperature after the next 7 minutes? The temperature of the surroundings is 10ºC. Assume that the Newton’s law of cooling holds good throughout the process. [MNR, 1992] [Ans. 28ºC] 46. The peak emission from a black body at a certain temperature occurs at a wavelength of 9000 Å. On increasing its temperature the total radiation emitted is increased 81 times. At the initial temperature when the peak radiation from the black body is incident on a metal surface, it does not cause any photo emission from the surface. After the increase of temperature, the peak radiation from the black body caused photo emission.
t.m e/
s: //
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43. A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed
er si ta rio s
iv
.m
//t
ht tp s:
π K r1r2 (T1 − T2 ) Ans. l
ita
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e/
s:
tp
ht
42. A tapered rod of length l and thermal conductivity K has its end radii r1 and r2 maintained at temperature T1 and T2 respectively. Calculate the rate of flow of heat through the rod.
K1 + 3K 2 Ans. 4
44. One end of a copper rod of uniform cross-sectional area and of length 1.5 m is in contact with ice and the other end with water at 100ºC. At what point along its length should a temperature of 200ºC be maintained so that in steady state the mass of ice melting is equal to that of steam produced in the same interval of time? Assume that whole system is insulated from surroundings. [MNR, 1992] [Ans. 1.396 m]
er s
//w
41. A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. Calculate the effective thermal conductivity of the system. [IIT, 1988]
w
w
.ja
m
ar a
na .c om
40. An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of 20ºC inside it when the outside temperature is –10ºC. The walls have three layers of different materials. The innermost layer is of wood of thickness 2.5 cm, the middle is of cement of thickness 1.0 cm and the outermost layer is brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1.0 watt m–1 (ºC)–1 respectively. [IIT, 1986] [Ans. 9 k W]
coaxially on a thin metal disc of mass 0.4 kg and of the same cross section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If thermal conductivity of the material of the cylinder is 10 watt/m-K and the specific heat of the material of the disc is 600 J/kg-K, how long will it take for the temperature of the disc to increase to 350 K? Assume, for the purpose of calculation, the thermal conductivity of the disc to be very high and system to be thermally insulated except for the upper face of the cylinder. [IIT, 1992] [Ans. 2 min. 46.35 sec]
rio s_ in fo
Boltzmann constant = 5.67 × 10–8 watt/ m2 (K)4. [MNR, 1986] [Ans. 0.177]
2.67
39. A body which has a surface area 5.00 cm2 and a temperature of 727ºC radiates 300 joules of energy each minute. What is the emissivity?
Kinetic Theory of Gases and Thermodynamics
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(b) temperature of the coil. (Given σ = 17 × 10–8) 3 [IIT, 2003] [Ans. (a) 595 watt/m2, (b) 147ºC]
49. The air density of Mount Everest is less than that at the sea level. It is found by mountaineers that for one trip lasting a few hours, the extra oxygen needed by them corresponds to 30,000 cc at sea level (pressure 1 atmosphere, temperature 27ºC). Assuming that the temperature around Mount Everest is –130ºC and that the oxygen cylinder has capacity of 5.2 litre, find the pressure at which oxygen be filled (at site) in the cylinder. [MNR, 1978] [Ans. 2.75ºC]
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52. One gram mole of oxygen at 27ºC and one atmospheric pressure is enclosed in a vessel. (a) Assuming the molecules to be moving with vrms, find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (b) The vessel is next thermally insulated and moved with a constant speed v0. It is then suddenly stopped. The process results in a rise of the tempera-
To
Ta = 27°C Hot Oil
T = 127°C
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51. An electric bulb of volume 250 cm3 was sealed off during manufacture at a pressure of 10–3 mm of Hg at 27ºC. Find the number of molecules in the bulb. [IIT, 1974] [Ans. 8.03 × 1015]
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48. Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity k = 0.149 J/(m ºC-sec), thickness t = 5, emissivity = 0.6. Temperature of the top of the lid is maintained at Tf = 127º. If the ambient temperature Ta = 27ºC, calculate
50. Modern vacuum pumps can bring down the pressure to P = 4.0 × 10–15 atm. at room temperature T = 300 K. Assuming that the gas exhausted is nitrogen, find (a) number of molecules per cc, and (b) average distance between the molecules at this temperature and pressure. [Ans. (a) 105 molecules/cc, (b) 0.2 mm]
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47. A thermocole vessel contains 0.5 kg of distilled water at 30ºC. A metal coil of area 5 × 10–3 m2, number of turns 100, mass 0.06 kg and resistance 1.6 Ω is lying horizontally at the bottom of the vessel. A uniform, time varying magnetic field is set up to pass vertically through the coil at time t = 0. The field is first increased from zero to 0.8 T at a constant rate between 0 and 0.2 s and then decreased to zero at the same rate between 0.2 and 0.4 s. This cycle is repeated 12000 times. Make sketches of the current through the coil and the power dissipated in the coil as functions of time for the first two cycles. Clearly indicate the magnitudes of the quantities on the axes. Assume that no heat is lost to the vessel or the surroundings. Determine the final temperature of the water under thermal equilibrium. Specific heat of the metal = 500 J kg–1 K–1 and the specific heat of water = 4200 J kg–1 K–1. Neglect the inductance of the coil. [IIT, 2000] [Ans. 5.63ºC]
(a) rate of heat loss per unit area due to radiation from the laid.
To bring these photo electrons to rest, a potential equivalent to the excitation energy between the n = 2 to n = 3 Bohr levels of hydrogen atom is required. Find the work function of the metal. [Roorkee, 1993] [Ans. 2.25 eV]
Kinetic Theory of Gases and Thermodynamics
2.68
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Kinetic Theory of Gases and Thermodynamics
59. A gas consists of rigid diatomic molecules and has a molar heat capacity C = R. If the volume of this gas is increased η = 3 times in a polytropic process, find how many times will the rate of collisions of molecules against the vessel’s wall be reduced as a result of this process. [Ans. 4.32]
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63. Two perfect gases at absolute temperature T1 and T2 are mixed. There is no loss of energy. Find the temperature of mixture if masses of molecules are m1 and m2 and the number of molecules in the gases are n1 and n2 respectively. [Roorkee, 1989]
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57. A vessel contains m = 15 g of oxygen at T = 300 K. Find the amount of heat that must be transferred to the gas to double the root mean square velocity of its molecules. [Ans. 8.764 kJ]
62. Find the temperature at which the most probable velocities of oxygen and nitrogen molecules differ by ∆v = 40 m/s. [Ans. 0.5 kK]
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56. An astronaut takes a cylinder of volume 10 litres containing nitrogen gas at a temperature of 27ºC and pressure 50 atmosphere. He makes a hole of area 1 sq. cm in this cylinder and places it in open space. Estimate the time it would take for the cylinder to become empty. (Boltzmann constant = 1.38 × 10–23 J/K) [Roorkee, 1985] [Ans. 0.387 sec]
61. Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0ºC and a pressure of 76 cm of Hg. One of the bulbs is then placed in water bath maintained at 62ºC. What is the new value of pressure inside the bulbs? [IIT, 1985; MNR, 1990] [Ans. 83.75 cm of Hg]
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55. By how many times has a gas consisting of a rigid diatomic molecules to be expanded adiabatically so as to reduce its root mean square velocity by η = 1.5 times? [Ans. 7.59]
60. A gas consisting of rigid diatomic molecules undergoes expansion in a polytropic process. Find the molar heat capacity of the gas in this process, if the rate of collision of the molecules with the walls of the vessel does not change. [Ans. C = 3R]
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54. A thermally insulated vessel containing gaseous nitrogen at temperature T = 300 K and moving with a velocity v0 = 100 m/s is suddenly stopped. By how much per cent and in what way will the gas pressure change? [Ans. 2.25%]
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58. A gas consists of rigid diatomic molecules and is initially at normal temperature and pressure. If the gas is compressed adiabatically to η = 5.0 times, what will be the mean kinetic energy of a rotating molecule in the final state? [Ans. 7.88 × 10–21 joule]
53. A polyatomic gas having N atoms per molecule is heated isobarically. If all the degrees of freedom of the molecules, viz., translational rotational and vibrational, are excited, find what fraction of heat transferred to the gas in this process is spent to perform the work of expansion. What is its value in case of a monatomic gas? 1 2 Ans. , 3 5 3N − 2
ture of the gas by 1ºC. Calculate the speed v0. [Ans. (a) 1.97 × 1025 (b) 34.35 m/s]
2.69
n1T1 + n2 T2 Ans. n1 + n2
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2.70
69. A 0.1 kg steel ball falls from a height of 10 m and bounces to a height of 7 m. (a) Why does it not bounce back to its original height? (b) If all the dissipated energy were absorbed by the ball as heat, how much will its temperature rise? (Specific heat of steel = 0.11 kilo cal/kg ºC, 1 cal = 4.2 joule). [Roorkee, 1978] [Ans. (b) 0.064ºC]
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72. Two different adiabatic paths for the same gas intersect two isothermals at T1 and T2 as shown in P – V diagram. How does Va /Vd compare with Vb / Vc ? [Roorkee, 1983] Va / Vd = 1 Ans. Vb / Vc
68. An insulated container containing monoatomic gas molar mass m is moving with a velocity v0. If the container is suddenly stopped, find the change in temperature.
mv02 Ans. 3R
73. n moles of an ideal gas is contained under a weightless, freely moving piston of a vertical cylinder at a temperature T0. The space over the piston opens into the atmosphere. Find the work that has to be done to increase isothermally the gas volume under the piston η times by slowly raising the piston. [Ans. nRT (η – 1 – loge η)]
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71. Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas is 105 N/m2 and 6 litre respectively. The final volume of the gas is 2 litre. Molar specific heat of the gas at 3 constant volume is R . 2 [IIT, 1982] [Ans. 972 joules]
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67. A cubical box of side 1 meter contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any collision with other atoms. Take R = 25/3 J./mol-K and k = 1.38 × 10–23 J/K. Find: (a) temperature of the gas. (b) average kinetic energy per atom. [Ans. (a) 160 K, (b) 3.31 × 10–21 J]
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66. The temperature of a gas is –68ºC. To what temperature should it be heated so that (a) the average kinetic energy of the molecules be doubled? (b) the root mean square velocity of the molecules be doubled? [MNR, 1995] [Ans. (a) 137ºC, (b) 547ºC]
70. Two moles of an ideal gas at a temperature T0 = 300 K were cooled isochorically so that the gas pressure reduced η = 2 times. Thereafter, as a result of isobaric process, the gas expanded till its temperature became T0 again. What is the total amount of heat absorbed by the gas in this process? [Ans. 2.49 kJ]
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65. 0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27ºC. How much heat has to be transferred to the gas to double the rms velocity of its molecules? [Roorkee, 1991] [Ans. 337.5 K]
64. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have? (b) Obtain the work done by the gas during the expansion as a function of initial pressure P and volume V. [IIT, 1990] [Ans. (a) γ = 1.4, (b) W = 1.25 PV]
Kinetic Theory of Gases and Thermodynamics
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78. An ideal gas has a specific heat at con5 stant pressure C p = R. The gas is kept 2 in a closed vessel of volume 0.0083 m3, at a temperature of the 300 K and a pressure of 1.6 × 106 N/m2. An amount of 2.49 × 104 joules of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas. [IIT, 1987] [Ans. 674.78 K; 3.6 × 106 N/m2]
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where a and α are constants. Find the work performed by the gas and the amount of heat transferred to it when its internal energy increases by ∆U. What is the molar heat capacity of the gas in this process? R R γ −1 Ans. 1 + α ∆U; γ − 1 + α
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80. An ideal monatomic gas is confined in a cylinder by a spring loaded piston of cross section 8 × 10–3 m2. Initially, the gas is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the spring is in its relaxed (unstretched, uncompressed) state. The gas is heated by a small electric heater until the piston moves out slowly by 0.1 m. Calculate the final temperature of the gas and the heat supplied (in joules) by the heater. The force constant of the spring is 8000 N/m, atmospheric pressure is 1.0 × 105 N/m2. The cylinder and the piston are thermally
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76. A heat conducting piston divides a thermally insulated cylinder containing an ideal gas at temperature T0 into two equal parts, each of volume V0. The piston is slowly displaced. Find the gas temperature as a function of the ratio η of the volumes of the greater and smaller sections. The ratio of the specific heats of the gas is γ. ( η + 1)2 Ans. T0 4 η
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79. The internal energy U of an ideal gas having adiabatic exponent γ varies with its volume V in a certain thermodynamic process as U = aVα,
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9 Ans. (a ) 4 T0 , (b)12.9T0 75. 70 calories of heat are required to raise the temperature of 2 moles of an ideal gas at constant pressure from 30ºC to 35ºC. Find the amount of heat required (in calories) to raise the temperature of the same gas through the same range (30ºC to 35ºC) at constant volume [IIT, 1985] [Ans. 50 calorie]
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77. A container is divided into two equal portions. One portion contains an ideal gas at pressure P and temperature T while the other portion is a perfect vacuum. If a hole is opened between two portions, find the change in energy and the change in temperature of the gas. [Roorkee, 1987] [Ans. No change]
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(a) final temperature of gas in each chamber and (b) work done by the gas in the right chamber.
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74. Two rectangular boxes shown in the figure has a partition that can slide without friction along the length of the box. Initially, each of the two chambers of the box has one mole of a monatomic ideal gas (γ = 5/3) at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands, pushing the partition until the final pressure in both chambers becomes 243 P0/32. Determine:
Kinetic Theory of Gases and Thermodynamics
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ar a
γR αR αR R +R+ = + , (b) (a ) C = γ − 1 P V γ − 1 P0 V 0 Increase in internal energy for one mole P = 0 ( V f − Vi ) Total work done by the gas γ −1 Vf , Heat transferred = P0 ( V f − Vi ) + α log e Vi Vf λ P0 to the gas = V f − Vi ) + α log e ( λ −1 Vi
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= RK = a constant 83. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume
86. A cycle consists of two isobaric and two adiabatic lines. Assuming the working substance to be an ideal gas with adiabatic exponent γ, find the efficiency of the cycle if the pressure changes n times within the cycle. [Ans. η = 1 – n(γ – 1)/γ] 87. A cycle consists of two isochoric and two isobaric lines. Assuming the working substance to be an ideal gas, having adiabatic exponent γ, find the efficiency
becomes 5.66 V while its temperature T falls to . 2 (a) How many degrees of freedom do the gas molecules have?
85. A diatomic gas is used as a working substance in a Carnot cycle. What is the efficiency of the cycle if as a result of adiabatic expansion, the gas volume increases n times? [Ans. η = 1 – n1 – γ]
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α ( γ − 1) R ⋅ T0 ( η − 1) ; PV γ ⋅ e PV Ans. W = α log e η − γ −1
α(γ −1) PV
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82. The molar heat capacity of an ideal gas, having an adiabatic exponent γ, varies α with temperature as C = where α is T a constant. Find the work performed by one mole of this gas during its heating from temperature T0 to a temperature η times higher, and the equation of the process in the variables P, V.
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81. A piston divides a closed gas cylinder into two parts. Initially, the piston is kept pressed such that one part has a pressure P and volume 5 V and the other part has pressure 8 P and volume V. The piston is now left free. Find the new pressures and volumes for the adiabatic and isothermal processes. For this gas, γ = 1.5. [Roorkee, 1990] [Ans. Under adiabatic change: P1 = 1.837 P, Under isothermal change: P′ = 2.17 P]
[Ans. 800 K, 717.01 J]
Find: (a) heat capacity of the gas as a function of its volume; (b) increase in internal energy of the gas, the work performed by it, and the amount of heat transferred to the gas in the process, if its volume increased from Vi to Vf.
Open atmosphere Rigid support
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84. The pressure P of one mole of an ideal gas having adiabatic exponent γ varies with its volume V as á P = P0 + , where P0 and α are constants. V
(b) Obtain the work done by the gas during the expansion as a function of initial pressure P and volume V. [Ans. (a) 5, (b) 1.25 PV]
Kinetic Theory of Gases and Thermodynamics
insulated. The piston is massless and there is no friction between the piston and the cylinder. Neglect heat loss through the lead wires of the heater. The heat capacity of the heat coil is negligible. Assume the spring to be massless. [IIT, 1989]
η − 1);PV γ ⋅ e
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Kinetic Theory of Gases and Thermodynamics
in
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D → A: heating at constant volume
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A → B: adiabatic expansion C → D: adiabatic compression
tively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation PV19/13 = constant, in adiabatic processes. (a) Find the number of gram moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at T = 300 K. (c) If T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adia1 batically to of its initial volume 5 V. Find the change in its adiabatic compressibility in terms of the given quantities. [IIT, 1995] [Ans. (a) 2, (b) Speed of sound in a gaseous mixture = 400.93 m/s (c) 0.17% (d) 8.278 × 10–5 V]
88. One mole of a monatomic gas is taken through the cycle shown in the given figure:
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The pressure and temperature at A, 90. The pressure in a monatomic gas increases linearly from 4 × 105 Nm–2 to 8 × 105 Nm–2 B etc. are denoted by PA, TA, PB, TB, when its volume increases from 0.2 m3 to etc., respectively. Given that 0.5 m3. Calculate the following: 2 1 TA = 1000 K, PB = PA and PC = PA , (a) work done by the gas, 3 3 (b) increase in the internal energy, calculate the following quantities: (c) amount of heat supplied, (a) the work done by the gas in the pro(d) molar heat capacity of the gas. cess A → B [Roorkee, 1995] (b) heat lost by the gas in the process 5 [Ans. (a) 1.8 × 10 joules, (b) 4.8 × 105 B → C. joules, (c) 6.6 × 105 joules, (d) 17.14 J/ (c) temperature TD. mol-K] 2/5 2 [IIT, 1993] 91. One mole of a diatomic ideal gas (γ = 1.4) Given : = 0.85 3 is taken through a cyclic process starting from point A. The process A → B is an [Ans. (a) 1860.48 joule, (b) 1571.27 joule, adiabatic compression, B → is isobaric (c) 499.72 K] expansion, C → D is an adiabatic expan89. A gaseous mixture enclosed in a vessel of sion and D → A is isochoric. The volume volume V consists of one gram mole of a ratios are VA/VB = 16 and VC/VB = 2 and 5 the temperature at A is TA = 300 K. Calgas A with γ(= C p / C v ) = and another 3 culate the temperature of the gas at the 7 gas B with γ = at a certain temperapoint B and D and find the efficiency of 5 the cycle. [IIT, 1997] ture T. The gram molecular weights of [Ans. TB = 909 K, TD = 791.33 K, 61.4%] the gases A and B are 4 and 32 respec
of the cycle if the absolute temperature of the gas rises α times both in the isochoric heating and in isobaric expansion. α+γ Ans. n = 1 − 1 + αγ
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volume ratios are VB = 2 and VD = 4. If VA VA the temperature TA at A is 27ºC, calculate:
V VB A
O
TA
TB T
(a) the temperature of the gas at point B, (b) heat absorbed or released by the gas in each process, (c) the total work done by the gas during the complete cycle. Express your answer in terms of the gas constant R. [IIT, 2001] [Ans. (a) 327ºC, (b) 900 R, (c) 600 R]
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93. A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the given figure. The
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92. A weightless piston divides a thermally insulated cylinder into two parts of volumes V and 3V. 2 moles of an ideal gas at pressure P = 2 atmosphere are confined to the part with volume V = 1 litre. The remainder of the cylinder is evacuated. Initially the gas is at room temperature. The piston is now released and the gas expands to fill the entire space of the cylinder. The piston is then pressed back to the initial position. Find the increase of internal energy in the process and final temperature of the gas. The ratio of the specific heat of the gas γ = 1.5. [Roorkee, 1999] [Ans. No change in the internal energy of the gas and its temperature. Final temperature of the gas is, equal to the room temperature]
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
2. An ideal gas expands according to the law pV2 = const. The molar heat capacity C is (a) CV + R (b) CV – R (c) CV + 2R (d) CV – 3R
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6. An empty pressure cooker of volume 10 liters contains air at atmospheric pressure 105 Pa and temperature of 27ºC. It contains a whistle which has area of 0.1 cm2 and mass of 100 gm. What should be the temp. of air inside so that the whistle is just lifted up? :KLVWOH
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5. The average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is (a) 75 J (b) 100 J (c) 150 J (d) 125 J
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1. Which of the following will have maximum total kinetic energy at temperature 300 K? (a) 1 kg, H2 1 1 (b) kg H 2 + kg He 2 2 1 3 (c) kg H 2 + kg He 2 4 (d) 1 kg, He
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ONLY ONE OPTION IS CORRECT
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6. Explain how it is possible that more than half of the molecules in an ideal gas have kinetic energies less than the average kinetic energy. Shouldn’t half have less and half have more?
7. Explain how an automobile airbag protects the passenger from injury. Why would the airbag be ineffective if the gas pressure inside is too low when the passenger comes into contact with it? What about if it is too high?
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5. Why does a helium weather balloon expand as it rises into the air? Assume the temperature remains constant.
4. Four particles have velocities 1, 0, 2, 3 m/s. The root mean square of the particles is (in m/s) (a) 3.5 (b) 3.5 (c) 1.5 (d) 14/3
4. Suppose we have two tanks, one containing helium gas and the other nitrogen gas. The two gases are at the same temperature and pressure. Which has the higher number density (or are they equal)? Which has the higher mass density (or are they equal)?
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3. Why must we use absolute temperature (temperature in kelvins) in the ideal gas (PV = NkT)? Explain how using the Celsius scale would give nonsensial results.
3. Two gases occupy two containers A and B the gas in A, of volume 0.10m3, exerts a pressure of 1.40 MPa and that in B of volume 0.15m3 exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa) (a) 0.70 (b) 0.98 (c) 1.40 (d) 210
2. One way to loosen the lid on a glass jar is to run it under hot water. How does that work?
1. Will the temperature of gas in a container increase when we put the container on a moving train? Explain.
CONCEPTUAL QUESTIONS
1
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B
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V0 ρgh 1 + p
17. When an ideal diatomic gas is heated at constant pressure, the fraction of the heat energy supplied which increases the internal energy of the gas is (a) 2/5 (b) 3/5 (c) 3/7 (d) 5/7
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13. The pressure and temperature of an ideal gas is a closed vessel are 720 kPa and 40ºC respectively. If 1/4h of the gas is released from the vessel and the temperature of the remaining gas is raised to 353ºC, the final pressure of the gas is
(b) T1 < T2 (d) T1 = T2
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16. An air bubble of volume v0 is released by a fish at a depth h in a lake. The bubble rises to the surface. Assume constant temperature and standard atmospheric pressure above the lake. The volume of the bubble just before touching the surface will be (density) of water is ρ (a) v0 (b) v0(ρgh/p)
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(a) T1 > T2 (c) T1 ≤ T2
12. The root mean square speed of the molecules of a diatomic gas is v; then what would be the new root mean square speed when the temperature is doubled and the molecules dissociate into atoms? (a) v (b) 2 (c) 2v (d) 4v
a.
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11. What is the average velocities of particles of a gas? 3RT (a) (b) zero M
15. Maxwell’s velocity distribution curve is given for two different temperature. For the given curves 1
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10. In a process, the density of gas remain constant. If the temperature is doubled, then the change in the pressure will be (a) 100% increase (b) 200% increase (c) 50% decrease (d) 25% decrease
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9. A mixture of ideal gases N2 and He are taken in the mass ratio of 14 : 1 respectively. Molar heat capacity of the mixture at constant pressure is (a) 6R/19 (b) 13R/6 (c) 6R/13 (d) 19R/6
(c)
14. Let V denote the root mean square speed of the molecules in an ideal diatomic gas at absolute temperature T. The mass of a molecule is ‘m’. Neglecting vibrational energy terms, which of these is false? (a) A molecule can have a speed greater than 2 v (b) V is proportional to T (c) The average rotational kinetic energy of a molecule is mv2/4 (d) The average kinetic energy of a molecule is 5mv2/6
8. A gas mixture consists of 2 moles of oxygen and 4 moles of a argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is (a) 4RT (b) 5RT (c) 15RT (d) 11RT
(b) 1080 kPa (d) 540 kPa
7. The specific heat of Ar at constant volume is 0.075 kg–1 K–1. Calculate the atomic weight (R = 2 cal mol–1 K–1) (a) 40 (b) 40.4 (c) 40.2 (d) 40.80
(a) 1440 kPa (c) 720 kPa
(b) 227ºC (d) 600ºC
(a) 400ºC (c) 327ºC
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18. The internal energy of monatomic and diatomic gases are respectively due to (a) Linear motion and rolling motion (b) rolling motion and linear motion (c) Linear motion and rotatory motion (d) rotatory motion and linear motion
1 p1 p2 + 2 T1 T2 p T − p2T1 (d) 1 2 T1 − T2
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28. The root-mean-square (rms) speed of oxygen molecules (O2) at a certain absolute temperature is v. If the temperature is doubled and the oxygen gas dissociates into atomic oxygen, the rms speed would be
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2v
(d) 2 2v
(c) 2v
(b)
(a) v
27. A gas at absolute temperature 300 K has pressure = 4 × 10–10N/m2. Boltzmann constant k = 1.38 × 10–23 J/K. The number of molecules per cm3 is of the order of (a) 100 (b) 105 (c) 108 (d) 1011
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26. Determine the absolute gas temperature at which the root mean square speed of helium molecules exceeds their most probable speed by 200 m/s. (a) 110.2 K (b) 90.2 K (c) 190.2 K (d) 100.2 K
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25. Two containers of equal volume contain the same gas at pressures p1 and p2 and absolute temperatures T1 and T2 respectively. On joining the vessels, the gas reaches a common pressure p and a common temperature T. The ratio p/T is equal to
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24. The velocity of sound in air is 332 m s–1 at NTP. Find the rms speed of air molecules at NTP. (γ = 1.41) (a) 484 ms–1 (b) 418 ms–1 –1 (c) 248 ms (d) 382 ms–1
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23. Two ideal gases at temperature T1 and T2 are mixed. There is no loss of energy. If the
22. Two identical containers joined by a small pipe initially contain the same gas at P0 pressure and absolute temperature T0. One container is now maintained at the same temperature while the other is heated to 2T0. The common pressure of the gases will be 4 3 (a) p0 (b) p0− 3 2 5 1 (c) p0 (d) p0 3 3
(d)
21. An ideal gas has volume V and pressure P. The total translational kinetic energy of the gas 3 (a) must be PV 2 3 (b) may be PV 2 3 (c) must be ≥ PV 2 3 (d) must be ≤ PV 2
n2T1 + n1T2 n1 + n2
20. The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled if the pressure on the surface is equal to h meter of mercury column and relative density of mercury is ρ then the depth of lake is (a) 2ρhm (b) 4ρhm (c) 8ρhm (d) 7ρhm
(c)
19. The root-mean-square (rms) speed of oxygen (O2) molecules at a certain absolute temperature is v. If the temperature is doubled and the oxygen gas dissociates into atomic oxygen, the rms speed would be (a) v (b) 2v (c) 2 v (d) 2 2v
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masses of molecules of the two gases are m1 and m2 and number of their molecules are n1 and n2 respectively, the temperature of the mixture will be T T T +T (a) 1 2 (b) 1 + 2 n1 n2 n1 + n2
Kinetic Theory of Gases and Thermodynamics
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37. The average translational energy and the rms speed of the molecules in a simple of oxygen gas at 300 K are 6.21 × 10–21 J and 484 m/s respectively. Assuming ideal gas behaviors, the corresponding values at 600 K are nearly (a) 12.42 × 10–21 J, 968 m/s (b) 8.78 × 10–21 J, 684 m/s (c) 6.21 × 10–21 J, 968 m/s (d) 12.42 × 10–21 J, 684 m/s
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36. A mixture of n1 moles of monatomic gas and C n2 moles of diatomic gas has p = γ = 1.5 Cv (a) n1 = n2 (b) 2n1 = n2 (c) n1 = 2n2 (d) 2n1 = 3n2
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34. Each molecule of a gas has f degrees of C freedom. The ratio p = γ for the gas is Cv
38. Two identical containers A and B have frictionless pistons. They contain the same volume of a ideal gas at the same temperature. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The changes in the pressure in A and B are found to be ∆p and 1.5 ∆p respectively. (a) 4mA = 9mB (b) 2mA = 3mB (c) 3mA = 2mB (d) 9mA = 4mB
33. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ration Cp = γ for the gas is Cv (a) 2 (b) 3/2 (c) 5/3 (d) 4/3
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32. When an air bubble rises from the bottom to the surface of a lake, its radius becomes double. Find the depth of the lake, given that the atmospheric pressure is equal to the pressure due to a column of water 10 m high. Assume constant temperature and disregard surface tension. (a) 30 m (b) 40 m (c) 70 m (d) 80 m
( f − 1) 3
35. N (< 100) molecules of a gas have velocities 1, 2, 3, ........N km/s respectively. Then (a) rms speed and average speed of molecules is same (b) ratio of rms speed of average speed is
31. A closed vessel is maintained at a constant temperature. It is first evacuated and then vapor is injected into it continuously. The pressure of the vapour in the vessel (a) increases continuously (b) first increases and then remains constant (c) first increases and then decreases (d) none of the above
1 f
30. A gas has volume V and pressure p. The total translational kinetic energy of all the molecules of the gas is (a) (3/2) pV (b) (3/2) pV only if the gas is diatomic (c) > (3/2) pV if the gas is diatomic (d) (3/2) pV in all cases
(d) 1 +
2 f
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29. The average translational kinetic energy of O2 (molar mass 32) at a particular temperature is 0.048 eV. The average translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is (a) 0.0015 (b) 0.003 (c) 0.048 (d) 0.768
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
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4. Hydrogen gas and oxygen gas have volume 1 cm3 each at NTP (a) Number of molecules is same in both the gases (b) The rms velocity of molecules of both the gases is the same (c) The internal energy of each gas is the same (d) The average velocity of molecules of each gas is the same.
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5. Pick the correct statement(s). (a) The rms translational speed for all ideal gas molecules at the same temperature is not the same but it depends on the mass. (b) Each particle in a gas has average translational kinetic energy and the equation 1 2 3 mvrms = kT establishes the relation2 2 ship between the average translational kinetic energy per particle and temperature of an ideal gas. It can be concluded that single particle has a temperature. (c) Temperature of an ideal gas is doubled from 100ºC to 200ºC. The average kinetic energy of each particle is also doubled. (d) It is possible for both the pressure and volume of monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change.
(a) The value of a V0 is 2N (b) The ratio Vavg/V0 is equal to 2/3 (c) The ratio Vrms/V0 is equal to 1/ 2 speed V (d) Three fourth of the total particle has a speed between 0.5V0 and V0.
9 VSHHG9
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3. Let v , Vrms and Vp respectively denote the mean speed, the root-mean-square speed, and the most probable speed of the molecules in an ideal monoatomic gas at absolute temperature T. The mass of a molecule is m. (a) No molecule can have speed greater than Vrms. (b) No molecule can have speed less than Vp . 2 (c) Vp < V < Vrms (d) The average kinetic energy of a mol3 ecule is mv 2p 4
1. Pick the correct statement(s). (a) The rms translational speed for all ideal-gas molecules at the same temperature is not the same but it depends on the mass. (b) Each particle in gas has average translational kinetic energy and the equation 1 3 mvrms = kT establishes the relation2 2 ship between the average translational kinetic energy per particle and temperature of an ideal gas. It can be concluded that single particle has a temperature. (c) Temperature of an ideal gas is doubled from 100ºC to 200ºC. The average kinetic energy of each particle is also doubled. (d) It is possible for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change.
ONE OR MORE THAN ONE CHOICE MAY BE CORRECT
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True. 1. Statement 1: Internal energy of a gas must V increases when its temperature is increased. Statement 2: Internal energy of a gas is proportional to the velocity of the vessel in which gas is contained.
7. The total kinetic energy of translatory motion of all the molecules of 5 litres of nitrogen exerting a pressure P is 3000J. (a) the total KE of 10 litres of N2 at pressure of 2 P is 3000 J (b) the total KE of 10 litres of He at pressure of 2 P is 3000 J (c) the total KE of 10 litres of O2 at pressure of 2 P is 20000 J (d) the total KE of 10 litres of Ne at pressure of 2 P is 12000 J
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6. Statement 1: The maxwell speed distribution graph is symmetric about the most probable speed. Statement 2: The root mean square speed of an ideal gas depends on the type of gas (monatomic, diatomic or polyatomic).
5. Statement 1: State variables (P, V and T) of any gas at low densities obey the equation PV = nRT. Statement 2: Real gases are good approximation of an ideal gas at low density.
7. Statement 1: The r.m.s. speed of an ideal gas depends on type of gas.
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3. Statement 1: Absolute zero temperature is not the temperature of zero energy. Statement 2: Only the translational kinetic energy of the molecules is represented by temperature. C 4. Statement 1: The ratio v for a monaCp tomic gas is more than for a diatomic gas. Statement 2: The molecules of a monatomic gas have more degrees of freedom than those of a diatomic gas.
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ASSERTION AND REASON QUESTIONS Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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10. If the atoms in a diatomic molecule can vibrate, the molecule has (a) 2 rotational degrees of freedom (b) 3 rotational degrees of freedom (c) 1 vibrational degrees of freedom (d) 2 vibrational degrees of freedom
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9. An actual gas deviates from perfect gas behaviour as its molecules (a) Have definite size (b) Attract each other (c) Show Brownian motion (d) Are not spherical
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8. According to the kinetic theory, 0K is that temperature at which (a) pressure of ideal gas is zero (b) volume of ideal gas is zero (c) internal energy of ideal gas is zero (d) ideal gas liquefies
2. Statement 1: Systems are in thermal equilibrium, when their temperature are same or average kinetic energy per molecule is same. Statement 2: Total kinetic energy or internal energy or total energy does not determine the direction of flow of heat.
a.
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6. Two vessels of the same volume contain the same gas at same temperature. If the pressure in the vessels be in the ratio of 1 : 2, then (a) the ratio of the average energy is 1 : 2 (b) the ratio of the root mean square velocity is 1 : 1 (c) the ratio of the average velocity is 1 : 2 (d) the ratio of number of molecules is 1 : 2
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
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Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II.
MATCH THE COLUMN TYPE QUESTIONS
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3. Column II give γ for gas mention in column II, match them correctly. Column I Column II 5 (a) monoatomic (p) = 1.67 3 7 (b) diatomic (q) = 1.4 5 4 = 1.33 (c) linear polyatomic (r) 3 3 (d) non-linear polyatomic (s) 7
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14. Statement 1: Two different gas molecules having same temperature will have the same KE. Statement 2: KE of gas molecules follows the Boltzmann’s law.
2. Column II gives value of Cp for gas mention in column I match them correctly. Column I Column II (a) Helium (p) 7R/2 (b) Oxygen (q) 5R/2 (c) Methane (r) 3R/2 (d) Air (s) 4R
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13. Statement 1: Degree of freedom of an ideal gas is three. Statement 2: We can boil water by convection inside earth’s satellite.
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12. Statement 1: The internal energy of a given sample of an ideal gas depends only its temperature according to kinetic theory of gases. Statement 2: The ideal gas molecules do not exert intermolecular forces.
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11. Statement 1: If temp. of gas increase than most probable speed will be increase. Statement 2: Mean free path of gas 1 λm = 2πd 2 n
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10. Statement 1: The pressure exerted by an enclosed ideal gas does not depends on the shape of the container. Statement 2: The pressure of an ideal gas depends on the number of moles, temperature and volume of the enclosure.
9. Statement 1: In Maxwell’s speed distribution graph, for a given amount of gas, the area under the graph increases as the temperature of the gas increases. Statement 2: Increase in temperature broadening the curve.
Column I Column II (a) In the con(p) Pressure of the gas tainer X is more. (b) In the con(q) RMS speed of gas tainer Y molecules is more (c) Since the (r) Average thermal number of energy of a molmolecules is ecule is the same as more that of a molecule in the other container (d) Since the (s) Internal energy of gas has a the system is more smaller (s) than that of the molecular system in the other mass container.
1. X and Y are two equal size containers. X contains 5 mol of H2 and Y has 10 mol of O2 at the same temperature. Assuming that the gases are ideal, match Columns-I and II:
8. Statement 1: All the poly-atomic gas molecules have six degrees of freedom. Statement 2: The number of degree of freedom depends only on the atomicity of gas.
Statement 2: Different degree of freedom is associated with gases of same nature of gas.
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PASSAGE BASED QUESTIONS PASSAGE-1 The rms speed of an oxygen molecule in a sample at 300 K is about 480 m/s. This does not mean
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m dN = 4πN 2πkT
3/2
v 2 e − mv
2
/ 2 kT
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7. It can be concluded that (a) average translational kinetic energy of a molecule in container X is more than that of a molecule in container Y (b) average translational kinetic energy of a molecule in container Y is more than that of a molecule in container X (c) average translational kinetic energy of a molecule in container X could be more or could also be less than that of a molecule in container Y (d) average translational kinetic energy of a molecule in container X is equal to that of a molecule in container Y
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2. The relation between T2 and T1 is given by (a) T2 > T1 (b) T1 < T2 (c) T1 = T2 (d) None
4. The order of magnitude between vp, vrms and vmean is (a) vp < vrms < vmean (b) vrms < vp < vmean (c) vp < vmean < vrms (d) None
8. It can be concluded that (a) internal energy of gas in container X is more than that of the gas in container Y (b) internal energy of gas in container X is less than that of the gas in container Y
3. Area under graph at T1 is A1 and under T2 is A2 then (a) A2 > A1 (b) A1 < A2 (c) A1 = A2 (d) None
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dN is dv (b) L–1T (d) None
1. The dimensional formula for
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2kT speed. Its value is given by v p = . m
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6. From the given information, we can conclude that (a) the molecules in container X are moving as fast as those in container Y (b) the molecules in container X are moving faster than those in container Y (c) the molecules in container Y are moving faster than those in container X (d) no possibility can be ruled out, i.e., molecules in X could be moving faster than these in Y or the molecules in Y could be moving faster than those in X
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We see that there are some molecules which have speeds many times greater has the mean speed. This fact helps in making nuclear fusion reactions in a laboratory. The speed vp at which dN/dv is maximum is called the most probable
X and Y are two equal size containers. X contains a gas ‘A’ at a temperature 75ºC and Y contains a gas ‘B’ at a temperature 40ºC. Each gas behaves as an ideal gas and specific heat at constant pressure for both gases has the same value. There is no further information about the systems.
dv
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PASSAGE-2
Figure shown plots of dN/dv against v at two different temperatures.
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5. On increasing temperature number of molecules corresponding to most probable speed will (a) Increase (b) Decrease (c) Remain same (d) None
that the speed of each molecule is 480 m/s. Many of the molecules have speed less than 480 m/s and many have speed more than 480 m/s. Maxwell derived an equation giving the distribution of molecules in different speeds. If dN represents the number of molecules with speeds between v and v + dv then
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Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
(c) internal energy of gas in container X could be more or could also be less than that of the gas in container Y (d) internal energy of gas in container X has the same value as that of the gas in container Y
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2. A thermally insulated vessel is divided into two parts by a heat-insulating piston which can move in the vessel without friction.
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5. A smooth vertical tube having two different sections is open from both ends and equipped with two pistons of different areas as shown in figure.
7. A vessel of volume V = 7.5 ℓ contains a mixture of ideal gases at a temperature T = 300 K. η1 = 0.10 mole of oxygen, η2 = 0.20 mole of nitrogen, and η3 = 0.30 mole of carbon dioxide. Assuming the gases to be ideal, find (a) the pressure of the mixture; (b) the mean molar mass M of the given mixture which enters its equation of state pV = (m/M)RT, where m is the mass of the mixture. 8. For silver, the molar specific heat at constant pressure in the range 50 K to 100 K is given by, Cp = (0.076 T – 0.00026 T2 – 0.15) cal/K where T is the temperature. If 2 mole of silver is heated from 50 to 100 K, calculate the heat required in calories?
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4. A vessel of volume V = 30 ℓ contains ideal gas at the temperature 0ºC. After a portion of the gas has been let out, the pressure in the vessel is decreased by ∆p = 0.78 atm (the temperature remaining constant). Find the mass of the released gas. The gas density under the normal conditions ρ = 1.3 g/ℓ.
6. Two identical vessels are connected by a tube with a valve letting the gas pass from one vessel into the other if the pressure difference ∆p ≥ 1.10 atm. Initially there was a vacuum in one vessel while the other contained ideal gas at a temperature t1 = 27ºC and pressure p1 = 1 atm. Then both vessels were heated to a temperature t2 = 107ºC. Up to what value will the pressure in the first vessel (which had vacuum initially) increase?
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3. Molar volume is the volume occupied by 1 mole of any ideal gas at STP. Show that it is 22.4 litres.
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The left part of the vessel contains one mole of an ideal monatomic gas, and the right part is empty. The piston is connected to the right wall of the vessel through a spring whose length in free state is equal to the length of the vessel (figure). Determine the heat capacity C of the system, neglecting the heat capacities of the vessel, piston and spring.
Each piston slides within a respective tube section. One mole of an ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional area of the upper piston is ∆S = 10 cm2 greater than that of the lower one. The combined mass of the two pistons is equal to m = 5 kg. The outside air pressure is p0 = 1 atm. By how many kelvins must the gas between the pistons be heated to shift the pistons through l = 5 cm?
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1. Two perfect diatomic gases at absolute temperature T1 and T2 are mixed. There is no loss of energy. Find the temperature of the mixture if masses of the molecules are m1 and m2 and the number of molecules in the gases are n1 and n2.
SUBJECTIVE QUESTIONS
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3. A cubical box of side 1 m contains helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 s, an atom travelling with the root mean square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall without any collision with other atoms. Take R = 25/3 J/mol-K and k = 1.38 × 10–23 J/K] [2002]
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12. One mole of an ideal monoatomic gas undergoes a process defined by U = a V where U is internal energy and V is its volume. The molar specific heat of the gas for * R. The numthis process is found to be 12 ber in the numerator is not readable. What may be this number?
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11. For the case of an ideal gas find the equation of the process (in the variables T, V) in which the molar heat capacity varies as (a) C = CV + αT; (b) C = CV + βV; (c) C = CV + aP, and α, β and a are constant
2. A gaseous mixture enclosed in a vessel of volume V consists of one gram mole of gas A with γ = (Cp/CV = 5/3) and another gas B with γ = 7/5 at a certain temperature T. The gram molecular weights of the gases A and B are 4 and 32 respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation, PV19/13 = constant in adiabatic process. (a) Find the number of gram moles of the gas B in the gaseous mixture. (b) Compute the speed of sound in the gaseous mixture at 300 K. (c) If T is raised by 1 K from 300 K, find the percentage change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 of its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. [1995]
10. One mole of an ideal gas whose adiabatic exponent equals γ undergoes a process p = p0 + α/V, where p0 and α are positive constants. Find (a) heat capacity of the gas as a function of its volume; (b) the internal energy increment of the gas, the work performed by it, and the amount of heat transferred to the gas, if its volume increased from V1 to V2.
9. A vertical cylinder of cross-sectional area S contains one mole of an ideal monatomic gas under a piston of mass M. At a certain instant, a heater which transmits to a gas an amount of heat q per unit time is switched on under the piston. Determine the established velocity v of the piston under the condition that the gas pressure under the piston is constant and equal to p0, and the gas under piston is thermally insulated.
Kinetic Theory of Gases and Thermodynamics
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(a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the box.
4. An insulated box containing a monoatomic gas of molar mass M moving with a speed v0 is suddenly stopped. Find the increment in gas temperature as a result of stopping the box. [2003] 5. Statement 1: The total translational kinetic energy of all the molecules of a given mass
1. A closed container of volume 0.2 m3 contains a mixture of neon and argon gases at a temperature of 27ºC and pressure of 1 × 105 Nm–2.The total mass of the mixture is 28 g If the molar masses of neon and argon are 20 and 40 g mol–1 respectively, find the masses of the individual gases in the container assuming them to be ideal. [Universal gas constant R = 8.314 J/mol-K] [1994]
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PREVIOUS YEAR ’ IIT-JEE QUESTIONS
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Kinetic Theory of Gases and Thermodynamics
(b) Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1 (c) Statement 1 is True, Statement 2 is False (d) Statement 1 is False, Statement 2 is True
of an ideal gas is 1.5 times the product of its pressure and its volume, because Statement 2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision. (a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement 1
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ONE OR MORE THAN ONE CHOICE MAY BE CORRECT 3. (c, d) 10. (a, d)
5. (a) 13. (d)
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2. (b) 10. (a)
7. 15. 23. 31.
(a) (b) (d) (b)
8. 16. 24. 32.
(d) (d) (a) (c)
6. (b, d)
7. (c, d)
7. (d)
8. (d)
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ASSERTION AND REASON QUESTIONS 1. (c) 9. (d)
(c) (c) (b) (d) (c)
4. (a, c, d ) 5. (a, d)
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6. 14. 22. 30. 38.
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(b) (c) (d) (c) (a)
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Kinetic Theory of Gases and Thermodynamics
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QUESTION BANK 2 (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?
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12. If you leave the refrigerator door open and the refrigerator runs continuously, does the kitchen get colder or warmer? Explain.
4. Is it possible to make a heat pump with a coefficient of performance equal to 1? Explain.
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CONCEPTUAL QUESTIONS
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2.88
Kinetic Theory of Gases and Thermodynamics
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12. If W1 is the work done in compressing an ideal gas from a given initial state through a certain volume isothermally and W2 is the work done in compressing the same gas from the same initial state through the same volume adiabatically, then
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containing 3 moles of Helium gas. There is a heater which gives out a power 100 cal to the nitrogen gas. A power of 30 cal is transferred to Helium through the bottom surface of the piston. The rate of increment of temperature of the nitrogen gas assuming that the piston moves slowly
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18. 0.5 mole of an ideal gas at constant temperature 27ºC kept inside a cylinder of length L and cross-section area A closed by a massless piston.
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Kinetic Theory of Gases and Thermodynamics
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ONE OR MORE THAN ONE CHOICE MAY BE CORRECT
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Kinetic Theory of Gases and Thermodynamics
Statement 2: Initial and final temperature is equal, therefore change in internal energy is zero.
2. Statement 1: If an ideal gas expands in vacuum in an insulated chamber, ∆Q, ∆U and ∆W all are zero. Statement 2: Temperature of the gas remains constant.
MATCH THE COLUMN TYPE QUESTIONS Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II. 1. Match the column.
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Column II (p) –P/V
Column I (a) Adiabatic bulk modulus
9. Statement 1: In cyclic process, initial and final state are same. Therefore net work done is zero.
7. Statement 1: Internal energy change is zero if the temp is constant, irrespective of the process being cyclic or noncyclic. Statement 2: dU = n CvdT for all process and is independent of path. 8. Statement 1: Adiabatic expansion is always accompanied by fall in temperature. Statement 2: In adiabatic process, volume is inversely proportional to temperature.
Column II (Symbols have usual meaning) (p) ∆U = 0 (q) Q = 0 (r) W = 0 (s) ∆U = nCV∆T
2. The P-V diagram of 0.2 mol of a diatomic ideal gas is shown in figure. Process BC is adiabatic γ = 1.4. Column I Column II (a) ∆QAB (J) (p) 602 (b) ∆WBC (J) (q) –644 (c) ∆UCA (J) (r) 1246 (d) ∆UBC (J) (s) –602
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Column I (Processes for ideal gases) (a) Isothermal (b) Isobaric (c) Isochoric (d) Adiabatic free expansions
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compared to that at constant pressure for doing work in expansion.
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Kinetic Theory of Gases and Thermodynamics
(b) Slope of P – V graph in isothermal process (c) Degree of freedom (d) The ratio of molar heat capacity at constant pressure to universal gas constant R
(q) 2/(γ – 1)
PASSAGE BASED QUESTIONS PASSAGE-1 One mole of an ideal gas is taken from an initial state P0, V0 and temperature T0 through the following activities. (γ = CP/CV, CP = 7 R/2, CV = 5 R/2) (i) Heating at constant volume to a temperature three times. (ii) Adiabatic expansion to a volume 2V0. (iii) Cooling at constant volume to a temperature one third. (iv) Adiabatic compression so that it is returned to its initial state.
(r) γP γ γ −1
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4. The temperature at D is (a) T0 21.4 (b) T0 20.4 (c) T0 2–1.4 (d) T0 2–0.4
7. Identify the gas filled in the container A and B. (a) N2, Ne (b) He, H2 (c) O2, Ar (d) Ar, O2
3. The total heat given to the system is (a) 1.25 RT0 (b) 3.75 RT0 (c) 2.25 RT0 (d) 2.275 RT0
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∆PA = 2.5 atm, ∆PB = 1.5 atm Required data for unknown gas: Mono (molar gas) Kr 84g
Xe Rd 131g 222g
O2 32g
Cl2 71g
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He Ne Ar 4g 20g 40g Dia (molar gas) H2 F2 N2 2g 19g 28g
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PASSAGE-2
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8. Total number of molecules in ‘A’ (Here NA = Avagadro number) 125 (a) NA (b) 3.125 NA 64 125 (c) NA (d) 31.25 NA 28
6. Identify the type of gas filled in container A and B respectively. (a) Mono, mono (b) Dia, dia (c) Mono, dia (d) Dia, mono
2. The pressure at point C is given by 3P 3P (a) 0.40 (b) 1.40 2 2 (c) 3P0 × 20.4 (d) 3P0 × 21.4
Kinetic Theory of Gases and Thermodynamics
PASSAGE-3 The PV diagram for a cyclic process performed on an ideal monoatomic gas is shown in figure. The curve AC is one fourth of circle. 3 1P
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(a) 6 RT0 (1 − η−1/3 )
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16. If initial temperature of system is T0 and it is compressed to η times its initial volume then work done by system is
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14. Which of the following statement is correct about this process? (a) It is an adiabatic process 1 (b) It is a process in which ∆Q = − ∆U 5 (c) It is a process in which dQ = dU (always) (d) It is a process in which dQ = –dU (always) 15. If this process is possible (a) its molar heat capacity is negative (b) its molar heat capacity is positive (c) its molar heat capacity is zero (d) molar heat capacity cann’t be defined until volume or pressure is constant.
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1. One mole of a certain ideal gas is contained under the weightless piston of a vertical cylinder at a temperature T. The space over the piston opens into the atmosphere of pressure P0. What work has to be done in lifting the piston slowly so that the volume of the gas under the piston increases isothermally n times?
12. Heat absorbed by the gas in process B to C is (a) 150 J (b) –150 J (c) 300 J (d) Can’t be calculated
SUBJECTIVE QUESTIONS
Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
Determine the ratio of the energy given out by the gas to its surroundings during the isochoric section of the cycle to the expansion work done during the isobaric section of the circle.
5. A horizontal cylindrical vessel of length 2l is separated by a thin heat-insulating piston into two equal parts each of which contains n moles of an ideal monatomic gas at a temperature T.
7. One mole of an ideal monoatomic gas undergoes the cyclic process as shown. Find out efficiency (in percent) of the cycle. [Take ln 2 = 0.7]
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The piston is connected to the end faces of the vessel by undeformed springs of rigidity k each (figure). When an amount of heat Q is supplied to the gas in the right part, the piston is displaced to the left by a distance x = l/2. Determine the amount of heat Q, given away at the temperature T to a thermostat with which the gas in the left part is in thermal contact all the time.
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P = (3/2) T1/2. It is found that its temperature increases by 100 K in this process. Now it is taken from state B to C through a process for which internal energy is related to volume as U = (1/2) V1/2. Find the total work performed by the gas (in Joule), if it is given that volume at B is 100 m3 and at C it is 1600 m3. [Use R = 8.3 J/mol-K]
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Kinetic Theory of Gases and Thermodynamics
the bubble rises to the surface its diameter increases n = 1.1 times. Find how deep is the pond at that spot. The atmospheric pressure is standard, the gas expansion is assumed to be isothermal.
9. A heat conducting piston can move freely inside a closed, thermally insulated cylinder with an ideal gas (γ = 5/3). At equilibrium the piston divides the cylinder into two equal parts, the temperature of gas being equal to 300 K.
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The piston is slowly displaced by an external agent. Find the temperature of gas when the volume of the greater section is seven times the volume of the smaller section.
(b) 3/5 (d) 5/7
2. An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. (i) How many degrees of freedom do gas molecules have? (ii) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [1990]
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(a) Find work done in each process (b) Find heat rejected by gas in one complete cycle. (c) Find the efficiency of the cycle.
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4. Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, calculate the following quantities in this process. [1992]
If p1 = 105 Pa, p0 = 3 × 105 Pa, p2 = 4 × 105 Pa, V2 – V1 = 10 ℓ, and segments 4-3 and 2-1 of the cycle are parallel to the V-axis.
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7. An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1 = 5960 J, Q2 = –5585 J, Q3 = –2980 J and Q4 = 3645 J, respectively. The corresponding quantities of work involved are W1 = 2200 J, W2 = –825 J, W3 = –1100 and W4 respectively. [1994]
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6. One mole of a monoatomic ideal gas is taken through the cycle shown in the figure. [1993] A → B : adiabatic expansion B → C : cooling at constant volume C → D : adiabatic compression D → A : heating at constant volume
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5. An ideal gas is taken from the state A (pressure P, volume V) to the state B (pressure P/2, volume 2V) along a straight line path in the P-V diagram. Select the correct statements from the following. [1993]
(a) The net change in the heat energy (b) The net work done (c) The net change in internal energy
The pressure and temperature at A, B, etc are denoted by PA, TA, PB, TB etc respectively. Given that TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA, calculate the following quantities: (a) The work done by the gas in the process A→B (b) The heat lost by the gas in the process B →C (c) The temperature TD. [Given: (2/3)2/5 = 0.85]
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(b) 2 mA = 3 mB (d) 9 mA = 4 mB
11. Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in
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10. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be ∆P and 1.5 ∆P respectively. Then [1998]
13. One mole of an ideal monoatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate [1998]
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Kinetic Theory of Gases and Thermodynamics
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(a) the temperature will decrease (b) the volume will increase (c) the pressure will remain constant (d) the temperature will increase 20. P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to [2001]
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Column II ∆W > 0 ∆W < 0 ∆Q > 0 ∆Q < 0
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27. P-V graph of cyclic process is shown for an ideal gas. Match the column correctly. [2006]
(a) change in temperature (b) work done and (c) change in internal energy [Given: specific heat 400 J/kg/ºC, coefficient of cubical expansion, γ = 9 × 10–5/ ºC, density ρ = 9000 kg/m3, atmospheric pressure = 105N/m2]
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Kinetic Theory of Gases and Thermodynamics
P0 Mg − 2 πR 2
(d)
P0 Mg + 2 πR 2
(c) An ideal monoatomic gas expands to twice its. (d) An ideal monoatomic gas expands such that its. Column II (p) The temperature of the gas decreases separated by a valve. Chamber I contains an ideal gas and the Chamber II has vacuum. The valve is opened.
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Kinetic Theory of Gases and Thermodynamics
ONLY ONE OPTION IS CORRECT 1. 9. 17. 25. 33.
(c) (c) (b) (a) (a)
2. 10. 18. 26. 34.
3. 11. 19. 27. 35.
(d) (a) (c) (c) (d)
4. 12. 20. 28. 36.
(a) (d) (b) (c) (a)
5. 13. 21. 29. 37.
(b) (b) (d) (d) (a)
6. 14. 22. 30. 38.
(b) (d) (c) (a) (a)
(d) (b) (d) (b) (b)
7. 15. 23. 31. 39.
(b) (a) (a) (c) (c)
8. 16. 24. 32.
(a) (d) (c) (c)
ONE OR MORE THAN ONE CHOICE MAY BE CORRECT 3. (a, c) 4. (a, b, d) 5. (a, c, d) 6. (b, d) 11. (b, c, d) 12. (a, c) 13. (a, d) 14. (c, d)
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5. We know the quantity of specific heat taken Q = ms ∆t
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3. In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperature becomes equal. The final temperature can be the mean temperature (T1 + T2)/2 only when thermal capacities of the two bodies are equal.
2. When a body is taken from one position to another, it spends mechanical energy to overcome friction. This energy is dissipated as heat. Now, if the body is allowed to go round to its path in reverse direction, its energy spent is not recovered but instead is further spent in the reverse path. Hence the process is not reversible.
CONCEPTUAL QUESTIONS
HINTS AND SOLUTIONS
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Kinetic Theory of Gases and Thermodynamics
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= nRdT – (nRT/P) dP = nRdT + nRdT = 2nR∆T = 1200 R
W = PdV = nRdT – VdP
∴ ∆U = 0 ⇒ ∆Q = ∆W = –RT0 ln 2 Process 3-4: Since V ∝ T, thus P = constant 5 ∴ ∆Q = RT0 ; ∆W = RT0 2 Process 4-1: Since temperature is constant
3 WAB = ∫ P d v = ∫ T 1/ 2 dv 2
4. Process A → B
3. 2γ+1 P0 (both cases)
Process 2-3: Since temperature is constant Thus process is isothermal ⇒ PV = constant
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2 2 = [40 − 10] = × 30 = 20 JW 3 3
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10. (a) The gas would pass from A to B. The change in pressure and volume will take place under adiabatic conditions. The final pressure in the two cylinders would be 0.5 atm. (b) The change in internal energy of the gas is zero. (c) The change in temperature will be zero. (d) The process is rapid, the intermediate states are non-equilibrium states and hence do not satisfy the gas equation. The intermediate states do not lie on the P-V-T surface.
Process B → C 1 U = V 1/2 2 1 3 1 RT = V 1/2 ⇒ 3 PV1/2 = 1 ∴ p = 3 V 2 2 1600 1 2 dv = V Now WBC = ∫ Pdv = ∫ 3 3 V 100
8. The climate of a Harbour town is more temperate because of the large amount of water vapours present in the air due to its proximity with sea.
SUBJECTIVE QUESTIONS
3 1 = ∫ T 1/2 × RT −1/ 2 dT 2 3 On solving, WAB = 50 R = 50 × 8.3 = 415 J
6. During driving the temperature of the air, inside tyres increases due to friction force between road and tyres. As volume of air in the tyre is constant according to Charle’s law. (P ∝ T)V pressure increases with increase in temperature.
Larger the value of specific heat of the coolent more is the heat absorbed by it.
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Thus, PV = constant ∴ ∆Q = ∆W = 2RT0 ln 2
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2.112
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= –RT0 – RT0 ln 2 + RT0 + 2RT ln 2 = RT0 ln 2 5 Heat supplied = RT0 + 2 RT0 ln 2 2 ∴ Efficiency RT0 ln 2 = (5 − 2) RT0 + 2 RT0 ln 2
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PREVIOUS YEARS’ IIT-JEE QUESTIONS
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Thus efficiency = (Net work done)/(Heat supplied)
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Kinetic Theory of Gases and Thermodynamics
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Kinetic Theory of Gases and Thermodynamics
27. (A) → (s), (B) → (p, r), (C) → (r), (D) → (q, s)
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Kinetic Theory of Gases and Thermodynamics
Q = ∆U + W
∆U = +ve (∆T = +ve)
W = +ve (As volume increases)
PV (d): T = as produced of P and V nR increases, so temperature increases
So, Q = +ve
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C H A P T E R
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3.1 WAVE MOTION
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Wave Mechanics
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In simple terms, we can say that wave motion involves transfer of disturbance (energy) from one point to the other with the particle of medium oscillating about their mean positions i.e., the particles of the medium do not themselves travel along with the wave. Instead, they oscillate back about some equilibrium position. As the wave passes by, only the disturbance is propagated. In this chapter, we will limit discussion to mechanical waves (elastic waves) which require a medium to travel. There are also electromagnetic waves which do not require any medium and can travel in vacuum. The idea of wave motion can be understood by a very simple example. If a stone is dropped in a pool of water, a disturbance is created when it enters the surface. The disturbance is not confined to that place alone but spreads out and eventually it reaches all the parts of water, and brings the particle with which it comes in contact, into motion. These particles then set into motion neighbouring particles. They in turn produce similar motion in other particles and in this fashion disturbance is transmitted from particle to particle. The particles oscillate over short paths about their initial positions and as a result a wave moves through the medium. The medium as a whole does not go in the directions of the motion of the wave.
3.2 TRANSVERSE WAVE In transverse waves, the particle of the medium oscillate perpendicular to the direction in which the wave travels. Travelling waves on tight rope are transverse waves. If one of the rope is rigidly fixed and the other end is given periodic up and down jerks, the disturbance propagates along the length of the rope but the particles of the rope oscillate up and down. Disturbance travels along the rope in form of crests (upward peaks) and troughs (downward peaks). :DYHPRWLRQ
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3.2
Wave Mechanics
Oscillation of Particles &
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T = trough
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Single line arrows show the oscillation of particles and the double line arrows show the directions of the motion C = crests
T = trough
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Transverse waves require that their should be a shearing force in the medium. Hence, transverse wave can be propagated only through medium which will support a shearing stress i.e., mainly solids.
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In longitudinal waves, the oscillation of the particles is parallel to the direction in which the wave travels. Disturbance travelling in a spring parallel to its length, a pressure variation propagating in a liquid, sound waves travelling in a medium are examples of longitudinal waves. Let us consider a stretched spring. If one end of the spring is suddenly given in and out oscillation parallel to length of the spring, the coils starts exerting force on each other and the compression and the expansion points travel along the length of the spring. The coils oscillate right and left parallel to the spring.
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R = rarefaction https://t.me/universitarios
Wave Mechanics
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The spring in the above example can be replaced by a long tube of air with a piston at the left end. The piston is set into oscillation along the length of the tube. ZDYHPRWLRQ RVFLOODWLRQRI PROHFXOHV
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Compression (crowding together of molecules) and rarefactions (spreading out of molecules away from each other) travel along the tube. The pressure at the compression point is higher and the pressure at a rarefaction point is lower. The molecules of air oscillate right and left i.e., parallel to the wave propagation. Longitudinal waves do not require shearing stress and, hence can travel in any elastic medium: solid, liquid and gases.
Wave Speed (c)
The speed of wave is the distance it covers in one second. It depends on medium. It should be carefully noted that wave speed is completely different from particle speed. Particle speed the speed of the vibrating particles in the medium. On the other hand wave speed is the speed with which (the disturbance/or wave) propagates in the medium.
Wave Frequency (ν) The frequency with which the particle of the medium (through which the wave passes oscillates is known as wave frequency. In transverse waves, frequency is the number of crests (or troughs) that pass through a point in one seconds. In longitudinal waves, frequency is the number of compressions (or rarefactions) that pass through a point in one second. https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
Time Period (T) The time period of the oscillation of the particle in the medium is the time period of the wave.
Amplitude (A) Amplitude of the wave is same as the amplitude of the oscillating particles.
Wave Length (λ) Wave length is the distance between two consecutive crests (or compression) in a wave. Wave length, wave speed and frequency are related as follows:
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When a wave passes through a medium, all particles oscillate with same frequency, but they reach the corresponding positions in their path at different time instants.
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For example, in the above figure, the particle at P is at its top extreme, the particle at Q, is passing through its mean positions, the particle at R is at its bottom extreme. These relative positions represents the phase of motion. (i) If two particles have same position and same velocity at all time instants, they are said to be in same phase (or in phase). (ii) Two particles are said to be in opposite phase (or exactly out of phase) if their displacement from mean position and their velocities are equal in magnitude but opposite in direction.
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Wave Mechanics
3.5
■ The distance between particles in same phase = 0, λ, 2 λ ... ■ The distance between particles in opposite phase = λ/2, 3λ/2,
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3.5 INTENSITy OF A WAVE (I) In a travelling wave, energy is transferred through the medium in the direction in which the wave travels. The transfer of energy per unit time per unit area perpendicular to the direction of motion of the wave is called as the intensity of the wave. If A is the amplitude, c is the speed, ν is the frequency, ρ is the density of the medium, I is the intensity, then I = 2π2 c ρ ν2 A2
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Note: Intensity is proportional to the square of the amplitude (I ∝ A2). As the wave propagates through the medium, its energy is shared by a larger number of particles and the energy through unit area per second decreases. Hence, the intensity and amplitude go on decreasing as we move from the source of the waves. For a point source giving out wave in all directions (three-dimensional wave), the variations of intensity with distance (r) from the source is given as ...(ii)
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3.6 WAVE EqUATION
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Wave Mechanics
■ φ0 depends upon the initial condition (at φ = 0) and is called initial phase. 2π x 2 xt − + φ0 is called the phase angle and represents the phase λ T of a particle located at x at time instant t. ■ The wave equation can also be written in the following equivalent forms:
■ The angle φ =
y (x, t) = A sin (kx – ω t + φ0) 2π y ( x, t ) = A sin ( x − ct ) + φ0 λ
ta
s_
rio
2π x1 2π x2 2π − = (∆ x) λ λ λ
.ja m
=
ar
an a.
in fo
co
m
2π where, k = angular wave number = , and λ ω = angular frequency. Phase difference between two particles located at x1 and x2 at the same time instant is 2π x1 2π t 2π x2 2π t ∆ φ = φ1 − φ2 = − + φ0 − − + φ0 T T λ λ
un iv
rio
∆ φ = 0, 2π, 4π .... for ∆x = 0, λ, 2λ ....
//w
⇒
s
er
w
w
si
where, ∆ x is the separation between the particles.
ta
si
er
un iv
∆ φ = π, 3 π,5 π, ....for ∆ x =
s:
//t
ht
tp
.m e/
s:
Hence, particle are in same phase if the phase difference is an even multiple of π. Similarly, we can see that λ 3λ , ,... 2 2
dy vp = dt x = constant
ht tp
s:
//t
ht
tp
.m e/
Hence particles are in opposite phase if the phase difference is an odd multiple of π. The velocity of the particle at any instant isgiven by
= – ωA cos (kx – ωt + φ0)
The equation of wave travelling towards left (negative x-axis) y (x, t) = A sin (kx + ωt + φ0).
(i) Find the frequency, wave length and the speed of sound wave in air. (ii) If the equilibrium pressure of air is 1.0 × 105 N/m2, what are the maximum and minimum pressure at a point as the wave passes through that point?
Example The equation of a sound wave in air is given by P = (0.01N/m2) sin [(1000s–1) t – (3.0 m–1) x]
3.6
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Wave Mechanics
3.7
(i) Comparing with the standard form of a travelling wave x P = P0 sin ω t − υ we see that, ω = 1000 s–1 ω 1000 Frequency, υ = = Hz = 160 Hz 2π 2π Also from the same comparison
ω = 3.0 m–1 c ω or c = 3.0 m −1
1000 s −1 = 333 m/sec. 3.0 m −1 c Wavelength, λ = ν
si ta rio
m .ja w
=
s_
an
ar
333m/sec. = 2.1 m 160 Hz
s
ta
s:
un
iv
//w
rio
w
er
(ii) The pressure amplitude is P0 = 0.01 N/m2. Hence, the maximum and minimum pressure at a point in the wave motion will be (1.01 × 105 ± 0.01) N/m2.
in
a.
=
fo
co m
)
) Solution
iv
un
//t
ht
tp
.m e/
er si
Example The density of air at NTP is 1.293 gm/litre and γ = 1.36. Calculate the frequency of a turning fork which emits sound of wavelength 66.5 cm at 17º C.
ht
∴
e/
γP0 1.36 × (1.013 × 106 ) = ρ0 0.001293
//t
s:
c1 =
.m
)
tp s:
) Solution Pressure at NTP = 1.013 × 106 dyne cm–2
tp
= 32640 cm s–1
ht
Let, ct is velocity of sound at 17º C (290 K).
ct T = c1 T0
But
∴
ct = 33640 cm s–1
or
ct 290 = 32640 273
or
ct = νλ
v=
33640 = 506 Hz 66.5
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Wave Mechanics
Example A transverse wave at wavelength 50 cm is travelling towards positive x-axis along a string whose linear density is 0.05 gm/cm. The tension in the string is 450 N. At = 0, the particle at x = 0 is passing through its mean position with upward velocity from an equation describing the wave. The amplitude of the wave is 2.5 cm. )
) Solution Let the wave be described by y (x, t) = A sin (kx – ωt + φ0)
2π 2π = = 4π and A = 2.5 cm λ 0.5 y (0, 0) = 0 A sin φ0 = 0 φ0 = 0, π
in f s_ rio
iv
iv er
.m e/
2πc 2π (300) = = 1200 π rad/sec. λ 0.5 Using all the quantities, the equation is ω = 2πν =
e/
tp
s:
un
//t
⇒
T 450 = = 300 m/sec. m 0.05 × 10−1
un
=
s:
ht
tp
//w
w
Velocity of transverse wave in the string
s
er si
φ0 = π
ta
w .ja
m
– Aω cos φ0 > 0 From equations (i) and (ii), we get
o
dy >0 dt (0,0)
ar a
We also have
si ta rio
∴
na .c om
∴ ⇒
where, k =
//t
ht
.m
y = 2.5 sin (4π x – 1200 πt + π) cm.
ht
tp s:
Example The sound level at a point is increased by 30 dB. By what factor is the pressure amplitude increased? )
) Solution The sound level in dB is I B = 10 log10 I0
If B1 and B2 are the sound levels and I1 and I2 are the intensities in the two cases, then
I I B2 – B1 = 10 log10 2 − log10 1 I0 I0
or or
3.8
I 30 = 10 log10 2 I1 I2 = 103 I1
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Wave Mechanics
3.9
As the intensity is proportional to the square of the pressure amplitude, we have
P2 I = 2 P2 I1 ≅ 32.
= 1000
3.7 SUPERPOSITION OF WAVES Two or more waves can propagate through the same medium simultaneously. At the point where the two waves overlap, the resultant displacement of the particle is the sum of displacement of the individual waves.
m
⇒ y = y1 + y2 + ....
in
s_
rio
m
ta
si
s
er
w w
si ta
un
iv
//w
Stationary Waves
rio
ar an
a.
fo
(i) Stationary waves: These are formed when two waves of same frequency travelling in opposite direction overlap. (ii) Beats: These are formed when two waves of slightly different frequency travelling in same direction overlap.
.ja
co
The two most important situations involving superposition are:
iv
//t .
y = y1 + y 2
un
m
ht tp
The resultant wave is described by
e/
er
e/
//t .m
y1 = A sin (kx – ωt) travelling along positive x-axis. y2 = A sin (kx + ωt) travelling towards negative x-axis.
s:
ht tp
s:
Let us consider equation of two transversewaves travelling along a tight string in opposite directions.
tp
s:
y = A sin (kx – ωt) + A sin (kx + ωt) y (x, t) = (2 A sin kx) cos ωt
ht
Hence, the particle at a location x is oscillating in simple harmonic motion. With angular frequency ω and amplitude 2 A sin kx. As the amplitude depends on location (x). Particles are oscillating with different amplitudes. The points where amplitude is minimum are called nodes and the points where amplitude is maximum are called antinodes.
Nodes: Amplitude = 0
⇒
⇒
⇒
2 A sin kx = 0 π 2π x = 0, , k k
x = 0, λ/2, λ, 3λ/2, 2λ
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Wave Mechanics
Antinodes Amplitude is maximum ⇒ sin kx = ± 1 π 3π ⇒ x= , ,.... 2k 2k λ 3λ 5λ , , ,.... 4 4 4 O $
1
[
$
O
O $
1
$
[ O O
m
ar a
[
1
na .c om
1
O
rio s_ in fo
⇒ x=
[
[ O
ht tp
s: //
er si ta rio s
t.m e/
7 W 7
W
un
W
//t ht tp s: 1
iv
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
■ Nodes are completely at rest. Antinodes are oscillating with maximum amplitude (2A). The points between a node and an antinode have amplitude between 0 and 2 A. ■ Separation between two consecutive nodes (co-ordinates) = λ/2 ■ Separation between a node and the next antinode = λ/4 ■ Nodes and antinodes are alternately placed.
ht
3.10
1
1
1
W 7 W 7
■ N1, N2 are nodes and, hence remain at rest. The segment of the string between N1 and N2 moves up and down. The figure shows the position of this segment at t = T T 3T T 0, , , , , where T is the time period of the wave. 8 4 4 2 ■ The portion between two consecutive nodes is often called as a loop. ■ It is clear from the figure that since nodes are at rest they do not transfer energy. In a stationary wave, energy is not transferred from one point to the other. https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
3.11
Vibrations of a Stretched String Consider a string of length l stretched between two fixed supports. When a wave is set up in the string, the disturbance travels in both directions while being reflected at each end. For a stationary wave to exist, the fixed ends must be nodes. 1
$
1
O
na .c om
As the distance between two consecutive nodes is
ar a
m
⇒ ⇒
rio s_ in fo
nλ if there are (n + 1 nodes in all) 2 2l c nc n T λ = ⇒ ⇒ ν = = ν= n λ 2l 2l m l=
λ , therefore 2
un iv
1 2l
er si ta rio s
tp
s:
where, ν0
T m
e/
//w
w
er s
w
ita
.ja
where, T = tension, and m = mass per unit length. Hence, the string can vibrate with those frequencies which are integral multiples of the fundamental frequency υ0
iv un
//t
ht
.m
The possible frequencies are: ν0, 2ν0, 3ν0 .....
ht tp
s: //
t.m e/
ht tp s:
The fundamental frequency is simply the lowest possible frequency for vibration in the string. The nth multiple of the fundamental is known as the nth harmonics. ν0 is the first harmonic, 2ν0 is the second harmonic, 3ν0 is the third harmonic and so on. The shape of the loops in the strings for these harmonics is shown below: Fundamental first harmonic = υ0 $
1
1
$ 2QHORRS
Second harmonic = 2υ0 (first overtone) https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics $
$
1
1
1
7ZRORRSV
$
$
$ 1
1
rio s_ in fo
1
ar a
1
na .c om
Third harmonic = 3υ0 (second harmonic)
ita
.ja
m
7KUHHORRS
er si ta rio s
w
er s
w
Vibration in an Air Column in a Tube
$
O
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
e/
s:
un iv
//w
1. Open at both ends: The open ends of the tube become antinodes because the particle at the open end can oscillate freely.
ht
$
ht tp
3.12
Let l = length of the tube (or air column). If there are (n + 1) antinodes in all, then l=
nλ 2
because distance between two consecutive antinodes is ⇒
ν=
λ . 2
c nc = λ 2l
where, c is speed of sound wave.
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Wave Mechanics
3.13
The possible vibration frequencies are integral multiples of the fundamental or the lowest frequency ν0 ν0 =
c 2l
Possible frequencies are: ν0, 2ν0, 3ν0 ......
$
$
$ 1
1
$
$
ar a
1 $
1
$
Second harmonic = 2 ν0
er s
un iv
Third harmonic = 3 ν0
er si ta rio s
Fundamental mode first harmonic = ν0
w w //w
$
ita
.ja
m
1
$ 1
rio s_ in fo
na .c om
1
e/
iv
.m
un
//t
ht
tp
s:
2. Closed at one end: The open end becomes antinode and the closed end beλ comes a node. The distance between a node and antinodes is . 4
t.m e/
ht tp s:
⇒ If there are n nodes and n antinodes, then
ht tp
s: //
$
O
1
l = (2n – 1)
λ 4
c (2n −1) c = λ 4l ⇒ The possible vibration frequencies are odd multiples of the fundamental or the lowest frequency v0. ν=
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Wave Mechanics
v0 =
c 4l
Possible frequencies are: v0, 3v0, 5v0 ..... $
$
$ 1 $
1
1 $
na .c om
$ $
Third harmonic = 3 ν0 (first overtone)
m
ar a
Fundamental mode first harmonic = ν0
rio s_ in fo
1
1
Fifth harmonic = 5 ν0 (second overtone)
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
er si ta rio s
e/
.m
tp
s:
un iv
//w
w
er s
w
ita
.ja
Note: There are only odd harmonics in a tube closed at one end. Beats: Beats are formed by the superposition of two waves of slightly different frequencies moving in the same direction. If two sources of sound having slightly different frequencies are placed in the same medium, the resultant effect heard at any fixed position will consists of alternate loud and weak sounds. Let us consider two waves of frequencies ν1 and ν2 and amplitude A. We will analyse the net effect of a convenient point (say x = 0) y1 = A sin 2π ν1 t y2 = A sin 2π ν2 t ⇒ y = y1 + y2 ⇒ y = A (sin A 2π ν1 t + sin 2π ν2t) y = [2A cos π (ν1 – ν2) t] sin π (ν1 + ν2) t
ht
3.14
Thus, the resultant wave can be represented as a travelling wave whose frequency ν + ν2 is 1 and amplitude is 2A cos π (ν1 – ν2) t. As the amplitude term contain t, 2 the amplitude (and, hence intensity or loudness in case of sound) varies periodically with time. For loud sounds: net amplitude = ± 2 A ⇒
cos π (ν1 – ν2) t = ± 1
⇒
π (ν1 – ν2) t = 0, π, 2π, 3π ....
⇒
t = 0,
1 2 , .... ν1 − ν 2 ν1 − ν 2
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Wave Mechanics
3.15
For weak sounds, proceeding similarly, we get t= ∴
1 3 , .... 2(ν1 − ν 2 ) 2(ν1 − ν 2 )
Interval between two loud (or weak) sounds =
1 ν1 − ν 2
the number of loud sound per second = ν1 – ν2 or beat per second ν1 – ν2 Example The vibrations of a string of length 60 cm fixed at both ends are represented by the equation
na .c om
πx y = 4 sin cos (96 π t) 15
ita
.ja
m
ar a
What is the maximum displacement of point at x = 5 cm? Where are the nodes located along the string? What is the velocity of the particle at x = 7.5 at t = 0.25 sec? Write equation of the component waves whose superposition gives the above wave.
er s
w
(i) (ii) (iii) (iv)
rio s_ in fo
where, x and y are in cm and t in second.
er si ta rio s
w
) Solution The given equation is represented by the stationary wave.
un iv
π 2π = 15 λ
⇒ λ = 30 cm
iv
k=
e/
A = 2 cm,
un
//t
ht
tp
s:
We get
.m
//w
Comparing this equation by the general equation y (x, t) = (2 A sin kx) cos ωt
ht tp s:
2π = 96 π, T
t.m e/
⇒ T =
s: //
(i) At x = 5 cm
1 sec. 48
ht tp
π ymax = 4 sin × 5 15 = 4 sin
π 4 3 = = 3.464 cm 3 2 λ 30 = = 15 cm 2 2 = distance of two consecutive nodes
(ii) Loop length =
1 $
1
1
FP FP
1
1 %
FP
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Wave Mechanics
πx (iii) y (x, t) = 4 sin cos (96 π t) 15 Velocity =
∂y πx ( x, t ) = 4sin (–1) sin (96 πt) 96 π dt 15
At x = 7.5 cm and t = 0.25 sec. π× 7.5 c(x, t) = – 4 sin sin (96 π × 0.25) 96 π 15 π = − 4sin . sin (24 π) 96 π 2 =0
na .c om
(The particle is at its maximum displacement.)
ita er si ta rio s
er s
π y2 = 2 sin x − 96 t 15
and
//t
iv
.m
e/
s: tp
π y1 = 2 sin x + 96 t 15
un iv
//w
w
w
.ja
Comparing the equation, we get
rio s_ in fo
m
ar a
(iv) Let two waves are y1 = A sin (kx + ω t) and y2 = A sin (kx – ω t) ∴ y = y1 + y2 = 2 A sin kx cos ω t
ht
un
Example A rifle shot is fired in a valley formed between two parallel mountains. The echo from one mountain is heard after 2 seconds and from the other is heard after 2 seconds.
s: //
t.m e/
ht tp s:
(i) What is the width of the valley? (ii) Is it possible to hear the subsequent echoes from the two mountains simultaneously at the same point? If so, after what time? Given c = 360 ms–1.
ht tp
3.16
) Solution
(i) Let the width of valley is x $
[±\
\ 3
%
[
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Wave Mechanics
3.17
Let the distance of point P where the man hears the shot be ‘y’ from A where the man hears the echo. Thus, total distance travelled by the first case is 2 y and the second case is 2 (x – y). So, y = c × 1 = 360 × 1 = 360 m (x – y) = c × 2 = 360 × 2 = 720 m ∴ x = (720 + 360) m = 1080 m (ii) The sound reflected from A reaches P after 2 seconds (first echo). This sound travels toward B and after reflection at B reaches 4 seconds later (third echo). That is after a total time of 6 seconds. The sound reflected from B reaches P after 4 seconds (second echo). This travels towards A and after reflection reaches at A 2 second later (third echo). This shows that the echo is heard after 6 seconds simultaneously.
na .c om
e/
rio s_ in fo
7 %
ht tp s:
un
//t
ht
G G[ O
.m
tp
$
7
[
iv
un iv
s:
7
er si ta rio s
ita
er s
w
//w
w
) Solution
.ja
m
ar a
Example How long will it take the sound waves to travel the distance l between the points A and B if the air temperature between them varies linearly from T1 and T2? The velocity of sound propagation in air is equal to v = α T , where, α is constant.
dx = v ⋅ dt or
l
t
0
0
ht tp
According to kinematics
...(i)
s: //
T −T T = T1 + 2 1 x l
t.m e/
Let the temperature is T at a distance x from the point A. Then,
∫ dx = ∫ v. dt
T −T where, v = α T = α T1 + 2 1 x l
∴
t
t
0
0
∫ dt = ∫
dx T −T α T1 + 2 1 x l
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Wave Mechanics
Put,
T −T T1 + 2 1 x = z , l
T −T ⇒ 2 1 dx = dz l
The limit of z → T1 to T2 T2
t
−
T
1
l dz l 2 z 2 dz t = ∫ dt = ∫ = α . z1/2 (T2 − T1 ) α T∫1 (T2 − T1 ) 0 T1
=
2l T2 − T 1 = α T2 − T1 α
2l
(
T2 + T1
)
.
Example
fo
in
an
a. co
m
The displacement of the medium in a sound wave is given by the equation: y1 = A cos (ax + bt), where A, a, b are positive constant. The wave is reflected by an obstacle situated at x = 0. The intensity of reflected wave is 0.64 times that of the incident wave.
si
ta
.ja
ta
si er un
e/
s:
tp
ht
2π m λ
b Hz 2π (ii) Equation of the reflected wave is y2 = A2 cos (–ax + bt)
⇒λ =
ω = b or 2π f = b
and
∴
2π = a, λ
//t .m
tp s ht
we get,
iv
2π y1 = A cos x + ωt λ
://
(i) Comparing the wave equation
rio
iv
un
)
ht tp
t.m e/
) Solution
s
er
w
w
s: //w
s_
rio
ar
m
(i) What are the wavelength and frequency of the incident wave? (ii) Write the equation of reflected wave. (iii) In the resultant wave formed after reflection, find the maximum and minimum values of the particles speed in the medium. (iv) Express the resultant wave as a super position of a standing wave and a travelling wave. What are the position of the antinodes of the standing wave? What is the direction of the propagation of the travelling wave?
3.18
f =
Here, x is negative because if incident wave has +ive x direction than reflected wave has –ve x direction y = A2 cos (–ax + bt) It is given in the problem that intensity of reflected wave is 0.64 times of incident wave I1 A12 1 = 2 = I 2 A 2 0.64
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Wave Mechanics
(a) When there is zero phase change: Y = yi + yr = A cos (ax + bt) + 0.8 A cos (–ax + bt) = 0.8 A [cos (ax + bt) + cos (–ax + bt)] + 0.2 A cos (ax + bt) = 2 × 0.8 A cos ax. cos bt + 0.2 A cos (ax + bt) This is the mixture of stationary wave (1st term) and a progressive wave (2nd term) progressive in (–x) direction. ∴ At ax = 0, π, 2π Y = 2 × 0.8 A cos bt + 0.2 A cos bt = 1.8 A cos bt with amplitude 1.8 A (maximum) π 3π 5π At ax = , , 2 2 2 y = 0.2 A cos bt, with amplitude 0.2 A (minimum) (b) y = resultant wave when π phase change = 0.8 A cos (ax + bt) – 0.8 cos (–ax + bt) + 0.2 A cos (ax + bt) = – 2 × 0.8 A sin ax. sin bt + 0.2 A cos (ax + bt) This is again a mixture of stationary and progressive wave. At ax = 0, π, 2π, 3π.....
or A2 = 0.8 A Thus y2 = ± 0.8 A cos (– ax + bt) Here, ± depends on whether there is zero or π phase change on reflection.
m
in fo
rio
iv
rs ita
un
e/
y = ± 0.2 A cos bt with ampli tude = 0.2 A (minimum) π 3π 5π ax = , , , .... 2 2 2 y = 2 (0.8 A) sin bt + 0.2 sin bt
un
e/
t.m
ht
= 1.8 = sin bt with amplitude = 1.8 A (maximum)
tp
s:
//t
iv e
.m
tp ht
At
s
er si
ta
.ja
s:
//w
w
w
m
rio
s_
ar an
a. co
3.19
s: //
tp
ht
π 3π 5π , , ..... 2a 2a 2a (c) Progressive wave travels towards the reflecting surface i.e., negative x-axis
(iii) vm = Amax ω = 1.8 Ab and vmin = Amin ω = 0.2 Ab (iv) (a) For resultant wave, y = 1.6 A sin (ax) cos (bt) + 0.2 A cos (ax + bt) or y = – 1.6 A sin (ax) sin (bt) + 0.2 A cos (ax + bt) (b) For zero phase changing, position of antinode are π 2π x = 0, , ,..... a a For π phase change, position of antinode are
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Wave Mechanics
)
Example A wire of density 9 cm–3 is stretched between two clamps 100 cm apart while subjected to an extension of 0.05 cm. What is the lowest frequency of transverse vibrations in the wire, assuming Y = 9 × 1011 dyne cm–2? stress ) Solution Young’s modulus, Y = strain T/A T l Y= = × ∆l / l A ∆l T Y∆l = A l
∴
9 × 1011 × 0.5 100 = 45 × 107 dyne cm–2 Mass per unit length of wire, m =A ρ
m in fo rio s_ si ta
s io
iv
iv
t.m
er
si
un
ta r
1 45 × 107 = 35.3 Hz. 2 × 100 9
e/
s: //w
=
45 × 107 9
er
1 2l
ar
=
T Aρ
.ja m
1 2l
ht tp
v=
w w
an
a. co
=
tp s
:// t.m
ht tp
e/
un
s: //
Example A string 25 cm long and having mass of 2.5 gm is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first over tone and the air in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in is air 320 ms–1. Find the tension in the string. )
ht
) Solution Since by decreasing the tension of the string, the number of beats
decreased, therefore vstring > vpipe
v 320 = = 200 Hz 4 l 4 × 40
vpipe =
1 T l m where, m = mass per unit length
Also,
Fundamental Frequency νstring – νpipe = number of beats = 8 ∴ ν string (1st ovetone) = 200 + 8 = 208Hz
3.20
ν string =
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Wave Mechanics
ν string =
∴
2.5 × 10−3 = 10−2 kg m−1 25 × 10−2
1 T = (208) 2 l2 m
or
3.21
∴
T = (208)2 × (25 × 10–2)2 × 10–2 = 27.04 N
m
Example A sonometer wire of length 100 cm carries a metal block hanging at its free end. When the block is in air, its resonance is in unison with a tuning fork. When the block is in water, the length of vibrating wire in unison with the same fork is 90 cm. Find the specific gravity of the block. )
a. co
) Solution Let w1 and w2 be the weight of the block in air and water respectively.
io s
ta r si
e/ u
tp s:
ni
//w
w
Squaring and dividing, we get w1 l12 = w2 l22
in fo
rs ita rio s_
w2 m
ve
1 2 l2
an
v=
w1 m
ar
1 2 l1
.ja m
v=
w
and
∴
In both the case, the frequency of vibration of the wire remain the same.
But
w1 weight in air = specific gravity = w1 − w2 loss of weight in water
e/ un
iv
//t .m //t
l12 l − l22 2 1
ht tp
s:
ht
.m
tp
s:
Specific gravity =
=
∴
er
w1 l2 = 21 2 w1 − w2 l1 − l2
ht
∴
1002 = 5.26 1002 − 102
Example Two wires of radii r and 2r respectively are welded together end to end. This combination is used as a sonometer wire kept under tension T. The welded point is midway between the two bridges. What would be the ratio of the number of loops formed in the wire such that the joint is a node when stationary vibrations are set up in the wires?
)
) Solution Let m1 and m2 be the masses at per unit length. ∴ m1 = A1 × ρ = πr2 × ρ and m2 = A2 × ρ = π (2r)2 × ρ https://t.me/universitarios_infohttps://www.jamarana.com
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3.22
∴
m1 1 = m2 4
Wave Mechanics
p 2l
T m1
and
q 2l
T m
Now ν1 = ν2 =
fo ita iv
t.m
er si ta
e/ un i
s
ve rs
w
w
w
s: //
tp
ht
un
e/
ht tp s:
//t
.m
tp
s: //
Consider a source of sound (S) producing sound wave of frequency ν. If an observer (O) hears the sound waves, he will record a frequency same as ν, if both the sources and the observers are stationary. If the source or observers or both are moving the apparent frequency as heard by the observer will be different from ν. Let us call this apparent frequency as ν′. Let Vs, V0 be the velocity of the source and the observer (or the components of their velocity along the line joining them). If c is the speed of sound the apparent frequency is given by c V0 ν' =ν c Vs
ht
3.8 DOPPLER’S EFFECT
rio
p 1 = . q 2
rio s_ in
or
.ja m
m1 p = q m2
or
.c o
p p m1 m1 T T m m
ar an a
∴
p p 2l 2l q = q = 2l 2l
m
where, ν1 and ν2 are the frequencies of the vibrations, p and q are the number of loops. Since, it is a continuous wire, therefore ν1 = ν2 (as it vibrates with the same frequency)
V0, Vs are positive, if they are directed from source to the observer. They are negative if they are directed from observer to source. In other words, we will assume that the direction from source to the observer is the positive direction. If the medium is moving relative to ground, the speed of sound will be taken as c + Vm (Vm is the speed of the medium). Again note that Vm will be positive if the medium is moving from source of the observer. So, in general the apparent frequency is
c Vm V0 ν' =ν c Vm Vs https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
3.23
If the sound reaches the observer after being reflected from a stationary surface and the medium is also stationary, the image of the source in the reflecting surface will become the source of the reflected sound. Example Given v = 1000 Hz, c = 330 ms–1.
(i) What are the wave length ahead of and behind the moving source (source moving away from the stationary observer) if Vs = 30 ms–1? (ii) What is the frequency as heard by the observer? (iii) If the source is at rest and the observer is moving away from the source at 30 ms–1, what is the frequency as heard by the observer?
fo
co
m
(i) For the waves in front of the source c − Vs λ= ν 330 − 30 = = 0.30 m 1000 For the waves behind the source
in
rio
ta si
s
er
iv
si
ta
un
er
e/
.m
//t
iv
c c + Vs
s:
un
tp
v′ = ν
rio
330 + 30 = 0.36 m 1000 (ii) Frequency as heard by the observer
ht
s_
an
m
c + Vs ν
//w w
λ=
s:
w
.ja
ar
a.
)
) Solution
.m
//t
(iii) Here
V0 = 30 ms–1, Vs = 0
s:
ht tp
e/
330 = 1000 × = 916.7 Hz 330 + 30
ht
tp
c − V0 v′ = ν c
330 − 30 = 1000 = 909.1 Hz 330
Example An ambulance travels down a highway at a speed of 33 ms–1. Its siren emits sound at a frequency of 400 Hz. Find the change in frequency measured by a passenger in a car travelling at 23 ms–1 in the opposite direction as the car approaches the ambulance and as the car moves away from the ambulance. c = 343 ms–1. )
) Solution Apparent frequency measured by the passenger, when ambulance and car approach each other.
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3.24
c + V0 ν1 = ν c − Vs
Wave Mechanics
343 + 23 = (400) 343 − 23
366 = 400 = 472.2 Hz 310
Apparent frequency measured by the passenger in the car when ambulance and car move away from each other, 343 − 23 = 400 343 + 33
na .c om
c − V0 ν2 = ν = 400 c + Vs
320 = 340.4 hz 376
ar a
rio s_ in fo
Change in frequency measured by the passenger ∆ν = ν1 – ν2
tp
e/
s:
WORKED OUT EXAMPLES
t.m e/
s: //
λ=
or
2 × 1011 8000
= 0.5 × 104 = 5000 m/sec. Equation of the stationary wave is y (x, t) = 2 Asin kx cos ωt Given, 2A = 2 × 10–6
1
$
1
$
1
$
1
$
∴
Since both ends are open, therefore start from antinodes at each side. Since distance between the consecutive λ antinode is , therefore 2
and
P
$
=
Y ρ
)
1
2 = 0.4 m 5
Velocity of sound in rod, c =
ht tp
) Solution $
λ 5 =1 2
iv un
//t
ht tp s:
ht
.m
1. A metallic rod of length 1 m is rigidly clamped at its mid-point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either side of the mid-point. The amplitude of an antinode is 2 × 10–6 m. Write the equation of motion at a point 2 cm from the mid-point and those of the constituent waves in the rod. (Young’s modulus = 2 × 1011 Nm–2, density = 8000 kg m–3)
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
= 472.2 – 340.4 = 131.8 Hz
2π 2π = = 5π λ 0.4 c ω = 2πv = 2π λ 5000 =2π× = 25000 π 0.4 k=
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3.25
Wave Mechanics
λ=
∴
Y = y1 + y 2
5π 2π = L λ
It gives, k =
2L 5
... (ii)
At x = 2cm y (x, t) = 2 × 10–6 sin (0.1 π) cos (25000 π t) The constituent wave is
Velocity of the particle in the string, c =
T Tension = µ Mass per unit length
The two wave travels with same velocity and frequency in opposite directions. y1 = A sin (kx – ω t) and y2 = A sin (kx – ω t) ∴ y1 = 10–6 sin (5 πx – 25000 πt) ∴ y2 = 10–6 sin (5 πx + 25000 πt)
m
= 2π
fo
na .
2 π 5π T = t L µ
ta
Time period =
2L µ . 5 T
er si
rio s
iv
∴
5π T t L µ
3. The following equations represent transverse waves: Z1 = A cos (kx – ω t) Z2 = A cos (kx + ω t)
un
e/ .m
Z3 = A cos (ky – ω t)
//t
y (x, t) = 2 A1 sin kx cos ωt
s:
Identify the combinations of the waves which will produce
tp
Z = Z1 + Z2
5π y0 (x) = A sin x L
(i) Standing waves
)
) Solution
But in the problem 2A1 = A y (x, t) = A sin (kx) cos (ωt) ∴ y0 (x) = A sin (kx) ...(i) On comparing this equation with the given equation, we get
[ /
(i) standing waves (ii) a wave travelling in the direction making an angle if 45º with the positive x-axis and positive y-axis. In each case, find the position at which the resultant intensity is always zero.
ht [
s_ in
s:
ht tp
)
) Solution The equation of stationary wave is
un
e/
//t
.m
ht tp
(ii) What is the period of motion?
[
(ii) Period of motion =
iv
//w
s:
w
5ð x y (x, 0) = y 0 (x) = A sin , L ∂y at (x, 0) = 0, where A is con∂x stant.
5π y (x, t) = A sin x cos L
er si ta rio
ar a
w
.ja
m
(i) Write down the simplest possible expression for the stationary wave y (x, t) describing the motion at all time t > 0 if the initial condition is
c λ
Hence, equation of the stationary wave is
co
2. Consider a vibrating string of length L under tension T and with mass per unit length µ. The string is fixed at both ends (x = 0 and x = L)
= 2π
T 5 5π T . = µ 2L L µ
ω = 2 πν
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= A cos (kx – ωt) + A cos (kx + ωt) = 2 A cos kx · cos ω t
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3.26
Wave Mechanics
No power is dissipated during the propagation of the wave-pulse. Calculate:
Amplitude of the resultant standing wave = 2 A cos kx Since the intensity is directly proportional to the square of the amplitude, therefore I ∝ A2 = 0 So, I=0 ∴ 2 A cos kx = 0 or cos kx = 0 2π or cos x=0 λ 2π Thus, x = (2π − 1) π / 2 λ λ ∴ x = (2n – 1) , where n = 1, 2, 3 .... 4 (ii) Wave travelling in the direction making an angle of 45º degrees with the positive x and positive y. So, resultant equation is the sum of the first and third equation. Z = Z1 + Z3
) Solution
)
o
in f
s_
T m
rio ta
cPQ =
80 × 4.8 = 80 m/sec. 0.06
cQR =
80 × 2.56 = 32 m/sec. 0.2
rio
s
si ta
Time, t = t1 + t2
4.8 2.56 + = (0.06 + 0.08) sec 80 32
er
∴
un iv
=
= 0.14 sec
e/
E 1 2 l 2 1 2 = ωm A = ω t 2 2 t mv A2 The conservation of power at the junction Q gives Pi = P r + Pt P=
m1 v1 (Ai2 – Ar2) = m2 v2 At2 ...(i)
or
4. A long wire PQR is made by joining two wires PQ and QR of equal radii, PQ has a length 4.8 m and mass 0.06 kg. QR has a length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave-pulse of amplitude 3.5 is sent along the wire PQ from the end P.
2π x − y π = (2n − 1) λ 2 2 λ x – y = ± (2n – 1) , where 2 n = 1, 2, 3, ...
(ii) The energy transferred by the wave pulse through the wire 1 = (ω2m) A2 l . 2 Then, average power,
t.m tp
ht
π x− y k = (2n − 1) 2 2
s: //
ht
∴
.m
tp s:
//t
ht
or
er si
iv
un
e/
s:
tp
Hence, if the intensity is zero, then
or
5
where, T = tension, and m = mass per unit length
k ( x − y) k ( x + y) .cos − ωt 2 2
x− y cos k =0 2
4 P NJ
(i) Velocity of the wave in the rod =
ar
m
.ja
w
//w w
= A cos (kx – ωt) + A cos (ky – ωt)
= 2 A cos
P
3P NJ
a.
an
P
co m
(i) The time taken by the wave pulse to reach the other end R of the wire. (ii) The amplitude of reflected and transmitted wave pulses as for the incident wave pulse crosses the joint Q.
The division of amplitude of Q gives Ai = Ar + At Substituting this in equation (i), we get m1 v1 (Ai2 – Ar2) = m2v2 (Ai – Ar)2
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/
Wave Mechanics
a.
s_ i
5λ 5 × 1.6 = (3.6 m) − m = 1.6m 4 4
L−
7λ 7 × 1.6 = (3.6 m) − m = 0.8m 4 4
ta
s
io
si
er
9λ 9 × 1.6 = (3.6 m) − m = 0m 4 4
un
iv
L−
e/
According to the Bernoulli’s equation, the flow of the fluid at any point is given by 1 2 p+ ρv + ρgh = constant 2 For the two points near the top and the bottom of the pipe filled with water, we get 1 1 p + ρv 2 + ρgH 0 = p0 + ρv02 2 2 Taking pressure to be identical at two points, we get
s: //t .m
ht tp
L−
iv
e/ un
:// t.m
tp s
ht
3λ 3 × 1.6 = (3.6 m) − m = 2.4m 4 4
si er
w
w
tp s: //w
ht
1 2 1 ρv + ρgH 0 = v02 2 2 1 2 1 2 or v + gH 0 = v0 …(i) 2 2 Also from the principle of continuity, we get vA = v0 a ...(ii) where, A and a are the areas of cross-section of the pipe and small hole, respectively. Eliminating v0 from equations (i) and (ii), we get
)
L−
rio
m
.ja
5. A 3.6 m long vertical pipe resonates with a source of frequency 212.5 Hz, when water level is at certain height in the pipe. Find the height of water level (from the bottom of pipe) at which resonance occur. Neglect end correction. Now, the pipe is filled to a H of 3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as the function of H. If the radii of the pipe and the hole are 2 × 10–2 m and 1 × 10–3 m respectively, calculate the time interval between the occurrance of first two resonance. Speed of sound in air is 340 m/s and g = 10 m/s2.
) Solution
λ 1.6 m = (3.6 m) − = 3.2m 4 4
Given: Ai = 3.5 cm, ∴ Ar = 0.6 × 3.5 = 2.1 cm Hence, At = Ai – Ar = (3.5 – 2.1) = 1.4 cm
L−
an
A r 1.5 = = 0.6 A i 3.5
ar
co m
1 + (A r / A i ) 0.2 / 2.56 = = 2.5 1 − (A r / A i ) 0.06 / 4.8
o
Substituting the value of m2 and m1, we get
nf
The condition of resonance is shown in the figure. v 340 m/sec. Wave length, λ = = =1.6 m ν 212.5/sec. The height of water level in the vertical pipe at which resonance occur are
ta r
Y2
1 + (A r / A i ) m2 = m1 1 − (A r / A i )
or
Y
O + 2
F , m
Ai + A r m2 = m1 Ai − A r
we get
or
$
O
O
/
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O + 2
Y
Since v =
A i + A r m2 v2 = A i − A r m1 v1
or
3.27
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3.28
1 2 1A v + gH 0 = v 2 2 a
2
t2 =
Wave Mechanics
1/2
2g H a2 v= 2 0 2 A −r In terms of radii, we get
=
or
2gH r v= 4 0 4 R −r Since, r < R, we ignore r4 in comparison to R4, therefore
0.00559 m /s 2 3.6 − 2.4 = 62.3 sec 0.00559
∆ t = t2 – t1 = (62.3 – 19.4) sec
Finally
1/2
= 42.9 sec
r2 2 gH 0 R2 dH v= , we get dt
4
H0 − H2
o
in f
s_
rio
si ta
rio
s
)
er
iv
(i) When the observer receives the sound, the source is closest to him. It means that frequency is emitted by the source sometimes before figure shows and the source approaches the stationary observer with velocity Vs cos θ. Hence, frequency noted by the observer
un e/
s: //
c v = v0 c − Vs cos θ
)
H = H 0 − 0.00559 m /s 2 t
ht
tp
t.m
tp
) Solution
...(iii)
ht
r2 1.0 × 10−3 2 g = 2 × 10 R2 2.0 × 10−2
(
(i) The frequency of sound received by the observer at the moment when the sound gets closest to him. (ii) The distance between the source and the observer at the moment when the observer receives a frequency υ = υ0.
si ta
er iv un e/ .m
)
= .01118 m /s 2 Hence, equation (iii) becomes
m co
ar m
w
//w w
ht
(
2g t
s:
r2 i.e., 2 H 0 − H = 2 R For the given data, we get
H0
t
dH r 2 = 2 2 g ∫ dt H R 0
s:
H
−∫
tp
−
.ja
dH r 2 = 2 2 g dt H R Integrating, we get or
//t
−dH r 2 = 2 2 gH dt R
a. an
v=
Setting
6. A source of sound with natural frequency υ0 = 1.8 kHz moves uniformly along a straight line separated from a stationary observer by a distance l = 250 m. The velocity of the source is equal to η = 0.80 fraction of velocity of sound. Calculate:
6 T
Since for the first resonance H1 = 3.2 m, therefore H 0 − H1 t1 = 0.00559 m /s 2
[
=
3.6 − 3.2 0.00559 m /s 2
= 19.4 sec
Since for the second resonance, H2 = 2.4 m, therefore https://t.me/universitarios_infohttps://www.jamarana.com
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Wave Mechanics
v″ = v
x l 2 + x2 = Vs c
But
c+u c c+u . =v c c−u c−u c −u λ″ = λ or c+u λ″ c − u or = λ c+u λ″ c−u 2u ∴ 1− = 1− = λ c+u c+u Hence, percentage change in the wavelength
...(i)
V0 c = v0 = c − η c cos θ 1 − η cos θ
3.29
x
Vs =η c
om
s_
an
s
rio
ita
ve rs
ni
∂y = −am ω cos kx sin ωt …(ii) ∂t So, the kinetic energy confined in the string element of length dx is given by or
1 m ∂ y dT = dx 2 l ∂ t
2
Wall will reflect the sound with same frequency v′. Thus, frequency noticed by a stationary observer after reflection from wall. c v″ = v′ c−u
e/ u
c+u c
v′ = v
.m
//t
tp s:
ht
)
) Solution Frequency of sound when it reaches the wall
(i) Let two waves y1 = A cos (ωt – kx) and y2 = A cos (ωt – kx), superimpose and as a result, we have standing wave (the resultant wave) in the string of the form y = 2 A cos kx cos ωt. According to the problem, 2A = am Hence, the standing wave excited in the string is y = am cos kx cos ω t ...(i)
un iv
s: //t
ht tp
7. A stationary source sends forth monochromatic sound. A wall approaches it with velocity u = 33 cm/s. The propagation velocity of sound in the medium is c = 330 m/sec. In what way and how much in per cent, does the wavelength of sound change on reflection from the wall?
) Solution
.m e/
ht
tp s:
The distance between the source and the observer at this time
(i) The maximum kinetic energy of the string. (ii) The mean kinetic energy of the string averaged over one oscillation period.
si ta
er
rio
m ar
w .ja
//w w
Vs t = lη.
in fo
8. A string of mass m is fixed at both ends. The fundamental tone oscillation are excited with circular frequency ω and maximum displacement amplitude am. Calculate:
a. c
(ii) When the source is right in front of O, the sound emitted by it will not be Doppler shifted because θ = 90º. This sound will be received at O at time l t= after the source has passed it. c Then the source will have moved ahead by a distance
= l 1 + η2 = 0.32 km
λ − λ″ λ 2u = × 100% = 0.2 % decrease c+u =
v frequency, v = 0 2 = 5 kHz 1− η
)
we get
or cos θ = η ...(ii) Hence, from the equation (i) and (ii),
l +x
=
2
2
∴
1m = dx am 2 ω2 cos 2 kx sin 2 ωt 2 l
Since, wall behaves as a source of frequency v′, therefore
=
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mam2 ω 2 2π x dx sin ωt cos 2 λ 2l
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3.30
Wave Mechanics
For the interference path difference ∆ = λ, 2λ, 3λ ... c 360 λ= = = 2m v 180 For the minimum distance, from equation (i)
Hence, the kinetic energy contained in the string corresponding to the fundamental tone
T = ∫ dT =
mam2 ω2 sin 2 ωt 2l
λ /2
∫ cos
2
2π x dx λ
2 x 2 22 + − x = λ = 2 string, l2 = λ Because, for the fundamental tone length of the 2 x2 2 λ or 4 4 + = x + 4x + 4 Because, for the fundamental tone length of the string, l = 4 2 ' Integrating, we get
0
o in f rio
[
ta
s rio ta
er si
un
iv
e/
10. A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a horn that has dominant frequency 800 Hz. What will be the apparent frequency heard by driver in the second car? Speed of sound in air = 330 m/sec.
ht
)
) Solution From the figure, path difference between two waves reaching the detector,
First car in source, Vs =
108 × 1000 = 30 m/sec 3600
Second car is observer, 72 × 1000 = 20 m/sec 3600 According to Doppler equation, when the source and observer are moving in the same direction, the apparent frequency V0 =
…(i)
) Solution
tp
9. A source emitting sound of frequency 180 Hz is placed in front of a wall at a distance 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and detector for which the detector detects the maximum sound. Speed of sound in air = 360 m/sec.
2 x = 2 22 + − x 2
16 + x2 = x2 + 4x + 4 4x = 12 x = 3 cm.
.m s: //t
1 < T > = mam 2 ω2 8
or or or
dt
ht tp
0
sin ωt dt
)
∫
un
iv
0 2 π/ ω
6
.m e/
∫
2
//t
=
2 π/ ω
s:
ht tp
1 2 2 mam ω 4
er si
.ja
//w
∫ T dt ∫ dt
s:
w w
m
1 ∴ Tmax = mam 2 ω2 4 (ii) Mean kinetic energy averaged over one oscillation period
s_
ar an
a. co
m
1 T = mam 2 ω2 sin 2 ωt 4 Since for maximum kinetic energy, Tm, sin2ωt = 1
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3.31
Wave Mechanics
c − V0 c − Vs
(c + Vm )(c + Vb ) − (c − Vm )(c − Vb ) = ν c 2 − Vb2
c+V = 2 ν Vb 2 m2 . c − Vb
11. A band playing music at a frequency v is moving towards a wall at a speed Vb. A motorist is following the band with a speed Vm. If c is the speed at sound, find an expression for the beat frequency heard by the motorist.
m
) Solution First overtone of the open organ pipe
s
rio
ta
si
iv er
un
.m
e/
lc =
Hence from equations (i) and (ii), we have 330 3 330 − × = ± 2.2 lo 4 3 4 330 − 3 = ± 2.2 lo
ht
or
330 = (300 ± 2.2) Hz lo lo =
330 330 or 332.2 327.8
= 0.993 m
or
or
or 1.0067 m
c + Vm c + Vm = ν − ν c − Vb c + Vb
s_
rio
si ta
c 330 3 = = m 4 × 110 4 × 110 4 Given in the problem | υo – υc | = 2.2 Hz
c + Vm ν″ = ν c − Vb The beat frequency heard by the motorist is ∆υ = ν″ – ν
…(ii)
where, l = length of the closed organ pipe. Fundamental frequency at closed organ pipe. c υ′c = = 110 Hz 4 lc
tp
s:
//t
ht
Thus, the frequency heard by the motorist is
c 3c = 4 lc 4 lc
e/ .m //t
tp
s:
ν . c − Vb
vc = 3 ⋅
er
iv
un
s:
tp ht
where, c = velocity of sound (ii) Frequency indirectly received after reflection from the wall. In this case, the motorist is moving towards reflected frequency which is
…(i)
in
m
//w
w
w
.ja
c + Vm ν′ = ν c + Vb
c c = 2lo lo
where, lo = length of the open organ pipe. First overtone of the closed organ pipe
ar
an
(i) Frequency directly received by the band. In this case, the motorist moves towards the band with velocity Vm and the band moves away from motorist with velocity Vb. The frequency heard by the motorist is
v0 = 2
frequencies:
)
a. co
)
) Solution The motorist receives following two
12. The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the length of the pipes. Velocity of sound c = 330 m/sec.
(330 − 20) 310 = 800 × = 827 Hz (330 − 30) 300
= 800
(330 − 20) 310 = 800 × = 827 Hz (330 − 30) 300
= 800
fo
ν′ = ν
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3.32
14. A column of air at 51º C and a tuningfork produce 4 beats per second when sounded together. As the temperature of the air column is decreased, the number of beats per second tends to decrease and when the temperature is 27º C, the two produce one beat per second. Find the frequency of the tuning-fork.
13. A source of sound moving along a circular orbit of radius 3 metres with an angular velocity of 10 rad s–1. A sound detector located away from the source is executing linear simple harmonic motion along the line BD (shown in the figure) with an amplitude BC = CD = 6 m. The frequency of oscillation of the detector is 5 per second. The source is at the point π A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and the minimum frequencies recorded by the detector (velocity of sound in air = 330 ms–1).
)
) Solution According to the given condition,
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15. A boat is travelling in a river with a speed 10 m/sec along the stream flowing with a speed 2 m/sec. From this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible.
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(i) What will be the frequency detected by receiver kept inside the river downstream? (ii) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/sec. in the opposite direction of the river stream. Determine the frequency of the sound detected by the receiver. Given: Temperature of the air and water = 90º C Density of the river water = 103 kg/m3
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s:
tp
ht
270 = 255 Hz 360
v = 75 Hz
un
//t
s:
tp
ht
un
∴
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s:
tp
ht
c − V0 330 − 60 = 340 × c + Vs 330 + 30
= 340 ×
Given that: v1 – v = 1.039 or 1.039 v1 – v = 4 × 1 or .039 v = 2.961
iv
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)
both approach each other and the minimum frequency occurs when both move away from each other. Linear velocity of the source, Vs = ωr = 10 × 3 = 30 m/sec. Maximum velocity of the detector at mean position Vo = ωA = (2 π υ) A where, A = amplitude of simple harmonic motion of detector 5 Vo = 2π× × 6 = 60m /sec ∴ π Since the time period of the source and detector are equal, therefore c + V0 330 + 60 vmax = v = 340 × c − Vs 330 − 30 390 = 340 × = 442 Hz 300 vmin = v
…(ii)
where, v1 is the frequency of air column at temperature 21º C
&
) Solution Maximum frequency occurs when
or
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P
P
v2 T2 324 = = v1 T1 300
or
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frequency of the air column is greater than the frequency of tuning-fork. Let frequency of the tuning fork is v ∴ v2 – v = 4 ...(i) where, v2 is the frequency of air column at the temperature 51º C.
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Cp,m Cv , m
9D
9V
for air = 1.4
The frequency heard by the receiver is c − V2 ν′ = V0 c − V1
(i) The velocity of sound in water is
)
) Solution
1/ 2
B 2.088 × 109 Pa = ρ 103 kg /m3
344 − 5 5 = 105 Hz = 1.030 × 10 Hz 344 − 15
= 1445 m /sec
c=
5
6
Bulk modulus of the water = 2.088 × 109 Pa Gas constant R = 8.31 J/mol – K Mean molar mass of air = 28.8 × 10–3 kg/mol
3.33
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ta rio
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Dr = Ar sin (k1 x + k2 V1 t)
γ RT M
Dt = At sin (k2 x – k2 V2 t) Ai = Ar + At
V − V2 (ii) Ar = 1 V1 + V2
k1 − k1 Ai = Ai k 2 + k1
(iii) Under what conditions is Ar negative?
The relative velocity of source with respect to the medium (i.e., air) is V1 = Vs – Va = (10 + 5) m/sec. = 15 m/sec
(i)
= 344 m/sec.
Show that:
16. A cord stretched to a tension FT consists of two sections whose linear densities are µ1 and µ2. Taking x = 0 to be the point (a knot) where they are joined, with µ1 referring to the section of cord to the left and µ2 that to the right. A sinusoidal wave Di = A1 sin (k1 x – k1 V1 t), starts at the left end of the cord. When it reaches the knot, part of it is reflected and part is transmitted. Let the equation of the reflected and transmitted waves be
(ii) The velocity of sound in air is
1/2
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= 1.007 × 105 Hz
(1.4)(8.314Jk −1mol−1 )(293K) = 28.8 × 10−3 kg mol−1
= 1.037 × 105 Hz
s: //t .m e/
ht tp
1445 + 2 = 105 × 1445 − 8
ht tp
s: //t
ht
.m
The relative velocity of receiver with respect to medium i.e., water = 2 m/sec. The frequency heard by the receiver is c + V0 v′ = v0 c − V1
c=
344 − 5 ν′ = (105 Hz) 344 − 17
∴
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9Z
e/
s: //w
5
w
9V
tp
6
Note: In the above solution, speed of 5 m/sec. of air is relative to that of the water. If the value is taken with respect to earth, then V1 = Vs + V2 – Va = 10 m/sec. + 2 m/sec. + 5 m/sec. = 17 m/sec. V2a = V2 – Va = 5 m/sec.
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c 1445 v= = = 105 Hz −3 λ 14.45 × 10 The relative velocity of source with respect to the medium (i.e., water) is V1 = Vs – Vw = (10 – 2) m/sec. = 8 m/sec
The frequency of sound in water is
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3.34 )
At x = 0, we have d dx (Di + D r ) = k1 Ai cos (– k1 V1 t) x =0 + k1 Ar cos (k1 V1 t) = (k1 Ai + k1 Ar) cos (k2 V1 t)
) Solution
k1 V1 = k2 V2
d = k2 At cos [– k2 V2 t] dx Dt x =0 = k2 At cos (k2 V2 t) Equating these two expressions and using the fact that k1 V1 = k2 V2, we get k1 (Ai + Ar) = k2 At Since Ai = Ar + At, we get k1 (Ai + Ar) = k2 (Ai – Ar)
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k1 V1 , we get V2
V1 − V2 V1 + V2
⇒
FT FT > µ2 µ2
µ2 < µ1
⇒
(b) 180º (d) 270º
(a) 90º (c) 45º
uestions
t – 1.57x) and another wave is y´ = 0.10 sin (314 t – 1.57 x + 1.57) The phase difference between the two wave is ∆φ = (314 t – 1.57x) ~ (314 t – 1.57x + 1.57) = 1.57 radians =
1.57 × 180 = 90° π
Another wave has the equation
where t, x and are in seconds, meters and centimeters respectively.
Then the phase difference between the two waves is
pe
t
) Explanation One wave is, y0 = 0.30 sin (413
y = 0.30 sin (314 t – 1.57x)
y´ = 0.10 sin (314 t – 1.57x + 1.57)
q
)
1. A simple harmonic wave has the equation
SOLVED OBJECTIVE y
x 2. The equation y = a cos2 2 πnt − 2 π α represents a wave with
ht tp
s: //
t.m e/
V2 > V1
un
iv
(iii) For Ar to be negative, we must have
.m
//t
d (Dt) = k2 At cos (k2x – k2 V2 t) dx
ht tp s:
k − k A r = Ai 2 1 k2 + k1
Putting k2 =
or
e/
s:
tp ht
Ai (k1 – k2) = – Ar (k1 + k2)
A r = Ai
d (Di + Dr) = k1 Ai cos (k1 x – k1 dx V1 t) + k1 Ar cos (k1 x + k1 V1 t)
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w
w
(ii) Since, the cord is continuous, slope of the cord at the boundary (x = 0) will be identical in either side of the knot.
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Since, ∴ or
sin (k1 V1 t) [– Ai + Ar] = sin (k2 V2 t) [– At] k1 V1 = k2 V2, we have – Ai + Ar = – At A i = Ar + A t
m
or
Since, x = 0 at the knot, we will have Ai (– k1 V1 t) + Ar sin (k1 V1 t) = At sin (– k2 V2 t)
Di + Dr = Dt
Since, the rope is continuous, the displacement at infinitesimal distance on either of the knot will have the same value. Hence,
(i) Since frequency of the wave will be same in both sections, therefore
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wave-
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5. When two simple harmonic motions of same periods, same amplitude, having 3π , and at right angles to each phase of 2 other super imposed, the resultant wave form is a (a) circle (b) parabola (c) ellipse (d) figure of eight
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motions of different amplitudes and same period are superimposed, the resulting motion is in general elliptical. When the phase difference is 0 or 180º, the motion will be a straight line. When the phase difference is 90 or 270º, the motion will be an ellipse whose axes coincide with coordinate axes. If the amplitudes are the same and the phase difference is 90º or 270º, the motion will be circular.
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1 2 19 (d) 18 (b)
18 19
(a)
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6. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered is f2. If the speed of sound is 340 m/s, then the ratio f1/f2 is
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) Explanation When two simple harmonic
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tp
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T m
p2 = 2p1 p 1 : p2 = 1 : 2
∴ or
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ht
of a SHM of amplitude y0 and frequency f is 2πfy0. The wave velocity is fλ. For 2π fy0 to be πy0 equal to 4fλ, λ has to be (since λ = a) 2 4. Two wires of radii r and 2r respectively are welded together end to end. The combination is used as a sonometer wire and is kept under a tension. The welded point is midway between the bridges. What should be the ratio of number of loops formed in the wires such that the joint is a node when stationary vibrations are set up in the wires?
p T = 2 πr 2 d 2l
an a
p1 2l
i.e.,
The maximum particle velocity is equal to four times the wave velocity if a is equal to (a wavelength) πyo πyo (a) (b) 4 2 (c) πy0 (d) 2πy0
p 2l
When the string is vibrating with p loops as the frequency of wave in both they will be the same,
wave-
3. A transverse wave is described by the equation
)
f =
) Explanation The maximum particle velocity
)
string is given by
wave-
1 2πx = a 1 + cos α 2 (2n)t − 2 λ /2 a Hence Amplitude is , frequency is 2n and 2 λ wavelength is 2
y = y0 sin 2π (ft – x/a).
(b) 1 : 4 (d) 1 : 1
) Explanation Frequency of transverse wave in a
.c
2πx y = a cos 2 2πnt − λ
∴
(a) 2 : 3 (c) 1 : 2
wave-
)
(a) amplitude a, frequency n and length λ (b) amplitude a, frequency 2n and length 2λ a (c) amplitude , frequency 2n and 2 length λ a (d) amplitude , frequency 2n and 2 λ length 2 1 2 2 ) Explanation From cos θ = (1 + cos θ) 2
3.35
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3.36
(a) (b) (c) (d)
340 ) Explanation f1 = f ;
340 f2 = f 340 − 17
) Explanation After 2 seconds, the pulses will be
)
7. A flat horizontal platform moves up and down in S.H.M. with an amplitude of 1 cm. A small object is placed on the platform. What is the maximum frequency the platform can have, if the object is not to separate from it during any part of the motion?
at the same location, giving zero displacement at each point. Thus, the energy will not be in the form of potential energy.
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v0′ =2 v0
10. The amplitude of a wave disturbance propagating in the positive x-direction is 1 given by y = at time t = 0 and by y 1 + x2 1 = at t = 2 seconds, 1 + ( x − 1) 2
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where x and y are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is (a) 1 ms–1 (b) 0.5 ms–1 –1 (c) 1.5 ms (d) 2 ms–1
v + v′0 6 = 5 v
1.2v = v + v0′
.m
tp s:
8. Two pulses in a stretched string whose centers are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be
⇒ ∴
ht
1 g 1 f = = 980 2π a 2π
and
//t
ht
tp
s:
//t
simple harmonic motion is the weight of the object in the platform. If A is the amplitude, we have mω2A = mg, where, ω = 2π f.
)
ve
⇒
.m
)
) Explanation Maximum restoring force of
v + v0 5.5 = 5 v 1.1 v = v + v0
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(d) 2π × 980 per second
FP
(b) 2 (d) 11/6
) Explanation From Doppler effect,
//w
980 per second 2π
(a) 242/252 (c) 5/6
w
(c)
ar
980 per second 2πn
9. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B, he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is
m
(b)
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980 per second 2π
w
(a)
f1 19 = f 2 18
)
340 − 34
zero purely kinetic purely potential partly kinetic and partly potential
Wave Mechanics
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3.37
Wave Mechanics
) Explanation Writing the general expression for
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(b) 80 m/s (d) 120 m/s
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7. What will be the ratio of amplitude of the reflected wave to the incident wave in case a sinusoidal wave produced in a stretched string gets reflected and losses 36% of its energy in getting reflected from the fixed end of the string? (a) 6 : 10 (b) 8 : 10 (c) 1 : 1 (d) None of these
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4. A 1 metre long sonometer wire is fixed at both ends and a bridge is placed at onetenth of length of wire from one of its ends. It is set into transverse vibrations. The velocity of the wave in the wire if the frequency is 400 Hz, is
6. The 4th overtone of an open organ pipe resonates with the first overtone of semi-open organ pipe. The ratio of the length of the open pipe to the semi-open pipe is (a) 5 : 3 (b) 4 : 1 (c) 10 : 3 (d) none of these
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(d)
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(b) 4π (d) none of these
3. If the speed of longitudinal waves equals 10 times the speed of the transverse waves in a stretched wire of material which has modulus of elasticity E, then the stress in the wire is (a) 10 E (b) 100 E
(a) 40 m/s (c) 100 m/s
2
5. A plane progressive wave of frequency 25 Hz, amplitude 2.5 × 10–5 m and initial phase zero moves along the negative x-direction with a velocity of 300 m/s. A and B two points 6 m apart on the line of propagation of the wave. At any instant the phase difference between A and B is φ. The maximum difference in the displacement at A and B is ∆. (a) φ = π (b) φ = 0 (c) ∆ = 0 (d) ∆ = 5 × 10–5 m
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2. A particle oscillates according to equation x = 10 sin (0. 5πt). The speed of the particle when it is at a distance 4 m from the mean position
E 10
1[ x − v(2) ]
OBJECTIVE T PE UESTIONS
1. A closed organ pipe resonates in its fundamental mode at a frequency of 200 Hz in O2 at a certain temperature. If the pipe contains 2 moles of O2 and, 3 moles of Ozone is now added to it, then what will be the fundamental frequency of same pipe at same temperature? (a) 268.23 (b) 172.7 (c) 149.45 (d) None of these
(c)
1
Comparing with the given equation we get 2v = 1 or v = 0.5 m/s
At t = 0
(a) 2π 3 (c) 2π
y=
At t = 2 s,
1 1 + ( x − vt ) 2
y=
1 1 + x2
y=
)
y in terms of x as
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3.38
11. Three coherent sources S1, S2, S3 are placed at the three vertices of a right angled, triangle as shown in the figure. A detector is placed at P (mid-point of hypotenuse). The intensity of light as detected by detector is (a) maximum (b) zero (c) neither maximum nor minimum (d) insufficient data
9. A transverse wave travels along a string of material of specific gravity 10. If maximum particle velocity is one-tenth of wave velocity v, then intensity of the wave in the string (in SI units) is (a) 0.05v3 (b) 5v3 (c) 50v3 (d) 5 × 105 v3
Q
(d)
W
W
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15. A string of length L is stretched along the x-axis and is rigidly clamped at its two ends. It undergoes transverse vibration. If n is an integer, which of the following relation may represent the shape of the spring at any time t?
(c)
Q
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14. Sound waves of frequency 600 Hz fall normally on a perfectly reflecting wall. The distance from the wall at which the air particles have the maximum amplitude of vibration is (speed of sound in air = 330 m/s) (a) 13.75 cm (b) 41.25 cm (c) 68.75 cm (d) all of the these
s: // ht tp
(b)
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Q
13. The frequency of a sonometer wire is f. The frequency becomes f/2 when the mass producing the tension is completely immersed in water and on immersing the mass in a certain liquid, frequency becomes f/3. The relative density of the liquid is (a) 1.32 (b) 1.03 (c) 1.41 (d) 1.18
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(b) 3 I (d) 5 I
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(a) 2 I (c) 4 I
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10. A railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside a railway track. The frequency (n) of the sound heard by the observer is plotted against time (t). Which of the following best represents the resulting curve?
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12. Two source of sound of the same frequency produce sound intensities I and 4I at a point P when used individually. If they are used together such that the sounds from them reach P with a phase 2π difference of , the intensity at P will be 3
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8. A source of sound of frequency 1020 Hz is moving towards a wall with a velocity 20 m/s. The position of a doctor D is such that the detector is in between the source and the wall, as shown in figure. If the speed of sound is 320 m/s, the number of beats recorded by the detector per second is (a) 128 (b) 68 (c) 60 (d) zero
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(a) (b) (c) (d)
nπx (a) y = A sin cos ωt L nπx (b) y = A sin sin ωt L
16. A wire having a mass per unit length m and tension T resonates at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. The length of the wire is
in f
s_
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π (c) y ( x, t ) = A sin ωt − kx − 2
//t
π (d) y ( x, t ) = A sin ωt + kx − 2
22. A man wants to produce sound of frequency 680 Hz using a closed organ pipe of length 34 cm. Suggest him. (a) break the close part of pipe. (b) apply more and more pressure to produce then open end. (c) make a hole at a distance 25 cm from open end. (d) closed the other part of pipe also because it is not possible.
s:
ht tp
er s
π (b) y ( x, t ) = A sin ωt + kx + 2
//t
ht
(b) 0.5 m/s (d) 50 m/s
23. A tuning fork of frequency 340 Hz is sounded above an organ pipe of length 120 cm. Water is now slowly poured in
19. A standing wave is produced in a vapour of a chemical of atomic mass (127 g mol–1) at 400 K. Nodes are found to be 9.57 cm apart when frequency of source was 1000 Hz. The vapour used for the experiment is (R = 8.31 JK–1 mol–1)
(a) 5 m/s (c) 15 m/s
rio
π (a) y ( x, t ) = A sin ωt − kx + 2
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s:
tp
18. A stationary observer receives sound waves from two tuning forks, one of which approaches and the other recedes with the same velocity. As this takes place, the observer hears beats with frequency 2 Hz. What is velocity of each tuning fork if their oscillation frequency v = 680 Hz and the velocity of sound in σy air is vs = 340 m/s?
s
ni ve
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s:
tp
ht
rs
w
17. An air column in a pipe which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 330 Hz if the length of the column is (a) 50 cm (b) 75 cm (c) 150 cm (d) 100 cm
21. A sinusoidal wave (longitudinal or transverse) is propagating through a medium in the direction of negative x-axis. The parameters of the waves are A, ω and k. The particle at x = λ/4 executes the motion y (t) = A sin ωt. Possible equation of the wave is
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m
(d)
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3 T 490 m
(c)
(b) 210
1 T 140 m
(a)
co m
20. When an open organ pipe resonates in its fundamental mode, then at the centre of the pipe, (a) the gas molecules undergo vibration of maximum amplitude (b) the gas molecules are at rest (c) the pressure of the gas is constant (d) the pressure of the gas undergoes maximum variation
nπx (d) y = A cos sin ωt L
monatomic diatomic polyatomic can not determine
nπx (c) y = A cos cos ωt L
3.39
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3.40
(c)
Q
24. An open pipe of sufficient length is dipping in water with a speed v vertically. If at any instant l is length of tube above water, then the rate at which fundamental frequency of pipe changes is [c is the speed of sound in air]
fo
s_ in
s
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28. If the velocity amplitude of a particle acted upon by a force f = f0 cos ωt along x-axis is given by 1
v0 =
2
aω − 2bω + 3c
b a
(b) b 2 = 6ac,
(a) b 2 = ac,
W
then the condition for the resonance and the angular resonance frequency respectively are
Q
(a)
ta
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tp
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e/
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s: //
25. A railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside the railway track. The frequency (n) of the sound heard by the observer is plotted against time (t). Which of the following best represents the resulting curve?
27. Two sound waves of same amplitude and same frequency of 100 Hz travel in opposite direction in nitrogen gas at 200 K. What is the approximate distance between two consecutive nodes of the standing wave so formed? (R = 8 J mol–1 K–1) (a) 2.8 m (b) 1.4 m (c) 0.7 m (d) 5.6 m
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ht
cv 4l 2 cv (d) 4v 2 t 2
tp s:
cv 2l 2 cv (c) 2V 2t 2 (a)
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26. A plane wave of sound travelling in air is incident upon a plane surface of a liquid. The angle of incidence is 60º. The speed of sound in air is 300 m/s and in the liquid it is 600 m/s. Assume Snell’s law to be valid for sound waves, the wave will (a) refract into liquid away from normal (b) refract into liquid towards the normal (c) reflect back into air (d) none of these
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a b
b ab (d) b 2 = 3ac, a bc 29. A motorcyclist going around a circular track of radius 50 m with a speed of 25 m/s, is at a point X with a speed of 25 m/s.
(c) b 2 = 4ac,
W
it. The minimum height of water column required for resonance is (speed of sound in air = 340 m/s) (a) 25 cm (b) 95 cm (c) 75 cm (d) 45 cm
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5. A wave is expressed by the equation y = 0.5 sin π (0.01 x – 3 t)
where y and x are in metres and t in seconds. Find the speed of propagation. [Roorkee, 1981] [Ans. 300 m/s]
3. The velocity of sound in a tube containing air at 27ºC and a pressure of 76 cm of mercury is 330 m/s. What will be its velocity when the pressure is increased to
2. An engine approaches a hill with a constant speed. When it is at a distance of 0.9 km, it blows a whistle whose echo is heard by the driver after 5 sec. If the speed of sound in air is 330 m/s, calculate the speed of the engine. [IIT, 1966] [Ans. 30 m/s]
4. A tuning fork whose frequency as given by the manufacturer is 512 Hz is being tested with an accurate oscillator. It is found that the fork produces a beat of 2 Hz when oscillator reads 514 Hz but produces a beat of 6 Hz when oscillator reads 510 Hz. Find the actual frequency of the fork. [Ans. 516 Hz]
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100 cm of mercury and the temperature is kept constant? [IIT, 1976] Ans. 330 m/s
//t
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8. (d) 16. (a) 24. (b)
ht (
7. (b) 15. (a,b) 23. (d)
(c) (d) (c) (a)
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6. 14. 22. 30.
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EXERCISE 1
1. The temperature between two points A and B, separated by a distance l, varies linearly from T1 to T2. If the velocity of sound propagation in air is equal to v = α T , where α is a constant. Find how long will it take the sound waves to travel from the point A to point B? 2l Ans. t = α T2 + T1
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ANSWERS
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30. When an observer is moving towards a stationary sound source with uniform velocity vo, the fractional change in frequency is found to be x. Now, if the same observer moves away from the stationary sound source with the same uniform speed v0, then the fractional change in frequency will be numerically (a) equal to x (b) greater than x (c) less than x (d) nothing can be predicted
A static siren at Y is emitting sound of frequency n. How many times (approximately) in an hour will the motor cyclist hear the sound of actual frequency Y? (a) 24 (b) 287 (c) 600 (d) 573
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Molar heat capacity at constant volume = Cv = 2.5 R
12. Calculate the speed of sound in oxygen from the following data:
6. A stone is dropped into a well and its splash is heard at the mouth of the well after an interval of 1.45 seconds. Find the depth of the well. Given that velocity of sound in air at room temperature is equal to 332 m/s. [MNR, 1981] [Ans. 9.9 m]
Molar heat capacity at constant pressure = Cp = 3.5 R
Pressure = P = 1.013 × 105 N/m2
7. The equation of a wave travelling in a string can be written as
Temperature = T = 273 K
Mass of 22.4 litre of oxygen at NTP = 32 g. Ans. 315.08 m/s] 13. The intensity of sound from a point source is 1.0 × 10–8 W/m2 at a distance of 10 m from the source. Find the intensity at a distance of 100 m from the source. [Ans. 1.0 × 10–10 W/m2]
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15. A string tied between x = 0 and x = l vibrates in fundamental mode. The amplitude A, tension T and mass per unit length µ are given. Find the total energy of the string. [IIT, 2003] A 2 π2 T ] [ Ans. E = 4l 16. A wire having a linear density of 0.05 gm/cc is stretched between two rigid supports with a tension of 4.5 × 107 dynes. It is observed that the wire resonates at a frequency of 420 cycles/sec. The next highest frequency at which the same wire resonates is 490 cycles/sec. Find the length of the wire. [IIT, 1971] [Ans. 214.28 cm]
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11. Two speakers connected to the same source of fixed frequency are placed 2.0 m apart in a box. A sensitive microphone placed at a distance of 4.0 m from their mid-point along the perpendicular bisector shows maximum response. The box is slowly rotated till the speakers are in a line with the microphone. The distance between the mid-point of the speakers and the microphone remains unchanged. Exactly 5 maximum responses are observed in the microphone in doing this. Calculate the wavelength of sound wave. [Roorkee, 1989] [Ans. λ = 0.4 m]
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10. The speed of sound in hydrogen is 1270 m/s. Calculate the speed of sound in a mixture of oxygen and hydrogen in which they are mixed in the ratio 1 : 3. [Ans. 582.72 m/s]
14. Two audio speakers, separated from one another by a certain distance, are driven by the same amplifier system. A listener, in line with the speakers and in between them is at a distance of 6 m from one speaker and 6.4 m from the other. If the frequency of the source is varied continuously from 1000 Hz to 4000 Hz, find the frequencies for which there is destructive interference at the place of the listener. The speed of sound in air is 320 m/s. [Ans. υ = 1200 Hz, 2000 Hz, 2800 Hz, 3600 Hz.]
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9. A progressive wave of frequency 550 Hz is travelling with a velocity of 360 m/s. How far apart are two points 60º out of phase? [MNR, 1993] [Ans. 10.9 cm]
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where x and y are in cm and t is in seconds. Find its frequency. [MNR, 1986] [Ans. 1 sec–1]
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8. The equation of a transverse wave is given by
y = 3 cos π (100 t – x). Find its wavelength. [MNR, 1985] [Ans. λ = 2 cm]
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24. A copper wire is held at the two ends by rigid supports. At 30ºC, the wire is just taut, with negligible tension. Find the speed of transverse waves in this wire at 10ºC (α = 1.7 × 10–5/ºC, Y = 1.3 × 1011 N/ m2, d = 9 × 103 kg/m3). [IIT, 1979] [Ans. 70.08 m/s.]
25. A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of vibration is found to be 320 Hz. The velocity of sound in air is 320 m/s. Estimate the diameter of the tube. One end of the tube is now closed. Calculate the lowest frequency of resonance for the tube. [IIT, 1980] [Ans. 163.27 Hz.]
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20. A tuning fork having frequency of 340 vib/sec is vibrated just above a cylindrical tube. The height of the tube is 120 cm. Water is slowly poured in it. What is the minimum height of water required for resonance? Velocity of sound in air = 340 m/s. [IIT, 1975] [Ans. 0.45 m.]
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[IIT, 1974]
23. AB is a cylinder of length 1.0 m fitted with a thin flexible diaphragm C at the middle and two other thin flexible diaphragms A and B at the ends. The portions AC and BC contain hydrogen and oxygen gases respectively. The diaphragms A and B are set into vibrations of same frequency. What is the minimum frequency of these vibrations for which the diaphragm C is a node. Under the conditions of the experiment, the velocity of sound in hydrogen is 1100 m/s and in oxygen is 300 m/s. [IIT, 1978] [Ans. 11 × 150 = 1650 Hz.]
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22. A column of air at 51ºC and a tuning fork produces 4 beats per second, when sounded together. As the temperature of air column is decreased, the number of beats per second tends to decrease and when the temperature is 16ºC, the two produce 1 beat per second. Find the frequency of the tuning fork. [IIT, 1977] [Ans. n = 50 Hz]
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19. A pipe of length 1.5 m closed at one end is filled with a gas and it resonates in its fundamental mode with a tuning fork. Another pipe of the same length but open at both ends is filled with air and it resonates in its fundamental with the same tuning fork. Calculate the velocity of sound at 0ºC in the gas, given that the velocity of sound in air is 360 m/s at 30ºC where the experiment is performed. [Ans. 683.42 m/s]
21. A wire of density 9 gm/cm3 is stretched between two clamps 1.00 m apart while subjected to an extension of 0.05 m. What is lowest the frequency of transverse vibrations in the wire? Young’s modulus Y = 9 × 1010 N/m2. [IIT, 1975] [Ans. 36.27 Hz]
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18. An underwater swimmer sends a sound signal to the surface. If it produces 5 beats/sec when compared with the fundamental tone of a pipe 20 cm length closed at one end, what is the wavelength of sound in water? (Velocity of sound in air = 360 m/s, velocity of sound in water = 1450 m/s) [IIT, 1973] [Ans. λ´ = 3.258 m or 3.187 m]
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17. A sonometer wire fixed at one end has a solid mass M hanging from its other end to produce tension in it. It is found that a 70 cm length of the wire produces a certain fundamental frequency when plucked. When the same mass M is hanging in water completely submerged, it is found that the length of the wire has to be changed by 5 cm in order that it will produce the same fundamental frequency. Calculate the density of the material of mass M hanging from the wire. [IIT, 1972] 3 3 [Ans. 7.26 × 10 kg/m ]
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26. A string 25 cm long and having a mass of 2.5 gm is under tension. A pipe closed at one end is 40 cm long. When the string is vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats/sec are heard.
It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string. [IIT, 1982] [Ans. T = 27.04 N]
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12. Water waves can travel in any direction when far offshore, but they always arrive at the beach nearly head on. Explain.
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1. A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s. (a) 1 m/s (b) 2 m/s (c) 0.5 m/s (d) 1.5 m/s
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8. Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, Explain.
3. If the equation of a progressive wave is t x π given by y = 4sin π − + then which 5 9 6 of the following is correct? (a) v = 5 cm/s (b) λ = 18 m (c) A = 0.04 cm (d) f = 50 Hz
7. A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, Explain.
6. Why is it possible to understand the words spoken by two people at the same time?
2. A note has a frequency of 128 Hz. The frequency of a note which is two octave higher than this is (a) 256 Hz (b) 320 Hz (c) 400 Hz (d) none of these
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4. In a sound wave, a displacement node is a pressure antinode and vice versa, Explain.
11. The shape of a pulse gets distorted during propagation in a dispersive medium.
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3. The piano strings that vibrate with the lowest frequencies consist of a steel wire around which a thick coil of copper wire is wrapped. Only the inner steel wire is under tension. What is the purpose of the copper coil?
10. The wavelength of the fundamental standing wave on a cello string depends on which of these quantities: length of the string, mass per unit length of the string, or tension? The wavelength of the sound wave resulting from the string’s vibration depends on which of the same three quantities?
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2. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all (a) y = 2 cos (3x) sin (10t) (b) y = 2 x − vt (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2t
9. Why is transverse wave sometimes called a shear wave?
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1. When connecting speakers to a stereo, it is important to connect them with the correct polarity so that, if the same electrical signal is sent, they both move in the same direction. If the wires going to one speaker are reversed, the listener hears a noticeably weaker bass (low frequencies). Explain what causes this and why low frequencies are affected more than high frequencies.
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3.46
(b)
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13. Equation of a stationary and travelling waves are as follows y1 = a sin kx cos ωt and y2 = a sin (ωt – kx). The phase difference π 3π between two points x1 = and x2 = 3k 2k is φ1 in the standing wave (y1) and is φ2 in
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14. In a sonometer wire, the tension is maintained suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m3. The fundamental frequency of the wire is 260 Hz. If the suspended mass is completely submerged
9. The extension in a string, obeying Hooke’s law, is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be (a) 1.22 v (b) 0.61 v (c) 1.50 v (d) 0.75 v
(a) 1 (c) 3/4
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8. A string fixed at both ends has consecutive standing wave modes for which the distances between adjacent nodes are 18 cm and 16 cm respectively. The minimum possible length of the string is (a) 144 cm (b) 152 cm (c) 176 cm (d) 200 cm
12. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is (a) 25 Kg (b) 5 Kg (c) 12.5 Kg (d) 1/25 Kg
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7. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is (a) 242/252 (b) 2 (c) 5/6 (d) 11/6
11. The intensity of a plane progressive wave of frequency 1 kHz is 10–10 watt/m2. If the density of air is 1.3 kg/m3 and the speed of sound is equal to 330 m/s, then pressure amplitude of the wave is (a) 3 × 10–5 N/m2 (b) 3 × 10–4 N/m2 (c) 3 × 10–6 N/m2 (d) 3 × 10–3 N/m2
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6. A sound wave of frequency 440 Hz is passing through air. An O2 molecule (mass = 5.3 × 10–26 kg) is set in oscillation with an amplitude of 10–6 m. Its speed at the centre of its oscillation is (a) 1.70 × 10–5 m/s (b) 17.0 × 10–5 m/s (c) 2.76 × 10–3 m/s (d) 2.77 × 10–5 m/s
(d) None of these
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5. The frequency of a man’s voice is 300 Hz and its wavelength is 1 meter. If the wavelength of a child’s voice is 1.5 m, then the frequency of the child’s voice is (a) 200 Hz (b) 150 Hz (c) 400 Hz (d) 350 Hz
10. If the source is moving towards right, wave front of sound waves get modified to
4. Two coherent sources of different intensities send waves which interfere. The ratio of the maximum intensity to the minimum intensity is 25. The intensities are in the ratio (a) 25 : 1 (b) 5 : 1 (c) 9 : 4 (d) 625 : 1
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in water, the fundamental frequency will become (a) 240 Hz (b) 220 Hz (c) 230 Hz (d) 280 Hz
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19. The equation of plane progressive wave motion is y = a sin 2π/λ (vt – x). Velocity of particle is
24. Two pulses is a stretched string whose centers are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 second, the total energy of the pulses will be
23. A man standing between two cliffs hears the first echo of a sound after 2 sec and the second echo 3 sec after the initial sound. If the speed of sound be 330 m/s, the distance between the two cliffs should be (a) 1650 m (b) 990 m (c) 825 m (d) 660 m
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18. A taut string at both ends vibrates in its nth overtone. The distance between adjacent Node and Antinode is found to be ‘d’. If the length of the string is L, then (a) L = 2 d (n + 1) (b) L = d (n + 1) (c) L = 2 dn (d) L = 2 d (n – 1)
21. The average density of earth’s crust 10 km beneath the surface is 2.7 gm/cm3. The speed of longitudinal seismic waves at that depth is 5.4 km/s. The bulk modulus of earth’s crust considering its behaviour as fluid at that depth is (a) 7.9 × 1010 Pa (b) 5.6 × 1010 Pa (c) 7.9 × 107 Pa (d) 1.46 × 107 Pa
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17. Two monoatomic ideal gases 1 and 2 of molecular masses m1 and m2 respectively are enclosed in separate containers kept at the same temperature. The ratio of the speed of sound in gas 1 to that in gas 2 is given by
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16. In the resonance tube experiment, the first resonance is heard when length of air column is ℓ1 and second resonance is heard when length of air column is ℓ2. What should be the minimum length of the tube so that third resonance can also be heard? (a) 2ℓ2 – ℓ1 (b) 2ℓ1 (c) 5ℓ1 (d) 7ℓ1
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20. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment, the displacement of the wire is y1 = A sin (πx/L) sin ωt and energy is E1 and in another experiment its displacement is y2 = A sin(2πx/L) sin 2ωt and energy is E2. Then (a) E2 = E1 (b) E2 = 2E1 (c) E2 = 4E1 (d) E2 = 16E1
15. The beat frequency produced by two tuning forks when sounded together is observed to be 4 Hz. One of the forks makes 384 vibrations per second. When the other fork is loaded with a small piece of wax, the beats disappear first. The frequency of the second tuning fork is (a) 388 Hz (b) 380 Hz (c) more than 388 Hz (d) less than 380 Hz
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28. Organ pipe P1 closed at one end vibrating in its first harmonic and another pipe P2 open at both ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is (a) 8/3 (b) 1/6 (c) 1/2 (d) 1/3
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26. Choose the correct option A string of length L, fixed at its both ends is vibrating in its 1st overtone mode. Consider two elements of the string of same small length at positions ℓ1 = 0.2 L and ℓ2 = 0.45 L from one end. If K1 and K2 are their respective maximum kinetic energies then (a) K1 = K2 (b) K1 > K2 (c) K1 < K2 (d) It is not possible to decide the relation
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25. A source S emits electromagnetic waves which are detected by the detector D kept at a distance of d/4 from the source. The waves directly received from S and those reflected by a layer at a distance d above the ground are found to be in phase. If the layer moves up by a distance h = 0.01d, the interfering waves are once again in phase. The wavelength of the waves from the source is about
(a) (b) (c) (d)
27. A source of frequency f is stationary and an observer starts moving towards it at t = 0 with constant small acceleration. Then the variation of observed frequency f registered by the observer with time is best represented as
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(b) 1/2 (d) 19/18
35. The amplitude of a wave disturbance propagating in the positive x-direction is given by 1 1 y= at t = 0 and y = at t 2 1+ x 2 + x2 − 2x = 2s, where x and y are in metre. Assuming that the shape of the wave disturbance does not change during the propagation, the speed of the wave is (a) 0.5 m/s (b) 1 m/s (c) 1.5 m/s (d) 2 m/s
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37. A racing car moving towards a cliff sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If V = the velocity of sound, the velocity of the car is (b) V/2 (a) V / 2 (c) V/3 (d) V/4
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39. A tube 12 m long and weighing 900 g is fastened to a fixed support at one end. The other end of tube is attached by a rope and it passes over a pulley and it supports a 5 kg block.
34. A train moves towards a stationary observer with speed 34 m/s. The train sounds whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s the frequency registered is f1. If the speed of sound is 340 m/s, then the ratio f1/f2 is
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33. At a certain instant, a stationary transverse wave is found to have maximum kinetic energy. The appearance of string at that instant is (a) sinusoidal shape with amplitude A/3 (b) sinusoidal shape with amplitude A/2 (c) sinusoidal shape with amplitude A (d) straight line
36. Two vibrating tuning forks produce progressive waves given by y1 = 4 sin (500πt) and y2 = 2 sin (506πt). These tuning forks are held near the ear of a person. The person will hear (a) 3 beats/s with intensity ratio between maxima and minima equal to 2 (b) 3 beats/s with intensity ratio between maxima and minima equal to 9 (c) 6 beats/s with intensity ratio between maxima and minima equal to 2 (d) 6 beats/s with intensity ratio between maxima and minima equal to 9
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32. Three coherent sonic sources emitting sound of single wavelength ‘λ’ are placed on the x-axis at points → (−λ 11/6, 0), (0, 0), (λ 11/6, 0). The intensity reaching a point (0, 5λ/6) from each source has the same value I0. Then the resultant intensity at this point due to the interference of the three waves will be (a) 6 I0 (b) 7 I0 (c) 4 I0 (d) 5 I0
31. In resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air columns is 19 cm. If the 400 Hz tuning fork is replaced by 1600 Hz tuning fork then to get resonance, the water level in the tube should be further lowered by (take end correct = 1 cm.) (a) 5 cm (b) 10 cm (c) 15 cm (d) 20 cm
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30. A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/sec. If sound travels at a speed of 330 m/sec, then number of beats per second heard by an observer between the wall and the source is (a) 7.7 Hz (b) 9 Hz (c) 4 Hz (d) None of these
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Wave Mechanics FP
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44. Two vibrating strings of the same material but length L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency v1 and the other with frequency v2. The ratio v1/v2 is given by (a) 2 (b) 4 (c) 8 (d) 1
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42. A wave pulse on a string has the dimension shown in figure. The wave speed is v = 1 cm/s. If point O is a free end. The shape of wave at time t = 3 s is FP
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41. Under similar conditions of temperature and pressure, in which of the following gases, the velocity of sound will be largest? (a) H2 (b) N2 (c) He (d) CO2
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43. Two waves are passing through a region in the same direction at the same time. If the equa2π tions of these waves are y1 = a sin (vt − x) λ 2π and y2 = b sin [vt − x) + x0 ] then the λ amplitude of the resulting wave for x0 = (λ/2) is (a) |a – b| (b) a + b (c) a 2 + b 2
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46. A source of frequency f gives 5 beat/s when sounded with source of frequency 200 Hz. The second harmonic of the source gives 10 beat/s when sounded with a source of frequency 420 Hz. The value of f is (a) 200 Hz (b) 210 Hz (c) 205 Hz (d) 195 Hz
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54. When a sound wave is reflected from a wall, the phase difference between the reflected and incident pressure wave is (a) 0 (b) π (c) π/2 (d) π/4
55. A wave travels uniformly in all directions from a point source in an isotropic medium. The displacement of the medium at any point distance r from the source may be
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50. A composite string is made up by joining two strings of different masses per unit length → µ and 4µ. The composite string is under the same tension. A transverse wave pulse Y = (6 mm) sin (5 t + 40 x), where ‘t’ is in seconds and ‘x’ is in metres, is sent along the lighter string towards the joint.
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53. A closed organ pipe has length ℓ. The air it is vibrating in 3rd overtone with maximum amplitude ‘a’. The amplitude at a distance of ℓ/7 from closed end of the pipe is equal to (a) a (b) a/2 a 3 (c) (d) zero 2
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49. An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is (a) 200 Hz (b) 300 Hz (c) 240 Hz (d) 480 Hz
48. When beats are produced by two progressive waves of nearly the same frequency, which one of the following is correct? (a) The particles vibrate simple harmonically, with the frequency equal to the difference in the component frequencies. (b) The amplitude of vibration at any point changes simple harmonically with a frequency equal to the difference in the frequencies of the two waves (c) The frequency of beats depends upon the position, where the observer is (d) The frequency of beats changes as the time progresses
Thus the periodic motion consists of three components.
51. A pipe open at the top end is held vertically with some of its lower portion dipped in water. At a certain depth of immersion, the air column of length (3/8) m in the pipe resonates with a tuning fork of frequency 680 Hz. The speed of sound in air is 340 m/s. The pipe is now raised up by a distance ‘a’ until it resonates in the “next overtone” with the same tuning fork. The value of ‘x’ is (a) 20 cm (b) 40 cm (c) 50 cm (d) 25 cm
47. The displacement y of a particle executing periodic motion is given by y = 4 cos2 (t) sin (1000t). This expression may be considered to be a result of the superposition of waves (a) two (b) three (c) four (d) five
The joint is at x = 0. The equation of the wave pulse reflected from the joint is (a) (2 mm) sin (5 t – 40 x) (b) (4 mm) sin (40 x – 5 t) (c) –(2 mm) sin (5 t – 40 x) (d) (2 mm) sin (5 t – 10x)
Wave Mechanics
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Wave Mechanics
beat frequency of 4 Hz is heard. For forks A and Y, the beat frequency is 7 Hz. For forks X and Y, the beat frequency is 3 Hz. (a) The frequency of X is 436 Hz (b) The frequency of Y is 447 Hz (c) The frequency of X is 444 Hz (d) The frequency of Y is 433 Hz
represented by (A is a constant representing strength of source) A (a) sin (kr − ωt ) r (b) A r sin (kr − ωt )
(c) Ar sin(kr – ωt) A /r sin (kr − ωt )
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5. Standing waves are produced on a stretched string of length L with fixed ends. When there is a node at a distance L/3 from one end, then (a) minimum and next higher number of nodes excluding the ends are 2, 5 respectively (b) minimum and next higher number of nodes excluding the ends are 2, 4 respectively (c) frequency produced may be V/3 L (d) frequency produced may be 3V/2 L [V = Velocity of waves in the string]
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1. In a resonance tube experiment, a close organ pipe of length 120 cm resonates when tune with a turning fork of frequency 340 Hz. If water is poured in the pipe then (given vair = 340 m/sec) (a) minimum length of water column to have the resonance is 45 cm. (b) the distance between two successive nodes is 50 cm. (c) the maximum length of water column to create the resonance is 95 cm. (d) None of these
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56. The wires X and Y of a guitar produce 4 beats per second. If the tension of Y is raised, then the number of beats becomes 2 per second. If the frequency of X is 300, then frequency of Y (a) 296 (b) 298 (c) 300 (d) 294
4. A transverse sinusoidal wave of amplitude a, wavelength λ and frequency f is traveling on a stretched string. The maximum speed of any point on the string is v/10 where v is the speed of propagation of the wave. If a = 10–3 m and v = 10 ms–1, then λ and f are given by (a) λ = 2π × 10–2 m (b) λ = 10–3 m 3 (c) f = 10 /(2π) Hz (d) f = 104 Hz
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12. Velocity of sound in air is 320 ms–1. A pipe closed at one end has a length of 1 m. Neglecting end correction, the air column in the pipe can resonate for sound of frequency.
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11. In a wave motion y = a sin(kx – ωt), y can represent (a) electric field (b) magnetic field (c) displacement (d) pressure
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10. y-x curve at an instant for a wave travelling along x-axis on a string is shown. Slope at the point A on the curve, as shown, is 53º.
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3.53
9. The vibration of a string fixed at both ends are described by Y = 2 sin (πx) sin (100πt) where Y is in mm, x is in cm, t in sec, then (a) Maximum displacement of the particle at x = 1/6 cm would be 1 mm. (b) velocity of the particle at x = 1/6 cm at time t = 1/600 sec will be 157 3 mm/s (c) If the length of the string be 10 cm, number of loop in it would be 5 (d) None of these
7. A string of length L is stretched along the x-axis and is rigidly clamped at its two ends. It undergoes transverse vibration. If n is an integer, which of the following relations may represents the shape of the string at any time t?
Wave Mechanics
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3.54
Q
ASSERTION AND REASON UESTIONS
Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
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4. Statement 1: When two coherent waves of intensity I1 and I2 are superimposed with a constant phase difference φ, then the intensity of resultant wave is given by I = I1 + I 2 + 2 I1 I 2 cos φ Statement 2: For two incoherent sources, resultant intensity is given by I = I1 + I2.
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a is reflected from a rigid wall b is reflected from a yielding surface c is reflected from a yielding surface d is reflected from a rigid wall
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2. Statement 1: An 80 dB sound has twice the intensity of a 40 dB sound. Statement 2: Loudness of a sound of a certain intensity I is defined as: L (in dB) = I 10 log10 I0 3. Statement 1: An open tube having fundamental frequency f generates the same frequency when half immersed in a dense liquid. Statement 2: On immersing the tube in a liquid, the open tube behaves as a closed pipe.
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1. Statement 1: All points on a wavefront vibrate in same phase with same frequency. Statement 2: Two sources are said to be coherent if they produce waves of same frequency with a constant phase difference.
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(a) Statement 1 is True, Statement 2 is True, Statement 2 is a correct explanation for Statement 1. (b) Statement 1 is True, Statement 2 is True, Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True.
13. In the figure, A is an incident pulse and a, b, c, d are the possible forms of the reflected pulse. Tick the correct answer.
(b) 240 Hz (d) 400 Hz
(a) 80 Hz (c) 320 Hz
5. Statement 1: A plane wave of sound travelling in air is incident upon a plane water surface. Angle of incidence is θ. If Snell’s law is valid for sound waves, it follows that
Wave Mechanics
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Wave Mechanics
Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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10. Statement 1: A person is standing near a railway track A train is moving on the track. As train is approaching the person, apparent frequency keeps on increasing and when train has passed the person then apparent frequency keeps on decreasing. Statement 2: When train is approach
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9. Statement 1: A transverse waves are produced in a very long string fixed at one end. Only progressive wave is observed near the free end. Statement 2: Energy of reflected wave does not reach the free end.
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1. For four sine waves, moving on a string along positive x-direction, displacementdistance curves (y-x curves) are shown at time t = 0. In the right column, expressions for y as function of distance x and time t for sinusoidal waves are given. All terms in the equations have general meaning. Correctly match y-x curves with corresponding equations.
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7. Statement 1: Like light; sound waves can be polarized. Statement 2: Polarization is a phenomenon related to all waves.
8. Statement 1: The intensity of a plane progressive wave does not change with change in distance from the source. Statement 2: The wavefronts associated with a plane progressive wave are planar.
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6. Statement 1: A tuning fork vibrated near the open end of the tube vibrate the air column above the water in it. At the surface of water in the tube, with reference to pressure, the status will be antinode. Statement 2: Displacement minima corresponds to pressure maxima.
sound will be refracted into water away from normal. Statement 2: From Snell’s law, angle of refraction is more than angle of incidence when wave travel from denser to rarer medium.
3.55
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3.56
Column II y = A cos (ωt – kx) y = – A cos (kx – ωt) y = A sin (ωt – kx) y = A sin (kx – ωt)
(c) y = 10 sin (2πx – 120t) + 10 cos (120t + 2πx) (d) y = 10 sin (2πx – 120t) + 8 cos (118t – 59/30πx)
2. A string of linear mass density 5 × 10–3 kg/m is stretched under a tension of 72 N between two rigid supports 60 cm apart. The string is vibrating in second overtone so that amplitude at one of its antinode is 0.25 cm (values in Column II are in CGS unit) Column II Column I 150 π (a) maximum (p) 2 velocity of a particle at antinode 2 (b) maximum (q) 90000π 2 velocity of a particle at 5 cm from any node (c) maximum (r) (150 π) acceleration of a particle at antinode (d) maximum (s) 90000 π2 acceleration of particle at 5 cm from any node
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5. A closed organ pipe P1 vibrating in its first overtone and another open pipe P2 vibrating in its third overtone are in resonance with the tuning fork of frequency 12 Hz. In column II, the beats frequency in Hz is given when produced in conditions provided in column I. Match column I and II.
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Column II Particles at every position are performing SHM Equation of travelling wave Equation of standing wave Equation of Beats
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4. Match columns I and II. Column I (a) y = 4 sin (5x – 4t) + 3 cos (4t – 5x + π/6)
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Column I (a) P1 resonate in 5th harmonic and P2 resonate in 3rd overtone (b) P1 resonate in 1st overtone and P2 resonate in 3rd harmonic (c) P1 resonate in 9th harmonic and P2 resonate in 7th overtone (d) P1 resonate in 2nd overtone and P2 resonate in 4th overtone
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Wave Mechanics
PASSAGE–1 A plane pressure pulse triangular in shape approaches a rigid wall along normal at a speed of 400 m/s. At time t = 0, situation is shown in the figure. The peak pressure is 100 P. By the wall pulse gets reflected and pressure near the wall
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Wave Mechanics
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11. The transverse of the particle at x = 13 cm and t = 0.015 s will be (a) –250 cm/s (b) –500 cm/s (c) 500 cm/s (d) –1000 cm/s
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10. The vertical displacement of the particle of the string at x = 7 cm and t = 0.01 s will be (a) 0.75 cm (b) 0.5 cm (c) 0.25 cm (d) zero
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Here f and x are in centimeters. The linear mass density of the string is 50 g/m and it is under a tension of 5 N.
The figure represents the instantaneous picture of a transverse harmonic wave travelling along the negative x-axis.
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A pulse is started at a time t = 0 along the +x direction on a long, taut string. The shape of the pulse at t = 0 is given by function f(x) with
4. Total impulse imparted by the pulse on the wall will be (a) 300 Ns (b) 150 Ns (c) 750 Ns (d) none of these
PASSAGE–2
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3.57
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3.58
PASSAGE–4
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16. If m2 < m1/2, then (a) vD > vB > vA > vC (b) vB > vA > vD = vC (c) vD > vA = vB > vC (d) vD > (vA = vB) > vC
13. The phenomena of beats can take place for (a) Only transverse waves (b) Only longitudinal waves (c) Both longitudinal and transverse waves (d) For sound waves only
17. If m1 > m2 > m1/2, then (a) vD > vC > vB > vA (b) vA = vB > vD > vC (c) vB > vA > vC > vD (d) vA = vB > vC > vD
14. The frequency of beats produced air when two sources of sound are activated, one emitting wavelength 32 cm, other 32.2 cm is (Take Vsound = 350 m/s)
In the figure, four strings are placed under tension by one or two suspended blocks, all of the same mass. Strings A, B and C have the same linear density m1, string D has linear mass density m2. The pulleys and strings are frictionless and massless. The velocity of waves in the strings are vA , vB, vC and vD respectively.
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12. If two sound sources of frequency difference 25 Hz are sounded together. Then which of the following is correct? (a) A normal human ear will hear 25 Hz beat frequency (b) A normal human ear will hear only 10 Hz beat frequency (c) A normal human ear-cannot hear this frequency (d) A normal human ear can hear maximum of the two frequency sounded together
PASSAGE–5
ω − ω2 ω1 + ω2 x = 0, y = 2 A cos 1 .t . t sin 2 2
Thus amplitude frequency is small and fluctuates slowly. A beat i.e., a maximum of intensity occurs, also intensity depends on square of amplitude. The beat frequency is given by ωbeat = |ω1 – ω2| Number of beats per second is called as beat frequency. A normal ear can detect only upto 15 Hz of frequency because of persistence of ear.
15. A tuning fork of unknown frequency makes 3 beats per second with a standard fork of frequency 384 Hz. The beat frequency decreases when wax is put on prongs of first fork and the frequency of this fork is (a) 381 Hz (b) 387 Hz (c) 384 Hz (d) None of above
Superposition of waves results in maximum and minimum of intensities such as in case of standing waves. This phenomenon is called as interference. Another type of super-position result in interference in time which is called as beats. In this case waves are analysed at a fixed point as a function of time. If the two waves are of nearby same frequency are superimposed, at a particular point, intensity of combined waves gives a periodic peak and fall. This phenomenon is beats. If ω1 and ω2 are the frequencies of two waves then by superimposed y = y1 + y2, we get at
(b) 18 (d) 10
(a) 14 (c) 7
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Wave Mechanics
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3.59
Wave Mechanics
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2. A straight line source of sound of length L = 10 m, emits a pulse of sound that travels radially outward from the source. What sound energy (in mW) is intercepted by an acoustic cylindrical detector of surface area 2.4 cm2, located at a perpendicular distance 7 m from the source. The waves reach perpendicularly at the surface of the detector. The total power emitted by the source in the form of sound is 2.2 × 104 W (Use π = 22/7)
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5. Sound from two coherent sources S1 and S2 are sent in phase, and detected at point P, equidistant from both the sources. Speed of sound in normal air is V0, but in some part in path S1, there is zone of hot air having temperature 4 times, the normal temperature, and width d. What should be minimum frequency of sound, so that minima can be found at P?
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4. Three successive resonance frequencies for a certain string are 75, 125 and 175 Hz (a) Find the ratios of each pair of successive resonance frequencies. (b) How can you tell that these frequencies are for a string fixed at one end only rather than for a string fixed at both ends? (c) What is the fundamental frequency? (d) Which harmonics are these resonance frequencies? (e) If the speed of transverse waves on this string is 400 m/s, find the length of the string.
1. In the figure shown a source of sound of y frequency 510 Hz moves with constant velocity vs = 20 m/s in the direction shown. The wind is blowing at a constant velocity vw = 20 m/s towards an observer who is at rest at point B. Find the frequency (in Hz) detected by the observer corresponding to the sound emitted by the source at initial position A. [Speed of sound relative to air = 330 m/s]
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SUBJECTIVE UESTIONS
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3.60
8. How long will it take sound waves to travel the distance ℓ between the points A and B if the air temperature between them varies linearly from T1 to T2? The velocity of sound propagation in air is equal to v = α T , where α is a constant.
11. A sound source of frequency f0 = 1300 Hz is dropped from a height H = 505 m above the ground. At the same time a detector is thrown upwards with a velocity v = 50 m/s along the same line. If the velocity of sound is v = 300 m/s, find the frequency (in Hz) detected by the detector at t = 5 s.
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PREVIOUS EARS’ IIT-JEE UESTIONS
9. In the given figure, a string of linear mass density 3 × 10–2 kg/m and length L = 1 m, is stretched by a force F = (3 – kt) N, where ‘k’ is a constant and ‘t’ is time in sec. At the time t = 0, a pulse is generated at the end P of the string. Find the value of k (in N/s) if the value of force becomes zero as the pulse reaches point Q.
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1. A band playing music at a frequency f is moving towards a wall at a speed vb. A motorist is following the band with a speed vm. If v is the speed of sound, obtain an expression for the beat frequency heard by the motorist.
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3. The air column is a pipe closed at one end is made to vibrate in its second overtone by tuning fork of frequency 440 Hz. The speed of sound in air is 330 m/s. End corrections may be neglected. Let P0 denote the mean pressure at any point in the pipe and ∆P0 the maximum amplitude of pressure variation. [1999]
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11. A boat is travelling in a river with a speed 10 m/s along the stream flowing with a speed 2 m/s. From this boat a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is 14.45 mm. Assume that attenuation of sound in water and air is negligible. [2001]
(a) What will be the frequency detected by a receiver kept inside the river downstream? (b) The transmitter and the receiver are now pulled up into air. The air is blowing with a speed 5 m/s in the direction opposite the river stream. Determine
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8. A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the train’s speed is reduced to 17 m/s, the frequency registered is f2. If the speed of sound is 340 m/s then the ratio f1/f2 is [2000]
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(b) magnetic field (d) pressure
(a) on a string clamped at both ends (b) on a string clamped at one end and free at the other (c) when incident wave gets reflected from a wall (d) when two identical waves with a phase difference of π are moving in the same direction
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10. A 3.6 m long pipe resonates with a source of frequency 212.5 Hz when water level is at certain heights in the pipe. Find the heights of waver level (from the bottom of the pipe) at which resonances occur. Neglect end correction. Now the pipe is filled to a height H (≈ 3.6 m). A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of H. If the radii of the pipe and the hole are 2 × 10–2 m and 1 × 10–3 m respectively. Calculate the time interval between the occurrence of first two resonances. Speed of sound in air is 340 m/s and g = 10 m/s2. [2000]
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9. Two vibrating strings of the same material but of length L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency v1 and the other with frequency v2. The ratio v1/v2 is given by [2000] (a) 2 (c) 8
0.8 represents a mov[(4 x + 5t ) 2 + 5] ing pulse where x and y are in metres and t is in second. Then [1999]
5. Y ( x, t ) =
(a) The time taken by the wave pulse to reach the other end R and (b) the amplitude of the reflected and transmitted wave pulse after the incident wave pulse crosses the joint Q.
(b) 1/2 (d) 19/18
(a) 18/19 (c) 2
length 2.56 m and mass 0.2 kg. The wire PQR is under a tension of 80 N. A sinusoidal wave pulse of amplitude 3.5 cm is sent along the wire PQ from the end P. No power is dissipated during the propagation of the wave pulse. Calculate [1999]
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Wave Mechanics
12. The ends of a stretched wire of length L are fixed at x = 0 and x = L. In one experiment the displacement of the wire
15. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by mass M. The wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is [2002]
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16. Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both end and is filled with a monoatomic gas of molar mass MA. Pipe B is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass MB. Both gases are at the same temperature. [2002]
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13. Two pulses in a stretched string, whose centres are initially 8 cm apart, are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds the total energy of the pulses will be [2001]
(a) E2 = E1 (c) E2 = 4E1
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(b) 2 (d) 11/6
(a) 242/252 (c) 5/6
a frequency of 5.5 kHz, while the train approaches the siren. During his return journey in a different train B, he records a frequency of 6 kHz while approaching the same siren. The ratio of the velocity of train B to that train A is [2002]
the frequency of the sound detected by the receiver. (Temperature of the air and water = 20ºC; Density of river water = 103 kg/m3; Bulk modulus of the water = 2.088 × 109 Pa; Gas constant R = 8.31 J/mol-K; Mean molecular mass of air = 28.8 × 10–3 kg/mol; Cp/Cv for air = 1.4)
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17. In a resonance tube experiment to determine the speed of sound (in air, a pipe of diameter 5 cm is used. The air column in pipe resonates with a tuning fork of frequency 480 Hz when the minimum) length
zero purely kinetic purely potential partly kinetic and partly potential
14. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records
(a) (b) (c) (d)
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Wave Mechanics
The frequency of sound recorded by an observer who is standing in air is [2004]
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24. An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct option. [2005] 5 3 (a) n = 3, f 2 = f1 (b) n = 3, f 2 = f1 4 4 5 3 (c) n = 5, f 2 = f1 (d) n = 5, f 2 = f1 4 4
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21. A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s and in air it is 300 m/s.
25. A tuning fork of 512 Hz is used to produce resonance in a resonance tube experiment. The level of water at first resonance is 30.7 cm and at second resonance is 63.2 cm. The error in calculating velocity of sound is [2005]
(b) 0.025 m (d) 0.024 m
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20. In the experiment for the determination of the speed of sound in air using the resonance column method, the length of the air column that resonates in the fundamental mode, with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate the end correction. [2003]
4 L ρ1 3 ρ2
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(b) 110 cm/s (d) 280 cm/s
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22. A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open organ pipe is [2004]
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19. A police car moving at 22 m/s chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle. If it is given that the motorcyclist does not observe any beats. [2003]
(a) 200 Hz (c) 120 Hz
18. A string of mass per unit length µ is clamped at both ends such that one end of the string is at x = 0 and the other is at x = ℓ. When string vibrates in fundamental mode amplitude of the mid-point of the string is a, and tension in the string is T. Find the total oscillation energy stored in the string. [2003]
of the air column is 16 cm. Find the speed of sound in air at room temperature. [2003]
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31. At x = 0, how many times does the net amplitude between 0 and 1 s (a) 46 (b) 42 (c) 50 (d) 100
27. A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m/s and 90 m/s2 respectively. Velocity of the wave is 20 m/s. Find the waveform. [2005]
32. In the experiment to determine the speed of sound using a resonance column. [2007]
26. An observer standing on a railway crossing receives frequency of 2.2 kHz and 1.8 kHz when the train approaches and recedes from the observer. Find the velocity of the train. [The speed of the sound in air is 300 m/s] [2005]
(a) Prongs of the tuning fork are kept in a vertical plane. (b) Prongs of the tuning fork are kept in a horizontal plane. (c) In one of the two resonances observed, the length of the resonating air columns is close to the wavelength of sound in air. (d) In one of the two resonances observed, the length of resonating air column is close to half of the wavelength of sound in air.
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(a) 340 m/s for passengers in A and 310 m/s for passengers in B.
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Two trains A and B are moving with speed 20 m/s and 30 m/s respectively in the same direction on the same straight track, with B ahead of A. The engines are at the front ends. The engines of train A blows a long whistle. Assume that the sound of the whistle is composed of components varying in frequency from f1 = 800 Hz to f2 = 1120 Hz as shown in the figure. The speed in the frequency (highest frequency lowest frequency) is thus 320 Hz. Speed of sound in still air is 340 m/s. [2007]
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Wave Mechanics
(b) 360 m/s for passengers in A and 310 m/s for passengers in B. (c) 310 m/s for passengers in A and 360 m/s for passengers in B. (d) 340 m/s for passengers in both the trains. 34. The distribution of sound intensity of the whistle as observed by the passengers in trains A is best separated by
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37. A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm is [2008]
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36. A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is [2008]
(a) 310 Hz (c) 350 Hz
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Y
ONL ONE OPTIONS IS CORRECT 1. 9. 17. 25. 33. 41. 49.
(c) (a) (b) (c) (d) (a) (a)
2. 10. 18. 26. 34. 42. 50.
3. 11. 19. 27. 35. 43. 51.
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4. w12. 20. 28. 36. 44. 52.
(b) (b) (d) (a) (a) (d) (d)
5. 13. 21. 29. 37. 45. 53.
(c) (a) (c) (b) (b) (d) (a)
6. 14. 22. 30. 38. 46. 54.
(a) (d) (a) (a) (c) (d) (a)
(c) (a) (d) (d) (d) (c) (a)
7. 15. 23. 31. 39. 47. 55.
(b) (a) (c) (c) (d) (b) (d)
8. 16. 24. 32. 40. 48. 56.
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8. For the propagation of transverse waves a medium should have young modulus and coefficient of rigidity. But gases do not have both factor so transverse waves can not propagate in gases. On the other hand propagation of longitudinal waves a medium should have bulk modulus. Hence longitudinal waves can propagate in all the three mediums.
11. In a dispersive medium the wavelength of a pulse gets changed and hence the shape of the pulse gets distoned.
5. Bats produce ultrasonic waves and receive them after reflection. During the time interval, between the production of ultrasonic and its reception back bats estimate the size direction etc. of the obstacle.
4. At the modes pressure is maximum. Therefore a displacement node is a pressure antinode.
7. Two notes of the same frequency differ in overtones and their relative strength. As a result, timber of the two notes are different and two notes can be distinguished.
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2. Imagine a cyclinder of radius 7 m and length 10 m. Intensity of sound at the surface of cylinder is same everywhere.
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Lets check at any point, say at x = 0, y = (10 cos t) sin (100t) at any point amplitude is changing sinusoidally. so this is equation of beats. (c) y = 10 sin (2πx – 120t) + 10 cos (120t + 2πx) = superposition of two coherent waves travelling in opposite direction. ⇒ equation of standing waves (d) y = 10 sin (2πx – 120t) + 8 cos (118t – 59/30πx) = superposition of two waves whose frequency are slightly different (ω1 = 120, ω2 = 118) ⇒ equation of beats
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4. (a) → p, q, (b) → s, (c) p, r, (d) → s
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3. (a) → r, (b) → q, (c) → p
∂y . Like for (a), y = 0 at x = 0 and t = 0. ∂t ∂y > 0 i.e., positive therefore it matches ∂t with (r).
T Vω = µ
f0 =
300 + 20 350 = × 510 = 525 Hz 330 + 20 − 10 340
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5. (b, c, d) 6. (a, b, c, d)
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Vair 330 = = 0.33 m f 1000 9
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m
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F cos α + N – mg cos α = 0 mg cos α mg sin α N = mg − cos α N = 1 + sin α 1 + sin α
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4. (b) the even harmonics are missing, (c) 25 Hz, (d) 3rd, 5th, 7th harmonic, (e) 4 m
d V =λ= ⇒ f (2n + 1) f
mg sin α 1 + sin α
1 K = T 10−4 (0.1) + 4 × 10−4 (0.5) 2
µ=
mg sin α – F sin α – f = 0
F= f =
1 K = T m12 (0.1) + m22 (0.5) 2
0.15 2 0.1 2 m dx + m 1∫ 2 ∫ dx 0.1 0
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8. (c)
27. y = (0.1 m) sin [(30 rad/s)t + (1.5 m–1)x + φ]
9. (d)
13. (b)
14. (b)
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37. (a) y = 5 cm and V = +ve
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30. (a)
12. (c)
29. (a)
11. (a) 1.0069 × 105 Hz, (b) 1.0304 × 105 Hz
28. (a)
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C H A P T E R
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4.1 REFLECTION
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Laws of Reflection
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Light (Geometrical Optics)
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(i) The angle of incident equals the angle of reflection, i.e.,
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(ii) The incident ray, the normal and the reflected ray lie in the same plane.
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Fermat’s Principle
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In going from one point to another, a light ray travels a path that requires equal or less time than the time required for neighbouring paths. In other words, the light ray takes the path that requires the minimum time.
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Concave mirror
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Light (Geometrical Optics)
Real/Virtual Object (Image)
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When a group of incident rays strike a surface, the point from which they diverge or the point at which they try to converge is known as object. After reflection we get a group of reflected rays. The point where reflected rays converge or the point from which the reflected rays appear to diverge is known as the image of the object. Objects and images may be real or virtual. If the light, after reflection, actually passes through the point where the image is located, the image is called as a real image, if the light only appears to come from the point where the image is called a virtual image. A real image will appear on a screen placed at its position while a virtual image will not.
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Similarly, if the object is a point where the light energy is actually present, it is a real object. If the light rays only tend to go towards the point where the object is located, it is a virtual object.
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Sign convention (i) All the distances will be measured from the pole (or the centre of the surface) as origin. Distance of the object from the surface is called the object distance (u) and that of the image will be called image distance (v). (ii) All real distances will be taken positive. Distances of all virtual points will be taken negative.
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Light (Geometrical Optics)
4A.3
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Convex mirror O: Pole of mirror OC = AC = BC = R = Radius of curvature
Focus If a group of rays parallel (and close) to the principal axis strike a concave mirror, the rays converges after reflection to a point which is known as focus (F). In case of a convex mirror, these incident rays will diverge away from focus (after reflection). The distance of focus from pole is called as focal length (f). (i) In a concave mirror, focus is a real point and, hence focal length is positive. (ii) In convex mirror, focus is a virtual point and, hence focal length is negative. (iii) The magnitude of focal length in spherical mirrors is half the radius of curvature, i.e., R f = . 2 https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
Mirror Equation (Magnification) Object distance, image distance and focal length in mirrors are related by the equation 1 1 1 + = u v f
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Ray Diagrams
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For a linear object, the ratio of image size (I) to the object size (O) is called magnification (M). It can also be proved that this ratio is equal to the ratio of v and u.
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The reflection of light rays and formation of images are shown with the help of ray diagram. Some typical incident rays and the corresponding reflected rays are shown below:
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■ A light ray parallel to principal axis passes through focus after reflection.
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■ A light ray passing through focus becomes parallel to principal axis after reflection.
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Light (Geometrical Optics)
4A.5
■ A light ray striking the Mirror normally i.e., collinear with C will be reflected along the same path.
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■ A light ray going towards focus becomes parallel to principal axis after reflection from a convex mirror.
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4.2 IMAGE FORMATION FOR A REAL LINEAR OBJECT
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Consider a real linear object AB and study with the help of rays diagrams images for various locations of AB.
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Light (Geometrical Optics)
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Object: at F Image: at infinity
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Object: between F and C Image: beyond C, real, inverted
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Object: beyond C Image: between F and C, real inverted, diminished.
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Object: between F and pole Image: behind the mirror, virtual erect and enlarged.
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Light (Geometrical Optics)
4A.7
Convex Mirror
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Some Important Results
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Object: vanywhere in front of mirror. Image: behind the mirror, between pole and F. Virtual, erect and diminished.
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(i) Reversibility of path: The path of a ray is completely reversible. If a light ray is turned back, it will take its original path. (ii) Whenever, a light ray strikes a reflecting surface normally, it comes back along the same path. (iii) In the case of plane mirror, focal length can be taken as infinite v = – u. We get a real image for virtual object and virtual image for a real object. The object and the image are equidistant from the mirror. (iv) If the incident ray on a plane mirror is kept fixed and the mirror is rotated through angle θ, the reflected ray gets rotated through an angle 2θ.
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(v) When a light ray is reflected, it deviates through an angle 180º – 22i,, where i = angle of incidence. Angle of deviation = D = angle I’ OR = 180 – (i + r) = 180 – 2i
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Light (Geometrical Optics)
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(vi) If two plane mirrors are inclined at an angle θ with each other, the total number of images of a point object placed between them is equal to the smallest integer 360° − 1 . greater than or the equal to θ (vii) The concave mirror is also called as converging mirror and the convex mirror is also called a diverging mirror.
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Example A 2.0 cm high object is placed on the principal axis of a concave mirror at a distance of 12 cm from the pole. If the image is inverted, real and 5.0 cm high, find the location of the image and the focal length of the mirror.
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Since the image is formed at 30 cm from the pole on the side of the object, we have 1 1 1 1 1 7 = + = + =− f v u −30cm −12cm 60cm https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
f =−
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4A.9
60cm = –8.6 cm. 7
Hence, focal length of the mirror is 8.6 cm. Example Rays of light are incident on a plane mirror at 45º. At what angle with the first should a second plane mirror be placed such that the rays emerge from the second mirror parallel to the first mirror?
) Solution The incident ray AB at an angle of incidence 45º is reflected along BC.
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This ray BC now suffers reflection at the second mirror at C and emerges along CD, where, ∠BCN′ = ∠N′CD (law of reflection). Since, ray CD is parallel to the first mirror. ∴ ∠M1CD = ∠BCO = θ Now in triangle BOC 2θ + 45º = 180º 135° or θ= = 67.5º 2
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Example An object is placed 25 cm from the surface of a convex mirror and a plane mirror is set so that the image formed in the two mirrors lie adjacent to each other. The plane mirror is then 20 cm from the object. What is the radius of curvature of the convex mirror?
) Solution Since, the image formed in plane mirror is as far behind the mirror as the object is in front of it. Image distance = |v| = (2d – u) = 40 – 25 = 15 cm (virtual)
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Light (Geometrical Optics)
| v | = –15 cm
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The focal length of a concave mirror is 30 cm. Find the two positions of the object in front of the mirror so, that the image is three times the size of the object.
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Light (Geometrical Optics)
4A.11
Case II: For virtual image: m = +3, since the image is erect. −v m= 2 =3 ∴ v2 = –3u2 u2 1 1 1 Now, = + f v2 u2 1 1 1 = + 30 3u2 u2 u2 = 20 cm
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Example Find the angle between two plane mirrors such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror, parallel to the first mirror.
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) Solution Let θ be the angle between the two plane mirrors.
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The ray AB parallel to OM2 is reflected along BC, where ∠ABM1 = ∠CBO = θ At C, the ray is reflected along CD, where CD is parallel to OM1. ∴ ∠M2CD = ∠BCO = θ Now in triangle BOC, 3θ = 180º ∴ θ = 60º Example A ray of light is incident on a plane reflecting surface at an angle of 30º to the normal. Find the deviation in the incident ray. What will be the deviation if the ray suffers a reflection again at a surface inclined at 60º to the first surface?
) Solution By the geometry
∠B′CM2 = 60º
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Light (Geometrical Optics)
∠M2CD = 60º
∠B′CD = 60º + 60º = 120º
∠DCN2 = ∠N2CB = 30º
∠ABN1 = ∠N1BC = 30º
∠CBA′ = 120º = D1
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The incident ray of light AB on the first mirror M1 at an angle of incidence of 30º is reflected along BC at an angle of reflection 30º. ∴ Deviation, D1 in the incident ray = 60º + 60º = 120º The reflected ray BC is incident on the second mirror M2.
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Since, ∠N1ON2 = 60º and ∠BCO = 60º ∠BCN2 = ∠N2CD = 30º.
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Example A rod of length 10 cm lies along the principal axis at a concave mirror of focal length 10 cm in such a way that the end closer to the pole is 20 cm away from it. Find the length of the image.
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Radius of curvature of the mirror is R = 2f = 20 cm. Thus, the nearer end B of the rod AB is at the centre of the curvature and hence, its image will be formed at B itself. We shall now locate the image of A.
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Light (Geometrical Optics)
4A.13
Here, u = –30 cm and f = –10 cm. We have, =
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v = –15 cm
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Thus, the image of A is formed at 15 cm from the pole. Hence length of the image is 5.0 cm.
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Example A convex mirror produces a magnification of 1/2 when an object is placed at a distance of 60 cm from it. Where should the object be placed so that the size of the image becomes 1/3rd that of the object?
) Solution When u1 = 60 cm, m1 = =
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Light (Geometrical Optics)
Example A plane mirror is placed at a distance of 50 cm from a concave mirror of focal length 16 cm. Where should a short object be placed between the mirrors and facing both the mirrors so that its virtual image in the plane mirror coincides with the real image in concave mirror? What is the ratio of the sizes of the two images?
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Let the object be at a distance x from plane mirror. ∴ Distance of object from concave mirror u = (50 – x) For the plane mirror, since object and image distances are equal, therefore
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where, u = (50 – x) v = (50 + x) f = 16 cm.
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1 1 1 = + 16 50 − x 50 + x
1 100 = 16 502 − x 2 or x2 = 502 – 16 × 100 = (2500 – 1600) cm2. = 900 cm2. ∴ x = 30 cm. Hence the object must be placed at a distance of 30 cm. from the plane mirror. or
AB A′B′ u 50 − x 50 − 30 20 1 = = = = = v 50 + x 50 + 30 80 4
Ratio of image sizes =
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Light (Geometrical Optics)
4A.15
The real image formed by the concave mirror is 4 times the size of virtual image formed by the plane mirror.
4.3 REFRACTION AT A PLANE SURFACE Refractive Index of a Medium (i) Refractive index of a medium 1 with respect to air µ1 = =
Speed of light in air Speed of light in medium 1 cair c1
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Refractive index of a medium with respect to air is also known as absolute refractive index of the medium. (ii) Refractive index of a medium 1 with respect to another medium 2 Speed of light in medium 2 c2 = 2 µ1 = Speed of light in medium 1 c1 µ1 2 µ1 = µ2
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Light (Geometrical Optics)
Rarer Medium and Denser Medium When two mediums are in contact, the one with the lesser refractive index is known as rarer medium and other with greater refractive index is known as denser medium.
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(i) When a light ray goes from the rarer medium to the denser medium, it bends towards normal. (ii) When a light ray goes from the denser medium to the rarer medium, it bends away from the normal. (iii) In both cases, the magnitude of the angle of deviation for the light ray is | i – r |.
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When a light ray goes from a denser medium to a rarer medium, the angle of refraction is greater than the angle of incidence.
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If the angle of incidence is increased, the angle of refraction may eventually becomes 90º. The angle of incidence for which the angle of refraction is 90º is called the optical angle for that interface. Let µr be the refractive index of the rarer medium and µd be the refractive index of the denser medium. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.17
Obviously, µr < µd From Snell’s Law: µ sin C = r sin 90° µ d
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If a ray of light travelling in a denser medium strikes a rarer medium at an angle of incidence i which is greater than the critical angle C, it gets totally reflected back into the same medium. This phenomenon is called a total internal reflection.
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4.4 REFRACTION THROUGH A GLASS SLAB When a light ray passes through a glass slab having parallel faces, it gets refracted twice and finally emerges out parallel to its initial direction. i.e., the angle of emergence (e) is equal to the angle of incidence (i). https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
Incident ray
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⇒ e=i Lateral displacement of the ray = d = distance between incident ray and emergent ray. It can be shown that
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When there is a layer of denser medium between a real object and an observer (in air), the object appears to be shifted towards the observer. If t is the thickness of the layer of denser medium, apparent shift is given as
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1 ∆t = 1 − t µ (i) A coin lying at the bottom of a vessel containing water upto depth t appears uplifted. Let, t = real depth of coin O. t ∴ Apparent depth = µ
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1 and Apparent Shift = OI = 1 − t µ (ii) (a) A glass slab of thickness t is placed between a real object O and an observer. The object appears to be shifted towards the slab by the distance
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Light (Geometrical Optics)
4A.19
(b) Whenever, a glass slab is placed in the path of converging rays, the point of convergence is shifted away from the slab by the distance 1 t 1 − µ W
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(iii) Multiple slabs: (Near normal rays) These relations hold also for images formed by successive refraction through a number of parallel sided slabs. It t1, t2, t3, ............ are the thicknesses of each of the slabs and µ1, µ2, µ3 ........... are the absolute refractive indexes of material of each slab, then object O viewed from rarer air medium is seen at I. Then, as seen from air
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AO = real depth, AI = apparent depth, OI = apparent shift. Example Consider the situation shown in the figure. Find the maximum angle θ for which the light suffers total internal reflection at the vertical surface.
) Solution For this case, optical angle θ '' = sin −1
1 4 = sin −1 1.25 5
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Light (Geometrical Optics)
4 5 π Since, θ '' = − θ ', we have 2 3 sin θ′ = cos θ′′ = 5
or sin θ '' =
From Shell’s Law sin θ = 1.25 sin θ′
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Maximum value of θ, for which the total reflection takes place at the vertical 3 surface, is sin −1 . 4 Example When a ray of light in air enters a liquid, it is found that the angle between the incident and the refracted rays is 150º and the angle between the reflected and refracted ray is 60º. Calculate the refractive index of the liquid.
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4A.20
) Solution If the liquid is denser than air, the refracted ray bends towards the normal. Given: ∠AOB = 150º ∴ ∠AOQ + 90º + r = 150º or 90º – i + 90º + r = 150º or i – r = 30º Also ∠BOC = 60º (Given) ∴ ∠BOP + ∠POC = 60º or 90º – r + 90º – i = 60º ∴ i + r = 120º
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Light (Geometrical Optics)
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4A.21
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Example
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sin 75° 0.9659 = = 1.366. sin 45° 0.7071
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=
er s
w
sin i sin r
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Now µ =
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m
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From equations (i) and (ii), we get 2i = 150º or i = 75º ∴ r = 45º
1
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Show that the lateral displacement of a ray of light passing obliquely through a parallel glass slab of thickness t and absolute refractive index µ is given by
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cos i sin i. t 1 − µ 2 − sin 2 i
) Solution
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Light (Geometrical Optics)
…(i)
In triangle BCE, x sin(i − r ) = BC ∴ x = BC sin (i – r) In triangle BCD,
BD = cos r BC
t [ BD = t ] cos r Substituting in equation (i), we get t x= sin(i − r ) cos r BC =
.c
…(ii)
s_ i
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sin i cos r cos i sin r x=t − cos r cos r
∴
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∴
ita rio s
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.m
µ 2 − sin 2 i µ
iv er
t.m
s: //
cos r =
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∴
si
sin 2 r µ2
e/
tp ht
Again, cos 2 r = 1 − sin 2 r = 1 −
…(iii)
ta
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sin i µ
er s
w
w
//w
sin r =
s:
∴
m
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sin i =µ sin r
Also,
rio
ar
Note: In the desired expression there is no term containing r. Therefore, we must eliminate the term sin r and cos r by finding their values in terms of i.
//t
Substituting the values of sin r and cos r in equation (ii), we get
ht
tp
s:
cos i sin i x = t sin i − µ µ 2 − sin 2 i µ ∴
4A.22
cos i x = t 1 − sin i 2 2 m − sin i
Example A layer of ice (µ = 1.33) lies on a glass plate (µ = 1.5). A ray of light makes an angle of incidence 60º on the surface of ice. Find the angle between this ray and the normal in glass.
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Light (Geometrical Optics)
4A.23
) Solution
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U ,FH %
na .c om
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At B:
sin r1 µ g = sin r2 µice
er s un iv
e/
r2 ≈ 35°
Example
iv un t.m e/
3 1 ⋅ 2 1.5
s: //
∴
sin r2 =
…(ii)
ht tp
or
ht tp s:
//t
ht
.m
tp
sin 60° = µg sin r2
…(i)
ita
.ja
w w
//w
s:
From equation (i) and (ii), we have
er si ta rio s
sin 60° µice = sin r1 1
m
At A:
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In the figure, the ray gets deflected twice from air to ice at A and from ice to glass at B. Find the angle r2.
A glass cube of refractive index 1.5 is 4 cm. on a side. A ray of light is incident on one face 1 cm from the edge at an angle of incidence 60º and directed towards the near side. Where and at what angle will the ray emerge from the cube?
) Solution Let the ray strike the cube at A. After refraction into the cube, it again strikes the other surface at B as shown. If r is the angle of refraction at A, the angle of incident at B is (90º – r). At point A, or
sin 60° 1.5 = sin r 1 sin r =
1 3
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Light (Geometrical Optics)
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∴ r = 35º Angle of incident at B, i = 90 – r = 55º
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sin θ = 1.5 sin r =
iv
t.m e/
1.5 3
un
.m
//t
ht tp s:
tp
e/
sin r 1 = sin θ 1.5
s:
At point C,
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
1 2 Critical angle for glass air interface = sin −−11 = sin −−11 ≈ 42° 1.5 3 As i > critical angle, the ray will be total internally reflected at B. Finally, the ray strike at C as shown.
ht
s: //
∴ θ = 60º Now CQ = BQ tan r = (PQ – BP) tan r = (PQ – AP cot r) tan r = PQ tan r – AP 4 = − 1 = 1.83 cm 2 The light ray emerges out from the opposite face at a distance of 1.83 cm from the edge at an angle of 60º.
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4A.24
Example A beaker contains water upto a height h1 and kerosene oil above water upto another height h2. Find the apparent shift in the position of the bottom of the beaker when viewed from above. Refractive index of water is µ1 and that of kerosene oil is µ2.
) Solution The apparent shift of the bottom due to the water is 1 ∆t1 = 1 − h1 µ 1 https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.25
And apparent shift of bottom due to kerosene oil is 1 ∆t2 = 1 − h2 µ2 1 1 Total shift = ∆t1 + ∆t2 = 1 − h1 + 1 − µ µ 1 2
∴
h2
ita
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m
ar a
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Example A concave mirror of radius 40 cm lies on a horizontal table and water is filled in it up to a height of 5.0 cm as shown in the figure. A small dust particle flows on the water surface at a point P vertically above the point of contact of the mirror with the table. Locate the image of the dust particle as seen from a point directly above it. The refractive index of water is 1.33.
er si ta rio s
s:
un iv
//w
w
er s
w
3
e/
iv
.m
Let us first locate the image formed by the concave mirror. Let us take vertically upwards as the negative axis. Then, R = 40 cm. The object distance, u = –5 cm. Using the mirror equation,
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s: //
t.m e/
ht tp s:
un
//t
ht
tp
) Solution The ray diagram is shown in the figure.
3
3 3
or
1 1 2 + = u v R 1 2 1 = −2 − 1 = 6 = − v R u 40 cm −5 cm 40 cm
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or
v = 6.67 cm
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Light (Geometrical Optics)
The positive sign shows that the image P1 is formed below the mirror and hence, it is virtual. These reflected rays are refracted at the water surface and go to the observer. Depth of the point P1 from the surface = 6.67 cm + 5.00 cm = 11.67. Due to refraction at the water surface, the image P1 will be shifted above by a distance 1 = (11.67 cm) 1 − = 2.92 cm 1.33 Thus, the final image is formed at a point (11.67 – 2.92) cm = 8.75 cm below the water surface.
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na .c om
Example The base of a tank is horizontal plane of glass 8 cm thick (µ = 1.6). Above this is a layer of liquid of thickness 4.5 cm (µ = 1.5) and upon this floats a layer of water 12 cm thick (µ = 4/3). Looking from above, what is the apparent position of the spot on the bottom tank and where should the eye be held to see this spot?
ar a
) Solution O is the object on the bottom and its image appears at I.
ita
.ja
m
(
ht tp s:
P
s: //
t t1 t2 + + 3 µ1 µ 2 µ3
t.m e/
2
ht tp
Now, AI =
P
iv
e/ .m
tp
//t
FP
,
P
un
s:
un iv
//w
FP
er si ta rio s
w
er s
w
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FP
ht
4A.26
8 4.5 12 + + 1.6 1.5 4 3 = 5 + 3 + 9 = 17 cm Apparent position of the spot = 24.5 – 17 = 7.5 cm above O. Position of eye: Since, the least distance of distinct vision is 25 cm, therefore the eye must be held at a distance = 25 cm from I or at a distance (25 – 17) = 8 cm above A. =
Example A slab of glass 20 cm thick and refractive index 1.5 is kept in front of a plane mirror and a pin is kept in front of it in air at a distance of 40 cm from the mirror. Find the position of the image as formed by the mirror. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.27
) Solution 0LUURU FP
W
, 2 FP
6ODE
FP
FP FP
rio s_ in fo
na .c om
The virtual image of pin at O is formed by the slab I1, where OI1 is the shift due to slab 1 2 OI1 = t 1 − = 20 1 − µ 3
ar a
and
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
This image I1 now acts as an object for the plane mirror at a distance = 40 – 6.67 = 33.33 cm. ∴ Distance of image from the plane mirror = 33.33 cm.
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ht tp
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4.5 REFRACTION THORUGH A PRISM
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A light ray striking at one face of a triangular glass prism gets refracted twice and emerges out from the other face as shown above. The angle between the emergent and the incident rays is called the angle of deviation (D). The angle between the two refracting faces involved is called the refracting angle (A) of the prism. From AXY, we have A + (90 – r1) + (90 – r2) = 180º ⇒ r1 + r2 = A ...(i) Deviation, D = (i – r1) + (e – r2) = (i + e) – (r1 + r2) = i + e – A ⇒ A+D=i+e ...(ii) https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
We also have two equations from Snell’s Law at X and Y sin r2 1 sin i = µ and = sin r1 sin e µ
Angle of Deviation It can be easily seen that if we reverse the emergent ray, it goes back along the same path. The angle of incidence and emergence get interchanged but the angle of deviation remains same.
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T
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t.m e/
T
un
ht tp s:
//t
iv
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$
er si ta rio s
w
er s
w
angle of incidence = θ1 angle of emergence = θ2 θ1 + θ2 = A + D
//w s: tp
rio s_ in fo
ita
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m
ar a
T
ht
4A.28
angle of incidence = θ2 angle of emergence = θ1 θ2 + θ1 = A + D
Hence, the same angle of deviation D is possible for two different angles of incidence: θ1 and θ2 where, θ1 + θ2 = A + D
Minimum Deviation The angle of deviation is minimum when the path of light ray through the prism is symmetrical. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.29
i.e., angle of incidence = angle of emergence i=e ⇒ r1 = r2 Hence, A + Dmin = 2 i Also, A = r1 + r2 = 2 r sin i =µ sin r1 A + D min ) 2 =µ A sin 2 $
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ita
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un iv
//w
w
er s
w
U U U
Grazing Incidence
s:
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we have,
sin(
er si ta rio s
From,
iv
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ht tp
s: //
t.m e/
ht tp s:
un
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ht
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When i = 90º the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass-air. This is known as grazing angle. When e = 90º, the emerged ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for the glass-air. This is known as grazing emergence. $
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Maximum Deviation The angle of deviation is same for both the above cases (grazing incidence and grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
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Dispersion Through a Prism
µred < µviolet because λred > λviolet
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i.e.,
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The refractive index of a medium is different for light rays of different wavelengths. More is the wavelength,
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er si ta rio s
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A ray of white light passing through a prism gets splits into two different colours because the deviation is more for violet and less for red. This phenomenon is known as dispersion.
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s: //
t.m e/
ht tp s:
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ht
For a prism with very small refracting angle A, the deviation D is given by: D ≈ (µ – 1) A
ht tp
4A.30
Hence, deviation of violet = Dv = (µv – 1) A and deviation of red = (µr – 1) A The angle between the red and violet rays is known as the angle of dispersion. Angle of dispersion, φ = Dv – Dr = (µv – µr) A Angle of mean deviation, D m =
Dv + D r 2
µ + µr = v − 1 A 2 Dispersion power of glass, w =
µv − µr φ = Dm µv + µr − 1 2
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Light (Geometrical Optics)
4A.31
Combination of Prisms
ar a
Condition for achromatism is φ1 = φ2 ⇒ (µ1v – µ1r) A1 = (µ2v – µ2r) A2
er si ta rio s
er s
un iv
µ + µ1r µ + µ2r = 1v − 1 A1 − 2 v − 1 A 2 2 2
e/
tp
s:
//w
w
w
Net deviation produced, Dnet = Dm1 – Dm2
ita
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m
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(i) If two prisms are placed opposite to each other (as shown) and their angles of dispersion are equal, the combination is known as achromatic combination. The dispersive effect of one prism is cancelled by the other.
iv
.m
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
(ii) If we want to have only dispersion and no deviation, we combine two prisms having same mean deviation. Dm1 = Dm2
µ + µ1r µ + µ2r ⇒ 1v − 1 A1 = 2 v − 1 A 2 2 2 Such a combination is known as non-deviation or direct-vision combination. The net angle of dispersion produced by the combination is Qnet = φ1 – φ2 = (µ1v – µ1r) A1 – (µ2v – µ2r) A2. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
Example A glass prism has a reflecting angle 60º and refractive index 1.5. A ray of light is incident at 48.6º. Find the angle of emergence and angle of deviation. )
) Solution Given: A = 60º, i = 48.6º, µ = 1.5 Now,
µ=
or
1.5 =
sin i sin r1 sin 48.6° sin r1
0.7501 = 0.5 1.5 ∴ r1 = 30º Also, r1 + r2 = A ∴ r2 = (60º – r1) = 60º – 30º = 30º Which shows that r = r1, i.e., the ray passes symmetrically through the prism. ∴ e = i = 48.6º D=i+e–A = 48.6º + 48.6º – 60º = 37.2º This deviation is minimum at i = e sin r1 =
o
s_ in f
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si ta
s rio
w
iv er
w
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m
ar
an
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ta
si
er
iv
t.m
tp
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//w
Example The refractive indexes of silicate flint glass for wavelength 400 nm and 700 nm are 1.66 and 1.61 respectively. Find the minimum angles of deviation of an equilateral prism made of this glass for light of wavelengths 400 nm and 700 nm.
ht
un
tp
s:
//t
ht
tp
A + Dm sin 2 µ= A sin 2
.m e/
)
s: //
) Solution The minimum angle of deviation Dm is given by
ht
D sin 30° + m D 2 = = 2sin 30° + m sin30° 2
For 400 nm light,
or or
or
D 1.66 = 2sin 30° + m 2
4A.32
D sin 30° + m 2 30° +
= 0.83
Dm = 56° 2 Dm = 52°
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Light (Geometrical Optics)
4A.33
For 700 nm light, D 1.61 = 2sin 30° + m 2 Solving, we get Dm = 48°. Example A prism of angle 60º deviate a ray of light through 40º for two angles of incidence which differ by 11º. What is refractive index of the glass of the prism?
) Solution The incident ray PQ is deviated through 40º and the principle of
rio s_ in fo
e/
5
L H
6
er si ta rio s
un iv
//w s:
U
ita
4
U
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Hence other angle of incidence causing same deviation of 40º is e = i2 where, i1 – i2 = 11º (Given) But i + e = A + D = 60º + 40º = 100º ∴ i1 + i2 = 100 ∴ i1 = 55º . 30′ and i2 = 44º . 30′
s: //
sin i1 sin i2 = sin r1 sin r2
ht tp
Now, µ =
t.m e/
ht tp s:
un
//t
iv
.m
tp ht
L
[
er s
3
w
w
.ja
m
ar a
na .c om
reversibility of ray through optical system. Shows that the ray SR would also suffer the same deviation of 40º.
where, r2 = 60° –r1 ∴
µ=
sin 55° ⋅ 30′ sin 44° ⋅ 30′ = sin r1 sin(60° − r1 )
…(i)
From equation (i), we first calculate r1, therefore 0.8241 = sin r1
0.7009 3 1 cos r1 − sin r1 2 2 Dividing both sides in denominator by sin r1, we get 0.8241 0.7009 = sin r1 3 1 sin r1 cot r1 − sin r1 2 2 sin r1 https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
1 0.7009 ( 3 cot r1 − r2 ) = 2 0.8241
or
3 cot r1 = 2.701
cot r1 = 1.559 r1 = 32.40′
or or
or
∴ µ=
sin 55° ⋅ 30′ 0.8241 = = 1.526 sin 32° ⋅ 40′ 0.5400
Example
om
A ray of white light incident at an angle of 35º on one face a crown glass of refracting angle 60º. Find the angle of dispersion and mean deviation produced by the prism. Given for the crown glass, µv = 1.538 and µr = 1.52. )
.c
) Solution The angle of prism is 60º and hence it is not a small angle prism,
⇒ rv 1 = 21.9º
∴ rv 2 = (60 – 21.9) = 38.1º
fo in s rio ta
un i
//w
ve
w
rs
w
s_
rio
ita
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m
ar an a
i = 35º, A = 60º. (i) For the violet colour at the first face, sin i sin 35° µv = = 1.538 = sin rv1 sin rv1
iv
un
//t
si
sin ev sin ev = 1.538 = sin rv2 sin 38.1°
er
e/
µv =
.m
tp s:
Again at the second face,
ht
e/
s:
tp s:
sin i sin 35° = 1.52 = sin γ r1 sin γ r1
µr =
ht
//t
.m
ht tp
⇒ ev = 71.6º ∴ Deviation for the violet ray, DV = (i + ev) – A = (35 + 71.6) – 60 = 46.6º. (ii) For the red colour at the first face,
γ r1 = 22.2°
Then, γ r2 = (60 − 22.2) = 37.8° Again at the second face, sin er sin er µr = = 1.52 = sin γ r1 sin 37.8°
⇒
⇒
4A.34
er = 68.7º
Deviation for the red ray, Dr = (i + er) – A = (35 + 68.7) – 6.0 = 43.7º
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Light (Geometrical Optics)
4A.35
Again of dispersion, φ = (Dv – Dr)
= (46.6 – 43.7) = 2.9º 46.6 + 43.7 Angle of the mean deviation = 2 = 45.15°.
om
Example Crown glass has µv = 1.538, µr = 1.52; quartz glass has µv = 1.557; µr = 1.542. Given a quratz prism of refracting angle 4º. Calculate (i) The angle of the crown prism to form a direct vision combination for white light. (ii) The angular width of the spectrum product by this combination.
an a. c
)
) Solution The prisms are small angled. For direct vision combination
ta
0.5495 × 4 = 4.16° 0.5290
si
s ita rio
w
A crown =
w
∴
un iv er
.ja
rio
m ar
s_
in fo
1.557 + 1.542 − 1 µ − ( 1) A crown 2 m quartz m quartz = 0.5495 crown = = A quartz (µ mm crown 1.538 + 1.520 0.5290 quartz crown − 1) − 1 2
ve rs
ni
//t .m
ht
tp
e/
s: //
w
Then, angular width of the spectrum produced by the combination = (µv – µr)quariz – (µv – µr)crown = |(0.015) 4 – (0.018) × 4.16 | = 0.01488º
s: //
ht
t.m
e/ u
tp s:
Note: The net angle of dispersion is the angular width of the spectrum and is the modulus of the difference between the angle of dispersion produced by the individual prisms.
ht
tp
Example A 60º glass prism has a refractive index of 1.5. Calculate: (i) The angle of incidence for minimum deviation. (ii) The minimum deviation. (iii) The angle of emergence of the light at maximum deviation. )
) Solution
(i) At minimum deviation, the ray passes symmetrically through the prism. A 60° r= = = 30° ∴ 2 2 sin i Now µ = sin r ∴ sin i = µ sin r = 1.5 sin 30° = 0.75 ∴ i = 48°.35′ https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
A+D sin 2 µ = (ii) A sin 2 60° + D sin 2 or 1.5 = sin 30° or sin
∴
na .c om
60 + D = 1.5sin 30° = 0.75 2 60 + D = 48.35° 2 D = 37.10°
or
m
ar a
sin 90° = µ = 1.5 sin r1 sin r1 =
1 = 0.6667 1.5
∴ r1 = 41.49′ and r2 = (A – r1) = 60º – 41.49º = 18.11º sin e Now, = µ = 1.5 sin r2
un iv
er si ta rio s
er s
w
w
//w
un
//t
sin e = 1.5 sin r2 = 1.5 sin 18.55º = 0.4681 or e = angle of emergence = 27.55º
ht tp
s: //
t.m e/
ht tp s:
∴
iv
.m
e/
s: tp
ita
∴
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rio s_ in fo
(iii) Maximum deviation occur when the light is incident at 90º on one face.
ht
4A.36
U
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Example Light falls at grazing incidence on one of the surfaces of a prism of refractive index µ and refracting angle A. Show that the angle of emergence is given by sin e =
(
µ 2 − 1 sin A − cos A
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)
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Light (Geometrical Optics)
4A.37
) Solution The ray PQ is incident at grazing incidence (Since, i = 90º) and emerges
along RS with angle of emergence e. Now because of the reversibility of the ray, through the optical system, the ray RQ would emerge along QP with angle of refraction = 90º. ∴ r1 = C (critical angle) But r1 + r2 = A ∴ r2 = (A – r1) = (A – C) Now for refraction at R,
sin e =µ sin r2
iv
…(iii)
t.m e/
un
//t
cos C =
er si ta rio s
un iv e/
µ2 − 1 µ Substituting in equation (i), we get
∴
ht tp s:
ht
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Also, cos2C = 1 – sin2C 1 = 1− 2 µ
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∴ sin e = µ sin (A – C) or sin e = µ (sin A cos C – cos A sin C) ...(i) Note that in the expression asked in the problem, since the terms sin C and cos C are absent, therefore, we must find the values of sin C and cos C in terms of µ and A. Since, C is the critical angle, therefore 1 sinCC= or sin = …(ii) µ
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4A.38
Light (Geometrical Optics)
4.6 REFRACTION AT A SPHERICAL SURFACE Consider a spherical surface of radius R separating two media with refractive indexes µ1 and µ2. The radius of curvature R will be taken positive when incident rays strike the convex side of the surface. Let u is the object distance and v is the image distance. $
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µ1 µ 2 µ 2 − µ1 + = u v R The corresponding equation for refraction at a plane surface will be µ1 µ 2 + = 0 (taking R = ∞) u v
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Derivation of equation Let ∠DOP = α, ∠DCP = β and ∠DIC = γ. For paraxial rays, D is close to P and α, i, r, β, and γ are all small. From triangle ODC α + β = i And from ∆DCI, r+γ=β Also from Shell’s Law,
...(i) ...(ii)
sin i µ 2 = sin r µ1
We can write sin e ≈ i and sin r ≈ r, so that the above equation becomes µ1i = µ2r Putting values of i and r from equation (i) and (ii) into equation (iii), we get µ1 (α + β) = µ2 (β – γ) or µ1 α + µ2 γ = (µ2 – µ1) β https://t.me/universitarios_infohttps://www.jamarana.com
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...(iii)
...(iv)
Light (Geometrical Optics)
4A.39
As α, β, and γ are small, hence from figure DP DP DP α≈ ,β= and γ = PO PC PI Putting these values in equation (iv), we get DP DP DP µ1 + µ 2 PI = (µ 2 − µ1 ) PC PO µ1 µ 2 µ 2 − µ1 + = PO PI PC
or where,
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R = radius of curvature. µ1 µ 2 µ 2 − µ1 + = u v R
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∴
PO = Object distance, PI = Image distance, and
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4.7 EXTENDED OBJECTS: LATERAL MAGNIFICATION
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The lateral or transverse magnification is defined as h m= 2 h1 where, h2 = height of the image and h1 = height of the object In the figure, OQ = + h1 and O′Q′ = – h2. h O′Q′ m= 2 =− h1 OQ Since the triangles OCQ and O′CQ′ are similar, therefore −O′Q′ −O′C −PO′ − PC m= =− = OQ OC PO + PC In the figure, PO = – u, PC = + R and PO′ = + v From equation (i), we have R −v v−R m=− = −u + R R − u https://t.me/universitarios_infohttps://www.jamarana.com
…(i)
…(ii) https://t.me/universitarios
Light (Geometrical Optics)
µ 2 µ1 µ 2 − µ1 − = v u R
Also or
µ 2u − µ1v µ 2 − µ1 = uv R
or
R=
…(Here, u = negative)
(µ 2 − µ1 )uv µ 2u − µ1v
This gives, R − v =
µ1v(v − u ) µ µ (v − µ ) and R − u = 2 µ 2u − µ1v µ 2u − µ1v µ1v µ 2u
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Example There is a small air bubble inside a glass sphere (µ = 1.5) of radius 10 cm. The bubble is 4.0 cm below the surface and is viewed normally from the outside (in figure). Find the apparent depth of the bubble.
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4A.40
surface when the light from the bubble goes from the glass to the air. Here, u = –4.0 cm, R = –10 cm, µ1 = 1.5 and µ2 = 1 We have µ 2 µ1 µ 2 − µ1 + = v u R or
1 1.5 1 − 1.5 + = v 4.0 cm −10 cm
or
1 0.5 1.5 = − v 10 cm 4.0 cm
or v = –3.0 cm Thus, the bubble will appear 3.0 cm below the surface. https://t.me/universitarios_infohttps://www.jamarana.com
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4A.41
Example A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 9 mm situated in water. Assuming the light rays to be paraxial: (i) Find the position of the image due to refraction at the first surface and the position of the final image. (ii) Draw a ray diagram showing the positions of both the images.
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1 1 − 1.33 = v 2 mm
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Thus,
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For the first refraction (water to air). µ1 = 1.33, µ2 = 1, u = ∞, R = + 2 mm
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µ 2 µ1 µ 2 − µ1 + = v u R
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The ray diagram is shown in the figure. The equation for refraction at a spherical surface is
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or v = – 6 mm The negative sign shows that the image I, is virtual and forms at 6 mm from the surface of the bubble on the water side. The refracted rays (which seen to come from I1) are incident on the farther surface of the bubble. For this refraction, µ1 = 1, The object distance,
From equation (i), or
µ2 = 1.33, R = – 2
u = (6 mm + 4 mm) = 10 mm
µ 2 µ1 µ 2 − µ1 + = v u R
1.33 1 1.33 − 1 + = v 10 −2
or
1.33 0.33 1 =− − v 2 10
or
v = –5 mm
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4A.42
Light (Geometrical Optics)
This minus sign shows that the image is formed on the air side at 5 mm from the refracting surface. Measuring from the centre of the bubble, the first image is formed at 8.0 mm from the centre and the second image is formed at 3.0 mm from the centre. Both images are formed on the side from which the incident rays are coming.
4.8 LENSES
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A lens is one of the most familiar optical device for human being, the lenses are made of a transparent material bounded by two spherical surfaces. The surface may be both convex, both concave or one convex and one concave. When the thickness of the lens is small as compared to the other dimension like object distance etc., we call it a thin lens.
Convex Lens
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It a beam of light parallel to principal axis strikes a convex lens, it converges after refraction to a point called as focus (F). Focus in a convex lens is a real point and, hence focal length (f) is positive. Due to the converging effect, convex lens is also called as a converging lens.
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Lenses have two foci, because light can strike a lens from both the sides. As there are two spherical surfaces, there are two centres of curvatures C1 and C2 and correspondingly two radii of curvature R1 and R2. The line joining C1 and C2 is called the principal axis of the lens. The centre P of the thin lens which lies on the principal axis, is called the optical centre.
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Light (Geometrical Optics)
4A.43
Ray Diagrams for Convex Lens Some important rays and their paths after refraction are shown below.
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Consider a real linear object AB placed at different locations in front of a convex lens.
Object beyond 2F: Image between F and 2F, real inverted and diminished.
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Object at 2F: Image at 2F, real, inverted and of same size as object.
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Light (Geometrical Optics)
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Object between 2F and F: Image beyond 2F, real, inverted and enlarged.
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Object at F: Image at infinity.
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Object between F and lens: Image of the same size as object, virtual, erect and enlarged.
Concave Lens
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4A.44
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Light (Geometrical Optics)
4A.45
When a beam of light parallel to principal axis strikes a concave lens, it diverges away from a point known as focus (F). As the focus is a virtual point, the focal length (f) of a concave lens is negative. A concave lens is also called as a diverging lens.
Ray Diagrams for Concave Lens
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Image formation for a real object AB in a concave lens is shown below:
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Object anywhere in front of lens: Image between F and lens on the same side, virtual, erect diminished.
Sign Conventions The optical centre is taken as the origin and the principal axis as the x-axis. The positive direction of the axis is generally taken along the incident rays. The quantities u, v, f, R1 and R2 represent the x-coordinate of the object, the image, the focus, first centre of curvature and second centre of curvature respectively. The table below shows the signs of u, v, f, R1 and R2 in certain cases shown in the figure. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
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4A.46
(d) 2
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Light (Geometrical Optics)
4A.47
Generally, the incident rays and hence the positive direction of the axis is taken from left to right. Heights measured upwards are taken to be positive and the heights measured downward are taken to be negative. With the usual choice of axis, f of a lens is positive for converging lens and is negative for a diverging lens.
4.9 LENS MAKER’S FORMULA AND LENS FORMULA $
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µ 2 µ1 µ 2 − µ1 + = v u R
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The general equation for refraction at a spherical surface is …(i)
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µ1 µ 2 µ 2 − µ1 + = u v1 R2
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For the second refraction, the light goes from the medium µ2 to medium µ1. From equation (ii), we get µ1 µ 2 µ1 − µ 2 + = v v1 R2
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…(iii)
From equations (ii) and (iii),we get 1 1 1 1 µ1 + = (µ 2 − µ1 ) − v u R1 R 2 or
1 1 1 µ2 1 + = − 1 − v u µ1 R1 R 2
…(iv)
If the object O is taken far away from the lens, the image is formed close to the focus. Thus, for u = ∞, v = f. ∴
1 1 µ2 1 = − 1 − f µ1 R R 2 1
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Light (Geometrical Optics)
If the refractive index of the material is µ and its placed in air, µ2 = µ and µ1 = 1, then 1 1 1 = ( µ − 1) − f R R 2 1
…(vi)
This is called Lens Maker’s formula because it tells what curvature will be needed to make a lens of desired focal length. Combining equations (iv) and (v), we get 1 1 1 + = u v f which is known as Lens formula.
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Magnifi Magnification, cation, M M ==
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Example (i) Find the focal length of a biconvex lens in air if the radius of its surfaces are 60 cm and 15 cm. Refractive index of glass is 1.5. (ii) Find the focal length of a concave lens in water (µ = 4/3) if the surface have radii equal to 40 cm and 30 cm. (µg = 1.5).
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(i) Consider a light ray going through the lens as shown. If strikes the convex side of 60 cm radius and concave side of 15 cm radius while coming out.
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4A.48
Here, R1 = + 60, R2 = –15 cm Now
1 1 µ2 1 = − 1 − f µ1 R1 R 2 1 1.5 1 = − 1 − 1 60 −15 1 5 1 = 0.5 + = 0.5 × 60 15 60
f = 24 cm.
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Light (Geometrical Optics)
4A.49
(ii) If we image a light ray going through the lens as shown.
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1.5 1 1 = − 1 − 4 −30 +40 3
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Using the formula,
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Example A point source S is placed at a distance of 15 cm from a converging lens of focal length 10 cm on its principal axis. Where should a diverging mirror of focal length 12 cm be placed, so that a real image is formed on the source itself?
) Solution
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Light (Geometrical Optics)
Equation for the lens is 1 1 1 + = u v f
Here, u = 15 cm and f = + 10 cm 1 1 1 ∴ + = 15 v 10
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1 1 1 2 = − = v 10 15 60 or v = 30 cm The positive sign of v shows that the image I1 is formed to the right of the lens in the figure. The diverging mirror is to be placed to the right in such a way that the light rays fall on the mirror perpendicularly. Then only the rays will retrace their path and form the final image on the object. Thus, the image I1 formed by the lens should be at the centre of curvature of the mirror. We have, LI1 = 30 cm. MI1 = R = 2f = 24 cm Hence, LM = LI – MI1 = 6 cm Thus, the mirror should be placed 6 cm to the right of the lens. or
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Example A convex lens in air produces a real image having the same size as object. When the object and the lens are immersed in a liquid, the real image formed is enlarged two times the object size. Find the refractive index of the liquid (Take µg = 1.5). )
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) Solution Combining the lens equation and Lens Maker’s formula, we have
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1 1 1 µ2 1 + = − 1 − u v µ1 R1 R 2
t.m
v =1 u v=u 1 1 1.52 1 1 + = − 1 − u u 1 R1 R 2
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tp s
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m=
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In air,
v =2 u v = 2u 1 1 1.5 1 1 + = − 1 − u 2u µ1 R1 R 2
...(i)
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∴
...(ii)
⇒
In liquid, m =
From equations (i) and (ii), we get 1 1 + u u = 1.5 − 1 1 1 1.5 + − 1 u 2u µ 1
4A.50
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Light (Geometrical Optics)
2×
or or
4A.51
0.5µ1 2 = 3 1.5 − µ1
µ1 =
12 11
Hence, refractive index of liquid is
12 11
Example
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A converging lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axis coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens as shown in the figure. It is found that the final beam comes out parallel to the principal axis. Find the separation of the mirror and the lens.
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1 1 1 = − v f u
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or
1 1 1 + = u v f
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Here, u = 12 cm and f = 15 cm.
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1 1 −1 − = 15 12 60 or v = –60 cm The negative sign shows that the image is formed to the left of the lens as suggested in the figure. The image I1 acts as the source for the mirror. The mirror forms an image I2 of the source I1. This image I2 then acts as the source for the lens and the final beam comes out parallel to the principal axis. Clearly, I2 must be at the focus of the lens. We have I1I2 = I1P + PI2 =
= 60 cm + 15 cm = 75 cm. Let, the distance of mirror from I2 is x cm. Then for the reflection from the mirror, u = MI1 = – (75 + x) cm, v = – x and f = –20 cm. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
1 1 1 + = u v f
From the relation,
1 1 1 + = x 75 + x 20 75 + 2 x 1 = (75 + x) x 20
or
x2 + 35x –1500 = 0
or
−35 ± (35) 2 + 4 × 1500 2 ∴ x = 25 or –60 As the negative sign has no physical meaning, only positive sign should be taken. Taking x = 25, the separation between the lens and the mirror = (15 + 25) = 40 cm x=
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Example A convex lens mounted on an optical bench forms an image of an object on a screen, the magnification being 2.5. The object and the screen are kept fixed and the lens is moved through a distance of 10 cm, when a sharp image is again formed on the screen, the magnification now being 0.4. What is the focal length of the lens?
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and
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u −f 1 = 1 2.5 f
∴
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According to the given condition, f m= = 2.5 u1 − f
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5 u1 − f + 10 = 2 f
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5 2 10 − = 2 5 f
or
12 10 = 10 f
or
∴
…(ii)
or
4A.52
f = 4.76 cm.
Example Photographs of the ground are taken from an aircraft flying at an altitude of 2000 metres, by a camera with a lens of focal length 50 cm. The size of the film https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.53
in the camera is 18 cm × 18 cm. What area of the ground can be photographed by this camera at any one time?
) Solution Since, the object (ground) is at a very large distance from the lens,
therefore, the image of the ground can be assumed to be formed at the focal length distance ‘f ’. Hence for u = 2000 m v = 0.5 m Let the length of ground photographed be x.
x=
2000 × 0.18 = 720 m. 0.5
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0.18 0.5 = x 2000
or
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m=
Now,
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∴ Area of ground photographed = 720 m × 720 m
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4.10 CONJUGATE POINTS AND PLANES
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A consequence of the reversibility of light rays is that the positions of a real object and its real image are interchangeable. Any pair of such real object point and its real image point on principal axis are called conjugate points. The planes passing through these points and perpendicular to the axis are called conjugate planes.
Power of a Lens The reciprocal of focal length expressed in metres is known as the power of lens. It is expressed in Dioptres. I P = (m −1 ) f
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4A.54
Light (Geometrical Optics)
4.11 THIN LENSES IN CONTACT
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2
/
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3
3
,
u = object distance for the first lens v = final image distance for the second lens. v1 = image distance of the first image O1 for the first lens. As the lenses are assumed to be thin, v1 is also object distance to the second lens. 1 1 1 Then, − = …(i) u v1 f1
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1 1 1 1 − = + v u f1 f 2
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From equations (i) and (ii), we get
…(ii)
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1 1 1 − = v v1 f 2
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And,
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Let
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If the combination is replaced by the single lens, then 1 1 1 − = v u F 1 1 1 From equations (i) and (ii), we get = + F f1 f 2 1 1 1 1 In general, = + + + ... F f1 f 2 f 3
Two Thin Lenses Separated by a Distance G $
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…(iii)
Light (Geometrical Optics)
4A.55
If we place a mirror at the end of a group of lenses, the combination behaves as a mirror. Any light ray entering the combination passes through each lens twice before reflection and the after reflection from mirror. The equivalent focal length of the combination behaving as a mirror is given by: 1 1 1 1 1 = 2 + + + ... + F f f f 2 3 1 fm where, f1, f2, f3, ... are the focal lengths of the lenses and fm is the focal length of mirror.
Displacement Method
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Consider a real object and a screen fixed at a distance D apart. A convex lens is placed between them. Real images are obtained on the screen for two positions of the lens. One at a distance x1 and the other at a distance x2 from the object. 6FUHHQ
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[
un iv
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w
er s
w
ita
.ja
m
ar a
'±[
ht tp
s: //
)
[
'±[D
'±[
7
Two object distances ((x1 and x2) are roots of the equation 1 1 1 + = x D−x f or
x2 – Dx + Df = 0
or
x=
∴
x1 =
D ± D 2 − 4Df 2 D − D 2 − 4Df D + D 2 − 4Df and x2 = 2 2
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Light (Geometrical Optics)
Under the condition D > 4f the displacement of lens is L = x2 – x1 D 2 − L2 4D The above formula is frequently used to calculate the focal length of lens by measuring D and L. L = D 2 − 4Df
Also i.e.,
f =
D−L D+L + =D 2 2 x1 = D – x2 and x2 = D – x1. x1 + x2 =
Hence, object and image distances are interchanged in the two positions of the lens. Let the magnifications be m1 and m2, then
⇒
m1m2 =
⇒
I1 I 2 ⋅ =1 O O
na .c om
m1 =
ar a
(D − x1 ) (D − x2 ) =1 x1 x2
rio s_ in fo
D − x1 D − x2 and m2 = x1 x2
⇒
ita
.ja
m
⇒ Object size = I1I 2
//t
iv
er si ta rio s
e/
.m
tp
s:
un iv
//w
w
er s
w
Example A thin equiconvex lens is placed on a horizontal plane mirror and a pin held 20 cm above the lens coincides in position with its one image. The space between the lens and mirror is filled with water (µ = 4/3) and then to coincide with its own image as before, the pin has to be raised until its distance from the lens is 27.5. Find the radius of curvature of lens.
ht
un
) Solution In the first case, we get the focal length of the convex lens
ht tp s:
s: //
t.m e/
∴ f1 = 20 cm In the second case, we get the focal length F of the combination of the convex lens and water plano-convex lens.
ht tp
4A.56
)
I FP
∴ F = 27.5 cm Let f2 = focal length of water lens ∴
1 1 1 = + F f1 f 2
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Light (Geometrical Optics)
4A.57
1 1 1 = + 27.5 20 f 2 −220 ∴ f2 = cm 3 For the liquid lens, 1 1 µ2 1 = − 1 − f 2 µ1 R1 R 2 −3 4 1 1 = −1 − (µ1 = 1) or 220 3 R1 ∞ − 220 ∴ R = cm (R1 = R) 9 ∴ Radius of curvature of the convex lens, 220 R= cm. 9 [When the image and object coincide it is known as auto-collimation].
rio s_ in fo
ar a
na .c om
or
ht tp
s: //
t.m e/
ht tp s:
un
iv
.m
e/
1 1 1 = ( a µ g − 1) − f R1 R 2
//t
ht
tp
s:
) Solution We know,
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
m
Example An equiconvex lens of refractive index 1.5, focal length 10 cm is held with its axis vertical and its lower surface immersed in water. The upper surface being in air. At what distance from the lens will a vertical beam of parallel light incident on the lens be focused? Refractive index of water is 4/3.
$LU
:DWHU 7
O
Here, R2 = –R1 ∴
1 2 = 0.5 × 10 R
or R = 10 cm.
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Light (Geometrical Optics)
Now for air-glass interface µ 2 µ1 µ 2 − µ1 + = v u R 1.5 1 1.5 − 1 + = v ∞ 10 ∴ v = 30 cm. For glass-water interface u′ = – v = –30 cm, and R = –10 cm. µ1 µ 2 µ 2 − µ1 Now, + = u v' R 2 4 2 4 − 3+ 3 = 3 3 or v ' −30 −10 ∴ v′ = 20 cm
rio s_ in fo
na .c om
∴
er si ta rio s
ht tp
s: //
t.m e/
ht tp s:
un
//t
iv
.m
tp
e/
s:
un iv
//w
w
er s
w
ita
.ja
m
ar a
Example The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in the figure. (i) Where should a pin be placed on the optic axis such that its image is formed at the same place? (ii) The concave part is filled with water of refractive index 4/3. Find the distance through which the pin should be moved, so, that the image of the pin coincides with the pin.
ht
4A.58
) Solution (i)
1 1 µg 1 = − 1 − f g µa R1 R 2 1 1 1 = (1.5 − 1) − = 20 60 60 Now
1 1 1 1 = + + F f g fm f g =
=
1 1 1 + + 60 R 2/2 60
2 2 1 1 1+ 3 4 + + + = = 60 20 30 10 30 30
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Light (Geometrical Optics)
∴
F = 7.5 cm
But F =
4A.59
u , since, at u object and image coincide. 2
1 4 1 1 − − 1 × = f w 3 60 180
fw = 180 cm. = Focal length of combination, F′ is given by
co
∴
m
or
1 1 1 µw = − 1 − fw µa 60 ∞
Now
∴ u = 2F = 2 × 7.5 = 1.5 cm (ii) Here, fg = 60 cm, fm = 10 cm
s_
an
rio
ar
2 1 2 + + fw fm f g
1 1 1 + + 90 10 30
=
1 + 9 + 3 13 = 90 90
s si er iv
un
//t s:
io ta r
iv
un
.m
tp ht
90 13
But,
F′=
u′ , since, at u′ object and image coincide. 2
∴
u′ = 2 F ′ =
.m
//t
tp
s:
tp
e/
F′=
∴
ht
ta
=
si
2 1 2 + + 180 10 60
er
=
e/
s:
//w
w w
.ja
m
=
in
a.
fo
1 1 1 1 1 1 = + + + + F′ f w f g f m f g f w
180 cm = 13.85 cm 13
ht
Therefore, the pin has to be moved towards the lens by (15 – 13.85) cm = 1.15 cm
Example A thin equiconvex lens of glass of refractive index µ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water (µ = 4/3). On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in the figure. The separation between lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics) P
P
na .c om
0LUURU
) Solution The rays coming from the object will suffer refraction at air-glass
un iv
//w
w
µa µ g µ g − µa + = u v′ R
.m
iv
P:
u
u
ht tp
2
u u u u u u
Y
t.m e/
PD
u u u u u u PuJ u u u u u u
s: //
ht tp s:
//t
Y P
…(i)
un
tp
e/
s:
For refraction of glass-water surface,
er si ta rio s
er s
w
ita
.ja
m
ar a
rio s_ in fo
surface. It may be supposed to from a virtual image in the direction of refracted light. This image acts as an object for the subsequent refraction at glass-water surface to form the image. If this image is formed beyond the mirror, the actual image will lie in front of mirror at a distance identical to that of image behind the mirror. For refraction at air-glass surface,
ht
4A.60
O
u u u u u
O
O
µ g µw µw − µ g + = −v′ v −R Adding equations (i) and (ii), we get
…(ii)
µa µw µ g − µa µw − µ g + = − u v R R
…(iii)
The radius of curvature R of the surface of equiconvex lenses may be determined by the formula. 1 1 1 = µg −1 − f R1 R 2
(
)
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Light (Geometrical Optics)
4A.61
Here, Rg = –R1 = R 1 2 = (1.5 − 1) 0.3 R
∴
or R = 0.5 × 2 × 0.3 = 0.3 m Substituting this value of R along with other data in equation (iii), we get 3 4 3 −1 − 1 4/3 2 + = − 3 2 0.9 v 0.3 0.3 m 1 1 4 − = 0.6 0.8 1.8
or
4/3 4 1 2 = − = v 1.8 0.9 1.8
or
v=
rio s_ in fo
na .c om
=
m
ar a
4 1.8 × = 1.2 m 3 2
e/
tp
s:
Example
er si ta rio s
un iv
//w
w
er s
w
ita
.ja
The distance v = 1.2 m lies 0.4 m behind the mirror. Hence, the image is formed 0.4 m in front of the mirror. This point is also 0.4 m on the right of the convex lens.
iv
t.m e/
ht tp s:
un
//t
ht
.m
A converging lens of 20 cm focal length is arranged coaxially with a diverging lens of focal length 8 cm. A point object lies on the same side as the converging lens and very far away on the axis.
) Solution
ht tp
s: //
(i) What is the smallest possible distance between the lenses if the combination is to form a real image? (ii) If the lenses are placed 6 cm apart, what is the position and nature of the final image of the object? (i)
I ±FP
I FP
) [ I±[ I
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Light (Geometrical Optics)
For convex lens: Since, u1 = ∞, the rays would converge toward its focus F1, but the concave lens diverges and produces the final image. For concave lens: The point F1 would act as the virtual object for concave lens u2 = – (f1 – x) = – (20 – x)
1 1 1 1 1 −(20 − x) + 8 =− + = = − v2 f 2 u2 8 (20 − x) 8(20 − x)
∴
∴ v2 =
8(20 − x) x − 12
ta rio
s
er s
w
w
Example
s_ i
56 cm (virtual) (from concave lens) 3
.ja m
=−
nf
8(20 − x) 8(20 − 6) = x − 12 6 − 12
ita rio
∴ v2 =
ar an a
o
.c om
For the image to be real, v2 ≥ 0 Hence the minimum value of x should be 12 cm. (ii) Given: x = 6 cm
er si
ht
tp s:
//t
ht
.m e/
tp s:
//t
un iv
tp
.m e/
s:
un iv
//w
A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and image again coincide at a distance 25 cm from the lens. Calculate the refractive index of liquid.
ht
4A.62
) Solution Using Lens Maker’s equation 1 1 1 − = ( µ − 1) R R f 2 1 For the bi-convex lens µ = µg, R1 = R and R2 = –R
(
)
1 1 1 2 µg −1 = µg −1 − = R f1 R ( −R)
(
)
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Light (Geometrical Optics)
4A.63
For the plano concave liquid lens µ = µw, R = –R and R2 = ∞ 1 1 1 − ( µ w − 1) = ( µ w − 1) − = f2 R ( − R) ∞
∴
The focal length of the combination is
=
(
)
2 µ g − 1 − ( µ w − 1) R
...(i)
1 1 1 = + f f1 f 2
s_
ita rio
f = 15 cm
s ita rio ve rs
e/ un i
1 1 1 = + f 15 ∞
t.m
or
ht tp
s:
we get
//w
w
w
1 1 1 = + f u v
ve rs
.ja
m
ar an
in f
a.
o
co
m
With the above combination of lenses, it is given that the object and its image coincide with each other. This is possible when the light starting from the object point and after passing through the combination of lenses emerges perpendicular to the mirror so that the light retraces its path and reach the same starting material. Hence, in such a case we will have u = 15 cm and v = ∞ Thus, from the lens equation
e/ u
s: //
ni
Using f = 15 cm and the values of µg (= 1.5) and µw = 4/3 in equation (i), we get
.m
ht tp
R = f [2 (µg – 1) – (µw – 1)]
tp s:
//t
4 = 15 [2 (1.5 − 1) − ( − 1)] 3
∴
or
ht
2 = (15 × ) cm = 10 cm. 3 For the second case R = f[2 (µg – 1) – (µ1 – 1)] 10 = 25 [2 (1.5 – 1) – (µ1 – 1)]
µ1 = 2 −
10 40 = = 1.6 25 25
Example An object is kept at a distance of 100 cm from a screen. A convex lens is place between them produces a real magnified image on the screen. If the lens is shifted 30 cm towards the screen, real image is again obtained. Find the focal length of the lens. Calculate the size of the object, the image size are 16 mm and 9 mm respectively. https://t.me/universitarios_infohttps://www.jamarana.com
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4A.64
Light (Geometrical Optics)
) Solution The situation is based on the displacement method. Here, D = 100 cm, L = 30 cm Now, f =
D 2 − L2 1002 − 302 91 = = cm. 4D 4 × 100 4
We must verify that, D = 100 cm > 4f Image sizes are I1 = 16 mm and I2 = 9 mm. Object sizes = I1I 2 = 16 × 9 = 12mm
er si ta rio s
er s
iv un
//t
ht
.m
tp
e/
un iv
(QODUJHG LPDJH
s:
//w
w
w
ita
.ja
m
Simple Magnifier
rio s_ in fo
ar a
4.12 OPTICAL INSTRUMENTS
na .c om
Object sizes = I1I 2 = 16 × 9 = 12mm
)
2EVHUYHU
t.m e/
ht tp s:
)
s: //
FP
ht tp
A simple magnifier (or magnifying glass) consists of a convex lens to be placed before an object that is to be examined. The object is adjusted between the focus and lens, so, that a virtual magnified image is seen by eye. For most distinct vision, the image is usually adjusted to be at a distance of 25 cm. Let, Magnification obtained = M ∴
M=
25 v− f v 25 = − 1 = − + 1 ≈ − f f f f
Thus, the image size is almost
25 times the object size. f
Microscope A compound microscope basically consisting of two converging lenses, the objective and the eye-piece, mounted on a common axis, in two coaxial tubes, one telescoping into another. The distance between the lenses can be altered at will. https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Geometrical Optics)
4A.65
( %
2
Y
$ IR
Y
IH
$
IH
$
X %
X
%
rio s_ in fo
m
Angle subtended by the final image(at eye) Angle subtenede by the object at least distance of distinct vision from the eye
ita
.ja
=
ar a
na .c om
AB is the object held slightly beyond the principal focus of the objective. A′B′ is the real image produced by the objective, which serves as the real object for the eye-piece. A ′B′ lies within the focal length of the eye-piece. A “B” is the final image (virtual with respect to A ′B′) produced by the eye-piece. The final image is inverted with respect to the original object AB. The eye-piece works as a simple microscope. Magnifying power of a compound Microscope
er si ta rio s
e/
s:
un iv
//w
w
er s
w
= Magnification by the objective × magnifying power of the eye-piece v v = 1 1 + 2 u1 f2
iv
s: //
Special Cases
t.m e/
ht tp s:
un
//t
ht
.m
tp
where, u1 = Distance of the object from the objective. v1 = Image distance of the image produced by the objective. v2 = Distance of the final image from the eye-piece. f2 = Focal length of the eye-piece.
ht tp
(i) The final image at the distance of distinct vision from the eye (eye-piece) v2 = D, the distance of distinct vision from the eye.
∴ Magnifying Power =
v1 D 1 + u1 f2
(ii) Final distance at infinite distance from the eye-piece, D >>> f2 Magnifying Power =
v1 D × u1 f 2
(iii) To get maximum magnification, the object is very nearly at the focal point of the objective.
∴ u1 = f1 Magnification Power =
v1 D × for D >>> f 2 u1 f 2
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Light (Geometrical Optics)
Telescope It is similar to microscope with the difference that in microscope the objective has smaller focal length and aperture, while in the telescope the eye-piece has smaller focal length and aperture. X
Y $
D
D
$
E
na .c om
%
Y
rio s_ in fo
%
m
ar a
The object is at infinite distance for an astronomical telescope. The image A′ B′ produced by the objective is in the focal plane. The distance of eye-piece is so adjusted that the final image is
er si ta rio s
w
er s
w
ita
.ja
(i) at the distance of distinct vision, or (ii) at infinity.
e/
iv
.m
un
t.m e/
ht tp s:
∴
A′B′ u v β Magnifying Power = = 2 = 1 A ′ B ′ α u2 v1
//t
tp
s:
un iv
//w
Magnifying power of the telescope is defined as the ratio of the visual subtended by the image at the eye to that subtended by the object, directly at the eye.
ht ∴
s: //
For astronomical telescope, v1 = f1, (the focal length of the objective). If the final image is also at infinity, u2 = f2 (the focal length of the eye-piece). Magnifying Power =
ht tp
4A.66
f1 f2
If the telescope is used to view nearer objects as in the case of field glasses, A′B′ A′B′ α= ,β= v1 u2 ∴
Magnifying Power =
Now, v1 = ∴
v1 u2
u1 f1 v f and u2 = 2 2 u1 − f1 v2 − f 2
u f v − f2 Magnifying power = 1 1 2 u1 − f1 v2 f 2
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Light (Geometrical Optics)
4A.67
Special Cases (i) Final image of at distance of distinct vision v2 = D u f D − f2 ∴ Magnifying power = 1 1 u1 − f1 Df 2 (ii) Final image at ∞, v2 = ∞ u f Magnifying power = 1 1 u1 − f1
f2 v2 u1 f1 1 × f 2 u1 − f1 f 2
1−
f1 D − f 2 f 2 D
ar a
Magnifying power =
ita
.ja
m
4.13 CHROMATIC ABERRATION
rio s_ in fo
na .c om
(iii) Object at infinity, final image at distinct vision. u1 = ∞, v2 = D
er si ta rio s
iv
ht tp
s: //
t.m e/
ht tp s:
un
//t
ht
.m
tp
e/
s:
un iv
//w
w
er s
w
The refractive index of glass is different for different wavelengths (colours). Thus, when a white light ray strikes a lens, smaller wavelengths get refracted more than the larger wavelengths.
Voilet V oilet
fv
Red Fv
Fr
fr
In the above figure, violet rays are focused closer to a converging lens than are red rays. This property of a lens to converge of different wavelengths to different foci is known as chromatic aberration. As
µred < µviolet
∴
fred > fviolet
The colour spread on the axis of lens = fr – fv ≈ wfm where, fm is the mean focal length (usually for yellow colour).
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Light (Geometrical Optics)
Achromatic Combination of Lenses Two lenses-one converging and one diverging can be combined in such a way that the combined lens will not have chromatic aberration. If w1, w2 are dispersive powers of lenses and f1, f2 are their focal lengths, chromatic aberration will be eliminated if w1 w2 + =0 f1 f2 Such a combination is known as an achromatic combination of lenses. The equivalent focal length (F) can be calculated by
Example
rio s_ in fo
na .c om
1 1 1 = + F f1 f 2
er s
w
ita
.ja
m
ar a
A microscope of an objective lens of focal length 0.9 cm placed 13 cm from an eye-piece of 5 cm focal length. An insect .05 cm long is placed 1 cm from the objective. Find the nature, position and the size of the final image.
er si ta rio s
un iv
u1 f1 1× 0.9 = 9 cm.. = u1 − f1 1 − 0.9
.m
Objective
O
t.m e/
ht tp s:
V1
u2
ht tp
s: //
1 cm l2
Eye-piece
un
//t
iv
v1 =
tp
s:
∴
e/
//w
w
) Solution For the objective: u1 = 1 cm, f1 = 0.9 cm
ht
4A.68
l1 13 cm
For eye-piece: Distance between image of object and eye-piece, u2 = 13 – 9 = 4 cm, f2 = 5 cm v2 =
u2 f 2 4×5 = = − 20 cm u2 − f 2 4−5
Size of image =
v1 v 9 9 × size of object =1 × size × 0.05 of =object 0.45 cm. = × 0.05 = 0.45 cm. u1 u1 1 1
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Light (Geometrical Optics)
4A.69
v2 −20 × (size of first image) = × 0.45 = −2.25 cm u2 4
Size of final image =
v2 −20 × (size of first image) = × 0.45 = −2.25 cm u2 4 Hence, the final image is virtual at a distance of 20 cm from eye-piece and 2.25 cm long.
na .c om
1. A man stands 2 m in front of a concave spherical mirror. He finds that his image is one and a half times greater than that in a flat mirror placed at the same distance from the face. Find the radius of the concave mirror.
ita
When h < < 2a, α and β are small.
ht tp
H 3 h = a + b 2 2a
∴
v=b v=b
⇒
u=a
And for small angles, tan α = α, tan β = β
H 3 ( a + b) = h 4a
I I
u=a
er si ta rio s
un
t.m e/
β
O
s: //
β
β 3 = α 2 (i.e., the image of the man on the concave 3 mirror is times greater than that on the flat 2 mirror.) h H tan α = and tan β = ∴ 2a a+b Given that,
iv
.m //t ht tp s:
hO
…(ii)
er s
un iv
e/
s:
tp
ht h
H b = h a
m
.ja
w
//w
)
w
) Solution When the man (object) sees his own image by two different mirrors, the ratio of the sizes equals the ratio of the angles subtended at the position of the object.
rio s_ in fo
1 1 1 = + …(i) f a b Now for concave mirror, the linear magnification of the mirror is For concave mirror
ar a
Solving equations (i), (ii) and (iii), we get f =
h
R = 2f = 6 m
∴
h I
α a a
a a
I
2. A prism of flint glass with an angle of refraction 30º is placed inside water. (i) At what angle should a ray of light fall on the face of the prism so that
O
α
h O
h
3 3 a = ×2 = 3 m 2 2
WORKED OUT EXAMPLES
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4A.70
Light (Geometrical Optics)
For the extreme rays r
θ
inside the prism, the ray is perpendicular to the bisector of the angle of the prism? (ii) Through what angle with the ray turn after passing through both faces of the prism?
P 2r
α
) Solution The ray inside the prism QR is
)
m
3 2
8 3
ta
rio
ta
cos α =
s
s_
sin θ 1 2 1 = × = µ 2 3 3
rio
si
er
iv
un .m e/
2r r = 0 sin(90 + θ − α) sin α
//t
C
r 1 = 2r 2
sin α =
r0 =
s:
sin i = µ sin 15º i = 18.7º
or ∴
⇒
iv un S
s: ht tp
B
Q
In ∆ OPQ, angle ∠OQP = 90 + θ – α Applying sin e rule
//t
ht
i2
.m e/
r2
Q r1
si er
w R
3 2
in
a. m ar
⇒
w //w s: tp
D
r0
sin θ sin α
µ=
.ja
A
P
cos θ =
an
sin i = sin r
i1
⇒
µ
fo
co
Also A = 30º ∴ r = 15º. µ
sin θ =
O
Deviation, D = i1 + i2 – A = 2i – A A = r1 + r2 = 2r,
α
θ
evidently parallel to the base of the prism BC. Hence, we can conclude that the ray is undergoing minimum deviation. ∴ i1 = i2 and r1 = r2 But
θ
sin α sin α ⋅ 2r = 2r cos (θ − α) cos θ cos α + sin θ sin α
ht tp
sin α sin α ⋅ 2r = 2r cos (θ − α) cos cos sin sin θ α + θ α 3. A vertical beam of cross-sectional radius 1 r is incident symmetrically on the curved × 2r 2 surface of a glass hemisphere (µ = 3/2) 3 = = 2r of radius 2r placed with its base on a 3 8 1 1 2 6 +1 × + × horizontal table. Find the radius of the 3 2 3 2 luminous spot formed on the table. = 0.678r r0 =
)
The path of beam within the hemisphere is as shown. Let, θ = angle of incidence α = angle of refraction
4. A ray of light parallel to the axis of a concave mirror is incident at an angle of 30º. This ray after reflection meets the axis at I. Find PI, where P is the pole of the mirror. Focal length of the mirror is 10 cm.
) Solution
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Light (Geometrical Optics)
) Solution
1
3 2 η (y) ky 2 + 1 , where k = 1.0 (metre)–3/2 The refractive index of air is 1.0 (i) Obtain a relation between the slop of the trajectory of the ray at a point B (x, y) in the medium and the incident angle at that point.
)
For the mirror R = 2f = 20 cm ∴ AC = 20 cm
A 30° 30°
P
30°
60°
I
F
Y
C
P(r1, y1)
AIR
B(x, y)
m
K
3
s_ rio
s
rio
ta
si
er
e/ s:
= 10(2 − 3 +
1 3
)=
)
.m
//t
Y
10 3
B 1m
(2 3 − 2)
i
(r, y)
= 10(2 − 3) +
10
ht tp
From equations (i) and (ii), we get PI = PK + KI
i + θ = 90º
Slop, m = tan θ = tan (90 – i) = cot i
tp
ht
…(ii)
= 20 − 10 3 = 10(2 − 3)
un iv
s:
(i) From the figure,
PK = PC – KC
Now,
ta un
.m e/
) Solution
//t
ht
KC = 10 3
∴
…(i)
3 KC KC cos 30° = = AC 20
X
(ii) Obtain an equation for the trajectory y(x) of the ray in the medium. (iii) Determine the co-ordinate (x1, y1) of the point P, where the ray intersects the upper surface of the slab-air boundary. (iv) Indicate the path of the ray subsequently.
si
er
10
KI =
In ∆AKC,
iv
tp s: //w
w
w .ja
AK 10 = KI KI
tan 60° =
In ∆AKI
AIR
A(0, 0)
m
∴
Medium
in
a.
fo
co
t = 1.0m
ar an
AK AK = AC 20 AK = 10 cm.
In ∆ AKC, sin 30º =
∴
4A.71
3
)=
10
(2 3 − 2) 3 = 8.45 cm
5. A ray of light travelling in air is incident at grazing angle (incident angle = 90º) on a long rectangular slabs of a transparent medium of thickness t = 1.0 cm. (seen the figure). The point if incidence is the origin A (0, 0). The medium has a variable index of refraction η (y) given by
O
(ii) As the light passes through different medium, hence by Snell’s law sin α1 η2 = sin α 2 η1
1
∴ η1 sin α1 = η2 sin α2 = η3 sin α3 =...
3
= 10(2 − 3 +
10
3) +
i.e., at any point in the medium with variable refractive index η (y) sin λy = C = Constant
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4A.72
Light (Geometrical Optics)
where, η(y) = refractive index at any point y. αy = angle of incidence at any point y. At y = 0, Refractive index, η1 = [k ⋅ (0) + 1]1/ 2 = 1
P
mR
R
)
) Solution 3
α3
α2
2 1
α1
(iii) At y = 1 m x = 4 m
or
1.5 0.5 = R (1.5m + 1) R (1.5 m + 1) 0.5 = 1.5 2 4 m= = 1.5 3
7. The x – y plane is the boundary between two transparent media. Medium 1 with z ≥ 0 has a refractive index 2 and medium 2 with z ≤ 0 has a refractive index 3 . A ray of light in medium 1 given by
6. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance of mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure.
or
or
(iv) Let αe be the angle of emergence in the air. Then by Snell’s Law ηe sin αe = η1 sin α1 = 1 where, ηe = refractive index of air = 1 ∴ sin αe = 1 or αe = 90º Thus, it emerges parallel to the incident ray.
x 256
ht
y=
4
(∵k = 1)
∴
ht
∴ 4y1/4 = x
tp s
1 1 1.5 − 1.0 + = ∞ mR v or v = – 1.5 mR Since for the refraction at the curved surface, the image I acts as an object for this refraction, we will have µ1 = 1.5, µ2 = 1.0, u = 1.5 mR + R v = ∞, R = – R 1.5 1.0 1.0 − 1.5 + = ∴ −R 1.5 mR + R ∞ ∴
At x = 0, y = 0, Therefore c = 0
y +1/ 4 = kx+c (+1/4)
For the refraction at the plane surface, we have µ1 = 1.0, µ2 = 1.5, u = mR, R = ∞
dy = k y 3/ 4 dx
:// w
µ1 µ 2 µ 2 − µ1 + = u v R
2
∴ y −3/ 4 dy = k dx ⋅
The refraction formula is
∴ m=
R
V
or ky + 1 = 1 + m 3/2
u
P
I
w tp w s: .ja //t m .m ht ar tp e an /u s: a. ni //t co v .m e m rs e/ i un ta rio iv er s_ si in ta fo rio s
since, α1 = 90º, therefore, C = 1 × 1 = 1 ∴ η (y) = sin αy = 1 Substituting the values, we get 1 [ky 3/ 2 + 1]1/ 2 ⋅ =1 1 + m2
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4A.73
Light (Geometrical Optics)
Since, the refracted ray lies in the plane of incident ray (the plane containing the vector A) and the normal to the incidence plane, the refracted vector may be express as A r = ai + b j + ck
the vector A = 6 3i + 8 3j − 10 k is incident on the plane of separation. Find the unit vector in the direction of the refracted ray in medium 2.
) Solution )
with the value of a and b are equal to the corresponding values in the incident vector, i.e., A r = 6 3 i + 8 3 j + ck The value of c can be determined from the fact that A r ⋅ (−k ) cos r = | Ar |
z
si
(6 3) 2 + (8 3) 2 + c 2
1 2
−c
o
2
=
ta rio s_
ar a
w
2
−c
=
.ja m
1
1
(6 3) 2 + (8 3) 2 + c 2
in f
Thus,
na .
r
x
θ
co m
i
y
1 2
=
−c 1
2 2 300 + c
−c
=
1
2 2 300 + c
ve r
tp
)
ht
s:
//t
ht
.m
tp
e/
s:
un i
//t
ht
.m
tp
e/
si
un i
ta rio s
ve r
w
//w
s:
The given diagram in the figure. c = ± 10 3 The incident ray represented by the vector A and the refracted wave represented by Ar shown on the plane (x – y) of incidence (z = 0). Selecting, c = − 10 3 (as the refracted rays The cosine of the angle between the vector proceeds downward like the incident ray) A and the positive z direction is We get A r = 6 3i + 8 3 j − 10 3k A⋅k cos θ = |A| The unit vector of Ar is −10 (6 3 i + 8 3 j − 10k ) ⋅ k 1 =− = =− 1 A 6 3 i + 8 3 j − 10 3 k 3 i + 4 j − 5k r 20 2 = = 1 (6 3) 2 + (8 3) 2 + (−10) 2 2 | Ar | 5 2 (6 3) 2 + (8 3) 2 + (10 3) 2 2 −10 (6 3 i + 8 3 j − 10k ) ⋅ k 1 =− = =− 1 A 6 3 i + 8 3 j − 10 3 k 3 i + 4 j − 5k r 20 2 = = 1 (6 3) 2 + (8 3) 2 + (−10) 2 2 | Ar | 5 2 (6 3) 2 + (8 3) 2 + (10 3) 2 2 ∴ θ = 120º 8. When a telescope is in normal adjustAngle of incidence, ∠i = 180º – 120º ment, the magnification produced is 30. = 60º. The eye-piece has a focal length of 5 cm. From Snell’s Law of refraction, Find the magnification if the final image is to be formed at the distance of distinct sin i µ 2 3 = = vision. What will be the length of the telesin r µ1 2 scope in that case? 2 2 1 = (sin 60°) = Hence, sin r = (sin i ) ) Solution In normal adjustment, final image is 3 3 2 f formed at infinity and magnification is o 1 2 fe Also, cos r = 1 − sin r = 2 https://t.me/universitarios_infohttps://www.jamarana.com
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4A.74
fo = 150 cm.
or
M=
) Solution
fo = 30 fe
∴
)
Light (Geometrical Optics)
v1
A'
For the final image at distinct vision (25 cm), the magnification is
A v1=D
25 + f o 25 + 5 M = + fo = 150 = +36 25 × 5 25 f e
B'
Length of telescope = fo + object distance for eye-piece ve f e ve − f e
B
u1
The object is within the focal distance
m
β=
s_
Hence, the magnification will be 36 and the length of telescope = 154.17 cm.
rio
Magnifying power =
β D 25 = 1+ = 1+ =6 α f 5
er
//t
11. A telescope has an objective of focal length 50 cm and an eye-piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 200 cm away from the objective. Calculate: (i) The separation between the objective and the eye-piece. (ii) The magnification produced.
10. A convex lens of focal length f is used in simple microscope. Find its magnifying power if the final image is at the distance of distinct vision. Find also the magnification and the distance of the object f = 5 cm, D = 25 cm.
25 × 6 =6 25
12 25 31 31 = 10.33 1+ = 2× = 6 6 6 3
ht
M=
125 25 = cm. 30 6
Magnification =
∴
s:
=
v1 f −25 × 5 = v1 − f −25 − 5
.m
e/ un
iv
Also, u1 =
tp
where, u = 6 cm, fo = 4 cm, fe = 6 cm, D = 256 uf o 6× 4 = = 12 cm ∴ v= 2 u − fo
si ta
un
e/
m
//t .
s:
tp
v D 1 + u fe
ht
M=
rio
iv
w
://
tp s
ht
)
produced by the objective × angular magnification produced by the eye-piece. ∴
s
er
w w
si ta
.ja
9. The focal lengths of the objective and the eye-piece of a compound microscope are power = β = 1 + D = 1 + 25 = 6 Magnifying 4 cm and 6 cm respectively. If an object α f 5 is placed at a distance of 6 cm in front of A′B′ v1 25 the objective, what is the magnification = = Magnification = AB u1 41 produced by the microscope?
m
ar
∴
) Solution Magnification = magnification
A′B′ AB AB = ; α= v′ u1 D
in
an
a.
co
(−25)(5) 25 = 150 + = 150 + = 154.17 cm. −25 − 5 6
fo
= fo +
β α
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Light (Geometrical Optics)
) Solution
)
)
) Solution I
E
fe
B'
Object and image will coincide if the rays strike the mirror normally and come back along the same path. Hence, the ray must diverges away from the centre of the curvature C of the mirror. The concave lens must produce a virtual image at C for the lens:
B''
rio
f =
rs
ita
uv 12 × (−7.5) = = − 20 cm. u+v 12 − 7.5 = –20 cm.
un
iv e
uv 12 × (−7.5) = = − 20 cm. u+v 12 − 7.5
13. An equiconvex lens of focal length 30 cm in air is kept stationary on the surface of water (µ = 4/3). A point object is placed at a height 90 cm above the centre of the lens in air. If the refractive index of glass is 1.5, find the position of the final image.
.m
e/
//t
s:
tp
v v′ 200 1 6 = × = × × 25 × =2 u u′ 3 200 25
12. An object placed at 12 cm from a concave lens coincide with its image if a concave mirror of focal length 15 cm is placed at a distance of 22.5 from the lens on the other side. Find the focal length of the lens.
)
) Solution As the medium is not same on both sides, we cannot use
1 1 1 + = u v f
ht tp s:
//t
ht
−25 × 5 25 v′ f ′ = = cm v′ − f ′ −25 − 5 6
Total magnification produced = mo × me
er e/ un
f =
.m
tp
ht
where l is the separation between the lenses. The image (final) distance, v′ = 25 cm from the eye-piece and f′ = 5 cm.
u = 12 cm v = – (30 – 22.5) = –7.5 cm.
iv
w
s: //
si ta
.ja
w w
∴
This serves as an object for the eye-piece. The distance between A′B′ and the eye-piece is 200 u′ = l − , 3
s_ i
an a
m
ar
∴ and
200 × 50 1000 200 = = cm 200 − 50 15 3
200 × 50 1000 200 = = = cm 200 − 50 15 3
u′ =
nf o
.c om
Let AB be the position of the object. Here, u = 200 cm, f = 50 cm Let the image A′ B′ formed by the objective is at a distance v from it. uf v= ∴ u− f =
C
O.I.
s
A'
fo A''
B
rio
Q
Let us first calculate the ratio of the curvature of lens. Let x = magnitude of radius of curvature of each side.
A
∴
4A.75
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4A.76
Light (Geometrical Optics)
Refraction at first surface of the lens. µ µ µ − µ1 Using the relation, 1 + 2 = 2 u v R 1 1.5 1.5 − 1 + = 90 v1 +30
O
or
1.5 1 1 3− 2 1 = − = = v1 60 90 180 180
∴
v1 = 180 × 1.5 = 270 cm.
Air
Glass
Refraction at second surface: Glass to water µ µ µ − µ1 Using the relation, 1 + 2 = 2 u v R 4 − 1.5 1.5 4/3 3 + = or −270 v2 −30
fo
in
ta
rio s ta si er
uestions
un
iv
pe
)
2h µ
2. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids L1 or L2 having refractive indices n1 and n2
1 (d) h 1 + µ
2h µ
= x2 − x1 =
2h µ −1
rio s_
si er iv un
ht tp
ht tp
s: //t .m e/
(either plane or spherical) does not depend on the medium The image of P will be formed at a distance h below the mirror If d = depth of the liquid in the tank, then d −h apparent depth of P = x1 = µ apparent depth of the image of P d +h = x2 = µ apparent distance between P and its image
h
(b)
(c)
q
) Explanation Image formation by a mirror
O
(a) 2µh
a.
w s: //t .m e/
tp s
:// w
w
ht
SOLVED OBJECTIVE y
1. A plane mirror is placed at the bottom of a tank containing a liquid of refractive indeed µ. P is a small object at a height h above the mirror. An observer O, vertically above P, outside the liquid-sees P and its image in the mirror. The apparent distance between these two will be
P
4 −0.5 1.5 = + 3v2 −90 270
or v2 = 120 cm. Hence, the final image is formed 120 cm below the water surface.
t
x = 30 cm.
or
or
.ja
1 1 1 = (1.5 − 1) − 30 x −x
or
m ar an
1 µ 1 1 = − 1 − f 1 R1 R 2
co
m
Water
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Light (Geometrical Optics)
respectively (n2 > n1 > 1). The lens will diverge a parallel beam of light if it is filled with (a) air and placed in air (b) air and immersed in L1 (c) L1 and immersed in L2 (d) L2 and immersed in L1
B
A
θ
θ p
θ
4A.77
C
s
rio
ta
si
v=
4/3 8 = 3/2 9
( f0 + fe ) fe f0
f v image size l = e = = u f 0 object size L
f0 L = = magnification of telescope in normal adjustment l fe
= magnification of telescope in nomal adjusment 5. A ray of light passes through four transparent media with refractive indices
−18 9
iv
un or
∴
f0 1 1 1 = − = v fe f0 + fe ( f0 + fe ) fe
θ ≤ sin
or
Magnification =
n sin θ = w sin 90° ng
⇒
e/
)
internally at the point P (say). Therefore critical angle ≅ θ. Applying Snell’s Law of refraction at the critical angle of incidence (θ) we get
sin θ =
1 1 1 − = v −( f 0 + f e ) f e
ht
(d) None of these
) Explanation The light ray is just reflected
⇒
of the objective and eyepiece respectively. For normal adjustment, distance from the objective to the eyepiece (tube length) = f0 + fe. Treating the line on objective as the object, and the eyepiece as the lens, µ = – (f0 + fe) and f = fe.
:// t.m
2 3
tp s
2 8 < sin θ > 3 9
L+1 L−l
er
e/
(b) sin θ <
(c)
(d)
) Explanation Let f0 and fe be the focal lengths
//t s: ht tp
8 9
(a) sin θ ≥
rio
ita rs un
iv e
A
.m
ht tp C
s_ in
an
ar
m
.ja w w θ
L −1 l
(c)
s:
//w
B
fo
a. co
m
4. In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is L L +1 (b) (a) l l
)
)
3. A glass prism of refractive index 1.56 is immersed in water (R.I. = 4/3). The beam of light incident normally on the face AB is totally reflected to reach the face BC, if
Hence (d) is the correct answer.
2 1 µ2 = − 1 − f µ1 R µ Here f is negative if 2 > 1, i.e., µ2 = n2 and µ1 µ1 = n1 Hence (d) is the correct answer.
) Explanation
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4A.78
µ1
µ4
µ3
30º at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is (a) 28 (b) 30 (c) 32 (d) 34
µ1, µ2, µ3 and µ4 as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have
Light (Geometrical Optics)
2√3 m
C
B
B 30°
in s_
ta
.ja
= 0.2 tan 30° =
s
rio
er
(0.2/ 3)
iv
8. A screen beaming a real image of magnification m1 formed by a convex lens is moved a distance x. The object is the moved until a new image of magnification m2 is formed on the screen. The focal length of the lens is x x (a) (b) m2 − m1 m2 − m1
e/
un
x
(d) None of these
) Explanation In first case,
⇒
1 + m1 =
q f
1 1 1 q + = and = m1 p q f p
7. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle
m1m2
)
…(ii)
f0 =m=5 fe where, m = magnification = 5 Solving equations (i) and (ii) we get 5 fe + fe = 36 ⇒ fe = 6 cm and f0 = 5fe = 30 cm
l = length of the tube = 36 cm
(c)
...(i)
.m
//t
s:
tp
ht
ta
un
e/
.m
//t
s:
tp
ht
)
where
= 30
si
iv
//w w
s:
tp
ht
2 3
=
) Explanation We know f0 + fe = 36
0.2
3 Hence required number of reflections
er
w
si
)
6. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eye piece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the length fe of the eye piece respectively are (a) f0 = 45 cm and fe = –9 cm (b) f0 = 50 cm and fe = 10 cm (c) f0 = 7.2 cm and fe = 5 cm (d) f0 = 30 cm and fe = 6 cm
rio
)
m
due to a single reflection
) Explanation From Snell’s law, µ sin θ = constant ⇒ µ1 = µ2
A
) Explanation Forward shift of point of incidence
ar
(b) µ2 = µ3 (d) µ4 = µ1
(a) µ1 = µ2 (c) µ3 = µ4
an
a.
A
fo
co
m
0.2 m
In the second case,
1 1 1 q+x + = and = m2 q + x p′ f p′
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…(i)
Light (Geometrical Optics)
q+x f Solving equations (i) and (ii), we get
…(ii)
m2 =
⇒
x f
O 3h
m2 − m1 =
x f = m2 − m1
r
i
r
2h
h
9. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is
i
10. A beam of light, consisting of red, green and blue colours, is incident on a right angled prison, as shown. The refractive indices of the material of the prison for the above red, green and blue wavelength are 1.39, 1.44 and 1.47 respectively. The prison will
2h
s_
si ta .m
tp
1
) Explanation The angle of incidence of all the
5
=
rays is 45º at the hypotencese. For a critical angle of 45º, the refractive index must be,
5h
2
)
h
h + (2h) 2
)
2
=
h 2
=
2
1
(sin 45°) −1 = 2 = 1.414
2
h +h
s
s: tp ht h
) Explanation sin i =
45°
(a) separate part of the red colour from the green and blue colours. (b) separate part of the blue colour from the red and green colours. (c) separate all the three colours from one another. (d) not separate even partially any colour from the other two colour.
//t
ht h
2h
90°
e/
un
s: //
iv
t.m
er
e/
si ta
un
iv
rio
w //w s: tp ht 3k
sin r =
45°
er
w
.ja
(d)
o
in f
a.
m
3 2
rio
3 2
ar an
5 2
(b)
(c)
5 2
(a)
co m
⇒
4A.79
2
or µ =
5
5 2
1 sin i = = µ sin r
For red light, µ = 1.39 < 1.414. Hence, its critical angle > 45º Therefore, red light will pass through the surface into air.
According to Shell’s law,
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4A.80
Light (Geometrical Optics) P
60°
r1
11. A ray incident at a point at an angle of incidence of 60º enters a glass spheres of R.I. n = 3 and is reflected and refracted at the further surface of the spheres. The angle between the reflected and refracted rays at this surface is (a) 50º (b) 60º (c) 90º (d) 40º
r2
α
s_
rio
rio s iv
un e/
.m
y
s:
µ2
tp
//t
ht
µ1
ht
uestions
er si
pe
t
tion of a ray in the two media are along the vectors r1 = aiˆ + bjˆ and r2 = ciˆ + djˆ , then
//t
s:
tp
q
ta
un
e/
.m
ht
tp
s:
//w
w
iv er
si
ta
w .ja
m
UNSOLVED OBJECTIVE y
1. A point source emits sound equally in all directions in a no-absorbing medium. The ratio of amplitudes of the waves by the source at P to that of Q is 2 : 3. Ratio of distance of P and Q from the source is (a) 2 : 3 (b) 3 : 2 (d) 3 : 2 (c) 2 : 3
fo
in
a.
Putting r2 = 30º we get i2 = 60º Reflection at Q r2 = r1 = 30º ∴ α = 180º – (r2 + i2) = 180º – (30º + 60º) = 90º
ar an
sin r2 1 = sin i2 3
Refraction at Q
co m
)
sin 60° = 3 sin r1 1 sin r1 = ⇒ 2 ⇒ r1 = 30º Since r2 = r1, therefore r2 = 30º
(a) µ1a = µ2b
µ1a
(b)
=
µ2 d
a +b c2 + d 2 2 2 (c) µ1(a + b ) = µ2 (c2 + d2) (d) None of these
2
2
4. A liquid of refractive index 1.33 is placed between two identical palno-convex lenses, with refractive index 1.50. Two possible arrangement P and Q are shown. The system is
3. Two thin slabs of refractive indices µ1 and µ2 are placed parallel to each other in the x-z plane. If the direction of propaga-
x
2. A point object is placed at a distance of 12 cm from a convex lens of focal length 10 cm. On the other side of the lens, a convex mirror is placed at a distance of 10 cm from the lens such that image formed by the combination coincides with the object itself. The focal length of the convex mirror is (a) 20 cm (b) 25 cm (c) 15 cm (d) 30 cm
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Light (Geometrical Optics)
4A.81
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move to the left move to the right remain at P move either to the left or right, depending upon focal lengths of the lenses
9. Half the surface of a transparent sphere of refractive index 2 is silvered. A narrow, parallel beam of light is incident on the unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from the sphere will be a (a) parallel beam (b) converging beam (c) slightly divergent beam (d) widely divergent beam
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6. Two bodies A and B are moving towards a plane mirror with speeds vA and vB respectively as shown in the figure. The speed of image of A respect to the body B is
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5. A ray of light on a slab of transparent material is party reflected from the surface and partly refracted into the slab. The reflected and refracted rays are mutually perpendicular. The incident ray makes on angle i with the normal to the slab. The refractive index of the slab is (a) tan–1(i) (b) cot–1(i) –1 (c) sin (i) (d) cos–1(i)
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of a convex mirror of radius of curvature 10 cm, in order to have an image of the object coincidet with it, is (a) 70 cm (b) 30 cm (c) 50 cm (d) 60 cm
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4A.82
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Light (Geometrical Optics)
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4A.83
close very close to each other in vacuum. A point source is placed at the center point C1 of first sphere. Image formed by paraxial rays when viewed through second sphere is
15. A graph is plotted between angle of deviation (δ) and angle of incidence (i) for a prism, where prism is being used to study dispersion. The plot will be like
Light (Geometrical Optics)
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16. A black spot B is marked very near to base of the prism (µ = 3/2) as shown in the figure. If the image of black spot appears on the surface of the prism, the angle of the prism A will be $
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18. A Young’s double slit experiment is conducted in water (µ1) as shown in the figure, and a glass plate of thickness t and refractive index µ2 is placed in the path of S2. The magnitude of the optical path difference at O, will be
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4A.84
FP
(b) 7.5 cm (d) none of these
23. If final image formed after two refractions through the lens and one reflection
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26. A man stands on a glass slab of height 1 and inside an elevator accelerated upwards with ‘a’. The bottom of the slab appears to have shifted with respect to the man by a distance (if the R.I. of the glass is µg) 1 (a) less than µg 1 (b) greater then µg 1 (c) equal to µg
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25. A beam of light coming from infinity is passing through a convex lens having radius of curvature R = 20 cm each; is focused at a certain distance from lens. Find the radius of convex of emergent wave front from lens: (µlens = 1.5, µ = 1) (a) 10 m (b) 5 m (c) 2 m (d) 1 m
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24. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiec is 36 cm. The final image is formed at infinity. The focal length f0 of the objective and te of the eyepiece respectively are (a) 45 cm and –9 cm (b) 50 cm and 10 cm (c) 7.2 cm and 5 cm (d) 30 cm and 6 cm
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21. On the hypotenuse of a right angled prism of refractive index 1.5, a drop of liquid is placed as shown. A ray of light falls normally on the AB. What should be the maximum value of refractive index of liquid, so that it gets totally reflected at face AC? $
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(d)
µR (µ + 1)
(c)
from the mirror forming at same point O, then will be
20. A sphere of radius R made of transparent material of refractive index µ works like a thin lens having its optical center coinciding with the center of the sphere. The focal length of this thin lens is µR µR (a) (b) (µ1 − 1) 2(µ − 1)
Light (Geometrical Optics)
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29. A ray of light incident on a transparent glass slab of refractive index 3 . If the reflected and refracted rays are mutually perpendicular, then angle of incidence is
28. A lens of refractive index µ is put in a liquid of refractive index µ′. If the focal length of the lens in air is f, its focal length in liquid will be
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7. (c) 15. (a) 23. (b)
8. (b) 16. (d) 24. (d)
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2. A person looking through a telescope just sees the point A on the rim of bottom of a cylindrical vessel when it is empty shown in the figure. When the vessel is filled with a liquid of refractive index 1.5, he observes a mark B at the centre of the vessel, without moving the telescope or the vessel. What is the height of the vessel if the diameter of its cross section is 10 cm? [IIT, 1977] [Ans. 8.45 cm] 3. In a river 2 m deep, a water level measuring post embedded into the river stands vertically with 1 m of it above the water
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EXERCISE 1
1. A point source of light is placed at the bottom of a vessel containing a liquid of 5 refractive index . A person is viewing 3 the source from above the surface. There is an opaque disc of radius 1 cm floating on the surface. The centre of the disc lies vertically above the source S. The liquid from the vessel is gradually drained out through a tap. What is the maximum height of the liquid for which the source cannot at all be seen from above? [IIT, 1970] [Ans. 1.33 cm]
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30. Images formed by an object when placed between two plane mirrors making an angle of 90º with one another lie on a (a) straight line (b) circle (c) ellipse (d) parabola
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(a) 60º (c) 30º
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(a)
ANSWERS
4A.85
27. A thin prism P1 of angle 4º, and made from a glass of refractive index 1.54, is combined with another thin prism P2 made from a glass of refractive index 1.72, to produce dispersion without deviation. The angle of P2 is (a) 5.33º (b) 4º (c) 3º (d) 2.6º
Light (Geometrical Optics)
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4A.86
Light (Geometrical Optics)
reflected ray and refracted ray are mutually perpendicular, what is the angle of incidence? Give calculations. [MNR, 1987] [Ans. 58.19′]
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7. A ray of light falls on a transparent glass slab of refractive index 1.62. If the
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11. A man standing on the edge of a water tank looks at a stone lying at its bottom. Find the position where he sees the image of the stone if the line of vision makes an angle θ with the normal to the water surface. Take the depth of water in the tank to be H. (µ 2 − 1) H sin 3 θ Ans. l = − (µ 2 − sin 2 θ)3/ 2
12. A convex lens of focal length 15 cm is placed in front of a convex mirror. Both are coaxial and the lens is 5 cm from the apex of the mirror. When an object is placed on the axis at a distance of 20 cm from the lens, it is found that the image coincides with the object. Calculate the radius of curvature of mirror. [Ans. 55 cm]
6. A ray of light is incident from a denser medium on to a rarer medium. If θ1C is the critical angle and θ1B is the Brewster’s angle of incidence, find the relative refractive index of the two mediums. Given sin θ1c that, = η = 1.28 [IIT, 1983] sin θ1B [Ans. 0.798]
[Ans. 2 ]
4
8 Ans. sin θ ≥ 9
10. Light is incident at an angle α on one planar end of a transparent cylindrical rod of refractive index n. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of the rod irrespective of the value of α. [IIT, 1992]
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9. A ray of light falls on a glass slab making an angle of 45º with the vertical shown in the figure. Find the refractive index of the glass if total internal reflection takes place at the vertical face AB. [Ans. µ = 1.22]
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8. Glycerine (refractive index 1.4) is poured into a large jar of radius 0.2 m. to a depth of 0.1 m. There is a small light source at the centre of the bottom of the jar. Find the area of the surface of glycerine through which the light passes. [Roorkee, 1998] [Ans. 0.0327 m2]
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5. A glass prism of refractive index 1.5 is 4 immersed in water (refractive index ). A 3 light beam incident normally on the face AB shown in the figure is totally reflected to reach the face BC if
4. Explain how a fish inside a pond full of water can see everything above the water near the pond. Show that the field of vision of the fish is concentrated within a cone of half angle 49º inside the water 4 (refractive index of water = and (sin 3 49º = 0.75). [Roorkee, 1980) [Ans. 49º]
surface. If the angle of inclination of sun above the horizon is 30º, calculate the length of the post on the bottom 4 surface of the river. µ for water = 3 [Roorkee, 1978] [Ans. 3.44 m]
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4A.87
18. A pin is placed 10 cm in front of convex lens of focal length 20 cm made of material of refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature 22 cm. Determine the position of the final image. Is the image real or virtual? [IIT, 1978] [Ans. 11 cm; real]
13. An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm at a distance of 15 cm from the lens. A plane mirror is placed inclined at 45º to the lens axis, 10 cm to the right of the lens. Find the position and size of the image formed by the lens and mirror combination. Trace the path of the rays forming the image. [IIT, 1972] [Ans. Both the points A and B are at a distance of 20 cm from the lens axis. Size of image = 8 cm]
19. The radius of curvature of the convex face of a plano convex lens is 12 cm and its µ = 1.5. (a) Find the focal length of the lens. The plane surface of the lens is now silvered. (b) At what distance from the lens will parallel rays incident on the convex surface converge? (c) When a point object is placed on the axis 20 cm from the lens, Calculate the image distance when the object is placed. [IIT, 1979] [Ans. (a) 24 cm, (b) 12 cm, (c) 30 cm]
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21. The convex surface of a thin concavo convex lens of glass of refractive index 1.5 has a radius of curvature 60 cm. The convex side is silvered and placed at horizontal surface. Find: (a) Where should a pin be placed on the optic axis such that its image is formed at the same place? (b) If the concave part is filled with water 4 of refractive index , find the distance 3 through which the pin should be moved
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20. An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. You may take the distance of the near surface of the slab from the mirror to be 1 cm. [IIT, 1980] [Ans. 7.56 cm]
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17. A concave mirror of radius of curvature 1 m is placed at the bottom of a tank of water. The mirror forms an image of the sun when it is directly overhead. Calculate the distance of the images from the mirror for different depths, 80 cm and 40 cm of the water in the tank. [IIT, 1968] [Ans. When the depth of water 80 cm, distance = 50 cm, When the depth of water 40 cm, distance = 47.5 cm]
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16. Photographs of the ground are taken from an airplane flying at an altitude of 2000 m by a camera with a lens of focal length 50 cm. The size of the film of the camera is 18 cm × 18 cm. What area of the ground can be photographed by this camera at any instant? [IIT, 1976] [Ans. 720 m × 720 m]
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15. An object is placed at 20 cm left of the convex lens of focal length 10 cm. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens, find the magnification of the final image. [IIT, 1974] [Ans. 1]
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14. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm it is found that there is no parallex formed by the two mirrors. What is the radius of curvature of the convex mirror? [IIT, 1973] [Ans. 25 cm]
Light (Geometrical Optics)
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4A.88
22. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. Find the power of combination. [Ans. –1.5 diopter]
sin
23. A convex lens of focal length 20 cm and a concave mirror with a radius of curvature 16 cm are placed at a distance of 25 cm from each other. A luminous object is placed at a distance of 25 cm from the lens. Where will be the image formed and what will be its magnification? [Roorkee, 1984] [Ans. In-between the lens and mirror and at a distance of 7.23 cm from the mirror. Magnification, –0.39]
θ b = 2 µ R2
2 2 2 1/2 2 2 1/2 (µ R − b ) − ( R − b )
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28. A parallel beam of light travelling in 4 water (refractive index = ) is refracted 3 by a spherical air bubble of radius 2 mm, situated in water. Assuming the light ray to be parallel, find the position of the image due to refraction at the first surface and position of final image? [Ans. Image is formed at a distance of 6 mm on the object side. Final image will be formed at a distance of 5 mm to the left of the second surface.]
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29. The distance between two point source of light is 24 cm. Find out where would you place a converging lens of focal length 9 cm so that the images of both sources are formed at the same point. [Roorkee, 1988] [Ans. Lens, should be placed at a distance of 18 cm or 6 cm from the source.]
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30. In a compound microscope the objective and eyepiece have focal lengths of 0.95 cm and 5 cm respectively and are kept at a distance of 20 cm. The last image is formed at a distance of 25 cm from the eyepiece. Calculate the position of the object and the total magnification. [Roorkee, 1988]
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27. Two lenses of power +12 and –2 diopters are placed in contact. What will be the focal length of combination? [MNR, 1987] [Ans. 10 cm]
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25. A convex lens of focal length 20 cm and a concave mirror with a radius of curvature of 16 cm are placed as shown in the figure, at a distance of 25 cm from each other. A luminous object is placed at a distance of 25 cm from the lens. Where will be the image formed and what would be its magnification? [Roorkee, 1986] FP
[Roorkee, 1987]
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24. A plano-convex lens has a thickness of 4 cm. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottommost point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face of 25 cm. Find the the lens is found to be 8 focal length of the lens. [IIT, 1984] [Ans. 75 cm]
26. A ray of light passes through a transparent sphere of refractive index µ and radius R. If b is the distance between the incident ray and a parallel diameter of sphere, show that the angle of deviation θ is given by expression
so that the image of the pin again coincides with the pin. [IIT, 1981] [Ans. (a) 15 cm, (b) 1.06 cm]
Light (Geometrical Optics)
I FP U FP
[Ans. 159.37 cm, –3.44]
95 Ans. − 94 cm from objective and left of it, 94
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4A.89
(a) the separation between the objective and eyepiece. (b) magnification produced. [IIT, 1988] [Ans. (a) 70.83 cm, (b) – 2
31. A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away from the objective. Calculate:
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Light (Geometrical Optics)
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B
15. Why does a diamond sparkle?
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18. Explain why the planets do not twinkle. 19. Why does the Sun appear reddish early in the morning?
20. Why does the sky appear dark instead of blue to an astronaut?
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21. Does short-sightedness (myopia) or longsightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
22. The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense does then a magnifying glass provide angular magnification?
23. In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back? 24. Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power.
10. A concave spherical mirror has a radius of curvature of 40 cm. Draw ray diagrams to locate the image (if one is formed) for an object at a distance of (a) 100 cm (b) 40 cm (c) 20 cm and (d) 10 cm from the mirror. For each case, state whether the image is real or virtual, erect or inverted, and enlarged, reduced, or the same size as the object.
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9. A truck uses a convex mirror as view finder whose radius of curvature is 2.0 m. A maruti car is coming behind the truck at a distance of 10 m. What will be the position of the image of the car and size of the image of the car when observed by the driver of the truck through the convex mirror?
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8. In what way is the word AMBULANCE printed infront of the hospital vans? Why is it printed this way?
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17. Why do stars twinkle?
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7. Will the focal length of the lens change when it is put in water?
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6. What do you mean by a focal plane?
16. Can you photograph a virtual image?
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5. A ray of light passes through the optical centre of a lens. Does it suffer any deviation?
14. Why do different coloured rays deviate differently in the prism?
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4. For driving a car, what type of mirror would you prefer to see the traffic at your back?
13. What are fibre cables?
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3. In case of a spherical mirror, do both the sides act as reflecting surfaces?
12. When a monochromatic light passes through a prism, will it show dispersion?
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2. If you want to see an enlarged image of your face, state whether you will use a concave mirror or a convex mirror.
11. In previous question instead of concave if convex mirror (same radius of curvature) is used what will be the answer.
1. A concave mirror forms a real image of a object. What change in the image do you expect if a fly (a) is sitting on the object. (b) is between object and mirror and (c) is sitting on the mirror?
Q
CONCEPTUAL UESTIONS
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Light (Geometrical Optics)
Question
4A.90
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Light (Geometrical Optics)
v sin θ (d) v tan θ (b)
(c) v cot θ
(b) 4 5 (d) 36/ 7
(a) 36 5 (c) 36 7
4. A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is
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12. A mango tree is at the bank of river and one of the branch of tree extends over the river. A tortoise lives in river. A mango falls just above the tortoise. The acceleration of the mango falling from tree appearing to the tortoise is (Refractive index of water is 4/3 and the tortoise is stationary) (a) g (b) 3 g/4 (c) 4 g/3 (d) None of these
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19. Liquid is filled in a vessel of height 2H/3. At the bottom of the vessel there is a spot P and a hole from which liquid is coming out. Let d be the distance of image of P from an eye at height H from bottom at an instant when level of liquid in the vessel is x. If we plot a graph between d and x it will be like
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30. A bird is flying up at an angle sin–1 (3/5) with the horizontal. A fish in a pond looks at that bird. When it is vertically above the fish. The angle at which the bird appears to fly (to the fish) is – [nwater = 4/3] (a) sin–1(3/5) (b) sin–1(4/5) (c) 45º (d) sin–1(9/16)
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34. The plane mirrors are placed symmetrically on disks revolving in different directions as shown. ‘S’ is the source and the beam from it reflects repeatedly from the mirrors. If the angular velocity of rotation of the disks is ω, then the angular velocity Ω of rotation of a beam that is consecutively reflected ‘n’ times from the mirrors is
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35. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of
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38. An air bubble is inside water. The refractive index of water is 4/3. At what distance from the air bubble should a point be placed so as form a real image at the same distance from the bubble? (a) 2R (b) 3R (c) 4R (d) the air bubble cannot form a real image of real object
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44. A pendulum of length ℓ is free to oscillate in vertical plane above point O. An observer is viewing the bob of the pendulum directly from above. The pendulum is performing small oscillations in water (refractive index is µ) about its equilibrium position. The equation of trajectory of bob as seen by observer is
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43. One face of a rectangular glass slab is silvered. The thickness of slab is 9 cm and refractive index for material of glass is 1.5. An object is held 8 cm from front face of the slab. The distance of image from silvered surface is
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40. A student holds a hand plane mirror to observe the back of her head while standing in front of and looking into a wall plane mirror. If she is standing 4 ft away from wall mirror in front of the wall mirror and she holds the hand mirror vertically 1 ft behind her head, she will see the back of her head approximately how far behind the wall mirror? (a) 6 ft (b) 5 ft (c) 4 ft (d) 3 ft
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39. Focal lengths of two lens are f and f´ and dispersive powers are ω0 and 2ω0. To form achromatic combination from these (a) f´ = 2f (b) f´ = –2f (c) f´ = f/2 (d) f´ = –f/2
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Light (Geometrical Optics)
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59. A planoconcave lens is placed on a paper on which a flower is drawn. How far above its actual position does the flower appear to be?
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55. The magnifying power of an astronomical telescope can be increased, if we (a) Increase the focal length of the objective (b) Increase the focal length of the eyepiece (c) Decrease the focal length of the objective (d) Decrease the focal length of the objective and at the same time increase the focal length of the eye-piece
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58. The figure shows a parallel slab of refractive index n2 which is surrounded by media of refractive indices n1 and n3. Light is incident on the slab at angle of incidence θ (≠ 0). The time taken by the ray to cross the slab is t1 if incidence is from n1 and it is t2 if the incidence is from n3. Then assuming that n2 > n1, n2 > n3 and n3 > n1, then value of t1/t2
53. A fly is sitting on the objective of a telescope pointed towards the moon. What effect is expected in photograph of the moon taken through the telescope? (a) The entire field of vision is blocked (b) There is an image of the fly on the photograph (c) There is not effect at all (d) There is a reduction in the intensity of the image
52. A concave mirror has a focal length 20 cm. The distance between the two positions of the object for which the image size is double of the object size is (a) 20 cm (b) 40 cm (c) 30 cm (d) 60 cm
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60. Prism angle of a prism is 10º. Their refractive index for red and violet colour is 1.51 and 1.52 respectively. Then dispersive power will be (a) 0.5 (b) 0.15 (c) 0.019 (d) 0.032
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72. Light travels through a glass plate of thickness t and having a refractive index µ. If c is the velocity of light in vacuum, the time taken by light to travel this thickness of glass is (a) tµc (b) tc/µ (c) t/µc (d) µt/c
66. A beam of monochromatic light is refracted from vacuum into a medium of refractive
71. The refractive index of water is 1.33. What will be the speed of light in water? (a) 3 × 108 m/s (b) 2.25 × 108 m/s 8 (c) 4 × 10 m/s (d) 1.33 × 108 m/s
70. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together, what will be their resulting power (a) +6.5D (b) –6.55D (c) +7.5D (d) –0.75D
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69. Focal length of a convex lens will be maximum for (a) blue light (b) yellow light (c) green light (d) red light
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68. The focal length of the convex lens depends upon (a) frequency of the light ray (b) wavelength of the light ray (c) both (a) and (b) (d) None of these
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63. A telescope consisting of an objective of focal length 60 cm and a single–lens eye piece of focal length 5 cm is focussed at a distant object in such a way that parallel rays emerge from the eye piece. If the object subtends an angle of 2º at the objective, then the angular width of the image will be (a) 10º (b) 24º (c) 50º (d) 1/6º
62. Diameter of the moon is 3.5 × 103 km and its distance from earth is 3.8 × 105 km. It is seen by a telescope whose objective and eyepiece have focal lengths 4 m and 10 cm respectively. The angular diameter of the image of the moon will be nearly (a) 5º (b) 10º (c) 20º (d) 25º
67. Time taken by the sunlight to pass through a window of thickness 4 mm whose refractive index is 1.5 is (a) 2 × 10–8 second (b) 2 × 108 second (c) 2 × 10–11 second (d) 2 × 1011 second
61. Prism angle and refractive index for a prism for a 60º and 1.414. Angle of minimum deviation will be (a) 15º (b) 30º (c) 45º (d) 60º
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74. A ray of light from air is incident in water then which property of light will not change in water (a) velocity (b) frequency (c) amplitude (d) colour
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1. The quantity quoted first in the options is plotted on y-axis against the quantity quoted latter in the option on x-axis. Then for which of the options will the graph have the form as shown?
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2. A parallel beam of light is incident normally on the flat surface of a hemisphere of radius 6 cm and refractive index 1.5, placed in air as shown in figure (i). Assume paraxial ray approximation.
76. Radius of curvature of convex mirror is 40 cm and the size of object is twice as that of image, then the image distance is (a) 10 cm (b) 20 cm (c) 40 cm (d) 30 cm
75. The focal length of convex lens is 30 cm and the size of image is quarter of the object, then the object distance is (a) 150 cm (b) 60 cm (c) 30 cm (d) 40 cm
77. The combined power of two lenses is contact is +10D. When they are separated by 20 cm, their power becomes +6.25D. The powers of these lenses are (a) –3.5D, +6.5D (b) –7.5D, +2.5D (c) +7.5D, +2.5D (d) +9.0D, +1.0D
4A.101
(a) The distance of object from focus of a convex mirror plotted against the distance of the image from the same point (b) pressure plotted against the temperature for an ideal gas undergoing process P V = constant. (c) electric field due to charged conducting sphere plotted against distance from the centre of the sphere. (d) the tension in a string plotted against its fundamental frequency
Light (Geometrical Optics)
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9. In case of three plane-mirrors meeting at a point to form a corner of a cube, if incident light suffers one reflection on each mirror (a) The emergent ray is antiparallel to incident one (b) The emergent ray is perpendicular to incident one (c) The emergent ray is in phase with incident one (d) The emergent ray is in opposite phase with incident one
4. A man of height 170 cm wants to see his complete image in a plane mirror (while standing). His eyes are at a height of 160 cm from the ground. (a) Minimum length of the mirror = 80 cm (b) Minimum length of the mirror = 85 cm (c) Bottom of the mirror should be at a height 80 cm (d) Bottom of the mirror should be at a height 85 cm
Light (Geometrical Optics)
5. When light passes from air to water which of the following changes? (a) Wavelength (b) Velocity (c) Frequency (d) Colour
10. A ray of light is incident normally on one face of 30º – 60º – 90º prism of refractive index 5/3 immersed in water of refractive index 4/3 as shown in figure.
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8. If two mirrors are inclined to each other at 90º, the image seen may be (a) One (b) Two (c) Three (d) Four
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4A.102
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Light (Geometrical Optics)
the reflected ray does not rotate the reflected ray rotates an angle θ the reflected ray rotates an angle 2θ the incident ray is fixed
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12. An object is placed in front of a converging lens at a distance equal to twice the focal length f1 of the lens. On the other side of the lens is a concave mirror of focal length f2 separated from the lens by a distance 2(f1 + f2). Light from the object passes rightward through the lens, reflects from the mirror, passes leftwards through the lens, and forms a final image of the object.
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13. Which of the following can form diminished, virtual and erect image of your face? (a) Converging mirror (b) Diverging mirror (c) Converging lens (d) Diverging lens
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16. In displacement method, the distance between object and screen is 96 cm. The
14. The radius of curvature of the left and right surface of the concave lens are 10 cm and 15 cm respectively. The radius of curvature of the mirror is 15 cm.
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(a) Height of B, observed by F (from itself) is equal to 5.30 m (b) Depth of F, observed by B (from itself) is equal to 6.60 m (c) Height of B, observed by F (from itself) is equal to 8.80 m (d) None of the above
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(a) The distance between the lens and the final image is equal to 2f1. (b) The distance between the lens and the final image is equal to 2(f1 + f2). (c) The final image is real, inverted and of same size as that of the object. (d) The final image is real, erect and of same size as that of the object.
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15. A fish, F in the pond is at a depth of 0.8 m from the water surface and is moving vertically upwards with velocity 2 m/s. At the same instant a bird B is at a height of 6 m from the water surface and is moving downwards with velocity 3 m/s. At this instant both are on the same vertical line as shown in the figure. Which of the following statements are correct?
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4A.103
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Light (Geometrical Optics)
(c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True. 1. Statement 1: Maximum possible angle of refraction in a medium is critical angle. Statement 2: The critical angle θc (or µ) depends upon pair of medium, colour of light (or λ) and temperature.
ratio of length of two images formed by a convex lens placed between them is 4.84. (a) Ratio of the length of object to the length of shorter image is 11/5 (b) Distance between the two positions of the lens is 36 cm (c) Focal length of the lens is 22.5 cm (d) Distance of the lens from the shorter image is 30 cm
4A.104
2. Statement 1: A point object is placed at a distance of 26 cm from a concave mirror of focal length 26 cm. The image will form at infinity. Statement 2: For above given system the 1 1 1 gives v = ∞. equation + = u v f
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4. Statement 1: A convex mirror is used as a driver’s mirror. Statement 2: Because convex mirror’s field of view is large and images formed are virtual, erect and diminished.
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6. Statement 1: In visible light µr < µv Statement 2: This follows from Cauchy’s B C formula µ = A + 2 + 4 λ λ
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5. Statement 1: Focal length of a lens for red colour is greater than its focal length for violet colour. 1 1 1 Statement 2: Because = (µ − 1) − f R1 R2 and µ1 < µv.
7. Statement 1: Refractive index of material of a prism depends on angle of prism A and angle of minimum deviation δm. Statement 2: Refractive index of a prism,
Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
Q
ASSERTION AND REASON UESTIONS
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(a) As ‘x’ is increased from zero, intensity continuously decreases (b) As ‘x’ is increased from zero, intensity first increases then decreases (c) Intensity at centre of screen for x = 90 cm and x = 110 cm is same (d) Radius of bright circle obtained on screen is equal to 1 cm for x = 200 cm
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3. Statement 1: When a concave mirror is held under water, its focal length will increase. Statement 2: The focal length of a concave mirror is independent of the medium in which it is placed.
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17. A point source of light S is placed on the axis of a lens of focal length 20 cm as shown. A screen is placed normal to the axis of lens at a distance ‘x’ from it. Treat all rays as paraxial.
A + δm sin 2 µ= sin( A/2)
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16. Statement 1: A virtual image can be photographed. Statement 2: Only real objects are photographed.
15. Statement 1: If a spherical mirror is dipped in water, its focal length remains unchanged. Statement 2: A laser light is focused by a converging lens. There will be a significant chromatic aberration.
17. Statement 1: The small object, to be seen in a microscope, is kept between F and 2F of its objective. Statement 2: In this case, the image formed by the objective is nearer to the eyepiece.
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19. Statement 1: Light rays retrace their length when their direction is reversed (Law of reversibility of light rays). Statement 2: For the refraction light, water is denser than air, but for the refraction of sound, water is rarer than air.
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20. Statement 1: If the angle between the two plane mirror is 72º and the object is asymmetrically placed between the two mirrors, then 5 images of the object will be formed.
13. Statement 1: As the temperature of a medium increases, the refractive index decreases. Statement 2: When a ray travels from vacuum to a medium, then µ is known as absolute refractive index of the medium. (µvacuum = 1)
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12. Statement 1: If both plane mirror and object are moved through a distance x, then the image moves through a distance 3x. Statement 2: When the object is fixed and plane mirror is moved through a distance x. Then the image is also moved through the distance x.
14. Statement 1: Critical angle is minimum for violet colour. Statement 2: Because critical angle
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11. Statement 1: When the object moves with a velocity v , its image in the plane mirror moves with a velocity of −2v . Statement 2: The minimum height of the mirror to be required to see the full image of man of height h is h/2.
18. Statement 1: A point source of light is placed at a distance of 2f from a converging lens of focal length f. The intensity of the other side of the lens is maximum at distance 2f. Statement 2: In chromatic aberration, the rays of different colours fail to converge at a point after going through a converging lens.
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10. Statement 1: Keeping a point object fixed, if a plane mirror is moved, the image will also move. Statement 2: In case of a plane mirror, distance of object and its image is equal from any point on the mirror.
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9. Statement 1: Two convex lenses joined together cannot produce an achromatic combination. Statement 2: The condition for achromaω ω tism is 1 + 2 = 0 where symbols have f1 f 2 their usual meaning.
4A.105
8. Statement 1: Spherical aberration of a lens can be reduced by blocking the central portion or peripheral portion of the lens. Statement 2: Spherical aberration arises on account of inability of the lens to focus central (paraxial) and peripheral (marginal) rays at the same point.
Light (Geometrical Optics)
1 1 θc = sin −1 and µ ∝ µ λ
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Statement 2: For given system of mirror, the total number of images formed due
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Light (Geometrical Optics)
to successive reflection is equal to either
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MATCH THE COLUMN T PE UESTIONS Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II
1. An executed object is moving in front of concave mirror as shown in figure. On L.H.S. various velocity of object and position is given. On R.H.S. some properties of image and its velocity is given. Column I Object
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2. Column I shows four diagrams of real object point O, image point I and principal axis have been sketched. Select the proper optical system from column II which can produce the required image. (Image may be real or virtual).
26. Statement 1: If P1 and P2 be the powers of two thin lenses located coaxially in a medium of refractive index µ at a distance d, then the power P of the combination is P = P1 + P2 – P1P2d/µ Statement 2: Because for above given system equivalent focal length is given by f1 f 2 1 and P = F= F f1 + f 2 − d /µ
(p) +ve velocity
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(a) +ve velocity and object is between focus and centre of curvature (b) –ve velocity and object is between focus and pole (c) –ve velocity and object is beyond centre of curvature (d) –ve velocity and object is virtual
25. Statement 1: Rainbow is an example of the dispersion of sunlight by the water droplets. Statement 2: Light of shorter wavelength is scattered much more than light of larger wavelength.
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24. Statement 1: A man wearing glasses of focal length +1 m cannot see beyond 1 m. Statement 2: A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black, the intensity of the image will decrease but the image will not be shifted upward or downward.
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23. Statement 1: When we seen an object, the image formed on the retina is real and inverted. Statement 2: If the magnification of a system is less than one, then the image formed in inverted.
Column II Image
22. Statement 1: All large telescopes use parabolic mirrors rather than spherical mirrors. Statement 2: A parabolic mirror brings parallel rays (rays parallel to its axis) to an exact focus at an point.
21. Statement 1: Optical fibres are used to transmit light without any appreciable loss in its intensity over distance of several kilometers. Statement 2: Optical fibres are very thick and all the light is passed through it without any loss.
360° 360° 360° is or − 1 accordingly as θ θ θ odd or even respectively.
Q
4A.106
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Light (Geometrical Optics)
4. Match the following:
(b)
(s) −10iˆ
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(a) VA′, A (b) VB′, B (c) VC ′, C (d) VD′, D
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6. Match the following: Column I (a) Inverted crown-flint glass prism (b) Achromatism (c) Hollow prism (d) Glass slab Column II 1 (p) Deviation ∝ dispersive power
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5. A convex lens (f) forms an image on a screen considering the object to be at the zero mark in a scale, match the following: Column I (a) Image (b) Additional lens in contact (c) Reduction in refractive index (d) Slicing the lens to have one plane and another convex surface Column II (p) Moves the image of infinite object further away (q) Not unique as lens is moved between object and source. (r) Virtual for screen position at a distance <4f from the object. (s) Object at d forms real image further nearer plano-convex lens.
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3. Four particles are moving with different velocities in front of stationary plane mirror (lying in y-z plane). At t = 0, velocity of A is v A = iˆ, velocity of B is vB = −iˆ + 3 ˆj , velocity of C is vC = 5iˆ + 6 ˆj , velocity of D is vD = 3iˆ − ˆj. Acceleration of particle A is a A = 2iˆ + ˆj and acceleration of par ticle C is aC = 2t ˆj. The particle B and D move with uniform velocity (Assume no collision to take place till t = 2 seconds). All quantities are in S.I. units. Relative velocity of image of object A with respect to object A is denoted by VA′, A . Velocity of images relative to corresponding objects are given to column I and their values are given in column II at t = 2 second. Match column I with corresponding values in column II. Column I Column II
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Column II Diverging lens Converging lens Concave mirror Convex mirror
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4A.107
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Light (Geometrical Optics)
4A.108
(r) Absence of chromatic aberration (s) Dispersion without deviation 7. Match the features of a telescope listed in Column I with the quantities that they depend upon listed in Column II.
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Column I Intensity of light received by lens Angular magnification Length of telescope Sharpness of image
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PASSAGE BASED UESTIONS Q
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8. The graphs given apply to convex lens of focal length f, producing a real image at a distance v from the optical centre when self luminuous object is at distance u from the optical centre. The magnitude of magnification is m. Identify the following graphs with the first named quantity being plotted along y-axis. Column I (a) v against u 1 1 (b) against v u (c) m against v v (d) (m + 1) against f
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PASSAGE–1
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The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature
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Light (Geometrical Optics)
PASSAGE–2
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Age, (years) 10
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1. Minimum focal length of eye lens of a normal person is (a) 25 cm (b) 2.5 cm (c) 25/9 cm (d) 25/11 cm
2. Maximum focal length of eye lens of normal person is (a) 25 cm (b) 2.5 cm (c) 25/9 cm (d) 25/11 cm
3. A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be (a) +1 (b) –1 (c) +3 (d) –3
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In the normal human eye, light from an object is refracted by the cornea-lens system at the front of the eye and produces a real image on the retina at the rear of the eye. For a given eye, its lens-toretina distance is fixed at about 2.5 cm. Most of the focusing of an image is done by the cornea, which has a fixed curvature that is convex with respect to incoming light. The importance of the lens is that its radius of curvature can be changed, allowing the lens to fine-tune the focus. The lens is surrounded by the ciliary muscle. Contraction of the muscle decreases tension on the lens. This allows the natural elasticity of the lens to produce an increase in the radius of curvature. When the muscle relaxes, the lens flattens out, decreasing its radius of curvature. Unfortunately, the lens loses elasticity with age and the ability to alter curvature decreases. The range over which clear vision is possible is bounded by the far point and the near point. In normal vision, the far point is infinity and the near point depends on the radius of curvature of the lens. For normal eyes the average near point for reading is 25 cm.
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decreases) and focal length decreases. For a clear vision, the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person. A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grownup person, minimum distance of object should be around 25 cm. A person suffering for eye defects uses spectacles (Eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina. The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example, power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3. For all the calculations required you can use the lens formula and lens maker’s formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.
4A.109
In the myopic (nearsighted) eye, the lens-toretina length, is too long and/or the radius of curvature of the cornea is too great. This causes rays from an object at infinity to focus at a point in front of the retina. The far point is closer than normal. A corrective, lens will put a virtual image of a distant object at the position of the actual far point of the eye. In the hyperopic (farsighted) eye, the lens-to-retina length is too short and/or the radius of the curvature of the cornea is not great enough. This causes rays from an object at infinity to focus at a point behind the retina. The
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Light (Geometrical Optics)
(d) relaxing the ciliary muscle and decreasing the radius of curvature
near point is farther away than normal. A corrective lens will put a virtual image of the close object at the position of the actual near point. The relation among the object (o) and image (i) distances from the eye and the focal length (f) of the lens is given by the lens-distance rule
PASSAGE–3 Figure shows a convex lens of focal length 15 cm. A point object is placed on the principle axis of lens at a distance 20 cm from it as shown. On the other side of lens two observer eyes O1 and O2 are situated at a distance 100 cm from the lens at some distance above and below the principal axis.
1 1 1 + + o i f When using this equation, all distances are given in centimeters. The power of corrective lenses is usually given in units called diopters. Power, in diopters, is the reciprocal of the focal length in meters 1 Pdropter = f meter
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9. After activity–I, for which observer the intensity of image will be reduced to half?
8. After activity–I, which of the following observer will not be able to see the image of object, if before this activity both were seeing the image? (a) O1 (b) O2 (c) both O1 and O2 (d) neither O1 nor O2
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5. An optometrist examined John’s eyes. The farthest object he can clearly focus on with his right eye is 50 cm away. What is the power of the contact lens required to correct the vision in his right eye? (a) –0.50 diopters (b) –2.0 diopters (c) +2.0 diopters (d) +5.0 diopters
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7. In initial setup before the activities, which of the following statement is correct? (a) Observer O1 will see a real image at 60 cm from the lens but observer O2 will not be able to see it. (b) Observer O2 will see a real image at 60 cm from the lens but observer O1 will not be able to see it. (c) Both the observers will see a real image at 60 cm from lens irrespective the positions of O1 and O2. (d) Both the observers may or may not be able to see the image at 60 cm from lens depending on the positions of O1 and O2.
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4. The lens system of the myopic eye is best described as (a) producing too much convergence (b) producing too little convergence (c) producing too much divergence (d) producing too little divergence
6. In a mildly hyperopic eye, the focal length of the eye’s natural lens can be corrected by (a) contracting the ciliary muscle and increasing the radius of curvature (b) contracting the ciliary muscle and decreasing the radius of curvature (c) relaxing the ciliary muscle and increasing the radius of curvature
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Activity I: Now half position of lens below principal axis is painted black. Activity II: After this, lens is cut along its horizontal diameter and upper part of it is raised up slightly in vertical direction.
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I. Converging lenses have positive focal lengths, and diverging lenses have negative focal lengths. II. Real images have positive distances from the lens, and virtual images have negative distances from the lens.
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4A.110
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4A.111
Light (Geometrical Optics)
PASSAGE–4 The laws governing the behaviour of the rays namely rectilinear propagation, laws of reflection and refraction can be summarized in one fundamental law known as Fermat’s principle. According to this principle, a ray of light travels from one point to another such that the time taken is at a stationary value (maximum or minimum). If c is the velocity of light in a vacuum, the velocity c in a medium of refractive index n is , hence n n time taken to travel a distance l is . If the light c passes through a number of media, the total time
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1. A 4 cm thick layer of water covers a 6 cm thick glass slab. A coin is placed at the bottom of the slab and is being observed from the air side along the normal to the surface. Find the apparent position of the coin from the surface.
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2. How long will the light take in travelling a distance of 500 metre in water? Given that µ for water is 4/3 and the velocity of light in vacuum is 3 × 1010 cm/sec. Also calculate equivalent path. 3. A ray of light travelling in air is incident at angle of incident 30º on one surface of a slab in which refractive index varies with y. The light travels along the curve y = 4x2 (y and x are in metre) in the slab. Find out the refractive index of the slab at y = 1/2 m in the slab.
11. The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is
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4. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
10. In the figure shown, the speed of image in cm/sec with respect to mirror is *, then find the value of ‘*’
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13. A concave mirror and a convex mirror are placed co-axially, their reflecting surfaces facing each other. Their focal lengths are 15 and 12 cm respectively. An object placed between them is 20 cm from the concave mirror. The image formed by it is at the object itself. Calculate the distance of the concave mirror from the object. 14. The powers of the objective and eye piece of a compound microscope is 100D and 20D respectively. The microscope is adjusted for normal observation (eye muscles least strained). The magnifying power of the microscope is 50. What will be the magnifying power if the distance between the objective and eye piece is increased by 2 cm and under normal observation adjustment?
11. The angle of crown glass (µ = 1.52) prism is 5º. What should be angle of flint glass (µ = 1.63) prism, so that the two prism together may be used in direct vision spectroscope?
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16. An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.
2. An eye specialist prescribes spectacles having combination of convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm. The power of this lens combination in diopters is [1997]
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3. A thin equiconvex lens of glass of refractive index µ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water µ = 4/3. On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the object formed by the system. [1997]
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4A.113
15. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.
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1. A thin plano-convex lens of focal length f is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half lens is 2. Find the focal length of the lens and separation between the halves. Draw the ray diagram for image formation. [1996]
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(a) is absent (b) is smaller if the curved surface of the lens faces the object (c) is smaller if the plane surface of the lens faces the object (d) is the same whichever side of the lens faces the object
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9. The x–y plane is the boundary between two transparent media. Medium-1 with z ≥ 0 has a refractive index 2 and medium-2 with z ≤ 0 has a refractive index 3. A ray of light in medium-1 given by vector A = 6 3 iˆ + 8 3 ˆj − 10kˆ is incident on the plane of separation. Find the unit vector in the direction of the separation. Find the unit vector in the direction of the refracted ray in medium-2. [1999]
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8. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a
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6. A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to [1998]
(a) real and will remain at C (b) real and located at a point between C and ∞ (c) virtual and located at a point between C and O (d) real and located at a point between C and O
5. A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be [1998]
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15. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axis PQ and RS parallel but separated in vertical direction by 0.6 cm as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. If A’ B’ is the image after refraction from the lens and the reflection from the mirror, find the distance of A’ B’ from the pole of the mirror and obtain its magnification. Also locate positions of A’ and B’ with respect to the optic axis RS. [2000]
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13. A diverging beam of light from a point source S having divergence angle α falls symmetrically on a glass slab as shown. The angles of incidence of the two extreme rays are equal. If the thickness of the glass slab is t and its refractive index is n, then the divergence angle of the emergent beam is [2000]
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16. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 (n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax, such that the ray comes out only from the surface CD, is given by [2000] $
18. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prism Q and R of identical shape and of the same material as P are now added as shown in the figure. The ray will suffer [2001]
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20. A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 4/3. It is found that when a point object is placed 15 cm above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image
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Light (Geometrical Optics)
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Light (Geometrical Optics)
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23. Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30º at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is
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24. The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image [2003] (a) (b) (c) (d)
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25. A ray of light is incident at the glass-water interface at an angle i, it emerges finally parallel to the surface of water, then the value of µg would be [2003]
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A light of wavelength 6600Å is incident on face AB such that angle of incidence is 60º. Find (a) the angle of emergence and (b) the minimum value of thickness of the coated film on the face AC for which the light emerging from the face has maximum intensity. [2003]
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37. Focal length of the plano-convex lens is 15 cm. The object is at A as shown in the figure. The plane side is silvered. The image is [2006]
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34. A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths? [2005]
35. AB and CD are two slabs. The medium between the slabs has refractive index 2. Find the minimum angle of incidence of Q, so that the ray is totally reflected by both the slabs. [2005]
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38. Match the following: Column I Column II (a) Angular magnifi- (p) Dispersion of cation lenses (q) f0 and fe
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43. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60º). In the position of minimum deviation, the angle of refraction will be [2008]
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41. In an experiment to determine the focal length f of a concave mirror by the u – v method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in the line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then
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Light (Geometrical Optics) 6
(c) greater for the red colour (d) equal but not 30º for both the colours
44. An optical component and an object S placed along its optic axis are given in Column I. The distance between the object and the component can be varied. The properties of images are given in Column II. Match all the properties of images from Column II with the appropriate components given in Column I. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. Column I
45. A light beam is travelling from Region IV (figure). The refractive index in regionals I, n n n II, III and IV are n0 , 0 , 0 and 0 respec2 6 8 tively. The angle of incidence θ for which the beam just misses entering region IV is [2008]
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4. 12. 20. 28. 36. 44. 52. 60. 68. 76.
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5. 13. 21. 29. 37. 45. 53. 61. 69. 77.
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6. 14. 22. 30. 38. 46. 54. 62. 70.
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7. 15. 23. 31. 39. 47. 55. 63. 71.
(a) (a) (a) (b) (b) (a) (a) (b) (b)
8. 16. 24. 32. 40. 48. 56. 64. 72.
(b) (a) (c) (b) (a) (c) (c) (d) (d)
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Light (Geometrical Optics)
4A.122
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ONE OR MORE THAN ONE CHOICE MA BE CORRECT 1. (a, b)
2. (a, c)
8. (a, b, c) 15. (b, c)
3. (a, b, c)
9. (a, d)
10. (a, c)
4. (b, c)
5. (a, b)
6. (a, b, c)
7. (a, c)
11. (c, d)
12. (a, c)
13. (b, d)
14. (a, c)
5. (a) 13. (b) 21. (c)
6. (a) 14. (a) 22. (a)
7. (d) 15. (c) 23. (c)
16. (a, b, d) 17. (b, c, d)
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ASSERTION AND REASON UESTIONS 2. 10. 18. 26.
(b) (a) (c) (b)
3. (d) 11. (b) 19. (b)
(a) (d) (b) (a)
4. (a) 12. (d) 20. (a)
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8. (a) 16. (c) 24. (b)
1. (a) → r, q (b) → p, r (c) → p, s (d) → p, r
4. (a) → q, (b) → p, (c) → r, (d) → p
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3. (a) → s, (b) → p, (c) → s, (d) → q
2. (a) → p, s (b) → q, r (c) → q (d) → q, r
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6. (a) → s, (b) → r, p (c) → q, (d) → q
7. (a) → q (b) → p (c) → r, s (d) → p
8. (a) → r, (b) → s, (c) → q, (d) → p
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PASSAGE BASED UESTIONS
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10. 4/3
11. (c)
12. (d)
13. (b)
14. (d)
15. 15 cm, –3/2 16. (a)
9.
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8. (a)
7. (a) 600 nm, (b) sin–1(3/4)
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5. (d)
4. (b)
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2. (b)
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PREVIOUS EARS’ IIT-JEE UESTIONS
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HINTS AND SOLUTIONS
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43. (a) At minimum deviation for any wavelength r1 = r2 = A/2, Because r1 + r2 = A
19. (i) 4º, (ii) –0.04º
4A.123
18. (c)
21. (b)
22. (c)
23. (b)
24. (b)
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44. (a) → (p, q, r, s), (b) → (q), (c) → (p, q, r, s), (d) → (p, q, r, s) uf f u (a) f < 0, v = and = = u − f 1 − f /4 u /f − 1 v m = − values of v may be positive, negau tive or infinity, also it can have values less than or greater than u. (a) → (p, q, r, s) (b) In this case f is positive.
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Light (Geometrical Optics)
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c h a p t e r
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4.1 Wave optics: n introduction
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Light (Wave Optics)
4.2 Huygens’ principle
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In geometrical optics, we have represented light as rays which travel in straight lines in a homogeneous medium. By doing this, we have studied a variety of phenomenon involving mirrors and lenses. The phenomenon like interference and diffraction can not be explained on the bases of particle nature of light. These phenomenon can only be explained on the basis of wave nature of light. This part of optics is called physical optics. The wave theory of light was presented by Christian Huygens in 1678. During that period, Newton’s corpuscular theory had satisfactorily explained the phenomenon of reflection, refraction and rectilinear propagation of light. So the scientists believed in the corpuscular theory; no one really believed in Huygen’s wave theory. The wave characteristics of light was not really accepted until the interference experiments of Young in 1801. It should be pointed out that Huygens did not know whether the light waves were longitudinal or transverse and also how they propagate through vacuum. It was then explained by Maxwell by introducing electromagnetic wave theory in nineteenth century.
(i) Each point of a given wavefront is a source of new disturbance which is called secondary disturbance. The wavelets originated from these points spread out in all directions with the speed of light. (ii) The envelope of these wavelets in the forward direction gives the shape and position of the new wavefront at any subsequent time. To understand this, consider a spherical wave front AB as shown in figure. Every point such as 1, 2, etc., on AB becomes the source of secondary spherical wavelets. After time t the radius of each wavelet will be ct, where c is the speed
Huygens principle provides a geometrical method which allows us to determine the shape of the wavefront at any time, if the shape of the wavefront at an earlier time is known. A wave front is the locus of the points which are in the same phase. Huygens’ principle can be stated as follows:
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Light (Wave Optics)
of the light. Thus from the points 1, 2, 3,.... etc., draw spheres of radii equal to ct. These spheres represent the secondary wavelets. According to Huygens, the common envelope A1B1 in forward direction gives the position of new wavefront (see fi g.). $ $
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In the time, the disturbance at A reaches C, the secondary waves from B will travel a distance BD such that BD = AC. With the point B as centre and radius equal to AC draw an arc. From the point C, draw the tangent CD. In triangles BAC and BDC; BC is common and BD = AC https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.3
∠BAC = ∠BDC = 90º The two triangles are congruent, and so ∠ABC = ∠BCD i = r.
∴ or
Thus angle of incidence is equal to angle of reflection. This proves the law of reflection.
proof of law of refraction
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Let xy is the interface between two media 1 and 2 of refractive indexes µ1 and µ2 respectively. Suppose v1 and v2 are the velocities of light in two media. The second medium is optically denser than first and so v2 < v1. AMB is the plane wavefront incident at an angle i. In the time disturbance at B reaches C, the secondary waves from A will travel a distance AD = v2t, where t is the time taken by the waves to travel the distance BC. Thus
With A as the centre and radius AD draw an arc. Then draw a tangent CD to the arc. CD represents the refracted wavefront r be the angle of refraction. We have BC v1t v1 = = . AD v2t v2
...(i)
In triangles ABC and ACD, we have sin i BC /AC = sin r AD /AC BC . AD From equation (i) and (ii), we have sin i v1 = sin r v2
...(ii)
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4B.4
Light (Wave Optics)
v1 µ 2 = , v2 µ1
Since
sin i µ 2 = sin r µ1
∴
or µ1 sin i = µ2 sin r. This proves the law of refraction, which is called Snell’s law.
4.3 interference
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When two or more coherent waves superimpose, the resultant intensity in the region of superposition is different from the intensity of individual waves. This modification in the distribution of intensity in the region of superposition is called interference.
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young’s double slit experiment (ydse)
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Thomas Young in 1801 devised an ingenious method of producing coherent sources. In this method, a single wavefront is divided into two; these two split wavefronts act as if they originated from two sources having a constant phase relationship and therefore, when they were allowed to interfere, a stationary interference pattern was obtained. In the experiment, light from a source S fell on a cardboard which contained two pinholes (or slits) S1 and S2 which were very close to one another. The spherical waves originating from S1 and S2 were coherent and so beautiful interference fringes or bands were obtained on the screen.
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Young’s double slits arrangement.
coherent sources Two sources of light are said to be coherent if they emit light waves of same frequency and having constant phase difference (may be zero). It means the two sources https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.5
ote
1. Two independent laser sources of equal wavelengths can be coherent. Because they can maintained the constant phase difference for long time.
2. Two ordinary sources can not maintain the constant phase difference so they can not be coherent and hence will not interfere.
nalytical treatment of interference
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must emit waves of the same wavelength. In practice, it is not possible two have two independent sources which are coherent and so for practical purposes, two virtual sources formed from a single source can act as coherent sources. Young’s double slits arrangement, Fresnel’s biprism method, Llyod’s mirror arrangement are the methods of producing two coherent sources from a single source.
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Consider a monochromatic source of light S emitting light waves of wavelength λ and two narrow slits S1 and S2. S1 and S2 are separated a distance d and equidistance from S. S1 and S2 then becomes two virtual coherent sources of light waves. Let φ is the phase difference between the two waves reaching at point P. The equation of wave for any fixed position (say screen at x = 0) can be written as: y = a sin (ωt – kx), where x = 0 and so, we
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...(1)
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a2 sin φ = R sin θ, we get
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This shows that the resultant wave at any point P is simple harmonic of amplitude R. The amplitude R can be obtained as: Squaring equations (i) and (ii), we have R2 = a12 + a22 + 2a1a2 cos φ.
...(2)
I = I1 + I2 + 2 I1 I 2 cos φ.
...(3)
As intensity I of wave is proportional to square of the amplitude, and so Also dividing equation (ii) by (i), we get a2 sin φ . a1 + a2 cos φ
...(4)
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Light (Wave Optics)
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In Young’s interference experiment, incident monochromatic light is diffracted by slit S, which then acts as a point source of light that emits semicircular wavefronts. As that light reaches screen B, it is diffracted by slits S1 and S2, which then act as two point sources of light. The light waves traveling from slits S1 and S2 overlap and undergo interference, forming an interference pattern of maxima and minima on viewing screen C. Depending on the phase difference φ between the two waves, the intensity of resulting wave may be minimum or maximum. Accordingly, there are two types of interference. These are:
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or
2 I max = Rmax = (a1 + a2 ) 2 .
...(5)
(ii) Destructive interference (dark point) The intensity I will be minimum, when cos or
φ = –1 φ = (2n – 1)π, n = 1, 2, 3,...
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Light (Wave Optics)
∆x = (2n − 1)
Also
4B.7
λ 2
Now
2 I min = Rmin = a12 + a22 − 2a1a2
or
2 I min = Rmin = (a1 − a2 ) 2
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Thus
2 I max Rmax (a + a ) 2 = 2 = 1 2 2. I min Rmin (a1 + a2 )
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Interference between two waves with φ = 0.
Interference between two waves with φ = π.
Special cases: When two identical waves interfere,
a1 = a2 = a
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I = a2 + a2 + 2aa cos φ = 2a2 (1 + cos φ) = 2a 2 × 2cos 2
or
φ I = I max cos 2 . 2
φ φ = 4a 2 cos 2 2 2
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Light (Wave Optics)
intensity distribution It has been obtained that intensity at bright points is 4a2 and at dark points is zero. According to law of conservation of energy, the energy of the interfering waves as a whole remains constant. Thus the energy from points of minimum intensity transfers to the points of maximum intensity. The intensity variation with phase difference is shown in fi gure. , D
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Consider two sources S1 and S2 emitting monochromatic light of wavelength λ. The separation between them is d. The interference fringes are obtained on a screen placed at a distance D from the sources. The fringes are of equal width and alternatively bright and dark. The centre to centre distance between two consecutive bright or dark fringes is called fringe width β. Consider a point P on the screen at a distance yn from the centre of the screen O. The angular position of the point P is θ from the centre of the sources (see fi g. 4.8). The path difference between the waves on arriving at point P, is S2P – S1P, which is equal to ∆x. From the figure, ∆x = d sin θ. For small θ, we can write sin θ ≃ tan θ. Thus
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4B.8
...(i)
(i) Bright fringes There will be bright fringe at P, when ∆x = nλ. Thus path difference dyn = nλ D n Dλ or yn = ; n = 0, 1, 2, d
.........(9)
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4B.9
Light (Wave Optics)
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(ii) Dark fringes There will be dark fringe at P, when ∆x = (2n – 1)
λ . Thus 2
dyn λ = (2n − 1) D 2 or yn =
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...(11)
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Light (Wave Optics)
Equation (11) represents the position of nth dark fringe. The (n – 1)th fringe will be at a distance 2(n − 1) − 1 Dλ yn−1 = d 2 ∴ Fringe width β = yn – yn – 1
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note The maximum path difference ∆xmax = d, when sin θ = 1. If n are the number of d brights fringes on one side of the central bright, then d = nλ or n = . Thus total λ number of fringes that can be on the screen are = 2 n + 1, including central fringe.
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Light (Wave Optics)
4B.11
Special case: If YDSE is performed in water, and observer is in air, then fringe width
Dλ water . d λ = air , µω
βwater =
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mportant points
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1. In YDSE, the central fringe is bright, and all the bright fringes are of same intensity. Colour of bright fringes are of the colour of incident light. 2. If slits are of equal size, the intensity of all the dark frings are zero. 3. If slits are of unequal size, then the intensity of dark fringe is not zero. 4. All the fringes are of equal width. 5. If sources have random phase difference, then there will be no interference. The intensity at any point will be I = a2 + a2 = 2a2. 6. If white light is used in the experiment, then the central fringe will be white, and other fringes are overlapped colour fringes.
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1. The sources must be coherent. 2. The separation between the slits should be small (order of mm), so that size of fringe is large enough to observe. 3. The amplitudes of interfering waves are equal or nearly equal, otherwise the intensities of bright and dark fringes are not differentiable.
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xample 1 Consider interference between two sources of intensities I and 4I. Obtain intensities at points where the phase difference is (i) π/2 and (ii) π.
)
) Solution
We know that resultant intensity
For φ = π/2;
(i)
I R = I + 4 I + 2 I × 4 I cos
π 2
= 5I
(ii)
I R = I1 + I 2 + 2 I1 I 2 cos φ
For φ = π;
I R = I + 4 I + 2 I × 4 I cos π = 5I – 4I = I
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Light (Wave Optics)
example 2 A parallel wavefront of monochromatic light is incident on double slit arrangement at an angle α as shown in fi gure. Find the position of zero order maxima from the centre of the screen. The distance of screen from the slits is D (d << D).
) Solution
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∆x = (BS2 + S2P) – (AS1 + S1P) = (S2P – S1P) – (AS1 – BS2)
.m
s:
tp
//w
w
Suppose the position of zero order maxima is at P at a distance yo from O. The path difference between two waves at P is
ht
4B.12
example 3 The two coherent sources of monochromatic light of wavelength λ are located at a separation λ. The two sources are placed on a horizontal line and screen is placed perpendicular to the line joining the sources (see figure). Find position of the farthest minima from the centre of the sources.
) Solution Suppose at P the farthest minima will occur. Let it subtends an angle θ at the centre of the sources.
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Light (Wave Optics)
4B.13
3
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Thus in between these two positions there is only one minima for which ∆x = λ . Thus 2 λ = λ cos θ 2 1 or cos θ = 2 ∴ θ = 60º
e/
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s: //
t.m e/
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//t
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tp ht
er si ta rio s
w
∆xm = 0; when cos θ = 0 or θ = 90º
s:
er s
w
and minimum path difference
ita
∆xmax = λ; when cos θ = 1 or θ = 0º
.ja
m
The maximum path difference can be
rio s_ in fo
∆x = S2P – S1P ≃ d cos θ = λ cos θ
ar a
na .c om
The path difference
example 4 Two slits in Young’s interference experiment have width in the ratio 1 : 4. Find the ratio of intensity at the maxima and minima in their interference.
) Solution The intensity of the wave is proportional to the area of the slit. Thus I1 b1 b1 1 = = = . If a1 and a2 are the amplitudes of the waves, then I 2 b2 b2 4 I1 a12 1 = = I 2 a2 2 4
∴
a1 1 = . a2 2
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Light (Wave Optics)
The ratio
I max (a1 + a2 ) 2 = I min (a1 − a2 ) 2 (1 + 2) 2 9 = . (1 − 2) 2 1
=
example 5 In double slit arrangement, the source S is not symmetrically placed from the slits. It is located as shown in the figure. Find the position of the zero order maxima from the centre of the screen. The separation between slits and screen is D (d << D).
w
un iv
//w s:
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tp
GVLQT
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6
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) Solution
ht
s: //
Suppose the position of zero order maxima is at P at a distance yo from O. The path difference between two waves at P is
ht tp
4B.14
∆x = (SS2 + S2P) – (SS1 + S1P) = –(SS1 – SS2) + (S2P – S1P) =−
(
)
2 d − d + d sin θ
For small θ, sin θ ≃ tan θ =
∴
∆x = −
(
)
2d −d +
yo . D
dyo D
For zero order maxima, ∆x = 0 or
∴
0=− yo =
(
(
)
2d −d +
)
dyo D
2 −1 D
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Light (Wave Optics)
4B.15
example 6 In an interference arrangement similar to Young’s double slit experiment, slits S1 and S2 are illuminated with coherent microwave sources each of frequency 1 MHz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity Iθ is measured as a function of θ where θ is defined as shown in figure. If Io is the maximum intensity, calculate Iθ for (a) θ = 0º (b) θ = 30º and (c) = 90º.
) Solution
T
T
m
ar a
G
un iv e/
un
t.m e/
s: //
∆x = d sin θ.
ht tp
iv
.m
//t
ht tp s:
c 3 × 108 = = 300 m f 106
The path difference
er si ta rio s
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w //w s: tp ht
The wavelength of microwave λ=
2
ita
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w
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6
rio s_ in fo
6
na .c om
3
The corresponding phase difference φ= =
2π 2π ∆x = (d sin θ) λ 300 2π (150sin θ) 300
= π sin θ. The resultant intensity in interference is given by
IR = I1 + I 2 + 2 I1 I 2 cos φ I1 + I 2 + 2 I1 I 2 cos(π sin θ).
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Light (Wave Optics)
For
I1 = I2 = I.
IR = 2I + 2I cos (π sin θ).
(a)
(b)
For θ = 0º, IR = 2I + 2I cos 0º = 4I I Given I0 = 4I; ∴ I = 0 . Ans. 4 For θ = 30º
IR = 2I + 2I cos (π sin 30º) = 2I + 2I cos
(c)
For θ = 90º
IR = 2I + 2I cos (π sin 90º)
ar a
I0 I0 = Ans. 4 2
na .c om
=2
π = 2I 2
rio s_ in fo
4B.16
= 2I – 2I = 0.
er s
w
ita
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m
4.4 displacement of fringes (fringes sHift and numBer of sHifted fringes)
er si ta rio s
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t.m e/
ht tp s:
un
∆x = S2P – [S1P + (µ – 1) t] = (S2P – S1P) – (µ – 1) t
//t
ht
.m
tp
e/
s:
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Suppose a transparent sheet of thickness t and refractive index µ is introduced in front of one of the slits of YDSE. The optical path of the light waves emerging from slit will increase by an amount (µ – 1) t. In the arrangement shown the optical path of S1P + (µ – 1) t. Thus path difference between waves at P
S2P – S1P = d sin θ
s: //
From the geometry of the figure
ht tp
For small angle θ, sin θ ≃ tan θ =
yn . D 3
W P
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Light (Wave Optics)
∴ ∆x =
4B.17
dyn − (µ − 1)t. D
For bright fringes the path difference ∆x = nλ. Thus dyn − (µ − 1)t = nλ; D or yn =
n = 0, 1, 2, ....
n Dλ D(µ −1)t + . d d
...(1)
In the absence of the sheet, the position of nth bright, yn = n
rio s_ in fo er si ta rio s '
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2
iv un
//t s: //
t.m e/
ht tp s:
6
.m
tp ht
ita er s
W P
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//w
6
un iv
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of fringes
Dλ . Thus displacement d
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ht tp
)ULQJH SDWWHUQZLWKRXWVKHHW
Displacement of the fringes.
∆=
D(µ − 1)t d
...(2)
The position of (n – 1)th order bright fringe yn −1 =
(n − 1) Dλ D(µ − 1)t + d d
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Light (Wave Optics)
The fringe width β = yn – yn – 1 or
β=
Dλ . d
This shows that when a transparent sheet is introduced in the path of the slit, the entire fringe pattern will shift towards that side but fringe width remains same. The number of fringe shifted N=
∆ D(µ − 1)t /d = β Dλ d
N=
or
(µ − 1)t . λ
...(3)
rio s_ in fo
ita un iv
e/
tp
2
un
P
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t.m e/
//t ht tp s:
iv
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T
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ht tp
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6
3
er si ta rio s
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W P
6
s:
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example 7 Two transparent sheets of thickness t1 and t2 and refractive indexes µ1 and µ2 are placed infront of the slits as shown in the fi gure. If D is the distance of the screen from the slits, then find the distance of zero order maxima from the centre of the screen. What is the condition that zero order maxima is formed at the centre O?
ht
4B.18
) Solution Suppose P is the position of zero order maxima. The distance of P from the centre O of the screen is y0. the optical path of light waves from source S1
x1 = S1P + (µ1 – 1) t1
The optical path of light waves from source S2
x2 = S2P + (µ2 – 1) t2
The path difference
∆x = x2 – x1 = (S2P – S1P) + (µ2 – 1) t2 – (µ1 – 1) t1
From the geometry,
S2P – S1P = d sin θ ≃d tan θ =
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dy0 . D
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Light (Wave Optics)
dy0 + (µ2 – 1) t2 – (µ1 – 1) t1 D
∆x =
∴
4B.19
For zero order maxima, ∆x = 0.
dy0 + (µ2 – 1) t2 – (µ1 – 1) t1 D
∴
0=
or
y0 =
D[(µ1 − 1)t1 − (µ 2 − 1)t2 ] . d
For zero order maxima at the centre O, y0 = 0
or
(µ1 – 1) t1 = (µ2 – 1) t2.
ita
.ja
4.5 fresnel’s Biprism
ar a
na .c om
0=
rio s_ in fo
D[(µ1 − 1)t1 − (µ 2 − 1)t2 ] d
∴
m
iv
er si ta rio s
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Dλ . d
6 G
ht tp
s: //
t.m e/
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β=
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Fresnel developed another arrangement to get interference phenomenon. He used two small angled prisms (prism angle about 0.5º) placed base to base. With this arrangement he produced two coherent sources by using a single source. The given fi gure shows a source S of monochromatic light and biprism arrangement. S1 and S2 are the virtual images of S as formed by two prisms. These sources are originated from the same source, and so they are coherent. If λ is the wavelength of light used, then fringe width is
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In the arrangement D = (a + b). The value of d can be determined by two methods. These are: https://t.me/universitarios_infohttps://www.jamarana.com
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4B.20
Light (Wave Optics)
(i) By displacement method In this method, a convex lens is used to form real images of the sources. If d1 and d2 are the separations between images in two positions of the lenses, then d = d1d 2 . (ii) If A is the angle of prism, then angle of deviation produced by any prism
δ = (µ – 1) A.
From the geometry of the figure, for small angle δ=
d /2 a
d 2a ∴ d = 2a (µ – 1) A
ar a
4.6 lloyd’s mirror arrangement
rio s_ in fo
na .c om
or (µ – 1) A =
er si ta rio s
iv
t.m e/
∴
ht tp s:
0 = I + I + 2 II cos φ φ = π rad.
un
//t
ht
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In Lloyd’s mirror arrangement, a plane mirror is used to produce another source, which together with real source constitutes two coherent sources. This set-up was developed by Lloyd in 1834. In this arrangement the interference pattern similar to Young’s double slit experiment is obtained on the screen. But the central fringe is dark instead of being bright. It means there is destructive interference at the centre O. If we assume that mirror is perfectly reflecting, then the intensity of light waves from S1 and S2 is equal, let it is I. Then for destructive interference (zero resultant intensity)
ht tp
s: //
It shows that the wave after reflecting from mirror undergone a phase change of π rad. This experiment proves, that a light wave after reflection from an optically denser medium undergoes a phase change of π rad.
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Light (Wave Optics)
4B.21
example 8 A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment: (i) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å. (ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm. nDλ . d
For third bright, n = 3
3Dλ 3 × (120 × 10−2 ) × 6500 × 10−10 = d 2 × 10−3 = 1.17 × 10–3 m.
y3 =
(ii)
Let n1th bright of 6500 Å concides with the n2th bright of 5200 Å, then
ar a
∴
4 5
un
=
t.m e/
5200 6500
ht tp
s: //
ht tp s:
=
iv
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n1 λ 2 = n2 λ1
er si ta rio s
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Dλ1 Dλ 2 = n2 d d
//t
or
n1
er s
w
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ita
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m
rio s_ in fo
yn =
na .c om
) Solution (i) For bright fringe,
O c
O c
\
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Light (Wave Optics)
The minimum of value of n1 and n2 are 4 and 5 respectively. Therefore y4 =
4Dλ1 4 × (120 × 10−2 ) × (6500 × 10−10 ) = d 2 × 10−3
= 1.56 × 10–3 m.
m
ar a
) Solution
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3
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iv
6
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er si ta rio s
w
er s
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ita
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6 '[
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example 9 A parallel beam of monochromatic light is used in a Young’s double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. The light is incident an angle θ with the normal to the plane of slits. Find the value of θ so that, there will be dark fringe at the centre P of the pattern.
ht
The path difference between two wavefronts at P
ht tp
4B.22
∆x = d sin θ
For dark fringe at P, ∆x =
λ 2
d sin θ =
λ 2
Therefore or
λ θ = sin −1 . 2d
example 10 A narrow slit S transmitting light of wavelength λ is placed a distance d above a large plane mirror as shown in the fi gure. The light coming directly from the slit and that coming after reflection interfere at a screen Σ placed at a https://t.me/universitarios_infohttps://www.jamarana.com
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4B.23
Light (Wave Optics)
distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e., just above O? (b) At what distance from O does the first maximum will occur?
) Solution
6
6 G
na .c om
G
ar a
6
E
rio s_ in fo
2
er si ta rio s
er s
un iv
//w
w
ita
β Dλ /2d Dλ = . = 2 2 4d
w
y=
.ja
m
(a) There will be dark fringe at O, therefore intensity of light at O will be zero. (b) It is clear from the figure, that the distance of first maximum from O
6
iv
ht tp
s: //
) Solution
t.m e/
ht tp s:
un
//t
ht
.m
tp
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s:
example 11 Two plane mirrors M1 and M2 are inclined to each other at an angle θ and an illuminated slit S is placed infront of them at a distance of 12 cm away from and parallel to the line of intersection of the mirror. An eye piece is mounted at a distance of 60 cm from the line of intersection of the mirrors. If the bandwidth obtained is 0.16 mm and wavelength of light used is 5460 Å. Find the value of θ.
0
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T
T 6
FP
0
6 FP
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Light (Wave Optics)
From the figure D = 60 + 12 = 72 cm Let separation between the sources formed by mirrors is d. We have β= or
0.16 × 10–3 =
Dλ d (72 × 10−2 ) × (5460 × 10−10 ) d
d = 2.46 × 10–3 m.
∴
d 0.24 2.46 × 10−3 θ= 0.24 = 10.25 × 10–3 rad.
For small angle tan θ ≃ θ =
na .c om
or
rio s_ in fo
er si ta rio s
un iv
//w
w
er s
w
ita
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m
ar a
example 12 S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis 50 with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO.
un
//t s: //
ht tp
/
t.m e/
ht tp s:
) Solution
iv
.m
tp
e/
s:
(i) If the third intensity maximum occurs at the point A on the screen, find the distance OA. (ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase decrease, or remains the same?
ht
4B.24
3
6
PP 3
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2
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P
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Light (Wave Optics)
4B.25
For any half u = –0.15 m f = 0.10 m We have,
1 1 1 − = v u f
1 1 1 − = v −0.15 0.10
or
v = 0.30 m.
∴
Now from similar triangles S S1 S2 and S P1 P2 we have
d = 1.5 × 10–3 m
ar a
3 × 1.0 × 500 × 10−9 1.5 × 10−3
un iv
er s
w //w
w
=
= 1 × 10–3 m
e/
s: tp
iv
.m
(ii) As we have seen d is proportional to P1 P2, that is gap between L1 and L2, therefore with decrease in gap, d decreases and hence OA increases, because 1 y∝ . d
t.m e/
ht tp s:
un
//t
ht
Dλ d
ita
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m
(i) For third maximum OA = y3 = 3
er si ta rio s
or
rio s_ in fo
na .c om
d (0.30 + 0.15) = −3 0.5 × 10 0.15
ht tp
s: //
example 13 Consider the arrangement shown in the fi gure. The distance D is large compared to the separation d between the slits. (a) Find the minimum value of d so that there is a dark fringe at O. (b) Suppose d has this value. Find the distance x at which the next bright fringe is formed. (c) Find the fringe width.
) Solution 3 [ G 2 '
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Light (Wave Optics)
(a) The path difference at O, ∆x = 2 D 2 + d 2 − 2 D λ 3λ For the dark fringe at O, ∆x = , , 2 2 For minimum value of d, λ 2 λ or ( D 2 + d 2 )1/2 − D = 4 2 D2 + d 2 − 2D =
1/2
m
Dλ . 2
er si ta rio s iv
.m
[
un '
s: //
'
3
2
6
t.m e/
ht tp s:
6
6 G
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w
er s
w
(b)
ita
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or d =
ar a
na .c om
d2 λ or D 1 + −D= 2 2 D 4 2 d λ or D + −D= 2D 4
rio s_ in fo
d2 λ or D 1 + 2 − D = 4 D
ht
Here the path difference ∆x = (SS1 + S1P) – (SS2 + S2P)
ht tp
4B.26
= ( D 2 + d 2 ) + ( x − d )2 + D 2 − D + D 2 + x 2 For the next bright fringe after first dark fringe, ∆x = λ
∴ D 2 + d 2 + ( x − d )2 + D 2 − D + D 2 + x 2 = λ 1/2
1/2
( x − d )2 d2 or D 1 + 2 + D 1 + D2 D
1/2 x2 − D + D 1 + 2 = λ D
d2 ( x − d )2 x2 + D + or D + − D + D + =λ 2D 2D 2D https://t.me/universitarios_infohttps://www.jamarana.com
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4B.27
d 2 + ( x − d )2 =λ 2D or d2 + x2 + d2 – 2xd – x2 = 2λ D
or
or 2d2 – 2xd = 2λD Dλ , we get 2
For d =
2
in fo rio s_
ar an a
Dλ = − Dλ 2
2x
or
2
ve rs ita rio
iv
ni
.m
:// t
Dλ . d
e/ u
tp s
(c) Fringe width β =
ht
Dλ =d 2
e/ un
://
w
which gives x =
er si ta
w .ja
w
m
Dλ 2 or 2 x 2 = (− Dλ)
s
Dλ Dλ − 2x = 2λ D 2 2
.c o
or 2
m
Dλ Dλ 2 = 2λD − 2 x 2 2
t.m
ht tp
s: //
ht
E
tp s
xample 14 Two coherent point sources S1 and S2 vibrating in phase emit light of wavelength λ. The separation between the sources is 2λ. Consider a line passing through S2 and perpendicular to the line S1S2. What is the smallest distance S2 where a minimum of intensity occurs? )
) Solution Path difference at P, ∆x = d 2 + x 2 − x
∆xmin = 0, when x → ∝, and maxima will occur. ∆xmax = 2λ, when x = 0, again maxima will occur.
3λ 2
d 2 + x2 − x =
3λ 2
or
or
or
For nearest minima ∆x =
3λ d 2 + x2 = + x 2 d 2 + x2 =
2
9λ 2 + x 2 + 3λx 4
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6 G [
6
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0D[LPD
0LQLPD
0D[LPD
0LQLPD
0D[LPD
s:
ita
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7λ . 12
er si ta rio s
x=
w
or
3λ
9λ 2 4
er s
ar a m w
.ja
=
(2λ) 2 −
rio s_ in fo
'[ O
na .c om
'[ O
or
[
3
'[ O
2 9λ 2 d − 4 x= 3λ
(a)
λD 2d
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example 15 Consider the situation shown in the fi gure. The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength λ. The separation between the slits is d. The light transmitted by the slits falls on a screen E1 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3. Another screen E2 is placed a further distance D away from E1. Find the ratio of the maximum to minimum intensity observed on E2 if z is equal to
ht
4B.28
(b)
λD d
(c)
λD . 4d
6
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6
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6
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'
(
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Light (Wave Optics)
4B.29
)
) Solution Light from sources S1 and S2 get interfered and thereafter S3 and S4
become new sources. At S3 the path difference between the lights coming from S1 and S2 is zero. Therefore they interfere constructively and so a3 = (a + a) = 2a
(a) At S4 the path difference ∆x = d sin θ ≃ d tan θ λD d× z 2d =d = D D λ = 2
s_
an
in
a.
I max (a3 + a4 ) 2 (2a + 0) 2 = =1 = 2a − 0 I min (a3 − a4 ) 2
rio
ar
iv un
tp
Corresponding phase difference =
ht
s:
//t
.m e/
tp
s:
//t
ht
er s
.m e/
tp
I max (2a + 2a ) 2 = =∞ I min (2a − 2a) 2
and
λD d dz 4d λ (c) ∆x = = = D D 4
A42 = a 2 + a 2 + 2aa cos
rio
ita
Now a4 = a + a = 2a
ht
un
iv
//w
Corresponding phase difference = 2π radian
s:
w
w
er si
ta
.ja m
λD d dz d (b) ∆x = = =λ D D
s
The ratio
fo
a4 = 0
∴
co m
Corresponding phase difference = π radian
or
λ radian 2
π = 2a 2 2
A4 = 2a
( (
) )
2 2
2a + 2a I max = I min 2a − 2a
=
(3.414) 2 (0.586) 2
= 34.
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Light (Wave Optics)
example 16 In a modified Young’s double slit experiment, a monochromatic uniform and 10 parallel beam of light of wavelength 6000 Å and intensity W/m2 is incident π normally on two circular apertune A and B of radii 0.001 m and 0.002 m respectively. A perfect transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength of 6000 Å is placed infront of aperture A (see fi gure).
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Calculate the power (in watts) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focus spot.
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The intensities of sources after emerging from the lenses are
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∆x = (µ – 1) t = (1.5 – 1) × 2000 × 10–10 = 10–7 m
and
φ=
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4B.30
2π 2π π × ∆x = × 10−7 = radian λ 6000 × 10−10 3
Now power received at F I + I A − I B + 2 I A I B cos φ −6 −6 = 10−6 + 4 × 10−6 +2 10 × 4 × 10 cos
π 3
= 7 × 10–6 W
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Light (Wave Optics)
4B.31
4.7 interference in tHin films Thin film may be an oil film spread over water, paint on glass, or air film between two glass plates etc. The phenomenon of interference was first explained by Young. It has been observed that interference in the case of thin film takes place due to (1) reflected light (2) and transmitted light.
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(1) Interference in reflected light Consider a thin film of transparent material of thickness t and refractive index µ is situated in air. A ray incident at A on the upper surface of the film is partly reflected and partly refracted along AB. At B it is partly reflected along BC and finally emerges out along BG. This process of reflection and refraction goes continuously at several points on the film (theoretically infinity).
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To get the path difference between the light waves from points A and C, drop a perpendicular at D on the reflected ray from A and at E on the ray BC. Thus optical path difference,
∆x = (AB + BC) in medium – AD in air = (AB + BE + EC) in medium – AD in air
We know that AB distance travels in medium of refractive index µ becomes equal to µAB in air, and so for BE and EC. Therefore
∆x = (AB + BE + EC)µ in air – AD in air
...(i)
By Snell’s law; µ=
sin i AD /AC AD = = ∴ AD = µ EC. sin r EC /AC EC
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...(ii)
Light (Wave Optics)
Substituting this value in equation (i), we get ∆x = (AB + BE)µ As AB = A′B, ∴ ∆x = (A’B + BE)µ = (A′E)µ
From the triangle AA′E, A′E = 2t cos r, and so ∆x = 2µt cos r
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λ 2µt cos r = (2n – 1) ; n = 1, 2, … 2 λ If we take 2µt cos r = (2n + 1) , then n will be 0, 1, 2, 2 (ii) The minima will occur when ∆x = (2n – 1) λ/2
∴ or
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Here n = 0 is discarded because path difference 2µt cos r can not be zero. When this condition is fulfilled the film will appear dark in reflected light.
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(i) The maxima will occur when ∆x = nλ.
...(1)
co
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The path difference in equation (iii) is the optical path difference. It has been proved that when light is reflected from optical denser medium (upper surface of the film), it undergoes an abrupt phase change of π rad; an equivalent path difference λ/2. Therefore the effective path difference becomes,
∆xe = 2µt cos r ±
...(iii)
∴ 2µt cos r = nλ; n = 1, 2, (ii) The minima will occur when ∆x = (2n – 1) λ/2. ∴ 2µt cos r = (2n – 1) λ/2; n = 1, 2,
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ote For normal incidence, angle of incidence i and hence r become zero and so cos r = 1, then 2µt cos r → 2µt. (2) Interference in transmitted light The optical path difference between the light waves originated from B and F can be calculated as in the case of reflected light. Thus ∆x = 2 µt cos r. In this case, the light rays emerge from B and F has no phase change because light emerges from F has gone two reflections, each with a phase change of π rad. Thus ∆xe = 2µt cos r + 0 = 2 µt cos r. (i) The maxima will occur when ∆x = nλ.
4B.32
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...(5)
Light (Wave Optics)
4B.33
Conclusion: The condition of maxima and minima for interference in reflected and transmitted light are opposite to each other. It means, if any colour is strongly reflected, then it will be absent in transmitted light.
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example 17 A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in fi gure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25% of the incident on it and transmits the remaining. Find the ratio of the minimum and the maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.
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I min = I1 + I 2 − 2 I1 I 2
∴
I min I1 + I 2 − 2 I1 I 2 = I max I1 + I 2 − 2 I1 I 2
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Light (Wave Optics)
colours of thin film When white light is incident on thin film, beautiful colours are observed. The incident light will split up due to reflection at the top and bottom of the film. The splitted rays interfere and gives bright colours. The interfering rays of only certain wavelengths will have a path difference satisfying the conditions of bright fringe. Hence only such wavelengths will have a path difference satisfying the conditions of bright fringe. Hence only such wavelengths will be present there. Other wavelengths will be present with diminished intensity.
2 µt cos r = nλ.
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interference in wedge shaped film Consider a thin film of refractive index µ is enclosed between the two surfaces inclined at an angle α. For small angle α, the effective path difference between two light waves originated from upper surface can be taken equal to 2 µt cos r. Thus for minima
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...(ii)
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and
xn – 1
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β = xn – xn – 1 =
[Q± [Q
λ 2µ tan α
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Light (Wave Optics)
4B.35
Newton’s rings: Newton’s rings are formed due to interference between the waves reflected from the top and bottom surfaces of the air film enclosed between the lens and the plate. From the property of circle, we have
r × r = t × (2R – t)
or
r2 = 2Rt – t2 ≃ 2Rt [Since t << R] t=
∴
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...(i)
Similarly, for dark ring: 2µt = nλ, n = 1, 2, 3
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Light (Wave Optics)
E
or
r2 = nλ 2R nλR r2 = µ Dn2 nλR = 4 µ
Dn2 =
or
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xample 18
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In Young’s double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron is introduced in the path of one of interfering waves. The mica-sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift on the introduction of mica-sheet. Calculate the wavelength of the monochromatic light used in the experiment.
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D (2 D)λ (µ − 1)t = c d (µ − 1)t ∴ λ= 2
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(2 D)λ . d According to the given condition β=
ht
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(1.6 − 4) × 1.964 × 10−6 2 = 5892 Å.
=
xample 19 A glass plate of refractive index 1.5 is coated with a thin layer of thickness of t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If the λ = 648 nm. Obtain the least value of t for which the rays interfere constructively. E
4B.36
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Light (Wave Optics)
4B.37
) Solution
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2µt = (2n – 1)λ/2, n = 1, 2, 3, λ ∴ t = (2n − 1) 4µ
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648 × 10−9 = 90 × 10–9 m 4 × 1.8
s:
=
λ λ = 4µ 4µ
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Condition of constructive interference in the situation shown in figure is
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example 20 A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in figure. Calculate the position of the central fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled upto OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid.
) Solution $ 6
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Light (Wave Optics)
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∆x = (SS1 + S1P) – (SS2 + S2P)
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or
d sin θ1 = d sin θ2
or
sin θ1 = sin θ2
0.4 + 2
2
=
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After solving, we get y = 0.02 m = 2 cm.
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4B.38
When water is filled in the vessel, it causes shift in position of fringe
or
∆ = 2 cm D (µ – 1)t = 0.02 d
0.10 (µ – 1) × 0.1 = 0.02 0.8 × 10−3 which gives
µ = 1.002.
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&
4B.39
Light (Wave Optics)
example 22 In a modified YDSE the region between screen and slits is immersed in a liquid 5 T whose refractive index varies with time as µ l = − until it reaches a steady 2 4 5 3 state value of . A glass plate of thickness 36 µm and refractive index is in4 2 troduced infront of one of the slits. (a) Find the position of central maxima as a function of time and the time when it is at point O, located symmetrically on the x-axis. (b) What is the speed of the central maxima when it is at O?
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) Solution (a)
∆x = S 2 liquid P + (µ g − µ )t – (S1P)liquid
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The optical path difference between lights start from S1 and S2
= (S2P – S1P)liquid + (µg – µℓ)t = µl(S2P – S1P)air + (µg – µℓ)t or
∆x = µl d
y + (µg – µℓ)t D
For central maxima, ∆x = 0 y 0 = µ d or + (µg – µℓ)t D which gives
y=
− D(µ g − µ )t d µ
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Light (Wave Optics)
3 5 T − D − − t 2 2 4 = 5 T d − 2 4 T D 1 − t D(4 − T )t 4 = = 5 T d (10 − T ) d − 2 4
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T D 1 − t 4 0= ⇒ T = 4 s. 5 T d − 2 4
dy 6 Dt = dt (10 − T ) 2 d Central maxima is at O at T = 4 s Speed of central maxima v =
ita
6 Dt 6 Dt Dt = = 2 (10 − 4) d 36d 6d
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example 23 A glass surface is coated by an oil film of uniform thickness 1.00 × 10–4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm – 750 nm) which are completely transmitted by the oil film under normal incidence.
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t.m e/
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) Solution Optical path difference for the light transmitted through oil is = 2µt cos r for normal incidence
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4B.40
2LO P *ODVV P
∴
r = 0, cos r → 1. ∆x = 2µt
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Light (Wave Optics)
4B.41
But at the interface between oil and glass will produce an additional path difference of λ/2. Therefore effective path difference ∆x = 2µt + λ/2 For constructive interference in transmitted light λ = nλ, n = 1, 2, …. 2 2µt = (2n – 1) λ/2
λ=
4µt (2n − 1)
=
4 × 1.25 × 1× 10−6 2n − 1
=
5 × 10−6 (2n − 1)
o in f
n = 4, λ =
5 × 10−6 m = 714.29 nm 7
n = 5, λ =
n = 6, λ =
ita
5 × 10−6 m = 555.55 nm 9
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n = 3, λ =
5 × 10−6 m = 454.54 nm 11
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E
xample 24 Two square surface of a transparent plastic block having a shape of as shown in fig. 4.41 have thickness t1 at one edge and t2 at the other. The refractive index of the material is 1.6. When viewed at normal incidence, using a light of wavelength 5000 Å, 20 fringes are observed. Find the difference (t2 – t1). )
) Solution
...(i) ...(ii)
2µt1 = n λ 2µt2 = (n + 20)λ
and
Let, at the thickness t1, nth number of fringe is formed, then at thickness t2, (n + 20)th number of fringe will form. For reflected light
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4B.42
Light (Wave Optics)
W
W
2µ(t2 – t1) = 20λ 10λ (t2 – t1) = µ 10 × (5000 × 10−10 ) = 1.6 = 3.125 × 10–6 m.
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4.8 diffraction
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Subtracting (i) from (ii), we get
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The bending of waves around the edges of an obstacle or aperture is called diffraction. This phenomenon was discovered by Grimaldi. The theorical explanation was first given by Fresnel. According to him, diffraction results from the superposition of secondary wavelets originating from the different parts of the same wavefront. For diffraction to occur, the size of the aperture should be small enough. It should be comparable to the wavelength of wave. In case when size of aperture is very large in comparison to the wavelength, then there is no appreciable diffraction will occur, and so infront of the aperture, there is uniform illumination on the screen (see fi g.).
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Light (Wave Optics)
4B.43
If an obstacle with a small gap is placed in almost like a point source. If the gap is large however, the diffraction is much more limited. Small, in this context, means that the size of the obstacle is comparable to the wavelength of the waves.
difference Between interference and diffraction
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1. Interference takes place between two or more wavefronts originating from coherent sources. While in diffraction superposition takes place between the secondary wavelets originating from the same wavefront.
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2. In the interference pattern the regions of minimum intensity are usually almost perfectly dark. While it is not so in diffraction pattern. 3. The fringe width in interference may or may not be equal, while in diffraction pattern fringe widths are never equal. 4. In interference all the maxima are of same intensity but in diffraction they are of varying intensity.
4.9 fraunHoffer diffraction at single slit In this type of diffraction, a plane wavefront falls on a slit and its response is seen on a plane screen placed at very large distance from the slit. Suppose a plane wavefront falls on a slit of width d. Every point of the exposed part of the wavefront acts as a source of secondary wavelets. These wavelets superpose (interfere) and produce diffraction.
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Light (Wave Optics)
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intensity distribution
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If the intensity of the principal maxima is I0, then the intensity of the first maxima I I and second order maxima are found to be 0 and 0 respectively. The diffraction 22 61 fringes are of decreasing width and decreasing intensity. In general, for any angular position θ, the intensity I is given by
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4B.44
2
sin α I (θ) = I 0 , α πa where α = sin θ. λ
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Light (Wave Optics)
4B.45
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principal maxima When wavefront strikes the slit, the secondary wavelets from all point in the slit travel about the same distance to reach the centre of the diffraction pattern and thus are in same phase there. So they interfere constructively and produce principal maxima. To understand higher order maximas, we can use a clever technique and the slit can be divided hypothetically into odd number of zones; three zones for first order maxima and five zones for second order maxima and so on.
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Light (Wave Optics)
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positions of first and Higher order minimas To understand first order minima, we can divide the slit hypothetically into two identical zones Z1 and Z2 each of width d/2. Take a wavelet from top of the zone Z1 and other from the top of the zone Z2. The path difference between them is d/2 sin θ. All other similar pairs will have the same path difference. If these wavelets interfere destructively, then they will produce minima. This will happen when d/2 sin θ is equal to λ/2. Thus for first order minima or
d/2 sin θ = λ/2 d sin θ = λ
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Light (Wave Optics)
4B.47
The position of first order minima,
sin θ =
λ . d
The principal maxima spread symmetrically about central line in angle 2θ. Thus angular width of principal maxima is ...(2)
λ 2θ = 2sin −1 . d
m
λ Linear width of principal maxima; β = (2θD) = 2D sin −1 . d For second order minima, we can divide the slit hypothetically into four zones. The wavelets of nearest two zones interfere destructively and again produce minima. Thus for second order minima, we have
in
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d sin θ = 2λ.
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or
d λ sin θ = , 4 2
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ositions of first and igher order maximas
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P
...(3)
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n = 1, 2,
d sin θ = nλ;
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In general the positions of minimas can be obtained as:
er
:// tp s
...(4)
or
d λ sin θ = 3 z 3λ . d sin θ = 2
ht
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tp
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s:
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To understand first order maxima, we can divide the slit hypothetically into three identical zones Z1, Z2, Z3 each of width d/3. The wavelets of two neighbouring zones Z1 and Z2 interfere destructively and cancel their mutual effect. The wavelets of remaining third zone will cause first order maxima. Thus for first order maxima the path difference between two wavelets, one from top of the zone Z1 and other from top of zone the Z2 is d/3 sin θ. For destructive interference between them
Similarly for second order maxima, we can get
a sin θ =
5λ . 2
In general, for higher order maxima, we can write; λ d sin θ = (2n + 1) ; n = 1, 2, 2
N
...(5)
ote
The technique used to explain the positions of minima and maxima can not be used to get the intensity of maxima. https://t.me/universitarios_infohttps://www.jamarana.com
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4B.48
Light (Wave Optics)
4.10 diffraction grating A diffraction grating consists of equally spaced, parallel slits. If a is the width of the slit and b is the width of the opaque portion between two adjacent slits, then (a + b) is the distance between the adjacent slits. This is known as grating element. If there are N parallel slits (rulings) in each centrimetre, then grating element is given by
D
ar a
Diffraction grating.
m
1 cm. N
ita
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(a + b) =
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E
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(a + b) sin θ = nλ;
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s:
n = 1, 2, ...
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w
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w
The grating equation can be written as:
iv un
//t
ht
.m
tp
The above equation gives the position of nth maxima.
s: //
t.m e/
ht tp s:
example 25 Plane microwaves are incident on a long slit having a width of 5.0 cm. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at θ = 30º.
sin θ = Given,
θ = 30º
∴ sin 30º = or
λ d
ht tp
) Solution For first diffraction minima,
λ 5
1 2 = 2.5 cm.
λ = 5×
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Light (Wave Optics)
4B.49
diffraction by a circular aperture When a plane wavefront falls on a circular hole, it is diffracted by the hole. The response, if received on a screen at a large distance, the pattern is a bright disc surrounded by alternate dark and bright rings of decreasing intensity as shown in fi gure. The position of first dark ring from the hole at an angle θ with the axis 1.22λ , d where λ → wavelength of the light used.
m
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er si ta rio s
ita er s
w //w
1.22λD (for small θ). d
un iv
w
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The radius of first dark ring R=Dθ=
5
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T
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G
na .c om
sin θ =
iv
//t
ht
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tp
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s:
Most of the light coming from the hole is concentrated within the first dark ring, this radius is also called radius of the diffraction disc.
s: //
t.m e/
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un
example 26 Light of wavelength 560 nm goes through a pinhole of diameter 0.20 mm and falls on a wall at a distance of 2.00 m. What will be the radius of the central bright spot formed on the wall?
R = 1.22 =
ht tp
) Solution The radius of central dark can be obtained as λD d
1.22 × ( 560 × 10−9 ) × 2 0.20 × 10−3
= 0.68 cm.
example 27 Angular width of central maximum in the Fraunhoffer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 Å. When the slit is illuminated by light of another wavelength, the angular width is decreases by 30%. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid.
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Light (Wave Optics)
) Solution Angular width of central maxima = 2λ ′ d According to given condition, we have
2λ . d
For the other light it is = 2λ′ 2λ = (0.70) d d
λ′ = 0.7 λ = 0.7 × 6000 Å = 4200 Å.
⇒
When the apparatus is immersed in liquid, the same decrease is observed in the angular width.
na .c om
λ (2λ air )/µ = d d 1 µ = = 1.43 0.7
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or
rio s_ in fo
Angular width in liquid = 2
er si ta rio s
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w
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w
ita
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m
example 28 A convex lens of diameter 8.0 cm is used to focus a parallel beam of light of wavelength 620 nm. If the light be focused at a distance of 20 cm from the lens, what would be the radius of the central bright spot formed?
e/
1.22λ d
iv t.m e/
ht tp s:
un
//t
sin θ =
.m
tp
s:
) Solution The angular spread of central bright is given by
ht
5
s: //
T
ht tp
4B.50
'
1.22 × 620 × 10−9 0.08 = 9.455 × 10–6 rad. =
Since θ is small, so we can take tan θ ≃ sin θ = 9.455 × 10–6 rad By geometry, we have, tan θ = ∴
R = 9.45 × 10–6 D
R D
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Light (Wave Optics)
or
4B.51
R = 9.45 × 10–6 × 0.20 = 1.89 × 10–6 m.
example 29 A wedge of angle 0.5º is illuminated with sodium light whose two lines corresponds to the wavelengths 5890 Å and 5896 Å. Find the distance from the apex at which the maxima due to the two wavelengths first coincide when observed in the reflected light. (The wedge contains air).
) Solution Let the thickness of the wedge at the point where the maximums of both
rio s_ in fo
λ 2
ita
[
W
er si ta rio s
w
w
2µt cos r = (2n + 1)
T
er s
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m
ar a
na .c om
coincides be t. For constructive interference in reflected light, we have
un iv
e/
iv
.m
un
//t
and
λ1 2 λ2 2t = (2n + 3) 2 2t = (2n + 1)
ht tp s:
ht
tp
s:
//w
For normal incidence and µair = 1, we have
∴ and
s: //
(2n + 1)λ1 (2n + 3)λ 2 = 2 2 (2n + 1) (5896) = (2n + 3) (5890)
ht tp
or
t.m e/
According to given condition, we can write
n = (1499) whole number
λ1 (2 × 1499 + 1) = × 5896 × 10–10 4 4 = 0.004 cm
t = (2n + 1)
Let x be the required distance from the apex, then
∴
t = tan θ ≃ θ x t x= θ 0.004 = = 4.58 cm. π 0.5 × 180
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Light (Wave Optics)
resolution limit and resolving power When we decrease the separation between the two point objects, a limit is reached when our eyes can not differentiate them separately, even they are not in contact. The minimum separation between two closely placed objects, the eyes can distinguish them separately is known as resolution limit (RL). The reciprocal of resolution limit is known as resolving power (RP). Thus RP =
1 . RL
rayleigh Criteria
iv
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s: //
t.m e/
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E -XVWUHVROYHG
ht tp s:
er si ta rio s
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w
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w
ita
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D 8QUHVROYHG
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According to Rayligh, two images of two objects are said to be just resolved when central maxima of one diffraction pattern falls on first minima of other (see figure).
ht
4B.52
F :HOOUHVROYHG
(i) The resolution limit of normal human eye is one minute (1´). (ii) The resolution limit of a telescope is given by angle θ, where 1.22λ θ= rad. d Here λ is the wavelength of light used and d is the diameter of objective lens of the telescope. (iii) The resolution limit of a microscope is given by 1.22λ x= metre, 2µ sin θ https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.53
where µ is the refractive index of medium between objects and lens; θ is the angle subtended by the objects at objective lens of the microscope. µ sin θ is called numerical aperture of the lens.
4.11 transverse nature of ligHt
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Interference and diffraction phenomenon prove the wave character of the light. These phenomenon do not give any idea; whether light waves are transverse or longitudinal. Polarisation proves the transverse character of light waves. Transverse character of light waves was known in the early nineteenth century; however the association of electric and magnetic vectors with light wave was known only after Maxwell. The electric field in a light wave propagating in free space is perpendicular to the direction of propagation. Natural light consists of many pulses emitted by different atoms; in general have electric field in different directions. Hence the resultant electric field at a point changes continuously and randomly. Such a light is called unpolarized light. If Eo is the amplitude of the resultant of all the waves, then it can be resolved into two components; Ex and Ey, where Ex = Eo cos θ and Ey = Eo sin θ
ht tp
The intensity of light due to x-component is given by I x = Ex2 = Eo2 cos 2 θ, and due to y-component. I y − Eo2 sin 2 θ. To get the average intensity of light, we have to find 2π 1 1 the average of cos2 θ and sin2 θ over one complete cycle; which is cos 2 θ = 2π ∫0 2 2π 1 1 and sin 2 θ = . 2π ∫0 2 Thus resultant intensity of the light I0 can be written as:
I0 = Ix + Iy = E02 cos 2 θ + E02 sin 2 θ
or
I0 =
E02 E02 + = E02 . 2 2
Thus unpolarized light can be assumed of two components; each of intensity
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I0 . 2
4B.54
Light (Wave Optics)
4.12 polariZation When electric field vector (or magnetic) vibrates only along one direction perpendicular to the direction of propagation, the light is said to be linearly polarized light and the phenomenon associated is called polarization. Thus when unpolarized light is incident on a polarizer, the emerging light becomes linearly polarized of I intensity 0 . 2 ,\
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Linearly polarized light.
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w
plane of polarization
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The plane which passes through direction of propagation and perpendicular to the plane of vibration is known as plane of polarization.
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4.13 polariZation By reflection: BreWster’s laW In 1808, Malus found that when natural light is incident of a glass slab, the reflected light becomes partially polarized. Brewter forwarded his study and found that when light is incident on glass slab at certain angle, called polarising angle, the reflected light becomes linearly polarized. This happens when reflected and refracted rays are perpendicular to each other. Consider a beam of natural light is incident on a glass slab at an angle ip as shown in fi gure. A part of it is reflected and a part is refracted. If r is the angle of refraction, then https://t.me/universitarios_infohttps://www.jamarana.com
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Light (Wave Optics)
4B.55
r + iP = 90º r = 90º – iP.
∴
By Snell’s law; sin i sin r sin iP = sin(90° − iP )
µ=
=
sin iP cos iP
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L3 L3
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ita
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,[
rio s_ in fo
µ = tan iP. Brewster’s law
∴
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s: //
The above relation is known as Brewster’s law. For ordinary glass µ = 1.5, iP = tan–1 (µ) = tan–1 (1.5) ≃ 57º. Thus when natural light is incident on glass slab, the reflected light becomes linearly polarized and refracted light becomes partially polarized. ∴
polarization by double refraction When unpolarized light (natural light) is incident on a calcite crystal, it splits into two rays; O-ray (ordinary ray) and E-ray (extra-ordinary ray). O-ray has same speed in all directions and hence obeys law of refraction. While E-ray has different speed in different directions in a medium and so does not obey law of refraction. These two rays are plane polarized and their planes of vibrations are mutually perpendicular. https://t.me/universitarios_infohttps://www.jamarana.com
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4B.56
Light (Wave Optics)
8QSRODUL]HG OLJKW
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4.14 malus’ laW
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Iθ = (E cos θ)2 = E2 cos2 θ Iθ = I cos2 θ. Malus’ law
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ita
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Consider a polarizer and a analyser placed at an angle θ. If E is the amplitude of the incident wave, then the amplitude of the wave emerging from analyser will be E cos θ and thus the intensity of the emerging beam will be given by;
e/
iv
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un t.m e/ ,\ , (
ht tp
8QSRODUL]HGOLJKW
,\
s: //
ht tp s:
//t
ht
tp
s:
Thus when plane polarized light is incident on an analyser, which is placed at an angle θ with the polarizer, the intensity of emerging light from analyser becomes I cos2 θ.
3RODUL]HU
T
,T
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In complete rotation of analyser, we will see two maximum and two zero intensities. For
θ = 0º, Iθ = I cos20º = I θ = 90º, Iθ = I cos290º = 0 θ = 180º, Iθ = I cos2 180º = I θ = 270º, Iθ = I cos2 270º = 0.
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Light (Wave Optics)
4B.57
example 30 Two polaroids are placed at 90º to each other and the transmitted intensity zero. What happens when one more polaroid is placed between these two bisecting the angle between them?
) Solution
rio s_ in fo
na .c om
er si ta rio s
w
er s
w
ita
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m
ar a
Figure shows the orientation of the polaroids. If I is the intensity of the light emerging from the first polaroid, then intensity of emerging light from second and third are I cos2 45º and (I cos2 45º).sin2 45° These are: I/2 and I/4 respectively.
e/
iv
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un
//t
ht
tp
s:
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example 31 Three nicols prisms are placed such that, first and third are mutually perpendicular. Unpolarized light is incident on first nicol’s prism, the intensity of light emerges from third nicol’s prism is 1/16 the intensity of incident light. Find the angle between first and third nicol’s prisms.
ht tp s:
) Solution Suppose angle between first and second nicol’s prisms is θ. Then the
ht tp
s: //
t.m e/
angle between second and third nicol’s prisms becomes 90º – θ. If I0 is the intensity of the incident light on the first prism, then intensity of emerging light I from this will be I = 0 . The intensity of light emerging from second and third 2 I I0 2 nicol’s prisms be cos θ and 0 cos 2 θ sin 2 θ respectively. 2 2 T
± T
Thus according to the given condition I0 I cos2 θ sin2 θ = 0 2 16 https://t.me/universitarios_infohttps://www.jamarana.com
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4B.58
1 8 1 (2 sin θ cos θ)2 = 2 1 sin2 (2θ) = 2 (sin θ cos θ)2 =
or
or
2θ = 45º
ita
iv
.m
un
t.m e/ s: //
= x2 + d 2 − x
er si ta rio s
er s
8 n − n 2 It is easily seen that the path difference decreases from ∆x ≃ 4 m to (∆x → 0 at x → ∞) Hence, n can have value n = 3, 2, 1. Going away from O towards the x-axis, 8 3 7 first maxima occurs at x = − = m (for 3 2 6 n = 3) 8 2 second maxima occurs at x = − = 3 m (for 2 2 n = 2) 8 1 third maxima occurs at x = − = 7.5 m (for 1 2 n = 1)
ht tp
ht tp s:
Path difference, ∆x = S2 P – S1 P
//t
ht
⇒
x 2 + 16 − x = n
Solving, we get x =
e/
tp
s:
un iv
//w
w
w
.ja
m
)
Let
= x2 + d 2 − x
Using d = 4 m, λ = 1 m, we have
) Solution Let the detector be at the point P (x, 0) S1 S2 = d = 4 m
rio s_ in fo
For constructive interference ∆x = nλ
ar a
1. Two point sources of radiation: S1 and S2 radiate waves of same frequency and are excited by the same oscillator. They are also in phase with each other. S1 is placed at orgin O, while S2 is placed at the point (0, 4) on y-axis. Find the points of maxima of received intensity if a detector is moved towards positive x-axis starting from origin. The wavelength of radiation is 1 m.
X
AMPLES
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θ = 22.5º.
or
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2. The distance between the coherent sources is 0.3 mm and the screen is 90 cm from the source. The second dark band is 0.3 cm from central bright fringe. Find the wavelength and the distance of the fourth bright fringe from CBF.
6
6 2
)
) Solution Here, d = 0.3 mm = 3 × 10–2 cm [
'
;
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1 2
sin 2θ =
or
or
na .c om
Light (Wave Optics)
D = 90 cm, x2 = 0.3 cm
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)
) Solution
λD 6 × 10−5 × 80 = = 0.096 cm d 0.05 (ii) Distance of the 10th bright fringe from CBF λD x10 = n × d = 10 × 0.096 = 0.96 cm Distance of the 10th dark fringe on the other side of CBF λD ′ = (2n − 1) × x10 d
∴ w=
rio
ita
rio
ita
iv e
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e/
)
.m
⇒
4. A biprism is placed at a distance of 5 cm in front of a narrow slit illuminated by monochromatic light of wave length 6 × 10–5 cm. The virtual images formed by the biprism are 0.05 cm apart. Find the fringe width and a screen 75 cm from the biprism. Also
d=
180 × 6000 × 10−10 = 3.44 ×10-4 m 0.1 π
6. A beam of light constituting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment. (i) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å.
s:
10 × 589.3 × 10−9 m × 100 × 10−2 12 × 10−3 = 4.9 × 10–4 m = 0.049 cm
= 1.872 cm.
5. In a Young’s double slit experiment, the angular width of a fringe formed on a distant screen is 0.1º. The wavelength of the light used is 6000 Å. What is the spacing between the slits? w λ ) Solution Angular width = = D d λ 0.1π ⇒ = d 180
=
rs
ht
tp
//t
tp
ht
nλD d Given xn = 12 mm, n = 10 λ = 589.3 mm,
s
rs
un
.m e/
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s:
)
distance between the central line
D = 100 cm nλD d = xn
1 = (2 × 10 − 1) × × 0.096 2 = 0.912 cm ∴ Required distance = x10 + x´10
iv e
w
w
s: //
tp
ht
) Solution For the nth maximum fringe, the xn =
in f
o
ar
w
3. A double slit experiment is performed with sodium (yellow) light of wavelength 589.3 mm and the interference pattern is observed an a screen 100 cm away. The length bright fringe has its centre at a distance of 12 mm from the central maximum. Find the separation between the slits.
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m
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a.
co
m
or
s_
2 λ = × 10−4 cm 3 2 or λ = ×104 Å 3 th (ii) For n bright fringes, λD xn = n d 2 90 −4 Putting n = 4, x4 = 4 × × 10 × 3 3 × 10−2 = 0.8 cm
3 90λ 0.3 = × 2 3 × 10−2
⇒
(i) Here, d = 0.05 λ = 6 × 10–5 m D = distance between the screen and virtual images = 75 + 5 = 80 cm
Putting n = 2,
λD 2d x2 = (2 × 2 – 1) × λ × 90 2 × 3 × 10−2
xn =(2n – 1) ×
4B.59
find the distance between the 10th bright band on one side and 10th dark band on the other side of the central bright fringe.
(i) For nth dark fringes,
Light (Wave Optics)
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4B.60
∴
Icenter = Imax (∆ φ = 0)
2π (Path λ difference) 2π λ π = × = λ 8 4
At the given point, ∆ φ =
I = I max cos 2
(ii) What is the least distance from the central maximum where the bright fringes due to both the wavelength coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.
Light (Wave Optics)
)
) Solution Given: D = 120 cm. d = 0.2 cm
∆φ π = I max cos 2 2 8
I 1 + (1/ 2 ) 2 +1 = = I max 2 2 2
∴
in
s_
an
ht
Hence, fourth bright fringe due to λ1 coincides with fifth bright fringe due to λ2 at distance
x=
si ta
un
iv er
si ta
rio
s
er
t.m e/
s: //
s:
m 4 = n 5
6500 m = 5200 n
P
$
tp
⇒
//t
tp
⇒ ⇒
Dλ1 Dλ 2 and x = n d d m λ1 = m λ2 x=m
ht
∴
.m
ht
tp
e/
un
s: //
iv
w
w
w
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m
rio
ar
fo
co
a.
3Dλ1 3 × 120 × 6500 × 10−8 = = 0.117 cm d 0.2 Dλ1 w1 = (ii) d Dλ 2 and w2 = d th Let the m bright fringe due to λ1 and nth bright fringe due to λ2 coincide at a distance x from central maximum.
8. In the figure shown, S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two are placed symmetrically about the central axis so with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO. (i) If the third intensity maximum occurs at the point A on the screen, find the distance OA. (ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease or remain the same?
m
Let λ1 = 6500 Å and λ2 = 5200 Å (i) Distance of nth bright fringe from central maximum nDλ = d ∴ Distance of third bright fringe
P
6
6FUHHQ
4Dλ1 d
P
)
) Solution At central maximum, constructive interference occurs.
)
) Solution
For L1 and L2: u1 = u2 = 0.15 m, and f = 0.1 m
1 1 1 + = u v f
or
1 1 1 1 1 = − = − v f u 0.1 0.15
∴
4 × 120 × 6500 × 10−8 = = 0.156 cm. 0.2 7. The path difference of a certain point on screen in a double-slit experiment is oneeight of the wavelength. Find the ratio of intensity at this point and the intensity at central maximum.
2
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Light (Wave Optics)
4B.61
6 [ / I 6
[
[
3
2 [ /
I
ita er s
er si ta rio s
un iv
G
t.m e/
un
iv
G
D FP
s: //
= 2 × 10 ×
∴
w=
=
λD d
π π (1.5 –1) = cm 90 9
5890 × 108 × 100 = 0.01687cm. π 9
10. In Young’s experiment, the source is red
)
) Solution Here, λ = 5890 × 10–8 cm
d 2a d = 2a × A (µ – 1)
tan δ ≅ δ =
∴
rio s_ in fo
ar a e/
ht tp
∴
(ii) If gap between L1 and L2 is reduced from its original value of 0.5 mm, then d will reduce and hence OA will increase. 9. A Fresnel’s biprism of angle 2º is illuminated by light of wavelength λ = 5890 Å by a slit source 10 cm from it. Calculate the fringe width on a screen 90 cm from biprism, if the R.I. of biprism is 1.5.
G
The minimum deviation of the thin prism is given as δ = A(µ – 1)
3 × 500 × 10−9 × 1 = 1 × 10–3 m. 1.5 × 10−3
D = 90 + 10 = 100 cm
6
$
6
.m
ht tp s:
=
µ = 1.5
π radians 90 The distance d is found as shown:
//t
ht
tp
s:
//w
w
w
Hence distance between two source S1 and S2 = 3 x = 1.5 mm (i) The maximum intensity on the screen is observed, when the distance of maximum from O, λD xn = n d Given: n = 3, D = 1.3 – 0.3 = 1m d = S1S2 = 1.5 mm, λ = 500 nm ∴ OA = x3
6
Angle of biprism, A = 2º =
m
⇒
.ja
v = 0.3 m v1 v2 = =2 u1 u2
na .c om
X X P
or
[
Y Y P
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4B.62
11. In Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index. 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity. It is further observed that what used to be the fifth maximum earlier, lies below the point P while the sixth minimum lies above P. Calculate the thickness of the glass plate. (Absorption of light by glass plate may get neglected.)
6
ta
si
ta rio er si
iv un e/
= (1.7 – 1.4) t = 0.3t
...(i)
Since, P lies in between the original fifth maximum and sixth minimum, this path difference will also be given as Path difference = 5λ + ∆x
...(ii)
Equating equations (i) and (ii), we get
The change in fringe width due to the change in wavelength D w = w2 – w1 = (λ g − λ r ) d = 285.71 × (5 × 10–7 – 7 × 10–7) m
s
er
Path difference = (µ2 – µ1) t
6 × 7 × 10−7 + 1 = 1.6 7 × 10−6
6FUHHQ
µg =
WKPLQLPXP
0.3t = 5λ + ∆x
tp s
ht
or
3
The optical path difference between the two rays reaching P from the two slits S1 and S2 after the insertion of glass plates of equal thickness and having refractive indexes µ1 and µ2 respectively is given as
ht
6 × 7 × 10–7 = (µg – 1) × 7 ×10–7
or
WKPLQLPXP
WP
:// t.m
tp
s:
Now, for green light (λ = 5 × 10–7 m), we have ∆x = (µg – 1) t
= – 5.71 × 10–5 m.
WP
iv
un
e/
ht
5 × 7 × 10−8 = 7 × 10–6 m 1.5 –1 It is also given that the central bright fringe shift by 10–3 m, hence, from the expression λD xn = n d 10−3 D xn = = 285.71 ∴ = d nλ 5 × 7 × 10−7 =
6
//t .m
tp
s:
//w
w
w
.ja
where, µr (=1.5) is the refractive index of glass plate for red light and t is the thickness of the plate. Given: ∆x = 5λ. ∆x ∴ t= µr − 1
s_
)
m
ar
) Solution
in
an
by the insertion of glass plate in the path of one of the interfering beam is given as ∆x = (µr – 1) t
rio
)
a.
) Solution The optical path difference introduced
fo
co m
light of wavelength 7 × 10–7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beam, the central bright fringe shift by 10–3 m to the position previously occupied by the fifth bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10–7 m, the central fringe shift to a position initially occupied by the sixth bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength.
Light (Wave Optics)
...(iii)
The path difference ∆x can be determined from the given intensity at P, which is
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) Solution
)
According to the expression I ∆φ = cos 2 I0 2 2π ∆x Here ∆φ = λ 2
4B.63
(ii) If the incident beam makes an angle of 30º with the x-axis (as in the dotted arrow in the figure), find the y co-ordinates of the first minima on either side of the central maximum.
I 3 = I0 4
Light (Wave Optics)
I 2π ∆x 3 = cos 2 . = I0 λ 2 4
∴
\ T
G
3 π∆x cos = λ 2 π∆x π or = λ 6 λ or ∆x = 6 Now, thickness of the glass plates t is given by
2
rio
ta
si
un iv er e/
Thus, the allowed values of m are + 1, 0, –1 and – 2. Hence, four minima will be observed on the screen; distance on the screen is given by
;
For m = 1, sin θ =
3 3 , y= m 4 7
For m = 0, sin θ =
1 1 , y= m 4 15
For m = –1, sin θ =
1 –1 m , y=– 4 15
For m = –2, sin θ =
3 –3 m , y=– 4 7
6FUHHQ
' P
y = D tan θ = 1.0 sin θ sin θ = 1.0 1 − sin 2 θ cos θ
G
1 λ 1 sin θ = m + = m + 2D 2 0.5 mm 1 1 = m+ 1.0 mm 2 2
ht
tp
s: //
t.m
ht tp s:
e/
.m
//t
ht
12. A coherent parallel beam of microwave of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 mm is as shown in the figure. (i) If the incident beam falls normally on the double slit apparatus, find the y co-ordinates of all the interference minima on the screen.
1 d sin θ = m + λ; m = 0, 1, 2, … 2 Only those values of m are allowed for which sin θ does not exceed ± 1. For the given data, we get
s
si ta
un iv er
w
w
tp
s:
//w
or
(i) The interference minima will occur on the screen provided
λ 6 1 31 t= × λ 0.3 6 1 31 = × × 5400Å 0.3 6 = 9.3 × 104 Å = 9.3 × 10–6 m
0.3t = 5λ +
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an a
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om
or
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4B.64
(ii) Find the deviation of the beams transmitted through the face AC. (iii) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and the lower beams, immediately after transmission from the face AC, are 4I and I respectively, find the resultant intensity at the focus.
$
an a.
s_
ta
si
rio s
ta
.m e/
4
3
tp s:
H
G
c
G c
4
&
%
= 0.8 × 10–14 m2
∴ b = (1.25 – 1.20) (4 × 10–7)2 = 0.05 × (4 × 10–7)2
0.8 × 10−14 25 × 10−14 = 1.20 + .032 = 1.232 Let, e = angle of emergence of the 5000 Å ray of light
(ii) µ5000 = 1.20 +
where, λ is in Å and b is a positive constant. The value of b is such that the condition for total reflection at the face AC is just satisfied for one wavelength and is not satisfied for the other. (i) Find the value of b.
$
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13. Two parallel beams of light P and Q (separation d) containing radiation of wavelengths 4000 Å and 5000 Å (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in the figure. The refractive index of the prism as a function of wavelength is given by the relation. b µ(λ) = 1.20 + 2 λ
1 1 = sin θ 0.8 b = 1.20 + (4 × 10−7 ) 2
(i) µ 4000 =
//t
ht tp
)
m
//t .
s:
First minimum below the central maximum occurs when –1 1 m = –2, sin θ = , y = – m 4 15
and smaller the critical angle. For just total internal reflection, θ = critical angle. Critical angle for 4000 Å = θ
er
e/ un
tp s:
1 1 m = –1, sin θ = , y = m 4 15
ht
&
) Solution Greater the µ, smaller the wavelength
iv
//w
w
w
1 1 This gives sin θ = , y = m 2 3 First minimum above the central maximum occurs when
%
rio
ar
For the central maximum, we have d sin θ – d sin α = mλ, where, m = 0
.ja m
1 3 = m+ 2 2
6LQT
G
fo
1 11 +m+ 2 22
m
=
co
1 0.5mm = sin 30° + m + 2 1.0 mm
in
1λ sin θ = sin α + m + 2d
si
1 d sin θ – d sin α = m + λ, 2 m = 0, 1, 2, … From given data, we get
er
(ii) For angle of incidence, α = 30º, the condition for minima is
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= EF sin θ (µ5000) in air
or sin e = 0.8 × 1.232 = 0.9856 ∴ e = 80.27º Alow, θ = sin–1 (0.8) = 53.13º Then δ = deviation for 5000 Å = 80.27º – 53.13º = 27.14º
sin e = EF sin θ in air sin θ = EF sin e (in air)
Hence, when the converging lens foci P and Q at its principal focus, the beams have equal paths, i.e., phase difference = zero, but different intensities IP and IQ respectively. ∴ Intensity at the focus
$ T *
E
IO = I P + IQ + 2 I P IQ cos θ
ar a m
N
&
ita
er si ta rio s
■ The formula I0 = I1 + I2 + 2 I1I 2 cos θ is applicable only if the interfering beams are of the same single frequency.
type questions )
) Explanation Due to introduction of glass
plate, additional path difference = (µ – 1) t. For the intensity to be same as before, (µ – 1) t = nλ For minimum value of t, n = 1 ⇒ t = 2λ
2. Two coherent monochromatic light beams of intensities I and 4I are superimposed. The maximum and minimum possible intensities in the resulting beam are
ECTIVE
O
BJ
1. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is 2λ (a) 2λ (b) 3 λ (c) (d) λ 3
SOLVED
ht tp
s: //
t.m e/
un
//t
ht tp s:
ote: ■ If the data means P is 4000 Å and Q is 5000 Å and the lens refracts the two beams: (i) They will not come to the same focus. (ii) The interference affect is not constant at any position.
er s
.m
ht
un iv
e/
s:
tp
//w
w
w
.ja
(iii) The two beams of 5000 Å (P and Q) emerging from the face AC, have same path FG in air, till they reach FG (a) The beam P passes KE in prism + EG in air up to G. (b) The beam Q passes (LG + GH) in prism to reach F. It can be shown that the two paths of P and Q up to FG are equal as shown below: KE (in glass) = DH (in prism) EG (in air) = EF cos α
=9I
%
= 4I + I + 2 4I × 1
)
iv
+
4
rio s_ in fo
G
na .c om
D
(
3
4B.65
= EF sin e (in air) HF (in glass) = EF sin θ (in prism)
sin e = 1.232 sin θ
∴
Light (Wave Optics)
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4B.66
5I and I 5I and 3I 9I and I 9I and 3I
(b) 1/2 (d) 16
(a) 1 (c) 4
) Explanation Two waves of a single source )
(a) (b) (c) (d)
Light (Wave Optics)
having an amplitude A interfere. The resulting amplitude Ar is given by A 2r = A12 + A 22 + 2A1A cos δ
)
) Explanation Intensity ∝ (Amplitude)2 i.e., I ∝ A2 When two waves (beams) of amplitude A1 and A2 superimpose, at maxima and minima, the amplitude of the resulting wave are (A1 + A2) and (A1 – A2) respectively. If the maximum and minimum possible intensities are Imax and Imin respectively, then
where A1 = A2 = A and δ = phase difference between the waves
When the maxima occurs at the centre, δ = 0 ∴ Ir = 4I …(i) Since the phase difference between two successive fringes is 2π, the phase difference between two points separated by a distance equal to one quarter of the distance between the two successive fringes,
in f
a. an
s_
si ta
s
si
4I =2 2I
er
I r1
I r2
iv
un
=
4. The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the distance between two fringes from the centre is
)
//t
s:
tp
ht
600 n2 = 12 = 18 400
.m
e/
5. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference π between the beams is at point A and π 2 at point B. Then the difference between the resultant intensities at A and B is (a) 2 I (b) 4 I (c) 5 I (d) 7 I π ) Explanation IA = I + 4I + 2 1.41 cos = 5I 2
s: //
tp
ht
…(ii)
ta
un
t.m e/
s:
tp
ht
)
rio
er iv
//w w
3. In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is (a) 12 (b) 18 (c) 24 (d) 30 Dλ and ) Explanation Fringe width, w = d n1w1 = n2w2 ∴
π/ 2 I r2 = 4 I cos 2 = 2I 2 From equations (i) and (ii), we get ⇒
Imax = 9I, and Imin = I
⇒
1 π δ = ( 2π ) = radians 4 2
rio
m ar
I max I min
w .ja
Now,
2
A1 + 1 A1 + A 2 A 2 9 = = =A A1 – A 2 1 − 1 1 A 2 2
A1 I 1 = = A2 4I 2
o
co
m
2
Imax ∝ (A1 + A2) and Imin ∝ (A1 – A2) 2
I r = I1 + I 2 + 2 I1I 2 cos δ
∴
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IB = I + 4I + 2 1.4I cos π = I IA – IB = 4I
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+
er si ta rio s
iv
un
7. In Young’s double slit experiment, the separation between two coherent sources S1 and S2 is d and the distance between the source and screen is D. In the interference pattern, it is found that exactly in front of one slit, there occurs a minimum. Then the possible wavelengths used in experiment are
t.m e/
ita
6. In Young’s double slit experiment, we get 60 fringes in the field of view of monochromatic light of wavelength 4000 Å. If we use monochromatic light of wavelength 600 Å, then the number of fringes obtained in the same field of view is (a) 60 (b) 90 (c) 40 (d) 1.5
un iv
e/
.m
//t
ht tp s:
d2 , d2 , d2 ... D 3D 5D
(b) λ =
d2 , d2 , d2 ... D 5D 9D
(c) λ =
d2 , d2 , d2 ... D 2D 3D
(d) λ =
d2 , d2 , d2 ... 3D 7D 11D
s: //
ht tp
(a) λ =
8. A parallel beam of light is incident on a thin film of air of uniform thickness.
5. A monochromatic light source of wavelength λ is placed at S. Three slits S1, S2 and S3 are equidistant from the source S and the point P on the screen. λ 2λ . S1 P – S 2 P = and S1 P – S 3 P = 6 6
If I be the intensity at P when only one slit is open, the intensity at P when all the three slits are open is (a) 3 I (b) 5 I (c) 8 I (d) zero
er s
tp
ht
) &
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ar a
.ja
w
w
s:
//w
E
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4. In a double slit experiment, the distance between the slits is d. The screen is at a distance D from the slits. If a bright fringe is formed opposite to a slit, its order is (λ is wavelength of light used) d D2 (a) (b) 2Dλ 2dλ 2 d2 d (c) (d) 2Dλ 2Dλ
QUESTIONS
4
m
3. If white light is used in a Young’s doubleslit experiment, (a) bright white fringe is formed at the centre of the screen (b) fringes of different colours are observed clearly only in the first order. (c) the first-order violet fringes are closer to the centre of the screen than the firstorder red fringes (d) all of these
*
D
G
x (b) I0 cos2 β I 2 πx (d) 0 cos 4 β
(
3
x (a) I0 cos β πx (c) I0 cos2 β
T
2. In a Young’s double-slit experiment, let β be the fringe width, and let I0 be the intensity at the centre bright fringe. At a distance x from the central bright fringe, the intensity will be
4B.67
$
TYPE
1. In Young’s double slit experiment, we get 60 fringes in the field of view of monochromatic light of wavelength 4000 Aº. If we use monochromatic light of wavelength 6000 Aº, then the number of fringes obtained in the same field of view is (a) 60 (b) 90 (c) 40 (d) 1.5
O
unsolved BJ
ECTIVE
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4B.68
fo
in
s_
rio
s
rs ita
ve
15. Microwaves of frequency 3 × 104 MHz and ultrasonic waves of wavelength 1cm are passed through a slit of width 2 cm. Diffraction will occur (a) only in the microwaves. (b) only in the ultrasonic waves. (c) in both, but the diffraction patterns will be different. (d) in both and the diffraction patterns will be identical.
( µ 2 − 1) t
16. Let S1 and S2 be the two slits in Young’s double slit experiment. If central maxima is observed at P and angle ∠S1PS2 = θ, then the fringe width for the light of wavelength λ will be
11. A double slit of separation 0.1 cm is illuminated by white light. A coloured interference pattern is formed on a screen 100 cm away. If a pin hole is located in this screen at a distance of 2 mm from the central fringe, the wavelengths in the visible spectrum which will be absent in the light transmitted through the pinhole are
e/ //t s: tp
ht
(d)
( µ 2 − µ1 ) t
(c)
.m
ht tp µ (b) 1 − 1 t µ 2
µ2 (a) − 1 t µ1
ta rio
si
un i
s:
2
W
14. In the Young’s experiment with sodium light, the slits are 0.589 m apart. What is the angular width of the forth maximum? Given that λ = 589 nm (a) sin–1(3 × 10–6) (b) sin–1(3 × 10–8) (c) sin–1(0.33 × 10–6) (d) sin–1 (0.33 × 10–8)
un
e/ .m //t
ht
w
w
s: //
tp
6
6
13. In a two slit experiment with white light, a white fringe is observed on a screen kept behind the slits. When the screen is moved away by 0.05m, this white fringe (a) does not move at all (b) gets displaced from its earlier position (c) becomes coloured (d) disappears
iv er
w
.ja
m
10. A Young’s double slit experiment is conducted in water (µ1 as shown in the figure, and a glass plate of thickness t and refractive index µ2 is placed in the path of S1. Find the magnitude of the optical path difference at ‘O’. :DWHU
12. A plane monochromatic light wave falls normally on a diaphragm with two narrow slits separated by 2.5 mm. The fringe pattern is formed on a screen 100 cm behind the diaphragm. By what distance will these fringes be displaced, when one of the slits is covered by a glass plate (µ = 1.5) of thickness 10 µm? (a) 2 mm (b) 1 mm (c) 3 mm (d) 4 mm
co m
ar an
3 λ (d) 1 4 λ2
a.
4 λ1 3 λ2
3 λ (c) 2 4 λ1
(b)
4 λ2 3 λ1
5714 Å and 4444 Å 6000 Å and 5000 Å 5500 Å and 4500 Å 5200 Å and 4200 Å
9. In Young’s double slit experiment, the 8th maximum with wavelength µ1 is at a distance d1 from the central maximum and the 6th minimum with a wavelength λ2 is at a distance d2. Then d1/d2 is equal to (a)
(a) (b) (c) (d)
Wavelengths of 7200 Å and 5400 Å are observed to be missing from the spectrum of reflected light viewed normally. The other wavelengths in the visible region missing in the reflected spectrum is (a) 6000 Å (b) 4320 Å (c) 5500 Å (d) 6500 Å
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8. (b)
14. (a)
15. (d)
16. (a)
un
7. (a)
t.m e/ 3. In a Young’s double slits experiment two narrow slits 0.8 mm apart are illuminated by the same source of light (λ = 5893 Å). How far apart are adjacent bright bands in the interference pattern observed on a screen 2 m away? [Roorkee, 1982] [Ans. 0.147 cm]
E
4. In Young’s double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron is introduced in the path of one of the interfering waves. The mica sheet is then removed
2. Two slits are separated by a distance of 0.03 cm. An interference pattern is produced on a screen 1.5 m away. The fourth bright fringe is at a distance of 1 cm from the central maximum. Determine the wavelength of light used. [Roorkee, 1980] [Ans. 5000 Å]
ht tp
1. In Young’s experiments for interference of light, the slits are 0.2 cm apart and illuminated by yellow light (λ = 5896 Å). What will be the fringe width on a screen placed 1 m from the slit? [Roorkee, 1977] [Ans. 0.295 mm]
6. (c)
1
ERCISE
s: //
practice X
(d) λ
er si ta rio s
13. (a)
2λ 3
ita er s
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(b)
ar a
m
5. (a)
.m
11. (a) 19. (a)
4. (c)
//t
10. (c) 18. (a)
λ 3
(c)
.ja
w w //w s:
3. (d)
ht tp s:
9. (b) 17. (b)
2. (c)
ht
1. (c)
tp
ANSWERS
(a) 2λ
18. In a double-slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in the interference pattern the intensity of
rio s_ in fo
na .c om
X (m1 + m2 )
(d)
X m1 + m2
19. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is
iv
λ 2θ
(d)
17. In the displacement method, a convex lens is placed in between an object and a screen. If the magnification in the two positions be m1 and m2 and the displacement of the lens between the two positions is X, then the focal length of the lens is X X (a) (b) m1 × m2 m1 − m2 (c)
4B.69
(a) both the maxima and the minima increase (b) the maxima increases and the minima has zero intensity (c) the maxima decreases and that of the minima increases (d) the maxima decreases and the minima has zero intensity.
(b) λθ
λ θ 2λ (c) θ (a)
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4B.70
o
in f
s
rio
ta
si
er
un iv
.m
ht tp
s: //t
.m
e/
//t
s:
tp
ht
ta r
si
er
iv
un
e/
s:
tp
ht
10. In a modified Young’s double slit experiment, a monochromatic, uniform and parallel beam of light of wavelength 6000 10 Å and intensity W/m2 is incident π normally on two circular apertures A and B of radii 0.001 m and 0.002 m, respectively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength of 6000 Å is placed in front of aperture A Calculate the power of the lens (in watt) received at the focal spot F. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot. [IIT, 1998] [Ans. 7 × 10–6 W]
io s_
an
ar
m
.ja
//w
w
w
m
a. co
7. A two-slit Young’s experiment is done with monochromatic light of wavelength 6000 Å. The slits are 2 mm apart and the fringes are observed on the screen placed 10 cm away from the slits. Now a transparent plate of thickness 0.5 mm is placed in front of one of the slits and it is found that the interference pattern shifts by 5 mm. What is the refractive index of the transparent plate? [Roorkee, 1985] [Ans. 1.2]
9. A Young’s double-slit arrangement produces interference fringes for sodium light (λ = 5890 Å) that are 0.20º apart. What is the angular fringe separation if the entire fringe separation is immersed in water? (Refractive index of water = 4/3) [Roorkee, 1988] [Ans. 0.15º]
8. Consider interference between two source intensities I and 4 I. Obtain intensities at points where the phase difference is
[Roorkee, 1987]
[Ans. (a) 5 I, (b) I]
5. In a Young’s double slit experiment, the angular width of a fringe formed on a distant screen is 0.1º. The wavelength of the light used is 6000 Å. What is the spacing between the slits? [Roorkee, 1983] [Ans. 3.44 × 10–4 m]
6. A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of slits and the screen is 120 cm. [IIT, 1985] [Ans. (a) 1.17 mm, (b) 1.56 mm]
(b) π
(a) π 2
and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or, minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of monochromatic light used in the experiment. [IIT, 1983] [Ans. 5892 Å]
11. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in the figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25 percent of the light incident on it and transmits the remaining. Find the ratio of the minimum and maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate. [IIT, 1990]
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4B.71
2
1
12. In a two-slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10–2 m towards the slits, the change in fringe width is 3 × 10–5 m. If the distance between the slits is 10–3 m, calculate the wavelength of light used. [Roorkee, 1992] [Ans. 6000 Å]
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16. A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength λ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. Write the condition for their constructive interference. If λ = 648 nm, obtain the least value of t for which the rays interfere constructively. [IIT, 2000] [Ans. 4µt = (2n – 1) λ, 90 nm]
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15. Unpolarized light of intensity 32 Wm– 2 passes through three polarisers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is 3 Wm–2, what is the angle between the transmission axes of the first two polarizers? At what angle will the transmitted intensity be maximum? [Roorkee, 1995] [Ans. 30º, 45º]
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13. In the given figure, S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 mm. The distance along the axis from S to L1 and L2 is 0.15 m, while that from L1 and L2 to O is 1.30 cm. The screen at O is normal to SO.
1.30 m
(a) If the third intensity maximum occurs at the point A on screen find the distance OA.
17. A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass pates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and
L2 0.15 m
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14. A monochromatic light of λ = 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. The interference pattern is seen on a screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10–6 m and refractive index µ = 1.5 is placed between one of the slits and the screen. Find the intensity at the centre of the screen if the intensity is I0 in the absence of the plate. Also find the lateral shift of the central maximum. [Roorkee, 1993] [Ans. 0, 1.5 mm]
(b) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease or remain the same? [IIT, 1993] [Ans. (a) 1.0 mm, (b) Increases]
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4B.72
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18. A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB (see figure). The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. (a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure) (c) If the intensity at point P corresponds to a maximum calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum. [IIT, 2002] [Ans. (a) circular, (b) 16, (c) 300 nm]
passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure below. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled up to OQ. The central bight fringe is found to be at Q. Calculate the refractive index of the liquid. [IIT, 2001]
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(a) the speed of light in vacuum, (b) the speed of light in a medium (say, glass or water), depend?
7. In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? 8. When a tiny circular obstacle is placed in the path of light from a distance source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
On which of these factors, if any, does
4. Let us list some of the factors, which could possibly influence the speed of wave propagation (i) nature of the source. (ii) direction of propagation. (iii) motion of the source and/or observer. (iv) wavelength. (v) intensity of the wave.
9. Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.
10. Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?
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6. In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
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3. Huygens’ principle leads to the laws of reflection and refraction, use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
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2. Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
5. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?
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1. What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations? (a) The screen is moved away from the plane of the slits (b) The (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength (c) The separation between the two slits is increased (d) The source slit is moved closer to the double-slit plane (e) The width of the source slit is increased (f) The monochromatic source is replaced by a source of white light.
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11. When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
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8. In a YDSE experiment if a slab whose refraction index can be varied is placed in front of one of the slits then the variation of resultant intensity at mid-point of screen with ‘µ’ will be best represented by (µ ≥ 1). [Assume slits of equal width and there is no absorption by slab; mid-point of screen is the point where waves interfere with zero phase difference in absence of slab]
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4. Two coherent light sources each of wavelength λ are separated by a distance 3λ. The maximum number of minima formed on line AB which runs from –∞ to +∞ is 3
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3. In Young’s slit experiment 10th order maximum is obtained at the point of observation in the interference pattern for λ = 7000 Å. If the source is replaced by another one of wavelength 5000 Å then the order of maximum at the same point will be (a) 12 th (b) 14 th (c) 16 th (d) 18 th
7. A beam of light parallel to central line AB is incident on the plane of slits in a YDSE experiment as shown. The number of minima obtained on the large screen is n1. Now if the beam is titled by some angle (≠ 90º) as shown in the figure, then the number of minima obtained is n2. Then
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6. A parallel beam of monochromatic light is incident on a narrow rectangular slit of width 1mm. When the diffraction pattern is seen on a screen placed at a distance of 2 m the width of principal maxima is found to be 2.5 mm. The wave length of light is (a) 6250 Å (b) 6200 Å (c) 5890 Å (d) 6000 Å
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5. When wave of wavelength 0.2 cm is made incident normally on a slit of width 0.004 m, then the semi-angular width of central maximum of diffraction pattern will be (a) 60º (b) 30º (c) 90º (d) 0º
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1. The two coherent sources of intensity that ratio 2 : 8 produce an interference pattern. The values of maximum and minimum intensities will be respectively. (a) Il and 9 Il (b) 9Il and Il (c) 2Il and 8Il (d) 8Il and 2Il 2. In young’s double slit experiment, the distance between two slits is made three times then the fringe width will become (a) 9 times (b) 1/9 times (c) 3 times (d) 1/3 times
12. The principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?
(b) 4 (d) 8
(a) 2 (c) 6
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14. In the figure shown, if a parallel beam of white light is incident on the plane of the slits then the distance of the nearest white spot on the screen from O is (assumed d << D, λ << d]
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9. A thin oil film of refractive index 1.2 floats on the surface of water (µ = 4/3). When a light of wavelength λ = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is (a) 10–7 m (b) 2 × 10–7 m (c) 3 × 10–7 m (d) 5 × 10–7 m
13. In Young’s double slit experiment the two slits are illuminated by light of wavelength 5890Å and the distance between the fringes obtained on the screen is 0.2º. If the whole apparatus is immersed in water then the angular fringe width will be, it the refractive index of water is 4/3. (a) 0.30º (b) 0.15º (c) 15º (d) 30º
10. Light of wavelength 6328 Å is incident normally on slit having a width of 0.2 mm. The width of the central maximum measured from minimum to minimum of diffraction pattern on a screen 9.0 metres away will be about
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(b) d/2 (d) d/6
(a) 0 (c) d/3
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12. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is (a) 2I (b) 4I (c) 5I (d) 7I
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11. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wave-length λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is (a) 2λ (b) 2λ/3 (c) λ/3 (d) λ
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(b) 0.18º (d) 0.09º
(a) 0.36º (c) 0.72º
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4B.76
15. The intensities of two sources are I and 9I respectively. If the phase difference between the waves emitted by them is π then the resultant intensity at the point of observation will be (a) 3I (b) 4I (c) 10I (d) 82I
19. When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The
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23. The equation of two light waves are y1 = 6 cos ωt, y2 = 8 cos (ωt + φ). The ratio of maximum to minimum intensities produced by the superposition of these waves will be (a) 49 : 1 (b) 1 : 49 (c) 1 : 7 (d) 7 : 1
18. When a mica sheet of thickness 7 microns and µ = 1.6 is placed in the path of one of interfering beams in the biprism experiment then the central fringe gets at the position of seventh bright fringe. The wavelength of light used will be (a) 4000 Å (b) 5000 Å (c) 6000 Å (d) 7000 Å
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22. The intensity ratio of two waves is 9 : 1. These waves produce the event of interference. The ratio of maximum to minimum intensity will be (a) 1 : 9 (b) 9 : 1 (c) 1 : 4 (d) 4 : 1
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17. In Fresnel’s biprism experiment, the width of 10 fringes is 2 cm which are formed at a distance of two 2 metre from the slit. If the wavelength of light is 5100 Å then the distance between two coherent sources will be (a) 5.1 × 10–4 m (b) 5.1 × 104 cm (c) 5.1 × 10–4 mm (d) 10.1 × 10–4 cm
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21. Consider interference between waves from two sources of intensities I and 4I. Find intensities at points where the phase difference is π. (a) I (b) 5I (c) 4I (d) 3I
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24. Interference fringes were produced using white light in a double slit arrangement. When a mica sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some a distance. This distance is equal to the width of 30 interference bands if light of wavelength 4800Å is used. The thickness (in 1 µm) of mica is
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20. From a medium of index of refraction n1, monochromatic light of wavelength λ is incident normally on a thin film of uniform thickness L (where L > 0.1λ) and index of refraction n2. The light transmitted by the film travels into a medium with refractive index n3. The value of minimum film thickness when maximum light is transmitted if (n1 < n2 < n3) is
16. A ray of light is incident on a thin film. As shown in figure M, N are two reflected rays and P, Q are two transmitted rays. Rays N and Q undergo a phase change of π. Correct ordering of the refracting indices is Q
thickness of the film will be, if the wavelength of light is 5890Å. (a) 6.544 × 10–4 cm (b) 6.544 × 10–4 m (c) 6.54 × 10–4 cm (d) 6.5 × 10–4 cm
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25. In a Young’s experiment, the separation between the slits is 0.10 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1.0 m away. Find the separation between the successive bright fringes. (a) 6.6 mm (b) 6.0 mm (c) 6 m (d) 6 cm
29. Figure shows two coherent sources S1 – S2 vibrating in same phase. AB is an irregular wire lying at a far distance from the sources λ S1 and S2. Let = 10−3 ∠BOA = 0.12º. d How many bright spots will be seen on the wire, including points A and B.
(b) 12 (d) 24
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32. In the figure shown in a YDSE, a parallel beam of light is incident on the slits from a medium of refractive index n1. The wavelength of light in this medium is λ1. A transparent slab of thickness t and refractive index is put infront of one slit. The medium between the screen and the plane of the slits is n2. The phase difference between the light waves reaching point O (symmetrical, relative to the slit) is
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28. In Young’s experiment, the wavelength of red light is 7.5 × 10–5 cm and that of blue light 5.0 × 10–5 cm. The value of n for which (n + 1)th the blue bright band coincides with nth red band is (a) 8 (b) 4 (c) 2 (d) 1
31. Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5 × 10–7m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between third dark fringe and the fifth bright fringe is equal to (a) .65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.87 mm
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27. Two waves originating from source S1 and S2 having zero phase difference and common wavelength λ will show completely destructive interference at a point P if (S1 P – S2 P) is (a) 5λ (b) 3λ/4 (c) 2λ (d) 11λ/2
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30. In Young’s slit experiment, carried out with lights of wavelength λ = 5000 Å, the distance between the slit is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum will be at x equal to (a) 1.67 cm (b) 1.5 cm (c) 0.5 cm (d) 5.0 cm
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26. A broad source of light (I = 680 nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.034 mm in diameter at the other end. The total number of bright fringes that appear over the 120 mm distance is
6
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4B.78
33. In an experiment, the two slits are 0.5 mm apart and the fringes are observed to 100 cm from the plane of the slits. The distance of the 11th bright fringe from the 1st bright fringe is 9.72 mm. Calculate the wavelength. (a) 4.85 × 10–5 cm (b) 4.85 × 10–5 m –7 (c) 4.86 × 10 cm (d) 4.86 × 10–5 cm
38. Interference fringes were produced in young’s double slit experiment using light of wave length 5000Å. When a film of material 2.5 × 10–3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe width. The refractive index of the material of the film is (a) 1.25 (b) 1.33 (c) 1.4 (d) 1.5
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42. Find the minimum thickness of a film which will strongly reflect the light of wavelength 589 nm. The refractive index of the material of the film is 1.25 (a) 118 nm (b) 120 nm (c) 118 m (d) 188 mm
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41. If the ratio of the intensity of two coherent I −I sources is 4 then the visibility m ax min I m ax + I min of the fringes is (a) 4 (b) 4/5 (c) 3/5 (d) 9
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40. The path difference between two interfering waves at a point on screen is 171.5 times the wavelength. If the path difference is 0.01029 cm. Find the wavelength. (a) 6000 × 10–10 cm (b) 6000 Å (c) 6000 × 10–8 mm (d) None
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37. A beam of light consisting of two wavelength 6500Å and 5200Å is used to obtain
39. The path difference between two interfering waves at a point on the screen is λ/6. The ratio of intensity at this point and that at the central bright fringe will be (Assume that intensity due to each slit in same) (a) 0.853 (b) 8.53 (c) 0.75 (d) 7.5
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36. The wavelength of a light beam is given by the equation x + 2y + 3z = c, (where c is arbitrary constant) then the angle made by the direction of light with the y-axis is 1 2 (b) sin −1 (a) cos −1 14 14 −1 2 −1 3 (c) cos (d) sin 14 14
35. Two light waves are given by E1 = 2 sin (100πt – kx + 30º) and E2 = 3 cos (200πt – k′x + 60º). The ratio of intensity of first wave to that of second wave is (a) 2/3 (b) 4/9 (c) 1/9 (d) 1/3
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43. In Young’s double slit interference experiment, the distance between two sources is 0.1/π mm. The distance of the screen from the source is 25 cm. Wavelength of light used is 5000Å. Then the angular position of the first dark fringe is (a) 0.10º (b) 0.15º (c) 0.30º (d) 0.45º
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52. Select the right option of the following. (a) Christian Hygens a contemporary of Newton established the wave theory of light by assuming that light waves were transverse (b) Maxwell provided the compelling theoretical evidence that light is transverse wave (c) Thomas Young experimentally proved the wave behaviour of light and Huygens assumption (d) All the statements given above, answers the question “what is light”
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53. Two slits separated by a distance of 1 mm, are illuminated with red light of wavelength 6.5 × 10–7m. The interference fringes are observed on a screen placed 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (a) 0.65 mm (b) 1.63 mm (c) 3.25 mm (d) 4.88 mm
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48. The initial phase difference between the light waves from the two coherent sources S1 and S2 is π. Then the central fringe will be (a) a bright fringe (b) a dark fringe
51. Huygen wave theory allows us to know (a) The wavelength of the wave (b) The velocity of the wave (c) The amplitude of the wave (d) The propagation of wave fronts
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47. Mark the correct statement(s). (a) An interference pattern can be formed on a screen by two independent phase locked laser sources (b) An interference pattern can not be formed by two independent monochromatic sources of same wavelength (c) Interference pattern using white light contains few coloured fringes also (d) All of the above
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46. In Young’s double slit experiment (a) only diffraction takes place (b) only interference takes place (c) both diffraction and interference takes place (d) neither interference nor diffraction takes place
50. Which one of the following phenomena is not explained by Huygen’s construction of wavefront? (a) Refraction (b) Reflection (c) Diffraction (d) Origin of spectra
49. Wavefront means (a) All particles in it have same phase (b) All particles have opposite phase of vibrations (c) Few particles are in same phase, rest are in opposite phase (d) None of these
44. In Young’s double slit experiment, if the width of one of the slits is increased, then (a) contrast of the fringes decreases (b) contrast of the fringes increases (c) contrast of the fringes remains the same (d) interference pattern is destroyed immediately 45. One important similarity between light waves and sound waves is that both (a) can travel through vacuum (b) are transverse wave (c) have same speed in air (d) can interfere
4B.79
(c) bright or dark depending upon the wavelength of light (d) neither bright, nor dark
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60. Two Nicol prisms are first crossed and then one of them is rotated through 60º. The percentage of incident light transmitted is
4. If one of the slits of a standard Young’s double slit experiment is covered by a thin parallel sided glass slab so that it transmits only one half the light intensity of the other, then (a) The fringe pattern will get shifted towards the covered slit (b) The fringe pattern will get shifted away from the covered slit
59. Two points separated by a distance of 0.1 mm can just be inspected in a microscope when light of wavelength 6000 Å is used. If the light of wavelength 4800 Å is used, this limit of resolution will become (a) 0.80 mm (b) 0.12 mm (c) 0.10 mm (d) 0.08 mm
3. The intensity of a plane progressive wave in loss free medium is (a) directly proportional to the square of amplitude of the wave (b) directly proportional to the velocity of the wave (c) directly proportional to the square of frequency of the wave (d) inversely proportional to the density of the medium
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58. The first diffraction minima due to a single slit diffraction is at θ = 30º for a light of wavelength 5000 Å. The width of the slit is (a) 5 × 10–5 cm (b) 10 × 10–5 cm –5 (c) 2.5 × 10 cm (d) 1.25 × 105 cm
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2. An interference pattern is formed on the screen, when light from two different monochromatic sources are allowed to interfere. Then, it is true that (a) frequencies of light from the two sources are equal to each other (b) the sources are coherent (c) the sources should be located in the same medium (d) the path difference should either be an even or, an odd multiple of λ/2, where λ is the wavelength of light
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57. In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength λ1 is found to be coincident with the third maximum at λ2. So (a) 3 λ1 = 0.3λ2 (b) 3λ1 = λ2 (c) λ1 = 3.5λ2 (d) 0.3λ1 = 3λ2
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55. A double slit is illuminated by two wavelength 450 nm and 600 nm. What is the lowest order at which the maxima of one wavelength coincides with the minima of the other wavelength? (a) 1 (b) 2 (c) 3 (d) 4 56. x1 and x2 are the separations between the two images of slit obtained on a fixed screen, in a biprism set-up, using a convex lens, the separation in the two interfering virtual sources is (a) x1/x2 (b) x2/x1
(b) 25.0 (d) 50
1. If the first minima in a Young’s slit experiment occurs directly in front of one of the slits, (distance between slit and screen D = 12 cm and distance between slits d = 5 cm) then the wavelength of the radiation used can be (a) 2 cm (b) 4 cm (c) 2/3 cm (d) 4/3 cm
(a) 1.25 (c) 37.5
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54. The central fringe of the interference pattern produced by light of wavelength 6000 Å is found to shift to the position of 4th bright fringe after a glass plate of refractive index 1.5 is introduced. The thickness of the glass plate would be (a) 4.8 µm (b) 8.23 µm (c) 14.98 µm (d) 3.78 µm
Light (Wave Optics)
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12. A person views the interference pattern, produced by two slits illuminated by white light, on placing a green filter in front of his eyes. Then (a) He will see sharply distinguishable dark and bright fringes (b) The fringes will not be sharp but are differentiable (c) On replacing the green filter by a blue filter, the fringes will be again sharp but closer than those by the green filter (d) On using both the filters simultaneously, the central bands will be maximum bright
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7. A parallel beam of light (λ = 5000 Å) is incident at an angle α = 30º with the normal to the slit plane in a Young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is I0. Point O is equidistant from S1 and S2. The distance between slits is 1mm.
(a) the intensity at O is 4I0 (b) the intensity at O is zero (c) the intensity at a point on the screen 1 m below O is I0 (d) the intensity at a point on the screen 1 m below O is zero
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8. White light is used to illuminate the two slits in Young’s double slit experiment. The separation between the slits is b and the screen is at a distance d (>> b) from the slits. At a point on the screen directly in front of any one of slits, certain wavelengths are missing. Some of these missing wavelengths are (a) λ = b2/d (b) λ = 2b2/d (c) λ = b2/3d (d) λ = 2b2/3d
6. Huygen’s principle of secondary wavelets may be used to (a) find the velocity of light in vacuum (b) explain the particle behavior of light (c) find the new position of a wavefront (d) explain Snell’s law.
4B.81
5. If instead of monochromatic light, white light is used in Young’s double slit experiment, then (a) a bright white fringe is formed at centre of screen (b) fringes of different colours are clearly observed only in the first order place (c) the first order red fringes are relatively close to the screen than the first order violet fringes. (d) the first order violet fringes are relatively close to the centre of the screen than the first order red fringes
(c) The bright fringes will become less bright and the dark ones will become more bright (d) The fringe width will remain unchanged
Light (Wave Optics)
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4B.82
14. Two sources are called coherent if they produce waves (a) of equal wavelength (b) of equal velocity (c) having same shape of wavefront (d) having a constant phase difference
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4. Statement 1: When two coherent waves of intensity I1 and I2 are superimposed with a constant phase difference φ, then the intensity of resultant wave is given by I = I1 + I2 + 2 I1 I 2 cos φ Statement 2: For two incoherent sources, resultant intensity is given by I = I1 + I2.
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5. Statement 1: In YDSE central maxima means the maxima formed with zero optical path difference. It may be formed anywhere on the screen. Statement 2: In an interference pattern, whatever energy disappears at the minimum, appears at the maximum. 6. Statement 1: Red light travels faster in glass than green light. Statement 2: The refractive index of glass is less for red light than for green light.
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3. Statement 1: All points on a wavefront vibrate in same phase with same frequency. Statement 2: Two sources are said to be coherent if they produce waves of same frequency with a constant phase difference.
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(a) Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.
2. Statement 1: In YDSE, number of bright fringe or dark fringe can not be unlimited. Statement 2: In YDSE, path difference between the superposing waves can not be more than the distance between the slits.
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16. If one of the slits of a standard YDSE apparatus is covered by a thin parallel sided glass slab so that it transmit only one half of the light intensity of the other, then (a) the fringe pattern will get shifted towards the covered slit (b) the fringe pattern will get shifted away from the covered slit (c) the fringe pattern will be less bright and the dark ones will be more bright (d) the fringe width will remain unchanged.
ASSERTION
1. Statement 1: In Young’s double slit experiment if wavelength of incident monochromatic light is just doubled, number of bright fringe on the screen will increase. Statement 2: Maximum number of bright lunge on the screen is inversely proportional to the wavelength of light used.
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15. To observe a stationary interference pattern formed by two light waves, it is not necessary that they must have (a) the same frequency (b) same amplitude (c) a constant phase difference (d) the same intensity
Direction: Each question contains STATEMENT 1 (Assertion) and STATEMENT 2 (Reason). Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
(b) Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1. (c) Statement 1 is True, Statement 2 is False. (d) Statement 1 is False, Statement 2 is True.
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13. In a YDSE apparatus, we use white light. Then (a) the fringe next to the central will be red (b) the central fringe will be white (c) the fringe next to the central will be violet (d) there will not be a completely dark fringe
Light (Wave Optics)
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and Imin = ( I1 − I 2 ) 2 15. Statement 1: In YDSE, if a thin film is introduced in front of the upper slit, then the fringe pattern shifts in the downward direction. Statement 2: In YDSE if the slit widths are unequal, the minima will be completely dark.
QUESTIONS
TYPE
COLUMN
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λ , d << D, ℓ0 = 4 W/m2, 2 Amplitude of each wave = 2 units, λ = 6000Å) (Given: BP0 – AP0 =
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1. In the figure shown A, B and C are the three slits each of which individually produces the same intensity ℓ0, at point P0 when they are illuminated by parallel beam of light of wavelength λ.
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12. Statement 1: The fringe visibility will be maximum when amplitude of light waves from two coherent sources is exactly same. Statement 2: Fringe visibility I −I = max min I max + I min 13. Statement 1: In YDSE, the fringe width (β) depends on the medium in which the experiment is carried out. λD Statement 2: Because β = and λ d depends on the medium.
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Each question contains statements given in two columns which have to be matched. Statements (a, b, c, d) in column I have to be matched with statements (p, q, r, s) in column II.
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11. Statement 1: Diffraction takes place for all types of waves mechanical or non-mechanical, transverse or longitudinal. Statement 2: Diffraction’s effect are perceptible only if wavelength of wave is comparable to dimensions of diffracting device.
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9. Statement 1: YDSE, the intensity at the maxima is observed to be I0. If one of the slits is closed, then the intensity at the location of the maxima reduces to I0/4. Statement 2: In YDSE, fringes with blue light are thicker than those for red light. 10. Statement 1: If white light is used in YDSE, then the central bright fringe will be white. Statement 2: Because all the wavelengths produce their zero order maxima at the same position.
Statement 2: In YDSE Imax = ( I1 + I 2 ) 2
8. Statement 1: According to Ray light, intensity of scattered light is inversely proportional to the fourth power of the wavelength of light. Statement 2: One can not see in fog because intensity of light decreases due to scattering.
maximum when both the sources have same intensity. Statement 2: In YDSE, the nth maxima always comes before the nth minima.
14. Statement 1: In YDSE, if I1 = 9I0 and I I2 = 4I0 then max = 25 I min
I max is I min
4B.83
7. Statement 1: In YDSE, the ratio
Light (Wave Optics)
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Match the quantities of column I with their values in column II. Column I Column II (a) Resultant intensity (p) 4 at P0 in (W/m2)
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Light (Wave Optics)
(d) All fringes are equally spaced
(s) Lloyd’s mirror experiment 4. In Young’s double slit experiment, if distance between slits is d, distance between slit and screen is D, wavelength of light used is λ. Then match column I and II. Column I (a) condition for bright fringe
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D (b) condition for dark (q) (µ − 1)t d fringe (c) displacement of (r) Path difference fringe when glass =nλ plate of thickness t is placed (d) Distance between (s) Path difference central maxima = (2n – 1)λ/2 and first dark fringe when glass plate of thickness t is used.
5. In a YDSE experiment d = 1 mm, D = 2000 mm, λ = 10–5 mm positions are measured from central maxima on screen Column 1 1 (a) Optical path difference at x = 2 (b) Optical path difference at x = 2000 mm (c) Path difference is almost (λ) (d) Maximum path difference
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3. Match the observation given in Column I with the experiments given in Column II where these observation occur.
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Column II (p) 1 mm (q) Just a distance below central maxima (.02 mm).
(r) (1/2 2) × 10−3 mm (s) Far away from central maxima
Column I Column II (a) Central fringes (p) Interference is bright and all in wedgefringes are equally shaped thin spaced film (b) Central fringes (q) Young’s is dark and all double slit fringes are equally experiment spaced (c) Central fringes is (r) Fraunhofer bright and fringes single slit are not equally diffraction spaced experiment
6. The arrangement of Lloyd’s mirror experiment is shown in figure. S is a point source of frequency 6 × 1014 Hz. A and B are the two ends of a mirror placed horizontally, and LOM represents the screen.
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(b) Resultant (q) zero amplitude at P0 (in unit) (c) If slit C is closed (r) 2 then resultant intensity at P0 (d) If slit B is (s) 2 2 closed resulting amplitude at P0 (in SI unit)
4B.84
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Column I The minimum height, above point O, fringes will form is (in mm) The maximum height, above point O, fringes will be formed is (in mm) Total no. of fringes are Fringe width (in µm) is
1. Initially, the distance of third minima from principal maxima will be
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4B.85
coming out of S1 and S2 are same and is I0. In this situation, the principal maximum is formed at the point P. At the point on screen where principal maximum is formed, phase difference between two interfering waves will be zero.
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Light (Wave Optics)
PASSAGE–2 The YDSE setup shown in figure, is lighted by a bright point source S, placed at a distance D from the plane of slits of equal widths on the perpendicular bisector of the slits.
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4B.86
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1. Central maxima will remain at the centre of screen 2. Fringe pattern will slowly shift in the direction opposite to motion of source 3. Fringe pattern will slowly shift in the direction same as that of motion of source 4. Fringe width changes
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1. Central maxima will remain at the centre of screen 2. Fringe pattern will slowly shift in the direction opposite to motion of source 3. Fringe pattern will slowly shift in the direction same as that of motion of source 4. Fringe width changes (a) only 1 (b) only 2 (c) only 3 (d) 1 and 4
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4. During activity – I, which of the following statement is/are true?
On the basis of the phenomenon of polarization, it was established that light is a form of transverse wave motion.
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Activity–IV: Width of one of the slit is slightly increased.
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Activity– II: The wavelength of light of source is continuously decreased. Activity–III: Source is moved towards the plane of slits slowly.
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Activity– I:
1. Central maxima will remain at the centre of screen 2. Fringe pattern will slowly shift in the direction opposite to motion of source 3. Fringe pattern will slowly shift in the direction same as that of motion of source 4. Fringe width changes (a) only 1 (b) only 2 (c) only 3 (d) 1 and 4
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6. During activity – III, which of the following statements are true?
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(a) only 1 (c) only 3
Light (Wave Optics)
Since light is an electromagnetic wave, it con sists of electric vector and magnetic vector E H varying periodically in phase, at right angles to each other and also perpendicular to the direction of propagation. In ordinary light, the electric vector vibrates in all planes with equal probability at right angles to the direction of propagation. Hence the ordinary light also called unpolarized light can be represented by a star as shown. Hence unpolarized light is symmetrical about the direction of propagation. By the phenomenon of reflection, light can be polarized. The polarized light has vibrations confined only to a single line in a plane perpendicular to the direction of propagation and is called plane polarized. To polarize light by reflection we make use of Brewster’s law which states that the tangent of the angle of
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The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.
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1. An interference pattern is obtained by using a Fresnel’s biprism. If the fringe width is 4 mm when air is the surrounding medium, then find the fringe width if water is the surrounding medium. Keeping the same source. Assume nglass = 1.5, nwater = 4/3, nair = 1.
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12. Ratio of maximum to minimum intensities at P (a) 2 : 1 (b) 4 : 1 (c) 8 : 1 (d) 16 : 1
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11. If the intensity at P be maximum then the minimum distance through which the reflecting surface MN should be displaced so that at P again becomes maximum? (a) 3 × 10–10m (b) 6 × 10–10m –10 (c) 1.5 × 10 m (d) 12 × 10–10 m
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9. An analyser examines two adjacent planepolarized beams A and B whose planes of vibration are mutually perpendicular. In one position of the analyser, beam B shows zero intensity. From this position, a rotation of 30º shows the two beams as matched in intensity. The intensity ration IA/IB of the two beams is (a) 3 (b) 1/3 (d) 5 (c) 1/ 2
10. The shape of the interference fringes on the screen is (a) circle (b) ellipse (c) parabola (d) straight line
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polarization (p) is numerically equal to the refractive index µ of the medium. The device that polarizes unpolarized light is called the polarizer and the device that analyses the polarized light is called the analyser. The amplitude of a plane polarized light incident on the analyser can be resolved into two components, one parallel to the plane of transmission of the analyser and the other perpendicular to it. The parallel component, namely, (a cosθ) is transmitted through the analyser. 7. At the angle of polarization, the angle of inclination between the reflected and refracted rays is (a) π/8 (b) π/6 (c) π/4 (d) π/2 8. If in an unpolarized light E = 2iˆ + 3 ˆj and H = 3iˆ − 2 ˆj , then the direction of propagation is given by (a) −13k (b) −10 ˆj ˆ ˆ (c) −6i + 6k (d) 5 ˆj − 4kˆ
4B.87
Light (Wave Optics)
Slit S1 is covered by a transparent sheet of thickness t1 = 2.5µm and S2 by another sheet of thickness t2 = 1.25µm as shown in figure. Both sheets are made of same material having refractive index µ = 1.40. Water is filled
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4B.88
The widths of S1 and S2 are w and 2w respectively. A thin transparent film of thickness 4 mm and R.I. 3/2 is placed infront of S1. It absorbs 50% light energy and transmits the remaining. The interference is observed on the screen. Point O is equidistant from S1 and S2. If the maximum intensity on the screen is I, then find (i) intensity at O (ii) Minimum intensity (iii) fringe width (iv) Distance of nearest maxima from O (v) Distance of central maxima from O (vi) intensity at 4 mm from O upwards.
3. In a Young’s double slit experiment, a parallel light beam containing wavelength λ1 = 4000Å and λ2 = 5600Å is incident on a diaphragm having two narrow slits. Separation between the slits is d = 2 mm. If difference between diaphragm and screen is D = 40 cm, calculate (i) distance of first black line from central bright fringe (ii) distance between two consecutive black lines.
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7. Two slits S1 and S2 are on a plane inclined at an angle of 45º with horizontal. The distance between the slits is 2 mm. A monochromatic point source S of wavelength, λ = 5000Å is placed at a distance 1/ 2 mm from the mid-point of slits as shown in figure. The screen is placed at a distance of
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4. In a YDSE, the distance between the slits and the screen is 100 cm. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width 0.25 mm. When the distance between the slits is increased by ∆d = 1.2 mm, the fringe width decreases to n = 2/3 of the original value. In the final position, a thin glass plate of refractive index 1.5 is kept in front of one of the slits and the shift of central maximum is observed to be 20 fringe width. Find the thickness of the plate and wavelength of the incident light.
6. (a) Thin glass plate of thickness t and refractive index µ is inserted between screen and one of the slits in a Young’s experiment. If the intensity at the centre of the screen is I, what was the intensity at the same point prior to the introduction of the sheet. (b) One slit of a Young’s experiment is covered by a glass plate (µ1 = 1.4) and the other by another glass plate (µ2 = 1.7) of the same thickness. The point of central maxima on the screen, before the plates were introduced is now occupied by the third bright fringe. Find the thickness of the plate, the wavelength of light used is 4000Å.
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in space between diaphragm and screen. A monochromatic light beam of wavelength λ = 5000 Å is incident normally on the diaphragm. Assuming intensity of interference pattern obtained on the screen, where C is foot of perpendicular bisector of S1S2. (Refractive index of water, µw = 4/3)
Light (Wave Optics)
2 m. Find the fringe width of interference pattern on the screen.
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9. In a Young’s double experiment, the upper slit is covered by a thin glass plate of refractive index µ1 = 1.4 while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index µ2 = 1.7. Interference pattern is observed using light of wavelength λ = 5400 Å. It is found that the point P on the screen where the central maxima fell before the glass plates were inserted, now has 3/4 the original intensity. It is further observed that what used to be the fifth maxima earlier, lies below the point P while the sixth minima lies above the point P. Neglecting absorption of light by glass plates, calculate thickness of the glass plates.
13. A beam of light consisting of two wavelengths, 6500Å and 5200Å is used to obtain interference fringe in a Young’s double slit experiment. (i) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 6500Å. (ii) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.
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11. A Young’s double slit arrangement produces interference fringes for sodium light (λ = 5890 Å) that are 0.20º apart. What is the angular fringe separation, if the entire arrangement is immersed in water. (R.I. of water = 4/3)
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10. A plane monochromatic light wave falls normally on a single slit. At a distance D = 4 m from it there is a screen showing a certain diffraction pattern. The slit width is decreased n = 2 times. Calculate the new distance D’ at which the screen should be positioned to obtain the diffraction pattern similar to previous one but diminished n times.
14. In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000Å and intensity (10/π) W/m2 is incident normally on two apertures A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000Å and refractive index 1.5 for the wavelength of 6000Å is placed in front of aperture A (see figure). Calculate the power (in W) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.
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The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
12. In Young’s double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves.
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4B.89
8. In a Young’s double slit experiment, when a lens is interposed at two points between slits and screen, the distance between the two images of the slits are 4 mm and 9 mm. When the screen is shifted by 10 cm, the number of fringes is reduced from 100 to 80 in a 0.4 cm. Find the wavelength of the light used in the experiment.
Light (Wave Optics)
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4B.90
apparatus is immersed in a liquid. Find refractive index of the liquid. [1996]
QUESTIONS
-J
EE
IIT
YEAR
s’
1. Consider Fraunhoffer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum, the phase difference (in radian) between the wavelets from the opposite edges of the slit is [1995]
4. A double-slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of the slits and screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300Å. [1996] (a) Calculate the fringe width. (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minima on the axis.
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6. In a Young’s experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400Å. It is found that the point P on the screen where the central maximum (n = 0) fall before the glass plates were inserted, now has ¾ the original intensity. It is further observed that what used to be the 5th maximum earlier, lies below the point P while the 6th minima
I(θ) = I0 / 2 for θ = 30º I(θ) = I0 / 4 for θ = 90º I(θ) = I0 for θ = 0º I(θ) is constant for all values of θ
3. Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000Å. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original
(a) (b) (c) (d)
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The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150 m. The intensity 1(θ) is measured as a function of θ, where θ is defined as shown. If I0 is the maximum intensity, then I(θ) for 0 ≤ θ ≤ 90º is given by [1995]
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5. In Young’s experiment, the source is red light of wavelength 7 × 10–7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10–3m to the position previously occupied by the 5th bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 × 10–7m, the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of the glass for the green light. Also estimate the change in fringe width due to the change in wavelength. [1997]
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2. In an interference arrangement similar to Young’s double-slit experiment, the slits S1 and S2 are illuminated with coherent microwave sources, each of frequency 106Hz.
(b) π/2 (d) π
(a) π/4 (c) 2π
PREVIOUS
Light (Wave Optics)
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7. A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on screen placed parallel to the plane of the slits at a distance of 1.0 m from it, as shown in the figure.
9. The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 mm separation. \ 6
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10. Yellow light is used in a single slit diffraction experiment with slit width of 0.6. mm. If yellow light is replaced by X-rays, then the observed pattern will reveal [1999]
(a) (b) (c) (d)
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(a) If the incident beam falls normally on the double slit apparatus, find the y-coordinates of all the interference minima on the screen. (b) If the incident beam makes an angle of 30º with the x-axis (as in the dotted arrow shown in the figure), find the y-coordinates of the first minima on either side of the central maximum. [1998]
8. A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is [1998]
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(a) Find the location of the central maximum (bright fringe with zero path difference) on the y-axis. (b) Find the light intensity at point O relative to the maximum fringe intensity. (c) Now, if 600 nm light is replaced by white light of range 400 to 700 nm, find the wavelengths of the light that form maxima exactly at point O. [All wavelengths in this problem are for the given medium of refractive index 4/3. Ignore dispersion.]
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The lower slit S2 is covered by a thin glass sheet of thickness 10.4 µm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown. [1999]
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(b) π/2 (d) 2π
(Absorption of light by glass plate may be neglected)
4B.91
(a) zero (c) π
lies above P. Calculate the thickness of the glass plate. [1997]
Light (Wave Optics)
that the central maximum is narrower more number of fringes less number of fringes no diffraction pattern
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4B.92
11. A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat plate as shown. [1999]
is changed to 400 nm, number of fringes observed in the same segment of the screen is given by [2001]
Light (Wave Optics)
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(b) 2λ/3 (d) λ
(a) 2λ (c) λ/3
17. A point source S emitting light of wavelength 600 nm is placed at a very small height h above the flat reflecting surface AB (see figure).
(b) 4I (d) 7I
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13. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between resultant intensities at A and B is [2001] (a) 2I (c) 5I
16. In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is [2002]
12. In a double slit experiment instead of taking slits of equal widths, one slit is made twice as wide as the other, then in the interference pattern [2000] (a) the intensities of both the maxima and the minima increases (b) the intensity of the maxima increases and the minima has zero intensity (c) the intensity of maxima decreases and that of minima increases (d) the intensity of maxima decreases and the minima has zero intensity
14. In young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light
15. A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line perpendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S, 40 cm below P and 2 m from the vessel, to illuminate the slits as shown in the figure below. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now, a liquid is poured into the vessel and filled up to OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid. [2001]
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The observed interference fringes from this combination shall be (a) straight (b) circular (c) equally spaced (d) having fringe spacing which increases as we go outwards
(b) 18 (d) 30
(a) 12 (c) 24
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distance between the plane of the slits and the screen is I m. The minimum distance between two successive regions of complete darkness is [2004]
The intensity of the reflected light is 36% of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. [2002]
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λ (b) sin −1 2d λ (d) sin −1 4d
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Directions (Questions 22 to 24): The figure shows a surface XY separating two transparent media, medium–1 and medium–2. The lines ab and cd represent wavefronts of a light wave travelling in medium–1 and incident on XY. The lines ef and gh represent wavefronts of the light wave in medium–2 after refraction. [2007]
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21. In Young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this point is
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20. In a Young’s double slit experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again? Take D/d = 103. Symbols have their usual meanings. [2004]
2
(b) 5.6 mm (d) 28 mm
(a) 4 mm (c) 14 mm
(a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure) (c) If the intensities at point P corresponds to a maximum, calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum. 18. In the adjacent diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition on θ for constructing interference at P between the ray BP and reflected ray OP [2003] 3λ (a) cos θ = 2d λ (b) cos θ = 4d λ (c) sec θ − cos θ = d 4λ (d) sec θ − cos θ = d
4B.93
Light (Wave Optics)
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19. In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the
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4B.94
25. In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [2008]
23. The phases of the light wave at c, d, e and f are φc, φd, φe and φf respectively. It is given that φc ≠ φf (a) φc cannot be equal to φd (b) φd can be equal to φe (c) (φd – φf) is equal to (φc – φe) (d) (φd – φc) is not equal to (φf – φe)
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(b) (b) (a) (b) (a) (d) (b)
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6. 14. 22. 30. 38. 46. 54.
(a) (d) (d) (b) (c) (c) (a)
7. 15. 23. 31. 39. 47. 55.
() (b) (a) (b) (c) (d) (c)
2. (a) 10. (a)
(c) (b) (d) (a) (b) (a) (d)
8. (a, c) 16. (a, c, d)
QUESTIONS
REASON
AND
ASSERTION
1. (a, c) 2. (a, b) 3. (a, b, c) 4. (a, c, d) 5. (a, b, d) 6. (c, d) 7. (a, c) 9. (a, c, d) 10. (a, b, d) 11. (b, c, d) 12. (a, c) 13. (b, c, d) 14. (a, b, c, d) 15. (b, d)
1. (d) 9. (d)
8. 16. 24. 32. 40. 48. 56.
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3. 11. 19. 27. 35. 43. 51. 59.
(d) (a) (c) (b) (d) (a) (d) (b)
AN
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2. 10. 18. 26. 34. 42. 50. 58.
(b) (b) (a) (b) (d) (b) (a) (c)
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1. 9. 17. 25. 33. 41. 49. 57.
ONE
24. Speed of the light is (a) the same in medium–1 and medium–2 (b) larger in medium–1 than in medium–2 (c) larger in medium–2 than in medium–1 (d) different at b and d
A
(a) If d = λ, the screen will contain only one maximum (b) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen (c) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase (d) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase
22. Light travels as a (a) parallel beam in each medium (b) convergent beam in each medium (c) divergent beam in each medium (d) divergent beam in one medium and convergent beam in the other medium.
Light (Wave Optics)
3. (b) 11. (b)
4. (b) 12. (a)
5. (b) 13. (a)
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6. (a) 14. (a)
7. (c) 15. (d)
8. (b)
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1. (a) → r, (b) → r, (c) → q, (d) → q
4. (a) → r, (b) → r, (c) → q, (d) → p 6. (a) → q, (b) → p, (c) → s, (d) → r
4. (b)
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19. (d)
20. 3.5 mm
21. (c) 22. (a)
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17. (a) circular, (b) 1/16, (c) 300 m
24. (b)
11. (a)
12. (a)
13. (b)
14. (b) 15. 2 cm above point Q on side CD, µ = 1.0016
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8. (a)
16. (a)
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7. (d)
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QUESTIONS
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1. (c)
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5. (d)
12. (d)
25. (a,b). For at least one maxima, sin θ = λ/d If λ = d, sin θ = 1 and y → ∞ So there will be only central maximum on the screen. If λ < d < 2λ, sin θ exists and y is finite so there will be two more maxima on screen. Intensity of dark fringe becomes zero when intensities of two slits are equal. Initial intensity at both the slits are unequal so there will be some brightness at dark fringe. Hence when intensity of both slits is made same the intensity at dark on screen shall decrease to zero. So both (c) and (d) are false.
3. (a) 11. (a)
INTS
2. (b) 10. (a)
s’
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ASSAGE
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3. (a) → q, (b) → s, (c) → r, (d) → p
1. (a) 9. (b)
2. (a) → r,s (b) → q, (c) → p, (d) → s
5. (a) → r, (b) → r, (c) → q, (d) → p,s
p
4B.95
QUESTIONS
COLUMN
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Light (Wave Optics)
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