Rahul Sardana- Optics And Modern Physics.pdf

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ADVANCED

JEE PHYSICS for both Main & Advanced levels of JEE

OPTICS & MODERN PHYSICS

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ADVANCED JEE PHYSICS for both Main & Advanced levels of JEE

OPTICS & MODERN PHYSICS

RAHUL SARDANA M.Sc (Hons). Physics

--7

. Ef·,1--

--=1~11; __I l

ELSEVIER A division ofRe¢ Elsevier India Pvt. Ltd.

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ADVANCED JEE PHYSICS, Optics & Modem Physics Rahul Sardana ELSEVIER A Division of Reed Elsevier India Private Limited. Copyright © 2014 Effective Testbooks India (ETI)

This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. Reasonable efforts have been made to publish reliable data and infonnation, but the author and the publisher cannot assume responSibility for the validity of all materials or for the consequences of their use. Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming and recording, or by any information storage or retrieval system, without prior remission in writing ~om the publishers. The consent of ETI does not extend to copying for general distribution, for promotion, for creating new works, or for resale. Specific permission must be obtained in writing from ET/ for such copying. The export rights of this book are vested solely with the publisher. ISBN: 978-93-81269-97-8

Published by Elsevier, a division of Reed Elsevier India Private Limited, under special arrangement with ETI. Registered Office: 305, Rohit House, 3 Tolstoy Marg, New Delhi-110 001. Corporate Office: 14th Floor, Building No. JOB, DLF Cyber City, Phase-II, Gurgaon-122002, Haryana, India .. Printed and bound at Rajkamal Electric Press, Kundli, Haryana.

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text overview, goals & focus

In the past few years, the UT-JEE has evolved itself as an examination designed to check out true scientific skills. The examination pattern wants us to see those little details which, others fail to see. Those details which tell us how much in depth we should know to describe as much as possible. Keeping the present day scenario in mind, this book is written for students, to allow them not only to learn the tools that "Optics & Modem Physics" provides but also to see why they work so nicely in explaining the beauty of ideas behind the subject. The central goal of this text is to help the students develop a thorough understanding of the principles of "Optics & Modem Physics". This book stresses on building a rock solid technical knowledge based on firm foundation of the fundamental principles than on a large collection of formulae. The primary philosophy of this book is to act as a guide who creates a careful, detailed groundwork for strong conceptual understanding and development of problem solving skills like mature and experienced physicists. features of book theory with illustrations

"Optics & Modem Physics" are important topics, and in this book I have tried to make these topics lively, clear and precise to the greatest levels. I have generally seen students not stressing on the theoretical details. They always feel that doing more numerical problems will solve their purpose. But let me tell you here, that numerical problems are just the special cases of the theoretical concepts. The entire Physics is based on a simple program "IF~ THEN~ ELSE". Try to follow this and see how you get to your ultimate goal i.e., IIT~JEE. So, keeping this in mind, the entire theory part of all the chapters has been kept elaborative, simple to understand with supportive Illustrations at all the places. DO NOT TRY TO ATTEMPT ILLUSTRATIONS WITHOUT GOING THROUGH THE THEORY. conceptual notes, remarks, words of advice, misconception removal

Throughout the text, the Conceptual Notes and Remarks are highlighted which focus on the principal ideas and concepts that a student must take care of. Places where students commonly develop a misconception have been supported by Misconception Removal, highlighted in grey and supported by Words of Advice. Throughout my teaching career of 18 years I have always found my students getting benefitted from these Conceptual Notes, Remarks, Words of Advice, Misconception Removals. All these are actually used to provide warnings to the. students about common errors and ways to avoid them. problem solving techniques

These techniques, highlighted in grey, always ensure that the students become capable of solving a variety of problems in an easy way. Wherever necessary, the text is supplemented with them for having a thorough understanding to the application processes. in chapter exercises (ice) : topic wise

After you study the theory and apply it to the Illustrations, its time you practice something on your own and that too topic wise. For this purpose I have created In Chapter Exercises (ICE) (except for Mathematical Physics). Each ICE has the name of the topic(s) clearly mentioned on it. Please note that ICE are based on simple, single . concept classification technique. They are fully solved, so that if you come across only problem, then you just refer to the solutions.

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C:::::

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solved problems

After you have gone through the entire Theory (with Illustrations) and all the supplements (ICE, Conceptual Notes, Remarks, Words of Advice, Misconception Removal, Problem Solving Techniques), its high time to do problems that are a true mix of concepts studied. This section has problems that involve multiple concept usage so that your brain is exposed to the ultimate throttle required to extract the best from you at !IT-JEE. practice exercise sets {fully solved)

Now comes the time when you are very much ready to do the practice as per the !IT-JEE pattern. This section contains all the variety of questions that have been asked in the !IT-JEE. In this section you will come across the following variety of questions. single correct choice type (SCCT)

Each question, in this section, has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. multi11le correct choice t}'pe (MCCT)

Each question, in this section, has four choices (A), (BJ, (C) and (D), out of which ONE OR MORE is/ are correct. reasoning based questions/ assertion-reason type (ART)

This section contains Reasoning type questions, also called Assertion-Reason type question, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. E;ach question contains STATEMENT 1-(Assertion) and STATEMENT 2 (Reason). You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 ls TRUE andSTATEMENT2 is FALSE. Bubble (DJ If STATEMENT 1 is FALSE butSTATEMENT2is TRUE. linked comprehension tyl)e (LCT) / paragraph type

This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (BJ, (C) and (D), out of which only one is correct. (F_or the sake of competitiveness there may be a few questions that may have more than one correct options). matrix match type (MMT) / column matching

Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:

If the correct matches are A -> p, s and t; B -> q and r; C -> p and q; and D -> s and t; then the correct darkening of bubbles will look like the following : P

q

r

s

t

A@@©©CD B@@©©CD c@@©©CD D@@©©CD

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t=:J

integer answer type questions (IATQ) / numerical type questions

In this section the answer to each of the question is a four digit integer, ranging from O to 9999. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question number X (say) is 6092, then the correct darkening of bubbles will look like the following: X.@e@@

CD CD

G)

CD

®@®• ®@®®

@@©@

®@®®

~~~~

®®®® @@e@ answers & solutions

Each· chapter contains answers followed by solutions to the problems. The solutions· are exhaustive with complete methods and reasons which will help you a lot to understand a particular concept. Short cuts are also included (wherever necessary) for-enhancing you problem solving skills. This book, I hope, will nourish you with the concepts involved such that you get a great rank at !IT-JEE. To conclude, I apologise in advance for the errors (if any) that may have inadvertently crept in the text. I would be grateful to the readers who bring errors of any kind to my attention. I truly welcome all comments, critiques and suggestions at [email protected]. PRAYING TO GOD FOR YOUR SUCCESS AT !IT-JEE, GOD BLESS YOU!

The Author RAHUL SARDANA M.Sc.(Hons.) Physics

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CHAPTER 1

RayOptics (a) Reflection at Plane & Curved Surfaces _ _ _ _ _ _ _ _ _ _ 1.1 (b) Refraction at Plane Surfaces

1.27

(c) Refraction at Curved Surface

1.55

Solved Practice Problem

1.96

Practice Exercise Sets :> Single Correct Choice Type Questions _ _ _ _ _ _ _ _ J, 108

=

Matrix Match Type Questions (Column Matching Type), _ _ _..,,,40

:i

Integer Answer Type Questions

.43

Answers to In Chapter Exercises (ICE) & Practice Exercise Set

.45

Solutions to In Chapter Exercises (ICE)

.48

Solutions to Practice Exercise Sets

.52

CHAPTER4

Atomic & Nuclear Physics

:>

Multiple Correct Choice Type Questions

1. 135

AtomicPhysics ..... _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 4_7

:>

Reasoning Based Questions (Assertion Reason Type)

1. 141

Nuclear Physics....

4.21 4.48

:>

Linked Comprehension Type Questions (Paragraph Type) .......... 1. 144

Solved Practice Problem

:> :>

Matrix Match Type Questions (Column Matching Type) .............. 1.154

Practice Exercise Sets

Integer AnswerTypeQuestiohs ___________ 1,160

:i

Single Correct Choice Type Question~--------4.58

Answers to In Chapter Exercises (ICE) & Practice Exercise Sets ____ 1,163

:i

Multiple Correct Choice Type Question

Solutions to In Chapter Exercises (ICE) _ _ _ _ _ _ _ _ _ _ _ 1, 169

:i

Reasoning Based Questions (Ass'ertion Reason Type) _ _ _ _ 4.78

Solutions to Practice Exercise Set

:,

Linked Comprehension Type Questions (Paragraph Type) ............. 4.81

:,

Matrix Match Type Questions (Column Matching Type) ................ .4.89

:,

Integer Answer Type Questions _ _ _ _ _ _ _ _ _ _ _4.92

1. 193

CHAPTER2

Answers to In Chapter Exercises (ICE) & Practice Exercise Set,_ ____4.94

Wave Optics WaveOptiCS-------------------~-1 Solved Practice Problems....

.41

Solutions to In Chapter Exercises (ICE) _ _ _ _ _ _ _ _ _ _ _4.99 Solutions to Practice Exercise Set

Practice Exercise Sets :i

Single Correct Choice Type Question,,_ _ _ _ _ _ _ _ _..,_.52

=

Multiple Correct Choice Type Question

.66

:i

Reasoning Based Questions (Assertion Reason Type)

.69

:i

Linked Comprehension Type Questions (Paragraph Type) ............. 2.71

:i

Matrix Match Type Questions (Column Matching Type) ................. 2.77

:i

Integer Answer Type Questions

.80

Answers to In Chapter Exercises (ICE) & Practice Exercise Set.~---~,.83 Solutions to In Chapter Exercises (ICE)

.86

Solutions to Practice Exercise Sets

.92

CHAPTER3

Dual Nature of Radiation & Matter Dual Nature ofRadiqtion &Matter _ _ _ _ _ _ _ _ _ _ _ _ _..,,. I Solved Practice Problem

4J3

.1 S

Practice Exercise Sets :i

Single Cofrect Choice Type Question•s.._ _ _ _ _ _ _ _ __,_2 7

o

Multiple Correct Choice Type Question

.33

:,

Reasoning Based Questions (Assertion Reason Type)

.36

:,

Linked Comprehension Type Questions (Paragraph Type) .............3.38-·

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I Ray Optics

Contents

- --··-

- --

- -- -

---- - - ------ - - ---

'

(a) REFLECTION AT PLANE & CURVED SURFACES ..................................................... 1.1 (b) REFRACTION AT PLANE SURFACES ......................................................................1.27 (c) REFRACTION AT CURVED SURFACES ...................................................................1.55 Solved Practice Problems ..........••....•...•.•....•...•••••.••••...••....••...••...•••..•••....••...•....•.•..••....•••..1.96

Practice ·Exercise Sets

:> Single Correct Choice Type.Questions .•.....••..............•........................................1.108 :,

Multiple Correct Choice Type Ques1ions .............................................................1.135

:> Reasoning Based Questi?ns (Assertion Reason Type)-...................................... 1.141 :,

Linked Comprehension Type Questions (Paragraph Type) ................................. 1.144

::> Matrix Match Type Questions (Column Matching Type) .........................••....•...... 1.154 ::> Integer Answer Type Questions ..................................................................•••.....1.160 Answers to In Chapter Exercises (ICE) & Practice Exe!cise Sets .................._................. 1.163 Solutions to In Chapter Exercises (lCE) ...........................................................................1.169 Solutions to Practice Exercise Sets ................................. ;...............................................1.193

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NATURE OF LIGHT: An Introduction

E

ight is a form of energy that makes object visible to our eyes or light is the form of energy that produces in us the sensation of sight. In Seventeenth century Newton and Descartes believed that light consisted of a stream of particles, called corpuscles. Huygens proposed wave theory of light and proposed that light is a disturbance in a medium called Ether. This theory could explain the

L

phenomena of interference, diffraction, etc. Thomas Young,

through . his double slit experiment, measured the wavelength of light. Maxwell suggested the electromagnetic theory of light. According to this theory, light consists of electric and magnetic fields,.in mutually perpendicular directions, and both are perpendicular to the direction of propagation. Heinrich Hertz produced in the laboratory the electromagnetic waves of short wavelengths. He showed

-=c B In 1905, Albert Einstein revived the. old corpuscular theory using Plank's Quantum· Hypothesis and through his photoelectric effect experiment showed that light consists of discrete energy packets, called photons. The energy of each photon is

E=hf=hc A,

So, in view of these developments, light must be regarded to have a dual nahlre i.e., it exhibits the characteristics of a · particle in some situations and that of a wave in other sihlations. So the question'' Is light a particle or a wave?" is

purely inappropriate to be -asked. At present, it is believed that light has dual nature, i.e., it has both the characters,

wave-like and particle-like.

that these electromagnetic waves possessed all the properties OPTICS : An Introduction

of light waves.

E

Optics is the study of the properties of light, its propagation through different media and its effects. In most of the sihlations, the light encounters objects of size much larger

than its wavelength. We can assume that light travels in straight lines called rays, disregarding its wave nature. This allows us to formulate the rules of optics in the language of geometry, as rays of light do not disturb each other on intersection. Such study is called geometrical (or ray) optics. It includes the working of mirrors, lenses, prisins, etc. When light passes through very narrow slits, or when it

Direction of

propagation

ii

passes around very small objects, we have to consider the wave nature of llght. This study is called wave (or physical) optics.

Light travels in vacuum with a velocity given by

1

. . 8'

c= ~=3x10 ms

Vµoto

-1

·

where µo and Eo are the permeability and permittivity of free space (vacuum). The magnitudes of electric and magnetic fields are related to the velocity of light by the relation · ··

DOMAINS OF OPTICS

The study of light can be categorized into three broad domains. a) Geometrical Optics (Ray Optics) b) Physical Optics (Wave Optics) 1.1

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Advanced JEE Physics

Optics & Modem Physics

c) Quantum Optics Please note that these domains are not strictly disjoint as the transitions between them are continuous and not sharp. However for convenience we consider them as distinct. These domains are distinguished as follows.

5.

The Laws of Refraction (discussed later). The Laws of Refraction govern the bending of light when the light goes from one medium to the other (rarer to denser or denser to rarer) medium.

BASIC TERMS & DEFINITIONS A.

GEOMETRICAL OPTICS (RAY OPTICS)

This branch involves the study of propagation of light based on the assumption that light travels in fixed straight line as it passes through a uniform medium and its direction is changed when met by a surface. of a different medium or if the optical properties of the medium are non uniform either in time or in space. The ray approximation is valid for the wavelength A. very small compared to the size of the obstacle (d) or the size of the opening through which the ray passes.

SOURCE

A body which emits light is called source. ·Tue source can be ·a point one or an extended one. A source is of two types. a) Self luminous : The source which possess light of its own. EXAMPLE : sun, electric arc, candle etc.

Non-luminous : It is a source of light which does not possess light of its own but acts as source of light by reflecting the light received by it.

b)

This approximation 1,. « d proves to be very good for the study of mirrors, lenses, prisms and associated optical instruments such as microscope, telescope, cameras etc. B.

PHYSICAL OPTICS (WAVE OPTICS)

This branch involves the study of propagation of light in the form of a wave and it deals with the phenomenon of interference, diffraction, polarization etc. This nature of light has to be taken when the light passes through very narrow slits or when it goes past very small objects. So this branch works effectively when 1,. » d . C.

QUANTUM OPTICS

This branch involves the study of propagation of light as a stream of particles called as Photons. This concept of light behaving as particles called photons is of utmost importance while studying the origin of spectra, photoelectric effect, concept of radiation pressure, Compton effect etc. FUNDAMENTAL LAWS OF GEOMETRICAL OPTICS

To a first approximation, we can consider the propagation of light disregarding its wave nature and assuming that light propagates in straight lines called rays. This allows us to formulate the laws of optics in the language of geometry. Thus, the bfanch of optics where the wave nature of light is neglected is called geometrical (or ray) optics. Geometrical optics is based ~m five fundamental laws. 1. Law of Rectilinear Propagation of Light. It states that light propagates in straight lines in homogenous media. 2. Law of Independence of Light Rays. It states that rays do not disturb each other upon intersection. 3. The Law of Reversibility of Light. According to this law, if a ray of light, after suffering a number of reflections and refractions, has its path reversed at any instant, then the ray retraces its path back to the source. 4. The Laws of Reflection. The Laws of Reflection govern the bouncing back of the incident ray after striking a surface to the medium from which it was coming.

=

EXAMPLE : moon, objects around us, book etc. REMARK(S) Sources are also classified as isotropic and non-isotropic. Isotropic sources give out light uniformly in all directions whereas non-isotropic sources do not give out light uniformly in 1-~ll_dir~~!i?n~. _

RAY

The straight line path aiong which the light travels between two points in-a homogeneous medium or in a pair of media is called a Ray: It is represented by an arrow head on a straight line, the arrow head represents the direction of propagation of light. A ray of light will always follow a path along which the time taken is the minimum.

-- - - - . - . ., '-Ray

I

REMARK(S)

_A ~ingle_!~~-~ann_?~ be_~~ol~.t~~ !r<:_rTI _a_~oun:E:. of light.

MEDIUM

Substance through which light propagates or tends to propagate is called a medium. It is of following three types. a) Transparent: It is a medium through which light can be propagated easily. EXAMPLE : glass, water etc.

b)

Translucent : It is a medium through which light is propagated partially.

c)

Opaque : It is a medium through which light cannot be propagated.

EXAMPLE : oil paper, ground glass etc.

EXAMPLE : wood, iron etc.

BEAM

A bundle or bunch of rays is called a beam. It is of following three types.

1.2

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Ray Optics : Reflection at Plane & Curved Surfaces

a)

Parallel beam : It is a beam in which all the rays constituting the beam move parallel to each other and diameter of beam remains same. A very narrow beam is called a Pencil of Light.

0: b)

.-----· - -----

. . -... - .

·1

CONCEPTUAL NOTE(S) Virtual object cannot be seen by human eye, because for an o. bject or an- image io be seen by ihe eyes, the rays received by

t~~~x~s_m~st_~~di~e~ging. ___ -·---------

-- -

"

IMAGES

Convergent beam : In this case diameter of beam decreases in the direction of ray.

An optical image is a point where reflected or refracted rays of light either intersect or appear to intersect. Thus, the image of an infinite object is actually an assembly of the image points corresponding to various parts or the points of ~e object. The images formed can again be real or virtual. A.

REAL IMAGES

If the rays after reflection or refraction actually converge (or meet) at a point then the image is said to be real and it can be obtained on a screen. c)

Divergent beam: It is a beam.in which all the rays meet at a point when produced backward and the diameter of beam goes on increasing as the rays proceed forward.

Real Image

B. OBJECTS

The object for a mirror can be real or virtual. A.

VIRTUAL IMAGES

However, if the rays do not actually converge but appear to diverge from a point (or appear to meet at a point), then the image so formed is said to be virtual image. A virtual image cannot be obtained on a screen.

REAL OBJECTS

If the rays from a point on an object actually diverge from it and fall on the mirror, the object is said to be real. In simple language the incident rays are diverging and the point of divergence is the position of the real object. The following diagrams support the arguments given. Virtual Image

0

0

a) b)

B.

VIRTUAL OBJECTS

If the rays incident on the mirror appear to converge to a point, then this point is said to be virtual point object for the mirror. In simple language the incident rays are converging and the point of convergence is the position of the virtual object. The following diagrams support the arguments given:

1

I'·......'::-.o

l

rl·--~::::»o /

CONCEPTUAL NOTE(S) The real images can be obtained on a suitably placed screen, but virtual images cannot be obtained on-a screen. Human eye cannot distinguish between the real image and the virtual image because in both the cases the rays are d!ve!gi~Q-_. _~ _ __________ _

REFLECTION OF LIGHT When light strikes the. surface on an object, some part of the light or the complete light is sent back into the same medium. This phenomenon is called as reflection. The surface, which reflects light, is called mirror. A mirror could be plane or curved. CONCEPTUAL NOTE(S) In reflection, the frequency, speed and wavelength remain unchanged, but a phase change may occur depending on the __nature_. of reflec~ing.surface.

1.3

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--- - - - - - ----I Advanced JEE Physics

--

Optics & Modern Physics To understand this theorer;n, let ·us· consider two points A and B in the same medium. Since, we know that between these two points light travels in a straight line, so the time taken by the light to go from A to B must logically be the. minimum.

The reflection from a denser medium causes an addition phase

1.

--

change of n or a path change of - (by Stoke's Law) while 2 reflection from rarer.medium does not c~Use.an; phase c,hange:_..

Diffused (irregular) reflection takes place from a rough surface where as Specular (regular) reflection takes place from an extraordinarily smooth surface. However, the Law's of Reflection are applicable for both kinds of surfaces,

b)

The angle of incidence i is the angle made by the incident ray with the normal. The angle of reflection r is the angle made by the reflected ray with the normal.

Y~. .o

0

I

i

Concave Surtace

i

b~oundary, afte~ reflecti:~~~:~~ces its p ~ , - - - - , , ..

I

!

'

, c)

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I

/

L) -{' \

Concave Mirror (b)

~

i

~ -

OB

=!:.( .Ja

2

+x' +~b' +(d-x)2 )

~=0

dx

_!._(.Ja 2 +x 2 ) +_!,_Ub 2 + (d-x)') = 0 dx _

(c)

g. as shown in ·the-figure .. I

'' '

AO

t=-+c C

.dx

j

!

X+I

C

Convex Mirror

I

i

-0-" -----+t+d -

= f A-+O f- fO-+B

t

I

=90°-i. and hence r ---t

b

Now,.according to Fermat's,Principle, t is MINIMUM, so

For grazing. i_ncidence, the incident ray grazes the refle~ting

i

i

- Let the incident light start from A , hit the mirror at O and get reflected to point B. Let the points A and B be at perpendicular distances a and b from the mirror and let A and B have a separation d between them as shown in figure. The time taken by the light to go from A to O to B is given by

1(

2

The angle made by the incident ray with the plane reflecting i

surface, so i-),

r,'

----d----.a

,, ' Surface-Js called- glancing· angle. ;Thus, the, glancing angle l (a)

B

' '' ''' '

'i

1-4- X

-~~~~~~,,) !

j ('---~---" Plane Mirror

!

a

If i = P·, then r =O . It means a ray incident no.rmally on a J

I

I

i

f

i

i

I b)

A

Convex Surface

SPECIAL CASES a}

LAWS OF REFLECTION USING FERMAT'S THEOREM

The incident-ray, the reflected-ray and the normal to the reflecting surface at the point of incidellce, all lie in the same plane. The angle of reflection is equal to the angle of incidence (i = r) .

Plane Surface

B

LIGHT PATH

Consider a plane mirror on which light is incident as shown.

LAWS OF REFLECTION

a)

A

.Ja +x

) 1( 2(d-x)(-1)) 0 +2 ~b'+(d-x)' = (d-x)

X 2

2

~b'+(d-x)'

From the figure, we observe that

.I

x · · an"d (d-~ sm· r ~--=s1nz ..Ja'+x2 )b'+(d-x) 2 ,

I

I

2x ..Ja'+x'

::::> ··::::>

sini = sinr i =r.

{The Law'of Reflection}

FERMAT'S PRINCIPLE OF LEAST TIME

According to this theorem, light will always follow the path taken between any two points by a ray of light is the path along which the time taken is the minimum. ·This principle is sometimes taken as the definition of a ray of light.

=

I PROBLEM SOLVING TRICK(S) I a)

Basic Problems in Optics : Most of the problems asked in

L_

optics. expect us fo find tl)e position and na!ure of the final imag·e formed by-certain optical systems for a given abject. The ..aptical_system_may b~ jul;;t a mirror, m~a lernr or a__

I I

1.4

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Ray Optics: Reflection at Plane & Curved'Surfaces combinatiOn of several reflecting and ~r9fractinQSurface5. b)

Basic Strategy for Solving the Problems : To handle these kinds of problems, first of all, we identify the sequence in which the reflection and refraction are taking place. The several events of reflection or refraction can be

named as Event 1, Event 2 and so on following the · sequence in which they occur. Now; the ihlage of Event t would be object for Event 2, image of Event 2 will be object of Event 3 and so on. This

way one can proceed to find the flnal Image.

I

ANGLE OF DEVIATION (6)

· Deviation (6) is defined as - the angle between the initial direction of the incident ray and the final direction of the reflected ray or the. emergent ray.

Illustration 1 Two plane mirrors are inclined to each other at an angle 0 . A ray of light is reflected first at one mirror and then at the other. Find the total deviation suffered by the ray.

Deviation produced in Reflection is 6 = 180° -(i + r) Since r=i

=>

- -- - - - - - - - - - - - - - - CONCEPTUAL NDTE(S) The same ls fol!nd to'hold·good for three plane mirrors arranged mutually perpetldicular to each other thus fohning the comer of a cube such that the light incident on this arrangement suffers one reflection from each of the mirrors so as to emerge out antiparallel to the incident light. This arrangement of three mutually perpen_dicular plane mirrors fonning the comer of a cube is . called the CORNER REFLECTOR.

6=180°-2i

The variation of deviation (6) with the angle of incidence (i) js shown in figure.

Solution a be the angle of in~idence for mirror M 1

p be the angle of incidence for mirror M, 61 be the deviation due to mirror M 1 and

8_= Jt

62 be thE! deviation due to mirror M 2 D

0

The deviation is minimum for grazing incidelice i.e., when

I I . I

i-+i,then 6=6rr,;0 =0°.

1

: PROBLEM SOLVING TRICK($) • :he deviation is maximu~ for normal incidence i.e., when

I ,a) I· b)

[ c)

1=0 then, o=o= =180 .

!

While dealing with the case of multiple reflections suffered by a ray, the net deviation suffered by the incidentray is the : algebraiC; sy.m of deviation due to.each single reflection. So, 6total

= L 6!ndMdual

,_

""""""

' DO NOT FORGET TO TAKE INTO ACCOUNT THE SENSE OF I ROTATION WHILE SUMMING UP THE DEVIATIONS DUE TO SINGLE REFLECTION. 1'

TWO IDENTICAL PERPENDICULAR PLANE MIRRORS

If two plane mirrors are inclined to each other at 90° , the emergent ray is always antiparallel to the incident ray if it suffers one reflection from each (as shown in figure) whatever be the angle of incidence.

Fi-om figure, we observe 61 =1t-2a, 62 =1t-2P Also ray is rotated in same sense i.e., anticlockwise, so Snet =~i'=Total deviation =31 +l\

=>

6=2it-2(a+P)

Now in !J.OBC , LOBC + LBCO + LCOB = 180°

=>

(90'-a)+(90°-p)+0=1so 0

=> =>

a+P=0 6=,2it-20=360°~20

1.5

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Optics & Modern Physics

Advanced JEE Physics Alternative Method: 5 = LBEC + LCEA + LAED

Illustration 2 Find the co-ordinates of the location of the image formed for an object kept at origin as shown in figure. y·

Now, LBEC = LAED (vertically opposite angle) =>

LBEC=180°-2(o:+p)

=>

LBEC = 180° -28

Also, LCEA=2o:+2P

=>

{·: 8=o:+P)

=>

LCEA=2(0::1-P)=28

5=(180°-28°)+28+(180°-28°) 5.=360°-28

=>

REFLECTION FROM A PLANE SURFACE OR PLANE MIRROR -

When a real object is placed in front of a plane mirror, the image is always erect, virtual and of same size as the object. It is at same distance behind the mirror as the object is in front of it.

Solution The first thing we obsenre is that the object is virtual, because the ray of light is converging on plane mirror. Also, the co-ordinates of object are (0, 0, 0) and the image co-

ordinates are the reflection of object coordinates in the mirror as shown in figure.

0

"4------

d -----t+f-- d ----+t

14-- 8

d-d-

(a) Point Object

(b) Extended Object

LATERAL INVERSION

The image formed by a plane mirror suffers lateralinversion. That is, in the image the left is tu.med to the right _ and vice-versa with respect to object. Howev.er, the plane mirror .does not turn up and down, as shown in figure.

0 cm -----+1

The image lies on norma~ of mirror at I . From MOP, we have

=> =>

sin(30°) = PO 8 PO=4cm

OI = 2(PO) = 8 cm So co-ordinates of I are X

= -8cos(60°) = -4 cm,

y=8sin(60°)=4.J3 cm and

Z=O Image

Objeet

Image Actually, the plane mirror reverses forward and back in three-dimensions (and not left into right). If we keep a righthanded coordinate system in front of a plane mirror, only the z-axis is reversed. So, a plane mirror changes right-handed co-ordinate system (or screw) to left-handed. y

Objeet

y

So, the co-ordinates of image are (-4, 4../3, 0) ; PROBLEM SOLVING TRICK(S)

j For _finding, the :loC:Btion of an image of a point object placed in i front of. a plane· mirror, we must see the perpendicular distance··of

l the object f_rom the mirror.

I

·

-,,.! 0

J----•x 0

z Right Handed

=

System

'

¾

•-' --

x'+----to·

Coi-rect OM=Ml

z' Left Handed System

1.6

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Incorrect OM=Ml

I

-

.---•

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Ray Optics: Reflection at Plane & Curved Surfaces

=

FIELD OF VIEW OF AN OBJECT'

Suppose a point object O is placed in front of a mirror, then a question arises in mmd whether this mirror will form the image of this object or not. The answer is yes, it will form. A mirror, irrespective of its size, forms the ini.ages of all objects lying in front of it. But every object.has its own field of view for a given mirror. Field of vi_ew is the region where diverging rays· from object or image are present. If our eyes are present in field of view then only we can see the object or an image as the case may be. Field of view of image is decided by rays which get reflected or refracted from the extremeties or the extreme ends of the mirror or a lens and depends on the location of the object in front of mirror or lens.

MINIMUM SIZE OF A PLANE MIRROR TO SEE A COMPLETE IMAGE CASE-I: To find the minimum size· bf mirror to see a full image we use the fact that light rays from extreme parts of object should reach eye after reflection from mirror. Let us consider following two situations a) The minimum size of mirror to see one~s full height is H where H is the height of man. To see full image 2 mirror is positioned in such a way so that rays from head and foot reach eye after reflection from mirror, as shown in the figure.

--r(x+y)

y

_J __

f y

i

Field of view of image

E Man.

b)

Field of view of object

CONCEPTUAL NOTE(S) It _has been observed that a convex mirror gives a wider field of view than a plane mirror. Therefore, the convex mirrors are us~d as rear view mirrors in vehicles. Though they make the estimation of distances more difficult but still they are preferred because for a large movement of the object vehicle there Is only a small movement of the image.

A ray starting from head (A) after reflecting from upper end of the mirror (F) reaches the eye at C . Similarly the ray starting from the foot (E) after reflecting from the lower end (G) also reaches the eye at C . In similar triangles ABF and BFC AB=BC=x (say) Similarly in triangles CDG and DGE , we have CD=DE=y (say)

Now, we observe that height of the man is

2( x + y)

and that

the length of mirror is (x + y) , i.e., the length of the mirror is half the height of the man. Please note that the mirror can be placed anywhere between the centre line BF (of AC) and DG (of CE); CONCEPTUAL NOTE(S)

a) Field

Field

of

view of

convex , mirror

of

0

view of a

plane

b)

In· order to see full image of the man, the mirror is positioned such that the lower edge of mirror Is at height half the eye level from the ground. Minimum size is independent of the·distance between man and ·mirror.

mirror

CASE-ff: The minimum length of the mirror required to see the full image of a wall behind the man who is standing at the

middle of the mirror and the wall is ; , where H is the ·height of wall. The ray diagram for this situation is shown in figure. 1.7

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Optics & Modern Physics

Advanced JEE Physics A

·

f 2x

. 360°

Ca'.culate · · -8

t

i 1

(x+y)

.

360' is an INTEGER

(X;cY)

f

8

n·

. 360' is a· FRACTION 8

•.,

•

2y

:t Wall

C

Man I+-- d

ltis an ODD• lnteg"ef i-~~

IMsan EVEN! Integer

Mirror d-+t

In mangles HBI and IBC let HI= IC= x . Now, in mangles HBI and ABF , we have

\

~

AF HI

FB BI

AF

2d

n is an Integral' Part of fraction

Object lies symmetrically ' on th~ angle

Object lies . UnsYmmetrically

bisector

-;-=a

~

AF=2x _Similarly if, CK= Kf = y·, then DG = 2y. Now, we o]:,serve

that height of the wall is 3(x+ y) while that of the mirror is

(x+y).

·

360' n='_(· - - 1')

: . a

,----~~. ....-:-.-:-----__,.---' • CONCEPTUAL NOTE(S! . -

a)

If an object is placed between two•parallel mirrors (8 = 0°), .

·the number of im?Q_;s formed Wm,be infinite, ~b)

NUMBEROF IMAGES IN INCLINED MIRRORS

Let 0 be the angle between two plane mirrors and' n be the number of images formed:

360 if 360 is odd 8 ' · 8 Then n= (360 ) . 360. [ -1 , if 1s even

8

Further when

~=[(3:

1

_ ),

:o

3

,,''

w

'

M,

J.

J

360 is odd, then ·0

360

Further if

?1

8

0

a'

All th8·. images. lie on a,_clrcle with radius equal to the distanc~, "-~etween ~e object_,;~ and th0 point· of intersection the mirrors c·. • '• The riu(nber ·of images formed may b8 different from the numberfOf Images see_n (which depends on .the· position ·of,. the observer). · · · ·

1

------

:

a

1~ 0=9no,.~,., l"

~------6~

,.,.'I-

.

"..~., M; "

.l12 I,. _______ ,! __ '.'° ___ -,

if.object lies symmemcally on the

liz

angle bisec:to! of tw~ mirrorS

if object lies tinsymmemcally ROTATION OF A.PLANE MIRROR

is a fraction, then the number of images

form~d will be integral part of the fr~ctlon e.g.

il

:o

3

is 4.8,

then, n = 4 . Following diagram shows _the process to calculate n . Net deviation produced by two plane mirrors.inclined at an angle 0 is 6=360°-28 Clearly 6 is independent of the angle of incidence of the ray oflight. · · · ·

When a mirror is rotated by an angle 0 (say anticlockwise), keeping the incident ray fixed, then· the reflected ray rotates . by 28 along the same sense, i.e., anticlockwise. y(or N) N (ory)-

'. ' ''' ' . '' ' j,•j

N' ..

R

t

R'.

'' '' \' :...._,.~ ,, . \

•· 'I

\ i-0

,a•

,,

i-0~

'

Initially

=

On Rotation of Mirror

1.8

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Ray Optics: Reflection at Plane & Curved Surfaces

Let I be the incident ray, N the normal and R the reflected ray, then on rotation, I remains as it is, N and R shift to N' and R'. From the two figures we can observe that the reflected ray earlier made an angle i with y-axis while after rotating the mirror it makes the angle (i-20). So, we conclude that the reflected ray has been rotated by an angle 20.

d$ + 2 da =O dt dt d$ =~2 da dt dt So, the angular speed of the reflected ray is double the angular speed of the mirror. Since, y = xtan~ dy =xsec'$d$ dt dt

CONCEPTUAL NOTE(S) If a plane mirror rotates with angular velocity ro , then the

reflected ray rotates with angular velocity 2ro (excluding rotation

Since l~;l=2oo

of ~-~~r ~ith n?r~al ~s_t~e ~is).

=>

Illustration 3

A plane mirror hinged at O is free to rotate in a vertical plane. The point O is at a distance x from a long screen placed in front of the mirror as shown in figure. A laser beam of light incident vertically downward is reflected by the mirror at O so that a bright spot is formed at the screen. At the instant shown, the angle of incidence is 8 and the mirror is rotating clockwise with constant angular velocity o, , Find the speed of the spot at this instant.

Normal (N)

,/ ,,

Screen

2

l'Jil=(xsec $)(2oo)

So, the speed of the spot is

1:1

= 2xoosec' $

VELOCITY OF IMAGE IN A PLANE MIRROR

To understand and interpret the moving images of moving objects in front of plane mirror, we must understand the following cases. CASE!: Object moving along the normal to the plane mirror which is at rest. All velocities measured w.r.t. grOund frame. y

8 ,'

0 ·

------------X

V V 0___.,_ ______ ,-------....._I

.- Mirror ··-::.._.;

Lx

00

Solution Let P be the bright spot, shown on the screen. Let the distance of point P from 0 1 be y at this instant shown in figure. Then according to the problem we need to calculate dy dt

(v.= 0)

Velocity of object with respect to mirror is Varn= vi Velocity of image with respect to mirror is

vhn =-vi Velocity of object with respect to image is V01 =V0 -V1 =(2v)i CASEII: Object moving parallel to the plane of mirror (at rest)

i

!-------:- -_____ J

From the figure 0+0+$=90°

... (1)

8+$+a=90°

... (2)

=> =>

a=0 $+20=90°

=>

$+2a=90°

Lx

I

l (v. = 0)

Velocity of object w.r.t. mirror is

Dom =vj 1.9

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Advanced JEE Physics

Optics & Modem Physics

Velocity of image w .r. t. mirror is

Solution

= vj

Vrm

Velocity of object w.r.t. image is iio1

=0

CASE III:

Object moving neither along the normal nor alohg the parallel to the plane mirror (at rest).

1

o~-----Vo,

------~ 1 Vr,,,

The component of velocity of image perpendicular to mirror is

Lx

½=2Vm - \1

0

°" (v,L =2<-2)-(6)=-10 ms-1 For component of velocity of image parallel to the mirror

(v. = 0)

(V,)11 =8 ms-1

Clearly; we observe this case to be a combination of Case .I and Case II. So, here -

(v 01 ), =2v0 x and (v01 ), =0 · , PROBLEM SOLVING TRICK(S) ; While solving problems that involve·the calculation of image of an I object w.r,t. any observer, then .

' Step 1 : Firstly, calculate the velocity of image ·w.r.t. mirror : keeping in rl'llnd·that ·

'

( Vlm ) along l'Ofrror

=>

J

={ijOm)along mirror

; Since, both the object and the image approach the mirror with i equal and ORposite speed, so we have

(Vimtormaltomlrror = -(Vam)no,ma]tomirror

VI-Vm=-(Va-Vm)

=>

V1 =2Vm-Vo

=>

1

V, =-./100+64 =-./164 ms- and 0=tan-'(¾)

Illustration 5

(v,.)11 =(vem)u

=>

(v,): +(V,),:

Velocity of image(½)=

A point object is moving with a speed of 10 ms-I in front of a mirror moving With a speed Of 3 ms-I as shown in figure. Find the velocity of image of the object with respect to mirror, object and gro~nd.

t·-·

; Step 2 : Then the velocity of image w.r.t. mirror is

Solution

V,m =.(~1m\ +(V1m)1. , However, velocity' of image w.r.t. any· other observer, say A is

: then given by f

I

,-<:ic/--

5./3 ms-'

VIA=V;-VA

t .

- ~ 5 ~.-, ----10 ms-

Illustration 4

Find the velocity of image of a,moving particle shown in figure.

10

ms-'.,{

v0 =(-s-./31-sJ) ms-1

Velocity of mirror, VM

=3i ms-I

For component of velocity perpendiCular to mirror, we have

(vlM), =-(VoM)_,_ =-(Vo-VM)

/_:\.53°

/ ,,

.

I

't

Velocity ofobject,

, /

1

3ms-'

=>

(v™)_,_ =-(-s-./31-31)=(5-./3+3)1 ms-1

For component of velocity parallel to mirror, we have

(iit~)1 -(v0M)1 = v0

-iiM

=

-5}-o =-5}

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Ray Optics: Reflection at Plane & Curved Smfaces Since,

=>

=>

(vlM)=(v,Mt +(v1M)

2. 11

(v,M)=(5,J3+3)i-5f

A ray of light travels from paint A to a point B after-being reflected from a plane mirror as shown in figure. From where should it strike the mirror?

B t

'' '' ''' '' '

(s,/3+3)1-sJ=v,-(31) A

=>

(s,/3+3)i+31-s]=v,

=>

v, =[(s,/3 +6)1-sJ] ms-'

=>

v,0 =(5,/3+6)1-sJ-(-s,131-sJ)

=>

v10 =(5,/3 +6+5,J3)i +(5-5)]

=>

v10 = (10,ffi + 6)1 ms-1

t

5cm

'

1+- 20

3.

20cm

cm ---+1

A plane mirror is inclined at an angle 8 =60° with horizontal surface. A particle is projected from point P on the ground (see figure) at t =0 with a velocity v at an angle ct with horizontal. The image of the particle is observed from the frame of the particle projected. Assuming the particle does not collide the mirror. Find the time when image will come momentarily at rest with respect to particle.

Illustration 6

A plane mirror in y-z plane moves with a velocity

-3i

as shown in figure. An object O starts moving with a

velocity 4/ + J- 4k . Find the velocity of the image.

V

0

a p

GROUND

4.

Two plane mirrors are inclined to each other 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. (a) Find the angle between the two mirrors. (b) Also calculate the total deviation produced in the incident ray due to the two reflections.

5.

Two plane mirrors M1 and M2 _ are inclined at angle 0 as

0•

Mirror (M)

Solution

e

Since the mirror is placed in y-z plane, so the y and z components of the velocity of the image remain the same as that of the object. However, perpendicular to the mirror, the velocity of approach of object towards the mirror is always equal and opposite to the velocity of approach of the image towards the mirror, so, we have

shown in figure. A ray of light 1, which is parallel to M1 strikes M2 and after two reflections, the ray 2 becomes parallel to M2 • Find the angle

M,

e. 2

(voM); =-(v,M), =>

(v 0 ),-(vM), =-(v,), +(vM),

=>

(v 1 ), =2(vM),-(vo).

=>

(v,), =2(-31l-41 =-lOi

So,

6.

v, =-10i + ]-4k

Calculate the deviation suffered by an incident ray in the situation shown in figure after it suffers three successive reflections.

M,

~

ICE I c;;l

1'

'

j

•.. 50°

· · ~~.

BASED ON REFLECTION AT PLANE SURFACES. (Solutions on page 1.169) :

1.

A ray of light travelling in the direction

i(i + .Jsj)

is incident

30°

M,

on a plane mirror. After refleciion, it travels along the direction

iU -.Jsj).

Find the angle of incidence.

7.

Two plane mirrors are placed parallel to each other and 40 cm apart. An object is placed 1O cm from one mirror. 1.11

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Advanced JEE Physics Find th8 distance frolTl the object tO the iesp0ctive image for each of the five images that are closest to the object. 1

a.

13.

A ray of liQht is incident on 8.11 arrangement of two plafle--: mirrors inclined at an angle 0 with each other. It suffers two ' reflections one from each mirror and finally moves in a , direction making angle u with the incident ray ( a is acute). , Find the angle a and show that it is independent of angle 1 of Incidence.

14.

A ray of light is incident at an angle of 30" with the horizontal. At what angle with horizontal must a plane mirror ! be placed in its path so that it becomes' vertically upwards after reflection? '

Find the number of images formed of an object O enclosed , by three mirrors AB , BC , AC having equal lengths ln ·

situation shown in figure.

A

6

!

0

, 9.

..

· ··c

s··

15.

A point l\>Ource of light S , placed at a distance L in front of , the centre of a mirror of width d, hangs vertically on a wall. A man Walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. Find the greatest distance over which he can see the image of the light · source in the mirror.

l }---------1

A ray of light is incident on a plane mirror along a vector j T+ J- k . The normal at the point of incidence is along

Find the smallest size of a looking glass which a man with a : face 24 cm x 16 cm should purchase that will enable him to see his·whole face completely, lf the (a) man is one eyed.

(b) man is two eyed. Given that the separation between his eyes is 8 cm . In what direction should A beam of light is to be sent from point A (shown in figure) contained in a mirror box for it to fall onto point B after being reflected once from each of the four walls. If the points A and B are in one plane perpendicular to the walls of the box (i.e., in the plane of the , drawing) then in what direction should the beam be sent , fromBtoA?

A small curved reflecting surface can be considered to be a part of a sphere. Hence, such surfaces are called spherical mirrors. Depending upon the surface silvered, these are of two type&--eoncave and convex, as shown in figure. Some important terms are described below. a) Pole or Vertex: Centre P of the surface of the mirror. b) Centre of Curvature : Centre C of the sphere. c) Radius of Curvature : Radius R of the sphere. d) Principal Axis : Line PC , joining the pole and the centre. e) Linear Aperture : Distance XY between the extremities of the mirror surface. Note that Sil).ce lenses are also made of spherical surfaces, the above terms also apply to lenses, except that the pole is replaced by a new term called as Optical Centre.

•A

Silvered

surface

•B

112.

ap surface

10 ms-1

/ ----od3-°~------0bject

Mirror

Lx

axis

y , , ',

R

A

/ y

' , Concave'-- .. ____ ... _.,.,. Convex mirror mirror

1

------- 30° --

_f_ri_!l_g!J)_aJ

/I~

Silvered

The object and the mirror move with velocity shown in , figure. Calculate the velocity of the image. 5ms

I

REFLECTION FROM CURVED SURFACES

i+----2L----+I

11.

16.

i + j . Find a unit vector along the reflected ray.

i.-L_..

' 10.

1

Two plane mirrors are inclined to each other at an angle of 70" . A ray is incident on one mirror at an angle 8 . The · ray reflected from this mirror falls on the second mirror from where it is reflected parallel to the first mirror. Find the value 9.

Important Terms and Definitions a) Centre of curvature : It is the centre of the sphere of which the mirror/lens is a part. b) Radius of curvature : It is the radius of the sphere of which the mirror/lens is a part. c) Pole : It is the geometrical centre of the spherical reflecting surface of which the mirror/lens is a part.

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· Ray Optics: Reflection at Plane & Curved Surfaces d) e)

f)

Principal axis (for a spherical mirror) : It is the straight line joining the centre of curvature to the pole. Focus : When a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on the surface of a mirror (lens), the reflected (refracted) beam either converges to a point or appears to diverge from a point on the principal axis. Th.is point is called the focus (F) . Focal length (for a mirror) : It is the distance between pole and the principal focus (F) .

Real image : If reflected (or refracted) rays converge to a point (i.e. intersect there), then the point is a real image. h) Virtual image: If reflected (or refracted) rays appear to diverge from a point, then the point is a virtual image. i) Real object : If the incident rays diverge from a point, · then the point is a real object. j) Virtual object : If incident rays converge and appear to intersect at a point behind the mirror (or lens), then the point is a virtual object. g)

SIGN CONVENTIONS FOR MIRRORS

While solving problems, we must follow a set of sign conventions given for convenience. According to this sign convention a)

Origin is placed at the pole ( P) .,

b) c)

All distances are to be measured from the pole ( P) . Distances measured in the direction of incident rays are taken as positive. Distances measured in a direction opposite to that of the incident rays are taken as negative. Distances above the principal axis are taken as positive. Distances below the principal axis are taken as negative. This sign convention is used to find the position and nature (virtual or real, erect or inverted) of the image formed by the mirror (or lens). Object distance is denoted by u , image distance by v , focal length by f and radius of curvature by R.

d) e) f)

g)

h) i)

PARAXIAL RAYS

Paraxial rays are the rays which are either parallel to the principal axis or make small angles with it i.e., these rays are nearly parallel to the principal axis, Our treatment for the spherical mirrors has been restricted to these rays and due to this we shall be considering the curved mirrors that have smaller aperture. However, for the sake of convenience, comfort and clarity, we shall be drawing the diagrams of larger size.

Note that generally we keep the object to the left of the mirror (or lens), so that the ray of light starting from object must go from left to the right i.e., towards positive direction of x-axis. Now· since the distances have to be measured from the pole consequently, u must always be negative, v is positive (for a virtual image) and negative (for a real image). / is positive (for a convex mirror) and negative (for a concave mirror). For both the mirrors and lenses. Magnification for a real image is negative i.e., mreal = 0

FOCUS AND FOCAL LENGTH

Magnification for a virtual image is positive i.e.,

When a narrow beam of light, parallel to the principal axis and close to it, is incident on the surface of a mirror (lens), the reflected (refracted) beam is found to converge to or appears to diverge from a point on the principal axis. Titis point is the focus also called Principal Focus in case of mirror(s). The plane passing through the focus and perpendicular to the prigcipal axis is called focal plane.

mvmua1

=EB CONCEPTUAL NOTE(S)

The convention that all distances measured along the ray of light are ,positive and all distances measured opposite to the ray of light are negative matches exactly with the Cartesian coordinate system, where we can simply place the origin at the pole P and say that all distances to the left of the pole are negative, all distances to the right of the pole are positive, all distances above the pole are positive and all distances below the pole are negative.

-

C

+

+ '

Incident Ray:

' ''P

+-----.~---+ Concave mirror

Focal length

(f)

Convex mirror

'' ''

is the distance of focus (F) from the pole

''

t'

(p) of the mirror or the optical centre for a lens. j)

For solving problems in which any of u, v, f(or R) is to be found, we must make ·sure that no convention should be applied on the quantity to be found. The

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Optics & Modern Physics

Advanced JEE Physics tmknown quantity will automatically take up its sign from which we shall make obvious conclusion. The diagrams show the application of sign convention to curved mirrors.

k)

.--, 1(±)1

·---·-------------c F

------,

:'...t:'

\~_~:

• _)

b)

F

C

,-i-·

'

' ce-~.J --r---

' :+--t ~'

'' '' ''8''

,-------

' f=.B.:'

·...~)

2 ,'

'\

• _)

frincipal Axis

e'

p•

F

C

-,---_-_-_-_-,----;::>+\·-_--~...--i,--i---],-+--

r------,

'f - R '' '1

-----l~_i_j r

I

-2 ,'

t§:}_~] '

'

:+- f ~

.,._

---

·--------- F----c

P.is Pole, F-isfocus and C is·Centre of Curvature 1

--- .......

', ..... _ ----r----c-·

I

...... :(£)1

(

',

p -------p

A ray of light passing through the focus (in case of concave mirror) or appearing to pass through the focus (in case convex mirror) is reflected parallel to the principle axis.

!+--- R

:... t:

'._-- ...

c)

',::-------~ ... -..,.... __

--- '..---F

C

A ray of light passing through the centre of curvature falls .normally on the mirror and is therefore reflected back along the same path i.e., retraces its path.

'8' ,_ -_ ! - .

R ----.:

----- ---- ........ __ _ F

RULES FOR OBTAINING IMAGE

These rules are based on the laws of reflection, i.e: the angle of incidence equals the angle of reflection, i = r and are used to find the location, nature (real or virtual, inverted or erect) and size of the image formed by a spherical mirror. Take any two rays coming from any given point on the object. Find out at which point these rays actually meet (or appear to meet) after reflection from the mirror. This point is the real (or virtual) image. In this way, taking one point after another on the object, the entire image can be constructed. a) A ray of light coming parallel to principal axis, after reflection passes through the focus (in case of co:ri.cave mirror) or appears to come from the focus (in case of convex mirror).

d)

C

Incident and reflected rays at the pole of a mirror are synunetrical about the principal axis. (Because for the pole principle axis acts as normal and by Laws of Reflection i = r ). So by observing the size of erect image in a mirror we can decide the nature of the mirror i.e., whether it is convex, concave or a plane mirror.

M

M

M"

M'

IMAGE FORMATION BY CONCAVE MIRROR

I

OBJECT POSITION_

At infinity

DIAGRAM

POSITION OF IMAGE

~

/ C

At the principal focus ( F)

,p

F~

or in the focal plane

~

=

NATURE OF IMAGE

1.14

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Real, inverted and

Beyond C

Between F and C

At C

At C

Real, inverted and of same size as the object

Between F and C

Beyond C

Real, inverted and magnifi~d

At F or in the focal plane

At infinity

Between F and P

dirn)nished

Real, inverted and highly , magnified

Behind the mirror

Virtual, erect and magnified

I

IMAGE FORMATION BY CONVEX MIRROR

I

OBJECT POSITION

For all positions of object

DIAGRAM

--0

F

C

POSITION OF IMAGE

NATURE AND SIZE OF IMAGE

Images formed between the Pole and the focus (F) .

AIWays forms a Virtual, Erect and Di_minished Image

p~

RELATION BETWEEN FOCAL LENGTH (!) AND RADIUS OF CURVATURE (R)

A ray parallel to the principal axis passes through the focus (as in concave mirror) or appears to pass through the focus (as in convex mirror). The normal to the mirror(s) •at the point of reflection i.e., A must pass through the centre of curvature. In triangle CAN , we have

tani= AN NC

For paraxial rays and_mirrors of small aperture, we have ~ieei=·AN NC In triangle FAN , we have

... (1)

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Optics & Modem Physics P=a+i

tan(2i) = AN NF

and y=r+P

CONCAVE MIRROR

--

f .........

..... - 1C

--

..... ...i1 2i

A,-

--!~

,F

' ''

--

'' ' '''

- --A f''.::--....

CONVEX MIRROR

~

·,,1

pp p(

-r-

' ... , .................. _ 2·' I

- ....I

F,

'' '

'

'' ' t,t-- f----...:

R

R

--- ,c

... (2)

From (1) and (2), we get

2 (AN)=AN NC NF 2 1 => - - - ... (3) NC NF Since, aperture is small, so N coincides with P, so we have NC e; PC and NF" PF For convex mirror, we have

=>

,1

' ''' ''' _ _ _ V ----+I ,

'' ' ''' '' '

Again for paraxial rays and mirror of small aperture, we have

tan(2i)" 2i = ~~

0

o+----R

i+------u-------->< Since by Laws of Reflection, we have i=r =>

a+y=2P

Applying paraxial ray approximation, we get AP AP AP tana=a=-, tanP.,P=- and tany=y=. PO PC PI => tana+tany=2tanp =>

AP+ AP = 2 (AP) PO PI PC

Using sign conventions, we have· PO=-u, PI=-v and PC=-R

1 1 2 --+--=--

PC=+R and PF=+f

=>

t=¾

R Since we know that / = - , so we get

(-u) (-v) (-R)

2

For concave mirror, we have 1 1 2 1 -+-=-=u v R f

PC=-R and PF=-f

f=~ 2

So, for a curved mirror of small aperture, focal length is half the radius of curvature. MIRROR FORMULA

Consider a point object O placed on the principal axis of a concave mirror. A ray of light, incident on the point A at an angle of incidence i on the mirror. makes an angle r with the normal as sho\\ n in the figure. From the Laws of Reflection we know that i = r . Further to find the location of the image let us take another ray along the principal axis so that it hits the mirror normally at the point P to reverse its path and meet the other ray at I . This point of intersection of the two rays happens to be the place where the image is formed. Since from geometry we know that in a triangle, external angle equals sum of internal opposite angles, so for triangle CAO and triangle CAI , we have 1.16

For Convex Mirror Similarly we can drive a formula for a convex mirror. Since from geometry we know that in a triangle, external angle equals sum of internal opposite angles, so for triangle CAO and triangle CAI, we have i = a.+y

For Concave Mirror

=

(Mirror Formula}

and P=r+y

0

p

' ' 13" -- --... _ ___c r--·-lr'-<_

~

--

1,

'' ' ''' o+--v-'

i+-----u--------

R----•

Since i = r , so we get -a+P=2y

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Applying paraxial ray approximation, we get -tana+tanP=2tany =,,

_AP+ AP PO PI

=

For spherical mirrors positive value of m means v and u are having opposite signs i.e. when u is negative v is positive and vice versa. So for a real object if the image formed is virtual, erect and three times the size of the real object then, we have m = +3.

2(AP) PC

Using sign conventions, we have PO=-u, PI=+v and PC=+R 1

1

2

=,,

---+-=(-u) v _R

=>

1 1 2 1 -+-=-="

1n=--. 3

[Mirror Formula)

I

R

V

Similarly for a real object if the image formed is real, inverted and one third the size of the real object then 1

Interestingly, the mirror formula is the same irrespective of the mirror used. NEWTON'S FORMULA

If instead of measuring the object distance and the image distance from the pole, the distances are measured from_ the focus, then we get a modified mirror formula. This modified mirror formula is called the Newton's Formula. Let ~ 1 be the distance of object from focus and

Solution The rough figure indicating the pole of the mirror, focus, and the given distances is shown. The sign convention is also given. Since, u =-15 cm (negative since it lies to the left of O )

f = -10 cm (negative since it lies to the left of O ) Since we have, from mirror formula that 1 1 1 -+-=-

x2 be the distance of image from focus, then u:::: f

Illustration 7 An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm . Describe the size, nature and position of the image formed.

+X1

v

and v=f+x2 =,,

According to the mirror formula, we have 1 1 1 -+-=v u f 1 1 1 ---+---=(/ +x1 )· (f +x,) f

=,,

(2/ +x1 +x,)f=(f +x1 )(f +x,)

=,,

2/ 2 +(x1 +x,)f = / 2 +(x1 + x,)f +x1x 2 X1X2

f

1111 1 11 -=---=-----=--+ v f u (-10) (-15) 10 15

Object

_____!____________ _

p

C

F

' ''

This i; known as Newton's formula.

: '

To have an idea of the relative size of the image and the object, we define linear magnification also called as lateral magnification as size of the image h m=-----~- 2 size of the object h1 For both concave and convex mirrors, it can be shown that V

m=--

u

. knowth 1 1 1 so we get Smcewe at-+-=V

u

f-v

u

f-u

I

I

+

0

+

Sign convention

i.-10 cm-.i'

LINEAR MAGNIFICATION OR LATERAL MAGNIFICATION OR TRANSVERSE MAGNIFICATION

f

5

150

150 V=--=-30cm

= f2

v

-15+ 10 150

5

=,,

=,,

=,,

u

1-4-------15 cm------+1'

The negative sign for v shows that the image lies to the left of O. Now, the magnification is given by V -30 m=--=--=-2 u -15 The negative sign for m indicates that the image is inverted, and hence real and is double the size of the object. Thus, we find that the image is real, inverted, twice the size of the object, and is formed 30 cm in front of the mirror. The ray diagram is shown in figure.

m=--=--=--

==================================== = 1.17

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--------------•--c

p F

'

(

O

C

~'-----------------¥;/: Image

14-1 O cm---:

...-15 cm--------ti-t i + - - - - - - 3 0 c m - - - - - -'

V 60 . . Ma gnif1cation, m=--=--=-2 . u 30

Illustration 8

A beam of light converges towards a point O , behind a convex mirror of focal length 20 cm . Find the nature and position of image if the point O is (a) 10 cm behind the mirror (b) 30 cm behind the mirror Solution (a) Here, in this case the object is virtual. So, for this we have u=+lOcm, /=+20cm

.. farmula, -+-=-,we 1 1 1 U smgrrurror get V

=>

1

1

1

V

f

U

1 20

U

1 +10

f

1-2 20

-1 20

· Hence, the image· formed will be virtual, inverted and enlarged, and at a distance of 60 cm behind the mirror. ------ - ·-- -----·---

__ _______ ,,

CONCEPTUAL NOTE(S) Note that for the real objects, a convex mirror always gives virtual and diminished image, but for virtual objects it gives real image if u < f and virtual image if u > f • LONGITUDINAL MAGNIFICATION

MAGNIFICATION

OR

AXIAL

When an object is placed with its length along the principal axis, then instead of defining the linear magnification we define the axial magnification. Mathematically we define axial magnification, for small objects as Size of image along principal axis m...i Size of object along principal axis

v=-20 cm

a 1cation, m =-- =- - - = ---

M gnif

.

.

(-20) 10

V

u

2

1 1 1 . Further since we know that -+-=- 1.e., V

U

f

Talcing the derivative of this equation with respect to u , we get

:Jv- )+ :u(u-')= :Jr') 1

=>

=> Hence, the image formed will be real, erect and enlarged, and at a distance of 20 cm in front of the mirror. (b) Again, in this case too the object is virtual. So, we have u =+30 cm, f =+20 cm .. farmula-+-=-, 1 1 1 we get U smgnurror

1

=

1

1

1

f

V

U

1

3-2

•

_

dv

-v 2 - - u

du

_2

=0

m-=::=-:: =-(f~uJ =-(ffvJ

AREAL MAGNIFICATION

Areal magnification is simply defined as the ratio of area of image to the area of object, so

m• .Area ofimage =

1

Area ofobject

A=!C=(_l_)' Ao u' f - u

=(f-v)' f

RELATION BETWEEN OBJECT AND IMAGE VELOCITY vfu2030 60 60 FOR CURVED MIRRORS => v=60 cm According to the mirror formula, we have 1.18 ==================================

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1

1

When the object lies between F and P , then the image becomes virtual i.e., u and· f are negative while v is positive. So from mirror formula we get,

-+-=v u f =>

v-1 + u-1 =1-1 = constant

1 v

Differentiating with respect to time, we get

1 (-u)

1

-+--=--

(-!)

-v-2 dv -~-2 du =O

dt

~

dt

~

~:=-(::)~; dudt

Here

... (1)

is the rate at which the object distance

u is

changing i.e., it is the object speed if the mirror is stationary. Similarly, dv is the rate at which v (distance between

dt

image and mirror) is changing i.e., it is image sPeed if the mirror is stationary. So if at a known values of v and u, the object speed is given, we can find the image speed from the above formula.

.1 =>

1

1

---=u V f

-u-'(~;)-v-'(!:)=o (-~:)=(::)(-~:)

Now, when u is further decreased, v also decreases to keep 1 · - constant. So, -du.th - is e rate at w hichb. o ~ect 1s f di approaching towards mirror and (- ~:) is rate at which the image is approaching towards the mirror.

------::;1'

---+

C

,;,-"

/

,

p

-----"'~,..,.,,,--+ P

/

'' I

+-

""'oa+,

'' ''' ' 1'

I

''t+-- V -----+I

'>+----u----+<

Let us take the example for a concave mirror. Suppose the object is moved from infinity towards focus, then since u is decreasing therefore,

Further in this case we observe that the image is always enlarged i.e., v > u . Therefore, image speed is more than the object speed. Thus, the above entire discussion can simply be concluded as follows. CONCLUSION

-(du) = rate of decrease of u dt . =>

(

!~) = rate of increase of

{object speed) {image speed}

v

Further, when the object lies between oo and C, then v < u , {from equation (1)) Hence, when the object is moved towards the mirror, its image (which is real) will recede from the mirror with speed less than the speed of object. When the object is at C, image is also at C ~

When an object is moved from -oo to F , the image (real) moves from F to -oo and then when the object is further moved from F to P image (now virtual) moves from +«.> to P .

~--------------CONCEPTUAL NOTE(S) ·When the object Is either at centre of curvature C ot at pole P , the two speeds are equal. When the object is at pole, then due to the small aperture of the mirror, it car) be assumed, as if the image is being formed by a plane mirror.

----------·

FINDING COORDINATES OF IMAGE OF A POINT

If the coordinates of a point object (-x 0 , -y0 ) with respect to the coordinate axes shown in figure are known to us and the coordinates of image be (x;, Yi) then for finding the

V=U

x-coordinate, we use the mirror formula, acc-0rding to which 1

Hence, when the object is at C speed of image is equal to the speed of object. When the object lies between C and F then v > u

1

1

-+-=v u f 1

1

1

-+-=X; Xo f

so, the image speed is more than the object speed. 1.19

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=___Eg_

X

=>

So, image coordinates are ( - ~O, -1) cm

Xo-f-

i

7

I

y

I PROBLEM SOLVING TRICK(S)

i\

a)

(-Xo, -yo)•

_

II

___.,\,________.x

,

bj

JP

..!.+..!.=! V U f

For finding the y-coordinate, we apply the concept of magnification (m) , according to which, we have

=>

I EXAMPLE ! mirror of focal

'I

I A point object is placed at

Place the object to the left of the mirror (or lens), so that sign convention matches with the familiar sign convention in the coordinate geometry. Both for concave as well as convex mirrors, use ,he same mirror formula i.e .

(-40, 1) cm in front of a concave

length 5 cm having Its pole at the origin (0, 0). Assuming the principal axts to be along x-axis, flrid

Substitute the numerical values of the given quantities with proper sign (+ve or -ve) as per sign convention. I Though the SI unit of distance' is metre, it may be more ' d) convenient in some problems to take the given distances, in cm rather than in m. But then your answer too will be in cril. e) Do not ,-give any sign to .the quantity to be determined. In your answer, the unknown quantity will be obtained with its proper sign. In addition to the above hints, if you remember the following facts, it will ·help you. a) Since the object is always placed to the left of the mirrc;>r-so, u is always negative.

b) c) d)

the pOsitlon of th.e image form~d.

e)

SOLUTION The si!uation discussed in the problem is shown in figure. y-axis

f)

' g)

I

i h)

7cm \p --'------l-------__.~---x-axis

Ii ")

(0,0)

,/

I

l 1, ' , '' ..,__4ocm-1>-1 '

U

: c) I

1

(\

and m=-~

For a concave mirror, f is negatiVe. For a convex mirror, f is positive. A real image is formed in front of the mirror, so for a real image v is neg·ative.

A virtu·a1 image is formed behind the mirror, so for a virtual : image v Is positive. 1 A real Image is always inverted, so for a real image h· IS negative. ·A virtual image is always erect, so for a virtual image h is positive. · For the real image of a real object and the virtual image 6f. a virtual object, m is negative. For the virtual-image of a real object and the real image of a virtual object, .m is positive.

GRAPH BETWEEN

!

VERSUS

V

1

1

40

v = ----,. cm 7

. h1 V S1nce, m·=-=-h0

h0

Hence, the mirror formula i.e.,

U

"'.~

1

-40

1

I

1

1

1

v

u

I

=>

-+-=-

But h0 =7 cm

=>

-=--+-

0

=

1

v u

=>

h, =.-1 cm

- --···- --·--

1.20

---

u

I

----=--

~=-! h 7 ---

! + ,! =.!. becomes V

(-~o}

h,

U

Let us first take the case of a concave mirror. Here, two cases are possible. Case .1 : 'When the Image formed is Real. When the image is real, i.e., object lies between F and infinity. In such a situation u, v and f are negative.

1

-+----v (-"-40) (~5) =>

!

_____

_\

1 v

1 u

1

I

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1

1

1

1

1 1

v+ C---u) =t ~

f

-=,-+v u I

Comparing with y = mx + c , the desired graph is a straight line_

Do not confuse here, the slope m with magnification. 1/v

of-slope m = ,1 and intercept on y-axis equal·to

i.

The graph i$

thus shown in figure.

1/1~.

1/v

[/

I~

4 ---""~"---1/u 1/f

/ 1/f

Case 2 : When the Image formed is Virtual. When the image is virtual, i.e., object lies between l' and P . Under such situation .u and / are negative whil~ v is positive. The'mirror formula thus becomes 1 1 -v1 --u f

,/

.. ~45° ~....cc=---1-----------1/u PROBLEM SOLVING TRICK(S) a)

Comparing it with y = mx + c the desired graph is a straight . . .· .. . 1 line with slope m ': 1 andintercept on y-axis is equal to -

1

As focal-length of a spherical mirror f

=~- dep·ends only on

the radius of mirror and is independent of wavelength of light and refractive index of medium so the 'focal length of a spherica! mir_ror in air or water' and for rec;J or blue light is same. This is also why the image _formed by mirrors do not show Chromatic aberration.

.

1/v b)

In case of'spherical mirror if R---+ oo (i.e., it becomes plane), f=

...J._ _...,<..,4,;5,__'_ _ _ 1/u

~ = oo , the mirror formula

/1/f

111' 11 -+-=- reduces to-+-= O i.e., v =-U

//,.,,

V

-1/f ,.

The graph is thus shown in figure. The two graphs can be .drawn in one single graph as in figure. 1/v

U

f

V

U

c)

i.e., image is· at same distance behind the mirror as the· object Is infront of it. This in turn verifies .the correctness. of mirror formula. Every part of a mirror forms complete image. If some portion of a mirror is obstructed (say covered with, black paper), then complete image will be formed _but intensity will be reduced.

d)

In case of spherical mirrors ii object distance

.(x,)

and

image distance (x2 ) are measured from focus instead of

pole, u=(l+x,} and

..:!. + .:!. = ! V

u I

reduces

to

simplificatiori gives x 1x 2 e) CONCEPTUAL NOTE(S) Please note that

and

.!V

.!U

and

..!

are actually the magnitudes of

V

U

(i.e., without sign)

f)

For a convex mirror, the image· forrh8d,,is always virtual, i.e., u is always negative while v and f- are always positive. Hence, the mirror for.mul~~-~co!'JlBS,, _ _ _ _ _ _ _ _ _ __

v=(i+x,) 1 - --

~o the mirror formula

+ -1-

=

!

(l+x,) {f+x,) I

which

on

= f2 • This result is called 'Newton's

formula;. If an object is moved .at constant.speed towards a concave mirror from Infinity to focus, the image will ·move (slower in the beginning and faster later on) away from·the mirror. This is because, during the time the. object.move~ from oo to C the image will move from F to. C· and when the object moves from C to F the image wUI move from C to co • At C the speed of object and image will be equal. concave mirror behaves as convex lens (both convergent) while convex mirror behaves as concave lens (both divergent). This is shown in figure.

1-------- ~------------------1.21

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(21, 21). This all'is shown'ln figure. V

''

·F

2f

Convex lens

Concave mirror

U=V

I'

'' '

-+-'' ,.._-------___

- ,--1--------'

(a) Convergent behaviour

IL-L-...L-----+U

I 21

Illustration 9

A thin rod of length l_ is placed along the principal 3

Concave lens

·convex mirror

g)

h)

axis of a collcave mirror of focal length f such that 'its image, which is real and elongated, just touches the rod. What is the -magnification ?

(b).Divergenf behaviour As convex mirror gives erect, Virtual and diminished image, field of view is increased. This is Why it is used as rear-view mirror'in vehicles. Concave mirrors give enlarged erect and virtual image (if object is between F and P ) so are used by dentists for examining teeth. Further due to their converging property concave mirrors are also used .as reflectors in automobiles head lights aniJ' search lights and by ENT surgeons in ophthalmoscope" For .real extended objects, ifihe image formed by a sinQle mirror Is· erect it is always virtual and in this situation if the size of the image is

Solution According to the problem, the image is real and enlarged, the object must have been placed between C and F . Since one end of the image ju~t touches one end of the object so, this end must lie on C .' Let AB be the object and A'B' be its image, such that A and A' both lie at C, as shown in figure.

A

i

smaller.than object equal to object larger than object • the mirror Is convex. • the mlrror·ls plane • the mirror Is concave

s·

VP I

F O m<+1

m=+1

3

end B of the object from the pole P is

m>+1

For real extended objects, if the image formed by a single mirror is inverted, it is always real (i.El., m is ~ve) and ,the mirror.is concave. In this situation If the size of image is

i

smaller than object • object is between ooand C • 8.nd l,mage between F-and-C

i

,.__ _ _ _ _ 21 - - - - - I . t i Now, as the length of the object AB is l_, so the distance of

, i)

i

equal 'to object larger than object • object is at C • object is between • and image is at C C and F • and Image between Candoo ·

u,=-(PA-f)=~(2/-f) =-(¾)t The distance of the image of end B , v, , is calculated by using the mirror formula, 1 1 1 -+-=-

v u

f

1 1 1 -+--=v, -~f f 3

-1 <m
j)

I•

m=-1

m>-1

J

.In case of spherical _mirrors if we plot a graph between u and v the graph will be a hyperbola as foi' u = f , v = oo

~ _ _and for u = oo, v

= f.

p

i+-1/3 _,

• '' '

B

TL--===~--tfc.c~=::ii ·------~---------------

5

v, =-2.f Therefore, the size of the image A'B' is A'B' =Iv, 1-1 v, .

A line u =:__v will c~t.!his _h.Y~~~~~ at

:::::::J 1.22

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(f)!)

.. . A'B' 2 3 Now, magnification, m = - AB = -( =-

2 Illustration 11

Illustration 10

A gun of mass m1 fires a bullet of mass m1 with a horizontal speed -v0 • The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Find the speed of separations of the bullet and the image at the instant just after the bullet is fired from the gun.

Find the location, size and the nature of the image of an object of height 2 mm kept between two mirrors (as shown in figure) after two successive reflections, considering the first reflection at the concave mirror and then at the convex mirror. M,(f,

=15 cm)

M,(f, = 20 cm)

Solution

m2Vo

:(

2mm

P,

Let v1 be the speed of gun (or mirror) just after the firing of bullet. By Law of Conservation of Linear Momentum, we have

ii

!

= m1V1

I

=>

.\\

'

' ' 14--20cm-+-1

... (1)

1

' ' ' •'-----50cm-----'

Now, du is the rate at which .distance between mirror and

dt

Solution

bullet is increasing, so du

-=V1

dt

... (2)

+Vo

f-u

Since, we know that.1dvl =(__j_)'ldul dt dt m,

m, Vo,..__ . . . .

Bullet

... (3)

---v,

As asked in the problem, let us. first consider the reflection at mirror M 1 . Before executing the mirror formula, we must keep two things in mind. 1. The incident light must go from the object to the mirror and we preferably take it parallel to the principal axis. 2. All distances have to be measured from the pole of the respective mirror for which reflection is being considered. 3. All distances measured along the incident ray are positive and all distances measured opposite to the incident ray are negative.

{

Since, at the instant just after the bullet is fired from the gun, the bullet is actually very close to the pole of the mirror, so u ---t O and hence we get at that instant

:: =(f~uJ

?

=m'=l

So, from (2) and (3), we get dv

du

dt = dt = V1 + Vo

,'~ . l'

I

I,_

I

I

: :~ :

I·

I

: ' I I

I

i+-20cm-+iI I

... (4)

I

' I I

.

' I I

1+-20cm....,. I , I

'4-----50cm---------~ ' ' ' 10cm

where dv is the rate at which distance between image (of dt bullet) and mirror is increasing. Therefore, speed of separation of bullet and image will be,

v, =2(v1 +v0 ) Substituting value of v1 from equation (1) we get

Figure is just representative and not to scale

Now, for reflection at concave mirror M 1 , the incident ray from the object goes to left of object and object distance is measured towards right of pole ,Pi , so u=-20cm Similarly, / 1 = -15 cm Now, according to mirror formula, we have 1.23

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Optics & Modern Physics

Advanced JEE Physics·

1 v

1 u

Solution The ray diagram for the situation is drawn in figure (but not to scale).

1

-+-==>

1

t

1

1

;+ (-20) = (-15)

=> v=-60 cm Negative sign with v means that it is formed to right of pole P1 at a distance of 60 cm from P1 , ( 10 cm· behind M, ). V

=>

.

(-60)

m, =--;;=- -20 =-3

So, image ( 11 ) formed is real, inverted and three times size

of object i.e., 6 mm. This image

(I1 )

formed now acts as object for the convex

mirror. Further, this image formed is 10 cm to the left ~f P, and the incident ray from the original object goes to the right

M, 20 cm --+1 30 cm ---+1 50 cm - - - - - - ; M i+------60cm--------<M

1+-

for reflection at M 2 to take place, so .

1+-

u=+l0 cm

Similarly, / = +20 cm Applying the mirror formula,.!.+.!:.=.!., we get V u I

For reflection at concave mirror M 1 , .we have

1

=>

1 1 -+-=v 10 20 v=-20 cm

u=-20 cm

'!, =-15 cm

(-20) u 10 So, image (I) formed is virtual, erect and two times the size of object (here 11 ). Hence the size of I is 12 mm . So, finally I is formed at 20 cm in front of convex mirror M2 , with size 12 mm ; virtual and erect.

Since, .!:_ + .!:_ = .!.

Illustration 12 Find the co-ordinates of image of point object P formed after two successive reflections in figure, considering the first reflection at concave mirror and then at convex mirror. y f.:,=20 cm

=>

=>

V

V

m2 =--=---=2

f,=15cm P

1 1 1 -+--=-v, (-20) (-15)

=>

v1 =-60 cm

So, magnification m., = _.'.J. = - -GO = -3 (Inverted) u -20

~O-------E-\--'--_.X 2mm

M,

M, t+-20 cm_.,.

A'P' =m, (AP)=3x2=6 mm

For reflection at convex mirror M2 , we have u=+l0cm

f, =+20 cm Since.!:.+.!:.=.!.

'

a

=>

V

u

1

1

I

1

-+-=v, 10 20 v 2 =-20 cm

gnif' . v, (-20) . 1cation, m2 = - - = - - - = 2 Agam, ma u 10

=>

C'P'=m,(CP')=2x8=16rnrn

So, the co-ordinate of image of point object P as measured from the origin O is (30 cm, -14 mm)

1+-----SOcm-------iM

=

I

=>

~

-+

u

1.24

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Ray Optics: Reflection at Plane & Cttrved Sttrfaces (b)

BASED ON REFLECTION AT CURVED SURFACES

(c)_

(Solutions on page 1.173)

1.

2.

3.

An object -of height 2.5 cm is placed at a 1.5 f from a concave mirror, where f is the magnitude of the fcical length of the mirror. The height of the object is perpendicular to the principal axis. Find the height of the image. Is the image erect or inverted?

the. ball· at time t =

X=O

8.

0

•

A

6.

B

• I State whether it is a convex mirror or a concave mirror. (b) ,Draw a ray diagram to locate.the mirror.and its focus. Write down the steps of construction of the ray diagram. Consider the possible two cases: (i) When distance of I from AB is more than the distance of O from AB and (ii)

When ·distance ·of o· from AB is more than the distance of I from AB

9.

Convex and concave mirrors have the same radii of curvature R. The distance between the mirrors is 2R. At what point on the common optical axis of the mirrors should a point source of light A be placed for the rays to converge at the point A after being reflected first on the convex and then on.the concave mirror?

1D.

An object ABED is placed In front of a concave mirror beyond centre'of curvature C as shown in figure. State the shape of the image.

A concave mirror has a radius of curvature of 24 cm . How far is an object from the mirror when the image formed is (a) virtual and 3 times the size of the object. (b) real and 3 times the size of the object and

real and

An Image I is, formed· of a point object 0 by a mirror whose principal axis is AB as shown-in figure.

(a)

Find the distance of Object from a concave mirror of focal length 1O cm so that image size Is four times the size ·o_t the object.

(c)

T is time period of

C x~axis+----------H

' \ 1 ~m __________ _

5.

, where

oscillation?

A mir/'or (in a laughing gallery) .forms an erect image fbur times enlarged, of a boy standing. 2.5 m away. Is the. mirror concave or convex? What is its·radiu~ of curvature?

A concave mirror forms the real Image .of a point source lying on the optical axis at a distance of 50 cm from the mirror. The focal length of the mirror is 25 cm . The mirror is cut' in two halves and these halves are drawn apart at a distance of 1 cm in a direction perpendicular to the optical axis. How will the images formed by the halves of the miriOr be arranged?

T

2

(! __+____________ _ 4.

to the image ofthe swinging bait At what point does the ball appear to coincide with Its image. What will be the lateral magnification of the .image of

¾the size of the object?

A thin flat glass plate is placed in front of a convex mirror. At what distance b' from the plate should a point source of light S be placed so that its image produced by the rays reflected from the front surface of the plate coincides with the,image formed by the rays reflected fi'om the mirror? the focal length of the mirror is f = 20.cm and-the distance from the plate to. the ·mirror a= 5 cm . How can the coinciderice of the images be established by direct observation?

-1-B_,_E_____---J\P A

,,.

D

C

F

I

An_ object is 30 cm from a spherical mirror, along the central axis. The absolute value of lateral magnification of an inverted image is

i.

Find the focal length of the mirror?

-a-----b----+t 12. 7.

A ball swings back and forth- In front of a concave mirror. The motion of the ball is described approximately by the equation x =fcos(rot), where f is the focal length of the mirror and x is measured along the axis of mirror. The origin is taken at the centre of curvature of the mirror. (a). Derive an exRression for the di_stance from the mirror

A thin-rod of length

½is placed along the principal axis of a

concave mirror of focal length f such that its image just ,__ _,,toLJchesJhl;! rod. Calc_ulate magnifica_te;io,..n~._______...,

1.25

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REFRACTION OF LIGHT AT PLANE SURFACES

T

he phenomenon of the bending of light rays as they travel from one medium to the other is called . Refraction. The surface separating two media is called an Interface. In other words, the phenomenon of bending of light rays at the boundary between two media is called refraction. Incident Ray N

A

: Normal Medium 1 (µ,)

i'

,

Interface

, ,

,

Q1

1r

~

~

~

1,Medium 2 (µJ

:

-!,

-!, I -t,

ABSOLUTE REFRACTIVE INDEX

-j

-t,

Refracted Ray

LAWS OF REFRACTION

a)

b)

REFRACTIVE INDEX (RI)

The refractive index of a medium is not determined by its density. It is,govemed by the velocity of light in the medium. The lesser the value of the velocity of light, the more is the refractive index of the medium, and the denser. is the medium. A medium having greater refractive index is called denser medium whereas the other medium is Called rarer medium.

1' -t>

B ~

is simply called as the Absolute Refractive Index of medium 2, expressed as µ 2 or simply µ .

The incident ray, the refracted ray and normal at the point of incidence to the surface separating the Mo media all lie in the same plane. Snell's Law For two media, the ratio of sine of angle of incidence i to the sine of the angle of refraction r is constant (for a beam of particular wavelength). For a given set of media this constant is called the refractive index of the medium 2 with respect to medium 1 (represented as 1µ2) i.e., sini µ -.- =constant= -2 = 1µ 2 smr µ1

The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium, speed of light in vacuum C >1 µ speed of light in medium V Absolute refractive index is more than one because the speed of light is maximum in vacuum/ air. RELATIVE REFRACTIVE INDEX

The relative refractive index of medium 2 with respect to medium 1 is denoted by 1 µ 2 and is given by

1

_

µ, _(

µ, -;,--( (SNELL'S LAW)

:J:J-_ v, v,

The relative refractive index of medium 1 with respect to medium 2 is denoted by 2 µ 1 and is given by

OR µ 1 sini=µ 2 sinr where µ 1 and µ 2 are Absolute Refractive Indices of Medium 1 and 2 respectively and 1 µ 2 is the refractive index of medium 2 with respect to medium 1. If medium 1 happens to be the vacu~m, then the constant

2

:J:J

µ, ( v, µ,=-=-(-=µ, v,

===================================== = 1.27

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Optics & Modern Physics

Advanced JEE Physics CONCEPJ"UAL NOTE(S) a)

1

The velocity of light in air Is nof.much different from that in vacuum. Hence, while defining the refractive index of a

me_dium we often take velOcity of light in air rather, than that in vacuum Medium

Refractive Index'"'

Water

.±=1.33 3

=> =>

sini < sinr

i'< r

REFRACTION : IMPORTANT POINTS

' a)

Whenever light goes froni. one medium to another, the frequency of light (/) remains unchanged. Since µ=

b)

V

, !=1.50

Glass

2

Relative refractive index can be less than one. If we calculate the refractive index afwater with respect to glass, then

where A-., and "modi= being wavelengths of light in air and medium respectively.

(34)

8 ' 9µ =·µw =-=-<1

• c)

(¾)

µ,

C

g

sini

µ2 sinr - µ

Refractive index is different for different wavelengths for a

---~~r_:ifmedia,because µ 1A1 =µi. 2 -

1

'' i'' 'r

"-1 -

"-

2

2

µ1"-'1 = µ2"-2

o

µ;\,

=constant

Also, we conclude that µ1V1 =µ2V2

Air

b)

Water

=>

µv = q:mstant

2

1

µ1 x µ 2 =1

'

2

I'

1 µ1=-

'µ,

c) According to Snell's Law µ 1 sini = µ 2 sinr

,.

si~i=~>l sinr µ 1 ·

b)

v1

c - v

(MODIFIED FORM OF SNELL'S LAW) From above we conclude that

µ,

=>

-

V1

When light passes from rarer to ,denser medium, it bends towards the normal i.e., a·light ray passing from air to ·water bends towards the normal as shown in the figure.

f',

v2

_ _ _ _ _ _ _ _ __

BENDING OF A LIGHT RAY

a)

Speed of light in vacuum Speed of light in medium

C

When light propagates through a series of parallel layers of different medium as shown in the figure, then the Snell's Law may be written as µ 1 sin0 1 = µ 2 sin0 2 = µ 3 sin03 = µ 4 sin04 = constant In.general, µsine= constant

sini >sinr

=> i > r When light passes from denser to rarer medium it bends away from the normal as i.e., a light ray ·passing from water to air bends away froffi the normal shown in the figure. Air µ,

Water

d)

If light is incident normal to a boundary (i.e. i = 0 ), then, it passes undeviated from the boundary as shown in the figure. µ,'

According to Snell's Law µ 1 sini=µ 2 sinr

=

sini =~
µ2* µ1 Condition for no refraction

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Ray Optics: Refraction at Plane Surfaces

e)

·If the refractive indices of the two media are equal as shown in figure, then also the light ray is not refracted and the boundary between the two media is not visible. This is why a transparent solid is invisible in a liquid of same refractive index.

C

V

n(1+;J 0

dt

dx

n,(1+;.)

=1500 ms-1

'

l'

' :® ' 0 •x x-no.t:---::-,-_..-:c--,x~-~a G)

Illustration 15

Therefore, the refractive index of water with respect,to air, for sound waves is µw

V

t = 5n0 a 4c

speed in air, v 1 = 330 ms-1

,

1,

C O

0

Note that for sound waves, speed in water, vw

''

fdt=~J(1+..::..)ax 2a

Condition for no refraction

fJ

-----T-------~

C

dt Jl:?=µ1 =µ

y

dx sowe h ave v=-,

. Smce,

= !!,_ = 330 = 0.22 Vw 1500

For the arrangement shown in the figure, a light ray is incident at an angle of 60° on the layer of water. Find the angle betweeh this ray and the normal to the glass.

''

6001

Air(µ,,= 1)

Thus, we find that for the refraction of sound waves, water is rarer than air.

Water (µ, = 4/3)

Illustration 13

A ray of light falls on a glass plate of refractive index n = -./3 . What is the angle of incidence of the ray if the angle behveen the reflected and refracted rays is 90° ? Solution

µ 0 sin(60°) =µ 1 sinr1

According to Snell's Law sini n=-sinr Since i+r=90° ~ r=90-i

-./3 =

sini sin(90-i)

Solution According to Snell's Law, we have

= µ2 sinr2

µ 0 sin(60°)

µ,

tani

i = tan-1 (-./3) = 60° Illustration 16 Illustration 14

A ray of light passes through a medium whose refractive index varies with distance as

n=n + ;a) .If 0(

1

ray enters the medium parallel to x-axis, what will be the time taken for ray to travel between x = 0 and x =a? Solution

tan(¾)=(~:~}an(i;r )Solution

The angle of deviation 6 is given by

Since, we know that µ = 5..

O=i-r According to Snell's Law,

V

=>

A ray of light goes .from air to medium of refractive index µ. If· i be the angle of incidence, r be the angle of refraction and 6 be the angle of devi~tion, then prove that

C

... (1)

sini smr

V=-

µ=-.-

µ

So, if v be the speed at a distance x from y-axis, then

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Optics & Modern Physics

Advanced JEE Physics 9=i tan9= tani

dy dx

AIR

.

. .. (1)

-=tani

MEDIUM(µ)

y

Applying componendo and dividendo, we get µ-1 sini-sinr µ+1 sini+sinr

=>

=>

µ-1 µ+1

µ-1 µ+1

tan

2cos

---7

c+r). sin c-r) 2 2 z

~ ----P(x, y)

a'

'-'-.....:C-"'--,,o+----~---'--•x

z

. (i+r) cos (i-r) 2sm Applying Snell's Law at O and P, we get µ0 sini0 =µpsinip

tan(%)

=>. (l)(sm90°) =( µ'x)sini

tanC;r)

1-r

µ+l tan (i+r) (2a) = (µ-1) 2

1 sini=[ ~

~J

0

Illustration 17 A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to the x -axis and width parallel to the y-axis. A ray of light is travelling along y-axis at origin. The refractive index µ of the medium varies as µ =

µ(\), where

tan i = --,==('=l=-='~")'==7

... (2)

µ;-(1-~)' From equations (1) and (2), we get

µ 0 and r(> 1) are

1- r constants. The refractive index of air is 1. y

dy =

--,==("'l=-=~=)-,,.dx µ;-(1-~)'

Integrating, we get

A

i l ~--o=<------~-x

d

X

fdy=f

d

o

Determine the x-coordinate of the point A, where the ray intersects the upper surface of the slab-air boundary. (b) Write down the refractive index of the medium at A . (c) Indicate the subsequent path of the ray in air. Solution (a) Refractive index is a function of x, i.e., the plane separating the two media is parallel to y-z plane or normal to this plane at any point is parallel to x-axis. Further refractive index increases as x is increases. So, the ray of light will bend towards normal and the path is shown in figure. Let at the point P(x, y) the angle of (a)

=>

o

1-~ r

µ~-(1-~)'

dx

x=r[l- µ;-(~+1µ;-1)']

(b) Atpoint A, l-~= µ;-(~+1µ;-1)'

{ (c)

·.-µ=~} 1-r

After A, medium is again air. Hence, from Snell's Law, angle of incidence will again become 90' or it will move parallel to y-axis as shown.

incidence be i . Then

=

1.30

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Ray Optics: Refraction at Plane Surfaces y

h=(o.2{½)( ~)=0.086 m A

Hence, height from the mirror is d = h + R = 0.1 + 0.086 = 0.186 m (b) Using the principle of reversibility of light, we get i=2r=60°

'--~o+---------'----+X

Illustration 18

A cylindrical glass rod of radius 0.1 m and refractive index ./3 lies on a horizontal plane mirror. A horizontal ray of light moving perpendicular to the axis of the rod is incident on it. (a) At what height from the mirror should the ray be incident so that it leaves the rod at a height of 0.1 m

T

Now, QU =coti=cot(60°)= TU

above the plane mirror? (b) At what distance a second similar rod, parallel to the first, be placed on the mirror, such thcit the emergent ray from the second rod is in line with the incident ray on the first rod?

~

v3

_TU _0.1

=>

Qu -

./3 - ./3

So, the desired distance is

OC = 2(0.1) + 2( ~) = 0.315 m

Illustration 19 An opaque sphere of radius R lies on a

•

horizontal plane. On the perpendicular through the point of contact there is a point sonrce of light a distance R above the

Solution

Let us first draw the ray diagram for the situation.

R

sphere.

I

(a)

R

Find the area of the shadow on the plane.

(b) A transparent liquid of refractive index ./3 is filled above the plane such that the sphere is just covered with the liquid. Show that new area of the shadow.

t-----R

Solution

(a)

(a) Since, PO= OQ =>

The situation is shown in the fignre g.:::\I'-.::.

,,--

LOPQ = LOQP = r (say)

....

Also, i=r+r=2r

In Af'OS, we have

=>

h = OPsini = 0.lsini h = 0.1sin2r

=> h=0.2sinrcos,r Applying Snell's Law at P, we get r,; __ sini 2sinrcosr 'V;) 3 = 2 cosr sinr sinr => r=30° Substituting in equation (1), we get

... (1)

Since, we observe that

sin0=0Q= R =1 OS 2R 2 1.31

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Advanced JEE Physics =>

A'= rcR'2 = 2rcR2

0=30°

Further, radius of shadow is given by

=>

r =MP= MStan30°

CONCEPT OF OPTICAL PATH LENGTH (OPL) AND REDUCED THICKNESS

~~(3Ri(l)=v'3R

If a distance L separates two buildings, then the measured distance has nothing to do with the medium between the buildings. If this separation is filled with water, then too the distance behveen the buildings is L . However the time taken by the light to travel between the buildings is different for different media between the buildings. This time difference is due to the interaction of the light with the molecules of the .medium which impede (slow down) the light's velocity and this cause the light to take more time to

So, area of the shadow is 2

A= 1t( v'3R) = 3itR 2

(b) The situation is again shown in the figure.

travel the same physical distance for different media. Due to this a,new concept of distance needs to be introduced

that accounts for the delay in the travelling time of the light in water (or a denser medium) in comparison to air (or a rarer medium). This new distance is called the Optica! Path Length (OPL) or Optical Path and takes .into account the slower velocity of light within a denser medium and it is simply the product of the distance with the refractive index

LIQUID

AB=AD+BD AB=AE+BD Further AB = R' + R tan i AC

=>

cosr= AB

i.e.1

OPL=µL

[by geometry}

Thus, light passing through a denser medium seems to travel

a longer distance than the light propagating in free

2R R'+Rtani

=>

R'=2Rsecr-Rtani Also, we observe that R' =BC +CM =2Rtanr+Rtani From equations (1) and (2), we get 2secr-tani = 2tanr + tani => secr-tanr=tani

. . . (1)

space/vacuum, during the same time intervals for both the media.

Let me illustrate this thing to you. For that let me take_ two media, one rarer of length L1 , refractive index µ1 and other

... (2)

denser of length L2 and refractive index µ 2 , as shown. 2

RARER(µ,)

DENSER(µ')

V1={f;

V2= ~2

---L,---..

---L,---..

Using the concepts of trigonometry, we get

,tani

1-sinr cosr

Time taken by light to travel a distance L, in rarer medium . h speed v = -C 1s • wit 1

µ,

=>

tani

t,=h=(L')=µ,L, V1

... (1)

C

The time taken by light to travel a distance L2 in denser

=> =>

:,

2i =!:__!..

2 2

medium with a speed v2

... (3)

=_..:... µ,

is

... (2)

Also, from Snell's Law, we get

sini=v'3sinr Solving the above equations, we get

... (4) Now if both times are equal, as said above, then

R' =.fi.R So the new area of shadow is

=

tl = t2 ~

µ1L1

1.32

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= µ2L2

... (3)

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Ray Optics: Refraction at Plane Surfaces Since, Optical Path Length (OPL) is the distance travelled by light in vacuum/air/rarer medium during the same time it travels a distance L2 in medium. So, from (3), we get

=µairLair = µmediumLmedium =µ2L2 Since, µair =1 , so, we get OPL = Lrur = µmediumLinedium

10--

d

and t, =

v: = (3x10')

6x10-15 s

1.8

µlLl

So, tmin (b)

{for same time in air and medium} Since light always travels slower in denser medium, s6 the OPL (the distance in air corresponding to same time in both) is always longer than the actual thickness L of the medium.

= 2 X 10-

15

s

The total number of wavelengths in a film of refractive index µ , thickness d is n = Optical Path Length Wavelength of Light ~

µd

n='/,.

So, total number of wavelengths, is

CONCEPTUAL NOTE(S)

n = µ,d, + µ,d, + µ,d,

Also, note that µ 1L1 is OPL in air/rarer medium and µ 2 L2 is OPL

'/,.

in denser medium. However for standard purposes OPL is the distance travelled by light in vacuum/air to travel a distance L in _ a medium during the same time in either air or medium. ______ _

So, from equation (3), we conclude that for a pair of media, Optical Path Length ) = (Optical Path Length) ( in Air/Rarer Medium in Denser Medium

1,,

'/,.

~

1 n=i:(µ 1d1 +µ 2 d2 +µ 3 d3 )

~

n=

~

n = 1000 (4.5) = 4500 = 75 600 600

10-xlO_, ((1.2)(1)+ (l)(l.5)+(1.8)(1)) 600

=µ2L2

=>

µ1 L1

~

L2- -µ,- L1

µ,

Since µ 1 < µ 2 , so we get

L2 < L1 Due to this reason, L2 is also called the Reduced Thickness.

So, in general, we get R~duced) = ( µra= ( Thickness µdenser

)L,=, = OPL in air

Illustration 21 A light ray enters the atmosphere of a planet and descends vertically 20 km to the surface. The index of refraction where the light enters the atmosphere is 1 and it increases linearly to the surface where it has a value 1.005. How long does it take the ray to traverse this path. Solution Since variation is linear, so we have

x-0 µ-1

--=

µdenser

Illustration 20 A light beam of wavelength 600 nm in air passes firstly through film 1 of thickness 1 µm and refractive index n1 =1.2 , then through an air film 2 of thickness

1.5 µm and finally through film 3 of thickness 1 µm and refractive index n3 = 1.8 . Which film does the light cross in the least time and what is that least time? (b) Calculate the total number of wavelengths (at any instant) across all three films together.

~

2xl0 4 -0 1.005-1

0.005x µ= 1 + 2xl0 4

Now, by definition, we have C

µ(x) = v(x)

~

v(x)

= µ[x)

(a)

where µ(x)=µ=l+ O.OOS: and c=3xl0 8 rns-1 2x10 X=O, µ= 1

i

Solution

(a) Since, 11

Similarly,

d,

v,

f2

10-' (3xl08) 1.2

d, C

x,µ

4x10-15 s

1.5x10-' 3xl08

2x10-15 s

X, =

2 X 10' m, µ= 1.005

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Advanced JEE Physics

=>

=> =>

3xl08

3xl0 8 0.005 1 + - -4x 2xl0

C

v(x) = µ(x)

X

di

1 di = --, [(1+2.5x10-'x)dx] 3xl0 t 1 2><104 3x 10

=> =>

4 )'

1 ..

1 .

C

V

=>

-smz=-smr

=>

sini C -=-=µ sinr v

[The Law of Refraction}

µ 1 sini =µ 2 sinr

0

1 =-1-[ 2 xl0' + (2xl0 3xl08

sini

VECTOR FORM OF SNELL'S LAW

fdl=--, f (1+2.5x10-'x)dx 0

d-x

Jb 2 +(d-x)'

3xl08 1+2.5x10-7 x

dx

~ini and

Using our knowledge of cross product of vectors, we have

(2.5x10-')] 2

ixfi=(l)(l)sini, 0 outwards

5

1=6.68x10- s

i

LAWS OF REFRACTION USING FERMAT'S PRINCIPLE

Consider a refracting surface / interface separating medium 1 from medium 2. Let the incident light start from A , in medium 1, hit the surface at O and get refracted to a point B, in medium 2. Let the points A and B be at perpendicular distances a and b_from the interface. Further, let A and B be at a separation d as shown in figure. The time taken by the light to go from A to O to B is t=tA...,.o+to-)oa

Medium 1 (µ 1) Medium

A

-n

(µ,)

'

and -ilxf=rxfi=(l)(l)sinr, 0 outwards So, from above we conclude that µ 1 (1 X fi) = µ 2 (f X fl)

=> =>

Ja' +x'

Jb' +(d-x)'

C

V

I---+~----

Now, according to Fermat's Theorem, t is MINIMUM, so In Medium 1

..'!!_=0

In Medium2

dx

REFRACTION THROUGH A COMPOSITE SLAB

'

.',o ''

(d-x)

X

Interface

I+----

=>

=>

Medium2

1

'r

''

=>

Medium 1 (vacuum)

d

' :' r'

b

!

B ---+I

J:..i( Ja' +x' )+J:..i( Jb' +(d-x)') C

di

1

V

2x

di

1 ( 2(d-x)(-1)) 0

2c Ja' +x' + 2v Jb +(d-x)' 2

1

X

cJa'+x'

1

(d-x)

vJb'+(d-x)'

From the figure, we observe that

=

Consider the refraction of light ray through a series of media as·shown in figure. The ray AB is incident on interface X1Y1 at an angle i . The ray is deviated in medium 2 along BC towards the normal. Then it falls on interface X2Y2 and is again deviated towards normal along CD. If the last medium is again Medium 1, the ray emerges parallel to the incident ray. Let r1 and r2 be angles of refraction in Medium 2 and Medium 3 respectively. Then from Snell's Law, sini µ 1 - . - = -2= µ2 ... (1) smr1 µ 1 sinr smr2

µ

2

- . -1= -3= µ3

... (2)

sinr2 µ 1 3 - .-. =-= µ1

... (3)

Slill

µ2 µ3

µ 1 = refractive index of medium 1 µ2

µ3

= refractive index of medium 2 = refractive index of medium 3

·= ==================================== 1.34

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Ray Optics: Refraction at Plane Sn,faces A

Solution 1

X,

f

Given, wµ 8 =2. and aµ,=~

8

Y,

2

t.

'

X,

3

t

~

~

I

X,

Y, Y, ~

Multiplying (1), (2) and (3), we get 1µ2x2µ3x3µ1 =1

=>

LATERAL SHIFT ON PASSING THROUGH A GLASS

tµ2x2µ3=1µ3

In general if a ray passes through a number of conposite parallel plate glass slabs, then 1 µ2 X 2µ3 X 3µ4 X 4µ5

= 1µ5

Illustration 22

A light beam passes from a parallel plate glass slab of refractive index µ 1 placed·in a medium of refractive index µ 1 • Show that the emerging beam is parallel to the incident beam.

SLAB

Consider a ray AO incident on the slab at an angle of incidence i through the glass slab EFGH of thickness t . After refraction the ray emerges parallel to the incident ray. Let PQ be perpendicular dropped from P on incident ray produced as OQ . The lateral displacement caused by plate, x = PQ = OPsin(i-r)

OM . (·1-r) x=--sm

~

Applying Snell's Law at A, we get µ 1 sini =µ 2 sinr1

~

1:!.= sinr1

~

µ2

/._-OP= OMl cosr

l

COST

Solution

COST

... (1)

sini

t . (.i-r ) x=--sm

t (smzcosr-cosismr . . .. ) x=-COST

~

x = t(sini- cositanr) A

Air(µ= 1)

µ,

µ,

Applying Snell's Law at B, we get µ 2 sinr2 = µ 1 sine

=>

1:!. = sinr2 µ2

sine

Air(µ= 1)

,e

''

' N'

... (2)

B

'

sini . Smce µ=-.-

From equation (1) and (2), we get i=e i.e., the emergent ray is parallel to incident ray.

smr

=> ::::::>

Illustration 23

Refractive index of glass with respect to water is 9/8 . Refractive index of glass with respect to air is 3/2 . Find the refractive index of water with respect to air.

. sini smr=--

µ

tanr

sini

~

1.35

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Optics & Modern Physics

Advanced JEE Physics

If i is very small then r is also very smalt hence, sin i ---+ i , and cos i ---+ 1, Then expression for lateral displacement takes the form.

b)

At near normal incidence (small" angle of incidence~ i )' apparent depth (d') is given by

Observer -;;;11. Medium 'I

:=n(1-¾)

f I

RARER RARER

APPARENT DEPTH

An object O placed in a medium of 'refractive index µ is observed from air at a small angle a to the normal to the interface (in figure, angle a is shown exaggerated for clarity) i.e., for near normal incidence.

d

l

'' 'A '' ''

'

B "'

,, '

fo·•"~,'

6X

',p

io '

Observer Medium

µrerawu

,

tanp

d =distance of abject from the' interface

µ=-=--,S--=-

surta'c8.'

--

·

\\,

·

..

Illustration 24 A fish rising vertically to the surface of water in a lake at a uniform speed of 3 ms-1 • It observes that a bird diving vertically towards the water at a uniform speed of

_!, find the 3

Solution Let x be the depth of the fish F below the surface of water, and y be the height of the bird B above the surface

d

Apparent depth, d' = -

µ

=d(1-¾)

at an instant.

To the fish, the bird will appear to be farther away, at an apparent height y' given by y

In case the object is seen through n number of slabs with different refractive indices, the total apparent shift is simply the sum of individual shifts, so Ax=Ax1 +Ax2 +&3 + .... +filn

=

Perp~ndicu1ar

actual speed of dive of the bird.

The apparent shift in normal direction (or the normal shift) in the position of the object is

a)

··,

v :;= v13locity of object perper:idicular to interface, relative to surface. v: = ve1acity at iffiage ta interface relative ta

9 ms-1 • If the refractive index of water is

Therefore, from Snell's Law, we get, smt sina. tana. d sinr sinP tanp d'

=,,

= real depth.~ = apparent

d' = distance of image from the interface depth. - · ,..

d d'

dX= d-d'

RI of medium of incidence obje_qt

Medium(µ)

P: ' ''' !'

Since angles a. and j3 are small, so sin a. ~ tan a. and sinp "tanp.

=:,

µrolaUvo

where µ ra· = ro live RI of medium of refraction observer

Air

H the object O is at a real depth d from the interface, its apparent depth d' can be calculated. From t,.s ABO and ABO', tancx

4

' d - iV d' =- - and v' =-·--

~

.,'

'' d'

'~·

DENSER

DENSER

( µ""'.'m•H•m.) µfishmedium =:,

ax=ai(1- :J~a,(1-:J+a,(i-:J+ . +a"(i- :J . CONCEPTUAL NOTE(S) If the medium in which the object is placed is rarer (µ,,< 1') and it is seen from the denser medium, the apparent shift calculated will be negative. If means· that the object apparently shifts away from the observer. If the shift comes out to be ,positive, the ima9e of the· object shifts towards the observer.

'

y

y = (;) =µy

The total apparent distance of the bird from the fish is s=x+y' =:>

s=x+µy

Differ~ntiating w.r.t. time t, we get C

ds dx dy dt ~ dt + µ dt

1.36

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Ray Optics: Refraction at Plane Surfaces

!f

•B'

SHIFT OF POINT OF CONVERGENCE OR DIVERGENCE

r

•B

y

If a glass slab of thickness t., refractive index µ is placed in the path of a convergent (or divergent) beam of light, the point of convergence (or divergence) gets shifted by

s

Air Water

X

~={1-¾)

i •F µ

Substituting the values, we get 9=3+(±) dy 3

dt

i+-t ....

1+-t-+1

Therefore, the actual speed of dive of the bird is given by

(B) Divergent beam

(A) Convergent beam

dy dt

=(9-3) (~) = 4.5 ms-' 4

,.

Illustration 25 A vessel is filled with a non-homogeneous liquid whose refractive index varies with the depth y from the

free ·surface of liquid as

µ·= ( 1

+;}

Calculate the

apparent depth as·seen by an observer from above, if H is the height to which the liquid is filled in the vessel.

Illustration 26 A point object O is placed in front of a concave mirror of focal length 10 cm . A glass slab of refractive

index µ =

¾and thickness 6-cm is inserted between-object

and. mirror, Find the position of final image when the distance x shown in figure is (a) 5 cm (b) 20cm 6cm

Solution Let us consider a thin layer of liquid of thickness dy at a distance y below the free surface of liquid. The apparent

•

0

depth of this layer having real depth dy is dH' = dy . µ Free Surface

r

Solution The normal shift produced by a glass slab is given by

y

H

l =>

dH;= dy

=>

dH'= dy ' H y) .' ' ·( , 1+-

~=(1-¾}=(1-¾}(6)=2 cm i.e., for the mirror the object is placed at a distance (32- ~) = 30 cm from it. Applying mirror formula i.e. 1 1 1 -+-=-, we get

µ

V

U

f

1 1 1 ---=--

ToW _apparent depth is obtained. by integrO:ting this expreSsion within appiopriate ~imifs. So, H

Hlog,(H +y)

H'=Hlog,2

-x-

1-4---32 cm-+i

30 10 => v=-15 cm (a) When x =5 cm The light falls on the slab after being reflected from the mirror as shown. But the slab will again shift it by a distance tu= 2 cm . Hence, 'the final real image is formed at a distance (15 + 2) = 17 cm from the mirror. v

================================== = 1.37

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Advanced JEE Physics

Optics & Modern Physics 4

Further, =:>

3

sin(45°) sinr

r=32°

=tanr =tan(32°) h-lO =0.62 h

Now EF GE

20

(b) When x = cm This time too the final image is at a distance 17 cm from the mirror but it i~ virtual as shown.

Solving this, we get

h=26.65 cm MULTISLABS

If a number of slabs (or immiscible liquids) of depth d1 , d,, d3 , •••• and refractive index µ 1 , µ 2 , µ 3 , •••• are placed one over the other, the real depth is d=d,+d,+d,+ .... The apparent depth is given as

Illustration 27

A cubical vessel with non-transparent walls is so located that the eye of an observer does not see its bottom but sees all of the wall CD. To what height should water be poured into the vessel for the observer to see an object F arranged at a distance of b =10 cm from comer D ? The face of the vessel is a = 40 cm and refractive index of water .

d' =E.!..+E1_+ d3 + ... µ1

µ3

Therefore, for the combination, the effective µ is

µ=I d'

4

IS-,

µ2

d,+~+~+...

L~

r(iJ

(i:)+(i:)+(i:)+...

If there are only two slabs, of equal thickness, d, =d, =d, d+d

3

--.-

d, f,,-===~==='~ __ t_ d2

!;}}jit;tir.1~!f~

--+· d3

µJ

Solution

1 . __ _ _ _ _ _.J. __

Since, the vessel is cubical, GE=ED=h (say)then EF=ED-FD

LGDE=45°

t_

and Illustration 28

Eye~A

The bottom of a tub has a black spot. A glass slab of thickness 4.5 cm is placed over it and then water is filled to the height of 8 cm above the glass slab. Looking from top, what shall be the apparent depth of the spot below the water surface? Also find the effective refractive index of the combination of glass slab and water layer. (Refractive index of glass is

But tan(45')=1=GE ED

cc,,

ED=GE=h

cc,,

EF=ED-FD=h-10

~ 2

and of water is

± ). 3

Solution

The apparent depth is given as d d 4.5 8

d,=µ:+µ:=rn)+(¾)=3+6=9cm

= :::::================================= 1.38

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Ray Optics: Refraction at Plane Surfaces

I

6.

I.

4 µ2=3

,.

3 µ1=2

\

surface of water. Where will the image of this source in the flat mirror-like bottom of a vessel be if the depth of the vessel full of water is d ? Refractive index of water is

!'"

n =~ . co·nsider only two steps.

4.5cm

•·

Spot

7.

µclfuctire

= !!,_ =

d, + d,

d,

d,

A plate with plane parallel faces having refractive index 1.8 rests on a plane mirror. A light ray is incident on the upper face of' the plate at 60° . How far from the entry point will the ray ·emerge after reflection by the mirror of the plate is 6 cm thick?

The effective refractive index is given as Real Depth µeffective Apparent Depth

=>

A point source of light is arranged at a height h above the

'''

= 4.5 + 8 = 12.5 = 1 _39

9

''' N:'

60~1

,M

9

IC:Eilli BASED ON GENERAL REFRACTION (Solutions on page 1•.176)

1.

An object lies 100 cm inside water. It is viewed from air nearly normally. Find the apparent depth of the object.

2.

A concave mirror is placed Inside water with its shining surface upwards and principal axis vertical as shown. Rays are jncident parallel to the principal axis of concave mirror. Find the position of final image.

!! water 4/3

~--

R=40cm

3.

4.

5.

f

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 PC?Sition· of the coin from the surface. Air

3cm h,

Glass

t

A ray of light falls onto a plane-parallel glass plate 1 cm thick at an angle of 60° . The refractive index of the glass Is

./3 ,

A 2 cm thick layer of water covers a 3 cm thick glass slab.

Water

~.

Some of the light is reflected and the rest, being refracted,, passes into the glass Is reflected from the bottom of the plate, refracted a ~econd time and emerges back into the air parallel to the first reflected ray. Determine the djstance £ between the rays .

The velocity of light in air is 3 x 108 ms-1 • If yellow light of wavelength 6000 A is passed-from air to glass of refractive index 1.5, determine the velocity, the wavelength and the colour .of light in glass.

f

happens to be 1 m above the water. Determine the length of the shadow of the pole at the bottom of the lake if the sunrays make an angle of 45° with the Water surface. The

9.

!

h,

A pole 4 m high is driven into the bottom of a lake and

refractive index of water is

A small object Is placed on the principal axis of a concave spherical mirror of radius 20 cm at a distance of 30 cm . By how .much will the position and size of the image alter, when a parallel-sided slab of glass of thickness 6 cm and refractive index 1.5 is Introduced between the centre of curvature and the object? The parallel sides are perpendicular to the principal axis.

f 2cm

8.

Air

30cm ~

MirrOr

.

10. A ray of light ·is refracted through a sphere whose ·material has a refractive index µ in such a way that it passes through the extremities of two radil which make an angle .~ with each other. prove that if a;. is the deviation Of the ray caused by its passage through the sphere, cos(~;")= µcos(%}

11. A vertical beam of light of cross-sectional radius r is incident symmetrically on the .curved surface of a glass hemisphere

(µ =%)

of radius 2r placed with its base on a

horizontal table. Find the radius of the luminous spot formed on the table.

12. A material having an index of refraction µ is surrounded by vacuum and i_s in the Shape of a quarter circle of radius R . A light ray parallel to the base of the material is incident from the left at a distance of 'L above the base .and emerges out of the material at an angle 8 • Determine an expression for a .

Coin

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Advanced JEE Physics

Optics & Modern Physics CRITICAL ANGLE

According to Snell's Law, we have sini smr

d

-.-= µ,

sin C sin 90

j 13. A circular disc of diameter d lies horizontally in~ide a

I

metallic hemispherical bowl of radius a. The disC. iS just visible to an eye looking over the .edge. The bowl is now filled with a liquid of refractive ·index µ. Now, the whole of the c;lisc is ju$t visible to the eye· in the same positioli. Show

that

,

d=2a(µ'-

V-

µdensl'r

· C =µrarer 1 -sm --=µderner

µdenser

· 1 ) µ2 +1

Eye

= µrarer

where µdenser is ~he refractive index of the denser medium w.r.t. the rarer medium. The lesser the value of µdenser, the greater is the critical angle C . For a given pair of media, since µ depends on the wavelength of light the critical angle ·also depends on the wavelength. The greater the wavelength, the greater will be the critical angle.

............ _ _ _ _ _ _ _ _ _ __

--, ---- .......... _

.............. _ ...........

Critical angle TOTAL INTERNAL REFLECTION (TIR)

Media Pili

When a ray of light goes from a denser to a rarer medium, it bends away from the normal. If the angle of incidence in the denser medium is increased the angle of refraction in the rarer medium also increases. At a particular angle of incidence in the denser medium (called as the Critical angle C ), the angle of refraction in the rarer medium is 90° (i.e., the refracted ray grazes the interface). This angle of refraction in the denser medium for which the refracted ray grazes the interface is called the critical angle for the pair of interface. ' Please note, that for small angles of incidence, both reflection and refraction occur, however we shall be neglecting the reflection at the interface as most of the light is refracted. However, when i > C, no part of light is refracted· and the entire light is reflected back to the denser medium itself. This phenomenon is called total internal reflection (TIR) and was first noted by Kepler in 1604.

0

Denser µ2

=

1.40

-1(-1-..· ·,) µdense'(

Water-Air

I

Glass-Air

I

Glass-Water

µd· _µ._4(3_4 _____ _

µ,

1

3

µ, µ,

3/2

3.

49°

µd=-=-=-

1- · 2

63°

EXAMPLES OF TOTAL INTERNAL REFLECTION

a)

Mirage : Mirage is an optical illusion ol;,served in deserts and roads on a hot day. When the ai_;- near the ground is hotter o (and hence rarer) than the air above, E '........ Denser j > 8~ there occurs a Rarer continuous decrease of refractive index of air towards the ' 'Earth ground. A ray of light from a point 0 of a tree is, therefore, refracted more and more away from the normal. Ultimately it gets totally reflected to reach the eye E. To tf1e observer it appeai;-s to come from I, which is the image of O . This image gives the impression of reflection from a pond of water. Looming : Similarly, in extremely cold regions (near polar regions), the refractive index decreases ·with height. Due to TIR (shown in figure), the image of a hut appears hanging in the air. This is called looming.

<:l1,,-_ ______

µz:> µ1

Images formed by TIR are much brighter than those formed by the mirrors (or lenses). Some loss of intensity always takes place, when light is reflected from a mirror (or refracted through a lens).

. C =sin

b)

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Ray Optics: Refraction at Plane Surfaces

rm ' ' ~-------~.._,...... \:,.,-/ Rarer

.......................... ! goo ~--·r--' 1-:c:c:__4r d

Earth-

c)

d)

e)

The µ of diamond is 2.5, for which C is only 24°. Diamonds are cut such that _i > C ; so TIR takes place again and again inside it The light coming out from few meticulously cut surfaces makes it sparkle. Air bubbles in water shine due to TIR The working of an optical fibre is due to multiple TIR inside it.

f)

Porro prisms used in periscopes or binoculars bend the ray due to TIR. Some examples are shown in figure. 45"

/

(a) Bending of rays by 180'

L-~-------l

µ

Fish in Glass Tank Similarly, if a source of light is kept in ·a pond, its light will come out only through a circular region. For any incident angle i greater than C, the light will be totally reflected back into the water, making corresponding region on the surface of water appear dark. Illustration 29 Light is incident making an angle 0 with the axis of a

transparent cylindrical fiber of refractive index. n =

/

i =45°

90'

'' ''

45'

~

as

shown in figure. Determine the maximum value of 8 so that the light entering the cylinder does not c~me out of. the curved surface. ·

t)-----~ - -(-)-

(b) Bending of rays by 90'

Solution The ray of light is incident at A and it just gets reflected totally at B . Therefore incident angle at B is equal

A' --+--
B'fr_.,__+-_

to the critical angle give~ as C = sin ~l ( ; ) (c) Erecting of image

FIELD OF VISION OF A FISH

A fish inside a pond does not see the outside world through the entire surface of water. The light from outside can reach the fish only through a circular patch, which forms a cone of half angle equal to the critical angle. If r is the radius of the circular patch, d is the depth of the fish and µ is the refractive index of water, then r=dtanC=dsinC =d sinC cosC .Jl-sin2 C

. . C =1 sm Smce, µ

~

d r= ~µ'-1

Snell's Law ofrefraction at A gives sin0 --=n sinr . sin0 ::::::,. smr=-n Since r+C=90°

... (1)

~ sinr=sin(90°-C)=cosC For a ray not to come through the curved surface, r,;90-C

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=>

2

sinr5.Jl-sin C

Optics & Modern Physics

5'11-:

2

... (2)

Eliminating sin r from (1) and (2), we get

=>

µ2>1+1

=:,

µ > Jz

So, the minimum value of refractive index is

sin0:<;)l-~

n

µml"

n

=:,

sine :5. .Jn' -1

=:,

sin2 0,; 1.25-1

=:,

sin2 0:5.0.25

=:,

. 0 51 sm 2 0:<;30°

=:,

0max = 3QO

=:,

( ·: maximum value of 0 can be 90°}

=Fz

Illustration 31 Light is incident at an angle a. 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 a ..

'

Illustration 30

A rectangular block of glass is placed on a printed page lying on a horizontal surface. Find the minimum value of the refractive index of glass for which the letters on the page are not visible from any of the vertical faces of the block.

Solution . . C =1 sin Smce, n In triangle ABN, r' + r + 90° = 180°

:s

Solution

-~()

Light will not emerge out from the vertical face BC, when De-------, C

Glass

=:,

r'=90-r

=:,

(r')mm = 90° -(r )=

and n

sin(i)max sin(rl-

sin90° (i = 90') sin(r)mu max

Then, sin(r)max =.!.=sinC n (r)=,=C

i > Critical Angle (C)

=>

sini > sinC

=>

. . 1 smz>-

=:,

Now, if minimum value of r' i.e., 90° -Sc is greater than

{·: sinC=;}

µ

(r')- ?:C

sin0=µsin(90°-i)=µcosi .

sin9

cosz=-µ

=:,

(90°-C)?:C

=:,

=:,

sin(90°-C) ?:sinC cotC ~ sinC cosC ~ 1

=:,

.Jn'-1 ?:l

=:,

"2 ~2

=:,

n?:Jz

=:,

sini=.Jl-cos 2 i =~1- sin'O

µ'

Therefore, the condition for no light to emerge from vertical face BC becomes,

1

>µ

=>

2

ec,

then obviously all values of r' will be greater than 8c i.e.,

total internal reflection will take place at face AB in all conditions. Therefore, the necessary condition is

Applying Snell's Law at O, we get 1sin8=µsinr =:,

(r')m1n = (90° -C)

Theref9re, minimum value of n is

J2

2

µ >l+sin 8

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Ray Optics: Refraction at Plane Su,faces Illustration 32 A point source of light is placed at a distance h below the surface of a large and deep lake. Show that the fraction f of light that escapes directly from water surface is

Solution Let the critical angles at 1 and 2 be C1 and C2 respectively. Then

~~-----~~ µ, =-,/2

;

i

independent of h and is given by, f

I

'

!

2

µ2=2

i

p

Solution Due to TIR, light will be reflected back into the water for i > C . So only that portion of incident light will escape which passes through the cone of angle 8 = 2C .

µ,=../3

I

T

2

C_· -1(b_)- · -,(j_)-45° 1 -Sll1

µ,

In

-Slfl

..,2

-

C A

1

1

and C2 =sin- (~:)=sin-

(

~)=60°

h

f

ForTIR, i>C2

s

Therefore, minim.urn angle of incidence, for total internal reflection to take place on both slabs must be 60° .

irnin =60°

+

So, the fraction of light escaping is given by

f = Area of Surface ACB

BASED ON TOTAL INTERNAL REFLECTION {TIR) (Solutions on page 1.179)

Total Area of Sphere ~

f= 21tR2 (1-cosC) 1-cosC 4nR2 2 Now, as f depends .on C, which depends only on µ, hence

1.

Light refracts from medium 1 into a thin layer of medium 2, crosses that layer and then is incident at the critical angle

on the interface between media-2 and 3 as shown in figure.

f is independent of h . Since, we know that

. C =1

SIIl

µ

cosC= ~µ'-l " ' µ'

=JI- µ'1

C

(a) (b)

1-J1- l

µ'

!=~~~

2

2.

Illustration 33 ' 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. A

Find. the angle e . If 8 is decreased, wiffthe l!ght be refracted to medium 3?

A ray of liQht enters into a glass slab from air as shown in figure. If refractive of glass slab varies with t, the thickness of the Slab measured from the top as µ =A - Bt where' A and B are constants. Find the maximum depth travelled by ray in the stab. Assume thickness of slab to be sufficiently large.

a

Air

Slab p

C

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Advanced JEE Physics 3.

A container contains water suj)to a height of 20 cm ,and·. there is a point Source of light at the benti'e of the .bott6m-of the cohtainer. A rubber ring of. radius ~ floats centr'ally On the water. The ceiling of the.l'oo·m is 2.m· above the Water

AIR

',,

surface. (a)

Find the radius of the shadow of the ring 'formed, on

theceilingif a=15cm. (b)

Find the maximum iJalue of

"··

8.

'

C

for which the shadOw of

n2·=·1.4

the ring iS fonrted on the cejflng.

Refr,active index,of w9-;er

" "· = 1.3 n3 = "1.32

=~--·

n..,=1.45

4.

ABCD· is,the plane of glasS'cube of ref~ctive in~ex µ. A" horizOnfal beam of light enters th_e·face AB at the graiing

incidence. (a)

Show that the angle e which any ray emerging from'

BC would make with noml_al to BC is given b·y sine= cot a, -~here a 'is the.critical 8ngle. ·

))---------' ----- )---1

o,._______, C (b)

..

,

Plot the deviation (S) versus the angle of incidence (I)

graph for a ray travelling frorii c1enser tO rarer medium.

6.

A point source of light s is placed at the bottom of a vessel containing a. liquid- of refracti~e i~dex ~ . A persOn is " 3 viewirig the source from, above 'the surface. There i_s an· opaque diSc Of' radius 1 cm fl08ting on the surfaC:e. The, center of the disc lieS verticially. above the.source S. The' liquid from. the vessel is gradually _Pralned out through a tap. What Is the maximum height of the liquid for which th~ source Cann_of at all seen from· above.

What Is the greatest valuidhat the·refractlon index of glass may have if any oFthe .light is io emerge from

.BC?

5.

o.

In figUre,. light refracts into material 2, crosses that material and is. then incident at the critical angle on the interface ,i · between materials 2 and 3.

11.

A rectangular glass block is placed on top of a sheet ,of_ paper on .which there is a small cross. When the paper is Soaked in alcohol and a' sodium. lamp is placed opposite to one vertical-of the block the cross can be·seen through th_e opposite vertical face Up to a. ·priint where the angle qf emergence of the light is 30° . If the refractive index of the . glass fs 1.5, find the refractive index of alcohol. Why can'.t the black cross be seen through the-face when the paper Is dry. ' ,,

12.

Rays·oflighlfall on the plane surlace of a half cylinder at an angle 45°, in -the plane perpe_ndicular to the. axis (sf2!eM·

, MEDIUM 1

01=1.6

MEDIUM

:i

n3 = ·1.:r

(a)

./2..

What is angle e?

figure). Refractive index of glass. is Dis~uSs ·, fh~" condition-that the rays do flat suffer tot8.! intemal'reflection. ·, .. " • -+·

(b), , If 0' is increased, is there refraction of light into material 3~ 7.

An isotropic point source is placed at a depth h· below tt}e · water surface. An opaque disc 9apable of floating on water

surface,.1s ·placed on th~ surface of water so that the bµlb is not visible from the slirface. Find the minimum radius of the disc fol° the bulb not to be viSible. Take ;refractive index of

water=µ., 8.

=

In figure, light begins from medium of- refractive ind8.x , n1 = 1.3', undergoes three refractions as it heads downward and a reflection and then a refraction· to reach the air. The initial angle 01 = 30° . Find the_ value:,s of the angles

13.

Find the maximum value of

-k·,

so that the beam of_ light·

inci_c;fenl normally .at the face A of a U shaped glaSs tube ~-~em=erges th!Q.~Qh B as shown- iri the figure. The refractive·

1.

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Ray Optics: Refraction at Plane Snrfaces index of glass is µ

between incident ray EF and emergent ray GH (produced backw~ds) is called angle of de':'iation D .

= ¾.

A

E A

H

B<-µ_ _ _ _ _ _ ___,C

8

In triangle OFG ,

D =(i-r1 ) + (e-r,)

PRISM

Prism is a traJl'iparent medium bounded by any number of surfaces in such a way that the surface on which light is incident and the surface from which light emerges are plane and non-parallel. Refracting angle of prism, or simply the angle of prism is the angle between the faces on which light is incident and from which light emerges. In all the prisms shown in figure above, angle A is the angle of prism.

Angle of deviation (D) is the angle between the incident ray and the emergent ray. Sometimes the angle of deviation is also denoted by 6 .

~

D=(i+e)-(r, +r,)

' ... (1)

Also in quadrilateral AFNG , A+ 90° + 0+90' = 360" ~ A+0=180' And in triangle FGN , r1 +r2 +8=180°

... (2) ... (3)

Comparing equations (2) and (3), we get A=r1 +r2

... (4)

From (1 ), we get

D=i+e-A i+e=A+D

... (5) If µ is the refractive index of material of prism, then from Snell's I.aw sini sine ... (6) µ=--=-sinrt sinr2 For a prism with small refracting angle, we have D=(µ-l)A ~

CONCEPTUAL NOTE(S) a)

Angle of deviation (D) means the angle between emergent and incident rays i.e., the angle through which incident ray turns in passing through a prism. It ls represented by D and is shown in figure. A

Please note that, for a glass-slab, the angle of prism is zero, and the incident ray emerges parallel to itself, i.e., there is no deviation. If µ of t!J.e prism material is same as that of its surroundings, no refraction takes place and light passes through undeviated. REFRACTION THROUGH A PRISM

A

Consider a monochromatic ray EF to be incident on the face AB of prism ABC of refracting angle A at angle of incidence i . The ray is refracted along FG , r1 being angle of refraction. The ray FG is incident on the face AC at angle of incidence r2 and is refracted in air along GH . Thus GH is the emergent ray and e is the angle of emergence. The angle

- - - _____ B

c_______ ,

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c)

Optics & Modern Physics

If the faces of a prism on which light is incident and· from which it emerges becomes parallel (as in figure), angle of prism will be zero and as incident ray will e·merge parallel to itself, deviation will also be zero i.e., the prism will act as

a slab. ' If µ of the material of the prism becomes equal to that of surroundings, no refraction at its faces will take place and light will pass through it undeviated. So, deviation is zero. i.e., D=O

---------- --- ------ ----

Thus, a ray of light will not emerge out of a prism (whatever be the angle of incidence) if A> 2C , that is, if µ > cosec(;). CONDITION FOR GRAZING EMERGENCE

A ray can enter a prism in such a way that the angle of emergence, e = 90° , as shown in the figure. A

CONDITION OF NO EMERGENCE

The light entering the prism at surface AB , will not be able to come out from the surface AC, if TIR takes place at this surface. For any angle of incidence, this condition will be satisfied, provided we have at surface AC,

(r,)mm > C Since, r1 + r2 = A

We can determine the angle of incidence i for such grazing emergence. We should have

~

'2 =A-r1

~

(r,)""' = A-(r1 ) -

'2 =C

... (1)

Since, for a prism, r1 + r2 = A

A

=> '

',, p

r,

i

r1 =A-r2 =A-C

Using Snell's Law,

r,

lsini=µsinr1 =µsin(A-C) ~

sini = µ(sinAcosC-cosAsinC)

~ sini = µ[(sinA),/1-sin 2 C -(cosA)(sinC)]

B

But (r1 )max is possible when i=imax =90° i.e., incident ray grazes the interface AB . Now, applying Snell's Law at AB, lxsini=µsinr1 ~

1

sin(90°) = µsinr

r1 =sm

~

'1 =C

... (2)

From equations (1) and (2), we get r1 +r2 >2C Since, r1 + r2 = _A Therefore, the condition becomes A>2C ~

sm(1)>sinC

~

sm

~

µ > cosec(

sini=sinAJµ 2 -l ~cosA

=>

i=sin-1 (sinAJµ 2 -1-cosA)

The light will emerge out of the. prism only if the angle of incidence i is greater than the above value.

. -1(1) ·µ

~

=>

... (3)

MAXIMUM DEVIATION

The angle of deviation D is maximum when the angle maximum, i.e., i = 90°.

is

Dmu = (i + e)-A = (90°+ e)-A Under such conditions of grazing incidence, r1 = C A

. (ZA) >µ1

1J B.'------------' C

=

And at the. second surface, 1.46

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Ray OpHcs: RefracHon at Plane Surfaces

emergent ray for the same incident ray, values of au the angles.

µsinr2 = lsine

=>

sine= µsinr2

A

.,

Since r1 +r2 =A ~

sine=µsin(A-r,)=µsin(A-C)

~

e = sin-' (µsin(A-C))

indicating the

,,,,,/

--+----i-----,;cr.'" /

Illustration 34 An isosceles glass prism has one of its faces coated . with silver. A ray of light is incident normally on the other face (which is equal to the silvered face). The ray of light is reflected twice on the same sized faces and then emerges through the base of the prism perpendicularly. Find angles of prism. Solution As the ray is incident normally at the face AB , so

90'

B

Solution

For total internal reflection to take place at surface AB, we have i>C

=>

'1 =0

.45'

C

sini>sinC

. . C =1 sm Smce, µ

E

~

sin45'>(t)

~

µ>Ji

~

µmin

=Ji

When the prism is immersed in water, the boundary AB now separates glass from water. B

0 C

Since, we know that r1 + r2 = A , so we get

... (1)

r2 =A=180'-20 Now, LDfE=180°-90°-2r2 ~

LDfE=180°-90°-360°+40

{·: r2 =180-20)

~ LDFE=40-270' Since, r, = 90° - LDFE ~

... (2) ... (3)

r,=360'-40

Again LBFG = 90° - 0 = 90° - r, =>

T3

=8

... (4)

From equations (3) and (4), we get 50=360° ~ 0=72' and 180°-20=36'

--

c =sm · -•(. µ,=, · -1 (Ji) · - -) =Sm --

~

C=70.12°

µ,_

1.33

Since i = 45° and also, we.observe that i
Hence, TIR will not take place. From Snell's Law, we get µ 1 sini = µ 2 sinr

So, the angles of .prism are 72°, 72° and 36°.

Ji sin(45') = 1.33sinr

Illustration 35 A ray of light incident normally on one of the faces of a right angle isosceles prism is found to be totally reflected as shown. What is the minimum value of the refractive index of the material of the prism? When prism is immersed in water (µ =1.33) trace the path of the

Ji( 1 )

sinr

T2

0.752 1.33 r = sin-• (0.752) = 48.75'

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Optics & Modern Physics

r2 >C

Illustration 36

A ray of light is falling on face AB of a tetrahedral of refractive index µ at angle of incidence i . The ray after getting internally reflected on face BC emerges from AD perpendicularly to the incident. beam. Find the range of µ and i.

=,.

sinr2 > sinC

=,.

sin(45°) > 1_ µ

=,.

->-

=,.

µ > .fi.

B

1

1 µ

.fi.

{--- sinC =;}

Further, we have µ =

sini

.

S1IlT1

sini sin(30°)

2sini

Since, µ > .fi. D

=,.

2 sin i > .fi.,

=O>

Sin I

. .

Solution Since, r1 + r2 = LB = 75°

... (1)

B

=>

1 > .fj_ i>45°

MINIMUM DEVIATION

The angle of deviation D varies with the angle of incidence i , as shown in the figure. D

__ J

D

From figure, we observe that e = i , because 1 and 2 are perpendicular

'' ''

i=e

It is found that D is minimum when i = e . Thus,

Dm1n =(i+e)-A=2i-A Using Snell's Law,

-----1 - ------1

·, e o/ I

c§):

lsini = µsinr1

2

and µsinr2 = lsine = sini

''

(Ray 1) 1- (Ray 2)

In quadrilateral QCDR, we have (90° -r,)+(90°+r,)+60° + 135° = 360° r, = 360°-60°-135°-(90°-r,)-90° T3

=>

µsinr1 =µsinr2

=,.

r1 = r,

Since, rt' + r2

... (2)

::::>

sini sine Further,µ=-.-=-.smr1 smr3

=A

2

..

{because i = e}

... (3)

Solving equations (1), (2) and (3), we get r2 = 45° and r1 = 30°

smi µ=--= sinr1

. (A+Dm;•) 2

Slil

. Sin

(A) 2

Note that if .the prism is equilateral or isosceles, then the ray inside the prism is parallel to its base.

Now, for TIR (total internal reflection) to take place at the face BC; we have

=

r (say)

A r=-

=r2 -15°

=>. r3 = r1

=

1.48

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Ray Optics: Refraction at Plane Surfaces A

A

s~-------~c Illustration 37 The angle of minimum deviation for a glass prism with refractive index ./3 equals the refracting angle of the prism. What is the angle of the prism? Solution Since we know that

./3

r1 = r3 = r (say)

=>.

6,+6Q+6, =120°

=>.

(i-r)+(180-2C)+(i-r)=120°

=>.

2i-2(60°-C)+180°-2C=120'

sin(1)

=>.

2i = 60°

=>.

cas(1)=

=>.

A=30° 2 A=60°

(say)

r=60°-C Given, OTotaI = 120°

sinA

./3 = 2cos( 1)

=i

=>.

=()min =AI so we get

=>.

=>.

=>.

=60°

Also, r2 ::::; C

sin(1) Bminimum

Since, r1 + r2 = r2 + r3

Similarly by symmetry, we have i1 =i,

(A+6m) s. m -µ= 2

Since

r1 = r3 = r i1=i2=i r2~c

=> i=30° Now, according to Snell's Law, we have sini sini µ = -1- = sinr1 sinr

~

But r1 +r2 =60° =>.

r+C=60° r=60'-C

=>.

µ

=>.

µ

=>.

Illustration 38 The path of a ray of light passing through an equilateral glass prism ABC is shown in the figure.

sin(30°) sin(60°-C) 0.5

./3 cosc--1.sinC 2

A

2

But sin C = -1. µ

BL.__ _ _-',L_ _ __:,.C

The ray of light is incident on face BC at an angle just greater than the critical angle for total internal reflection to take place. The total angle of deviation after the refraction at face AC is 120° . Calculate the refractive index of the glass. Solution The ray diagram is drawn for the sake of convenience.

=>.

./3Jµ'-1 =2

=>.

3(µ'-1)=4

=>.

3µ 2 =7

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=>

µ=~

=>

µ=1.52

Optics & Modern Physics B µ=A+,_ 2

'

WHITE LIGHT

White light consists of infinite number of continuous wavelengths (colours) ranging from 4000 A to 7800 A. For convenience it is divided into seven colours. Violet, Indigo, Blue, Green, Yellow, Orange, Red called as 'VIBGYOR' pattern. The Violet having least wavelength (maximum frequency) and Red having maximum wavelength (minimum frequency). VARIATION OF REFRACTIVE INDEX WITH COLOUR (CAUCHY'S FORMULA)

The refractive index

(µ)

of a medium varies with

wavelength (:\,) according to Cauchy's formula

where A and B are constants. The smaller the value of A, the larger is the value of µ . Thus, µ is maximum for violet colour and minimum for red. The deviation of a ray depends on µ it is larger for higher µ . Hence, violet suffers ~e maximum deviation and red the minimum. If light from sodium lamp falls on a prism then it disperses (breaks) into two lines called D1 ( 5890 A) and D, ( 5896 A) lines. Thus we observe that a prism causes deviation as well as dispersion. If Dv, DR and '!)y are the deviations caused by prism for violet, red and mean yellow rays, then for prism with small refracting angle (A) , we have Angular Dispersion D=Dv -D, =(µv -µ,)A DISPERSIVE POWER OF A PRISM

The ratio of angular dispersion to the mean deviation is called dispersive power, so Dispersive Power is

B C µ=A+,_,+,,,+ ...

where A , B and C are constants. From above we observe that refractive index decreases with increase of wavelength. It is maximum for violet and minimum for red colour and due to this variation of. the refractive index with the wavelength or the colour, a composite beam of light entering a prism splits into constituent colours.

0)

Angular Dispersion Mean Deviation

D Dy

Dv -D, Dy

where Dy is the deviation of mean light i.e., yellow light, whose wavelength is considered as mean of all the wavelengths present. Further for a prism of small refracting angle A , we have

D=(µ-l)A So, we have

DISPERSION

Dv=(µv-1)A, D,=(µ,-1)A and Dr=(µr-1)A

It has been observed that when a beam of composite light (consisting of several wavelengths) passes though a prism, it splits into its constituent colours. This phenomenon is called dispersion. The band of colours thus obtained on a screen is called the spectrum. If white light is used, seven colours are obtained as shown in the figure. The sequence of colours is VIBGYOR, from bottom to top.

So the dispersive power ro becomes o,

(µv-µ,)A (µy-1)A

(µv-µ,) µy-1

~

µ-1

where dµ:=µv-µ, and µ=µy The dispersive power co has no units and no dimensions. Its value depends on the material of the prism. COMBINATION OF TWO PRISMS

Red

Orange Yellow

Green Blue Indigo

Violet The dispersion of light takes place because the refractive index µ of the medium depends on the wavelength of light as given by Cauchy's formula, according to which

From a single prism, it is not possible to get deviation without dispersion, or to get dispersion P' without deviation. However, two small angled prisms may be p combined to produce Dispersion without Deviation or Deviation without Dispersion. The prism placement for both is shown here. The placement remains the same. It is just that we are to decide the relation between their refractive indices such that required condition may be achieved.

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Ray Optics: Refraction at Plane Surfaces A.

This give~

DISPERSION WITHOUT DEVIATION

Two prisms can be combined in such a way that the deviation of the mean ray produced by one is equal and opposite to that produced by the other. Such a combination is called a direct vision pris'm.

A'=(µv-µ')A µ~-µ~

Also from (1) we get

(µv -µ,)A=(µ~ -µ~)A'

A

=>

'White

1""""'~~~R7-c1~:....

ligh::,I

=>

V

or

A''=

(.!:.=!.)A. µ'-1

The net dispersion produced is

=(Dv -D,)-(D~ -D~)

Net Dispersion =(µv -µ,)A-(µ~ -µ~)A' . Net_ ·)=(µv-µ•J(µ-l)A-(µ~-µ~)(µ'-l)A' ( D1spers10n µ -1 µ' - 1 Net Dispersion = roD- ro'D' where co and ro' are dispersive powers of prisms P and P'.

o

B.

µ-1

D-D' =(µ-l)A-(µ'

So, in this arrangement of prisms, the mean deviation ·(o) caused by one prism is cancelled by the mean deviation (D') caused by the other prism i.e. D-D'=0

(µ-l)A-(µ' ~ l)A' =0

µ-1.. coD =ro'D'

.

is the condition for Deviation without Dispersion. The net mean deviation is

Crown ~-----~A,'

=>

(µv -µ,)(µ-l)A=(µ~ -µ~)(µ' -l)A'

-i)A'

Illustration 39 The refractive indices of the crown glass for blue and red light are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and.1.73 respectively. An isosceles prism of angle 6° is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident prism. (a) Determine the angle of the flint glass prism. (b) Calculate the net dispersion of the combined system. Solution (a) When angle of prism is small and angle. of incidence is also small, the deviation is given by o= (µ -1 )A

· Net deviation by the two prisms-is.zero, when deviation due to one cancels the deviation due to the other. So, 01 -0,=0

DEVIATION WITHOUT DISPERSION (ACHROMATIC PRISM)

... (1)

It is possible to combine two prisms of different .materials in such a way that each cancels the dispersion due to the other, Thus, the net dispersion is zero but a deviation is produced. So, in this arrangement of prisms, the dispersion (Dv -DR) caused by one. prism is cancelled by dispersion ( D~ -

Here, µ 1 and µ 2 are the refractive indices for crown . and flint glasses respectively, where

o;)

produced by the other prism. /

Crown

µ,

1.51 + 1.49 2

l 5 d · an µ,

1.77 + 1.73 2

l 75 ·

Angle of prism !or crown glass is A, = 6° Substituting this values in equation (1), we get (1.5-1)(6°)-(1.75-l)A, =0 This gives A2 =4°

i.e., (Dv-D,)-(D~-D~)=0 or

(µv-µ,)A-(µ~-µ~)A'=0

Hence, angle of flint glass prism is 4°

... (1)

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COLOURS OF OBJECTS AND COLOUR TRIANGLE

Flint

The colours of objects are due to·a number of phenomena.

6,+62=0

(b) Net dispersion due to the two prisms is given by Net Dispersion=(µ~ -µ,,)A,-(µ,,

-µ,,)A,

=> Net Dispersion = (1.51-1.49)(6°)-(1.77 -1.73)( 4°) =-0.04° ~

Net dispersion =--0.04°

Illustration 40 A ray of light is incident on a prism ABC of

refractive index

COLOUR TRIANGLE

If, Red (R), Green (G) and Blue (B) are primary colours. If

-./3 as shown in figure. B

The colours of opaque bodies are due to Selective Reflection. For example grass appears green because when white light is incident on grass, it absorbs all colours except green which is reflected. Black appears black because it absorbs all colours falling an it an reflects nothing. Similarly white appears white because it reflects all colours falling on it and absorbs nothing. · The colours of transparent bodies are due to Selective Transmission. For example a glass appears blue, because it absorbs all colours except blue, which it transmits. The colours of sky, rising and setting sun are due to scattering while the colours of soap bubble and kerosene oil film are due to interference.

P denotes Peacock Blue also called Cyan, M denotes Magenta, Y denotes Yellow and W denotes White, then from colour triangle we observe that

D 60'

R

E Find the angle of incidence for which the deviation of light ray by the prism ABC is minimum. (b) By what angle the second prism must be rotated, so that the final ray ·suffer net minimum deviation. (a)

At minimum deviation, we have r1 = r2 = 30° According to Snell's Law, we have

sini

µ=-.-1 smrt ~

.Js

sini1 sin(30°)

. . =2 .rs

~

Stnl1

~

i1 =60°

RAYLEIGH LAW

(b) In the position shown, net deviation suffered by the ray of light should be minimum. Therefore, the second prism should be rotated by 60° (anticlockwise). B. Dl'CT=---..--,E 60° 60° 60°

60'

A

B

R+G+B=W R+G=Y G+B=P R+B=M B+Y=W R+P=W G+M=W

Solution (a)

G

6_0'

60'

According to Lord Rayleigh, intensity (I) of scattered light is inversely proportional to the fourth power of the wavelength 1,. • So, l Ioc1,. 4

It can also be concluded that the amplitude (a) of the scattered light is inversely proportional to the square of the wavelength. 1 So, aoc ..2 {·.-Joca 2 } 1

C

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Ray Optics: Refraction at Plane Surfaces COLOUR OF THE SKY

When light from the sun travels through earth's atmosphere, it gets scattered by the large number of molecules of various

gases. It is found that the amount of scattering by molecules, called Rayleigh scattering, is inversely proportional to the fourth power of the wavelength. Thus light of shorter wavelength is scattered much more than the light of longer wavelength. Since blue colour has relatively shorter wavelength, it predominates the sky and hence sky appears bluish.

r

ii. the system is placed in wale. r what will be the net deviation? Refractive index of water .

A ray of light incident.on the face of a prism is refracted and escapes through- an adjacent face. What ·is the maximum permissible angle of refraction of th~ prism, If it is made of glass with a- refractive index of µ = 1.5 .

5.

In an isosceles prism of angle 45° , it is found that when the angle of incidence is same as the prism angle and the emergent ray grazes the emergent surface. (a) Find the refractive Index of the material of the prism. (b) For what angle of incidence the angle of deviation will be minimum? -

I

I

6.

A prism of fli_nt glass

(a)

BASED ON PRISM (Solutions on page 1.183) The path of a ray undergoing refraction in an equilateral prism is shown in figure. The ray suffers refraction at the faqe AB ,and the refracted ray is incident on the face AC

at an angle slightly greater than the critical angle and hence, totally reflected. After refraction- at the face BC the emergent ray makes an angle of 30° with normal at BC at" the point of emergence. Find the A

(a) (b)

2.

3.

(µ =¾) with an angle of refraction 9

i) .

At what angle should a ray of light fall on the face of the prism so that inside the prism the ray is perpendicular to the bisector of, the angle of the prism.

(b)

Through what angle will the ray tum after passing through both face5: of the prism?

7,

Light rays from ~ source are incident on a glass prism of_ index of refraction µ and angle of _priSm q.. At near normal incidence, calculate the angle of deviation of the emerging rays.

8.

One face of a prism with a refractive angle of 30° is coated with silver. A ray incident on another face at an angle of 45° is refracted and reflected from the silver coated face and retraces its path. What is the refractive index of the prism?

9.

A ray of light is incident at an angle of 60° at one face of a prism having refracting angle 30° . The ray emerging out of the prism makes ah angle of 30° with the incident ray. Find the angle of emergence and calculate the refractive index of

the material of the prism.

corresponding angle of incidence i. refractive index of the prism.

10.

A beam of light enters .a glass prism at an angle a and emerges into the air at an angle J3 . Having passed through th_e prismj the beam is deflected from the original direction by an angle 6 . Find the refracting angle of the prism and the refractive index of the materiatof the prism. In a prism of refractive index µ = 1.5 and refracting angle 60°, the condition for minimum deviation is fulfilled. If face AC is polished

60° µ= 1.5

i)

in a direction perpendicular to the other face.

Prove ttiat the ray will fail to emerge from the other face if cotA < cotC-1 , where C is critical angle for the material of prism.

11.

The index of refraction for violet light in silica flint glass is

1

I

1.66 and that for red light is 1.62. Find the angular dispersion of visible light passing through a prism of apex angle 60° , if the angle of incidence is 50° ·:

12.

A light ray is passing through a prism with refracting angle A= 90° .and· refractive index µ = 1.3 . Find the minimum and maximum-angle of deviation.

13.

A ray of li9ht. is incident at an angle of 60° on the face of a prism h~ving refracting angle 30° . The ray emerging out of the prism makes an angle 30° with the incident ray. Find the angle of emergence of the ray.

s~------~c Find the net deviation,~-~-----------~

A ray of light is incident upon one face of a prism (angle of prism <

A

L_(?)

3

'

30° i_S placed inside water ( µw =

1.

=i

4.

COLOUR OF CLOUDS

Large particles like water droplets and dust do not have this selective scattering power. They scatter all wavelengths almost equally. Hence clouds appear to the white.

b)

,_

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Advanced JEE Physics 14.

The .refracting angle of a glass prism is 30° ., A ray is incident onto .one of the faces- perpendicular to it. Find the angle 8 between- the incident~ray and the ray that leaves

the prism. The refractive indeX Of .glass is n =1.5.

15.

·

The refractive index of the mB.terial of a prism is,1.6 fora

strikil"lg at ,the first face again and emerging from it. in a direction making an angle of 4¢< with the reversed direction_ of the inCident beam. The refracted beam is found to have undergone a deviation of 1° from- the original direction. Ca,lcula,te.the refractive index of the glass and the angle ·of the prism.

certain monochromatic ray . .What should be the max;finum angle of incidence of this ray on the priSm so that no· total internal reflection occurs when the ray leaves ·the prism? The refracting angle of the prism is 45° .

16.

A ray of white light falls onto the side surface of an isosceles prism at such an angle that the red ray leaves the prism normally-to the second face. Find the deflection of the red and violet rays from the initial direction if the refraction a11gl8 of the prism is 45° . The refractive indices of the prism material for red and violet rays are 1.37 and 1.42, resp~ctively.

17.

A parallel beam of light falls normally on the first face of a prism 'of small refracting ang18 .. At the second face it is _Rartly_Jrahsmitte_Q_anQ..p_artly..reflected,Jhe_mflected beam~

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SINGLE REFRACTING SURFACE

b)

spherical surface which separates two media of different refrac~ve index is called a single refracting . surface. The copvexity or concavity of the surface is decided by looking at it from rarer medium as shown in figure.

A N

',,, ,,,.x_____--:

µ,

µ, I Denser!

Rarer

x~N~----

µ,

/

p

C

'' '

l

V ,._R-+< ' Convex Refrac!ing Surface

µ,

,

Rarer

C,

' 'l'

p

'

YL--------'

',._R-+i ' Concave Refracting Surface

SOME TERMS C6NNECTED WITH SINGLE REFRACTING SURFACE

1.

2. 3.

4. 5.

d)

I

Denser'/

/

c)

The distances measured against the incident ray are taken as negative. The. distances measured along the incident ray are taken as positive. All transverse measurements done above the principal ax.is are taken as positive while. the ones done below the principal axis are taken as negative.

Pole (p) : It is a point which bulges out most (in case of convex surface) or is depressed most (in case of concave surface) as seen from the rarer medium. Centre of Curvature (C): It is the centre of the sphere of which the surface forms a part.

ASSUMPTIONS

'While obtaining some relations, in ray optics, we make some assumptions given below. All those formulae will hold good only if these conditions are satisfied. a) The object/ source is considered to be point object/ squrce placed on principal axis. b) TI1e aperture of the surface/lens is small. c) Rays of light make smaller angles with the principal axis i.e., are paraxial in nature. REFRACTION OF LIGHT AT CURVED SURFACES

For the curved surfaces the same law of refraction are applicable. When a light-ray enters a denser medium, it bends· towards the normal. The figures show six situations. The shaded region is denser.

''

Radius of Curvature (R): It is the radius of the sphere of which the surface forms a part. Aperture (XY) : The diameter of the refracting surface is called the aperture of the surface. Principal axis : The line joining the· pole and centre of curvature and extended on either side of the surface is called the principal axis.

''

,

Real'

-~------

C

µ,

Denser

l

- i

'

, ,'

Real -- ______ _..._ '' ------µ, I 0 ~11 C Denser

(B)

(A)

In Figs. (A) and (B), the object O is kept relatively far from the refracting surface, and the image formed is real.

SIGN CONVENTIONS

Following sign conventions must be used while dealing with ray diagrams. a) All the distances will be measured from the pole of the surface.

\

Virtual

-·--------' Cµ2 !!J•1 Denser

(C)

1 :

fl:?

Denser

(D)

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AM

In Figs. (C) and (D), the object is nearer the refracting

AM AM

paatanP=-=- and MI PI

- _____.,.............."'-,

Virtual 0 µ,

I

AM

a.::::tana.=--=-MO PO'

surface, and the image is virtual.

C

AM

AM

yaatany=--=-MC PC Therefore (5) becomes

enser

(E)

(F)

In Figs. (E) and (F), the refraction always directs the ray away from the central axis, and hence virtual images are formed. Note the major difference from the images formed due to reflection from a spherical mirror. Here, real images are formed on the other side of the refracting surface, and virtual images are formed on the same side as the object. REFRACTION AT CONVEX SURFACE

CASE 1 : When the object lies in the rarer medium and the

=>

µ,(~~)+µ,(":)=(µ,-µ,)(~)

=>

l1_ + ~ = µ2 - µ1

PI PC Since PO=-u, PI=+v, PC=+R so we get 12+µ2 =µ2-µl -u V R

If the object is in air, then µ 1 = 1 and µ 2 =µ,so we get 1 µ' µ-1 -+-=--u V R

image formed is real.

Consider a spherical surface of radius R separating the two media 1 and 2 (µ, > µ 1 ) . A point object O is placed on the principal axis to the left of the pole P · at a considerable distance from it. The incident ray from O falls on point A and is refracted according to

. ... (1) Since the rays are paraxial, so the angle a is small and hence the angles i and r will also be small. Thus, applying such

paraxial approximation, then sin i ~ i and sin r ;;; r , so we get from (1)

... (2)

PO

CASE 2 : When the object lies in the rarer medium and the

image formed is virtual. Consider a spherical surface of radius R separating the two media 1 and 2 (µ 2 - µ 1 ) . A point object O is placed on the principal axis to the left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (1) Since the rays ~e paraxial, so the angle a is small and hence the angles i and r will also be small. Thus, applying such paraxial approximation, then sini i and sinr r, so from (1), we have

=

=

... (2)

µ,

0

I

''' '' y ,.__R_., '

(Real)

µ,

I

I

Denser

'' ''

,.__ U - - - , - - - V ---+I

!+- u ~

Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC,

i=a+y and r=r+p

... (3) ... (4)

Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ 1 (a+r)=µ 2 (r-P)

=>

µ 1a+µ 2 P=(µ 2 -µ 1 )r

I

'1

R ----+1

l+--V----+i

Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC,

i"=a+y

... (3)

and r=P+r

... (4)

Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ, (a+y) = µ 2 (P+r)

... (5)

µ 1a-µ 2P=(µ·2 -µ 1 )r

Now, since the aperture of the refracting surface is small, so

=>

M and P are very close to each other and hence, We have

Now, since the aperture of the refracting surface is small, so M and P are very close to each other and hence we have

=

1.56

... (5)

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Ray Optics: Refraction at Curved S111faces AM AM a:=:tana=--ee-MO PO'

AM AM yeetany=--ee-MC PC

AM AM peetanP=-ee- and MI PI AM AM yaatany=--ee-MC PC

µ,( ,:)+ µ,( ,:)=(µ, -µ,)( ~ )

.l:L+l:L= µ2-µt PO PI PC Since PO =-n, PI =+v, PC =-R so we get

µ,( :)-µ,( ,:)=(µ, -µ,)( ~~)

12+12 = µ1 -µ2

_&_-~= µ2-µ1

PO PI PC Since PO=-u, PI=-v, PC=+R so we get

11+ µ2 = µ2 -µ1 -u

V

R

CASE 3 : When the object lies in the denser medium and the image formed is real. Consider a spherical surface of radius R separating the two media 1 and 2 (µ, > µ,) . A point object O is placed on the

principal axis to the left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (1) Since the rays are paraxial, so the angle a is small and hence

the angles i and r will also be small. Thus, applying such paraxial approximation, then sin i =i and sin r =r , so from (1), we have ... (2)

-u V R Simply replace subscript 2 by 1 and 1 by 2 in the formula derived in Case 1 or Case 2. CASE 4 : When the object lies in the denser medium and the image formed is virtual. Consider a spherical surface ·of radius R separating the two media 1 and 2 (µ 2 > µ,) . A point object O is placed on the principal axis to the left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (1) Since the rays are paraxial, so the angle a is small and hence

the angles i and r will also be small. Thus, applying such

paraxial approximation, then sini::: i and sinr =r, so from

(1), we have ... (2)

µ, Denser

C

-

R __,___ V---+!

Using the geometrical property that an exterior angle of a

14----u----.a

triangle is equal to the sum of the two internal opposite

Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC y=a.+i ... (3) and r=P+r

... (4)

Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get

µ,a.+ µ,p = (µ, -µ,)y

... (5)

Now, since the aperture of the refracting surface is small, so M and P are very close to each other and hence we have

AM AM a:::tana.=--aa-MO PO' AM AM peetanP=-aa- and MI PI

angles, we get from triangles AOC and AIC a.=i+y

... (3)

and P=r+y ... (4) Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ, ( u-r) = µ1 (P-r) ~

µ, ( r- a)=µ, (P +r) ~

p

µ,a.-µ,P=(µ,-µ,)y

... (5)

Now, since the aperture of the refracting surface is small, so M and P are. very close to each other and hence we have

AM AM a:=tana=--aa-MO PO' AM AM paatanP=-ee- and MI PI AM AM 1 "' tany= MC" PC

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µ,( :)-µ,( ~)=(µ, -µ,)( ~ )

...!:L_..!::1= µ2-µ1

~

-u

h+ µ2 = µ2 -µ~

µ2

-u

R

V

-1:1_ _ ..e..t = µ1 - µ2

PO PI PC Since PO=-u, PI=-v, PC=-R .soweg~t

PO PI PC Since PO =-u, PI =-v, PC =-R sowe get

11 +11 = µ1 -

µ,( :)-µ,( ~)=(µ, -µ,)( ~ )

R

V

REFRACTION AT CONCAVE SURFACE

CASE 2: When the object lies in the denser medium.

For a concave. refracting surface the image formed is always

Consider a spherical surface of radius R separating the two media 1 and 2 (µ 2 > µ 1 ). A point object O is placed on the principal axis to the· left of the pole P . The incident ray from 0 falls on point A and is refracted according to ... (}) µ 2 sini = µ 1 sinr

virtual irrespective of the placement of the object. CASE 1: When the object lies in the rarer medium. Consider a spherical surface of radius R separating the two media 1 and 2 .(µ 2 > µ 1 ) • A point object O is placed on the

principal axis to the left of the pole P .' The incident ray from 0 falls on point A and is refracted according to

: .. (1) Since the rays are paraxial, so the angle a is small and hence the angles i and r will also be small. Thus, applying such paraxial approximation, then sin(:: i and sinr::: r, so from (1), we have .... (2) µ,

µ,

Rarer

Denser

Since the rays are paraxial, so the angle a. is small and hence the angles i and r will also be small. Thus, applying such paraxial approximation, then sini =i and sinr.:= r, so from,

(1), we have ... (2) -------A µ, Rarer

0

/ f;~; 0

l

C

·R--

C

'--------'B

Using the geometrical property that an exterior angle of a triangle. is equal to the sum of the two internal opposite angles, we get from triangles AOC and AIC

y=a+i

... (3)

and r=P+r

... (4)

Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ,(y-a)=µ,(y-p) ~

µ,a-µ,P=(µ, -µ,)y

... (5)

Using the geometrical property that an exterior angle of a triangle is equal to the sum of the two' internal opposite angles, we get from triangles AOC and .AIC

i=a+y

... (3)

and r=P+y

... (4)

Substituting the value of i and r from Equations (3) and (4) in Equation (2), we get µ, (a +y) = µ, (P +y) ~

AM AM a=:tana=--ee-MO PO'

AM AM a=:tana=--ee-MO PO'

AM AM peetanP=-ee-·- and MI PI

AM AM · peetanP=-ee- and MI PI AM AM yeetany=--ee-MC PC

AM AM yeetany=·MC ee PC

1.58

... (5)

Now, since the aperture of the refracting surface is small, so M and P are very close to each other and hence ~e have ·

Now, since the aperture of the refracting surface is smalt so M and P a_re very close to each other and hence we have

=

µ,a-µ,p = (µ, -µ,)y

~

µ,( :)-µ,( ~)=(µ, -µ,)( ~ )

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Ray Optics: Refraction at Curved Surfaces

=>

J:L_l:.!.= µ1-µ2 PO

PI

Since PO=-u, PI=-v, PC=+R so we get

AB=utani=ui and A'B'=vtanr~vr

CONCEPTUAL NOTE(S)

b)

For both ·convex and concave spherical surfaces, the refraction formulae are same,, only proper signs of u , v and R are to be used. For refraction from rarer to denser medium, the refraction formula is

-u

R

V

For refraction from denser to rarer medium, we Interchange µ 1 and µ 2 and obtain the refraction formula, J:g__+&=~-µ2 -u .v R

d)

Using Equations (1), (2) & (3), we get m= ~:·

(µ 1 = 1) and the ,denser medium

has refractive index µ (i.e., µ 2

=µ ), then

-u

V

R

.2._+l'.= µ-1 e)

u

µ2 - µ1 R

The factor

u µ,

H m is positive, the image is erect and virtual.

,------------------ -·-- - - CONCEPTUAL NOTE(S) 11+ µ2 = ~ -µ1

is equally applicable to R plane refradting surfaces i.e., surfaces for which R -i- oo . Let us . Real Depth denve µ -'-'--'-"'--'""'-"--~ using this. Apparent Depth

for refraction from medium:to air, we have -v

m=~l::1

=>

The refraction formula

for refraction from air to medium, we have

_!___+!'.= µ-1 (ii)

=:; =(:)(;,)

H m is negative, the image is inverted and real.

If the rarer medium is air

(i)

... (2) ... (3)

The magnification is defined as height of image A'B' m height of object AB

.&.+ µ2 = µ2-µ, ci)

... (1)

µ,

Now, in MBP and M'B'P, we have

12+1:!.=µl-µ2 -u V R

a)

~=11

=>

PC

Applying

R is called power of the spherical

refracting surface. It gives a measure of the degree to which the refracting surface can converge or diyerge the rays of light passing through It. For air-medium interface, the power is

.Hi.+ µ 2

=µ

-u

V

µ1 R with proper sign conventions and values, we get

-u

2 -

V

_µ_+.!.= 1-µ =0 -(-d) V oo d V=-µ

:::)

I

I

P= µ-1

R

R->oo

i

TRANSVERSE MAGNIFICATION

Instead of a point object O let us now, keep an extended object AB at point O such that its image A'B' will be formed at point I. The distance x(=-u) and y(=v) are related by the above formula.

~------~

I

B

!~~ 0 i.e., image of object O is formed at a distance

,

app

A

~ µ

on same side.

::. dactuar µ Real Depth

s-1 ,-u

!

f+ve

d

So d

A"!

I

2

µ = Apparent Depth

V

A ray from point B of the object is incident at point P and is refracted, in accordance with Snell's Law, such that µ 1 sini=µ 2 sinr

Illustration 41

In figure, a fish watcher watches a fish through a 3 cm thick glass wall of a fish tank. The watcher is in level

{applying paraxial approximation)

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with the fish; .the index of refraction of the glass is .

that of the wateris

.

.

_ll_

5

For face CD , we have

and

,8

_2_ - ~ =0

!.

AI,

3

To the fish, how far away does the watcher appear to be? · (b) To the watcher, how far aw'ay does the fish appear to be?

-11.16

(a)

:..-a cm

o

AI, =-(11.16)(¾)=--6.975 cm

o

FI, =8+6:975- •

o

Ff; = 14.975 cm

6.8cm

'' '' ·---------Observer

'

Illustration 42

There are two objects 0 1 and 0 2 at an identical distance of 20 ·cm on the two sides of the pole of a

Solution

spherical_ ·concave refracti.ng J,oundary of radius 60,cm. The indices ·of• refraction of the llledia on two sides of the I ,, ' I ,.

(a)

boundai; are 1 and

OA=3cm So,

AI; =(n, )(OA)

o

AI, =(¾)c3l=4.8 cm

the object (a) o, 'wheri see;,_ from (b)

E

C

r·

O, when see!' from· 0 1 •

The formula for-refraction from a curved boun.dary ls

:f.,

_&_.!:':!_= µ,-µ, v u' R

-4

.'

F

(a)

From the ray diagram drawri, we get,

G - ---+- +ve

D

' . ' µ =1, µ =4 · u1 =-20cm, R=--60cm, 1 2 3

For refraction at EG(R--> oo), using

n2 _!::!_= n2 -11i V u R 4· 8. 5

µ!

=1

0

-

60 cm _____.._'

FI, =6.5+6.8=13.3 cm

For face Ef, we have

8·

v=-24 cm

4

,.

(b)

BI, =-(6.8>(¾)(¾)=-8.16 c_m

C

'.

'

Keeping the object 0 2 on the left of the· pole P as shown, here, we get

E

··, i

.

• µ, -(4/3)

_,:_'_________

.

, ~

O, I P D

G

4 3

u1 =-20cm, R=+60cm, µ 1 ,=-, µ2 =1

O•--------A . ''·',B. --- ----- "7 ~F

=

i'

Thus, the-object 0 1 , will appear at a distance of 24 cm from P towards C .

...2._ - _3_ = 0 BI, --6.8

o

µ,= (4/3)

20cm:

-(4.8+3) 1+--

(b)

... ~·

-.::=::-_-,:-.:---~c~ --r o,_ P---~----------~.

BI, =-(7.8>(¼)(¾)=-6.5 cm

So,

o, .

Solution

___ J..: __ B, 'f.

3 BI,

(¾} respecfo'.ely. Find the location of

-- -- -.. ____ _ -------O,

C

+ve....,_ -----1 60 cm ---+1 20cm

1.60

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Ray Optics: Refraction at Curved Surfaces

1 (¾) 1-(¾)

Solution

(a)

; - (-20) =----;;a

Applying the formula for the refraction at the curved boundary i.e., '

=>

v=-16.36 cm Thus, the object 0 2 will appear at a distance of 16.36 cm from P towards 0 2 •

µ2 _l:!_ = µ2 -µ1 V u R

For, refraction at the first surface, the pole is P1 and we observe that

Illustration 43

An object of height 1 cm is kept at a distance of 40 cm fr~m a concave spherical surface having radius of curvature R = 20 cm , separating air and glass having refractive index µ = 1.33 . Find the location, height and the nature of the image formed. 1

1 (¾) 1-(¾) ; - (-oo)

V=-3R the left of pole P1 •

C

Titis image acts as an object for the refraction at the secon.d surface, with pole P2 • For this refraction, we have

' 20cmi I'

,..........

i+-40

cm-+-:'

4-

U=-(3R+2R)=-SR, R=-R, µ1 =1, µ 2 =3

Solution According to Cartesian sign convention, u=-40cm, R=-20cm

v=-~R 2

1.33 _ _ 1_= 1.33-1 V (-40) (-20)

Thus, the final image 12 is at a distance

v=-32cm

Magnification, m _h2 h,

µ,v µ,u

±_1

±

l, __1_=_3_ V -SR -R

Applying the formulaµ,_&_=µ,-µ, ,we get V u R

=>

(+R)

Thus, the first image 11 is formed at a distance of 3R to

µ,=1.33

emf

=

lx(-32) 1.33x(-40)

(b)

0.60

towards left. The ray diagram is shown in figure.

So, height of image, h, = mh., = 0.6 x 1 = 0.6 cm

_

The positive sign of magnification indicates that the image is virtual and erect.

¾) is

P, µ=1

µ=4/3 (a)

µ=413

Find the position of the image due to refraction at the first surface and the position of the final image. (b) Draw the ray diagram showing the position of the two images.

I

I

.

--- ............... . .

--------------------

o+---3R---a>t+--2R_.,

refracted by a spherical air bubble of radius R situated in water.

P,

I

---------

---....r~... --- ------«:{_. ...... ------•-----!; ----- ...........12___ .......... P1 P2

Illustration 44

R

------7

__,~-=--~?"" -

_!µ~~;'.'.4/'_:3 _ _________ ...

-

A parallel beam of light travelling in water ( µ =

SR from P2 2

---5R/2---M

Illustration 45 A solid glass with radius R and an index of refraction 1.5 is silvered over one lielllisphere. A small object is located on the axis of the sphere at a distance 2R to the left of the vertex of the unsilvered hemisphere. Find the position of final image after all refr~ctions and reflections have taken place. Solution The ray of light first gets refracted then reflected and then again refracted. For first refraction and then reflection

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the ray of light travels from left to right while for the last refraction it travels from right to left. Hence, the sign

C

conventio:h. will change accordingly.

A'-'------~D

Solution

Applying Snell's Law at face AB, we get

First Refraction : µ2 _!:!_= µ2-µl V u R conventions, we get 1.5 1 1.5-1 -;;;-- (-2R) = ~

Using

=>

V1

appropri9:te

1

=>

with

appropriate

sign

R

on face CD, we get

R

1.514

../2

1.514-../2

OE

oo

0.4

Illustration 47

A glass sphere of radius R =10 cm having refractive µ2 V

_

1:2_ = µ 2 - µ 1 u R

index µ8 with reversed sign

convention, we get 1 1.5 1-1.5

----=--

=>

= µ, - µ,

Solving this equation, we get OE"6 m

2

Again using

v,

1

Now applying, µ, - µ, V u

=-2 Third Reflection : Vz

.

smr=2

=> r=30° i.e., ray becomes parallel to AD inside the block.

-+-=-Vi 00 R

=>

(1)sin45° = (../2)sinr

sign

~oo

Second Reflection : 1 1 1 2 Using -+-=-=v u f R conventions, we get

1

with

-1.5R

=!

2

is kept inside water. A point object O is

placed at 20 cm from A as shown in figure. Find the position and nature of the image when seen from other side of the sphere. Also draw the ray diagram. Given refractive index of water is µw

-R

-

=_! . 3

v,=-2R A

0 M- 20

0

B

cm---.i

Solution R/2 1+-1,SR~

i.e., final image is formed at the vertex of the silvered face i.e., at the pole of silvered/ curved surface.

A ray of light starting from O gets refracted twice. The ray of light is travelling in a direction from left to right. Hence, the distances measured in this direction are taken positive. Applying

11+&= µ2 -µ1 , twice with appropriate signs at the Illustration 46 Figure shows an irregular block of material of refractive index ../2 . A ray of light strikes the face AB as shoWII. in tlie figure. After refraction it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at

-u

E . Find the distance OE upto two places of decimal.

=

1.62

R

V

two refracting surfaces, we get

M

p 2

0

I

2

B -----+- +ve

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Ray Optics : Ref.-actio11 at Curved Sm/aces

µ

(¼) + (¾) (¾-¼) -(-20) ~

10

AI1

Now, the first image 111 acts as an object for the second surface, so~ we have BI1 = u =-(30+20)=-50 cm

-u

= µ, -µ,

V

R

~

, we get

u, =-(3R-v,)

-10

~

i.e., the final image 12 is virtual and is formed at a distance 100 cm (towards left) from B . The ray diagram is as shown. N

-----::::::-r,

o

A

C

8

"4-20 cm--+1 1+--- 30 cm -----+1 1+-------100 c m - - - - - - + 1

u2 = -(3R ( 2:

u, =-(4µ-9)R 2µ-3

For refraction at curved surface 52 , we have 1

µ

(1-µ)

v,

-(4µ-9)R 2µ-3

(%)

v,

Illustration 48 A ·glass rod· has ends as shown in figure. The refractive index of glass is µ. The object O is at.a distance 2R from the surface of larger radius of curvature. The distance between apexes of ends is 3R. R

(4µ-9) >0 (10µ-9)(µ-2)

... (4)

The equation (4) is satisfied when Origin for first refraction

Origin for refraction at second surface

A

B

'' S' :

C

I

D

S'2 ,._ 2R _ _ R.,_3R/2___.. 1

R/2

Case I: (4µ-9)>0

R/2

9 µ>4

..-2R - ; - - - 3R ____..,

Find the distance of image formed of the point object from right hand vertex. (b) What is the condition to be satisfied if the image is to be real? Solution For refraction at curved surface 51 ,

... (3)

The image will be real if v 2 is positive, i.e.,

-J;?'4~~~ r';l~

(a)

... (2)

(4µ-9) (10µ-9)(µ-2)

0

air

3 ))

On solving the above expression for v 2 , we get

Following points should be kept in mind while drawing the ray diagram. (i) At P the ray travels from tarer to a denser medium. Hence, it will bend towards normal PC . At M, it travels from a denser to a rarer medium, hence, moves away from the normal MC. (ii) The ray PM when extended backwards meets the principal axis at I1 and the ray MN when extended meets the principal axis at I2 •

O

... (1)

2µ-3

vertex/pole of surface 52 • Object distance for second refraction is

BI, =-100 cm

r,

v,·= 2µR

The first image acts as object for refraction at second surface S2 • The origin of our Cartesian coordinate system is now at

(¾) + (¼) = ¼-¾ -(-50) .BI,

(µ-1)

µ 2µ-3 -=-v, 2R

AI1 =-30 cm

Again applying .l:L+ µ,

1

;,-- (-2R) =-R-

and (10µ-9)(µ-2)>0 ~

0.9<µ<2

there is no common solution for this condition and hence this is rejected. Case II: 4µ-9<0 1.63

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Optics & Modern Physics

Advanced JEE Physics

Solution Since parallel rays after passing thr9ugh a lens must converge (or appear to converge) atthe point. So this point is the place where focus is located and the final image is also formed at the focus. For refractio:ri at first surface, we get ·

9 4

=>

µ<-

(10µ-9)(µ-2)<0

=>

µ>2 OR

µ<0.9

Hence the common resul_t is

i

< µ < 2.25

1:1__1:1_ = µ2 -µ1 V1 -00 +R

Illustration 49 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- 111 for which a ray from P will emerge parallel to the table as shown in figure. ~ +ve

... (1)

I

--

1.-.-.-.-

., ...... .-

,-,

-->---!-"<'"-

:'

'

' ' !+V2 ---+: ,..'s - - - V , - - - - + !

For refraction at 2nd surface, we get 1+ mR +t+- R----+1

·

µ3 _ µ2·,=:µ3 -µ2 · Vi V1+R

Solution

Applying µ,

Adding equations (1) and (2), we get

~11= µ,-µ,, firstly at the plane surface

u. R and then at the curved surface.

.

V

µ3 =µ3-µl v, R

For the plane surface, we get 1.5

1

_ 1.5-1 _ 0

... (2)

=>

{·: R-->oo)

-µ,R

Vz=---

µ3 -µ1

Al, - (-mR) ----;;;--

Hence, focallength of the given lens system is

=>

AI,= -(l.5mR)

For the curved surface, since the final image is formed infinity, so we get

1

1.5 -(l.5mR+R)

00

=>

1.5 0.5 (l.5m+l)R. R

=>

3=1.5m+l

=>

2

=>

m=3.

3

-·1=...l!L

at

1-1.5 -R

µ3 -µ1

· Illustration 51 A parallel beam of light travelling in water having

refractive index·¾ is refracted by a spherical air bubble of

m,,,.2 4

Illustration 50 In the figure, light is incident on the thin lens as shown. Th~ radius of curvature_ for both the surface is R . Determine the focal length of this system.

radius 2 nun situated in water. Assuming the light rays to be paraxial. (i) F;_;,d the position of the image due to refraction at the first surface and the position of the final image; (ii). Draw a r_ay diagram showing.the positions.of.both the images. Solution (i) To get the desired result(s), we shall be applying

µ 2 - µ1 V

u

µ 1 , one by one on two spherical surfaces. .

4

1-±

oo

+2

.!a - 1. =-'-1. V1

==>

1.64

R

F~r first re-~actio~ at AP1B, ~~ have

µ,

=

=µ2 -

1 v,

1

-=--

6

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Ray Optics: Refraction at Curved Surfaces

-

=>

V1

====-

~~

sphere of refractive index 1.5 and of radius r is silvered to make the inner side reflecting. An object is placed at the axis of the sphere at a distance 3r from the centre of the

=-6 mm A

sphere. The light from the object is refracted at the

P,

P,

unsilvered part, then reflected from the silvered part and

again refracted at the unsilvered part. Locate the final image formed.

~\

B

So, the first image will be formed at 6 mm towards left

--•------------

of P,

0

For second refraction at AP2B, the distance of first

-~

image I, from P, will be 6mm+4mm=10mm (towards left). So, we get

±_1 _Q_ __l_ _ 3_ v,

-10

4 1 1 4 --------3v, 6 10 15

=>

Vi=-5 mm

Consider the figure shown. A hemispherical cavity of radius R is carved out from a sphere (µ =1.5) of radius 2R such that principal axis of cavity coincides with that of sphere. One side of sphere is silvered as shown. Find the value of x for which the image of an object at O is formed at 0 itself.

-2

=>

/

,.__ _ _ 3r _ _ _....,

3.

±

-----•------l

(ii) The ray diagram is shown in figure· Q,/

..... -I,

----+-

+ve

.--------------··· ...~x::·..,·~.: .' ~

I,

C

2mm 2mm

i.-

6 mm-->...--<---.!

4.

t
a)

At P and Q both normal will pass through C.

b)

At P ray of light is travelling from a denser medium (water) to rarer medium (air) therefore, ray of light will bend away from the normal and on extending meet at 11 • Similarly at 0. ray of light bends towards the normal.

c)

A glass sphere has a radius of 5 cm and a refractive index of 1.6. A paperweight is constructed by slicing through the sphere on a plane that is 2 cm from the centre of the sphere and perpendicular to a radius of the sphere that passes through the center of the circle formed by the intersection of the plane and the sphere. The paperweight is placed on a table and viewed from directly above by an observer who is 8 cm from the tabletop, as shown in figure. When viewed through the paperweight, how far away does the tabletop appear to the observer?

Observern

Both the images 11 and 12 are virtual.

. -~ -- I BASED ON REFRACTION AT CURVED SURFACES

Bern

~---- _____ • ________ICE_V_I _

3cm

(Solutions on page 1.187}

1.

A spherical convex surface separates object and image space of refractive index 1 and

the surface is 1O cm, find its power. 1+-40cm~

5.

F fl, '-µ,-'-------

2.

5cm

.±3 . If radius of curvature of

One end of a long glass rod having refractive index µ = 1.5 is formed into the shape of a convex surface of radius 6 cm. An object is located in air along the axis of the rod, at a distance of 1O cm from the end of the rod. (a) How far apart are the object and the image formed by the glass rod? (b) For what range of distances from the end of the rod must the object be located in order to produce a virtual image?

A hemispherical portion of the surface of a solid glass

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,fi.

An o~ject is at a distance of d =·2.5 cm from the surface of

a glass sphere with a radius R = 1O cm . Find the,position of

j

the image produced by the sphere. The refractive index of the glass is µ =1.5.

1 1

7.

l

A glass hemk:fphere of radius 10 cm and refractive. index µ =1.5 is Silvered over its curved ,surface. There is an air bubbl0 Jn the glass 5 ems fr9m the _plane surface along the axis. Find the position of the Images of this bubble seen _by. observer lo(lking along thB- axis ·into the flat surface of the

hemisphere. 8.

a

and cancavo-convex lenses. Sometimes, converging lens is represented symbolically by a double headed arrow as shown.

00 ~

Equiconvex

Biconvex

~

I

PlanoConcavo-convex Symbolic convex (Convex meniscus) representation

A hollow sphere of glass of refractive index µ has a small mark on its interior surface which is observed from a point outside the sphere on 'the sid9 opposite the center. The inner cavity is concentric with external surface and the thick11ess of the glass is everywhere equal to the. radius of the inner surface. prove that the ·inark will appear nearer

than it really is, by a distance (µ -l)R , where R is the

(3µ-1)

radius,of the inner surface.

~-----------------'

CONCAVE OR DIVERGING LENSES

Diverging lenses convert a parallel beam of rays into a divergent beam. Diverging lenses are concave, i.e. such that the thickness at. their edges is larger than the thickness at the middle. A concave l_ens is thinner at ·the centre. They include concave-concave~ piano-concave and convexo-concave lenses. Sometimes, a diverging lens is represented symbolically by a line with inverted arrows at its two ends.

THIN SPHERICAL LENSES

A lens is a piece of transparent material with two refraCting surfaces, at least one of them being curved. It may hav~ one surface plane. A spherical lens has spherical surfaces as bounds. If the thickness of the lens is small (compared to the radius of curvature of spherical surfaces, the object distance, the image distance, etc.), it is said to be thin. There are two types of lenses : a) convex or converging lenses, b) concave or diverging lenses. NAMING CONVENTION FOR LENSES

While naming a lens, the surface· with larger radius- of curvature is named first. The lens has a nature of the surface that has the smaller radius of curvature. EXAMPLE :

A lens with one surface plane and the other surface convex will be named as Plano-Convex irrespective of its placement and this lens will have converging . nature (the saine as the nature of the surface having smaller radius of curvature).

Similarly a Convexo-Concave lens will have a diverging nature and ConcaVo-Convex lens will have a converging nature. To summarise, we can say that the first name of a bifocal lens is derived from the name of the surface with bigger radius of curvature and the last name of.the lens is derived froirt the nature of the lens.

Equiconcave

BiPlanoConvexo-concave Symbolic concave concave ~ (Concave meniscus) representation

OPTICAL CENTRE OF LENS

The central portion of a lens (both convex and concave) behaves as a flat slab. Optical centre O is a' point through which any ray passes ~deviated. Face 1

---..

Face2

Face 1 ~-~ Face 2

---..:

/

c,

c,

Principal

axis Corwexlens

/

C,

C,

_Principal axis Concave Lens

PRINCIPAL AXIS OF A LENS

The line joining the centres of curvature C1 C1 is called the principal axis of the lens.

CONVEX OR CONVERGING LENSES

Converging lenses convert a parallel beam of incident rays into a convergent beam. Converging lenses are convex, i.e. such that the thickness at the middle is larger than . the thickness of edges. A cohvex lens is thicker in the centre than at its edges. They include convexo-convex, piano-convex,

=

1.66

PRINCIPAL FOCUS

A lens has two focal points. The first focal point F, is a point object on the principal axis for which the image is at infinity.

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, /

F,

F,

----:.~,. F,

2. The second focal point f 2 is a point image on the principal axis for which the object is at infinity.

F,

--

-- ........

3.

Focal length (/) is the distance between O and the second focus F2

' '''

F,

A ray passing through the optical centre goes through the lens undeviated.

F, F,

F,

/

F,

F,

F,

A ray passing through the focus (in case of a convex lens) or appearing to pass through the focus (in case of a concave lens) is rendered parallel to the principal axis after refraction through the lens.

•

Aperture is the effective diameter of the light transmitting area of the lens. The intensity of the image formed by the lens,

',

F,

F,

F,

',, F,

2

I oc (Aperture )

Converging and diverging action of a lens is due to the fact that a lens may be thought of a combination of small prisms, as shown in figure. A parallel beam of light, when incident on a convex lens, converges to a point called focus F .

,,

Any two of the above three rays can be used to obtain the location of the image.

,

Eiiiii:::--- . ',·

Converging action

Convex Lens

Diverging action

A concave lens diverges a parallel beam of light. It appears

A

to be diverging from a point F, called focus.

For thin lenses, we neecl not consider refraction of light at the two surfaces separately. Instead, we say that the light-ray is bent (towards the principal axis in case of convex lens, and away from the principal axis in case of concave lens) when it passes through a thin lens. RULES FOR OBTAINING IMAGES IN LENSES

1.

B

Concave Lens

THIN LENS FORMULA FOR A CONVEX LENS

A ray parallel to the principal axis, after refraction through the lens, converges to the focus (in case of a convex lens) or appears to diverge from the focus (in case of a concave lens).

Assumptions used in the derivation of lens formula a) The lens used is thin. b) The aperture of the lens is small. c) The incident and refracted rays make small angles with the principal axis.

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d)

The object is a small object placed on the principal axis. CASE 1 : When a real image is formed Consider an object AB placed perpendicular to the principal axis of a thin convex lens between its F' and C' . A real, _ inverted and magnified image A'B' is formed beyond C on the other side of the lens, as shown in figure.

-v--

Since OM= i\.B =:>·

A'B'

B'F

--=-

OF. From (1) and (2), we get . OB' B'F OB'+OF OB OF OF AB

Since, M'B'O alld AABO are similar, A'B' .OB'

AB

... (1)

OB

A'B'.- FB'

--=-·

MO OF Since MO = AB,

-v -v+ f -u=-f-

FB'

--=AB

OF From (1) and (2), we get OB' .FB' OB'-OF -=OB Of· OF

... (2)

.!.=.!._I. f

vf=-uv+uf

V

U

Thin lens fomtuia for a Concave Leris Let O be the optical centre and F be the principal focus of concave lens of focal length / . AB is an object placed perpendicular io its principal axis. A virtual, erect and diminished image A'B' is formed due to refraction through the lens. ·

.!'_=V-f -u f UV =Uf-vf

uv=uf-vf

Dividing both sides by, uvf , we again get

Object distance, OB= -u Image distance, OB'= +v Focal length, OF=+ f

=:-

-vf =uv-uf =:>

Using new-Cartesian sign convention, we get

=:>

Using new Cartesian sign converttion,

Object d~tarice BO= -u Image distance OB'= -v Focallength OF=+/

Also M'B'F artd llMOf are sll?lllar,·

A'B'

... (2)

Dividing both sides by uvf , we get

1

1

1

-=--V U

!

Ca~e 2 : When a virtual image is formed When an object AB is placed between the optical centre 0 and the focus F of a conVex lens, the image A'B' formed by the convex lens is virtual, erect and magnified. as shown in figure. · ·since, triangles A'B'O and ABO are s~ar, so Wf: have A'B' OB' -=... (1) AB OB Also, triangles A'B'F and MOF are similar, so we have A'B'

B'F

OM

OF

1

Since, M'B'O and MBO are similar A1B1 OB' So, AB= OB

... (1)

Also, M'B'F arid ~OF are similar

A'B' FB' So,--=OM ·OF Since OM= AB, therefore

•

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Ray.Optics: Refraction at Curved Surfaces A'B' FB' --=AB OF From (1) and (2), we get OB' FB' OF-OB' OB OF OF Using new Cartesian sign convention, we get OB=-u, OB'=-v, OF=-/

... (2)

= -f +v -f

=>

-v -u

=>

vf=uf-uv

=>

uv=uf-vf

Dividing both sides by uvf , we again get 1

1

1

I

V

u

IMAGE FORMATION BY CONVEX LENS OBJECT POSITION

At infinity

DIAGRAM

F

Between F and 2F

At F

2F

2F

F

Beyond 2F

At 2F

POSITION OF IMAGE

2F

2F

2F

2F

NATURE AND SIZE OF IMAGE

At the principal Focus (F) or in the focal plane

Real, inverted and extremely diminished

Between F and 2F

Real, inverted and diminished

At 2F

Real, inverted and of same size as the object

Beyond 2F

Real, .inverted and magnified

At infinity

Real, inverted and highly magnified

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Advanced JEE Physics OBJECT POSITION

DIAGRAM

Between F and optical centre

POSITION OF IMAGE

On the same side as the object

2F 2F

F

NATURE AND SIZE OF IMAGE

Virtual, erect and magnified

IMAGE FORMATION BY A CONCAVE LENS

I

OBJECT POSITION

DIAGRAM

For all positions of object

POSITION OF IMAGE

hnages formed between the optical centre of the lens and the focus (F)

2F 2F

NATURE AND SIZE OF IMAGE

Always forms a Virtual, Erect and Diminished Image

Illustration 52

'' ' o~-~-~-~-~---~-~-.ri:~t'_-_-_________ : ____ ~!3__ _

A point object O is placed at a distance of 0.3 m from a convex lens of focal length 0.2 m . It then cut into two halves each of which is displaced by 0.0005 m as shown in figure. Find the position of the image. If more than one image is· formed, find their number and the distance between them.

'

'

''

,_,__ V = 0.6 m -------+I

Since the triangles OL,,L2 and 01112 are similar. So, we have

j_

0

2 X 0,0005

ffi

I,I, =OB= u+v OA u

L,L,

I,I, L,L, 1+-

0.3 m --+1

=,,

Solution Each part will work as a separate lens and will form its own image. For any part, we have u = ..:..Q.3 rn, / = +0.2 m .

1 1 1 Therefore, from lens formula, - - - = - , we have V

1 V

=,,

U

f

1 1 0.3' 0.2

v=0.6m

So, each part forms a real image. of the point object O at 0.6 m from the lens, as shown in figure.

= 0.3 + 0.6 0.3

0.9 =3 0.3

I,I, = 3(L,L,) = 3(2x 0.0005) = 0.003 m

Illustration 53

An object is placed 45 cm from a converging lens of focal length 30 cm . A mirror of radius 40 cm is to be placed on the·other si«:fe of lens so that the object coincides with its image. Find the position of the mirror if it is · (a) convex? (b) concave? Solution

The object and image will coincide only if the light ray retraces its path and it will happen only when the ray strikes the mirror normally. In -other words the centre of the curvature of the mirror and the rays incident on the mirror are collinear.

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Ray Optics : RefractioH at C11rued S11rfaces (a)

The rays after refraction from lens must be directed towards the centre of curvature of mirror at C . If x is the separation, then for the lens u=--45cm, v=x+40, /=30cm

r_ _ _ t----c

0, I

1+- 45

=>

V1

=120 cm

In the second case, shift due to the glass slab is given by

fil=(1-f;c}=(1-/8 }=4 cm i.e., u will now become ( 40-4) = 36 cm, so now we have

1 1 1 -+-=v, 36 30 =>

v2 =180 cm

Therefore, the screen will have to be shifted 60 cm away from the lens.

cm-+t---- x - - - 4 0 cm+i

1 1 1 Using lens formula - - - =V

U

f

Illustration 55

Find the distance of an object from a convex lens of focal length 10 cm if the image formed is two times the size of object. Focal length of the lens is 10 cm .

1 1 1 x+40 --45 30 45 3 X= ( 0) -40=50cm 45-30

----=-

Solution

(b) In case of concave mirror, the refracted rays from lens meet at C, the centre of curvature (C) of the mirror.

A convex lens forms both type of images, real as well as virtual. Since, nature of the image is not mentioned here, so we will have to consider both the cases. CASE I : When image is real In this case v is positive and u is negative with

lvl =2lul, so if u=-x then v=2x and f =10 cm

0, l

. . . 111 Sb u stitutmgm ---=-,we get V

1+-45cm-+1+-----x----~ Using lens formula

1 1 1 ---=- where u=-45cm, V

U

f

v=x-40, /=30cm,weget 1 x,-40

1 --45

1 30

x- 40 = 45x30 45-30 x=90+40=130 cm

1 1 1 -+-=2x X 10 3 1 2x 10 => x=l5cm x = 15 cm, means object lies between F and 2F. CASE II : When image is virtual In this case v and u both are negative. So if u=-y then v=-2y and f=lOcm V

A lens with a focal length f = 30 cm placed at a

1 2y

Through what distance should the screen be shifted for the image of the object to remain distinct? The refractive index of the glass of the plate is µ = 1.8 .

1

1

1

U

f

1 1 1 --+-=-2y y 10

distance of a= 40 cm from the object produces a sharp image of an object on the screen. A plane parallel plate with thickness of d =9 cm is placed between the lens and the object perpendicular to the optical axis of the lens.

-+-=v, 40 30

f

. . . 1 1 1 Su bstitutmg m, - - - = - , we get

Illustration 54

Solution In the first case,

U

1 10

y=5cm y = 5 cm, means object lies between F and P . LINEAR MAGNIFICATION (m)

The linear magnification (also called lateral or transverse magnification) m produced by a lens is defined as the ratio of the height of image to the height of the object. So, 1.71

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Advanced JEE Physics Height of Image Height of Object

m

I 0

{·: f

Since triangles ABC and A'B'C are similar, so

A'B'

CA'

AB

CA

( dv)=(~)du dt u' dt

AB=O CA=-U

Substituting the values in equation (1 ), we get Magnitude of rate of change of position of image is

...CA'=+v

=>

I

V

0

-u

- dv = 0.09 rns-1 dt

--=-

- - . u ----,--v -

I

V

0

u

V

Lateral magnification, m =u Rate of change of lateral magnification is given by dv du u---v- (--0.4)(0_09)-(1.2)(0.01) dm dt dt u' dt (0.4)2 -

ni=-=-

Please note that for both the lens and mirror we have

mreal = NEGATIVE i.e. mroal < 0 mvirtual

= POSITIVE

i.e. mv1rtua1

;> 0

dm

Linear/Transverse/Lateral Mcmn_lfication proauced by a lens'is' I 0

V f u f+u

where I. is size ·of image perpendicular to Principal Axis ~nd O is size of object perpendlcular to principal Axis. Axial Magnification :· Axial magnification iS the ratio of the size of image along the principal ·axis to the size of the

object along the ·pril1cipal axis. SO m axiai

·= Size of Image along Principal Axis Size of <;)bject along Principal Axis

dv

v2

m?XJa1 = du =u2=m c)

dv du

2

Areal Magnification : Areal fl'lagnificatiDn is the ratio of ' the area of_ imaJ~e to the area·of object. m . _ Area of Image_ areat Area of Object

A1

2

V

mareai =A='if"=m

dt => Magnitude of rate·of change of lateral magnification is dm . dt 0.3 per seco~d

=

f-v· f

·m----------

b)

'

- =--0.3 per second

CONCEPTUAL NOTE($)

a)

.. ,(1)

Further, substituting proper values in lens formula, we get 1 1 ·1 -+-={': u=--OA m __ ,/=0.3 rn} V 0.4 0.3 => V=1.2 rn

Using Conventions, we get A'B'=-l

=>

= cons_tant}

Illustration 57 Two thin convex·lenses of focal lengths f, and

are

separated by a horizontal distance d (where d < f, , d < f, ) and their priitcipal axes are separated .by a vertical distance b as shown in the figure. Taking the centre of the first lens ( 0) as the origin of co-ordinate system and considering a . parallel beam of light coming from theieft, find the X and - y-coordinates of the focal point of'this lens system. y

.A 1 A0

L, I

L, I

2

tb

0

0

C

Illustration 56 An ·Object is approaching a thin convex lens of focal length 0.3 m with a speed of 0.01 ms-1 • -Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distail~e of 0.4 m from the lens. Solution 0

1 -1 1 "h . . the 1ens formµ 1a •. Differentiating - - = - wit respect V

to time, we get

f,

u

I

-

11/

I

X

.

' d

Solution

Fo:r; the refraction through the firs~ lens, . We have u-+oo,so

v, = I, Since, d < f, , the first image (formed by L,) lies to the right of second lens L2 , so

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u, =+(!, -d) 1

1

V

U

Applying Lens Formula - - -

1

1

This is called Newton's formula. where x1 ~ F1 O = . distance of object froni. first principal focus

=-1 , we get

1\

f

x2 = F2 I = distance of image from second principal focus F2

1

v,-(11 -d)=f, LENS MAKER'S FORMULA FOR THIN LENS

f2U1 -d)

V

Consider a thin Jens having its optical centre at C and let 0 be the point object situated on its· principal axis as shown in figure. Light starting from O strikes the first surface of t\te

f,+f,-d

2

~ x=v,+d f,f,+d(f,-d)

lens at A 1 and heads towards I1 , however, refraction takes place at the second surface, thereby giving a final real image at I.

f, + f,-d Magnification for second Jens is given by

!, !, + f,-d

m v, u,

µ,

The image due to the second Jens is formed below its principal axis and is of the size mb . So, the y coordinate of the focal point system is given by y=b-mb

,~

~

y=b

,..___ u-~--1--i--+---

Consider refraction at the first surface only. Let a 1 ,

So, the coordinates of the focal point of this system are

(11--d)b] ~(x, y )-_rl1,1,+a(f1 -a) ~ ~-~~~,

f, + f, -d

and

y1 be the angles which the incident ray (OA1 ), refracted ray According the general law of refraction, applied at A,, we get µ 1 sin i1 =µ 2 sin r1

If the distapce of object and image are not measured from optical centre ( C) , but from ·first and second principal foci respectively, and if x 1 is the distance of the object from the first focus x2 is the· distance of the image from the second

is the focal length of the lens, then we have

u=-(f +x,),_ v= f +x,

P,

( A1I 1 ) and normal ( A1C1 ) make with the principal axis.

f,+f,-d

NEWTON'S FORMULA

f

I,

i.-R,+--<-t

f,b

f,+f,-d ' (!, -d)b y= f,+f,-d

focus and_ if

--

0

Since the angles are smal), so sini1 ~ i1 ·and sinr1 =r1 ~

µ1i1 = µ 2 r1

•••

(1)

·In A.4.iC1O, we have it =rt +at

In M 1C1I,, we have y1 = r1 + P =>

rt=Yt-P1

Substituting for i1 and r, in equation (1 ), we get ~

0

µ1 (r1 +a 1 )=µ 2 (r1 -P,) µ,p, +µ,a,=(µ, -µ,)y,

Since the angles are small, so they can be replaced by thej.r tangents. ,.___U--<M+--V----+I

According to the lens formula, we have

1 1 1 · ---=-

v

~

u

f

1

1

1

f +x,

-(f +x,)

f

=>

µ 2 tanj3 1 + µ1 tana 1 =(µ 2 -µ~)tanyt

=>

µ2 M1J~

( )(A,M M10 = µ2 -.µ1 .M1C1. (A,M,) + µ1 (A,M,)

1)

For· a thin lens, Mt lies close to C . Therefore, all ,the distances measured from M1 can be replaced by those measured from C . Hence we have

... (2)

x,x, =('

==================================== = 1.73

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Now consider r~fraction at the second surface. The ray A 1A, which is the refracted ray for first surface becomes the incident ray for second surface. Applying general law of refraction at A, (light going from denser to rarer medium),

l.=('µ,-1)(1-_1-) I R, R, If the first medium is air, then 1 µ 2 =µ,so we have

we get

µ 2 sin i2 ::::::,.

µ2i2

l. =(µ-1)(1-_1-) I R, R,

=µ 1 sin r2

= µ1r2

... (3)

In M,C2I1 we have i, = r2 + P1 In M,C,I, we have r,

This formula is called Lens Maker's Formula .

I

PROBLEM SOLVING TRICK(S) I a) A concave lens forms virtual, erect and diminished image. b) A convex lens may form real and virtual images. The real image Is inverted, it may be diminishe9 or magnified while virtual Image (armed by convex lens is erect and enlarged. c) Different Media 0'1 either side of Lens

=r, + P,

I

Substituting for i, and r, in equation (2), we get

=>

... (6)

µ,(r, +P1)=11t(Y, +P,) µ1P,-µ,P1 =(µ,-µ1)r,

1f a lens of refractive index µ 2 has different media an either

Since angles ·are small so replacing the angles by their tangents, we get 1

side, the medium of object space has refractive index ,µ1

µ1tanp, -µ 2 tanp1=(µ, - µ1)tany2

and that of image space has refractive index µ 3 , then focal length f of lens is

(A,M,) ( J(AM2C2 M,) Mi M2I1 = (A,M,) 2

::::::,.

µ1

µ2

µ3 =µ2-~+µ3-~ f R, R,

~2 - µ1

Since the lens is thin, M 2 lies close to C , so we get

b. - l'L = µ, - l't CI

CI1

I I

CC,

As a special case if we put µ 3

=µ 1 ,

we get the Lens

Maker's Formula.

... (4) Illustration 58

Adding equations (2) and (4), we get

A piano-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 bottom most · 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 is found to 25 be cm . Find the focal length of the lens. Assume

12_ + l::L +l:!_ _-12. = µ2 - µ1 + µ2 - µ1 CI1 CO CI CI1 CC1 CC,

µ,(c~ +;I)=(µ,-µ,i(c~, +c~J

8

;I c~ +;I =(l,, -l)( c~, +c~J

thickness to be negligible while finding its focal length.

c~ + =(~:- 1)( c~1 +c~J

Solution

When placed on a horizontal table with curved surface in contact with it~ In this case refraction of the rays starting from O takes place from a plane surface as shown in Figure I.

Applying sign convention, we have CO=-u, CI=+v, CC1 =+R1 , CC2 =-R2

v

:u +~=(l,, -1)(~ +-~J 1 1 ;---;;-=(l,, -1) (1R, - R,1)

f !

4cm

... (5)

Since, focal length of a convex lens is defined as the distance of that point from the centre of lens where a beam coming parallel to principal axis comes to focus after refraction through the lens, so when U-)-oo we have p=f

'C7 •

0 Figure I

So, we can use Apparent Depth

Real Depth µ

Substituting in equation (5), we get

=

1.74

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Ray Optics: Refraction at Curved Surfaces

=>

4

Solution

3

(a)

µ=-

Image of object will coincide with it, if the ray of light after refraction from the concave surface falls normally on concave mirror so formed by silvering the convex surface i.e., image after refraction from concave surface should be formed at centre of curvature of concave mirror or at a distance of 20 cm on same side of the combination. Let x be the distance of pin from the given optical system.

When the plane surface is in contact with the horizontal table. In this case refraction takes place from a spherical surface as shown in Figure IL

4lm0

Applying,

1.5 1 1.5-1 -20 -x -60

0

---=

Hence, applying

4

1

-µ 1

X

,we get

R

u

V

1

=µ 2

3

~ - --4

1

=>

4

-3

1

1 8

3

1

8

40 120 120 120

X=-=15 an 8

When the concave part is filled with water, then before striking with the concave surface, the ray is first refracted from a plane surface. So, let x be the distance of pin, then the plane surface will form its image at a 4 distance of liapp = µh i.e., x from it.

(b)

= -R

8

=>

II

V

v=-20 cm, U=-x, R=-60 cm

Figure II ~-µ 1

&,_& = µ, R- µ, , we get for

1

-=---=-

3R 3 25 75 => R=25 cm Now, to find the focal length we will use the lens maker formula:

3

Now, using

1:2. _ ~ = µ 2 - µ 1 with proper signs, we u

V

R

get

..!:.=(µ-1)(2.._2..) f

R,

R,

15 . -20

7=(¾- 1)(~- -~5)=;5

4

4 1.5--

3

-(¼x)

3

=-=w

1 -3 1 -26 -=-+-=-

/=75 cm

40 360 360 360 X= =-13.84 an -26 x

Illustration 59 The convex surface of a thin concavo-convex lens of glass of the 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.

'' '' '

LENS IMMERSED IN A LIQUID

If a lens (made of glass) of refractive index µ 8 is immersed in a liquid of refractive index µ 1 , then its focal length in liquid, f, is given by

'

:s:±2:, (a)

..!:.=('µ,-1 )(2-_2..) f,

If

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 of refractive index 4/3 , find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.

1 0

R,

R,

is the focal length of lens of air, then

]__=( oµ,-1 )(2-_2..)

lo

R,

R,

!,=[~:=: ]lo Now three cases arise which are discussed here. 1.75

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µ, > µ1 , then f,

and /, are of same sign and f, > f,

.

a)

If

b)

That is the nature of lens remains unchanged, but its focal length increases and hence power of lens decreases. In other words the convergent lens becomes less convergent and divergent lens becomes less divergent. If µ g =µ 1 , then f, ~ co and the lens behaves as a simple .

c)

glass plate. If µ, < µ, , then f, and /, have opposite signs and the

nature of lens changes i.e. a convergent lens becomes divergent and vice versa.

1

1 1 V u f 1· 1 1 ----=D-u -u f

---=-

~

~

1 1 1 --+-=D-u u f

~

Df=(D-u)u

~

u 2 -Du+Df=0

~

u

,-u --1+---VF,

zw

-~-+--+-t-------t ~ 0 0

(/)

I+----- D----~

2

D±~D -4JD 2

For u to be mathematically real,

Illustration 60

D 2 -4JD?:0

A lens has a power of +5 dioptre in air. Calculate _its power if it is completely immersed in water? Given µg =·~.

~

D?:4/

2

and µw=

4

3

.

PROBLEM SOLVING TRICK(S)

ISo, if the,object and the screen are placed at a distance less than j

Solution

Let /, and fw be the focal lengths of the lens in air and water respec_tively, then

P =_l

U=V=-

2

Since lens has power +5 D in air, so

i.e. the lens is placed exactly between the object and the screen. CaseII:For D>4/

1

f, =p=s=0.2 m=20 cm Using Lens Maker's formula, we get

_l =(µ,

f,

Similarly,

.

~

-1 )(_!_ __!_) RI R,

... (1)

We get two different position of lens (L1 and L2 ) for which the image of object.on the screen is distinct and·clear.

_!_=(.t2-1)(_!_ __!_) !. µ. R, R,

pw = ;:

=(µ,-µ.)(;!

41 ~

D

fw

1

_2>:

Casel:For D=4/

and P10 =µ·

afa

4f, then a virtual image will be formed. Hence, for a real image

I to be formed D

-;J

14-U,---1--------V,----------<>I

... (2)

Dividing-_equation (2) by equation (1), we get,

P. P,

(µ,-µ.) (µ,-1)

1 3

p =p'=+~D w 3 3 DISPLACEMENT METHOD

Consider an object and a screen fixed at a distance D apart. Let a lens of focal length f be placed between the object and the screen. From figure we observe that u+v=D

~

v=D-u

Also from Lens formula

L, First Position of Lens L, Second Position_ol the Same Lens (shownin grey) I

,i NOTE:. /

!· I

_

Do not Develop· a'rriisconceptl0:rfthat there are two lenses, intact the same lens is displace~Jhrough x from position L1 to L,.

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The object distances for these two positions are given by D-JD'-4JD

... (1)

2

.'

D+JD'-4JD

u,

i.e. size of the object (O) is the geometric mean of the sizes of the image for two position oflens L, and L2 • Also,

m,-m

... (2)

'2

4Dx , m1-m2=-D, -x

Since u+v=D, so

D+JD'-4JD

v,

D-JD -4JD

(D:;x')

2

v,

... (4)

2 We observe that·

!

u, =V2 = u(say)

... (5)

v1 =u, =v(say)

... (6)

x=v1 -u,=JD'-4JD x =D'-4JD

f=--

=>

.. ,(15)

m1-m2

Further if m, = m , then

m, = ..!_ m

f =...!!!!_ m 2 -1

... (16)

Finally, we observe that

I

D

•• X

.... (7)

2

"" I

X

m, -m, =-

Let the Jens be displaced through x, then we observ~ fro\11 figure that ,

=>

D-x D+x

X

... (3)

2

D+x D-x

=-----

2

.. '.(S)

m1

m2

=(D+x)'

... (17)

D-x

Using (7) in (1), (2), (3) and (4), we get U1 =V2

'D-x

... (9)

=-2-=U

D+x

... (10)

V1=U2=:=-2-=V

If m1 is the magnification for the first position of lens i.e. ~1 , then

m, _b,_5.._!:,_ D+x 0

u1

u

... (ll)

D-x

If m2 is the magnification for the second position,of the Lens i.e. L2 , then I 0

v u2

u v1

u

D-x

v

D+x

2 1 m, = 2- = ==-=--

... (12)

From (11) and (12), we observe that

,m,m,=1 ,

... (13)

SO, if magnification for position L1 , is m , then maghification . . L2 1s . _;_ 1 . for position m Also from (13), we get 00 02

=>

0=.,JY;i;

Dear

Students,

you

must

J(eep

in

mind

that

actually

"Displacement Method" is not in the syllabus, but the Examiner generally· askS the problems not in its name ·but' by its concept_ e.g. an examiner's mind may fabricate a problem not having the name Displ~cement method but then the problem must be having a clue which may state D > 4f or the lens is displaced to get'two 1 real images on Screen and stuff like that. So, you are advised not I to overlook the.topic as this is very important (not by name) but'\ by the concept involved.

Illustration 61

A thin converging lens of focal length f is moved behveen a candle and a screen. The distance behveen the candle and th~ screen is J?(> 4/). Show that for two different positions of the lens, two different images can be obtained on the screen. H the ratio of dimensions of the image is p, fi~d the value of

(P +½)- _

Solution Let x be the separation between two positions of the lens for which a real image is formed on ·the sc;reen. Then, v+u=D ... (1) and v-u=x ... (2)

b_ I, =1 =>

PROBLEM SOLVING TRICK(S)

=1112 ... (14)

.

D-x

D+x

Solvmg we get u = - - and v = - 2 .2

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D+x D-x

Now, m1 =...1..=--

I D-x and m, -..1.=-- 0 D+x =>

.!i_=(D+xJ =~

=>

D+x=..ffi D-x

I,

=>

5u(u-4)-5(u-4)=0

=>

(Su-5)(u-4)=0

=> u=1m and u=4m Both the values are real, so this m~ans that there exist two positions of lens that form images of object on the screen.

D-x

(b)

V

m=u

(5-4) (5-1) 1-4 ~ Hence, both the images are real and inverted, the first has magnification -0.25 and the second -4 . Also, we observe that

m, = -(--) = -0.25 and m, = _(_)_ = -4

=>

x=(j:~)D D'

Now f=D'-x' , 4D

=

l'

._ffi - 1 - ~-

v)'

D

m,m, =(-0.25)(-4)=1

4D

Illustration 63

For two positions of a converging lens between an object and a screen which are 96 cm apart, two real images are formed. The ratio of the lengths of the two images is 4. Calculate the focal length of the lens.

1 (D7-2)' -2

~+Jl=

Solution

Since, m1 =4

Illustration 62 An object is 5 m to the left of a flat screen. A

converging lens for which the focal length is / = 0.8 m is placed between object and screen. (a) Show that two lens positions exist that form images on the screen and determine how far are these positions from the object? (b) How do the two images differ from each other?

m,

(~~:)' =4

D+x = . D-x

Substituting D = 96 cm, we get

96+x 96-x

Solution

(a)

Using the lens formula .!. _.!_=.!.,we get V

u

I

f=O.Bm

B

1+--

f = 21.33 cm

u ---- 5 - u

1 u

5

5-u+u (5-u)u

5 4

--+-=-

The power of a lens P is actually the measure of its ability to deviate the incident rays towards axis. The greater the curvature of the two surfaces (i.e., the shorter the focal length / ), the greater is the lens action. The shorter the focal length of a lens the more it converges or diverges the light, as shown in figure.

4

=>

20 = 25u-Su' 5u2 -25u + 20·= 0

=>

5u 2 -20u-5u+20=0

1.78

c::===================================

=>

=

1 5-u

x=32 cm D2-x2 Since, f = ~

POWER OF A LENS

1 1 1 ----=5-u -u 0.8 =>

=2

=>

=>

A

2

I,

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The power ofa lens placed in air is actually the reciprocal of the focal length of the lens in metre and is given by P=

1 f(in metre)

100 f(in cm)

Since/=½

P.,=L

=:>

/med

where µ is the refractive index of the medium and /med is the focal length of the lens in that medium. As a convention, the power of a converging lens (or convex lens) (with focal length positive) is taken to be positive. The power of a diverging lens (or concave \ens) (with focal length negative) is taken to be negative. AlsO we must note that for a mirror, power is defined as

p _ __::i__

100 /(incm)

/(inmetre)

Thus a convex lens and concave mirror have converging nature and hence they have poSitive power, whereas the concave lens and convex mirror have diverging nature and hence have negative power.

Nature of Lens

Mirror

Focal Length (f)

Power 1 plens

pmlrrcr

Concave mirror

Convex lens

Convex mirror

-ve

+ve

+ve

= f,

=-r1

+ve

Converging / Diverging

Converging

+ve

Converging

-ve

-ve

-ve

Ray Diagram

Diverging

Diverging

1 /=-=0.25m=25cm 4

Illustration 65

An object 4 cm high is placed at a distance of 10 cm from a convex lens of focal length 20 cm . Find the position, nature and size of the image. Also find the power of the lens. Solution

Here, u = -10 cm (the object assumed to be kept to the left of optical centre) f = +20 cm (positive for a convex lens)

h,_ = +4 cm (object kept above the principal axis) Using the lens formula, we get 1 1 1 ---==:,

v

u

1

1 -10

V

=:,

' Concave lens

Here P1 =6D, P2 =-2D

Power of the combination is given by P=P1 +P, =6-2=4 D

SI wlit of power is dioptre (D) . Power of a lens placed in a medium is defined as m

Solution

~ ~

=;t

=f-

Illustration 64

A convergent lens of power 6 D is combined with a diverging lens of -2 D . Find the power and focal length of the combination.

f

1 20

v=-20 cm

That is, the· image is 20 cm from the lens, on the same side as the object. Hence, the image is virtual. The linear magnification, h V m=i=-

h, " So, size of the image is

2 h, =h 1 (!:.)=4x - 0 =8 cm u -10 The positive sign indicates that the image is erect (and virtual).

Since, the power of the lens is given by p

1 /(in m)

1 =+5 D +0.2

Illustration 66 A converging lens forms a five folds magnified image of an object. The screen is moved towards the object by a distance d = 0.5 m and the lens is shifted so that the image has the same size as the object. Find the lens power and the initial distance behveen the object and the screen. Solution In the first case image is five times magnified. Hence

lvl=5lul 1.79

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Advanced JEE Physics In the second case image and object are of equal size. Hence

Here too, P11 P21 .••.. are to be substituted with proper signs

attached.

lvl=lul

The magnification of the combination is

M=m 1 xm 2 x ... =t'rm j,.]

In many optical instruments, the combination of lenses in contact are uSed so as to improve the performance of ·the

1+-X-+++-----Sx-------.l

instrument.

CASE-I

'' ' d=O.sm:'

1+---Y----<---Y-----++ CASE-II

From the two figures, we get

Consider two lenses of focal lengths I, and

6x=2y+d

=>

I,

kept in

contact. Let a point object O be placed at a distance u from

... (1)

6x-2y=0.5

the combination. The first image (say I1 ) after refraction

Using the lens formula for both the cases, we get for

from the fjrst lens is formed at a distance v1 (whatever may

CASE-I,

be the sign of v,) from the combination. This image I, acts an object for the second lens and let v be the distance of the final image from the combination. Applying the lens 1 1 1 formula---=-, we get

1 5x

1 -x

I

1

6 5x

I

as

---=-

1

... (2)

-=-

For the first lens, _!_ _ _! = _!_

CASE-II,

1

I

u

V

1

1

y -y

I

V1

---=-

=>

2

1

y

I

... (1)

/1

U

1

1

V

Vt

and for the second lens, - - -

' ... (3)

-=-

Solving these three equations, we get x = 0.1875 m and

=-1

/2

... (2)

Adding equations, (1) and (2), we get 1 1 1 1 1 ---=-+-=- (say)

vuf,l,I

I= 0.15625 m

Therefore, initial distance between the object and the screen

where,

I

is the equivalent focal length of the combination.

is 6x=l.125 m

1

1

1

F

fi,

I,

-=-+-

1 1 Power of the lens, P = - = - - - D = 6.4 D I 0.15625

TWO THIN LENSES SEPARATED BY A DISTANCE LENSES IN CONTACT

If two or more lenses of focal lengths

Iv I,, ... are placed in

contact, then their equivalent focal length / is given by

.!=_!_+_!_+ ... =

I I, I,

Where / 1 ,

/ 2 , ••••

.

are to be substituted with proper signs

The power of combination

combination is

1

1

F

I, I, Iii,

1

X

-=-+----

i;.!. ,., J;,

attached.

If two thin lenses of focal lengths Iv I, are placed at a distance x apart, then equivalent focal length · of

or Power for the combination is

P=P1 +P2-XP1P2 The net magnification of the combination is still remains

m=m1xm2

P=P,+P,+ ... =:tP; i~t

= ================================== 1.80

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Ray Optics: Refraction at Curved Surfaces Illustration 67

Consider a co-axial system of two thin convex lenses of focal length / each separated by a distance d . Draw ray diagrams for image formation corresponding to an object at infinity placed on the principal axis in the following cases: (a) d < f (b) d =f (c) f < d < 2/ (d) d = 2/ and (e) d > 2/. Indicate the nature of the combination (concave, convex or plane) in each case. Solution The formula

1

(d) When d = 2/: The incident parallel beam emerges out as a parallel beam but inverted. The combination behaves as a plane glass slab, which inverts the beam.

-r-

(e)

d

1

1

F

/1 /, f,f,

-=-+----

When d > 2 /: The incident parallel beam emerges out as a convergent beam. The combination behaves as a convergent or convex lens.

-r-

is valid only for small values of d compared to / 1 and /,. Therefore, we cannot use this formula in the given cases. However, we can draw the ray diagram to decide the nature. of the combination. (a) When d < f : The ray diagram is shown in figure. The out-coming rays are convergent. Obviously, the combination is a convex lens with F < f.

Illustration 68

··-····--...

----------·

---F--....., (b) When d = f : The incident parallel beam converges to a point and then passes without any more deviation. The combination behaves like a convex lens of F = f .

Two equi-convex lenses of focal lengths 30 cm and 70 cm , made of material of refractive index =1.5 , are held in contact coaxially by a rubber band round their edges. A liquid of refractive index 1.3 is introduced in the space between the lenses filling it completely. Find the position of the image of a luminous point object placed on the axis of the combination lens at a distance of 90 cm from it. Solution According to Lens Maker's Formula, we have

l_)

J_= (1.5-l)(J_ __ 30

=>

-R1

R1 =30cm

Similarly, radius of curvature of the second lens is 70 cm . Since 1 1 1 1 -=-+-+... (1)

----F---(c)

R1

F

When / < d < 2/: The incident parallel beam emerges

/1 f, f,

Here, / 1 =30 cm,/, =70 cm

out as a divergent beam. the combination behaves as a divergent or concave lens.

Now / 3 is calculated again using the Lens Maker's Formula, so we get ,.___F----+1

1 .!:..=(1.3-1)(---J_) /, -30 70 1.81

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Optics & Modern Physics

Advanced JEE Physics

=>

I, =-70 cm

=> F =30 cm {from equation (1)) According to Lens formula, applied on the combination of lenses, we have

1 1 1

---=-

v u

1

F

1

; - (-90)

1

= 30

v=45 cm LENSES WITH ONE SILVERED SURFACE

P=-2_=3_-~ f /, Im where f, is focal jength of lens and Im is focal length of spherical'rnirror formed due to silvering of surface. To have · a fundamental understanding of this we Can understand the silvering of lenses using the following' arguments. A ray incident on a lens with its backside· silvered will be refracted through the lens twice and will be reflected from the mirror once, as shown. a) Light from object O passes through lens ·10 form image I,.

When one face of a lens is silvered as shown in figure it acts like a lens-mirror combination.

b)

The image ! 1 acts as an object (virtual) for the curved mirror to form image 12 •

c)

The image I2 acts as an object (virtual) for the lens· to form the firtal image I •

=

Lenses with one face silvered act like lens-mirror combination

It is obvious from ilie ray diagram as shown in figur~ that the incident ray of light is refracted through the lens twice (i.e., once when light is incident on the lens and second time when reflected by the mirror) and reflected from the m!rror once.

+

+

The silvered lens acts like a mirror with equivalent focal ' length F , given by The combination acts like a mirror whose effective power is given by P,,e1 =2Pi+Pm whei-e Pi is the power of the lens and Pm is the power of the mirror. Since for a mirror we have

p =-~

m Im

and for a lens, we have

P, =2_=(µ-1)(~-~) f, R, R2 So, the combination acts like a mirror having net focal length given by

1 Fnet=-pnel

111121

------+----F /, Im f, f, Im where I, is focal length of lens and Im is focal length of spherical mirror formed due to silvering of surface. SIGN CONVENTION While using the above formula, we make use of tti_e following sign conventions.

a)

f is positive for converging (convex) lens and- concave mirror.

b)

f Is negative for diverging (concave) lens and convex mirror.

For example, for a plano-convex lens, from Lens Maker's Formula we get

2_ = (µ-1)(2--2-) = µ-1 f,

Rao

R

= ================================== 1.82

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Ray Optics: Refraction at Curved Surfaces R µ-1

=>

/,=-

a)

when plane surface is silvered, fm-+

0.03 m and when viewed normally through the curved surface it appears to be 0.036 m . If the actual thickness is 0.045 m , find the

co

(a) (b) (c) (d)

Since we know that

1

2 f,

1 "'

2(µ-1)

--=---

F

=> . b)

R

R

F

refractive index of the material of the lens. radius of curvature of lens. focal length if its plane surface is silvered. focal length when the curved surface is silvered.

Solution

2(µ-1)

when convex surface is silvered, then in general we know the relation between radius of curvature and the focal length is given by

(a)

Since, µ =>

µ=

Real Depth Apparent Depth

d-. dapp

R

= 0.045 =1.5 0.03

fm=z

Since we know that

1

2

2

--=-+F f, R

=>

2(µ-1)

R

2 2µ +-=-

R

R

-

F=-~ 2µ

0.045 m

Illustration 69

The plane surface of a plano•convex lens of focal length 60 cm is silver plated. A point object is placed at a distance 20 cm from the convex face of lens. Find the position and nature of the final image formed.

(b)

Using1 .!::,__b_=µ,-µ, V U R

=>

Solution

1 1.5 (-0.036) (-0.045) R=0.09 m=9 cm

weget I

1-1.5 (-R)

Since, P = 2Pi + Pm 1

=>

2

1

-I'= t,- Im where, Ji = +60 cm => =>

and fm -+ co

1 2 1 1 =---= F60oo30 F=-30 cm

(c)

The problem is reduced to a simple case where a point object is placed in front of a concave (converging) mirror of focal

But f," --> oo

length 30 cm . Using mirror. formula i.e.,

=>

1 1 1 -+-=v u f 1

1 1 v -20 -30 v=60 cm

-+-=-

1 F

R1 =+9 cm

2

I,

where, 2.=(µ-l)(J__l_) I, R, "'

where u=-20cm and /=-30cm =>

If the plane surface is silvered, then 1 2 1 -=-+F f, f.,

1+-- 20

cm -----+f

=> The image is virtual and erect

=>

1

2(µ-1)

F

R1

1 2(1.5-1) F +9 => F=9 cm The nature is given by applying negative sign to the final result. So, this will behave as a concave mirror. =>

Illustration 70 The greatest thickness of a planoconvex lens when

viewed normally through the plane surface appears to be 1.83

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Optics & Modem Physics

Advanced JEE Physics (d) When curved surface is silvered then R, ----> oo , R, = -9 cm 1 F

2 f,

1

-=---

fm

=>

_! = 2(µ-1)(~-_!_)-~

=>

.!=2(µ-1)+3.

=>

_!=2µ

~

F

,oo-9-9

9

F

F

9

9

1 1.5x2

-=--

+9 /=3cm F

=>

DEFECTS OF IMAGES : ABERRATIONS

The simple theory of image formation developed for mirrors and lenses suffers from various approximations. As a result, the actual images formed 'contain several defects. These defects can be broadly divided in two categories. a) Monochromatic Aberration The defects, which arise when light of a single colour, is used, are called monochromatic aberrations. b) Chromatic Aberration The index of refraction of a transparent medium differs for different wavelengths of the light used. The defects arising from such a variation of the refractive index are termed as chromatic aberrations.

shows spherical aberration for a concave mirror for an object at infinity. The rays parallel to the principal axis are incident on the spherical surface of the co_ncave mirror. The rays close to the principal axis (Paraxial Rays) are focused at the geometrical focus F of the mirror. The rays farthest from the principal axis (Marginal Rays) are focused at a point F' somewhat closer to the mirror. The intermediate rays focus at different points between F and F'. Also, the rays reflected from a small portion away from the pole meet at a point off the axis. Thus, a three- dimensional blurred image is formed. The intersection of this image with the plane of figure is called the Caustic Curve. If a screen is placed perpendicular to the principal axis, a disc image is formed on the screen. As the screen is moved parallel to itself, the disc becomes smallest at one position. This disc is closest to the ideal image and its periphery is called the Circle of Least Confusion. The magnitude of spherical aberration may be measured from the distance FF' between the point where the paraxial rays converge and the point where. the marginal rays converge. The parallel rays may be brought to focus at one point if a parabolic mirror is used. Also, if a point source is placed at the focus of a parabolic mirror, the reflected rays will be very nearly parallel. The reflectors used in automobile headlights are made parabolic and the bulb is placed at the focus. The light beam is then nearly parallel and goes up to large distance.

MONOCHROMATIC ABERRATIONS F

A.

SPHERICAL ABERRATION

Throughout the discussion of lenses and mirrors with spherical surfaces, it has been assumed that the aperhlre of the lens or the mirror is small and the light rays of interest make small angles with the principal axis. Only then, it is possible to have a point image of a point.object.

The rays reflect or refract from points at different distances from the principal axis. In general, they meet each other at different points. Thus, the image of a point object is a blurred surface. Such a defect is called Spherical Aberration. Figure

=

A lens too produces a blurred disc type image of a point object (due to finite aperture of lens). Figure shows the sihlation for a convex and a concave lens for the rays coming parallei'to the principal axis. We see from the figure that the marginal rays deviate a bit strongly and hence, they meet at a point different from that given by geometrical optics formulae. Also, in the situation shown, the spherical aberration is opposite for convex and concave lens. The point FM, where the marginal rays meet, is to the left of the focus for convex lens and is to the right of the focus for the concave lens. M-+---+.

M--+-----\--1

P-+--1-...__·,

P------A-;.,r

P-+--~rl-/

P----,---L

M-+----¥

M --+------f---¾.

1.84

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Ray Optics: Refraction at Curoed Surfaces The magnitude of spherical aberration for a lens depends on the radii of curvahrre and the object distance. For minimum spherical aberration the ratio of radii of curvahrre of lens is R1 R1

further above P1 • The image seen on the screen. thus have a comet-like appearance. Image of P

2µ 2 -µ-4 µ(2µ+1)

However, it cannot be reduced to zero for a single lens which forms a real image of a real object. A simple method to reduce spherical aberration is to use a stop before and in front of the lens. A stop is an opaque sheet with a small circular opening in it. It only allows a narrow pencil of rays to go through the lens hence reducing the aberration. However, this method reduces the intensity of the image as most of the light is cut off. Otherwise, the spherical aberration is less if the total deviation of the rays is distributed over the two surfaces of the lens. Example for this is a planoconvex lens forming the image of a distant object. If the plane surface faces the incident rays, the spherical aberration is much larger than that in the case when the curved surface faces the incident rays. In the former case, the total deviation occurs at a single surface whereas it is distributed at both the surfaces in the latter case.

R

Coma can be reduced by properly designing the radii of curvature of lens surfaces. It can also be reduced by using appropriate stops placed at appropriate distance from the lens. C.

d=f,-f, B.

COMA

CURVATURE

So far we have considered the image formed by a lens on a plane. However, it must be kept in mind that the best image may not be formed along a plane. For a point object placed off the axis, the image is spread both along and perpendicular to the principal axis. The best image is, in general, obtained not on a plane but on a curved surface. This defect is known as curvature. It is intrinsically related to astigmatism. The astigmatism or the curvature may be reduced by using proper stops placed at proper locations along the axis. E.

It has been observed that if a point cbject is placed on the principal axis of a lens and the image is received on a screen perpendicular to the principal axis, the image has a shape of a disc because of spherical aberration. The basic reason is that the rays passing through different regions of the lens meet the principal axis at different points. If the point object is placed away from the principal' axis and the image is received on a screen perpendicular to the axis, the shape of the image is like a comet. This defect is called Coma. the lens fails to converge all the rays passing at different distances from the axis at a single point. The paraxial rays form an image of P at P' . The rays passing through the shaded zone forms a circular image on the screen above P' . The rays through outer zones of the lens form bigger circles placed

ASTIGMATISM

Spherical aberration and coma refer to the spreading of the image of a point object in a plane perpendicular to the principal axis. The image is also spread along the principal axis. Consider a point object placed at a point off the axis of a converging lens. A screen is placed perpendicular to the axis and is moved along the axis. At a certain distance, an approximate line image is focused. If the screen is moved further away, the shape of the image changes but it remains on the screen for quite a distance moved by the screen. The spreading of image along the principal axis is known as Astigmatism (you must not confuse this with a defect of vision having the same name). D.

The spherical aberration can also be reduced by using a combination of convex and concave lenses. A suitable combination can reduce the spherical aberration by compensation of positive and negative aberrations. If two thin lenses are separated by a distance d, then condition for minimum spherical aberration is

~,

DISTORTION

It is the defect arising when extended objects are imaged. Different portions of the object are, in general, at different distances from the axis. The relation between the object distance and the image distance is not linear and hence, the magnification is not the same for all portions of the extended object. Hence a line object is not imaged into a line but into a curve and shown.

(b)

(c)

Object (a) and its distorted Images (b) & (c)

1.85

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Optics & Modern Physics

Advanced JEE Physics CHROMATIC ABERRATION

B.

The inability of a lens to form the white image of a white object is called chromatic aberration. In this case the lens forms coloured images of a white object. The chromatic aberration arises .due to the fact that the focal length of a lens depends upo~ the refractive index of material of the lens. The lens has different refractive indices for· different colours or wavelengths in accordance with Cauchy's formula given by

This is the spread of images perpendicular to principal axis and is given by

B

µ=A+, ). Accordingly, the refractive index is maximum for violet

(). =4000A)

I,-Iv

u

u

)0 =(rov' )0 u

f

u

ACHROMATISM & ACHROMATIC DOUBLET

ro,+ro'=O f, f, f, =- ro, f, ro,

and minimum for red (). =7800A) . Since

1

foc-µ-1

where ro1 and ro 2 are dispersive powers of materials of

Hence focal length of a lens is maximum for red and minimum for violet

=>

= v,O _ VvO =(v,-vv

The lens system free from chromatic aberration is called achromatic combination. This is obtained by using two lenses of different materials and different focal lengths and process is called, to Achromatise which satisfies the relation

7=(µ-1i(~ -~) =>

LATERAL CHROMATIC ABERRATION

lenses for focal length a)

freo > J,,,;,,~.

Figllre_ represents the chromatic aberration caused by a lens in the image of an object AB of size O . FR and Fv are second principal foci for red and violet

b)

colours respectively. The images of object AB are of different sizes and of different colours between AvBv and ARBR . The chromatic aberration is of two types. >+---V,---+< 14--.Vv~

f,

and

f,

respectively.

and ro, are always positive, therefore f,/ f, must be negative. This means the combinatj.on mu~t have one lens convergent and other divergent. For the achromatic combination (also called Achromatic Doublet) to be convergent, the powCr of convex lens must be greater or the focal length of convex lens must be smaller than that of concave lens. As dispersive power for crown glass is less than that for flint glass, therefore the convex lens must be made of crown glass while concave lens must be made of flint glass. Condition for minimum chromatic aberration obtained by two thin lenses of same medium separated by a distance d is As

ro1

d~f.,+f, 2

DEFECTS OF EYE

A,

A normal eye has nearer point at D ( 25 cm) called distance

of distinct vision and far point at c.o.

Chromatic Aberration

A. A.

AXIAL CHROMATIC ABERRATION

This is the spread of images along the principal axis and is given by VR-Vv

SHORT-SIGHTEDNESS OR MYOPIA

A short-sighted eye can see only nearer objects. It is due to elongation of eye-ball. It is corrected by using a concave lens, whose focal length is ~qua! to the far point of defective eye.

rov' =1

~\_

where ro is dispersive power, v is distance of image from lens for mean (yellow) colour and f is mean focal length of lens. If object is at infinity, then axial chromatic aberration, f,-fv=rof

1.86

Corrective DEFECTIVE EYE Image is not Created on the Retina

Lens DEFECT CORRECTED

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Ray Optics: Refractio11 at Curved Surfaces B.

LONG-SIGHTEDNESS OR HYPERMETROPIA

A long sighted eye can see only farther objects. It is due to contraction of eye-ball. It is corrected by using convex lens. This lens brings the nearer point of defective eye at a distance which equals to the distance of distance of distinct vision D( = 25 cm).

~\_

MAGNIFYING POWER

-~Tu I

l

Corrective DEFECTIVE EYE Image Js Created beyond the Retina

C.

When a small object is placed between optical centre and focus of a convex lens, its virtual, erect and magnified image is formed on the same side of the lens. The lens is held close to eye and the distance of the object is adjusted, till the image is formed at the least distance of distinct vision from the eye. For a normal eye, the least distance of distinct vision is 25cm.

Lens DEFECT CORRECTED

Case I : When image is formed at D It is defined as the ratio of the angle subtended by the image at the eye and the angle subtended by the object seen directly, when both lie at the least distance of distinct vision. It is also called angular magnification produced by the simple microscope. It is denoted b)' M. By definition, magnifying power of the simple microscope is given by

PRESBYOPIA

A presbyopic eye can see objects only within a definite range. This defect is corrected by using bifocal lenses. D.

M=Q_ a

In particle, angles a and p are small. Therefore, angles a and p can be replaced by their tangents i.e.

ASTIGMATISM

It arises due to distortion in spherical shape in cornea. This defect is corrected by using cylindrical lenses. OPTICAL INSTRUMENTS

An optical instrument is a device which is constructed by a suitable combination of mirrors, prisms and lenses. The principle of working of an optical instrument in based on the laws of reflection and refraction of light The common types of optical instrument are a) Projection instruments: These are used to project on the screen a real, inverted and magnified image of an opaque or transparent object so as to be viewed by a large audience. The object is, however, so fitted that its image is seen in erect form. An eye, a photographic camera, a projection lantern, an episcope, an epidiascope, an over~head projector, a film projector, etc., are examples of projection instruments. b) Microscopes: These are used to see very small objects in magnified form which otherwise cannot be seen distinctly when placed close to the naked eye.

·M = tanp =CA'=~ tana CA u If f is focal length of the lens acting as simple microscope, then 1 1 1 ---=v u f =>

~=1-~ u f

=>

M=l-~ f

B"

... ____________ _

Q

A'

F"

EXAMPLE A simple microscope and a compound microscope.

c)

-u-

Telescope: These are used to see astronomical and distant objects in magnified form which, otherwise cannot be seen clearly with the naked eye. EXAMPLE An astronomical telescope, a Galilean telescope, a terrestrial telescope, a reflecting telescope, etc.

4

1+--f~ - - - D ----+<

Since, image is formed at distance of distinct vision, so according to new Cartesian sign convention. v=-D and f=+f

SIMPLE MICROSCOPE (MAGNIFYING GLASS)

A convex lens of short focal length can be used to see magnified image of a small object and is called a magnifying glass or a simple microscope.

F

=>

D

M=l+/

1.87

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Optics & Modern Physics

Advanced JEE Physics

b)

a)

" CONCEPTUAL NOT!=(S) .· From above it follows that lesser'is the focal length of the convex. lens used as simple microscope, greater is the·

value of the magnifying power obtained. Further, the positive value of magnifying power of a simple microscope tells that .image formed is erect and hence virtuat.

c)

b)

Case II: When image is formed at infinity It is defined as the ratio of the angle formed by the image (situated at infinity) at the eye to the angle formed by object at the eye, when situated at least distance of distinct vision. It is denoted by M .

d) e) f)

In science laboratories, a magnifying glass is used to see slides and to read the vernier scales attached to the instruments. The use of magnifying glass enables us to place the object close to eye, making it appear bright and yet clearly visible. In position AB, object lies close to the eye. In absence of lens, the object will not be clearly · visible. It is also used by astrologers to read the fate lines of the hand. Used by Biology students to see slides. Used by detective department to match finger prints.

Illustration 71

A man with normal near point (25 cm) reads a book with small print using a magnifying glass (a thin convex lens) of focal length 5 cm. Find the (a) closest and farthest distance at which he can read the book when viewing through the magnifying glass. (b) maximum and minimum maguifying power possible using the above simple microscope.

I+-- u = f-----+-i i.----D----

Draw a line· A'B' = AB and perpendicular to principal axis at a distance CA'= D (least distance of distinct vision). Joint B' C . Then LB' CA' = a is the angle formed by object at the eye, when situated at distance D . The angle formed by the image situated at infinity at the eye is same as the angle formed by the object AB at the eye. Thus, LBCA = P is the angle formed by the image at the eye,

Solution (a) For a normal eye, far and near points are co and 25 cm, respectively. So, we have Vmm

=-25 cm

, 1 1 1 Using lens formula,---=-

v

u

f

U=-/-

(f)-1

By definition, M=f_

-> --<0 and

Vmax

So, u will be minimum, when v is minimum i.e.,

ex

Vmm

In practice, angles a. and 13 are small. Therefore, angles a. and p can be replaced by their tangents i.e. M= tanp =CA'=~ tancx CF u Applying new cartesian sign conventiOns CF;,,-/ and CA'=-D

=-25 cril

(u)m1n

{-: A'B'=AB)

5

-(15)- 1

~25 =--4.17 cm 6

And u will be maximum, when v is maximum i.e., vmax-),

co 5

(u)m~=-(s) =-5cm - -1

M=D

f

00

(b) Since magnifying power for a lens is CONCEPTUAL NOTE($)

V

It folloWs that magnifying power of the simple microscope is one less, when the image is formed at.inffnity. However, the vi0w(.ng Of the image is more comfortable.

USES

a)

Jewellers and watch makers make use of convex lens of short focal length to obtain a magnified view of the fine jewellery work and the small components of the watches.

m=-

u Magnifying power will be minimum, when u maximum i.e., um',.,.,; = -5 cm

=>

D

is

-25

(mlm1n =-=-=5 f , -5

m will be maximum, when u is minimum i.e., Umin

=-

25

6

= --4.17 cm

= ================================== 1.88

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Ray Optics: Refraction at Curved Surfaces ~

-25

(mlm~ =~=6 6

COMPOUND MICROSCOPE

A compound microscope is used to see extremely small objects. It consists of two lenses. A lens of short aperture and short focal length facing the object is called object lens and another lens of short focal length but large aperture is called eye lens. The two lenses are placed coaxially at the two ends of a tube. To focus over an object, the distance of the object lens from the object is adjusted with the help of rack and pinion arrangement. When a small object is placed just outside the focus of the object lens, its real, inverted and magnified image is produced on the other side of the lens between F and 2F . The image produced by object lens acts as object for the eye lens. The distance of object from the object lens is so adjusted that the final image is formed at the least distance of distinct vision from the eye. - Let AB be an object placed just outside the focus F0 of the object lens. Its virtual image A'B' is formed on the other side of the lens. The image A'B' lies between focus F~ and optical centre C' of the eye lens and it acts as object for the eye lens. Using the rack and pinion arrangement, the distance between object lens and the object AB is adjusted, till it virtual and magnified image A"B" is formed on the same side at the least distance of distinct vision.

Since the angles u and p are small, they can be replaced by their tangents·i.e. tanp A'B" M=--=-tanu AB \

A"B' A'Q AB ) ·:tanP=-- and t a n u = - = - (asA"Q=AB) C'A" C'A" C'A"

Multiplying and dividing by A'B', we have A"B" A'B' A'B' A"B" M=--X--=--X-AB A'B' AB A'B' v 0 A'B' M gnif" . where m0 =-=--= a 1cation produced by Object u0 AB lens.

and me=

:e =~::~=Magnification produced by Eye lens. '

. Further, for the eye lens, 1 1 1 ---=Ve

le

Ue

~

~

m =l-~

'

I,

Applying new Cartesian sign conventions we get

v, =-D and MAGNIFYING POWER

Case I : When image i~ formed at D It is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object seen directly, when both are placed at the least distance of distinct vision. EYE LENS OBJECT LENS

8

a

,,..,,,.,,.9,

I

1

.,,,.~

__ ,,. - - - - -

B"t.,..--

_,.!-i+---

'

I,

So, finally we get M=m 0 xm, = Vo(1+ Uo

D) fe

In practice, the focal length of the object lens is very short and the object AB is placed just outside the focus of object lens. Hence u, = lo (in magnitude) So that the final image is formed at least distance of distinct vision from the eye, the image A'B' must lie between optical centre and focus of the eye lens. Since, the focal length of the eye lens is also small, the distance of the image A'B' from the object lens is nearly equal to the length ( L) of the microscope tube i.e. v0 = L

'

!''

~

I, =+I,

D m =1+-

u.,--.i

i+----v, _ __,., l+----1,---.i f+-----D----><

Let LA" C'B" = p be the angle subtended by the final image at the eye. Cut A"Q equal to AB and join QC'. Then, LA"C'Q = u, the angle subtended by the object at the eye, when situated at the least distance of distinct vision. ·By definition, magnifying power of the compound microscope,

According to new Cartesian sign conventions we have Uo ';::!,-lo and Vo ';::!,L Substituting for u0 and M=-.!:..(1+

lo

V0

we get

D)

I,

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a)

Optics & Modern Physics

CONCEPTUAL NOTE(S) From the above expression, it follows that a compound microscope will have large magnifying power, if both the

object lens and the eye .lens are ot·small. focal length. In practice, focal length of object lens is smaller than that of

eye lens i.e.

t <~ .

Further the· negative value of

magnifying power of compound microscope tells that. final image fanned is Inverted. b)

In

practice,

to · eliminate

chromatic

aberration,

a.

combina,ion of two lenses in coritact is used. It is called

objective. c)

In place Of

an

~ye lens,_ a combination of two lenses at

certain distahce apart satisfying certain co·nditions (to minimize chromatic-and spherical aberrations) is used. It.is called eye piece.

Case II : When image is formed at infinity The magnifying power of a compound microscope,

M=m0 xme If u0 is distance of object from the object lens and v0 , then distance of image from it, then as obtained eailier,

ASTRONOMICAL TELESCOPE (REFRACTING TYPE)

An astronomical telescope is used to see heavenly objects. It produces a virtual and inverted image. As such bodies are round, the inverted image does_not"affect the observation. An astronomical refracting telescope consists of two- lens systems, Th~J~ns system facing the object is called objective. It has large aperture and is of large focal length (/0 ) • The other lens system is called eye-piece. It has small aperture and is of short focal length (I, ) . The objective and the eye-piece are mounted coaxially ill two- metallic tubes. The tube holding the eye-piece can be made to slide into the tube holding the objective with the help of rack and pinion arrangement.·

The objective forms the real and inverted image of the distant object in its focal plane. The position of the eye-piece is adjusted, till the final' image is formed at least distance of distinct vision. In case, pdsition of the eye-piece is adjusted such that final image iS formed at infinity, the telescope ·is said to be in normal adjustment. The_ image formation is discussed in the following n:vo cases. ·

V

m, =--'1.

u,

Further, as in the case of a simple microscope we have D

m=-

' f, where f, is focal length of the eyeJens. Therefore, magnifying power of a compound microscope is given by

v,

D

Uo

fe

M=-x-

Illustration 72

A compound microscope has a magnifying power 30. The focal length of its eye-piece is 5 cm . Assuming the final image ·to be at the least distance of distinct .vision (25 cm), find the magnification produced by the objective.

Case I :. _When final 'image is formed at infinity (Normal adjustment) When a parallel beam of light rays from the distant object falls on the objective, its real and -inverted image A'B' is formed on the other side of the objective. If the position of eye-piece is adjusted, so that the image A'B' lies at its focus, then the final highly magnified image will be fanned at infinity. Under such a situation i.e. in normal adjustment tne magnifying power of a telescope is defined as the ratio of the angle subtended by the image at the eye as seen through the telescope to the angle subtended by the object seen directly, when both the object and the image lie at infinity. It is also called angular magnification of the telescope and is denoted ' by M. OB CTIVE J~E-

,"-

Solution

For a compound microscope, we have

m

'

=(1+ D)=l+ f,

25 5

=6

From equatio;.,, (1), we get -30=m0 x6 ;:::>

y

f(\Ql,\

As the object is at a very large distance, the angle subtended by it at the eye is practically the same as that subtended by it at the objective. Thus, LA'CB' = a. may be considered as the angle subtended by object at the eye. Let LA'C'B' =p .

M=Ia.

6

1.90

A'*~ :,;:~:..::

$.\~

Then

30 m0 =--=-5

Negative sign implies that image formed by the objective is inverted.

=

"-

::>•' ! + - - - - !,.------- f,+l

.,: .. (1)

Since the final image is formed at least distance of distinct vision, the magnification of eye-piece is

C

EVE PIECE

Since the angles a. and Therefore,.

p are small,

a. " tan a. and

f= tan p.

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Using new Cartesian sign conventions we get CA'=+fo

For eye lens, we have

C'A'=-fe

1

1

Ve

Ue le

1

---=-

{·: distance of A'B' from object lens is along the incident light} =CO>

1 1 1 -=--u, v, f,

=CO>

_!_=_.!.( 1-f,

{·: distance of A'B' from eye lens is against incident light} M=fo

=CO>

Ue

f,

fe

Ve

M=- 1•(1-f,

J J

-·--·--------------------· --- - - - - , CONCEPTUAL NOTE(S)

=CO>

a)

Applying new Cartesian sign conventions we get

It follows that the magnifying power of a telescope in normal adjustment will be large, if objective is of large focal length and the eye·piece -is of short focal length.

b)

Further, when telescope is in normal adjustment, the distance between the two lenses is equal to sum of their focal lengths ( +

t t ).

c)

Further, the negative value of the magnifying power of the. _telescope.tells that final image formed is inverted and real.

Case II : When final image is formed at least distance of .distinct vision When a parallel beam of light rays from the distant object falls on the objective, its real and inverted image A'B' is formed on the other side of the objective. The position of eye-piece is adjusted so that the final image NB' is formed at least distance o( distinct vision. Under such a situation the magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the image formed at the least distance of distinct vision to angle subtended at the eye by the object lying at infinity, when seen directly. g

OBJECTIVE

EYE PIECE

::; l.:------LJ >-

oa: (.) L1J lL al 0

ct

C

I+--

t4-- fa -->-t D ------,--tot

Again, as the object is at a very large distance, the angle a subtended by it at the objective is practically the same as that subtended by it at the eye. Therefore, if LA'C'B' = P, then

Again, as angle a and p are small, they can be replaced by their tangents,

M = tanp = CA' tana C'A' Since, CA'= lo and C'A' = ue

f,=+f,, v,=-D, f,=+f, =:>

M=-f'(1+f')

f,

D

Therefore, a refracting telescope will have large magnifying power, if the object lens is of large focal length and eye lens ·is of short focal length. Further, the negative value of magnifying power of the telescope tells that the final image formed is inverted and real. Out of the two adjustments discussed, this adjustment gives a higher magnification, since the fact_or ( 1 + -{; ) is gre~ter than one. Also, a telescope does not increase the size of object, but it forms an image nearer so that the angle of vision is increased and hence it appears to us as if the bigger image of object is formed. Illustration 73 The objective of a telescope is a convex lens of focal length 100 cm . Its eye-piece is also a convex lens of focal length 5 cm . Determine the magnifying power of the telescope for normal adjustment.

M=f' Here, =CO>

M=I a

=:,

·Ve

Solution For normal adjustment, the magnifying power of a telescope is given by

L~;-;;Ei;; =-B'' ,

le

f, Ip =100 cm, le =5 cm

M= 100 =20 5

TERRESTRIAL TELESCOPE

A terrestrial telescope is used to observe obj~cts on earth. An astronomical telescope is used. to view heavenly objects since the inversion .of their images does not produce any complication..While viewing earthly objects we would prefer to have their images erect and hence, astronomical telescope is not suitable in such cases. By usinl? an additional convex

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lens O (of focal length f ) in between 0 1 and 0 2 of an astronomical telescope, we can have the final erect image. The lens O is called erecting lens, while the improved version of the telescope is calle.d Terrestrial Telescope. Rays from the distant object get refracted through the objective 0 1 , giving a real inverted image A1B1 • The erecting lens O is so adjusted that its distance from

Ai.B1

a

o,

a

is

equal to twice its (erecting lens) focal length. An image A,B, having same size as that of A 1B11 inverted w.r.t. A1 B1 and hence erect w.r.t. the object is obtained at a distance 2/ on other side of O . A2 B2 acts_ as an object for lens at 0 2 and

J+--0---+I

Parallel beam of incident rays from infinity are focussed by the objective 0 1 • An inverted image A 1B1 (shown in grey) would have been formed after refraction through 0 1 • Before the rays meet at A1 , a concave lens ( at 0 2 ) intercepts them.

finally an erect and magnified image is obtained after refraction through 0 2 • If the distance 0 2 B2 is equal to focal

The beam diverges and the final erect image A,B2 is

length f, of the eye lens 0 2 , final image is formed at infinity and the telescope is said to be in normal adjustment as in Figure 1.

formed at the distance of distinct vision. If O2 B1 is equal to

a

obtained. The distance 0 2B1 is so adjusted that final image is the focal length f, of eye lens at 0 2 final image is formed at infinity and the telescope is said to be set in normal adjustment. In such a case the length of the tube is equal to the difference between the focal lengths of two lenses. The field of view of this telescope is small because of the use of concave lens. When set in normal adjustment, its magi:iifying power M iS given by

A1B1

FIGURE 1

M =I= tan~= B10 2 = B,01

If the distance O2 B2 is less than fe then corresponding to a certain value of this distance, a virtual and magnified image is obtained at the distance of distinct vision as shown in Figure 2.

a.

tancx

A,B,

B,O,

B101

=>

M

F

f

Focal length of objective Focal length of eyeiens

RESOLVING POWER

The resolving power of an optical instrument is defined as the reciprocal of smallest angular separation between two neighbouring objects whose images are just distinctly formed by the instrument. The smallest angular separation is called the Limit of Resolution.

a

1+2f . . . 2f+I


~

A.

1+---D---+1

Limit of Resolution is given by

FIGURE2

Since the sizes of

AzB2

LIMIT OF RESOLUTION FOR MICROSCOPE

S=-"/,,-

and A1 B1 are same, ~troduction of

2µsin8

the erecting lens O has nor produced any change in its magnifying power but, has helped in getting the final image erect only. It may also be noted that the use of erecting lens 0 results in an increase (equal to four times the focal length of erecting lens) in the length of the tube of telescope.

where µsin0 is called numerical aperture and µ is refractive index of medium in which it is placed. For high resolution of microscope a beam of electrons is used which has wavelength of the order of 1 A .

GALILEO'S TELESCOPE

B.

Instead of using a combination of two, lens 0 1 and 0 2 for

If a is aperture or diameter of telescope and A. the wavelength, then resolving limit is

getting an erect image, Galileo used only one concave lenses to get the final erect image.

=

LIMIT OF RESOLUTION FOR TELESCOPE

d8oc~ a

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E.

1 For spherical aperture d0 = ·22 '- · .

a

,,

Resolving power oc !!..

Resolving power of telescope or microscope has no concern with focal lengths of lenses.

INVERSE SQUARE LAW OBEYED BY ILLUMlNANCE

When radiant energy falls normally· on a surface, the illuminance E of the surface is inversely proportioflal to the square of distance of surface point from source i.e. 1 Eoc-;, ... (1)

PHOTOMETRY F. A.

RADIANT FLUX

It is the radiant energy emitted by a body per second in all directions including all wavelengths. Its unit is watt.

B.

LUMINOUS FLUX

The radiant energy emitted by a body per second in the visible region (i.e., between wavelengths from 4000A to 7800A) is called luininous flux. Its unit is lumen. C.

LUMINOUS INTENSITY OF A LIGHT SOURCE

The luminous intensity of a light source in any direction may be defined as the luminous flux per unit solid angle in that direction. Its unit is lumen/ steradian or candela. Luminous Intensity I= ,ill M (in. lumen/steradian or candela)

If the light source is isotropic, then luminous flux is uniform in all directions, so that total luminous flux is given by F =4111 .(since total solid angle for all directions is 41t) If we plot the ratio luminous flux/radiant flux against 685 wavelength of radiation, we get a graph as shown in figure. The graph indicates that the ratio luminous flux/radiant flux is maximum for sssoA i.e. for yellow colour. 1his indicates that "-,--~~"-~-1,. our eye is most sensitive for 4000A 5550A 7800A 5550A i.e. yellow colour. The maxirnti.m value of ratio luminous flux/radiant flux is 685 lumen/watt. Thus when the r~tio luminolls flux/radiant flux is 685 lumen/watt, the luminous efficiency is said to be 100%. The tungsten filament bulb converts 2-3% electrical energy into visible light energy, while fluorescent tube converts 8-9% electrical energy into visible light energy.

LAMBERT'S COSINE LAW OBEYED BY ILLUMINANCE

When radiant energy falls obiiquely o~~.a, surface, the illuminance E of surface is directly proportional to .the cosine of angle made by normal to the surface with the direction of incident radiation. So, ... (2)

E oc cose Combining (1) and (2), we get

E=Icos8 . .. (3) r' :rotal Luminous Energy falling on a surface is given by Q=EAt where E = illurninance of the surface A=areaand t = exposure time. The total luminous energy required to be incident on a given type of camera film is constant. For a box type camera Time of exposure oc ( ~ )' where

d= diameter

f G.

=focal length of camera lens.

THE LUMINANCE

(L) OR BRIGHTNESS

Luminance of a surface is the luminous flux reflected by unit area of the surface normally.

=>

Luminance= Illuminance x Reflection Coefficient.

H.

PRINCIPLE OF PHOTOMETRY

If two sources of light of illuminating power 11 and 12 are placed at distances r1 and r2 from the screen, then the screen will be equally illuminated due to two sources when [1

12

?"=?" 1 2 ICE Vil

D.

ILLUMlNANCE OF A SURFACE (E)

BASED ON LENS FORMULA

It may be defined as the luminous flux falling per unit area on the surface. Its unit is Jumenm-2 or lux.

E=M t,S

,.

{Solutions on page 1.189)

I I

The distance between two point sources of light is 24 cm . Find out where would you place a conv~rging lens of focal i length 9 cm, so that the images of both the sources are ! formed attt:Le same ppint. ~ -·-··- --~- ·: ______ j

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lens paralieltoihe major optical-aXiSOfthe system to leave the lens, remaining parallel to th8 optical BJCiS, after being reflected from the mirror? Find the' image of the object produced by the given opfical system.

n object !s ·moved along the, ·prfncipal axis of a convex

ns. An image. three times the !:;ize of the .object is ob.taine_d hen the object is at a.distance ·of 16 cm from the lens ·and 'a' distance of 8 cm from the lens. Find the focal leng~h of the'lens. · ·

3.

13.

luminous point object. The separation between the object and the lens Is 3\ . A glass slab of thickness t is placed

The radius of curvature of. the. convex surface of a planqconvex lens is 10 cm and its focal length-is 30.cm-. What should be the refractive index of its material?

4.

One face of an equi-convex lens

{µ = 1.5)

A _convex lens of focal length \ is placed infront of a

between object and the lens. A real image of the o!)ject !S formed at the shortest possible distance from_ the object. (a) Find the refractive Index of the slab. is (b) If a concave lens of very large focal length

of focal length

t

eo· cin is silvered. Does it behave like a concave mirror or

placed in contact with the convex lens, find the shifting .of the image,

ca~vex mirror? Also determine the equivalent focal l~nQth of the,mlrror.

5.

A bicorivex lens made of glass with a refractive index of

14.

medium' (a) with a.refractive index of µt = 1.5

between -the ,lenses is d =30 cm .. An object is placed at a

·

distance of u1 = 30 cm from !he first. lens. At what distance from thEi second !_ens will the image be obtained?

(b) · with a refractive index ofµ, ~1.7 15. 6.

A biconvex thin tens is prepared from glass of refractive index

%. The two bounding surface5; have equal

radii of.

25 cm each. One of th? surfaces is silvered .from outside to make it reflecting. Where should an object be placed•before this lens.so th~t the image coincides with the object. 7.

A converging lens of focal length 5 cm is placed in contact with· a diverging lens of focal .leng,h 10 cm. Find the combined focal length of the system_.

8.

A~ biconvex lens of refractive index 1.5 has a focal length of ~ = 1o cm . One of the lens surfaces having a radius ·of

/

curvature of R=10 cm Is coated with silver. Determine the position of the image if ~the object is at a distance- of u == 15 cm from the lens.

10.

A concave spherical mirror with a radius of curvature of 0:2 m is filled with water. Calculate'the focal length of this

,

~

is placed lnfront of, a

17.

The distance between an object and a divergent lens is m times.,greater than the focal length ,of the lens. How many times will'the image be smaller than the object?

18.

An image I is formed of point object O by a lens w_hose optic axis·is .AB as shown In figure.

0•

• I

.(a) (b)

'

I

State whether it is a convex ler:iS or concave? Draw a ray'diagram to locate 1he le_ns and Its focus.

19. Two thin lenses having focal lengths \

t = 6 cm

=7 cm

and

are placed at a distance d,= 3 cm apart. whaUS

the distance of the focus of the ~ystem from. the second lens? Assume th·e system to ,be a centred one. 20.

Two glasses with refractive in\'.flc~s of µ 1 =1.5 and µ 2 ,=1.7 are used to make two identical,double convex lenses. (a) Find the ratio betwee"n their focal lengths. (b) How will each of these lenses act on a ray parallel 1o Its. optical axis if the lenses are submerged into transparent liquid with a refraclive·index of 1.6? "

a

shift In.the position of the image . At what •distance from a biconvex lens of focal length f.= 1 m shou[d a concave spheri~I mirror with a radius of curvature· of R = 1 m be placed Jar a beam incident on the

B

A

luminous point P so that the distance of the .point P from len~ is greate_r than focal length and the image formed is at the shortest possible-distance. If now a concave lens of very large. focal length t. be placed in contact with first, find the

.12.

4µ-2

Three convergent thin lenses of focal lengths 4a , a and 4a respectively are placed 'in order_ along the axis so that the distance between· consecutive lenses is 4a . Prov~ that this combination simply inverts every small object _on the 8.Xis without change of magnitude orposition.

.± .

A convex lens of focal length

,

16.

system? Given that refractive index of water is ' ' 3 11.

If r be the ,radius of curvature of each fBce of thin converging lens whose one face is silvered and µ. is the refractive index of _lens material, prove that the l_ens is equivalent to a concave mirror of focal length _r_.

A convex lens is held 45 cm above the bottqrn of an empty tank. The image of a point at the-bottom of a tank is formed 36-cm above the lens. Now a liquid is poured into the-tank to a depth of 40 cm . It is found that the distance of the image of the same point on the bottom of the tank is 48 cm above the lens. Find the" refractive index of the liquid.

9.

An Opti98.I system consists of two convergent lenses with, =1O cm . The distance focal lengths \ = 20 cm and

t

µ =1.6 .has a focal length of f =1O cm in air. Calculate the focal length of this Jens if It is placed into a transparent

21.

A parallel beam of light is incident on a system consisting of three thitUenses WithyCOmJTIOQ- opJLcal axis._ The focal

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t = +9 cm ,

i =+1O cm , t =-20 cm

respectively. The distance between the

first and the second lenses is 15 cm and between the

silvered I8ns. Find, in terms of x 1 and x 2 , the

second and the third 5 cm . Find the position of the point at which the beam converges when· it leaves the system of

(a) (b) (c)

lenses. 22.

surface the Image of the object pin is formed at the position of the ob_ject. pin. placed at a distance of x 2 from the focal length of lens radius of curvature of the curved surface and index of refraction of the medium of lens.

--------~

Consider a piano-concave lens with one of the radii of curvature r made up of a transparent material whose refractive index varies with intensity .(I) of incident light as

µ =.µ 0 + al , where a> 0 and O< µ 0 <

3

2

. Calculate

the

intensity' when the focal length Is equal .to two times the radius of. curvature r . 23.

Paraxiat rays are incident on surfaces of a thin equiconvex

glass lens of refractive index µ and having radius of curvaturE',I R . If the final image is formed after n internal reflectlons, calculate distance of this image from pole of the lens. 24.

When the plane surface of a piano-convex lens is silvered. it is found that the image of the object pin is formed at the p9sition of the object pin placed at a dis!ance of x1 from the silvered lens. When the same l~n?,j?,_silvered on th8 CU'-'Y§lsl

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Problem 1 A plane mirror is mo~ng with a uniform speed of 5 ms-1 along negatiVe x-direction and an ·observer P is moving with a velocity of 10 ms-• along +x direction. Calculate the velocity of image of an object O, moving with a velocity of 10,J2, ms-• as shown in the figure, as observed by the observer. Also firid its magnitude and direction. ·

Further, _parallel tci ~e mfrror,.i.e., along y-axis, we have

(v,), =(vo), =10] Since

So, absolute velocity of the image is

y

v, = -20/ + 10] Now i\p.=V1 -iip

i\, =-20/ + 10]-101 => v,, = -30/ + 10] =>

10./2 ms- 1

04__ __ _

=> If

5ms

v, = (v, ), +(v, ),

tanP=

Solution

Let v0 be the velocity of the object O, v, be the velocity of the observer P, vM be the velocity of the mirror and ·v, be the velocity of image (Assume all these velocities w.r.t. ground), then

!

- 10,J2, Vo= ,J2, ('l + J')

!

Ly

I0

v0 =10(/ + ])

X

VP =lOi VM =-5i

5ms 1

=>

30ms 1

Jv"l=-.'900+100=10Mms-•

p is the angle made by

1

-J+-----'~L.:"f--.-x

VIP with -x axis, then

10 30

P = tan-•

(½): with -x axis

Problem 2 Consider the situation shown in figure. The elevator is going up with an acceleration of 2 ins_, and the· focal length of the mirror is 12 cm . All the surfaces are smooth and the pulley is light. The mass pulley system is released from rest (w.r.t. the elevator) at t = 0 ..when the distance of . B frQm the mirror is 42 cm . Find the distance between th~ iroage of the block B and the mirror at t = 0.2 s . Take g=10 ms-•.

(v,M), = -(VoM);, Where !he axis perpendicular to the, mirror is .the x~axis.

=>

(v,), -(vM), =-(vo). +(vM).

=>

(v,). =2(vM), -(v0 ),

=>

(v,), =2(-s1l-107

=>

(v, ), =-.207

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Solution Let us assume thatthe acceleration ofblocks A and B to be a w.r.t. lift and a, be the acceleration of lift. G;msider block A , as seen from the reference frame attached

to the lift (a non-inertial frame), we get N=mg+maL T=ma

... (1)

... (2)

N·

m

T

Problem 3 A fixed cylindrical tank of height H = 4 m and radius R = 3 m is filled up with a liquid. An observer observes through a telescope fitted at the top of the wall of the tank and inclined at 0 = 45° with the vertical. When the tank is completely filled with liquid, he notices an insect, which is at the center of the bottom of the tank. At t = 0 , he opens a cork of radius r = 3 cm at the bottom of tank. The insect moves in such a way that it is visible for a certain time. Determine

---+a

Free Body Diagram of A

Now, consider block B, as seen from the reference frame attache~ to the lift! we have mg+ma~ -T=ma ... (3)

H=4m

Cork

T

Insect

(a) the refractive index of the liquid (b) the velocity of insect as a function of time. Solution (a) At t=0 .

3

smr=5

Free Body Diagram Of B

On adding equations (2) and (3), we get

µ = sinr =

a= g+a, = 10+2 =6 ms-'

2

5

Slilt

3../2

2

'' :.-,

~

So, distance fallen by block (B) is x =½at'

~Eye

'2

=>

x=2-x6x(0.2) 2

=>

x=0.12 m=12 cm

4m

Now, consider reflection at convex mirror, we have u=-(42-12) =-40 cm

3m

J=+12cin · .

1

1

v

u .f

(b) Let at time t , insect be at a distance x from centre of the tank. Since,

1

Smee-+-=-

=>

X

---1. = tan r

h

1 1 1 -+--=v (-40) 12 (40)(12)

480

40+12

52

=>

V

=>

v=9.23 cm

4

3

=>

x1 =-h 4

So,

x=(H-h)+x, -3

=>

x=4-h+~h-3

=>

X=l--

3 4

Therefore, the distance between the image of block (B) and mirror is 9.23 cm

=-3

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Advanced JEE Physics ... (1) From Equation of Continuity, we have

Solution·

(a)

For refraction at first half )ens, using lens formula, 1 1 1 ---=-,we get V U f

-A,(~~)=A,"2gh

1

1 1 -20 15 v=60 cm V

(- dh) = A,.fii dt

=>

Jii

=>

A,

V 60 -3 . . M agnif1cation, m=-=-=

')' ' -J.:!1:.=" 3 xrn,- ,l2x9.Bf dt h

H

(

Jii

,c{3)

u -20 The image formed by first half lens is shown in Figure-1

o

B

~5'

:

F _________ A,_

i

A

H-h

C,

:45°

H-h

'

r,

*

mirror. Mirror forms its virtual image A2 C2 . So, 2

B,

dt

So, speed of insect is

=>

=¼(-~~)

'' :' '' ' •A,

A convex lens of focal length 15 cm is split into two halves and the two halves are placed at a separation of 120 cm . Between these two halves of the convex lens, a plane mirror is placed horizontally and at a distance of 4 mm below the principal axis of the lens halves. An object of length 2 mm is placed at a distance of 20 cm from one half lens as shown in figure. f=15cm

!. ~

2mm

1+----

For the second half of the lens, using lens formula 1 1 1 ---=-,we get V U f 1

1 v -60 v=+20

1 15

---=-

=>

V 20 1 m=-=-=-u -60 3

f=15cm

~

½

So, length of final image A,B, = A2 B2 = 2 mm . However, point B2 is 2 mm below the optic :axis of

20 cm ,_.....,...__ ___._4 ~m

second half lens. Hence, its image B3 is formed

120 cm -------H

Find the position and size of the final image. (b) Trace the path of rays forming the image.

I mm 3

above the principal axis. Similarly, point A, .is 8 mm below the principal axis. Hence, its image is

1.98

4mm

Figure·2

Problem 4

·

2mm

C,

V=1.1x10--<(2-2.21x10-4t) ms-1

(a)

mm

length of A,B1 (i.e., C1 B1 ) acts as virtual object for mirror. Real image C,B, is formed of this part. Image formed by plane mirror is shown in Figure-2.

2

- dh =4.42x10--<(2-2.21x10-4 t)

V= ~:

2mm O1A1=60cm

4 mm length of A1 B1 (i.e.,A1C1 ) acts as real object for

Substituting H = 4 m, we get h = (2-2.21x10--!t)

AB=2 mm A1B1 =6 mm A0 1 = 20 cm 4mm 0 1F= 15 cm

Now, the point B1 is 6 mm below the principal axis of the lenses. Plane mirror is 4 mm below it. Hence,

h

''

B,

Figure-1

___ _ 1

~ mm 3

above it.

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Ray Optics Therefore, image is at a distance of 20 cm behind the second half lens and at a distance of

I3 mm

above the ::::::>

principal axis. The size of image is 2 mm and is inverted as compared to the given object. Image formed by second half lens is shown in Figure-3.

O'

--------•-·

---+----------

O"

B,

(b)

1

2

V1

=-50 cm (-50) (-12.5)

V

m,=-;;-=

-4

So, the image formed by the mirror is at a distance of 50 cm from the mirror to the left of it. It is inverted and four times larger. Step 2 : Image formed by lens The image formed by mirror acts as an object for the lens and the image formed by the mirror is at a distance of 25 cm to the left of lens. Using the lens formula, 1 1 1 - - - = - we get V

A,

1

-+--=v, -12.5 -20

Figure-3

f

U

1

The ray diagram for the final image is shown in Figure-4.

v, V2

.

1 25

1 -16.7

=-50.3 cm

V -50.3 and m1 =-=--=-2.012 u 25 Net magnification is given by m=m1 xm2 ~s

Hence, the final image is at a distance 25.3 cm to the right of the mirror, virtual, upright enlarged and approximately 8 times. Positions of the two images are shown in figure.

,, ,,,, ,, ,, ,,,,

'

,, ,'

B"

•''

A,,V '

Figure-4 A'

Problem 5 An object lies midway between the lens and a mirror. The mirror's radius of curvature is 20 cm and the lens has a focal length of -16.7 cm. Considering that the rays that leave the object travel first towards the mirror, locate the final image formed by this system. Is this image real or virtual. Is it upright or inverted? What is the overall magnification?

\/ I

I

li

H---

25 cm -----...i

B' !+-- 25 cm

\(

B

l\

A

'' ' '' ''

A"

--+1<---M<<---t- 25.3 cm 12.5 cm 12.5 cm

-...i

Problem 6 A thin piano-convex Jens of focal length f is split into two halves. One of the halves is shifted along the optical axis as shown in figure. 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 two halves. Draw the ray diagram for image formation.

Solution Step 1 : Image formed by mirror

0

.. f l 1 1 1 2 U sing mirror ormu a-+-=-=-, we get V u f R

I--''-'-'_ _ ___,

\

'.

\J

1.Sm

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Advanced JEE Physics y

·solution For both the halves, position of object and image is same, however the only difference is of magnification.

--

Magnlfication for one of the halves is given as 2(> 1). This can be for the first one, because for this, [vi> [u[ . Therefore, magnification, [ml=

l~I

> 1. So, for the first half, we have

=> [v[=2[u[ Let u=-x,then v=+2x

f

co-ordinate axes. Solution

(a)

For concave lens (L,) 1 v =>

1 1 1 -=--+---! 1.2-d -(0.6+d)

__

.. .....-...-f"'o.4m

----4M

.. ":'I "... .. --

----1-

-----------+-----------.·

f ----+t, ... Ct, a ,,. \

,

/,

,..,...--_ni;d-

).,, \V-~

;

'

f sin a h=(fsina.)m ~ -

'

i

\~

... (2)

l+cosa.

Hence the x-co-ordinate of image 12 is given by

IB

II

x=2f-lvlcosa.-hsina. =>

x=2f- fcos'a.

=>

x=2f-

=>

x=f(2cosa.+1) cosa+l

l+cosa

!

l

t+--:" 0.6 m ---,..i+-- 0.6 m--+++- 0.6 m

/sin 2 a., l+cosa.

f

l+cosa.,

Problem 7

Two thin lenses of same focal length f are arranged with their principal axes inclined at an angle a as shown in figure. The separation between the optical centers of the lenses is 2/ . A point object lies on the principal axis of the convex lens at a large distance to the left of convex lens. =1.100

,,-

'' ''

...----

,'\

fsina\

,....,...- ---.. ,.,,. j 2 '(A,, A,)

I ,/ ,/ .,, ------- .... ------\ 2 ,,:,,.,,-,~~-::::----B

L,

So, height of I, from the principal axis of L, is

1+--d----..

,/ I

... (1)

I

A -------+--------//, I ...... -- .; B, ,/'.1 ; ... ,......... ! A

( fcosa.)

v=- l+cosa.

r'f------

The ray diagram is as follows:

.,.... ...

-1 f

L,

12 and magnlfication for the first half is m1 = !'. = ( · ) = -2 u - 0.6

f=0.4m ./

1 -fcosa.

The magnification is given by V 1 m u 1 + cos a.

1 1 1 -=--+ 0.4 1.2-d (0.6+d) Solving this, we get d = 0.6 m Magnlfication for the second half will be V 0.6 1 m, =-;; = -(1.2) = -2

'

'''

Find the co-ordinates of the final image formed by the system of lenses taking O as the origin of the

= 0.4 m

I

''

(b) Draw the ray diagram.

For the second half, we have

~

0'

- - - - 21 - - - + <

(a)

and [u[+lvl=1.8m => 3x=1.8 m => x=0.6m Hence, u=-0.6m and v=+1.2m .11111 1 Usmg - = - - - = - - - - = ! V U 1.2 -0.6 0.4 =>

+ '' ''

0

''

' 1------1 2 ',

I

.,..,. , .,.

a.1/ ,

IVI

h'',~//

-'.,,

Similarly, y co-ordinate of image I,

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Ray Optics . 1 1 1 v, - (15-5) = 20

y = _:(lvlsincx-hcoscx) On substituting the_values of lvl and h from (1) and (2),weget y=O.

=>

So, the coordinates of the final irhage·are

[

20

v2 = - cm 3

/;;:::J

6.67 cm

V 20/3 2 and m2=-=--=u 10 3 About Final Image Net magnification is given by

t(2coscx+l), o] cosa.+1

(b) Ray diagram is shown in figure

m=m1m2

=-31

i.e., height of the image is 3 ~ ½ = 1 cm Since, the net magnification is-negative, so the final image is inverted.

Further y -coordinate of a point of the image will be,

Y1 =myo-m2.6.

CONCEPTUAL NOTE(S)

... (1)

The Y-co-ordinate of 12 is zero is very obvious because a ray of

with respect to the principal axis of L1

light starting from 11 and passing ihrgugh O' will suffer no

So, y -coordinate of image of A is

deviation. Hence, 12 must be forined an this line itself [e.,

y,. =(-½)
Y=O.

·-------------

y -coordinate of image of B is

·Problem 8

Two thin lenses f, = 10 cm and f, = 20 cm are separated by a distance d = 5 cm . Their optical centres are displaced a distance 6. = 0.5 cm . A linear object of size 3 cm placed at 30 cm from the optical centre of left lens.

0

Find the naturt; position and size of final image.

y,. =(-½)<3)-(¾)(½)=-¼ cm

Lx y

-t--- 0-- --------------

.B

A

0

Thus final image is as shown in figure.

-1--- -- ------------ :: ::i,:::!f B

L,

L,

B'

l

1/3 cm

·

f

413cm

!

1

6.67 cm

L,

L,

Solution

Problem 9 A prism of refractive index n1 and another prism of

Step 1 : Refraction from the first lens L,

refractive index n2 ·are stuck together with a gap as shown

Using the lens formula, .!._.!.=.!_,we get V U f

in the figure. The angles of the prism are as shown, n1 and n, depend on 1,, , the wavelength of light according to the

1 v,

=>

1 -30

1 10

v1 =15 cm

relations

given

by

n1 -- l "20 +10·8'J...2x 10'

and

145 + 1.80/..,2x 10' where 1,, is in nm. n2=.

V 15 1 and m, = - - - = - u -30 2

D

Step 2 : Refraction from th'e second lens L2

Again using the lens formula, .!. _ .!. =· .!_, we get V U f

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Optics & Modern Physics

Advanced JEE Physics (a)

Calculate the wavelength 1.. 0 for which rays incident at any angle on the interface BC pass through without bending at that interface. (b) For light of wavelength i..,, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum. Solution

Since,

-~ = 1.20 + 10·8,.,x 10'

ff

and

_

145 + nz-.

4 . On 3

opening at one end of a tank filled with water µ =-

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. Find the position (relative to the lens) of the image of the object formed by the system.

1.80x10 4 ').}

~

where, A is in nm. (a) The incident ray will not deviate at BC only if n1 = n,

=>

1 _2 + 10.s:10• Ao

=>

9x;O' =0.2S i.., 1 ,_ = 3x10 0 0.5 1..0 =600 run

=> => (b)

0.9 m - - ~ 0.8 m

---+-1

•-----------

1 _45 + 1.so:10• (i..=i..,) A.o Solution

· 1 (1 1) , ApplyingLensMaker'sFormula, -=(µ-1) --f R, R, we get

The given system happens to be a part of an equilateral prism of prism angle 60° as shown in figure.

__!_ =(~ 0.3

=>

2

-1)(!R --.!..) -R

{·: R, =Rand R2 =-R}

R=0.3

Now applying, !!:1__ µ 1 = µ 2 -µ 1 at air glass surface, we get V

=>

(say}

2

surface. Therefore, applying !!:1__!:!_ = µ 2 -µ 1 at glass water surface,

Since according to Snell's Law, we have

V

sini smr

u

R

we get

=-.-

4 3

sini=n'1 sin(30')

. Smee, ni

v1 =2.7 m

This image 11 will act as the virtual object for glass water

60'

f1=f2=-=30°=r

=>

R

So, the first image 11 will be formed at 2.7 m from the lens.

At minimum deviation, we have

n1

u

, =1.2 + 10.Sxl0' 2

i..,

where A. 0 = 600 run

=>

. ·-{l.2 + 10.SxlO'}(l)3 Slilt- -1.5 - -_ -

=>

, , -1(3) t=sm -

(600)1

2

2

v,

=>

3 2 2.7

(¼)-(¾) -0.3

v, =1.2 m

So, the second image 12 is formed at 1.2 m from the lens1or

0.4 m from the plane mirror.

4

4

Problem 10 A thin equiconvex lens of glass of refractive index

This iITI4ge 12 will act as a virtual object for mirror. Therefore, third real image 13 will be formed at a distance of 0.4 m in front of the mirror after reflection from it. Now this image acts as a real object for water-glass interface. Hence applying,

& - 1:2 = µ, R- µ, , we get V

U

µ = ~ and of focal length 0.3 m in air is sealed into an 2

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Ray Optics 4 3

(¾)-(¼)

-(0.8-0.4)

0.3

3 2

=>

Applying Snell's Law at face AB, we get sini sinr1

1 n=--

v, =-0.54 m

So, the fourth image is formed to the right of the lens at a distance of 0.54 m from it. Now finally applying the same formula for glass-air surface, we get

1-(I) l_--- = - -2-=-0.9 m I2

v, -0.54

=>

sini1 =nsinr1

=>

i1 =sin-1 (nsinr1 )

Substituting value of r1 , we get i1 =sin-1 {nsin(45°-C)}

=>

-0.3

i1 = sin-1 [ n(sin45°cosC -cos45°sinC)]

=>

Hence, the position of final image is 0.9 m relative to the

lens (rightwards) i.e., the image is formed 0.1 m behind the

=>

mirror.

. =sm. -1[ .J2n [ y1--;,:--; ~ n,J]

11

Problem 11

A right angle prism ( 45° -90° -45°) of refractive index n has a plane of refractive index n1 (n 1
Therefore, required angles of incidence ( 4) at face AB

for which the ray strikes at AC at critical angle is given by 1

i1 = sin-

A

[

}i.(Jn'-n; -n, )]

(ii) The ray will pass undeviated through face AC when ___fl_

n1 =nor r2 =0°

i.e., ray falls normally on face AC Since it is given that rl-i < n , so the option 11i = n is 8

ruled out, hence

C

(i)

Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle. (ii) Assuming n = 1.352 , calculate the angle of incidence at AB for which the refracted ray passes through the . diagonal face undeviated.

Solution

(i)

Critical angle C at face AC will be given by . C =Sm

=>

-1(11,) n

'2 =00

=>

r1 =A-r2 =45°-0°=45°

Now applying Snell's Law at face AB, we get sini1 n=-sinr1

=>

1.352

sini, sin(45')

=> · sini1 =(1.352)(1)

sinC =:i

n

=>

sin i1 = 0.956

=>

i, = sin-1 (0.956) "73°

Therefore, required angle of incidence is

4 = 73°.

Problem 12

B

Now, it is given that r2

=>

C

=C

r1 =A-r, =(45'-C)

A ray of light travelling in air is incident at grazing angle (incident angle = 90°) on a long rectangular slab of a transparent medium of thickness t = 1.0 m . The point of incidence is the origin A(O, 0). The medium has a variable index of refraction

n(y)

~ven by

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Optics & Modern Physics

Advanced JEE Physics n(y) = [kt/I'+ 1]1/' where k = 1.0(meterr'''

dy

The refractive index of air is 1.0 =,,

---K~- .

,,..<' I

'

or

4y114 =x

(c)

a

-'--7==----"-=-----,-,-------+X Air A(0,0)

=x

At the point of intersection on the upper surface, y=lm =o>

Obtain a relation between the slope of the trajectory of the ray at a point B(x, y) in the medium and the

X

4

= (256)1/ = 4 ffi

So the co-ordinates are ( 4 m, 1 m) (d) As nA siniA =np sinip and as nA

incident angle at the point. (b)

Obtain an equation for the traje~tory y(x) of the ray in the medium.

(c)

Determine the co-ordinates

(x1 , y 1 )

... (3)

0

=,,

,~ B(x, y), 1 Medium

_,,.,,.,"

y_,1'dy = dx

The required equation ·of trajectory is 4y 114

,

t=1.0 m

Thei-efore, ip

=iA =90°

= np =1

i.e., the ray will emerge parallel

to the boundary at P i.e., at grazing emergence.

of the point P,

where the ray intersects three upper surface, of the slab-air boundary. . (d) Indicate the path of the ray subsequently.

Problem 13

A thin .biconvex lens of refractive index

~ 2

is placed

i+8=90°, 8=90°-i,

on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with

Slope of tangent = tan8 = tan(90° -i) = coti

water of refractive index _! . It is found that when a point

tan8 =

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 again coincide at a distance 25 cm from the lens. Calculate the refractive index of the liquid.

Solution

(b)

fy_, 14dy =fdx 0

P(x,, y,)

Air

(a)

y'1' or

y

y

(a)

=

dx

3

!

dy

.

...(1)

-=cotz

dx

Applying Snell's Law at A and B nAsiniA =n 8sini8

1,,/liiu~,,,,mr

nA=1 because y=O

Solution

Let R be the radius of curvature of both the surfaces of the equi-convex lens, then in the first case, the situation is shown in figure. µ2=~

r\~-~Le,;s,~';?ol.!\,, -- ", -. -:.2··"Water I

!

/)/if f I I l l ) J f l l / 7 ) 7 ) )

sin iA

na

=1

because iA

=90°

Let f, be the focal length of equi-convex lens of refractive index µ 1 and· / 2 be the focal length of piano-concave lens (made of water) of refractive index µ 2 • The focal length of the combined lens system is given by

Grazing incidence

=~Ky312 + 1 =~y3/2 + 1,

because K = 1.0 (ni

t

12

~=..!:..+..!:..=(µ, -1)(~-J:...)+(µ, -1i(J:...-~) F !, I, R -R -R <X)

(1 )(1) = J(y' 1' + l) sini =,,

. .

. 1

Sllll=~--

Jy3f2 + 1

=,,

coti = ~y312 or y314

... (2)

=,,

½=(¾-1)(¾)+(¼-1)(-½)=½- 3~ = 3~

=,,

F=3R 2

Equating equations (1) and (2), we get

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'. RayOpHcs

Now, image coincides with the object when ray of light retrace$ its path i.e., it is,fall_s normally on the _plane mirror. This is possible only'when object li~s at centre of curvature of theJens system.

A

p •

B

Q

-. -----------·S

(I) 0

f

F=U=15cm

1+-- 30

I

f=15 cm

=>

3

=>

"

Solution

... ~ =>

cm ---+t+ 20 cm -+1

Rays coming from object AB first refract from the lens and then reflect from the mirror.

[ ·: Distance of object is 15 cm)

For refraction from the lens, we have

u=-20 cm, [=+15 cm

R =15 cm 2

A

R=lO cm

-r- ----r---0.6cm

_ of new lens system. Then,

:, =(µ, -1)(!- -~)+(µ-1)(_~ -

Optic Axis

B,

In the second case, let µ be the refractive index of the liquid filled between lens and mirror and, let f' be the focal length

3

!)

of Lens

r

B

------------Optic Ax.is of Mirror

cm

i A, ..,.._ 30 cm ------ao cm_.,.. 20 cm -11-1

1 =(-' 1)'(2)_(µ-1)= 1_µ-1={2-µ) F'2

R

F' =~ = ___!Q_ 2-µ 2-µ

RRR

1. 1 1 Applying lens formula, ---=-,we get V U f

R

{·: R=lO cm}

Now, the image coincides with the object when it is placed at 25 cm distance.

=> '=>

2-µ

=>

25µ=40

=> =>

v=+60 cm

V +60 m1 =-=--=-3 u (-20) So, the first image formed by the lens will be 60 cm from it (or 30 cm from the mirror) towards left and 3 times magnified but inverted. Length of first image A 1B1 would be

___!Q_ = 25 50-25µ=10

=>

and linear magnification is given by

F'=25

=>

1 1 1 :;;- (-20) = 15

40 µ=-=1.6 25 µ=1.6

A 1B1 = 1.2 x 3 = 3.6 cm (inver(ed). For reflection from mirror, we have

Image formed by lens (A1B1 ) will behave like a virtual

Problem 14

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 ls 30 cm. An upright object AB 'of height 1.2 cm is placed on tlie optic axis .PQ of the lens at a distance of 20 cm fr~m the lens. If A' B' is the image after refraction from the lens and the reflection from the mUTor, find the distance of A'B' from the pole of the mirror and obtain· its magnification. Also l~cate positions of A' and B' with respect to the optic axis RS .

object for the mirror at a distance of 30 cm from it as shown. Therefore u = +30 cm , f = -30 cm .

1 1 1 Applying mirror formula, -+-=-,we get V

1

1

U

f

1

-+-=-v 30 30

=>

V=-15 cm

and linear magnification is given by V (-15) 1 m =--=---=+, u +30 2

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Optics & Modern Physics

Advanced JEE Physics So, the final image A'B' will be located at a distance of 15 cm from the mirror (towards right) and since magnification is

+.!., 2

length of final image would be

(n2 > n1 ) at an angle of incidence O as shown in the figure.

A'B' = 1.8 cm

Since the p"oint B1 is 0.6 cm above the optic axis of mirror,

therefore, its image

B'

would be

(0.6)(½) =0.3 cm

above

optic axis. Similarly, point A 1 is 3 cm below the optic axis, therefore,

its image A' will be 3 x

Monochromatic light is incident on a plane interface AB between two media of refractive indices tt 1 and n2

A'B'=3.6x½=1.8 cm =>

Problem 15

½=1.5 cm below the optic axis as

The angle 0 1s infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (as shown in the figure), show that for any value of n3 all light will ultimately be reflected back again into medium II. Consider separately the cases (a) n3 < n1 and (b)

n, > n1 Medium I (n,) D ------------ E : Medium Ill :

shown.

I

G:

(nJ

e,' ''' '' '

Optic Axis of Lens

Optic Axis of Mirror

1.5 cm

A'B'= 1.8 m

15cmj

(n,)

Solution At interface AB, 0 is infinitesimally greater (slightly greater) than the critical angle for interface, so

Net magnification of the image is given by

m=m1

Medium II

'' '

A'

~

lF

A--~~-~--~---B

xm =(-3i(+½)=-%

1

2

0>sin-

(::)

(a) When n3 < n1

A'B'=(m)(AB)=(-%)<1.2)=-1.il cm

C.ONCEPTUAL NOTE(S) If the co·ordinates of the object (X 0 , Y0 ) are generally known to us with reference to the pole of an optical instrument (whether it, is a lens 'or a mirror), the corresponding co·ordinateS of image

(X,.~)

are found as follows.

X1 is obtained using OR

.!+.!. := ! V

U

f

(for a mirror)

1 1 1 ;;-u = f (for a lens)

Here, v is actually X1 and· u is X0 le., the above formula can be written as

Similarly,

Yi

Here, I is

.!_ ± ...!.. =! X1

X0

Yi

Yi = mY0 •

( :: )

< sin-

( ~)

(b)

Hence, critical angle for Medium III and Medium II will be less than the critical angle for Medium II and Medium I. So, if TIR is taking place between Medium I and Medium II, then TIR will definitely take place between Medium 1 and Medium III. When n3 > n1 , then Mo further cases may arise. Case I: n1 < n3 < n2

='fj

and O -is Y0 i.e., the above formula can -be

Y,

1.106

·

Is obtained from m

written as ·m =~ or

=

f

1

1

sin-

·

In this case there will be no TIR between Medium I and Medium III but TIR will take place between Medium III and Medium II. This is because : Ray of light first enters from Medium II to Medium III i.e., from-denser to rarer. So,

i>8

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Ray Optics Medium I

Medium Ill

Medium II

Medium I

,.'

q

p'

e,'

Medium Ill

i>9

I'

'

i <8

n,

Applying Snell's Law at P , we get n2 sine= n3 sini

Since, sin 8 slightly greater than

~

So, sin i will be slightly greater than "'- x n, =

n,

n3 "2

sini=(::}ine

n, ~

However " 1 is sin C for Medium I and Medium ill

n,

Since, sin8 is slightly greater than !!,_, so

n,

sini is.slightly greater

th~ "

2

n3

However, 1?-i

n,

interface, so sini > sinC for Mediwn I and Medium Ill interface.

x5..= 1Z-i 1½ n3

~

is nothing but sinC for Medium I,

i >(C),,111

Therefore, TIR will again take place between Medium I and Medium III and the ray will be reflected back.

Medium III interface, so sini is slightly greater than sinC for Medium I, Medium III interface. ~

CONCEPTUAL NOTE(S)

The Cases I and II for n3 > "1 can be explained by single

i>(C)1>m

Hence, TIR will now take place on Medium I and Medium III interface and the ray will be reflected back to Medium IIL Case II: f1i < n2 < n3

equation only. But two cases are deliberately formed for b~tter understanding of refraction, Snell's Law and total internal reflection (TIR).

1bis time while moving from Medium II to Medium III, ray of light will bend towards normal. Again applying Snell's Law at P , we get n2 sin0·= n3 sini ~

sini = " 2 sin9

n,

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,

1bis section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

. 1.

.

-

A transparent hemisphere has 'a radius of curvature 8 cm and an index of refraction of 1.6. A small object 0 is placed on the axis halfway between the plane surface and the Spherical surface i.e. 4 cm from each. The distance between the two images when- viewed along t,he axis from the two sides of the hemisphere is approximately

0

2mm

1

1+- 20

cm --+1.

(A) the distance between the.images is 2 mm ./

(B)

µ= 1.6

the distance between the images is 4 mm

(C) the distance between the two images formed by

--+--

· suchalensis-6min

·

(D) only one image will be formed by the lens (A) 7.5cm (C) 2.5cm 2.

3.

(B) 8.5 cm (D) 13.5 cm

A square wire of side 3.0 cm is placed 25 cm in front of a concave mirror of focal length 10 cm with its centre on the axis of the mirror and its plane normal to the axis. The area enclosed by the image of the wire is (A) 7.5 cm' (B) 6.0 cm2 (C) 4.0 cm' (D) 3.0 cm' An object is placed at a distance 2/ from the pole of a convex mirror of focal length f . The linear magnification is 1 2 (B) (A) 3 3

(C)

3 4

(D) 1

5.

Ah object is placed at 26 cm from a convex mirror of · -focal length 20 cm. The distance of the image from the pole of the mirror is (A) infinity (B) 10 cm (C) 15 cm (D) 40 cm

6.

A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm . When a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness t of the slab is, (A) 5 cm (B) 10 cm (C) 15 cm (D) 20 cm

7.

Light is incident normally on face AB of a prism as showri in figure. A liquid of refractive index µ is placed on face AC of the prism. The prism is made of glass of refractive index

4.

=

A convex lens of focal length 10 cm is painted black at the middle portion as shown in figure. An object is placed at a distance of 20 cm from the lens. Then

1.108

i. The limits of 2

µ for which

total internal reflection takes place at the face AC is

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I

I

Liquid

(C) concave, 15 cm (D) convex, 15 cm 14. A boy of height 1.5 m with his eye level at 1.4 m stands before a plane mirror of length 0.75 m fixed on the well. The height of the lower edge of the mirror above the floor is 0.8 m. Then (A) the boy will see his full image. (B) the boy cannot see his hair. (C) · the boy cannot see his feet. (D) the boy cannot see both his hair and feet.

~ c 90° B

-J3

(A) µ<-. 2

3-/3

(C) µ<-48.

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 , there is no parallax between the images formed by the two mirrors. The radius of curvature of the convex mirror is (A) 60cm (B) 50 cm (C) 30 cm (D) 25 cm

9.

A concave lens forms.the image of an object such that the distance between the obje_ct and image is 10 cm and the magnification produced is of the lens will be (A) 10 cm (C) 6.2 cm

¼. The focal length

(B) 8.6 cm (D) 4.4 cm

10. For a concave mirror, the magnification of a real image was found to be twice as great when the object was 15 cm from the mirror as it was when the object was 20 cm from the mirror. The focal length of the mirror is (A) 5,0 cm (B) 7.5 q:n (C) 10 cm (D) 12.5 cm

11. The image formed by a convex mirror of focal length 20 cm is half the size of the object. The distance of the object from the mirror is (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm 12. A concave mirror of focal length_ f in vacuum is placed in a medium of refractive index 2. Its focal length in the medium is

L

(B)

f

(C) 2/

(D)

4/

(A)

2

15. A horizontal ray of light passes through a prism of µ =1.5 whose apex angle is 4 ° and then strikes a vertical mirror M as shown. For the ray, after reflection to become horizontal, the mirror must be rotated through an angle of (A) 1° (B) 2° (C) 30 . (D)

M

16. A man of height 1.6 m wishes to see his full image in a plane mirror placed at a distance of 2 m. The minimum length of the mirror should be (A) 0.4 m (B) 0.8 m (C) 1.6 m (D) 2.4 m

17. A plane mirror reflects a beam of light to form a real image. The incident beam is (A) parallel (B) convergent (C) divergent (D) any one of the above 18. A plane mirror is approaching you at 10 cms-1 • You

can see your image in it. The image will approach ·you with a speed (A) 5 cms-1 (B) 10 cms-1

(C) 15 cms-1

(D) 20 cms-1

19. An object is placed at A(DA> f) , where,

f

is the focal

length of the lens. The image is formed at B . A perpendicular is erected at O and C is chosen such that LBCA = 90°. Then the value of f (in tenns of a, bandc)is

13. A spherical mirror forms an ere~t image three times the size of the object. If the distance between the object and the image is 80 cm, the nature and the focal length of the mirror are (A) concave, 30 cm (B) convex, 30 cm

C

:

0

:

' ' :' :' :' :' ': . - a -a.:+-- b ........:'

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OpHcs & Modem Physics

Advanced JEE Physics (a+ b)3 (A) -c-,-

(C)

c' a+b

(a+b)c (a+c) a' (D) - - a+b+c

(B)

20. An observer moves towards a plane mirror with' a speed of 2 ms-1 . The speed of the image with respect to the observer is (B) 2ms-1 (A) 1 ms01 (D) 8 ms-1 (C) 4ms-1 21. A concave mirror of focal length / produces a real image n times the size of the object. The distance. of the objet from the mirror is

(A) (n-1)/ (C)

( n:1

)t

(B)

Cn+1)f

(D) ( n:1

)t

22. Two plane mirrors are arranged at right angles to each other as shown in figure. A ray of light is incident on the horizontal mirror at an angle e. The value of 0 for which the ray emerges parallel to the incoming ray after reflection from the vertical mirror is

(A) 30' (C) 60'

(;r

(B)

45° (D) all of the above

23. A convex mirror of focal length / produces an image

of the size of the object. The distance of the object

from the mirror is

l_

(A) nf

(B)

(C) (n+l)f

(D) (n-1)/

n

24. A real image formed by a concave mirror is 4.5 times the size of the object. If the mirror is 20 cm from the object, its focal length is

90 · (A) -cm 11

(C)

150 cm 11

(B)

120 cm 11 180 (D) -cm 11

25. An object is placed 10 cm in front of a convex mirror of focal length 20 cm. The distance of the image from the mirror is

(A) 10 cm

(B)

3

. (C) 10 cm

20 cm 3

(D)

40

-cm 3

26. A concave mirror forms the image of an object on a screen. If the lower half of the mirror is covered with an opaque card, the effect would be (A) to make the image less bright. (B) to make the lower half of the image disappear. (C) to make the upper half of the image disappear. (D) to make the image blurred. 27. Two plane mirrors are inclined at 70°. A ray incident On one mirror at angle 0, after reflection falls on the second mirror and is reflected from there parallel to the first mirror. e is (B) 50° (A) 45° (D) 60° (C) 55° 28. A man stands in a room with his eyes at the centre of the room. The height of the ceiling is H . The length of the shortest plane mirror, fixed on the wall in front of the man, so that the man can see the full image of the wall behind him is H (A) 2H (B)

(C)

3

2

H

(D) H

3

4

29. An object is ·placed between two parallel mirrors. The number of images formed is (A) 2 (B) 4 (C) 8 (D) infinite 30. A bulb is placed between two plane mirrors inclined at an angle of 60° . The number of images formed is (A)5. (B)6 (C) 4 (D) 3

31. Two plane mirrors are placed perpendicular to each other. A ray strikes one mirror and after reflection falls on the second mirror. The ray after ,reflection froni. the second mirror: will be (A) perpendicular to the original ray. (B) parallel to the original ray. (C) at 45° to the original ray. (D) can be at any angle to the original ray. 32. A real image is formed by a convex lens, then it is brought in contact with a concave lens such that again a real image is formed. This image will (A) remain in its original position (B) shift towards the lens system (C) shift away from the lens system (D) shift to infinity 33. Plane mirrors A and B are kept at an angle 0 with respect-to each other. Light falls on A, is reflected, then falls on B and is reflected. The emergent ray is opposite to the incident direction. Then the angle 8 is equal to

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(B) -450 (D) 90°

(A) 30° (C) 60°

34, A point source of light B · is

(D)

f J

placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall. A man t walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown.

1:

The gr~atest distance over which

4i. An object is moving towards a concave mirror of focal B

l

I+- L --.i M---

2L -------+l

he .can see the image of the light

(B)

2

(C) 2d

d •

f

that instant; is

(A) 4 ems·' towards the mirror

(Bj

9 ems·' towards the mirror

(C) 4 ems·' away from the mirror

42. A ray of light passes through an equilateral prism such that the angle .of emergence is equal to the angle of

(D) 3d

35. A thin rod of length

length 24 cm. When it is at a distance of 60 cm from the mirror its speed is 9 ems-' . The_ speed of its image at

(D) 9 cms·1 away from the mirror

source in the mirror is

(A) d

incidence and each is equal to ( lies along the axis of a concave

. nui;or of focal length f . One end of its image touches an end of the rod. The length of the image is

(A)° f

(B)

j_

(C) 2f

(D)

j_

r.

¾

of the angle of

prism. The angle of deviation is (A) 45° (B) 39° (C) 20° (D) 30° 43. An object is placed in front of a concave mirror of focal

2

length f as shown in figure. The correct shape of the image is represented by

4

36. How many images will be formed if two mirrors are fitted on adjacent walls and one mirror oi:,. ceiling? (A) 5 (B) 7 (C) 11 (D) 2

a

(A) x + y + z = constant (C) y = constant

C

I+- X > 2f

37. The wavefront that represents the light waves travelling in vacuum along the y-axis is

----+(

(B)_ x = constant (D) z =constant

38. A boy stands straight in front of a mirror at a distance of 30 cm from it. He sees his erect image whose height is

.!:.5

.-./3 '

(A)

(B)

(C)

(D)

of his-real height. The mirror he is using is .

(A) plane

(C) concave

(B) convex (D) piano-concave

39. The image of an object placed in front of a concave

mirror of focal length 12 cm is formed at a point which is 10 cm more distant from the mirror that the object. The magnification.of the image is

(A) 1.5 (C) 2.5

(B) . 2 (D) 3

44. The index of refraction of diamond is 2.0. Velocity of

40. The minimum value of the refractive index for a 90° - 45° - 45° - prism which is used to deviate a beam

through 90° by total internal reflection is (A)

~

-./3

(B) ·..fi.

light in diamond in cms·1 is approximately (A) 6xl0 10 • (B) 3xl010 10 (C) 2 X 10 (D) 1.5 X 1010 45. A plane mirror is placed at origin parallel of y-axis, facing the positive x-axis. An. object starts from

'

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Optics & Modem Physics

Advanced JEE Physics (2,0,0) m

with a velocity of (21+2])ms-1 • The

relative velocity of image with respect to object is along

(A) positive x-axis (C) negative x-axis

(B) positive y-axis (D) negative y-axis

angle of refraction, then the angle of incidence is

47.

(B)

(C) 2cos-'(%)

(D) 2sin-'(%)

is less than the critical angle. 51. The critical angle of light going from medium A into medium B is 0. The speed of light in medium A is v . The speed of light in medium B is

A point of source of light is placed at the bottom of a

~. 3

for which the source cannot be seen at all from above is

3

(C)

2 -cm 3

(B)

(A)

A

person is viewing 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. The 'liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid

(A) -cm 2

is less than the critical angle.

(C) rarer to denser medium and the angle of incidence is greater than the critical angle. (D) rarer to denser medium and the angle of incidence

sin-'(%)

vessel containing a liquid of refractive index

\

when the ray goes from (A) denser to rarer medium and the angle of incidence

is greater than the critical,angle. (B) denser to rarer medium and the angle of incidence

46. A ray of light passes from vacuum into a medium of refractive index n . If the angle of incidence is twice the

(A) cos-'(%)

50. Total internal reflection of a ray of light is possible

4

(C)

V

(B)

sine V

(D) vtan0

lane

52. Glass has refractive index

index

vsin0

3

2

. and water has refractive

±. If the speed of light in glass is 2.00 x 10 3

8

-cm

the speed of light in water in ms-1 is

3

(A) 1.50xl08 (C) 2.25xl08

3

(D) - cm 4

48. A beam of light consisting of red, green and blue f=Olours is incident on a right-angled prism as shown. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44

ms-1 ,

(B) 1.78xl08 (D) 2.67xl08

53. Two sides of an isosceles right prism are coated with a reflecting coating. A ray of light falls on the hypotenuse at an arbitrary angle i . The value of i for which the ray leaving the prism is parallel to the incident ray is

and 1.47 respectively. The prism will

.

,,,/

I/

1.

µ=-

../3

(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 colours.

49. A thin prism P1 with angle 4° and made from glass of refractive index is 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P2 is

(A) 5.33° (C) 30

(A) 30° 60° (C) 450

(B)

(B) 4° (D) 2.6°

(D) any arbitrary angle from 0 < i < 2: 2· 54. A diver in a lake wants to signal his distress to a person sitting on the edge of the lake flashing his water proof torch. He should direct the beam (A) vertically upwards. (B) horizontally. (C) at an angle to the vertical which is slightly less than the critical angle. (D) at an angle to the vertical which is slightly more than the critical angle.

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Ray Optics 55. Critical angle of light passing from a glass to water is

(Refractive Index of Glass = 1.5, Refractive Index of Water =4/3)

minimum for

(A) red colour (C) yellow colour

(B) green colour (D) violet colour

56. Mirage is observed in a desert due to the phenomenon of (A) interference (B) total internal reflection (C) scattering (D) double refraction 57.

(,Jz +1) cm

(C)

(D) 4 cm

2v'2 cm

(C)

sin-1(

(

4

(C) 16cm

58. Total internal reflection can occur when light tends to pass from (A) a denser to a rarer medium.

(B) a rarer to a denser medium. (C) one medium to another of different refractive index irrespective of which medium has greater refractive index. (D) one medium to another of equal refractive index. 59. A diverging beam of light from a point source S having divergence angle a 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 the refractive index n , then the divergence angle of the emergent beam is

(B)

(D)

. -1( - 1 ) ../3 . -1( -9../3) sm sm

16

hit it with a stick. Taking aim the boy holds the stick in the air at an angle of 45° . At what distance from the stone will the stick hit the bottom, if the depth is 32 cm (given 'µw = 4/3 ) (A) 8 cm (B) 12cm

focus of a convex lens of focal length f are 1 cm each, then f is

(B)

1) f)

3

sin-

62. A stone lies at the bottom of a stream. A boy wants to

The distances of an object and its virtual image from the

(A) (2+"2) cm

1

(A)

(D)

12v'2 cm

63. When the surface of the lake is calm, a fish submerged in water will see the entire out-side world within inverted cone whose apex is situated at the eye of the fish and the cone subtends an angle of (A) 10° (B) 60° (C) 98° (D) 30° 64.

A ray of light strikes a glass slab of thickness t . It emerges on the opposite face, parallel to the incident ray but laterally displaced. The lateral displacement is l!.x.

(A)

l!.x = 0

(C)

l!.x=-cosr

tsini

(B) (D)

l!.x l!.x

= tsin(i-r )cosr tsin(i-r) cosr

65. In cold countries the phenomenon of looming (i.e. ship appears in the sky) takes place because (A) refractive index of air decreases with height. (B) refractive index of air increases with height. (C) refractive index does not change with height.

s

"

(D) refractive index becomes infinity at the surface.

"ft

n

66. If D is the deviation of a normally falling light beam on a thin prism of angle A and Ois the dispersive power

:t (A) ZERO (C)

. -1(1) n

sm

of the same prism then

(B)

a

(D)

2 sm · - ·n

(A) (B) (C) (D)

1(1)

D is independent of A . D is independent of refractive index. 8 is independent of refractive index. 8 is independent of A .

60. A ray incident at an angle of incidence 60° enters a

glass sphere of refractive index µ = ../3 . This ray is reflected and refracted at the farther surface of the sphere. The angle between reflected and refracted rays at this surface is

(A) 40° (C) 70°

67. For an equilateral prism, it is observed that when a ray strikes grazingly at one face it emerges grazingly at the other. Its refractive index will be

(A)

(B) 60° (D) 90°

f

(B)

l

(C) 2

61. A water film is formed on a glass block. A light ray is incident on water film from air at an angle 60°. What is the angle of incidence on glass block?

68. A rectangular block of glass (refractive index 3/2 ) is kept in water (refractive index 4/3 ). The critical angle for total internal reflection is

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Advanced JEE Physics (A) sin-1 (

emerges as QR. The condition of minimum deviation is satisfied in the prisms

%) for a ray of light passing from glass to

water.

(B)

sin-1 (

%) for a ray of light passing from water to

B

glass.

\

C

__

._.

1

(C) sin- ( ¾) for a ray of light passing from water to

0:

glass.

(A) A and C (C) AandB

. -1(8)- for a ray of light passing from glass to

(D) sm

69. The refractive index of a given piece of transparent quartz is greatest for (A) red light (B) violet light (C) green light (D) yeliow light 70. A well cut diamond appears bright because (A) it emits light (B) it is radioactive (C) of total internal reflection (D) of dispersion 71. The maximum refracting angle of ·a prism of refractive index 2is

(B) 45' (D) 90'

A monochromatic beam of light passes from a denser to a rarer medium. As a result its

(A) velocity increases (B) velocity decreases (C) frequency decreases (D) frequency increases 74.

77. A beam of white light is incident on a hollow prism of glass. Then (A) the light emerging from prism gives no spectrum. (B) the light emerging from prism gives spectrum but the bending of ali colours is away from base. (C) the light emerging from prism gives spectrum, all the colours bend towards base, the violet most and red theleast. (D) the light emerging from prism gives spectrum, all the colours bend towards base, the violet the least and :red the most. 78.

72. When light passes from one medium to another, the physical quantity that remains unchanged is (A) velocity (B) wavelength (C) frequency (D) None of these 73.

(B) Band C

(D) in ali prisms A, Band C

9

air.

(A) 30' (C) 60'

R

When a ray of light enters a glass slab from air (A) its wavelength decreases. (B) its wavelength increases. (C) its frequency increases. (D) neither its wavelength nor its frequency changes.

75. A number of images of a candle flame are seen in a thick mirror (A) the first image is the brightest. (B) the second image is the brightest. (C) the last image is the brightest. (D) all images are equally bright.

76. Three glass prisms A, B and C of same refractive index are placed in contact with each other as shown in figure with no air gap between the prisms. Monochromatic ray

of light OP passes through the .prism assembly and

If the critical angle for the medium of a prism is C and the angle of prism is A , then there will be no emergent ray when (A) A<2C (C) A>2C

(B) A=2C (D) AS2C

79. The angle of a prism is 60° . What is the angle of incidence for minimum deviation? The refractive index

of the material of the prism is -/'i. . (A) 45° (B) 60°

(C)

30°

(D) sin-'(¾)

80. A ray of light is incident at angle i on one surface of a prism of smali angle A and emerges normally from the opposite surface. H the refractive index of the material of the prism is µ, the angle of incidence i is nearly equal to A (A) A (B) µ 2µ µA

(C)

µA

(D)

2

81. If 'µ i represents the refractive index when a ray of light goes from medium i to medium j, then the product 2

µ 1 x 3µ 2 x 4µ 3 is equal to

(A)

3

µ1

(B)

3

µ2

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(C)

1

'µ,

(D)

'µ,

82. An air bubble inside a glass slab ( µ =

¾) appears to be

6 cm deep when viewed from one side and 4 cm deep when viewed from the opposite side. The thickness of the slab is (A) 10 cm (B) 6.67 cm (C) 15 cm (D) None of the above 83. The refracting angle of a prism is A and the refractive

index of the material of the prism is cot(

1).

The angle

of minimum deviation is

(A) 180°-3A (C) 90°-A

(B) 180°+2A (D) 180°-2A

84. The angle of a prism is 30°. The rays incident at 60° at one refracting face suffer a deviation of 30°. The angle of emergence is (A) 0° (B) 30° (C) 60° (D) 90° 85. A

ray falls on a prism ABC(AB=BC) and travels as sho'Wn in the figure. The minimum refractive index of the prism material should be 4 (A)

(B)

J'i.

(C)

(D)

../3

3

3 2

A

(A) d(µ, +µ,)

(B)

a(_!_+_!_J µ1 µ2

(D)

~(~ +

d

(C) 2(µ,+µ,)

:J

formed at the same place? (A) 6 cm from S1 (B) C

87. A point source of light is placed 4 m below the surface 3

91. A vessel of depth d· is half filled with a liquid of refractive index µ 1 and the other half is filled with a liquid of refractive index µ 2 • The apparent depth of the vessel, when looked at normally, is

should a convex lens of focal length 9 cm be placed in between them so that the images of both sources are

B

~.

90. A convex lens forms a real image three times larger than the object on a screen. The object and screen are moved until the image becomes twice the size of the object. If the shift of the object is 6 cm then the screen has to be shifted by (A)' 9 cm (B) 18 cm (C) 36cm (D) 72cm

92. Two point sources S1 and S2 are 24 cm apart. Where

86. Critical angle is minimum when a light ray passes from (A) air to glass (B) glass to air (C) glass to water (D) water to glass

of a liquid of refractive index

89. The path of a refracted ray of light in a prism is parallel to the base of the prism only when the (A) light is of a particular wavelength. (B) ray is incident normally at one face. (C) ray undergoes minimum deviation. (D) prism is made of a particular type of glass.

(C) 12 cm from S1

93. Light travels ·through a glass plate of thickness t and having refractive index n . If c is the velocity of light in vacuum, the time taken by light to travel this thickness of glass is

(A)

t nc

(B)

(C)

nt

(D)

C

The minimum

diameter of a disc, which should be placed over the source, on the surface of the liquid to cut off all light coming out of water, is (A) oo (B) 6m (C) 4m (D) 3m 88. A man standing in a swimming pool looks at a stone lying at the bottom. The depth of the swimming pool is h . At what distance from the surface of water is the image of the stone formed? Line of vision is normal. Refractive index of water is n . n (A) h (B) n h (C) h (D) Im

10 cm from S1 (D) 15 cm from S1

n 2c n 2t C

94. A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is

± and the fish is 12 cm below the 3

surface of water, the radius of the circle in cm is (A) 36../5 (B) 36ft 36 (D) 4../5 (C) ,ff 95. A diver inside water sees the setting sun at (A) 41 ° to the horizon (B) 49° to the horizon (C) 0° to the horizon· (D) 45° to the horizon

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Optics & Modem Physics .

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i3

96. If the refractive index of water is

and that of glass is

~ , then the critical angle of incidence for light tending

3 to go from glass to water is (A)

(C)

¾) . -1(4) sm 1

sin-

(

5

1

(B)

sin-

(D)

Sill

(

¾)

. -1( 32) -

97. Two media A and. B of refractive indices µ 1 =1.5 and µ 2 = 2 are separated by x-z plane. A ray of light travels from A to B . The incident ray and the reflected

ray are represented by unit vectors U1 = ai + b] and

incidence at which minimum deviation will be obtained is

(-/2) 3

(A)

• -1 Sill

(B)

.30°

(C)

60°

(D) 45°

103. Light is incident at an angle ex on one planar end of a transparent cylindrical rod of refractive index n . The least value of n for which the light entering the rod will not emerge from the curved surface of rod, irrespective of value of a is 1 (B) (A)

-/2

-/2 1

(C)

./3

(D)

./3

U2 =ci+d}. Then

a

3 C 4 b 3 -=d 4

(A) -=-

(B)

(C)

(D)

a C

b d

104. For a prism the refractive index ( µ) is related to

4 3 4 3

wavelength (;\.) as µ = A +

98. The speed of light in medium A is 2.0xl0 8 ms-1 and

that in medium B is 2.4 x 10 8 ms-1 . The critical angle of

incidence for light tending to go from medium A to medium Bis

(A)

. -1( 125 )

Sill

-

1

(B)

sin-

(D)

Sill

(

¾)

. -1(3)

(B) 0.80 (D) 1.44

(C) 1.20

(A) _!![_

(B)

__f_

__f_

(D)

_!![_

4

2.5 x 108 ms-1 • The refractive index of the liquid is

100. A ray of light travelling inside a rectangular glass block of refractive index is incident on the glass-air surface at an angle of incidence of 45° . The refractive index of air is 1. The ray will (A) emerge into air without any deviation. (B) be reflected back into glass. (C) be absorbed. (D) emerge into air with an angle of refraction equal to 90°.

-/2

101. A fish in water sees an object which is 24 cm above the surface of water. The height of the object above the surface of water that will appear to the fish is (B) 32 cm (A) 24 cm (C) 18 cm (D) 48 cm 102. The angle of minimum deviation equals the angle of prism A of an equilateral glass prism. The angle of

(B) B is large (D) A and µ are large

105. An infinitely long rod lies along the axis of a concave mirror of focal length f . The near end of the rod is at a distance u > f from the mirror. The length of the image of the rod is

u+f

-

99. The speed of light in glass of refr~ctive index 1.5 is 2 x 108 ms-1 • In a certain liquid the speed of light is (A) 0.64

large if (A) A is large (C) µ is large

~ . The dispersive power is

).

(C)

u-f

u+f

u-f

106. Two transpar_ent slabs have the same thickness as shown in figure. One is made of material X of refractive index 1.5. The other is made of two materials Y and Z having thicknesses io the ratio 1 : 2. The refractive index of Z is 1.6. If a monochoromatic parallel beam passing through the slabs has the same number of wavelengths inside both, the refractive index of Y is 1+--t-i

---~LB .........c ....... 1.6

(A) 1.1 (C) 1.3

(B) 1.2 (D) 1.4

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., ·Say Optics 107. A curved mirror of focal length f (in vacuum) is placed in a medium of refractive index 2. Its new focal length in the medium is f' . (A) f' < f (B) f' > f (C) f'= f (D) f'e;,f

(C) 30 ,

(D) 25

114. A prism having an apex angle 4° and refractive index · 1.5 is located in frorit of a vertical plane mirror as

shown in figure. The total angle through which the ray is deviated after reflection from the mirror is given by

108. If

t 0 is the absolute permittivity of free space, µ 0 is absolute permeability of free space, e is the permittivity of medium, µ is permeability of medium and n is the

refractive index of medium then,

(A)

(B)

n=J µoto µe (A) 176°

(C) 178°

(C) n=Jµ~,

109. The critical angle of glass water

(µw = ¼) is 0

2 •

(µ, = f) is 0

1

115. A slab of glass of thickness 3 cm and refractive index and that of

(A) 9 cm (C) 11 cm

(A) less than 01 (B) less than 82 (C) between 0, and 0, . (D) greater than 02

angle 0 falls on M, and is then reflected parallel to M 1 for (A) 0=45° (B) 0=50' (C) 0 =55° (D) 0=60° 111. An object is placed at 20 cm from a convex mirror of

focal length 20 cm. The distance of the image from the pole of the mirror is (A) infinite (B) 10 cm (C) 15cm (D) 40cm

pole of a concave . mirror of focal length f . The diameter of the image of the sun formed by the mirror

2f0 D

(D) D0

2f0

(A)

2nd

(C)

2d n

(C)

113. An object is placed in front of a convex mirror. ai a distance of 50 cm. A plane mirror is introduced covering the lower hall of the convex mirror. If the distance.. between the Object and the plane ·mirror is 30 cm , there is no parallax between the images formed by the two mirrors. The radius of curvature of the convex mirror (in cm) is {A) 60 (B) 50

; ill

(B)

d

l

.

2d

n-1 d (D) -(l+n) n

117. A ray of light enters an anisotropic medium from vacuum at grazing incidence. If 0 is the angle made by the reflected ray inside the medium with the interface and n ( 0) is the refractive index ofthe medium then, (A) n(0)sin0=1 (B) n(0)cos0=1

112. The s~ (diameter D ) subtends an angle 0 radian at the

(C)

(B) 10 cm (D) 12cm

116. A tank contains a transparent p liquid of refractive index n the bottom of which is made of a mirror as shown. An object O lies at a height d above the mirror. A person P vertically above the object sees O and its image in the mirror and finds the apparent separation to be

110. Two plane ~ors M1 and .M 2 are inclined to eaCh other at 70°. A ray incident on the mirror M1 at an

(B)

2

axis of the concave mirror. If the radius of mirror is 10 cm, the distance at which an object must be placed from the mirror so that the image coincides-with.the object is

The critical angle for water-glass

f0

~

is placed with its face perpendicular to the principal

interface is

is (A)

(B) 4' (D) 2'

n_(0) =l sm8

(D)

n(0) =l cos8

118. · A person runs with a speed u towards a bicycle. moving away from him with speed v. The person approaches his image in the mirror fixed at the rear of bicycle with a speed of (B) u-2v (A) u -v (D) 2(u - v) (C) 2u-v ~

119. Light travels through a glass plate of thickness t · having refractive index µ. If c is the velocity of light in

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Optics & Modem Physics

Advanced JEE Physics vacuum, the time taken by the light to travel this thickness of glass is (A)

(C)

I

(B)

µc µI

(D)

µ'c

µ't C

C

120. A real image I is formed by a converging lens L on its optic axis. On introduction of a rectangular glass slab of thickness d and refractive index µ between the image and lens the image displaces it by (A) d(µ-1) away from L

(B)

(C)

(D)

a(1-;) d(1-;)

Ji= 10·cm.

The distance between the two lenses, so that rays after refraction from both the lenses pass undeviated is

away from L, towards L f,

the wavelength changes from 6000

A

to 4000

A. The

critical angle for the interface will be

(C)

126. Parallel beam of light is incident on the system of two convex lenses of focal length / 1 = 20 cm and

d(µ-1) towards L

121. When-a ray is refracted from one medium to another,

(A)

125. The light on reflection from a plane mirror can give a real image when (A) the convergent rays are incident on the mirror. (B) the divergent rays are incident on the mirror. (C) an object is placed very close to the mirror. (D) an object is placed very far away from the mirror.

(¾) . -1(2)-3 sm 1

cos-

(B)

sin-'(1)

(D)

cos-' ( 1 )

123. The image of an object, formed by a piano-convex lens at a distance of 8 m behind the lens, is real and is onethird the size of the object. The wavelength of light

~ 3

f,

(B) 40 cm (D) 90 cm

127. The plane faces of two identical piano convex lenses, each with focal length f are pressed against each other using an optical glue to form a usual convex lens. The distance from the optical centre at which an object must be placed to obtain the image same as the size of object is

122. A boy stands straight in front of a mirror at a distance of 30 cm away from it. He sees his erect image whose height is one fifth of the original height. The mirror used by him is (A) plane (B) convex (C) concave (D) piano concave

inside the lens is

(A) 30 cm (C) 60 cm

times the wavelength in free space.

(A) f_

(B) f_ 2

4

(C)

f

(D)

2/

128. A parallel beam of light incident on a concave lens of focal length 10 cm emerges as a parallel beam from a convex lens placed coaxially, the separation between the lenses being 10 cm. The focal length of the convex lens incmis (A) 10 (B) 20 (C) 15 (D) 30 129. A ray oflight is incident on a glass sphere of refractive

The radius of the cured surface of the lens is (A) 1 m (B) 2m (C) 3 m (D) 6m

index

%. The angle of incidence for which a ray t.hat

enters the sphere· does not come out of the sphere is 124. A diverging beam of light from s a · point source S having a, falls divergence angle symmetrically on a glass slab as shown. The angles of incidence n of two extreme rays are equa( If t the thickness of the glass slab is t and the refractive index n , then the divergence angle of the emergent beam is (A) ZERO (B) a

1

(C) sin-'(;;)

(D) 2sin-'(;)

(A) tan-1 (C)

45°

(¾)

(B)

sin-1

(¾)

(D) 90°

130. A thin prism P1 of angle 4° and made from glass of refractive index 1.54, is combined with another thin prism P2 made from a glass with refractive index 1.72, to produce dispersion without deviation. The angle of P2 is (A) 5.33° (C) 3°

(B) 4° (D) 2.6°

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Ray Optics 131. A transparent sphere of. radius R made of material of refractive index

~ 2

incidence i . It is observed that no ray emerges from the other face. For this the minimum value of i should be

is kept in air. The distance from the

(A) µsinA-cosA (BJ sin-'(sinA-µcosA)

centre of the sphere must a point object be placed so as

to form a real image at the same distance from the

(C) sin-1 [~µ 2 -lsinA-cosA]

sphere is

(A) R (C) 3R

(B) 2R (D) 4R

(D) ~µ 2 -1 sin A- cos A

132. An air bubble in water is to be placed in a way such that

a real image is obtained at the same distance from bubble. Taking µw,..,

=¼we have the distance of object

138. As the position of an object (u) reflected from a concave mirror is varied, the position of the image ( v) also varies. By allowing the u to change from O to +co, the graph between v versus u will be

from the air bubble as (A) R (B) 2R

V

(C) 3R (D) An air bubble is incapable to form a real image. 1

(A)

133. An object is placed at a distance of 12 cm from a convex lens on its principal axis and a virtual image of certain size is formed. On moving the object 8 cm away from the lens, a real image of the same size as that of virtual

deep when viewed from one side and 4 cm deep when viewed from the other side. Assuming µg)

~·

=~ , the thickness of slab is (B)

(C) 15cm

(D) 20cm

placed

at

a

surface of the sphere when a point source of light lies at

the surface of the sphere. The refractive index of the sphere is

distance

of

3 2

-cm

20

cm

and

(B)

(D)

(C) 2

135. On two sides of an oily paper screen, two bulbs A and B

are

(D) k:--t----+u

139. A parallel beam of light emerges from the opposite

(A)

20 3

(A) 10cm

V

(C)

..

2

(B)

V

image is formed. The focal length of the lens in cm is (A) 15 (B) 16 (C) 17 (D) 18 134. An air bubble inside a glass slab appears to be 6 cm

f-+---•u

5 3 5 2

140. Two spherical mirrors M 1 and

30 cm, so that equal intensity is obtained on both sides

M2 ,

of screen. If PA and P, be the powers of the bulbs A

concave having same radius of curvature R are arranged 2R coaxially at a distance (consider their pole separation to

and B respectively then PA is

P,

(A) 0.44 (C) 1.5

(B) 2.25 (D) 0.67

one convex and other

be 2R). A bead of radius a is

136. An achromatic combination pair of a telescope objective

will be (A) lenses of f = -50 cm and power +2 D (B) lenses of powers 3 D and -5 D respectively (C) lenses of f = +20 cm and power -4.5 D (D) lenses of f = +40 cm and power +2 D

placed at the pole of the convex M1 mirror as shown. The ratio of the sizes of the first three images of the bead is

(A) 1 : 2: 3

(C)

137. A ray of light enters the face of a glass prism of refracting angle A, refractive index µ at an angle of

1 1 1 -·-:3 ·11 41

M,

(B) 1 : :1:_: :1:_

2 3

(D) 3: 11: 41

141. Light is incident at an angle a. on one planar end of a transparent cylindrical rod of refractive index n . The least value of n for which the light entering the rod will

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Advqnced JEE Physics not·en,.1;rie.-fromthe curved surfaCe of.rod, irrespective of :value of a. is ..

(A) .

}i

(C)

.J3

1

'

.(B)

.Ji.

(D)

.J3

142. A ray of light is incident on one face of prism with refracting angle A (< 90°) . Tl_te incident ray is normal to the other face of the prism, If e is the critical angle for prism-air interface, then the ray_ will emerge from this face only if (A) cote< cotA + 1 · (B) cote> cotA+ 1 (C). cotA < cote +.1 (D) cot A> cote+ 1

146. Rays-of light from a luminous object are brought to focus at a point A. The rays are intercepted, before meeting at A by a convex lens- of focal length 20 cm placed at 24 cm from A ,and are forced to meet at B. Then AB e!J_uals (in cm) , (A) 12 (B) 24 (D) 48 . (C) 6

147. A point object is placed at a distance of 0.3 m from a convex lens of focal length 0.2 m cut into two equal . halves, each of wliich is displaced by .0.0005 m, as shown 'in figure. If. e, and e, be their optical centres ' then;

143. The image of point P when viewed from top of the glass Slabs is

µ=1.5

I

µ= 1.5

(A)

1.5cm

or, C2 along principal ·a~is. (B) two images are formed; one at a distance of 0.6 m · and other at a distance of 1.2 m from e, or e,

1.5cm

cm

along principal axis. (C) an image is formed at a distance of0.12 m from e,

(B) 0.5 cm above P (D) 1 cm above P

144. An isosceles prism has refracting angle A . Its one face is·si!vered (other than the base). A ray of light falling normally on the face not silvered emerges through the base of the prism normal to it. (A) A=45' . (B) A=90' (C) A=36' (D) A=72' 145. Two identical thin isosceles prisms of refracting angle · A and refractive index µ'are placed with their bases touching each other and this systeni can collec.ti_vely act as a crude converging lens. A parallel beam of light is incident on this system as shown. The focal length of this so. called ~onverging lens "is

f 1 _·

2h~--+--'+--+----'ri!I

.

(C) f

h

(µ-1)A

!

(B)

f=

(D)

f=_E_ µ-1

e1

1

P•

(A)

an ,itnag~ is f~rm~d at ~- distance ·of 0.6 m from

1·.scm

2

(A) 2cmabove P (C) 0.5 cm below P

0

·or c;

al~ng principal axis. .

'.

(D) two images are formed at a distance. of.0.6 m from C1 or C2 along principal axis at a separation of 0.003m.

148. All of the following statements are correct except that (A) the magnification produced by a convex mirror is always less than one . . ·(B) a virtual, erect, same sized. image can be obtained by using the plane mirror. (C) a virtual, erect, magnified image can be formed by using thE:! _concave mirror. (D) a' real, inverted, same sized image can be formed by using .a convex mirro,r. 149. A ray of ligltt undergoes deviation of 30° when incident on an equilateral prism of refractive index .Ji. . The angle made by the ray inside the prism with the base of the prism is (B) 15° (A) 0° · (DY 45° (C) 30° 150. A convex lens of focal length f forms an image of ·a heavenly body. The area · of the image formed is proportional to (B) f' (A) f' ,

'(C)

!'

(D)

f'

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Ray Optics 151. A piano convex lens has a thickness of 4 cm. When placed on a horizontal table with curved surface -in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens .is inverted such that the plane face is in contact with tl~e table, the apparent depth of the centre of plane face is 25 · found to be cm. The focal length of the lens is

8

(A) 50cm (C) 100cm

(B) 75 cm (D) 150cm

152. If an object is placed between two .parallel mirrors, an infinite number of images are formed. If the mirrors are

at a distance 2b and an object is placed at the middle of the two mirrors, the distance of the nth image from the object is 1 (A) nb (B) -nb

(A)' n1 =n, =n,

1+~,"=n, +n,

(C)

(B)

n1 =n, ,en,

(D)

l+·n;=n;+ni

156. Four lenses are made from same type of glass. The · · radius of curvature of each face is given . Out of .these, the 1ens having the greatest po~itiv:e. power is (A) 10 cm convex and 15 cm convex. (B) 20 cm convex and 30 cm concave. (C) 15 cm convex and plane. (D) 5 cm convex and 10 cm concave. 157. The sides of an isosceles right angled prism are silvered.

A ray of light falls on the hypotenuse of the prism at an angle $0 as shown. The ray leaving the prism will

2

(C)

1 (D) -nb

2nb

4

153. A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform acceleration of 21t rads' 2 • The reflected ray, at the end of -1 s must 4

have turned through (A) 90° (C) 22.5°

(B) 45° (D) 11.25°

154. In the situation shown in figure, water

(µw = ¾)

is

filled in a beaker upto a height of 10 cm . A plane mirror is fixed at a height of 5 cm from the surface of water. Distance of image from the mirror after reflection from it of an. object' 0 at the bottom of the beaker is '

(A) (B) (C) (D)

graze the face AC. emerge normally to the face AC. be parallel to the incident ray. make an angle of 30° :with incident ray.

158. A right angled prism .(45°-90°-45°), of refractive index n has a plate of refractive index n1 ( n1 < n) cemented to its diagonal face. The assembly is in air. A ray is incident on AB as shown. If the ray strikes the diagonal face AC at critical angle then (A) sini=(~)

(B)

.. (n)

smz=

(C) sini = (D) sini

0

(A) 7.5cm (C) 12.5 cm

(B) 10cm (D) 15 cm

155. Three right angled prisms of refractive indices n,, n, and n3· are fixed together 'Using an optical glue as shown in figlJ!e. If a ray passes ,through the prisms without sufferfug any deviation, then

A

n,

~

F¥I -fli ~n

2

B

-11i

J'i.

159. An astronomical telescope has an angula:i; magnification of magnitude 5 for distant objects. The separation between ~he. objective and the eye piece is 36 cm and the final image is formed at infinity. The focal length f, of the objective and the focal length f, of the·eye piece are (A) f, = 45 crri and f, = -9 cm (B) f, =50 cm and f, =10 cm (C) f, = 7.2 cm and f, =5 cm (D) f,=30cm and f,=6cm

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Optics & Modern Physics

Advanced JEE Physics y

160. The slab of a material of refractive index 2 shown in

---

figure has a curved surface APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices

as given in figure. An object O is placed at a distance of 15 cm from pole P as shown. The distance of the final

---

image of O from P, as viewed from the left is

(A) x=

(B) 30 cm (D) 50cm

(A) 20 cm (C) 40 cm

image which is real and elongated just touches the rod. The magnification is

{A)

(C)

4 3 3 2

(B)

5 3

(µ. =¼), is

20 cm deep. If the coin

(B) 1.5cm (D) 2cm

163. A ray of light undergoes deviation of 30° when incident on an equilateral prism of refractive index

,Ji .

The_ angle made by the ray inside the prism with the base of the prism is (A) 30°

(C)

60°

(C)

X

d----->l

f,f, , y=!J. f, + f, f, (!, +d) , y =6.-2 f,+f,-d f,+f, f,f, +d(f, -d) y t>.(f, -d) ~~~~ f,+f,-d f,+f,-d f,f, +d(f, -d) y=O f,+f,-d

165. Spherical aberration in a thin lens can be reduced by (A) using a monochromatic light. (B) using a doublet combination. (C) using a circular annular mask over the lens. (D) increasing the size of the lens.

plane surface of a glass slab. The angle of incidence is

is viewed directly from above, the apparent diameter of the coin is

(A) 1.67 cm (C) 2.67cm

X

,_~tJ.,,__.. x

166. A circular beam of light of diameter d = 2 cm falls on a

(D) None of above

162. A 2 cm diameter coin lies flat at the bottom of a bowl

in which the water

(B)

(D) x

161. A thin rod of length .!:_ f is placed along the optic axis 3 of a concave mirror of focal length f such that its

.. J..

---+-l~o+-----<

(B) 45° (D) 0°

164. Two thin convex lenses of focal lengths / 1 and /, are separated by a horizontal distance d (where d < f, and d < f, ) and their centres are displaced by a vertical separation 8. as shown. Taking the origin of coordinates 0, at the centre of the first lens, the x and y coordinates of the focal point of this lens system, for a

parallel beam of rays coming from the left, are given by

60° and refractiVe index of glass is µ =~ . The diameter 2 of the refracted beam is

(A) 2.52 cm (C) 3.26 cm

(B) 3 cm (D) 4cm

167. Two thin lenses, wheri in contact, produce a combination of power +10 D . When they are

0.25 m apart, the power reduces to +6 D . The focal lengths of the lenses (in m) are (A) 0.125 and 0.5 (B) 0.125 and 0.125 (C) 0.5 and 0.75 (D) 0.125 and 0.75 168. A real image of a distinct object is formed by a piano convex lens on its principal axis. Spherical aberration {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. 169. A concave mirror is placed on a horizontal table, wi"Ut

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

1.122

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Ray Optics C . If the mirror is now filled with water, the image will

be (A) (B) (C) (D)

real and will remain at C . real and located at a point between C and oo. virtual and located at a point between C and O . real and located at a point between C and O .

170. 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 the point 0 and PO - OQ . The distance PO is equal to (A) SR (B) 3R (C) 2R (D) 1.5R 171. 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

(A) (B) (C) (D)

convergent lens of focal length 3.5R . convergent lens of focal length 3.0R . divergent lens of focal length 3.5R . divergent lens of focal length 3.0R .

A

60'

µ=2

(C)

(D)

µ=2

µ=s/3

175. A rectangular glass slab ABCD of refractive index n1 , is immersed

(C) 1

(D)

-- ---------- -----

:,~~A.... 1--- -

r::~::::

n, in water of refractive index n2 (n1 > n2 ). A -,:,:,:::g -- _,-, - _- __-c __-_-- __ ray of light is incident ~-------------- ·----------- -------at the surface AB of the slab as shown. The maximum value of the angle of incidence amax, such that the ray comes out only from the other surface CD is given by 1

1

1

(B)

sin- {~cos(sin-

(C)

sin-

1

1

(B)

3

(B)

(D) sinP,~ ;+- mR ----+-+- R --+i

8

(A) µ = -/2

1

distance of mR from it. The value of m for which a ray from P will emerge parallel to the table is

1 3

C

---------------

(A) sin- {::cos( sin-

172. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a

(A)

-

( ::

( ::

J)}

(:J)}

J

(::J

I

2 3

4 3

173. 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 respectively ( n2 > rLi > 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 L, . (D) L, and immersed in L1 •

174. A ray of light falls on a transparent sphere of refractive index µ, having centre at C as shown in figure. The ray emerges from the sphere parallel to line AB, then

176. A ray of light passes through four transparent media with refractive in_dices µ 1 , µ2, µ3 and 8 µ4 as shown in figure. The c surface of all media are A µ 1 112 ~ µ4 parallel. If the emergent ray ~~~~-~== CD is parallel to the incident ray AB, we must have (A) µ 1 = µ 2 (B) µ 2 = µ 3 (C) µ, = µ, (D) µ, = µ,

177. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of same material as P are now added as shown in figure. The ray will now suffer

(A) greater deviation. (B) no deviation. (C) same deviation as before.

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(D) total internal reflection.

182. Critical anglefor a prism is 36°. The maximum angle of

178. Two identical equiconvex lenses of focal length f ,

(µ, =%) ·are kept in contact. The space between the two lenses is filled with water (µw =½).

made of glass

The focal length of the combination is 3 / (A) j_ (B) 2 4

(C)

f

(D)

4/

(A) (C)

~

(B)

#.

184. A prism, having refractive index

~2h-----+1

#

'

(C) R

[i ~

185. -Two plane mirrors are inclined at angle 8 as shown in figure. If a ray parallel to OB strikes the other mirror at P and finally emerges parallel to , OA after two refle'ctio~, then 8 equals

(D) 3 2

180. Which one of the following spherical lenses does not exhibit dispersion ? (A) R·

183. A screen is placed 90·cm from an object. The image of the object on the screen is formed by a convex lens at two different positions separated. from each other by 20 cm . The lens has a focal length of (A) 32.1 cm (B) 15.8 cm (C) 21.4 cm (D) 10.7 cm

.Ji. and refracting angle 30°, has one of-the refracting surfaces polished. A beam of light incident on the other r<efracting surface will retrace its path if the angle of incidence is . (A) 0° (B) 30° (C) 45~ (D) 60°

3

179. An observer can see through a pin hole, the top· of.. a thin rod of height h , placed as shown in figure. The beaker's height is 3h and its radius is h . When the beaker is filled with a liquid upto a height 2h, he·can see the lower end of the rod.. Then the refractive index of liquid must be

prism for which the emergent ray is possible is (B) 36°· (A) 18° (C) 72° (D) 144°

R,

R

(B)

_R_ ~ -

(D) H

-~

""

""

(A) 30° (C) 60°

(B) 45° (D) 90°

186. An object is placed 20 cm in front of a block of glass 10 cm thick having its farther side ,silvered. Th,e image is

formed 23.2 cm behind the silvered face. The refractive index 9£ glass is (B). 1.46 (A) 1.41 (C) 1.51 (D) 1.61 187. Two parallel rays are travelling in a medium of

181. Two plane mirrors A and B are alligned parallel-to each· other, as,shown in figure. A ligll.t ray is incident atan angle of 30° at a point just inside one end of A: The

plane of incidence coincides with the plane .of figure. The maximum number of- times the ray undergoes reflections (including the first one) before it emerge_s out is

. .m d·ex µ 1 =-4 refractive

3

. However, .

one of h t e rays

passes through a parallel glass slab of thickness t and refractive index µ 2 =~. The path diff~rence between 2 · the two rays due to the glass slab is t t

(A)

8

(B)

(C)

~-

(D)

'6°

3! 2

188. A ray of light entering from air to glass (refractive index (A) 28

=

(C) 32 1.124

(B) 30. (D) 34

1.5) is partly reflected and partly refracted. If the incident and the reflected rays are at -right angles to each other, the angle of refraction is

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(A) sin-'(

Ji)

(B)

(C) 'sin,'(

1)

(D) sm

' -1(./2.) 3

S!Il

195. One side of a glass slab of refractive index 1.5 is silvered as shm--vn. A ray of light is incident on the other side at angle of incidence i = 45° . The deviation of the ray of

-

. -1( -./31 )

light from its initial path when it comes out of the slab is

''

189. A beam of light is converging towards a point on a screen. A plane parallel sided plate of glass of thickness t and refractive index µ is introduced in the path of the beam. The convergence point is shifted by

1-;) { 1-;)

(A) {

away

(B)

(C)

nearer

(D)

1+;) '{ 1+;) {

away

µ = 1,5

nearer

190. The distance of an' object from the first focus -of an equiconvex lens is 10 cm 'and the distance of its real image from the second focus is 40 cm . The foca.l length of the lens is (A) 10cm (B) 20 cm (C) 25cm (D) .40 cm 191. A beam of monochromatic Hght is incident .on one face

of an equilateral prism, the angle of incidence being 55°. If the angle of emergence is 46° then the angle of minimum deviation is (A) 41° (C) > 41°

(B)

'' 450,''

< 41°

(D) ;, 41°

192. When a ray of light is refracted by a prism such that ihe angle of deviation is minimum, then (A) the angle of emergence is equal to the angle of incidence. (B) the angle of emergence is greater than the angle of incidence. (C) the angle of emergence is smaller than the angle of incidence. (D) the sum of the angle of incidence and the angle of emergence is equal to 90° .

(A) 45' (C) 120°

(B)

90°

(D) 180°

196. A piano-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 bottom-most 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 25 cm. The focal length of the lens is is found to be 8 (A) 50 cm (B) 75 cm (C) 100cm (D) 150cm 197. If J 8 and JR are the focal lengths of a convex lens for

blue and red lights respectively and F, and FR are the respective values for a concave lens, then (A) J, > JR and F, >FR ,.(B) f, <JR and F, >FR

(C) J, > JR and F,
a ray of light inside the liquid is best represented by

193. The image of a square hole in a screen illuminated by light is obtained on another screen with the help of a converging lens. The distance of the hole from the lens is 40 cm . If the area of the image is nine times that of the hole, the focal length of the lens is (A) 30 cm (B) · 50 cm (C) 60 cm (D) 75 cm

194. A short linear object of length b lies along the axis of a concave mirror of focal length J at a distance u from the pole of the mirror. The size of the image is approximately equal to

(A) (C)

bH

(B)

b~u~J

(D)

b(u;JJ b(u~ J )'

J and the distance of an object from the principal focus is x . The ratio of

199. The focal length of a convex lens is

the size of the real image to the size of the object is

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(A)

l

(B)

X

(C)

(C)

X

I

f

(D)

4/

205. In PROBLEM 204, if the lens is cut along PQ and RS simultaneously, the focal length of each part will be

(D) _!_

f+x

2/

f+x

200. Focal length of a convex mirror is 10 cm (A) image of an object placed at 20 cm is also at

20cm (B) image of an object placed at 10 cm is at infinity (C) both (A) and (B) are correct (D) both (A) and (B) are incorrect 201. An object is placed at a distance x1 from the principal focus of a lens and its real image is formed at a distance

x, from the principal focus, The focal- length of the lens

(A) l_ 2(C) 2/

(B)

I

(D) 4/

206. The layered lens as shown is made of two types of trarJSparent materials-one indicated by horizontal lines and the other by vertical lines. The number of images formed of an object will be (A) 1 (B) 2 (C) 3 (D) 6

is

(B)

(A) (C)

207. The distance between an object and its real image formed by a convex lens cannot be (A) greater than 2/ (B) less than 2/ (C) greater than 4/ (D) less than 4/

x, x, 2

.X1 +X2

(D)

2

~x, x, ·

202. The plane faces of two identical planoconvex lenses, each having focal length of 40 cm , are pressed against each·other to form a usual convex lens. The distance in cm from this lens, at which an object must be placed to obtain a real 'image with magnification unity is (A) 10 (B) 20 (C) 40 (D) 80

3 . .JS 2 placed at a distance of 10 cm from a thin convex lens of

203. ·A plane refracting surface of refractive index -

focal length 30 cm . The parallel rays incident on lens

208. Two thin symmetrical lenses of different nature and of different material have equal radii of curvature R = 15 cm are placed close together and immersed in water

(µ. =¼). The focal len~ of the system in water

is 30 cm . The difference between refractive indices of the two lenses is

(A)

(C)

1 2 ,

1 4

(B) (D)

1 3 3

4

will converge at a distance of

209. A needle of length 5 cm, placed 45 cm from a lens forms an image on a screen placed 90 cm on the other side of

i+-

1Ocm --+1

the lens. The type of lens and its focal length are (A) convex, 30 cm (B) concave, 30 cm (C) convex, 60 cm (D) concave, 60 cm

~J

210. In PROBLEM 209, the nature and size of the image are

(A) 30 cm from the lens.. (B) 25 cm from the lens. (C) 20 cm from the lens. (D) 40 cm from the lens. 204. The figure shows an equiconvex)ens of focal length / . If the lens is cut along PQ, the focal length of each half will be p

(A) real, 20 cm (C) virtual, 20 cm

(B) real, 10 cm (D) virtual, 10 cm

211. An object is placed 50 cm in front of a convex surface of radius 20 cm. If the surface separates air from glass of refractive index 1.5, the distance of the image from. the lens and its nature are (A) 30 cm, real (B) 30 cm, virtual (C) 300 cm, real (D) 300 cm, virtual

R ---+,>+---S ·

Q

(A)

l_ 2

(B)

I

212. One of the refracting surfaces of a prism of angle 30° is .silvered. A ray of light incident at an angle of 60° retraces its path. The refractive index of the material of prism is '

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-Jz

(B)

(C) fj

3 2

(B) 25 cm

(C) 20cm (D) cannot be estimated with given data

(D) 2

213. A slab of glass of refractive index 1.5 and thickness 3 crri is placed with the faces perpendicular to the principal axis of a concave mirror. If the radius of curvature of the mirror is 10 cm, the distance at which an object must be placed from the mirror so that the image coincides with the object is (A) 9cm (B) .10 cm (C) 11cm (D) 12cm

214. Figure represents a convergent lens placed inside a cell filled with a liquid. The lens has a focal length +20 err\ when in air and its material has refractive index 1.50 . If the liquid has a refractive index 1.60, the focal length of the lens in the new system is

t

(C)

-80 cm -160 cm

215. A point object O is placed on the principal axis of a convex lens of focal length 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. If the eye is placed 60 cm to the right of the lens at a distance h below the principal axis, then the maximum value of h to see the image will be (A) 0 cm (B) 5 cm (C) 2.5 cm (D) 10 cm 216. For two positions of a lens, the images are obtained on a fixed screen. If the size of object is 2 cm and the size of diminished image is 0.5 cm, the size of the other image will be (A) 1 cm (B) 4cm (C) 8 cm (D) 16cm 217. The. medium on both sides of lens is air. The distances of object O , image I from first and second foci F1 and

F, are shown in figure. The focal length of lens is

- F,

0

16 cm

L

(A) 16cm

incident on a plane mirror. After reflection, it travels

along the direction

(A) 30° (C) 60°

.!(1 - /3J) . The angle of incidence is 2

(B) 45' (D) 75'

219. A bi-convex lens is formed with two thin piano convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this bi-convex lens, for the object distance of 40 cm , the image distance will be

(A) -280 cm (C) 21.5 cm

(B) +80 cm (D) -24 cm

F,

½(1 + /3]} is

R = 14 cm

Lens

(A)

218. A ray of light ravelling in the direction

2 5cm

(B) 40 cm (D) 13.3 cm

220. An equiconvex lens, having radius of curvature 33 cm, is placed on a horizontal plane mirror and a pin held 20 cm above the lens coincides with its image. Now the space between the lens and the mirror is filled with a liquid. In order to coincide with the image the pin has to be raised by 5 cm. The refractive index of the liquid is (A) 1.33 (B) 1.53 (C) 2.33 (D) 2.66 221. A real image is formed by a convex lens. If we put a concave lens in contact with it, the combination again forms a real image. The new image . (A) is closer to the lens system. (B) is farther from the lens system. (C) is at the original position. (D) may be anywhere depending on the focal length of the concave lens. 222. 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). 60 cm (B) 40 cm (C) 30 cm (D) 20 cm

223. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of the combination in dioptre is (A) -1.5 (B) -6.5

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(D) +6.67

(C) +6.5

3/

224. Two lenses of powers1 +12 D and -2 D are in contact. The focal length of the combination is (A) 10 cm (B) 12.5 cm (C) 16'6 cm (D) 8.33 cm 225. A point object O is placed at a distance of 20' cm from a convex lens of focal length 10 cm as shown in 0 figure. The distance x from the lens where a concave mirr(?r o,f focal . length 60 cm has to, be placed so th~t final image coincides with the obje.ct is (A) 10 CI_ll (B) 20 cm , (C) 40cm (D) final image can never coincide with the object under the conditions provided.

I

(D) f, 3:

4'2

229. When a ray C>f light goes from air to a glass slab, then (A) its wavelength increases (B). its wavelength decreases , (C) . its frequency increases (D) neither.its wavelength nor its frequency changes 230. In the displacement method, a convex lens is placed in between ,an object .µ1d a screen. If the magnification in the two positions are m1 and m2 ( m1 > m, ) , and the distance between the two positions of the lens is x , the

focal length of the Jens is (A)

(C)

X --

(B)

m1 +m2 X

(D)

(m, +m,)'

x m1-m2 X

(m, -m,)'

231. A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at

226. A lens forms a sharp image on a screen. On inserting a

parallel sided glass slab between the lens and the screen, it is _found necessary to move the screen a distance d away from the lens in order for the image to

be sharp again. If the refractive index of the material of the slab is n , the thickness of the slab is d (A) nd (B) n · nd (C) (n-1)d (D) n-1 n

two different locations separated by 20 cm. The focal · length of the Jens is (A) 10.7cm (B) 21.4cm (D) 32.1 cm (C) 15.8cm 232. A light ray travelling in glass medium is incident on glass-air interference at an angle of i_J;i.cidence 8 . The

reflected (R) and transmitted (T) int~nsities, both as function of 0, are plotted. The.correct sketch is ,Intensity

227. A plano convex glass

Jens

(µ, =¾)

100o/o -----------, -·

of

radius of curvature R =10 cm is placed at

T

,' ,,

(A)

___R..

0

--~_,.,..'

o~----9LO'-+ 0

a distance of y from a

concave lens .of focal ,..._ X - - - Y ___,.. length 20 cm . The distance x of a point object O from the piano convex lens so that the position· of final image is independent of y'is (B) 30cm (A) 20cm (C) 40cm (D) 60cm

, Intensity 100%

T

(B)

228. A thin lens has focal length /, and its aperture has

diameter D .

Ii forms

an image of intensity

central part of the_ aperture, of diameter

i'.

If the

~ , is blocked

by an opaque paper, the focal. length of the lens and the intensity of image will become

(A) 1.128

fr,½

Intensity 100%

T

(C)

__ f:!_.-

(B)

o~--.L-~9~0--0

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T

(B)

,,- -----

a virtual, erect, same-sized image can be obtained using a plane mirror.

(C) a virtual, erect, magnified image can be formed using a concave mirror.

(D)

(D) a real, inverted, same-sized image can be formed using a convex mirror.

R --

O!=--:.:-::-c_-_ _...:::::90=,=---+. e 239. When an object is at distances x and y from a IellS, 'a

233. A lens is placed between the source of light and a wall. It forms images of area A1 and Ai on the wall for its two different positions. The area of the source of light is (A)

~A,A,

(C)

(B)

(D)

A,; A,

(C)

2.+2.)-1

( A,

the curved surface, then the system behaves like a

(C) rµ

(B)

r µ-1

(D)

r(µ-1)

..{xy

(D)

x+y 2

240. If the central portion of a convex lens is

A,

234. The plane face of a plano-convex lens is silvered. If µ be the refractive index and r the radius of curvature of concave mirror of radius r (A) µ

real image and a virtual image is formed respectively

having· same magnification. The focal length of the lens is given by (A) x-y (B) x+y

wrapped in black paper as shown in the figure, ~ (A) no image will be formed by the remaining portion of the lens. · (B) full image will be formed, but it will be less bright. (C) the central portion of the image will be missing. (D) there will be two images, each producecl by one of the exposed portions of the lens.

241. A plane mirror made of glass slab 235. A ray of light falls on the surface of a spherical ·paper

position of the final image is

deviation of the emergent ray from the direction of the incident ray is

(A)

(B)

2(a-p)

(a-P)

(D)

p-a

(C)

2

is 2.5 cm

thick and silvered at the back. A point object is placed 5 cm in front of the unsilvered face of the mirror. The

weight making an angle a with the normal and is refracted in the medium at an angle p. The angle of

(A) (a-P)

(µ, =1.5)

(B)

16 3

cm from unsilvered face

25

(C)

cm from unsilvered face 3 12 cm from unsilvered face

(D)

14 cm from unsilvered face

236. The magnification of an object placed in front of a

convex lens of focal length 20 cm is +2 . To obtain a magnification of -2, the object has. to be moved by a distance equal to (A) 40 cm

(C) 20cm

242. The distance between an object and the screen is 100 cm. A lens proQ.uces an image on the screen when

placed at either of two position~ 40 cm apart. The power of the lens is approximately (A) 4.25 D (B) 4.50 D (C) 4,75 D (D) 5.0 D

30 cm (D) 10cm (B)

. 237. A concave mirror is placed on a horizontal table with its axis directed vertically upward. Let ( 0) 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

(A) real and will remain at C . (B) real and located at a point between C and oo . (C) virtual ru,d located at a point between C and 0. (D) real and located at a point between C and 0.

243. A real image of an object is formed ·by a convex lens at

the bottom of an empty beaker. The beaker is now filled with a liquid of refractive index 1.4 to a depth of 7 cm. In order to get the image again at the bottom, the beaker should be moved 0

238. All of the following statements,are correct except

(A) the magnification produced by a convex mirror is always less than one.

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(A) 10

(B) 20

(C) 30

(D) 40

250. An object is placed at a distance of 12 cm from a convex lens on its principal axis and a virtual image of certain

244. The convex surface of a thin concavo-convex lens (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. At what distance from the lens should a pin be placed on the optic axis such that its image is formed at

size is formed. If the object is moved 8 cm away from the lens, a real image of the same size as that of the virtual image is formed. The focal length of the lens in cmis (A) 15 (B) 16 (D) 19 (C) 18

the same place?

(A) 15 cm (C) 22.5 cm

(B) 7.5 cm (D) 30 cm

245. In PROBLEM 244, if the concave part is filled with

water (refractive index

± ), the distance from the lens at 3,

251. A thin converging lens of refractive index 1.5 has a power of +0.5 D . When this lens is immersed in a liquid, it acts as a diverging lens of focal length 100 cm. The refractive index of the liquid is (A)

4

(B)

3

(C)

5 3

() D

2

which the pin should be placed to form the image at the same place is

90 cm 13 . 135 (C) -cm 13

(A)

(B)

45

-cm 13

2

252. The distance between an object and its real image

180

(D) -cm 13

246. An equiconvex lens of glass

3

(µ, =1.5)

of focal length

10 cm, silvered on one side behaves like a

formed by a lens is D . If the magnification is m , the focal length of the lens is mD (A) (m-l)D (B)

m

(C)

(A). convex mirror of focal length 5 cm (B) convex mirror of focal length 20 cm (C) concave mirror of focal length 2.5 cm (D) concave mirror of focal length 10 cm 247. A piano-convex lens of focal length 30 cm has its plane surface silvered. An object is placed 40 cm from the lens

(m-l)D

(D)

m'

m+l mD

(m+l)'

253. A piano convex lens of focal length 16 cm, is to be made of glass of refractive index 1.5. The radius of curvature of the curved surface should be (A) 8 cm (B) 12 cm (C) 16 cm (D) 24 cm

on the convex side. The distance of the image from the

lens is (A) 18 cm (C) 30 cm

254. A real image of a point object O was formed by an

(B) 24cm (D) 40 cm

248. Refraction takes place at a concave spherical boundary ' 3 separating glass air medium. If µg = , then for the

2

image to be real, the object distance (A) is independent of the radius of curvature of the refracting surface (B) should be greater than the radius of curvature of the refracting surface (C) should be greater than two times the radius of curvature of the refr~cting surface (D) should be greater than three times the radius of curvature of the refracting surface 249. A parallel beam of light incident on a concave lens of focal length 10 cm emerges as a parallel bearu from a convex lens placed coaxially~ the distance between the lenses being 10 cm. The focal length of the convex lens incm is.

equi-convex lens of focal length f and the magnification was found to be unity. Now the lens is cut into two symmetrical pieces as shown by the dotted line and the right part is removed. The position of the image formed by the remaining part is at

0

f

(A)

f

(C) i_ 2

(B)

2/

(D) Infinity

255. A convex lens, made of a material of refractive index 1.5 and having a focal length of 10 cm is immersed in a liquid of refractive index 3.0. The lens will behave as a

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' lens of focal length 10 cm. (A) converging (B) diverging lens of focal length 10 cm. (C) converging lens of focal length

10 3

(C) 12cm

cm.

(D) diverging lens of focal length 30 cm. 256. A point source S is placed at a height h from the bottom of a vessel of height H (< h). The vessel is . polished at the base. If the water is gradually filled in the vessel at a constant rat'e a m 3s-1 , the distance d of image of the source from the bottom of the vessel varies with time. t as

S•

260. The maximum and minimum distances between a convex lens and an object, for the magnification of a real image to be greater than one are (A) 2/ and f (B) f and zero (C) oo and 2/ (D) 4/ and 2/ 261. A point object is placed on the optic axis of a convex lens of focal length f at a distance of 2f to the left of it. The diameter of the lens is d . An observer has his eye at a distance of 3 f to the right of the lens and a distance h below the optic axis. The maximum value of h to see the image is d (A) d (B)

r

i

• h

4

H

(C)

(A)

(q

t=, l=, 1=, l=, 0)

(D)

257. A convex lens of glass has power P in air. If it is immersed in water its power will be (A) more than P (B) less than P (C) p (D) more than P for some colours and less than P for others 258. A biconvex lens, made of a material of refractive index 1.5, has radius of curvature of each side equal to 0.5 m. The power of the lens is

(A) 0.5 D (C) 1.5 D

(D) 6cm

(B) 1.0 D

(D) · 2.0 D

259. A convex lens forms a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object or the screen, again real image is formed on the screen which is 16 cm long. The length of the object is (A) 8 cm (B) 10 cm

d

2

3

(D) d

262. A convex lens is immersed in a liquid of refractive index greater than that of glass. It will behave as a (A) convergent lens (B) divergent lens (C) plane glass (D) homogeneous liquid

263. If the top half of a convex lens is covered with black paper, (A) the bottom half of the image will disappear. (B) the top half of the image will disappear. (C) the magnification will be reduced to half. (D) the intensity will be reduced to half. 264. In displacement method, the lengths of images in the two positions of the lens between the object and the screen are 9 cm and 4 cm respectively. The length of the object must be · (A) 6.25 cm (B) 1.5 cm (C) 6 cm (D) 36 cm

265. A convex lens of focal length 15 cm is placed on a plane mirror. An object is placed 20 cm from the lens. The image is formed (A) 12 cm in front of the mirror (B) 60 cm behind the mirror (C) 60 cm in front of the mirror (D) 30 cm in front of the mirror 266. A convex lens of focal length 40 cm is held coaxially 12 cm above a concave mirror of focal length 18 cm. An object held x cm above the lens gives rise to an image coincident with it. The x is equal to (A) 12cm (B) 15 cm (C) 18cm (D) 30cm

0 xcm

12cm

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267. Two thin lenses of powers 2 D and 3 D are placed in contact. An object is placed at a distance of 30 cm from the combination. The distance in cm of -the image from the combination is

(A) 30 (C) 50 .

(A) f_

(ll) 40 (D) 60

lenses is

(C)

lf,.-f,I

(ll)

h + !,

f,f,

(D)

I,!,

It, - !,I

(D) dispersion

270. A plane mirror is Rlaced at the bottom of a tank containing a liquid ofrefractive index µ . A small object P lie~ at a height h above the mirror. An observer O , vertically above P , outside the liquid, observe P and its image in the mirror. The apparent distance between these two will be

o, ''' '' ' 'I ' Ph

(D)

3

(D)

f

3/ 2

274. An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length f, of

275. A piano convex lens .of radius of curvature R fits exactly into a plano concave lens such that their plane surfaces are parallel to each other. If the lenses are made of differen~ materials of refractive indices µ1 and µ 2 , then focal length of the combination is given by (A)

(ll)

R 2-(µ, +µ,)

(C)

ZR

R 2(µ,-µ,)

(D)

R

276. A compound microscope has an objective of focal length 2.0 cm and an eye piece of focal length 6.25 cm separated by 15 cm. If the final image is formed at the least distance of distinct vision (25 cm), the distance of the object from the objective is (A) 1.5 cm (B) 2.5 cm (C) 3.0 cm (D) 4.0 cm

:~

(B)

(C)

2/

the objective and f, of the eyepiece are (A) f 0 =45cm and f>,-9cm (ll) f 0 =50cm and /,=10cm (C) f,=7.2cm and /,=5cm (D) / 0 =30cm and /,=6cm

h+f,

269. Chromatic aberration in a lens is caused by (A) reflection (ll) interference

(C) diffraction

(B)

2

268. Two convex lenses of focal lengths f, and f, are mounted coaxially separated by a distance. If the power of the combination is zero, the distance be~~en the

(A)

concave mirror of focal length f : To obtain a real image of same magnification, the object has' to moved 1;>y a distance

2h .µ-1 2h µ

277. In PROBLEM 276, the magnifying power of the

271. A person can see clearly between 1 m and 2 m .. His corrective lenses should be (A) bifocals with pow~r -0.5 .D and additional +3.5 D (B) bifocals with power ~1.0 D and additional +3.0D (C) concave with power LO D (D) convex with power 0.5 D

microscope is

(A) 10 (C) 20

(B) 15 (D) 30

278. A point object is placed at a distance of 20 cm from a

glass slab (µ, =

¾) half immersed in water (µ. = ¾) as

shown in figure. The distance between two images when se~n-from the other side of the slab is

272. Spherical aberration in a thin lens can be reduced by (A) using a monochromatic light. (B) using a doublet combination. (C) using a circular. annular mask over the lens. (D) increasing the size of the lens.

9cm

l+-+i

--:-

0

273. A virtual image of an object is formed with a magnification of 2, when the object is placed infront of

a

1+20 cm+i

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(A) 1 cm (C) 4cm

(B) 2cm (D). 6cm

279. A compound microscope has .a magnification of 30. The focal length of the eye-piece is 5 cm. If the final image is formed at the least distance of distinct vision ( 25 cm),

possible magnification, we choose the lenses of focal lengths (A) 100 cm, 0.3 cm (B) 10 cm, 0.3 cm (C) 10 cm, 4 cm (D). 100 cm, 4 cm 287. The angular magnification of a telescope which contains

an objective of focal length / 1 and eyepiece of focal length /, is

the magnification produced by the objective is (A) 5 (B) 7.5 (C) 10 (D) 15

(AJ 280. The least distance of distinct VisiOn is 25 cm. The focal length of a convex lens is 5 cm . It can act as a simple

microscope of magnifying power (AJ 4 (B) 5 (C) 6 (D) None of these 281. An astronomical telescope has an eye piece of focal length 5 cm. If the angular magnification of normal adjustment is 10, the distance between the objective and the eye.piece is (A) 45cm (B) 50 cm (C) 55cm (D) 110 cm 282. The focal lengths of the objective and the eyepiece of an

astronomical telescope are 100 cm and 20 cm respectively. Its magnifying power in normal adjustment is (A) 5 (B) 2 (C) 25 (D) 4 283. Two convex lenses of focal lengths 0.3 m and 5 cm are used to make a telescope. The distance kept between

them is equal to (A) 0.35 m (C) 5.3m

(C)

4 3

(DJ

/1 + /, I, f,f, !1 + I,

288. An achromatic combination is to be made using a convex and a concave lens. Tfle two lenses should have (AJ their power equal. (B) their refractive indices equal. (C) their dispersive powers equal. (DJ the product of their powers and dispersive powers equal. · 289. For a thin equiconvex lens, the optics axis coincides with the x-axis.and the optical centre coincides with the origin. The co-ordinates of a point object and its image

are (--40, 1) cm and (50,-2) cm respectively. Lens is located at (A) x=0 (B) X=-lOcm (D) x~-30cm (C) x=+20 cm 290. The near point of a person is 50 cm and the far point is for seeing distant objects are respectively

285. The angle of incidence for an equilateral prism is 60° . The refractive index of prism so that the ray inside the prism is parallel to the base of the prism is 9 .. 8

(BJ

1.5 m. The spectacles required for reading purpose and

(B) 5.3 cm (D) 0.15m

284. To have larger magnification by a telescope (A) the objective should be of large focal length and the eyepiece should be of small focal length (B) both the objective and the eyepiece should be of large focal lengths (C) , both the objective and the eyepiece should be of small focal lengths (D) the objective should be of small focal length and the eyepiece should be of large focal length

(A)

(C)

I, !1 /1 I,

(B). ..fi.

(D) ,/3

286. Four convergent lenses have focal lengths 100 cm, 10 cm, 4 cm and 0.3 cm. For a telescope with maxirnuffi

(A) +2 D, -(¾) D

(B)

+(¾) D, -2 D

(C) -2 D, +(¾) D

(D) -(¾) D, +2 D

291. Astigmatism for a human eye can be removed by using

(A) concave lens (C) cylindrical lens

(B) convex lens (D) prismatic lens

292. A hollow convex lens of glass behaves like a

(A) plane mirror (C) convex lens

(B) concave lens (D) glass plate

293. The far point of a myopic eye is 250 cm. The correcting lens should be a (A) diverging lens of focal length 250 cm. (B) converging lens of focal length 250 cm. (C) diverging lens of focal length.125 cm. (D) converging lens of focal length 125 cm. 294. A person cannot see clearly beyond 50 cm. The power Of the lens required to correct his vision is (A) -0.5 D (B) +0.5 D

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(C)

-2 D

(D) +2 D

295. A ray travelling in negative x-direction is directed towards positive y-direction after being reflected from a surface at point P . The reflecting surface is represented by the equation x2 + y2 = a2 • Then co-ordinates of point

296. A person cannot see clearly objects at a distance less than 100 cm,. The power of the spectacles required to see clearly objects at 25 cm is (A) +1 D (B) +3 D

(C)

+4 D

(D) +2 D

297. An object is kept at a distance.of 16 cm from a thin lens and the image formed is real. If the object is kept at a

P are

distance of 6 cm from the same lens the hnage formed

is virtual. If the size of the images formed are equal, the focal length of the lens will be · (A) 8 cm (B) 5 cm (C) 11 cm (D) ../% cm 298. A person can see clearly objects lying between 25 cm and 2 m from his eye. His vision can be corrected by

(A) (a, 0)

=

(B)

(0.6a, O.Ba)

c;, Jz)

using spectacles of power

(A) +0.25 D (C) -0.25 D

(B)

+0.5D

(D)

-0.5D

(C)

( 0.Ba, 0.6a)

1.134

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(D)

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This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/ are.correct.

1.

The x-y plane is the boundary between two transparent media. Medium 1 with z 2:: 0 has a refractive index ..fi. and medium 2 with z < 0 has a

·../3. A ray of light in medium 1 given A= 6../31 + s../3J-10k is incident on the

refractive index by the vector

6, (C) -=µ 61 4.

A ray of light from a denser medium strikes a rar~r medium at angle of incidence i . The reflected and the refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r' respectively. The critical angle is (A) sin-1 ( tanr) (B) sin-1 ( tani) 1 (C) sin- (tanr') (D) tan-1 (sini)

5.

A single converging lens is used as a simple microscope. In the position of maximwn magnification.

plane of separation. The refracted ray makes angle r with +z axis and incident ray makes an angle i with -z axis. Then, (A) i = 120° (B) i =60° (C) r=45° (D) r =135°

2.

A ray of light travels from a medium of refractive index µ to air. Its angle of incidence in the medium is i, measured from the normal to the boundary, and its angle of deviation is a. The curve that best represents the plot of deviation 6 (along y-axis) with angle of incidence i (along x-axis) is

Select the correct statement(s). (A) the object is placed at the focus of the lens. (B) the object is placed between the lens and its focus. (C) the image is formed at infinity. (D) the object and the image subtend the same angle at the eye. ·

il 6,

(A)

6.

o, 0'-""':.._4-_.,,__.. 0

'

'

,,

3.

(D)

'' 6 ~ il 1 I J!. 0

ilkut.

A light of wavelength 6000 A in air enters a medium of refractive index 1.5 . Inside the medium, its frequency is v and its wavelength is 'A. . (A) v=5x10 14 Hz (B) v=7.5x1014 Hz (C)

1. = 4000

A.

(D) ,_ = 9000

A.

il

il

(C)

(D)

0

In PROBLEM 2, (A) 9=sin-'(¾)

2

i

'

il ,

.

0

'' '

''

0

7.

(A) may form a real image (B) must form a real image (C) may form a virtual image (D) may be a parallel beam

n

-2

If a converging beam of light is incident on a concave mirror, the reflected light

i

8.

Two points P and Q lie on either side of an axis XY as shown. It is desired to produce an image of P at Q

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using a spherical mirror, with XY as the optic axis. The mirror must be p

•

(A)

b

C

(B)

b

C

(C)

(D)

a

ab C

X+----------•Y

14. A lens ofrocal length / is placed in between an object

• Q

and screen fixed at a distance D . The lens forms two real images of object on the screen for two of its different positions, a distance x apart. The two real images have magnifications m1 and m2 respectively

(A) converging (B) di verging (C) positioned to the left of P (D) positioned to the right of Q 9.

(m 1 >m,).

An object and a screen are fixed at a distance d apart. When a lens of focal length / is moved between the object and the screen, sharp images of the object are formed on the screen for two positions of the lens. The magnifications produced at these two positions are M 1 and M2 • (A)

d>2f

(B)

d>4f

(C)

MM =1

(D)

IM,1-IM,1=7

1

2

d

10. Resolving power of an electron n,,.icroscope is R, and that of optical microscope is R0 .

(A) R, > R, (C) R, =R,

(B) (D)

R, < R0 Data Insufficient

(A)

X

!=-m1 -mi

(C)

D2-x2

! =4D --

(B)

m1m2 =1

(D)

D?.4/

15. A planet is observed by , an astronomical refracting telescope having an objective of focal length 16 m and an eye piece of focal length 2 cm . (A) The distance between objective and eye piece is 16.02 m. (B) The angular magnification of the planet is -800. (C) The image of the planet is inverted. (D) The objective is larger than the eye piece.

16. A parallel beam of white light falls on a combination of

11. In PROBLEM 10, the correct argument for the correct selected option is that (A) electrons have greater wavelength than visible light. (B) electrons have lesser wavelength than visible light. (C) resolving power is inversely proportional to the wavelength of the wave used for detecting an object by the microscope. (D) resolving power is inversely proportional to the square of the wavelength of the wave used for detecting an object by the microscope. 12. The distance between two point objects P and Q is 32 cm . A convex lens of focal length 15 cm is placed between them so that the images of both the objects are formed at the same place. The distance of P from the lens could be (A) 20cm (B) 18 cm (C) 16 cm (D) 12cm

a concave and a convex lens, both of same material. Their focal lengths are 15 cm and 30 cm respectively for the mean wavelength in white light. On the other side of the lens system, one sees (A) a coloured pattern with violet at the outer edge. (B) a coloured pattern with red at the outer edge. (C) white light again. (D) that it is unable for the lens to converge the rays at a point. 17. Consider a ray of light going from A to B. Let the ray traverse, in going from A to B, distances S 11 s2 ,s 3 , ................... sm in media of indices n1 , n2 , n3 , ••...•..••.•. nm respectively.

(A) Total time offiight

1 t=-Ln,s, m

C i=l

(B) Total time offiight t = _!

f s,

C i=l

(C) Optical path length is (O.P.L.) =

f n,s, i=l

13. The graph shows the variation of magnification m produced by a convex lens with the image distance v . The focal length of the lens is

B

f

m

(D) For inhomogeneous media the O.P.L. = n(s)ds A

and the ray travels along 'Stationary Pathways'.

L,.__

-'--+ V

_.L._ _ _ _

,.__ a,---c-----

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Ray Optics 18. A point object is placed ai 30 cm from a convex glass lens

(µ, =¾)

on the lens returns as a parallel beam from the arrangement. Select the correct statement(s).

of focal length 20 cm . For the final

(A) The beam diameters of the incident and reflected beains must be the· same.

image of object to be formed at infinity, which of the following is/ are correct? ' (A) A concave lens of focal length 60 cm is placed in

CBl a= 211,Hl,I


contact with the convex lens

(D) If the entire arrangement is immersed in water, the

(B) A convex lens of focal length 60 cm is placed at a distance of 30 cm from the convex lens. (C) The entire convex lens system is immersed in a

liquid of refractive index

i3

(D) The ~ntire convex lens system is immersed in a

conditions will remain unaltered. 24. An astronomical telescope and a Galilean telescope use identical objective lenses. They have the same magnification, when both are in normal adjustment. The eyepiece of the astronomical telescope has a focal

length I . Select the correct statement(s). (AJ The tube lengths of the two telescopes differ by I . (BJ The tube lengths of the two telescopes differ by

liquid of refractive index !!_ 8

19. A converging lens is used to for_m an image on a screen. When the upper half of the lens is covered by an opaque screen (A) half.of the image will disappear (B) complete image will be formed (C) intensity of the image will increase (D) intensity of the image will decrease

21. (C) The Galilean telescope has shorter tube length. (D) The Galilean telescope has longer tube length. 25. Two plane mirrors M 1 and M 2 are placed parallel to each other 20 cm apart. A luminous point object 10 1 is placed between them at 5 cm from Mt as shown.

20. For a mirror, the linear rnagnificatioh ·is +2 . The conclusion(s) that can be drawn from this information

,._...20 cm---1>1

is/are (A) The mirror is concave

(B) The mirror can be convex or concave but it cannot

i

be plane (C) The object lies between pole and focus (D) The object lies·beyond focus

I

M,!

I= lo

(C)

V

' µ

= Vo µ

(B) (D)

M,

(A) The distances (in cm) of three nearest images mirror Mt are 5, 35 and 45 respectively. (BJ The distances (in cm) of three nearest images mirror M, are 5, 35 and 45 respectively. (C) The distances (in cm) of three nearest images mirror M 1 are 15, 25 and 55 respectively. (D)_ The distances (in cm) of three nearest images

21. A ray of light has speed v0 frequency lo and wavelength A.0 in vacuum. When this ray of light-enters in a medium of refractive index µ , corresponding values are v , I and "- . Then (A)

.............. !··--·---··· 0 j

from from from from

mirror M 2 are 15, 25 and 55 respectively.

I= lo

22. For which of the pairs of u and I for curved mirror(s), the,image formed is smaller in size.

26.

In the case of hypermetropia (A) the image of a near object is formed behind' the retina.

(BJ the image of a distant object is formed in front of

(A) u=-45cm, l=-l0cm (B) u=-10cm, l=20cm (C) u=-60cm, l=30cm (D) u=-20cm, l=-:-30cm

the retina. (C) a concave lens should be used for correction. (D) a convex lens should be used for correction. 27.

23. A di~erging lens of focal leng\h I, is placed in front of and coaxially with a concav~ mirror of focal length / 2 •

Their separation is d . A parallel beam of light incident

Which of the following produce a virtual image longer in size than the object?

(A) Concave lens (CJ Concave mirror

(BJ Convex lens (D) Convex mirror

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--C

A concave mirror has focal length 15 cm. Where should

an object be placed in front of the mirror so that the image formed is three times the size of the object?

(A) 7.5 cm (C) 17.5 cm

(B) 10 cm (D) 20 cm

29. A concave mirror of focal length f forms an image 2 times the size of object. The object distance from the mirror is

(A)

j_

(B)

3/

(D)

4

(C)

2

30.

4/ 3

µ = 1.5

(A) appear to meet after extending the refracted rays backwards. (B) actually meet at some point. (C) meet (or appear to meet) at a distance of 60 cm from the spherical surface. (D) meet (or appear to meet) at a distance of 30 cm from the spherical surface.

j_

35. The focal length of a lens in air and refractive index are f and µ respectively. The focal length changes to / 1

2

A point object P moves towards a convex mirror with

when the lens is immersed i:r:t a liquid of refractive

a constant speed V, along its optic axis. The speed of

index

.the image

(A) is always less than V. may be less than, equal to or greater than V, depending on the pcisition of P . (C) increases as P comes closer to the mirror. (D) decrease as P comes closer to the mirror. · (B)

!:

and it becomes / 2 when the lens is immersed 2 ' in a liquid of refractive index 2µ . Then

w

h

2(µ-1)

f

(B)

(D)

(C)

2{µ-1)

h

f

µ-1

!,=!

31. A bird flies down vertically towards a water surface. To a fish inside the water, vertically below the bird, the

36. Two

thin

lenses,

when

in

contact,

produce

a

combination of power +10 dioptre. When they are 0.25 m apart, the power is reduced to +6 dioptre. The respective powers of the lenses in dioptre, are (A) 1 and 9 (B) 2 and 8 (C) 4 and 6 (D) 5 each

bird will appear to (A) be closer than its actual distance. (B) be farther away than its actual distance. (C) move slower than its actual speed. (D) move faster than its actual speed. 32. There are three optical media 1, 2 and 3 with their refractive indices µ 1 > µ 2 > µ 3 • S~lect the correct statement(s) {A) When a ray of light travels from 3 to 1 no TIR will take place. (B) Critical angle between 1 and 2 is less than the critical angle between 1 and 3. (C) Critical angle between 1 and 2 is more than the critical angle between 1 and 3.

(D) Chances of TIR are more when ray of light travels from 1 to 3 as compare to the case when it travel from 1 to 2.

..Ji. . Select the correct alternative(s). (A) Minimum deviation from this prism can be 30° (B) Minimum deviation from this prism can be 45° (C) At angle of incidence 45°, deviation is minimum _(D) At angle of incidence 60°, deviation is minimum

37. A solid, transparent sphere has a small, opaque dot at its centre. When observed from outside, the apparent

position of the dot will be (A) independent of the refractive index of the sphere. (B) closer to the eye than its actual position. (C) farther away from the eye than its actual position. (D) the same as its actual position. 38. For a concave mirror (A) virtual image is always larger in size

(B) real image is always smaller in size (C) real image is always larger in size (D) real image may be smaller or larger in size

33. An equilateral prism has a refractive index

34. Parallel rays of light are falling on a convex spherical surface of radius of curvature R = 20 cm and refractive index µ = 1.5 as shown. After refraction from the spherical surface, the parallel rays

39. During refraction, ray of light passes undeviated, then (A) medium on both sides is same (B) angle of incidence is 90° (C) angle of incidence is 0° (D) medium on other side is rarer

40. A ray of light travelling in a transparent medium falls on a surface separating the medlUm from air at an angle of incidence 45° . The ray undergoes total internal reflection. If n is the refractive index of the medium

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Ray Optics with respect to air, select the possible value(s) of n from the following , · (A) 13 (B) 1.4 (C) 1.5 (D) 1.6 41. A convex lens made of glass ( µ,

=

¾) has focal length

f in air, The image of an object placed in front of it is real,

inverted

and

magnified.

Now

arrang~ment is immersed in water

the

(µ"' =¾)

whole without

changing the distance between object and lens, then (A) the,new focal length becomes 4/

(B) the new focal length becomes J_ .

4

(C) the new image formed will be virtual and magnified.

(D) the new image formed will be real and diminished. 42. . A thin, symmetric double-convex lens of power P is cut into ,three parts A , B and C as shown. The power of A

(D) If the_ entire arrangement is.immersed in water, the conditions will remain unaltered.

45. Check the wrong statement(s) (A) A concave mirror can give a virtual image. · (B) A concave mirror can give a diminished virtual image. (C) A convex mirror can give a real image. (D) A convex mirror can give a diminished virtual image. 46. ,When lights of different colours move through water, they must have different , (A) wavelengths (B) frequencies (C) velocities (D) amplitudes 47. A thin concavo-convex lens has two surfaces of radii of curvature R and 2R . The material of the lens has a refractive index µ . When kept in air, the focal length of the lens (A) will depend on the direction from which light is incident on it. (B) will be the same, irrespective of the direction from which light is incident on it 2 (C) will be equal to R .

µ-1

(D) will be equal to B

(A) Ais·P

(C) B

lS -

_2

48. A convex; mirror is used to form .an image of a real object. The image , (A) always lies between the pole and the focus. (B) is diminished in size. (C) is erect. (D) is.real.

A is 2P

(D) B

. p

IS -

' 4

43. A watch glass having uniform thickness and having

average radius of curvature of its two surfaces much larger than its thickness is placed in the path of a beam of parallel light. The beam will (A) be completely unaffected. (B) converge slighUy. (C) · diverge slighUy. (D) converge or diverge slightly depending on' whether the beam ·is incident from the concave or the convex side.

44. A converging lens of focal length / 1 is placed in front of and coaxially with a convex mirror of focal length f, . Their separation is d . Aparallel beam of light incident. on · the lens returns as a parallel beam from the . arrangement. Select the correct'statement(s) .. (A) The beam diameters of tne incident and reflected beams must be the same.

(Bl

d =t, -211,I

(q

d :c1,

-lt,1

µ-1

C

(B)

. p

~.

·

.,.

49. Which of the following form(s) the virtual and erect image for all positions of object ? (A) concave _mirror (B) convex lens (C) convex mirror (D) concave lens 50. A ray of light is incident on a prism of refracting angle A . C is the critical angle for the material of ihe prism with respect to the surrounding material (say air/vacuum). . (A) An emergent ray will be there for all values of C . (B) An emergent ray will be there only for A < 2C . (C) A-ray incident at an angle i can pass through the prism if sini> sin(A-C) for C
A thin plane-convex lens of focal length f is split into two equal halves. One of the halves is shifted along the optical axis as shown. The separation between object and image planes is 1.8 m and the magnification of 1.139=

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image formed by one of the half lens is 2: The separation between two halves is d .

(D)

f

=½[1- ~1- :, ]

53. An object is placed at a distance 2/ from the pole of a curved mirror of focal length f (A) The linear magnification is 1 for both. types of curved mirror.

(B) The linear magnification is 1 for a concave mirror. ,+----1.8 m-----+<

(A) /=0.4m (C) d=D.6m

(C) The linear magnification is

f =0.6 m (D) d=0.4m

(B)

!3

for a convex mirror.

(D) Data Insufficient.

52. A point source of light is placed at a distance h below the surface of a large and deep lake. If f is the fraction of light energy that escapes directly from water surface and µ is refractive index of water then, (A) f varies as a function of h (B) f is independent of value of h 1 (C) f 2~µ 2 -1

54. If a convergent beam of light passes through a

diverging lens, the result (A) may be a convergent beam. (B) may be a divergent beam. (C) may be a parallel beam. (D) must be a parallel be~.

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This sectiqn contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENTl is TRUE and STATEMENT 2 is FALSE. Bubble (D)·lf STATEMENTl is FALSE but STATEMENT 2 is TRUE. 1.

Statement-I:

5.

Statement-I:

A parallel beam of light traveling in air can be displaced laterally by a parallel transparent slab by distance more than the thickness of the plate.

The images formed due to total internal reflections are much brighter than those formed by mirrors or lenses.

Statement-2:

There is no loss of intensity in total internal reflection.

The lateral displacement of light traveling in air increases with rise j.n value of refractive index of slab.

Statement-2: 6.

Statement-I: A bird in air is diving vertically with speed v0 over a

2.

3.

Statement-I: Even in absolutely clear water, a diver cannot see very clearly.

tank filled with water and having flat silvered bottom serving as plane mirror, it observes velocity of its image in silvered bottom of tank as 2v0 upward relative to

Statement-2:

itself.

Velocity of light is reduced in water.

Statement-2:

Statement-I:

Bird and its image in bottom mirror are always equidistant from bottom mirror.

Spherical aberration of a lens can be reduced by blocking the central portion or peripheral portion of the lens.

4.

7.

Statement-1:

Statement-2:

We cannot produce a real image by plane or convex mirrors under any circumstances.

Spherical aberration arises ol1 account of inability of the lens to- focus central and peripheral rays at the same point.

The focal length of a convex mirror is always taken as positive.

Statement-I:

. 8.

For total internal reflection, angle of incident in denser medium must be greater than critical angle for the pair 1 of media in contact.

Statement-2: 1

Statement-I: If a light ray is incident on aily one of the Mo mirrors inclined at 90° with each other, then finally the r emergent ray is antiparallel with incident ray.

Statement-2:

µ = sinC , where the symbols have their standard

meaning.

Statement-2:

Finally, the reflected and initially incident rays are in same phase when successively reflected from two perpendicularly inclined mirrors.

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Advanced JEE Physics 9.

Statement-1: The formula connecting u , v and f for a spherical mirror is valid only for mirrors whose sizes are very

15.

small compared to their radii of curvature.

Statement-2:

Statement-2:

A convex mirror has a much larger field of view than a plane mirror or a concave mirror.

Laws of reflection are strictly valid for plane surfaces, but not.for large spherical surfaces.

10. Statement-I:

The focal length of the mirror is f and distance of the object from the focus is u , the magnification of the mirror is

16. Statement-1: When a ray of light enters glass from air, its frequency decreases.

Statement-2: The velocity of light in glass is less than that in air.

L. u

17. Statement-1: A ray incident along normal to the mirror retraces its

Statement-2: Magnification=

size of image size of object

path.

Statement-2: In reflection, angle of incidence is always equal to angle of reflection.

11. Statement-1:

Although the surfaces of the lens used in goggles are curved, it does not have any power.

12.

18.

Statement-1:

Statement-2: In case of goggles, the lenses are concavo-convex and

A concave mirror of focal length in air is used in a

both the surface of lens have equal radii of curvature.

mirror in medium becomes double.

Statement-1:

The radius of curvature of a mirror is double of the

medium of refractive index 2. Then the focal length of Statement-2:

A convex lens behaves as a concave lens when placed in a medium of refractive index greater than the refractive index of its .material.

focal length. 19.

real image of the object. If both the object and mirror are immersed in water, there is no change in position of the image.

normalr i.e., the convex lens would diverge the rays and behave as collcave.

Statement-2: The formation of image by reflection does not depend on surrounding medium, so there is no change in position of image.

Statement-1: The minimum distance between an object and its real image formed by a convex lens is 2 f . 20.

Statement-2:

Statement-1: Light from an object falls on a concave mirror forming a

Statement-2: Light in that case will travel through the convex Jens from denser to rarer medium. It will bend away from

13.

Statement-1: For observing traffic at our back, we prefer to use a convex mirror.

Statement-1: A convex lens can be convergent in one medium and divergent in other medium.

The distance between an object and its real image is minimum when its magnification is one.

Statement-2: 14. Statement-1:

The lens formula

.!. =.!. _.!, f

V

indicates that focal length

U

of a lens depends on distances of object and image from the lens.

Statement-2: The formula does indicate but when u is changed v also changes, so that /. of a particle lens remains constant.

=

1.142

In denser medium, convex lens is convergent and in rarer medium, convex lens is divergent. 21.

Statement-1: For a prism of refracting angle 60° and refractive index

../2.

minimum deviation is 30° .

Statement-2: At minimum deviation, r1 = r2

= A =30° 2

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Statement-1:

Statement-2:

There exist two angles of incidence for the same magnitude of deviation (except minimum deviation) by a prism kept in air.

Dispersion of light is. the phenomenon of splitting of a beam of white light into its constituent colours .

•

27.

Statement-2:

Convex mirror always form a virtual i:rhage.

For a prism kept in air, a ray is incident on first surface

Statement-2:

and emerges out of second surface (of prism) along the previous emergent ray, then this ray emerges out of first surface along the previous incident ray. This principle is called the Principle of Reversibility of Light.

Focal length of a mirror is half of the radius of curvature.

28.

23. Statement-1: A plane convex lens is silvered from plane surface. It

Statement-1: A fish inside a pond will see a person standing outside taller than he is actually.

can act as a. diverging mirror.

Statement-2:

Statement-2:

Light rays from person converges into eyes of fish on entering water from air.

Focal length of concave mirror is independent of mediuni. 29. 24.

Statement-1:

Statement-1: Maximum distance of image formed by convex mirror from pole of mirror equals 'f' for all the objects (real/virtual).

Statement-2: Convex mirrors forms virtual images for objects placed in front_ of mirror. 25. Statement-1: We cannot produce a real image by plane or convex

Statement-1: Optical fibre has thin glass core coated by glass of small refractive index and is used to send light signals.

Statement-2: All the rays of light entering the fibre are totally reflected even at very small angles of incidence.

30.

Statement-1: The mirror used in search light are parabolic and not concave spherical.

Statement-2:

mirror under any circumstances.

In a concave spherical mirror the image formed is

Statement-2:

always virtual.

Reflection Law is valid for plane mirror as well ,as convex mirror. 26. Statement-1: There is no dispersion of light refracted through a rectangular glass slab. ·

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This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (13), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may h~ve more than one c·orrect options) Comprehension 1 A ray of light is incident at 45° on the face AB of an equilateral prism ABC which has the face AC silvered. Based on the information provided answer the following questions.

.

A

(A)

(C) 4.

\

~

\

\

The refractive index µ of the material of th~ prism so that when the ray falls on face BC (after reflecting from AC) it makes an angle 60° with it is

(A) ,,'3 (C) 2

2.

(B)

.J2

(D) 1.5

The total deviation, when the ray of light finally emerges from BC is (A) 120° (B) 180° (C) 150° (D) 90°

Comprehension 2 A convex lens of focal length 20 cm and a concave lens . of focal length 10 cm are placed 20 .cm apart. In between them an object is placed at distance x from the convex lens. Based on the information provided answer the following questions.

3.

The value of x (in cm) so that images formed by both the lenses coincide is

20,,'3-1

,/3

(D) 10( ,,'3 - 1)

,/3

The linear magnification produced by convex lens and concave lens individually is (A) ( ,,'3 + 1) and

C

(B)

20

\

B

1.

20( ,/3 -1)

1

1

(13)

,,'3 and

(C)

( ,/3 + 1) and ( ,/3 - 1)

(D) ,/3 and (2,/3 -3) Comprehension 3 A telescope is an optical instrument used to increase the visual angle of distant objects such as stars, planets etc. Ah astronomical telescope consists of two converging lenses. The one facing the object is called objective and the lens close to the eye is called an eyepiece. It can be adjusted by displacing relative to the objective. The angular magnification is defined as the ratio of focal length of objective and eyepiece. One can see the image with unstrained eye if it forms at infinity. An astronomical telescope has an objective of focal length 50 cm and a magnification of 20. Based on above information, answer the following questions .

5.

Focal length of the eyepiece is (A) 2.5 cm (B) 5 cm (C) 7.5 cm (D) None of these

6.

To view remote object by an unstrained normal eye, separation between two lenses will be

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(A) 55 cm (C) 60 cm 7.

(B) 57.5 cm (D) 52.5 cm

(D) As frequency of light increases then refractive index of glass increases

If object is at a distance 600 m from the telescope to see

Comprehension 5

the image with unstrained eye separation between two lenses should be (in cm)

The 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 the lens two observer eyes 0 1 and 0 2 are situated at a distance 100 cm from the lens at some distance above and below the principle axis.

(A) 46.65

(B) 47.65.

(C) 49.96

(D) 49.65

Comprehension 4 Speed of light in a medium of refractive index n is

f=15cm

given by ~ where c is speed of light in vacuum refractive n

index of a medium depends on wavelength (;\.). As wavelength increases refractive index decreases. It is also given A.red rel="nofollow"> Aorange > /!.yellow Based on above information, answer the following questions. 8.

9.

In glass (A) orange light travels faster than yellow light (B) yellow light travels faster than orange light (C) yellow light travels faster than red light (D) orange light travels faster than red light The quantity that remains unchanged if light enters from water to glass is (A) Wavelength and colour (B) Refractive index and frequency (C) Frequency and velocity (D) Colour and frequency

10. The phenomenon that happens because of variation of wavelength is (A) Aberration (B) Dispersion (C) Total internal reflection (D) Bending of light 11. Which of the following statement is true? (A) Time taken (t) by yellow light to travel distance X0

in refractive index n can be t::;;; nxo c,

(B) Time taken (t) by yellow light to travel distance · factivem r " " d ex n can b e t ~ nx,x0 .mre c

(C) Since wavelength of yellow light increases in refractive index n its frequency must decreases. (D) None of the above /

12. Which of the following statement is false? (A) Light is a electromagnetic wave (B) Speed of light of each colour is same in vacuum (C) Time to cover distance x0 in a medium is same for

0 20cm

1+----

100 cm

------t-1

Now half position of lens below principle axis is painted black Based on above information, answer the following questions. 13. In initial setup (before painting the lens) which of the following statement is correct. (A) Observer 0 1 will see a real image at 60 cm from the lens but observer 0 2 will not be able to see it (B) Observer 0 2 will see a real image at 60 cm from the lens but observer 0 1 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 0 1 and 0 2 (D) Both the observers may or may not be able to see the image at 60 cm from lens depending on the positions of 0 1 and 0 2 14. After painting the lens, 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) 0 1 (B) 0 2

(C) Both 0 1 and 0 2

(D) Neither 0 1 nor 0 2

15. After painting the lens, for which observer the intensity of image will be reduced to half? (A) For 0 1 (B) For 0 2 (C) Both for 0 1 and 0 2 (D) Neither for 0 1 nor for 0 2 Comprehension 6

The figure shows a ray incident at an angle i =2: . If the 3

plot drawn shows the variation of

Ir - ii

versus

each colour

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where r is the angle of refraction, then based on above information, answer the following questions. Ir-ii

(C)

J, < J.

(D) None of these

20. If angle of incidence is 60°, then focal length of this

marginal ray is 92 -----------------µ, µ,

(A)

e,

R

(B)

R

2

(D) 0

(C) 2R

21. The total deviation suffered by the ray falling on mirror at an angle of incidence equal to 60° is

16. The value of k1 is (A)

(C) 17.

2

Comprehension 8 The XY plane is the boundary between two transparent media. Medium-I with z ';?_ 0 has a refractive

(B) 1

../3 1 ../3

(D) ../3

index ..ff. and medium-II with z:, 0 has a refractive index ../3. A ray of light in medium-I given by A = 6$; + B../3] - lOk is incident on the plane of separation.

2

The value of 0 1 is (A)

(C)

1t

(B)

3 1t

(D)

6

1t

Based on the above facts, answer the.following questions.

2

22. The vector representing the incident ray has a magnitude of (A) 5 units (B) 10 units (C) 15 units (D) 20 units

ZERO

18. The value of k2 is

(A) 1 1

(C)

(B) 2

-2

23. The angle of incidence is

(D) None of these

Comprehension 7 Spherical aberration in spherical mirrors is a defect

24.

which is due to dependence of focal length f on angle of incidence 0 as shown in figure is given by R 2 where R is radius of Curvature of mirror and 0 is the angle of incidence. The rays which are closed to principal axis are

J=R--sec0

are called marginal rays. As a result of above dependence different rays are brought to focus at different points and the image of a point object is not a point. B ,-

_,,,,-1!

(A) 30° (C) 60°

(B) 45° (D) 90°

The angle of refraction is (A) 30° (C) 60'

(B) 45° (D) 90'

25. The refracted ray is represented by the vector given by

(A) _6../3i -8../3 j + 10../3 k

(B)

8../3]

(C) -10../3k (D) 6$; + s../3]- 10../3/;

called paraxial rays and the rays far away from principal axis

26. The vector representing the refracted ray has a magnitude of (A) ,/6 units (B) 10 units

(C) 10,/6 units

~P.'.cri'.'..'.nc~ipt'.'.a~l.'.'.ax~is:'.,ac,''...'_'_~----+:\.P'..co~le (P)

C

27. Based on above information, answer the following questions.

J,

and

J.

(A)

J, =J.

(B)

J, > J.

4 ,

1 •

(A)

5..fi. i+ 5..fi_ I - ..fi. k

(B)

3/ +4]-5k

(C)

5..fi., - 5..fi_ I+ ..fi. k

(D)

2i -3k

represent the focal length of paraxial and

marginal rays respectively, then correct relationship is

(D) 20,/6 units

The unit vector along refracted ray is

3 ,

19. If

(B) 90° (D) Cannot be determined

(A) 60° (C) 30°

3 ,

4 ,

1 •

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Ray Optics Comprehension 9 The lens governing the behaviour of the rays namely rectilinear propagation, laws of reflection and refraction can be summarised in one fundamental law known as Fermat's Principle. According to this principle a ray of light travels from one point to another $UCh that the time taken is at a stationary value (maximum or minimum). If c is the .velocity of light in a vacuum, the velocity in a medium of

refractive index n is

.£., n

hence time taken to travel a

Comprehension 10 Consider an equiconvex lens of radius R , made of a material of refractive index µ. Its focal length is / 1 when any one face is silvered. Now consider another plano-convex lens of radius R, made of same material having focal length, / 2 when no face is silvered, / 3 when plane face is silvered

and / 4 when curved surface is silvered. Based on above information, answer the following questions. 31.

distance I is nl . If the light passes through a number of C

media, the total time taken is ( ~)

L nl or ~ fndl

index varies continuously. Now,

Lnl

is the total optical

28. If refractive index of a slab varies as m = 1 + x 2 where x is measured from one end, then optical path length of a slab of thickness 1 m is

(B)

32.

(C) lm

(C) 33.

3

34.

2R --(2µ-1)

(D)

2R 2(2µ + 1)

equals R 2µ-1 R (D) µ+1

(B)

R µ-1

equals R (A) 2µ

(B)

2R µ

R 2(µ-1)

(D) 2Rµ

f. equals · (A)

R µ

(B)

(C) 2Rµ

B

(B)

f,

(C)

- m

29. The optical path length followed by ray from point A to B, given that laws of reflection are obeyed as shown in figure is A

f,

(A) R

4 (D) None of these

3

R 2(2µ-1)

(C)

path, so that Fermat's Principle states then the path of a ray

4

equals R (A) (2µ-1)

if refractive

is such that the optical path in at a stationary value. This principle is obviously in agreement with the fact that the ray are straight lines in a homogeneous isotropic medium. It is found that it also agrees with the classical laws of reflection and refraction. Based on above information, answer the following questions.

(A) - m

f,

(D)

2R µ

R 2µ

Comprehension 11 SITUATION-I

(A) Maximum (C) Constant

Two identical piano-convex lenses L1 and L2 having radii of curvature R = 20 cm and refractive indices µ, =1.4 and µ 2 =1.5 are placed as shown in the figure .

(B) Minimum (D) None of these

...... ...... ...... ...... ...... ...... ...... ......

30. The optical path length followed by ray from point A to B, given that laws of reflection are obeyed as shown in figure is A

SITUATION-II Now, the second piano-convex lens is shifted vertically

p

(A) Maximum (C) Constant

---+- ---------

B

(B) Minimum (D) None of these

downward by a small distance of 4.5 mm and the extended parts of L1 and L2 are blackened as shown in figure.

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Advanced JEE Physics ---+ ---+. ---+ ---+

~----,----,--

-

~

---+ ---+ ---+ ---+

Principal axis~ ----,--_--- of lens Li

38. The nature of final image of the object when x = 2R is (A) Erect and.magnified (B) Inverted and magnified -(q· Erect.and same size (D)· Inverted and same size

L,

Based on above information, answer the following questions.

39. Jt, is observed that for x = R , . a ray starting from 0 strikes the spherical surface .at grazing incidence. Th_e angle with the normal at which the ray emerges from

the plane S1.!I'face is

35. In SITUATION-I, the position of the image of the parallel beam· of light relative to the common principal axis is

100 cm 3

(A) 1~0 cm

(B)

(C)

200 (D) - c m 9

200 cm 3

36. In SITUATION-II, the new position of the image of the parallel beam is 2 0 ~ cm in front of the lens 2 mm below the (A)

(B)

Comprehension 13 The refractive indices of'the crown glass for blue and red lights are 1.51 and 1.49 respectively and those of the flint glass are 1.77- and ·1.73 respectively. An isosceles prism of angle 6~ is made of crown glass .. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crOwn·glass prism such

that there is no deviation of the incident light. Based on the

principal axis of L, . l~O cm behind the lens 2 mm below the principal

40.

2 0 ~ qn

The refractive.index·of crown glass f9r yellow,colour is .

behind the lens 2.5 mm below the

2 0 ~ cm in front of the lens 2.5 mm below the

41. The refractive index of flint glass for yellow colour is (A) 1.70 (B) 1.72 (C) 1:73 (D) 1.75 42. The refracting angle of flint glase;prisni is (A) +2° (B) +4° . (C) -20 (D) -40

principal axis of L, . Comprehension 12 A small object O is placed in air at the principal axis at a distance x from the pole of the curved .surf~ce of a transparent hemisphere having refractive index 2 ,and radius R as shown. Based on above information, answer th~ following questions. .

n=2

0

43. The net dispersion produced by the combined system is (A) 0,02° (B) , -0.02° (C) +0.04° (D) -0.04° Comprehension 14 An equilateral prism AllC is placed in air with its base side BC lying horizontally along x-axis as shown in figure. A ray of light represented by equation ,J3z + x =10 is incident a:t a point P ·on- the face AB of prism. Based on above information, answer the following questions. z • A

,_ X

--+--R---+<

37. The value of x, for which the final image of the object at O will be virtual is " • (A) 2R (B) 3R

(C)

!

3

(B) 1.49 (D) 1.59

(A) ,1.51 (C) 1.50

principal axis of L,. (b)

0°

(D) 60°

above facts, answer the following questions.

axis of_ L,.

(C)

(B)

(A) 90° (C) . 30°

(D) l.5R

" /

BJLJ-6_0_'---'---'(2__ ,,,.,. X · 0,0,0)

'/

y"/

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44. The value of µ, for which the ray.grazes the face AC is (A) -~ . (B) 4 · . ·2 ".' 3· ·2 .ff, (C) .fj (D) 2

the eyepiece as. the first image? Assume that the first image distanc~ is _d; from the eyepiece. (A) d, < f. ' . (B) d, =J,°

(C) J,
'

(D). d,>2/,

49. Two compound microscopes A ·!ID~. B were compared.

45. The direction of the finally refracted ray for µ = ~ is 2

(A) parallel to x-axis (C) parallel to y-axis ·

(B) parallel to z-axis, (D) parallel to face AB

46. The equation of ray emerging out of prism, if the bottom BC is silvered is (A) z + .fSx =10 \C) z+.fSx ;;20

(B) .f3z+x=10 (D) X + Z =10.fS

Comprehension 15 The schematic diagram of a compound microscope is shown in the adjacent figure. Its main components are two convex lenses: one acts as the main· magnifying lens and iS refer,ed to ~s the objective, and another lens called the eyepiece. The two lenses act independently of each other when bending light rays. ·

' 'Eyepiece .

Both had objectives and eyepieces with the same magnification but 4, gave an overall magnification that was greater than that of B . Which of the following is a possible explanation? ,. . . (A) The distance between object'and eyepiece in A is greater thaii. the cOITesponding distance in B. (B) The dis.lance between object and eyepiece in A is le~s than the corresponding distance in B . (q, The eyepiece anq. objective positions were reversed in A. ' (D) The eyepiece and objective positions were reversed in B. Comprehension 16 A ray of light travelling in air is incident at grazing angle on a long rectangular slab of a transparent medium of thickness t = 1.0 m . The point of incidence O is the origin (0, 0). The.medium has a variable index of refraci;ion µ(y)

1

given by µ(y)=[Ky'1'+1]1/2 where K=1.0(m)"1'. The

Objective

refractive index of air is 1.0. V

0

= focal point of objective J, = focal point of eyepiece / 0

Llght from the object (0) first passes through the objective and enlarged, inverted first image is formed. The eyepiece then magnifies thisimage. Usually the magnification of the eyepiece is fixed (either x 10 or x 15) and three rotating objective lenses are used : x 10, x· 40 and x 60. Angular magnification is defined as the angle subtended by the final image at the eye to the angle subtended by the object placed at least distance of distinct vision (,,, 25 cm) when viewed by the naked eye. Based on above information, answer th'e following questions. 47. The type of image that would have to be produced by the objective is

(A) (B) (C) (D)

"

Either virtua} or real Virtual Real It depends on the focal length of the lens.

48. Where would the first image have to be produced by the objective relative to the eyepiece such that a second, enlarged image would be generated on the same side 6£

Based on the above facts, answer the following questions.

50. The relation between the slope of the trajectory of the ray at the point B( x, y) in the medium and the. angle of incidence (i) at that point is given by (A) tan0.= sini (B) tane = 2sini (C) tane = coti.. (D). , tane = 2coti 51. The equation for the trajectory y(x) of the ray in.the medium is x' (A) y=-

(B)

(C)

(D) y= 256

16 x' y= 16

x'

y=16 · ·x•

52. The co-ordinates ( x, y) •of the point where the ray ' intersects the upper surface of the· slab-air boundary are

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Advanced JEE Physics (A) (1,l)m

(B)

(2, 1) m

(C) (3, 1) m

(D)

(4, 1) m

57. The position at which the image is formed is {A) 30 cm, right oflens (B) 40 cm, left of lens (C) 60 cm, right of lens (D) 70 cm, left of lens

53. The ray finally emerges {A) parallel to the incident ray (B) perpendicular to the incident ray (C) at an angle of 30° to the incident ray (D) ·at an angle of 45° to the incident ray

58. The total number of images arrangement is/ are (B) 2 (A) 1

(D) 6

(C) 4

Comprehension 17 The convex surface of a thin concavo-convex lens 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 horizontal surface as shown in the figure.

generated by the

59. The spacing between the images so formed is (A) 0.1 cm (B) 03 cm

of of of a

(C) 0.5 cm

(D) 1 cm

Comprehension 19

A thin biconvex lens of refractive index

~ 2

is placed on

a horizontal plane mirror as shown in figure.

'' '' ~

r=60cm

The space between the lens and the mirror is then filled with

20cm

----

---

Based on above information, answer the following questions. 54. The focal length of the combination has the magnitude

{A) 1.5 cm (C) 7.5 cm

(B) 15 cm (D) 8.6 cm

water of refractive index

! . It is found that when a point 3

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 again coincide at a distance 25 cm from the lens. Based on the above facts, answer the following questions.

55. The combination behaves like (A) a convex mirror (B) a concave mirror (C) a convex lens (D) a concave lens 56. A small object is placed on the principal axis of the combination, at a distance of 30 cm in front of the

60. The radius of curvature of both the surfaces of the convex lens is R , then the focal length of the convex lens is

(A) R

mirror. The magnification of the image is

(A)

1

3· (B)

3

(D)

(C) 5

(C)

4

1 4

Comprehension 18 A point object O is placed at a distance of 0.3 m from a convex lens (focal length 0.2 m) cut into two halves each of

which is displaced by 0.0005 m as shown in the figure. f =20m _________ 0

fi _ _

t____ _

2 X 0.0005

R 4

(B)

(D)

R 2

R 8

61. When .the space between the lens and mirror is filled with water, the focal length of water concave lens is (A) R (B) 2R (C) 3R (D) 4R

62. The radius of curvature R of each surface of convex lens is (A) 2cm (B) 5 cm (D) 15cm (C) 10cm 63. The focal length of the liquid concave lens is

20 (A) - cm 3 50 (C) -cm

ITT

---------v ' . ·i. ___ t____ ' _ l

3

40 - cm 3 70 (D) - cm (B)

3

l-4--30cm _.,.

Based on above information, answer the following qu'7stions.

64. The refractive index of the liquid is (A) 1.1 (B) 1.2 (C) 1.4 (D) 1.6

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Ray,Optics Comprehension 20 The diagram shows an equilateral prism. Th.e medium on one side of the prism has refractive index µ 1 • ·The

refractive index of the prism is µ =

Ji ·

68. The x -coordinate of the focal point of this lens system is (A)

The diagram shows

variation of magnitude of angle of deviation with respect to µ 1 • Consider the light ray to be incident normally on the first face.

(C)

;;;

·~ P,

6

"0 ;ii

P,

O> C

-,:

0

(C)

(B)

-J3 ../3

(D)

2

(CJ

U1 + I, +d)

k,

(C)

4 3 8

67. Value of

4

../3 8

5

3 10 3

(B)

!1+f,-d

2(/1 + d) U1 + !, -d)

(/1 -d)d

U1 + !, -d)

d

f

Medium-1 h

e/4/

y

Medi~m-2 _

Based <;>n the above facts, answer the following qu.estions. 70. Light travels as a (A) parallel beam in each medium

30'

(D) 90'

Comprehension 21 Two thin convex lenses of focal lengths / 1 and

f, _are

separated by a horizontal distance d ( d < / 1 and d < f,) and their centers are displaced l?y a vertical separation as shown in the.figure. Take the o_rigin of coordinates O at the center

of first lens. For a parallel beam of light coming from the left, as shown in figure. y

---

(D)

b

../3

p1 -P, is

(A) 20' (C) 60°

(B)

a~/

(D)

3

2d(/1+d)- Ji!,

Comprehension 22 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.

X

(B)

f,f, f1+f,-d

fl,

k,

66. Value of k1 is

(A)

(/1 + d)d 2(/1 -d)d

'' ''

65. Value of k, is

6

(D)

U1 + I, -d)

____ 1____

Based on above information, answer the following questions;

(A)

d(/1 -d) !1+f,-d

(A) ----i-- - -,-----------

'' '

(B)

f,+f,-d

69. The y -coordinate of the focal point of this lens system is

C

.Q

d(/1 ~d)+ f,f,

(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 71. The phases of the light wave at c , d , e and f are $,, $, and $1 respectively. It is given that $, (A) $, cannot be equal to $, (B) $, can be equal to $,

* $j

t,

(C) ($,-t1 ) is equal to($,-$,) (D) (t,-$.) isnotequalto

---- ... __ J

($ 1 -$,)

72. Speed of 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

~._.-c+--1-------+-1-----'./J.-• X

0

----d---•

Based on above·inforrnation, answerlhe following.questions.

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Optics & Modern Physics

Advanced JEE Physics Comprehension 23 A thin piano-convex lens of focal length f is split into·

two halves. One of the halves is shifted along the optical axis. The separation between the object and image planes is

77. The distance from itself at which the fish sees the image . of eye by viewing in the mirror is

WaG+~

~

HG+ID

1.8 m. The magnification of the image formed by one of the ·- · half lenses is 2. 78. The distance from itself at which the eye sees the, image

of the fish by directly observing the fish is

(A)

(C)

i+----1.8 m - - - . ,

2H(1+ 21J H(1+ 21J

(B) (D)

2H(½+¾) H(½+¾)

Based on above information, answer the following questions.

79. The distance from itself at which the eye sees the image of the fish by viewing in the mirror is

73. The focal length of the lens used is, (A) 0.4 m (B) 0.6 m (C) 1 m (D) 2m

(A)

74. The separation between the two halves of the thin plano-conveX lens is (B) ,0.4m· (A) 0.2m (D) 0.8m (C) 0.6m

75. The magnification for the second half lens is (A) 0.5 (B) -0.5 (C) 0.4 (D) -2 Comprehension 24 Consider a beaker filled with water (of refractive index

(C)

2(H +~) 2µ H(½+;)

(B) (D)

2

medium is given by the relation, n

2H

above information, answer the following questions.

and permeability of the medium respectively.

significantly different optical behaviour, without violating

t

in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in metci-m";terials.

76. The distance from itself at which the fish will see the image of the eye by direct observation is

(C)

is the speed of electromagnetic waves in vacuum, v its speed in the medium, e, and·µ, are the relative per~ttivity

negative, one most choose the _negative root of n . $uch negative refractive index materials can now be artificially prepared _and. are called· meta-materials. They exhibit

H/2

H(µ_+½) H(~+ 1)

where _c

any physical laws. Since n is negative, it results in a change

H

(A)

=(~)=±~erµ, ,

positive n for the medium. When both e, and µ, are

2H

i

H(l+ :J

In normal materials, both e, and µ, .~e p9_s!tive, imply~g

l

,;(/ E

+~2µ ··

Comprehension 25 Most materials have the refractive index, n > 1 . So, when a light ray from air enters a naturally occurring materiat then by Snell's law, sin 81 = n2 , it is understood sin0 2 n1 that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the

µ ) to a height H . A fish F is at a height H from the transparent base of the beaker which lies on a surface that happens to be a mirror. An observer whose eye E iS at a from the base of beaker is also there. Based on height

H

(B) (D)

~{1+~) ~ (µ+½)

80. Choose the correct statement.

(A) The speed oflight in the meta-material is v =clnl (B) The speed of light in the meta-material is v =l:I

(C) The speed of light in the meta-materials is v =c (D) The wavelength of the light in the meta-material (\,) is given by '-m=A_,lnl,where '-.,

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Each question in this section contains statements given in .two columns, which have to be matched. The statements llJ. COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A--> p, s and t; B --> q and r; C--> p and q; and D --> s and t; then the correct darkening of bubbles will look like the following : P Q r s t

A@@(D@@) B@@@®CD c@@©®CD D@@(D@@ 1.

For a real object, match the magnification situations in COLUMN-I, with their respective matches in COLUMN-II.

r·

COI.UMN-I

CJOLUMNcII

(A) mO (C) (D)

2.

(p) (q) (r) (s) (t)

lml
Plane mirror Convex- mirror Concave mirror Convex.lens Concave lens

Four particles are moving with different velocities in front of stationary plane mirror that lies in the y-z plane. At t = O, velocity of

A

is

v, =i, velocity of B i; v, =-i +3],' velocity of

C is

J

Ve =5i +6], velocity of D is v0 =3i -] . The iic = 2i + J, whereas.the particle B and D move

acceleration of particle A is iiA= 2i + and acceleration of particle C is with uniform velocity. Assume no Collision to take place till t = 2 s, all quailtities·to be in SI units, the relative velocity of image of object A with respect to object A to be denoted by vA"A • If velocity of images relative. to corresponding objects are given in COLUMN-I and their values at t = 2 s are given in COLUMN-II , then match the quantities in COLUMN-I with the corresponding values in COLUMN-II. A D

•

• •B C

•

y

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COLUMN-II

(A)

VA'A

(p)

(B)

VB'B

(q)

(C)

VC'C

(r)

(D) VD'D

3.

21 -61 -121+4] -101

(s) (t)

Perpendicular to the plane of mirror

The COLUMN-I shows some probable directions of velocity of images formed due to system shown in COLUMN-II. Match the quantities of COLUMN-I with the respective possibilities shown in COLUMN-II. COLUMN-I

COLUMN-II

y

(A)

+· +· -+· -+· y

(B)

-

0

(p)

Principle axis

----+~---

Real point object

0

(q)

Principle axis

~ Real point object

y

(C)

(r)

.........0

Principle axis Real point object

y

(D)

(s)

1-

o/

Real point object

0

(t)

.,__ Principle axis ----++---\

4.

Virtual point object

A bird in air is diving vertically over a tank with a speed of 6 cms-1 . The base of the tank is silvered. A fish in the tank is rising upward along the same line with a speed of 8 cms-1 • Taking

µwater

=±, match the quantities in COLUMN-I with 3

their respective values in COLUMN-II.

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Advanced JEE Physics

COLUMN-II

COLUMN-I 1

(A) Speed of the image of fish, in cms- as seen by the bird directly

(B) Speed of the image of fish, in cms-1 formed after reflection from the mirror as seen by the bird (C) Speed of image of bird, in cms-1 relative to the fish looking upwards

(p) 16 (q) 0 (r) 12 (s) 8

(D) Speed of image of bird, in cms-1 relative to the fish looking downwards in the mirror 5.

Match the descriptions in COLUMN-I with corresponding plot(s) in COLUMN-II.

i

COLUMN-I

COLUMN-II

(A) In convex mirror, when object is real and image is virtual

l~I

·

(B) In convex mirror, when object is virtual and image is real. (C) In concave mirror, when object is real and image is virtual. (D) In concave mirror, when object is real or virtual and image is real.

(p)

1+1

l~I (q)

l+I

IJI

...c.-··_--····-1-···- - -1

l~I (r)

IJI

- + - - , L - - - -1

l+I __....-

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-1~1- . (s)

_,,/_IJI

Iii . . . .-

6.

If (µ 1 , i..10 v1 ) and (µ,, i..,, v,) are the refractive indices, wavelengths and speeds of two light waves respectively, then match the entries of COLUMN-I with the entries ofCOLUMN-U.

COLUMN-I

7.

COLUMN-II

(A) µ1 > µ2

(p)

V1

< V2

(B)

µ1 < µ2·

(q)

Vl

> Vz

(C)

µ1

µ2

(r)

A-1

= A.2

(D) µ1 =µ2

(s)

A.1 < A.2

-;t;

Match the descriptions in COLUMN-I to corresponding details in COLUMN-IL

I

COLUMN-I

COLUMN-II

(p) (q) (r) (s)

(A) In refraction from a rarer to a denser medium. (B) In refraction. (C) In reflection from a denser medium. (D) In reflection.

Speed of wave does not change. Wavelength must be decreased. Frequency does not change. The reflected ray suffers an additional path change of

,.

2

· 8.

Light rays are incident on devices which may cause either reflection or refraction or both. The nature of the incident light and the devices are described in COLUMN-I. Some possible results of this on the rays are. given in COLUMN-II.

[

COLUMN-I

COLUMN-II

(p) (q) (r) (s)

(A) A ray of white light passes from an optically denser medium to an optically rarer medium. (B) A parallel beam of monochromatic light passes symmetrically through a glass lens.

Divergent beam Total internal reflection Lateral shift Dispersion

(C) A ray of white light is incident at an angle on a thick glass sheet.

(D) A ray of white light ls incident on one face of an equivalent glass prism.

9.

For a real object, match the descriptions in COLUMN-I to the corresponding details in COLUMN-II. \ COLUMNsI . COLUMN-II (A) Convex mirror (p) Virtual image (B) Concave mirror

(C) Convex lens (D) Concave lens

(q) (r) (s)

Real image Enlarged image D~shed image

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Optics & Modem Physics

Advanced JEE Physics

10. 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.

I

COLUMN-II

COLUMN-I

(p) (q) (r) (s)

~ s

(A)

Real image Virtual image Magnified image Image at infinity

b 7 s

(B)

~

(C)

s

•

I

s

(D)

\ 11. Match the details of COLUMN-I with the respective name and nature described in COLUMN-II. COLUMN-I

/ j (A)

COLUMN-II

(p) (q) (r) (s)

R ft .

Converging Concavo-convex Convexo-cOncave

Diverging

\,,

(B)

(C)

R

2R

µ

R

R

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(D)

R µ 2R

12 Match the following

I

COLUMN-I

COLUMN-II

(A) Concave mirror, virtual object (B) Convex mirror, virtual object (C) Convex lens, real object (D) Concave lens, real object

(p) Real image (q) Virtual image (r) (s)

Magnified image Diminished image

13. For a concave mirror of focal length 20 cm, match the object distances in COLUMN-I to the corresponding details of images formed in COLUMN-II.

I

COLUMN-I

COLUMNcII

(A) 10cm (B) 30cm (C) 40cm (D) 50cm

(p) Magnified, inverted and real (q) Equal size, inverted and real (r) Smaller, inverted and real (si. Magnified, erect and virtual

14. A point object is placed in front of a plane mirror as :shown and moving with velocity 3 ms-1 towards mirror. Mirror is moving with speed 2 ms-1 towards object, then

2ms'--1 ms-'

/

3 ....._.

l

COLUMN-I

(A) Speed of image w.r.t. ground (B) Speed of image w.r.t. mirror (C) Speed of image w.r.t. object (D) Speed of mirror w.r.t. object

COLUMN-II

(p)

10 ms-

(q)

sms-1

(r)

14ms·1 7 ms-1

(s)

I

1

15. A right angled prism of refractive index µ, is placed in a rectangular block of refractive index µ 2 , which is s~ounded by a medium of refractive index µ 3 , as shown in the 8 ,-,.+-rl-----'li;:E==---+.- g figure. A ray of light e enters the rectangular block at normal incidence. Depending upon the relationships between µ 1 , µ 2 and µ 3 , it takes one of the four possible paths h ef, eg, eh, or ei. Match the paths in COLUMN-I with conditions of refractive indices in COLUMN-II and select the correct answer usin the codes ·ven below. COLUMN-I COLUMN-II

(A) e--> f

(p)

µ,>,ff.µ,

(B)

e-->g

(q)

µ2>µ1 and µ2>µ3

(C)

e-->h

(r)

µl =µ2

(D)

e-).

(s)

µ, < µ1 < ,ff.µ, and µ, > µ 3

i

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In this section the answer to each of the question is a four digit integer, ranging from 0 to 9999. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question numbe_r X (say) is 6092, then the correct darkening of bubbles will look like the following :

X.@e@@ (DG)G)G)

®®®· ®®®®

@@)@@

®@®®

~~~~ @@®®

®@e® 1.

distance, in cm, from the concave mirror should a flat mirror be placed for the rays to converge again at the point S having been reflected from the concave mirror and then from the flat one? Will the position of the point where the rays meet change if they are first reflected from the flat mirror? The radius of the concave mirror is 80 cm .

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 30° at a point just inside one end of A . The plane of incidence coincides with the plane of the figure. Find the maximum number of times the ray undergoes reflections (including the first one) before it emerges out. ,...__ _ 2-$ m---.. B

5.

A concave mirror forms on a screen a real image of thrice the linear dimensions of the object. Object and screen are moved until the image is twice the size of the object. If the shift of the object is 6 cm, find the shift of the screen and the focal length of the mirror (both in cm).

6.

A fish is rising up vertically inside a pond with velocity 4 cms-1 and notices ~ bird, which is diving vertically downward and its velodty appears to be 16 cms-1 (to the fish). What is the actual velocity of the diving bird, in cms-1 , if refractive index of water is· 4/3 .

7.

A portion of a straight glass rod of diameter 4 cm and refractive index 1.5 is bent into an arc of radius R cm and a parallel beam of light is incident on it as shown in figure. Find the smallest R, in cm, which permits all the light to pass around the arc.

f i

0.2m

A

2.

Where should an object be placed, in cm, in front of a concave mirror of focal length 30 cm so the i.Irtage size is 5 times the object size?

3.

A rod of length 20 cm is placed along the optical axis of a concave mirror of focal length 30 cm . One end of the rod is at the centre of curvature and the other end lies between F and C . Fin~ the magnitude of the linear magnification of the rod

4.

A point source of light S is placed on the major optical axis of concave mirror at a distance of 60 cm . At what

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Ray Optics 13. A thin· converging lens of focal length f

placed along y-axis such that its optical centre coincides with the origin. A small light source S is placed at (-2; 0.1) m . 1; plane mirror.inclined' at an angle 0,

R

8.

A man of height 2 m is standing on level road where because of temperature variation the refractive index of air is varying as µ =-~1 + ay , where y is height. from

(where tan 0 = 0.3 ) is placed as shown in figure, such that y co-ordinate of final image is 0.3 m . Find the distance d , in metre. Also find the x co-ordinate of final image, in metre. y

•''

road. If a =2xl0.. m-1 • Then find the maximum distance, in km, till which he can see on the road. 9.

A ray of light falls on a glass sphere of refractive index ./3 such that the directions of the incident ray and emergent ray. when produced meet the surface at the same point on the surface. Draw the ray diagram and

''' '' '

''

s

•

:

10. A spherical ball of transparent material has index of refraction µ : A narrow beam of light AB is aimed as shown. What must the index· of refraction be in. order that the light is focussed at the point C on the opposite end of the diameter from where the light entered? Given that x <
-----

11. The pe:rpendh;ular faces of a right isosceles prism are coated with silver. The rays incident at an arbitrary angle on the hypotenuse face emerge from the prism after suffering a deviation of@) degree. Find x .

s, ~0:L

·-----------r

12. A , transparent solid sphere of radius 2 cm · and density p floats in · 10tcm a transparent liquid of ---------, density 2p kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature 8 cm and is silvered to make it concave mirror as shown in the figure. When an object is placed at a distance of 10 cm directly above the centre of the sphere its final 1ima.ge coincides with it. F~d h (as •t-.. /\ shown in figure)fthe height of the liquid surface in the beaker, in cm, from the apex of the bottom. Consider the paraxial rays only. The refractive index of the

14. An object is placed 12 cm to the left of a diverging lens of focal length --
16. The figure shows an arrangement of an equii:onvex lens of focal length 20 cm and a concave mirror of radius of curvature 80 cm . A point object O is placed on the principal axis at a distance 40 cm from the lens Such that the final image is also formed at the position of the object. Find the distance d, in cm . Also draw the ray diagram.·

I

2

,·, '' '' d _ _ _ _.,''

15. Determine the position of the image, in cm, produced by an optical system consisting of a concave mirror with a focal length of 10 cm and a convergent lens with a focal length of 20 cm . The distance from the mirror to the lens is 30 cm and from the lens .to the object 40 is cm. Consider only hvo steps.

R

~

----x

--0-,+.....;--~------~f",

A

I----+--....---""-IC

sphere is

,f

'0

find the value of angle of incidence, in degree.

fX

:= 1.5 m is

and that of the liquid is

± . 3

I,

I,

1+-----100 cm----+< 1-4------

40 cm ---+1+- 30 cm-+1

1+----aocm----+1 17. A converging beam of rays is incident on a diverging lens. After passing through the lens the rays intersect at a point 15 cm from the lens. If the lens is removed, the point where the rays meet, move 5 cm closer to the

==================================== = 1.161

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Advanced JEE Physics

Optics & Modern Physics

mounting that holds the lens. Find the focal length of the lens, in cm. 18. A lens with a focal length of

f =30 cm

produces on a

22. A lens with a focal length of 16 cm produces a sharp image of an object in two positions which are 60 cm apart. Find the distance, in cm, from the object to the screen.

screen a sharp image of an object that is at a distance of _ a = 40 cm from the lens. A plane-parallel glass plate having µ = 1.8 and a thickness of d = 9 cm is placed between the lens and the object perpendicular to the optical axis of the lens. Through what distance, in cm, should the screen be shifted for the image of the object to remain distinct?

23. An intense beam parallel to the principal axis is incident on a convex lens. Multiple extra images F1 , F2 , . . . are

formed due to feeble internal reflections, called flare spots as shown in the figure. The rardii of curvature of the lens are 30 cm and 60 cm and the refractive index is 1.5. Find the position of the first flare spot, in cm.

19. The height of a candle flame is 5 cm. A lens produces an image of this flame 15 cm high on a screen. Without touching the lens, the candle is moved over a distance of £"' 1.5 cm away from the lens and a sharp image of the flame 10 cm high is obtained again after shifting the screen. Calculate the focal length of the lens, in cm. 20. The focal length of a convex lens in air is 10 cm . Find

Principal

axis

its focal length, in cm, in water. Given that µ 8 = ~ and

24. A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm . Where should a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses. Give your answer in cm.

21. A converging beam of rays passes through a round aperture in a screen as shown in figure. The apex of the beam A is at a distance of 15 cm from the screen. How will the distance from the focus of the rays to the screen change, in cm~ if a convergent lens is inserted in the aperture with a focal length of 30 cm ? Plot the path of the rays after the lens is~itted.

25. One side of radius of curvature R, =120 cm of a

2

,'''

------------,'' _i''______ _

''',''

A

--L--.

''

'' r 1 5 c m -'

convex lens of material of refractive index µ =1.5 and focal length f, =40 cm is silvered. It is placed on a horizontal surface with silvered surface in contact with it. Another convex lens of focal length / 2 = 20 cm is fixed coaxial d = 10 cm above the first lens. A luminous point object O on the axis gives rise to an image coincident with it. Find its height, in cm, above the upper lens.

26. A source of light is located from a convergent lens of focal length f = 30 cm at a distance double the focal length of the convergent lens. At what distance from the lens should a flat mirror be. placed so that the rays reflected from the mirror are parallel after passing through the lens for the second time? Give your answer incm.

= ==================================== 1.162

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...

•

BASED;ONREFLECTJON AT PLANE SURFACES

ICE I ?: :;,," ,, J

.1.

30°

9.

3d

2.

4cm

10.

(a)

12 cmx8 cm

(b)

12cmx4cm

3.

4.

ucoso.(tann- tan8) g

(a)

60°

(b)

240°CCW

s.

60°

6.

100° (CW)

7.

20cm, 60cm, 80cm, 100cm and 140cm

12 --------------·-·-8.

•

j

---

ICEII

12.

-5(1+-/a)l +5]

13.

a=20

14.

30°

15.

50°

16.

• ') 1 ("i+j+k --

-/3

-- --------- -- -·· -~- -------------------'

BASED ON REFLECTIONA(CURVED SURFACES>\

0

1.

5 cm inverted

2~

Concave, 6.67 m

4.

7.5 cm, 12.5 cm

5.

(a)

8cm

(b)

16cm

(c)_

48cm

6.

15 cm

7.

(a)

2+cosrot) ( 1+cosrot 1

(b)

X =0

(c)

m->ro

8.

(a)

Concave .

9.

1 ( ../3 + )R from the convex mirror 2

11.

10cm

====================================1.163

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Optics & Modern P_hysics BASED ON GENERAi! REF.RAC'FION

1.

75 cm

7.

6.6cm

2.

7.5-cm

8.

2.88m

3.

.6.V=0.55 cm

9.

1 -cm

4.

2 x 10•

5.

3.5cm

6.

!h+d

I

l"fl2 z1.1

m,

ms-• , 4000 A yellow

.f3

_ 11. 0.7r

12.

sin-•(~, (~µ'R' - L' -,!R' ~ L' ))

3

BASED ON ir0FAI! INiliERN.A REF.l!EC'TION (iTIR) 1.

(a)_

54.34°

(b)

Yes

8.

i(A- -:) (a)

2,81 m

(b)

0:23,m

4.

(b)

-J2

6.

(a)

26.8°

(b)

Yes

3.

7.

• 1.

2.

5.

6:

=

40.54°

(b)

26.6°

9.

67.3°

10.

- _cm

11.

-J2

4 3

12. ·OPi~ 13.

h

2

~µ'-f

BASED ON li!RISM (a)

30°'

7.

sin-'(µsina).-a

(b)

l

8.

-J2

9.

0''

11.

6v-6R =4.59,

12.

22Q

A=<:t+~-6

. µ=sinp 3.

(a)

. 1+'( . _st" , smf}sm

)+cot(a+P-6))'-

e1+.fi-6,

.

I

56°

157.2°

13.

0'

(b)

128.4°

14.

19°

15.

10.1 °

16.

6"8
17.

2,.10

-./5

(b)

58.8°

(a)

34.2'

(b)

8.4°

•

.f3

(a)

(a)

0

1.164c::::=================================

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Ray Optics BASED ON REFRACTION AT .CURVED SURFAC!=S 1.

2.5 D

2.

Final image is formed at pole of the mirror

3.

x z 0.75R

.' 4.

5.

(a)

80 cm

(b)

U<12cm

6.

Final image is formed at 65 cm from first face on the same side of the object.

7.

3.33 cm, lnfi~ity

7.42 cm

I

BASED. ON LENS FORMU!cfilj 1.

6 cm from either of the object

14.

7.5cm

. 2.

12 cm

17.

(m + 1) times smaller

'3.

4 3

18.

(a)

19.

2.4cm

20.

(a)

21.

Rays will become parallel to the optic axis.

22.

1=3-2µ0

4.

Concave mirror of focal length 15 cm

. 5.

. 6.

(a)

90cm

(b)

102cm

12.5 cm in front of the silvered lens

7.

10cm

8.

2.14 cm

9.

1.37

23. 24.

1.4

2a R

2(µn+µ-1) (a)

x,

(b)

~

(c)

_x_,_

7.5cm

. 10.

4i'

11.

t

12.

2m, 1 m

13.

(a)

t t-i

(~

No Shift

•

Convex

X1 -X 2

X1 -X 2

ANSWERS

,

L__

SINGLE CORRECT CHOICE TYPE QUESTIONS.I

1.

C

2.

C

3.

A

4.

D

5.

B

6.

C

7.

C

8.

D

9.

D

10.

C

11.

B

12.

B

13.

A

14.

C

15.

B

16.

B

17.

B

18.

D

19.

C

20.

C

21.

C

22.

D

23.

D

24.

D

25.

B

26.

A

31.

B

27.

--- - - -------

- - - ---- - - -

-

32.

B

-----C

-

28.

C

33.

D

.

-

---

-

29.

D

30.

A

34.

D

35.

B

36.

B

37.

C

38.

B

39.

A

40.

B

41.

C

42.

D

43.

B

44.

D

45.

C

----------------

------

---

- -- -- -

46.

C

47.

B

48.

A

49.

C

50.

A

51.

A

52.

C

53.

D

54.

C

55.

D

56.

B

57.

B

58.

A

59.

B

60.

D

..

I

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Optics & Modern Physics

Advanced JEE Physics

I

61.

A

62.

B

63.

C

64.

D

65.

A

66.

D

67.

C

68.

A

69.

B

70.

C

71.

C

72.

C

73.

A

74.

A

75.

B

76.

C

n.

A

78.

C

79.

A,

80.

C

81.

C

82.

C

83.

D

84.

A

85.

B

86.

B

87.

B

88.

A

89.

C

90.

C

91.

D

92.

A

93.

C

94.

C

95.

A

96.

C

97.

B

98.

B

99.

C

100.

D

101.

C

102.

C

103,

B

104.

B

105.

C

106.

C

107.

C

108.

B

·109.

D

110.

B

111.

B

112.

A

113.

D

114.

C

115.

C,

116.

C

117.

B

118.

D

1.19.

C

120.

C

121.

C

122.

B

123.

C

124.

B

125.

A

126.

A

127.

C

128.

B

129.

D

130.

C

131.

C

132.

D

133.

B

134.

C

135.

A

136.

C

137.

C

138.

D

139.

C

140.

C

141.

B

142.

A

143.

D

144.

C

145.

C

146.

A

147.

D

148.

D

149.

A

150.

C

151.

B

152,.

C

153.

C

154.

C

155.

D

156. ·A

157.

C

158.

D

159.

D

160.

B

161.

c·

162.

D

163.

D

164.

C

165.

B

166.

C

167.

A

168.

B

169.

D

170.

A

174.

C

175.

A

C

171.

A

172.

D

173.

D

-116.

D

1n.

C

178.

B

179.

B

180.

181.

B

182.

C

183.

C

184.

C

185.

C

186.

C

187.

A

188.

B

189.

A

190.

B

191.

B

192:

A

193.

A

194.

D

195.

B

196.

B

197.

D

' 198.

D

199.

A

200.

D

201.

D

202.

C

,203.

D

204.

C

205.

C

206.

B

207.

D

208.

B

209.

A

210.

B

211.

C

212.

C

213.

C

214.

C

215.

C

216.

C

217.

C

218.

A

219.

B

220.

A

221.

B

222.

D

223.

A

224.

A

225.

B

226.

D

227.

A

228.

D

229.

B

230.

B

231.

B

232.

C

233,

B

234.

B

235.

B

236.

C

237.

D

238.

D

239.

D

240.

B

241.

B

242.

C

243.

A

·244:

A

245.

D

246.

C

247.

B

248.

D.

249.

B

250.

B

251.

C

252. r,D

253.

A

254.

D

255.

B

256.

C

257.

B

258.

D

259.

A

260.

A

261.

A·

262.

B

263.

D

264.

C

265.

A

266.

B

267.

D

268.

B

269.

D

270.

D

271.

A

272.

C·

273,

C

274.

A

275.

D

276.

B

2n.

C

278.

B

279.

A

280.

C

2111.

C

282.

A

283;

A

284.

A

285.

D

=1.166

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·I J

J

I

I ·1

Rahul Sardana with www.puucho.com

Ray Optics 286.

A

287.

C

288.

D

289.

B

290.

A

291.

C

292.

D

293.

A

294.

C

295.

D

296.

B

297.

C

· 298.

D

l~h:a ·iiiM~iJ:1i1PiilE[c0RREc;r,cHR1cE~Ti~E~au]~~10Nsl

lfNSWERS 1.

BC

2.

A ACD

6.

AC

7.

11.

BC

12.

16.

BD

21. 26.

3.

AD

4.

AB BCD

8.

AC

9.

AD

13.

BC

14.

· 17.

ACD

18.

AD

19.

BCD

22.

ABC

23.

AB

24.

AD

27.

BC

28.

BD

29.

5.

BD

10.

A

ABCD

15.

ABCD

BD

20.

AC

BC

25.

AD

CD

30.

AC B

31.

BD

32.

ACD

33.

AC

34.

BC

35.

36.

B

37.

AD

38.

AD

39.

AC

40.

CD

41.

AC

.42.

AC

43.

C

44.

AB

45.

BC

46.

ABC

47.

BC

48.

ABC

49.

CD

50.

BC

51.

AC

52.

BD

53.

BC

54.

ABC

(ANSWERS

:=J

a

ljlEASQ.l;llNG!BASElil!QUESTl0NS$

I

1.

D

2.

B

3.

A

4.

B

5.

A

6.

C

7.

D

8.

B

9.

C

10.

D

11.

A

12.

A

13.

D

14.

D

c·15.

A

16.

B

17.

A

18.

D

19.

A

20.

C

21.

B

22.

A

23.

B

24.

D

25.

D

26.

B

27.

D

28.

C

29.

C

30.

C

1.

B

2.

C

3.

B

4.

D

5.

A

6.

D

7.

C

8.

A

9.

D

10.

B

11.

D

12.

C

13.

D

14.

A

15.

D

16.

D

17.

C

18.

A

19.

C

20.

D

21.

A

22.

D

23.

C

24.

B

25.

D

A

28.

A

29.

B

30.

A

26.

C

27.

31.

C

32.

C

33.

B

34.

D

35.

D

36.

C

37.

C

38.

D

39.

C

40.

C

41.

D

42.

D

43.

D

44.

C

45.

B

46.

B

47.

C

48.

A

49.

A

50.

C

51.

D

52.

D

53.

A

54.

C

55.

B

56.

A

57.

C

58.

B

59.

B

60.

A

61.

C

62.

C

63.

C

64.

D

65.

B

66.

C

67.

B

68.

A

69.

D

70.

A

-1 1.167=

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Optics & Modern Physics

Advanced JEE Physics 71.

C

72.

B

73.

A

74.

76.

A

77.

B

78.

C

79.

81.

C

C ---

"]

--------

_ 75._D - - - - - ·

80.

D

B

' - - - - - ' - - - - - - - - - - - - - - - - - - ---··---·-··

I-

l.

~:

I

~~-~- r, s, I)

--- 2.

~:: :\ - - --- -- 3 ' ·

: :

~== ~~~

q, r) -- ---·

~

~==-~~)-·

5

·

~:%

q)

-

C-> (q: r, s, t) C _, (s, I) C-> (q, r, s) C-> (p) C-> (s) D-> (p, r, s) . D-> (t) D-> (p) D-> (p, q) • ' - - - - - , . - - - - - ~ _ . , D ->_(g,J) - - - - - ·-·-----6. A-> (p, s) 7. A-> (cj, r) 8. A-> (p, q, s) 9. A-> (p, s) 10. A-> (p, q, r, s) B-> (q) B-> (r) B-> (p, q) B-> (p, q, r, s) B-> (q) C -,· (p, q, s) C-> (p, r, s) C-> (r) C-> (p, q, r, s) C-> (p, q, r, s) r---',-D +(r). _______ D~(p, r) _________ ., D-> (p, q, s~)_ _ _ _ _D_-> (p, s~)_ _ _ _ _ _D_->~(p~,~q,~r~,s) 11. A->(P) 12. A->(p,s). 13. A->(S) 14c A->(S) 15. A->(q) -·] 1 ! B-> (p) B-> (p, q, r, s) B ->.(p) B-> (q) B-> (r) ! C-> (r, s) C-> (p, q, r, s) C-> (q) C-> (p) C-> (s) ~i'--~D~->~(g,_p) D -'>_(g,_s) D-> (r) . .P ->_(g)_ D ±(R).-'-

•' '"~IN'TEGER ANSW~R' TY~E,QUESTION$'· 0

-. 24, 36 3. 3 4. 90 2. I 1. 30.' . . 8. 2 6. 7. 12 5. 36 9 11. 180 12. 15 10. 2 60 I9,------------------- - 15. - 5- - 16. 30 13. 5, 4 14. 8 19_._9_ _ _ _ _ _ _ _ _2_0._:..~? _______7__,· 18. 60 111. 30 · _____ , _____ _J

21.

! 25.

5

22.

100

10

26.

45

1.168

23.

12

24.

25

--------

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1.

The angle between the incident ray and the reflected ray is 180-20, so, we have

i+.,/3]

i-.JJf

'

-2-

=>

:::::)

-2-

:'

X=4cm

So point of incidence of light from A should be at 4 cm from D on mirror.

'

''' '' e'e

3.

The image will be momentarily at rest when. the particle moves parallel to the mirror. Let at the time t the particle has a velocity v parallel to the mirror.

\180~-28

'' cos(18O°-20) =

( i+/31) 2

•

' ''. V

(i--.13]) 2

1i+;13j11i-;311 =>

(1-3) -cos(20) = - 4-

1

=>

-cos(20) = _ _! 2 1 cos(20) =

:::::) :::::)

20=60° 0=30°

=>

ucosa

vsin0=usina-gt and

vcos0 =ucosa

:::::,.

V=--

•.• (2)

COS0

From (1) and (2)

2

= usina-gt ( ucosa)sin0 case . =>

2.

ucosa

... (1)

1

Drawing the ray diagram and using the Law of Reflection, we get

B

= ucosa(tana-tanB)

g

4.

20cm

A .semi i

'

i=r :::::,. sini =sinr So we can say that ~DO and .6.OBC are similar

... (1)

From the figure, we observe 'that 30=180°

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Optics & Modern Physics

Advanced JEE Physics =>

-..--------" 3" ,,

0=60°

,, I

So, S, =180°-2(30°)=120° (CCW) I

ands, =180°-2(30°)=120° (CCW)

I

=>

S = 240° (CCW)

I

I

S = 180° + 8 = 240° (CCW) or 120° (CW)

I

\

1

\

I

I

I

\

2.' 1

I

I

\

"\

\\

,

0+0+0=180°

3 ,,

,•

\

I

'

\

\

,, J

I 1

\

'

•

5"

,,"1' ...... _-..!_..- \.

I~ 6Q0 1\ \

/

\

I

\_.--1'-_--...f ,

V

I',

I

I

\

\ \

\2' \ \ \

\

-----1---) ,-r•.-,,-::r,,c :, ,

r

I

\,: I \ J I r,5 1 , r3 1 I \. I' 5, / \ I' I 'f-. ... _._.--{ )-. ... _. __ -( I 1

1

0=60°

\

\

\

1',, ',0.,

\J

~

\

'

Bi\

t

\

\

\,4" \

\';).

}.----.._

I

\

we observe that

I

\

~---+---7 ----~---~ I \ \ 1"

/___ Tr----

Various angles made are as shown in figure. In triangle ABC,

=>

I

),,. - - - ... ../ ,, \ /'

I

Alternatively from the figure, we observe that

5.

2",f

I

I

So, total deviation 6 =61 + 62

\

I

,

,,

,,

1

4 ' .} _______ 4 ' ' \/ _______

1

So, the number of images formed by the combination is given by

N=6-1=-5

Combination of Mirrors

6.

From the figure, we observe that . S, =180°-2(50°)=100° (CCW)

Images Formed

AB & BC

1,2,3,4,5

AC & BC

1', 2', 3', 4', 5'

AB & AC

1", 2", 3", 4", 5"

These images along with the obJect must lie on a circle as shown in figure with an angular separation of

s, =180°-2(20°)=140° (CW)

360° = 360° = 720 N 5

s, =180°-2(10°)=160° (CW)

Similarly, the other two combination of mirrors also form 5 images each but we find from symmetry that 5 and 5' , 1 and 5" , 1 and 1" coincide. So the total number of images formed by three mirrors AB , BC and AC is

N'=(5)(3)-3=12 9.

• M,

The ray diagram is shown in figure. We observe that

Hl=AB=d

So, total deviation S = 100°(CCW) + 140°(CW)+ 160°(CW) => S = 100° (CW) or 260° (CCW)

DS=CD=_cl_ 2

G 7.

The image is formed as far behind the mirror as the object is in front of it. Also in:iage formed by mirror 1 i.e., 11 , acts as object for mirror 2, so 1; is formed 50 cm behind the mirror 2 as

H

shown.

I',

I", 20

I, 60

10

10

I'

I,

0 30

30

20

'

I"

60

''

_________ .., __________ I

'

B

:E

'

F J

2 Taking all distances to be in cm and plotting them as shown (but not to scale), we get 011 = 20 cm 012 = 60 cm

01;

= BO cm

O1;'=100cm

=>

GH=2CD=2(%)=d

Similarly, IJ =d

01; = BO cm O1;=140cm

So, the respective distances are 20 cm , 60 cm , 80 cm , 100 cm and 140 cm 8.

Also, AH = 2AD

10.

=>

GJ=GH+HI+IJ=d+d+d=3d

(a)

For a one eyed man, the required size will be half the each dimension of the face i.e., 12 cm x 8 cm

(b)

For a two eyed man, the

Let us first consider the mirrors AB and BC , for which we have

360° = 6 60°

Smallest length of the mirror

=>

= Half the length of face

(Smallest l~ngth) = _!_ x 24 = 12 cm ' of-the Mirror 2

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Ray Optics The smallest breadth of the mirror is calculated by using the fact that the rays from extreme part of face should reach one of the eyes after reflection from the mirror. The common overlapping portion is then the required breadth of the mirror. The ray

=>

13.

diagram is shown in figure.

v,=-5(1+.JsH+s]

Let AB be the incident ray and angle of incidence at the mirror M1 be i, then

p ----------------- ..."""P' M ,...,,..........

--

--

:'

A

'' '' '

'' '''

'-.,

I

----------------- '.:JQ'

From figure, we get MM'= .!.pa __!_E,E 2 2 ' Smallest Breadth)= _!_( 16 _ 8) = 4 cm of the Mirror 2 So, the shortest size of mirror is 12 cm by 4 cm. 11.

=> =>

=>

the image of B2 in mirror ac and B4 is the image of 8 3 in point of intersection of ab with line AB 4

•

•

ac with 8 2

,

~

Point C is the

14.

E with 8 1 and F with 8 .

B4,,

:' ', ............ : .............

!

LDCB=20-2i LCDB=180°-20 cr.=20

The angle between incident and emergent ray is 28 and it is independent of the angle of incidence i .

Let us now draw line

B 3 C from 8 3 and connect point D at which this line intersects

0

Using the Laws of Reflection, we get

Let us first find the image of point 8 in mirror bd (shown in figure). Let us then construct image B1 in mirror cd. Also, 8 3 is

mirror ab . Let us connect points A and 8 4

LCBO = go -i LBCO = 180° -0 -(goo -il LBCO = go -0+i 0

(

Suppose that a plane mirror is kept horizontal as shown in figure. The reflected ray will make an angle of 30° with horizontal, or an angle of 60° with the vertical.

Incident ray

. . . . . . . . ... ..,.,,

I

I

83

-----

,.,.,.

I

:

,,...

'

F

....... _,,"

_,,....

C

I

-

, __ -

---·B ,.,.1 1

,,/'

1 l,-'_,, - - - - - - - - - - - - - - - - - - - - - - - - - - -

··-

-- -

To make the reflected ray to go vertically upwards, the mirror is required to be rotated about O counterclockwise by 60°. To achieve this, therefore, the plane mirror is required to rotate about O by half the angle, i.e., by 30° , as shown in figure.

J 1

-- ...

0

I

--E',-d--l ..,

Reflected ray

~i/

. As b' lL-"-----------~~',C q_... I I

''' :

~ B2

It can be stated that the line ACDEFB is the sought path of the

Reflected ray

beam. Further, we observe that since, B3 CB4 is an isosceles triangle, CD is the reflection of beam AC . Similarly, we can show that DE is the reflection of CD and so on. This solution of the problem is not unique, as the beam will not necessarily always be sent initially to mirror ab .

Incident ray

12. Along x-direction i.e., perpendicular to the mirror, we have 30'

Relative Velocity of)= -(Relative Velocity of) ( Image w.r.t. mirror Object w.r.t. mirror

=>

V1-Vm=-(vo-vm)

=>

v, -(-5cos30°)=-(10cos60°-(-5cos30°))

=>

v, =-5(1+./3) ms-'

15.

In the direction parallel to the surface of mirror, i.e., along y-direction we have V1 =Vo

::::)

v,

Ray AB is incident on mirror OP at an angle e. The reflected ray BC is incident on second mirror OQ. Finally, the reflected ray CD is parallel to OP. Since CD and OP are parallel, and CO cuts them,

=> =>

LQCD = LCOP = 70° LDCN = gQ 0 - LQCD = go 0 LNCB=LDCN=20°

-

70° = 20'

= 10sin(60°) = 5 ms-1

So, velocity of the image

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Optics & Modern Physics

Advanced JEE Physics Q

16.

Reflection of a ray of light is just like an elastic collision of a ball with· a horizontal ground. The component of the incident ray

along the inside normal gets reversed while the component perpendicular to it remain!=i unchanged. So, the component of incident ray vector A= i +} -k parallel to normal, i.e., i + J

A

gets reversed while perpendicular to it, i.e.,

-k

remains

unchanged. So, the reflected ray is written as,

R=-1-]-k A unit vector along the reflected ray will be,

>l.!::::3.:::==::::';;:== p B

Further, LOCB = 90° - LNCB = 90° - 20° = 70° Now, in .6.COB , we ~ave LCBO = 180°-(LCOB+LOCB) = 180'-(70'+ 70°) = 40° => 0 = LNBC = 90' - LCBO = 90° - 40'= 50°

• Fi

-1-]-k

' =Fi =_Ja__,a_

=>

f=- ~(i+j+i<)

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1.

f . S1nce, m=-f-u Now, f=-f and U=-1.5f,so

:::::i-

~

:::::,.

-f m=-0.5 f

-f m=---f+1.5f m=-2

So, m=-~=-1 u

4.

Since m=~=-2 h, :::::,.

h; =-2h0 =-5 cm

f . Smce m=-f-u

inverted.

Given : u = -25 m , m = +4 (since the image is erect). Now, the magnification is given by

=>

.-4=_-=!Q_

-10-u u=-12.5cm Please note that here, lul > !ti and we know that in case of a concave mirror, image is real when object lies beyond F . Case 2 (When image is virtual) : So, m=+4 1 Since m=-1-u

V

m=--

=>

u

=> v=-mu=-(+4)x(-2.5)=10 m Using the mirror formula, we get 1 1 1 -+--=10 -2.5 f

f1 = 0.1-0.4 = -0.3 1 I= ---=-_1_()_ m

=>

0.3 3 Since f is negative so, the mirror is concave. The radius of curvature of the mirror is given by

· ( 310)

=:>

3

According to the mirror formula, we have

5.

.!.+.!.=! V

U

f

M,

"'~"':~~4==::~-~=='j,: 1::: -------------~r-- !oscm

optic-;;;;~1;:,,~-,\

~

!

M, i-.--------

50 cm

-------Joi

4=_-=!Q_

-10-u U=-7.5cm

Please, note that here, lul
=- 20 m=-6.67m

R=2f=2 -

3.

Given, f = -1 O cm . Since a concave mirror can fomi real as well as virtual image and since the nature of image is not given in the question. So we will consider two possible cases. Case 1 (when Image is real) : So, m = --4

The image is 5 cm long. The minus sign shows that it is 2.

---=V 50 -25 V=-50cm

Since f=·~=-12 cm Let the object be placed at a distance u from the pole. Since, we know that magnification m is given by f m=-f-u So, now we apply this formula to these situations one by one. a) Here, we have m = +3

=> => => =>

3=~ -12-u --4=-12-u 12+U=4 u=-8cm

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Advanced JEE Physics

Optics & Modern Physics

Here, m = -3

The following steps of construction for drawing the ray diagrams are used.

b)

-3 = ~ -12-u 4=-12-u

=>

=> c)

Here, m=-

=> 6.

(i)

=>

From I or O drop a perpendicular on principal axis,

such that CJ= CD or OC =CD.

U:=c-16 cm

1

(ii)

Draw a line joining D and O or D and I so that it. meets the principal axis at P . The point P will be the pole of the mirror as a ray reflected from the pole is always symmetrical about principal axis.

(iii)

From O draw a line parallel to principal axis towards the mirror so that it meets the mirror at M . Join M to I , so that it intersects the principal axis at F . F is the focus of the mirror as any ray parallel to principal axis after reflection from the mirror intersects the principal axis at the focus.

3 =>

36=-12-u

LI=-4Bcm

Distance of image formed by the plane mirror is (b-a) i.e., (b - 5) cm and distance of object from mirror is (b+a) i.e., (b + 5) cm . Using mirror formula,

1

..!. + ..!. =! V

U

f

we get

9.

1

(b-5)-(b+S) = 20

Let the point A be at a distance x from the convex mirror as shown in mirror, then assuming the origin to be placed at the pole of convex mirror, we get

Solving this equation, we get

1

b=15cm

,r'- :.-:;,

The coincidence of the images can be established by observing the changes in the relative position of the images when the eye is moved away from the optical axis of the mirror.

When the images are at various distance from the eye the images will be displaced with respect to each other. When the images are at the same distance, they will coincide irrespective of the placement of the eye.

7.

(a)

[ :

,

'' '' ' 1-4--2R ------+1'

At any instant t, .we have

2

~ l :

t

:

I

,

• Incident Ray

Incident ., , Ray ,

A

'

"4----X---+-: ,

----2R---~I

' .'

i+-'-1,_-x_.......v,-t,,t'

u = -(2f + x) = -(2f + f cos rot)

. hem1rror . formua, I -+-=-,we 1 1 1 get Usmg,t V U f -1

For convex mirror, we have

1

1

2

v 2f+fcoscot ::::,

V=-(2+coscot)t 1+coscot i.e., distance of image from mirror at time any instant t is

For concave mirror, we have 1 1

2 + cos oot)t ( 1+coscot Ball coincides with its image at centre of curvature, i.e., at

(b)

-(2R- x) -( 2R+ ~ )

2x+R Solving this equation, we get

X=O (c)

At t =

T

2

x=(1\F3)R and X=-( v'3/ 1)R

, we have

Ignoring the negative value, as we have already used a negative sign with x, so the object should be placed at a distance

rot= re =>

x = fcos(x) = -f

x = ( Js +

So, u = -f i.e., ball is at focus. So, its image is formed at infinity, so m -), co

8.

(a)

Since the image is on the opposite .side of the principal axis, the mirror is concave. Because convex mirror·always forms a virtual and erect image. The ray diagrams for two different cases are shown in figure.

(b)

M,' A p,'

0

-- -... --

0

2

10.

1

'M

A

:P

C'

D

---

,-' I Case2

'' ' ''

1

1

1 1 f U

u-f

-=---=-V

B

1

-v-u=-1

'

' ''

1)R from the convex mirror.

Object is placed beyond C. Hence, the image will be real and it will lie between C and F. Further u , v and f all are negative, hence the mirror formula becomes

1

''t----

Case 1

=

D, ,-r

''

'' '

2 R

Uf

f V=--f • 1-u

B

Now since, uAs > uEo

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D' A' -c---,--r---F~---mp

B' E' ...,....__ VMJ -----+I

14----Vm----+l

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Ray Optics

/·: m=-t) Therefore, shape of the image will be as shown in figure. Also note that v AB < uAB and .VEo < u~0 , 11.

lmABI < 1 and

=>

So, magnification, m

fmE 0 ] < 1

Length of Image Length of Object

_51_(-21)

3'

Since the image is inverted, so the hlirror is concave.

=>

Now, u =-30 cm

5f ~(-21) 2 3 Case JI : When the other end lies beyond C

1

=>

v u f

=,

1

Since .:!.+.:!.=!,soweget V U f

1 1 1 -+--=v (-i) -f

Since image touches the rod, the rod ·must be placed with one

end at centre of curvature. However, two cases arise here. Case I : When the other end lies between C and F

=>

71

V=--

4

For A, we have

U=-(2t-¾)=-~ f=-f

So, magnification, m = vA -vc UA -Uc

Since, .!.+.!=!,so we get 1

U

1

_

f =-f

1

(-15) + (-30) f=-10 cm

V

~2;__ _

U=-(2t+¾)=-~f

. 1 1· 1 Smee-+-==> =>

VA -Ve

For A, we have

v=~=-15 cm 2

1

m

UA - Uc

V

m=-2=-u

12.

51 2

V=-

f

1

A'

F

p

=>

m

-~-(-21) 4 _ 71 -(-21)

3

=-4

3

-+---v (-:')- (-0

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1.

· Magnification, m2 = ...:.'!...;,,. 4).55_ .. u .· J

d"=-d-= 100 =75 cm nm,attve 4/3

So, Liv;,, 0.55 cm

1 2.

The incident rays will pass undeviclted through_ the water surface and strike the mirror parallel -to its principal axis. Therefore for the mirror, object is at oo . Its image A (in figure) will be formed

at focus which is ·20 cm from tl]e mirror. Now for·the interlace

m2 i::l1.1 m1

4•

between water and air, d =10 cm .

.

The refractive index of glass, V C µ·=~=..;_ vglass

:::) Air 10 crh

'' ' ''

-...-:~

R =40 cm d'

l

The colour remains yellow, as the colour depends on the frequency and not on the wavelength.

10 ·7 5 =(~:)=(4;3f . cm

l)sing equation, the total apparent shift is

AX =h,(1- ~,)+h,(1-

Case I : When No Slab Is Inserted According-to mirror formula, we have

1 1 1 -+-=v u I

=> - Ax=2(1..:

~

6.

.. . V 1 Magm11cat1on, m1=--=--,-=-0.5 u 2 Case II : When Slab Is Inserted

When we consider only two st8ps, then the ray of light starting from object O first gets refracted and then reflected. Distance of image 11 formed after refraction from the plane surface is given by

Again, applying the mirror formula, we get

1

v, -(30-2)

=

1

-10

O• X=nh+d=.±h+d

,1

Shitt=(1-¾}=(1- 5 )6=2 cm

~

4J3)+s(1- 3J2)=1.5cm

-10

V1 =-15cm

1

µ:)

Thus, h=h,+h2 -filC=2+3-1.5=3.5 cm

1 1 1 -+-=-30

µ=v~T

A= i.., = 6000 =4000A µ 1.5

d

V1

c. A. 0

. S ince,

30cm

5.

3.

v=E.= 3x10a =2x10a nis-1 µ 1~ .

Since "the frequency of light rerriains the same when it passes from one medium to another, so we have C=fAo and V=fA

water

4/3

v·

3

Therefore, distance of image 12 formed by plane mirror will be

.±h+d 3

v 2 =-15.55cm

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Ray Optics 7

_

_ = sin(60°) · 18 s1nr => r =28.76"

=>

AB=(2)(1)(~)=

1

So, ·the distance between rays 1 and 2 is given by BE=ABsin(30°)= P

~

cm

N.

' I' ' ''' r' '

'--

10.

Total deviation suffered by the ray is given by ()Total

=>

=Sp + C\:i

a=0-r)+0-r)

"

. 1-r=2 Further, in ~OPQ , we have =>

0

Mirror

--'

Since, MP= POtanr

... (1)

r+r+P=180°

=> MP= (6)tan(28.76°) => MP=3.3cm So, MN =2MP =6.6 cm 8.

cm

=>

r=90°-.!l. 2

..• (2)

/

From Snell's Law, we have sin(45°) 3 sinr

i=

=>

sinr=¾sin(45°)

=>

r=32°

A

From equation (1),.we get

i=r+i=90°+(a;P)

.•. (3)

According to Snell's Law, we have sini µ=-.smr.

=> F

=>

o+----£--->t

sin[90°+(";~)]

cos(P;")

sin( 90"-%)

cos(%)

µ

cos(

p; a)= µcos~

Since, EF =ECtanr

=> EF=(3)tan32°=1.88 m Length of shadow at the bottom of the lake is £=DF=DE+EF=2.88 m

9.

· From Snell's Law, we have =:i-

11.

. . r sm11= 2r

::::)

i1 =30°

sin(60°) smr

,l3 =-·-.-

''

r=30°

E

r

l

60°160° 30° A• :r I

1

D

''' '' '' ' r'r C

Since; AB= 2(AD) =2(DCtanr)

''

' ''

2

I

1 cm

1

=2.

''

' ' \ ~,. ' ,

C Applying Snell's _Law at P , we get 3 sini1 2= sini2 => i2 =19.5° Now, Applying Sine Law (Lami's Theorem), on ACPR , we get 2r CR {·.- LPCR = 60°} sin(180° -60° -19.5°) sin(19.5") => CR=0.7r

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Optics & Modern Physics

Advanced JEE Physics

12.

Let LAOM=0 Then by symmetry, LAIB= e

Since, sin i =~

=>

i = sin-

1

V Eye

(~)

According to Snell's Law applied at A , we have

sini µ= sinr

1

__,.

B

. _,(sini) . -'(- L) r1 =sin =sm µ µR

Now from geometry of figure, we have

LA0!=2LIBA =>

LIBA = i:(90° - 0) Lr= LIBA = 90'- 9 2

Also Li=goo-0+0=9O0+0 2 2

Deviation suffered by the ray is

S= i-r =sin- (ij)-sin- (µLR) 1

•.• (2)

Now, Snell's Law gives

1

1

This is also the angle r2 , so we have

. (90°+0)

sini

sin - 2 -

µ= sinr

s. , n - -

. _,(L) . '( µRL) R ~sm-

(90°-0) 2

r2 =sin

0}' =µ'

cos-+sine2 2 ( 0 . 0

From the knowledge of inverse trigonometry, we have

cos--sm-

sin-'(C)-sin-'(D) = sin-,(cJ1-D' -0J1-C')

2

Ci']

Ct3 __!:__ Af-µ2Fr µRf-f\2

r2 =sin-,[!:_

... (1)

2

1+sin0

2

1-sin0=µ '-1

sin0=~ µ +1

Now, again applying Snell's Law at B , we get

sin0 smr2

µ=-.-

Since, sin

8 = sin-1 (µsinr2 )

=>

a=~ 2a

1 d =2a(µ:- ) µ +1

0=sin-'(µLJ1- L' _!:_J1- L') R µ 2 R2 R R2 =>

13.

0 = sin-

1

( ~2

2

2

2

~µ R -L

- ~ 2

2

2

../R -L

)

In the figure, let AB be the disc and O be the centre of the bowl.

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1.

(a)

Crltical angle between 2 and 3, is given by

1+--R--+1 l+X+I

sinC=~ 1.8 Now, applying Snell's Law, we get

1.6sinB = 1.8sinC = (1.8)(-1..2) =1.3 1.8 13 8 =Sm . -'(ffi ) .o::: 54 •34° (b)

2.

If 9 is decreased, the angle of incidence at the interface between 2 and· 3 gets decreased or i < C, so the light will refract into medium 3.

20cm

''

l

'''

µ=½

The path of ray is curved as shown in figure. As it travels successively into denser layers, it bends away from normal and TIR takes place at depth where angle of incidence approaches

So, the radius of shadow is R = ( 15 + B~O) cm

.::

~

R=

2

(b)

'' 6001'

,p

=>

max

=.!_(AB

J

a=. = 1sin(90°) 3 ~a!ax + 202

a

=>

16a!ax

Slab

=>

7a~~ = 9(20')

=>

a= = (

sin(i)

4.

(a)

= 9a!ax + 9 (202 )

~)<20

cm) = 0.23 m

At interlace AB , applying Snell's Law. we get

~ = (A-Bt,,,..) t

cm=2.81m

±[

Applying Snell's Law at interfaces P and Q , we get 1sin(60°) = µ 8

845

3

For shadow to be formed, angle of incidence must be less than critical angle. Using Snell's Law, we get

I\,.,

___ _!____

i

i•

1sin(i) = µsin a

,/3)

.

1 µ

stna=-

2

... (1)

At interface BC , applying Snell's Law again, we get

3.

(a)

Using Snell"s Law, we get µsini

µsin(90- a)= 1(sin0)

= sinr

:::::,

sin0 = µcos a

..• (2)

From equation (1) and (2), we get sine= cola ¾(,'15;~20' )=1(,/x':200') Solving, we get

800 x=-cm

(b)

For emergence from BC , we must have 90-as:C At grazing incidence we have

3

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Optics & Modern Physics

Advanced JEE Physics 6

a=C ~

90-u~C

:::::,

2c;;::so

:::::,

c~45°

=>

sin-

=>

_!>_1_

=>

1

µ -

A

B

',a '' '

0

(¾)~45°

n-2C

J2.

0 C n/2 Deviation versus angle of Incidence graph when TI_R is taking place

D '------~-__, C

µ~,/2.

So, the greatest value of refractive index is

µ-=h

CONCEPTUAL NOTE(S)

When ray is travelling from rarer to denser medium then-deviation

5.

Case I:

is given by

When angle of incidence (i) is less than critical angle C

15 = i - sin-~ (1µ 2 sini)

i.e., i
6

1Lsfrf1(1~) --------2 Rarer medium (µ,) Denser medium (µJ

0

Since, 8 =Deviation =r - i From Snell's Law, we get µ 1 sini = µ 2 sinr :::::,

1

r.= sin-

(

l!. 2

---(1) 6_

(a)

Critical angle between 2 and 3

sinC=g=~ 1_4 7

~:i)

Substituting in equation (1 ), we get 1_6 1.4

0 - .,(sini.) u=Sln - , - -I-·

µ, This is a non·linear function and graph is given below 6

;-,~-------------, ! .

Applying, Snell's Law, at P , we get 1_6sin0 =1.4sin(9O° -C) =1.4cosc

j

C

1_4J1-

Deviation versus angle of incidence graph when TIA is not taking place. · Case II: When the angle of incidence i is greater than the critical angle C,i.e., i>C In this case TIR will take place as shown, so deviation is 6=n-2i ___ (2)

/

/

0

''

(b)

=>

sine

~

Si::::26.8°

1-6

36 49

O.45

As 8 is increased, i1 will increase or i2 will decrease or

i2 < C and hence the light will refract in medium 3.

' Rarer medium Denser mediu'm

This is a linear function and so the graph is given below

7.

As shown in figure, the light from the source will not emerge out~ of water if i=C.

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·Ray Optics !'!=tanC=_:i_ h . 4

i ! h

' ,i}c

,,

h=±R=± cm 3 3

i l

s

i=C

11.

~µ -1 sinC=

1µ

''

C

s

Since r, + r2

~= tanc

4

=90° , so

2 (r,)., =i-sin·'( ; )

=>

R =htanC

=>

R=-h~µ' -1

Again, using Snell's Law, we get µ

(a)

Applying Snell's Law, we get 1.3 sin 81 =(1) sin 05

=>

=>

(b)

13 sin 0, = (1.3)sin(30°) = ; = 0.65

::::>

=>

1.a(.!.) 2

=>

1

=>

=>

p > 47.3°

For this to happen, we have

a <90°-47.3° => a<42.7° Applying Snell's Law at A , we get 0 < sin-1 (µsin42.7°)

=> =>

~

(-

For TIR to take place at B , we have P>C

smr2

GJa,;s -- 3_0~-

=..!.

1- 4µ2

9

9

4µ2 =~

9 µ

9

Alcohol

= -./2

Layer

i,

.I ''

Now, when the paper is dry then µ = 1

1-) = 47.3° C=sin- (.!.) = sinµ 1.36 1

= sin~ 3 oo)

·i

=>

sin0 4 = - --=0.49 1.45 84 ~ 26.6°

Critical angle,

9

=>

=> 85 r::::40.54° Applying Snell's Law, we get 1.3sin01 =1.45sin9 4 =>

3

:c

2 sinr, =~sin(90°) = µ 1.5 3 2 (r,J.~ = sin·' ( ; )

In llSAB,

9.

5

At the maximum, the ray can enter the glass at the grazing angle, so (Dmax = 90° . According to Snajl's Law µ 9 sinr, =µsini

C 2

' ' c,'

'' '' '' ''

h

Therefore, minimum radius R corresponds to the situation when

8.

'

R-+<

I+-

1

r1 =sin- (~}·:42°

Since r1 + r2 = 90°, so we get

8

r2

)

0 < sin·' [1.36sin(42.7°)] e < 67.3° So, the maximum val~e of 0 for TIR to take place at B is 67.3°.

= 48°

Critical angle at glass air interface, is given by

1 C = sin·' (--) ~ 42° 1.5 So, we observe that as r2 > C , so it can't be seen.

12.

According to Snell's Law, we have

-./2 = sin(45°) sinr r=30° The critical angle C is given by ~

1

C=sin- (¾)=45°

:::==================================== = 1.181

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Optics & Modern Physics

Advanced JEE Physics 13.

Incident angle i is least fqr ray AP and this angle should be

greater than the critical angle C i>C sini > sinC 1 R => - - > R+r µ

i.e., =>

=> Applying, Sine Law (i.e., Snell's Law) in t.OPM , we get

=>

OP

OM

sinC

sin(90° + r)

OP

3R > 2R+2r

=> =>

R

R 2 -->R+r 3

{R = radius)

R>2r r 1 -
A

(~)=cosr 2

=>

OP=~

As we move away from

O , angle

PMO

will increase.

Therefore, OP "f> ~ . Same is the case on left side of O .

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1".

The ray diagram for the situation discussed is shown in figure.

=>

A

2.

µ=~

r,+r2 =A Since i+e=A+c5 => 6=a+p-A Further applying Snell's Law at incident surface and emergent surface, we get µ= sina. and sinr2 =! sinr1 sinf3 µ sin a= sinf3 sinr1 sinr2

(a)

From the figure, we observe that r1+r2 =r2 +r3 = 60° ~

r1 =ra

Applying Snell's Law at the faces AB and BC , we get sIn1 sinr1

sin(30°) sinr3

µ=-=---

Since r1 = r3 ~

(b)

sina sin(A-r,)

sinp sinr2

sin(A-r,) sinr2

sina sinf3

sinAcosr.; sinr2

cosAsinr2 sinr2

. sina smAcotr2 =--+cosA sinp

i=30°

Since. r1 + r2 =A= 60° ~

r1 =60°-r2 :=60-C

Further, µ

µ

sina· sinp

. sin~ +cotA sInf3smA

(·: r, = C} 5

sini sin(30°) sinr1 sin(60° -C)

Sinceµ= !np =sinpcosecr2 smr2 µ =sin13:J1 + cot2 r~

0.5 · sin(60°)cosC-cos(60°)sinC

. µ=sInp 1+

Since sin C = ! µ

cosc=J1-

(

sma . - . +cotA s1nf3smA

)'

Since A=a+f3-c5

1 µ'

1 µ(..J3)J1 - 2 _ _! =0.5 2 µ 2 ( ~)~µ'-1 =1

. µ=smp 1+

3.

(

)'

sma . . (. ) +cot(a+p-6) smpsm a+P,-<5

At minimum deviation, we have

r=t=30° 2

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Optics & Modem Physics

Advanced JEE Physics A µ

sini sinr,

sin(45°) sin(45°-C)

µ

sin(45°) · sin(45°)cosC-cos(45°)sinC

Since sine=..! µ

Applying Snail's Law at AB , we get

(a)

15=~ · sin(30°)

=>

i =48.6°

• Since, 6roiai

(b)

=Op + 60 + 6R

=>

BT~, =0-r)+(1B0°-2r)+(i-r)

=>

()Total

=>

°"roiat =157 .2o

(b}

=180°+2i-4r-

µ~1-

=>

µ' -1 =4

=>

µ=-15

1 -1=1 µ'

(sin45° = cos45°)

At minimum deviation, we,have

i = 58:8°

=> 6.

(a)

Applying Snell's law ai D , we·get

(1)sini =(¾)sin30° ..

..~ ~, i' =34.2~ => 6ro1a1=180°+2i'-4r

=>

i =34:2°

A

OTotal =128.4°

Since, the condition for no emergence is A>2C 1

=>

A>2sin-

(¾) 1

A> 2s1~-1 ( --) 1.5

>83.62°

B~-~-----~c ·

Therefor~, Ama:,,. ="83.62°, for escaping of the ray, thr~ugh the

5.

=>

2

. r. sini Since, v5 - .. ( ) sin 22.5°

isini' = ¾sin(30°)

4.

cosC=~1- ;

r, =f2 =i=22,5°

Again applying Snell's ,Law for water-glass interface, we get

=>

=>

adjacent face.

Total deviation suffered bythe.ray·;s_ 6=60 +6E =260

T~e situation is shown in figure

=>

p

(b)

7.

6 = 20-30°) = 8.4°

At near normal incidence, i 1:::: r1 = 0° Since r1 + r2 ~

=A

r2 =Cl

From Snell's Law apRlied at the face from where the refracted ray emerges, we get _ sine µ=-.-' s1nr2 ~

e = sin-1 (µsina)

Now, deviation O= i+e-A =sin-1 (µsina)-a

(a)

~

Since, e = 90° Also, r2 =C=sin-

1

(¾)

8.

O=sin-1 (µsina)-a

{·: i = 0°}

For the ray to retrace its path, it must be incident normally to the face AC .:So, we have · "

r2 =Oo

Now, i=A=45° and r, =A-r2 =45°-C • Applying Snell's Law at AB , we get

= ==================================== L184

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Ray Optics A

sinA sinr1

=>

sinC

sinr1 = sinAsinC

... (3)

The ·ray does not ·emerge from the other face AC, when

r2 >C Since, r1 + r2

Since r1 +r2 =A

=> _r1 =A=30° From Snell's Law, we have µ= ~ini =../2 smr1 9.

From

11.

A-r, >C

=>

r1
= rel="nofollow">

sinr, < sin(A-C)

=> => =>

sinAsinC < sinAcosC-sinCcosA 1
For Violet Light: According to Snell's Law, applied at the plane of incidence, we get sini µ=-.-

the. statement of the problem, we gather the information

that

Slnr1

i=60°, A:::30°, 6=30°, Since, B=i+e-A

=>

=A

=>

=>

e=6-i+A=30°-60°+30°=0° A

=>

1.

= sin(so sinr1

66

0 )

r1 =27.5°

Since, r1 + r2 =A =:>

r2 =A-r1 = 32.5°

Applying Snell's Law at the plane of emergence, we get sine µ=-.s1nr2 .

i.e., the emergent ray is _perpendicula~ to the face through which it emerges. Further, r2 = O and r, + r2 = A

=>

1 _66 =

=>

e=63.1°

=:>

Sv=53.1°

For Red Light : According to Snell's Law, applied at the plane of incidence, we get

From Snell's Law applied at face AC , we get µ= ~ini =>/3

1.6 = sin(50°)

2

smr1 10.

sine sin(32.5°)

Since, 6v =Ci+e)-A

{as e = O}

r1 =A=30°

=>

:::::)

Fron:i geometry, we observe that the angle of incidence at the face AB is A . Applying Snell's Law at face AB , we get

sinA µ=-.smr1

f1

sinr1

=28.2°

=A

Since, r1 + r2

... (1)

r2 = A-r1 =31.8°

=:>

Applying Snell's Law at the plane of emergence, we get

A

1 _62 = :::::)

sine sin(31.8°)

8=58.6°

Since, SR

= (i + e)-A =48.6°

So, angular dispersion is given by Sv -6R

B c __ _ _ _ __,C If C is the critical angle of th_e prism, then 1 µ=sine From (1) and (2), we get

12.

=4.5°

For minimum deviation, we have . (A+o.) S in--

... (2)

µ=

2 sin(i)

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Advanced JEE Physics

Optics & Modern Physics

:::::,

=> 1.3sin(45°)=sin(45°+B.) Solving, this we get 6m =22° For maximum deviation, we have the emergent ray to be grazing on the surface of emergence. So, e=90°, r2 =C and r, =90°-r2 =90°-C =>

137=~ · sin(45°) i = 75.6°

=> => 6red=i+e-A=30.6° For violet ray, we have

6max =i+e-A

We can find i by using µ

sini = -.-

s1nr1

. C 1 and s1r:-i =-

µ

1.42 = sin(75.6°)

. . . 8 ubstituting

s1nr1

the values, we get 6max ::= 550

13.

14.

Since, r, + r2

= 0°

r, =43°

=>

r2 =A-r1 =2°

f:

r1 +r2 =A}

1.4 2 = s!ne

smr2

17.

=A

=>

Again, applying Snell's Law at the emerging face, for violet rays, we get

Given that, i = 60° , A = 30° and B = 30° Since, 6=i+e-A Substituting the values we get, e = 0° Now; e = 0°, means that the emergent ray is normal to the face through which it emerges.

Given that A= 30° and i = 0°, so r,

r, =A=45°

Now, according to Snell's Law, we have

=>

0=2.84°

=>

6v1o1ei=i+e-A=33.4°

As the angles are small we can take,

sine;::: e

=> r2 =A=30° Further applying Snell's Law at the plane of emergence, we get

A

1 _5 =s!ne s1nr2 Substituting the values, we get e=-49° =:>

15.

6=i+e-A=19°

For no total internal reflection, when the ray leaves the prism, r2 =C But sinC =

1 .!µ = -1.6 -

1 r, =C = sin-•(--) = 38.7° 1.6

Further r1 + r2 =A= 45°

=>

r1 =45°-38.7°=6.3°

Now, i = sin-• (µsinr,) = sin-' [1.6 x sin(6.3°)] = 10.1° 16.

C Applying Snell's Law for the two emerging rays at AC and AB , we get sin(A+1°) sin(4°) µ sinA sin(2A)

=>

A+ 1° 4° µ~~A-=2A

=>

A=1° and µ=2

For reQ ray, we have e=0°=r2

Since, r, + r2

=A

= c:================================== 1.186

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1.

Let us see where do the parallel rays converge (or diverge) on the principal axis. Let us call it the focus and the corresponding

f • Using

length the focal length

J:.g_ = 11-a V u

= µ2 -

µ1

Now applying .&__,& = µ 2 -µ, with the idea that 12 is formed

R

=>

appropriate values and signs, we get

4 3

1

3

4.

-R

Applying, µ 2

=1.5 , we get

1

P = 2.5 dioptre= 2.5 D

1.5

v,-

1

1.5-1 (-2r) = - -

i !E

.& = µ 2 R- µ, , we get

-

V

U

1.6

1-1.6

PI - (-3)

For first refraction at the unsilvered surface, we have

5cm

=>

PI =-2.42 cm

=>

EI= (5 + 2.42) cm

=>

EI= 7.42 cm

(a)

Applying, µ 2

'•

&T

= --::S

0

V1 --tco

i.e., rays become parallel to the principal axis. Hence the image formed by the curved mirror will lie at the focus of the mirror i.e., a distance

i

5.

=>

v,

-(-~r 1.5

µ,

= µ 2 R- µ,

, we get

U

1.5

1

1.5-1

V

-1

6

V=-90 cm

So, the distance between object and its image is 80 cm

For second refraction at the unsilvered surface, we have

1

-

V

from pole of mirror. {from pole of the mirror}

=)

µ-1

f=40cm=0.4m

1 1 P (in dioptre) = - ( -- ) = - f metre 0.4

:=:,.

1

-(2Rµ+x +R)

x :::::,Q.75R

Hence,

2.

µ

-(R+x)

Solving this equation for µ

Since, the rays are converging, its power should be positive.

=>

R

i_1

1-~= +10 =:-

u

V

at C, because light falls normally on the mirror.

with

(b)

1-1.5

Again applying µ 2

1.5

(-r)

-

.& = µ 2 -

=>

1

R

u

V

µ1

1.5-1

V - <-u) = - 6 -

V3=-2R

i.e., final image is formed at pole of the mirror. 3.

For the 'image of object O to be formed at O , the light should v is negative when

fall normally on mirror. First image 11 (after refraction from the

::::>

plane surface) will be formed at a distance of 2R+x from plane

12

u

>1

U<12cm

µ

surface ' because dapp.

= daC\1./al µ

.

6.

Applying, µ 2 V

1.5

_

,& = µ 2 - µ, twice, we get u R

1

1.5-1

:V, - (-2.5) = ~ ====================================1.187

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Advanced JEE Physics ~

-----.. +ve

30 v, = --=, cm

Further,

::::::,

Optics & Modern Physics

,1

1.5

v,

-(20+3;)

1-1.5 -10

v2 =-85cm

i.e., final image is formed at 65 cm from first face on the same

side of the object. 7.

First image will be formed by direct rays 1 and 2, etc. DO 5 D11 =-=-=3.33 cm · µ 1.5

Applying, µ 2 µ

Second image will be formed by reflected rays 3 and 4, etc. Object is placed at the focus of the mirror. Hence, 12 is formed

µ,

= µ 2 R- µ,

twice, we get

U

1

· µ-1

AI, - (-2R) = -R =>

at infinity.

-

V

AI ~ 2µR 1

1-2µ

1 µ 1-µ BI, (AI, -R) -R Solving this equation, we get 2R(4µ-1) BI, 3µ-1 So, the distance between the final image and the object is d= 3R 2R(4µ-1) (µ-1)R 3µ-1 (3µ-1) Further,

+

5cm

*:t

5cm

8.

·p

We have to see the image of O from the other side

= c:::::==========::::;:====================== 1.188

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1.

When the images of both the sources are formed at the same point, then v will be same for both (in value). However for one case the image will be real and for the other case it will be virtual. So,

For 0 1 ,

·

1

1 1 we have-+-=V

X

1 1 1 For 0 2 , we have--+--=V 24-X 9

... (2)

--=-

~

f=12 cm

Note that you would get the same answer by considering the othercase(m=-3,for u=-16cm)

... (1)_

9

-1+3 24

=>

3.

The Lens Maker's Formula is given by

f=9cm

!=(µ-1) f

Here, f =30 cm, R1 =10 cm, R2 =oo

o......______...oz

~= (µ-1)

=>

30µ-30=10

=>

µ~-

30

~X----S--24-x-t-t Adding equations (1) and (2), we get

x

24-x

9

4.

= 6 cm

= µ-1 10

4

3

1 1 2 -+--~SOiving we get x

(_!__.!) 10 ro

=>

o - - - - - - - - - - - - < •

0,

(_!_-~) RI R2

Applying Lens Maker's Formula,

!=(µ-1)(_!_-~)

Hence, the lens should be kept at a distance of 6 cm from

f

either of the object.

R1

R2

Here, µ=1.5, f=60 cm, R1 =+R, R2 =-R

2.

Here the image formed can be virtual as well as real, so the

=>

6~=(1.5-1)(¾+¾)

~

R=60 cm

value of m should be +3 in one case (virtual image) and -3 in

the other (rea! image). Magnification of +3 can be obtained

Therefore,· the focal length of the spherical silvered surface, is given by

only when the object is placed within F {I.e., for smaller value of the object distance). The magnification of -3 is obtained when the object is kept between F and 2F (i.e., for greater value of object distance). So, m=+3,for u=-Bcm Therefore, ·from the definition of magnification, we have

R 60 \,, =-=-=+30 cm 2 2 (Positive, because it is a converging mirror)· The equivalent focal length of the lens-mirror combination is then given by

V

m=-

1 2 2 1 -F=l-t:"= 60 - (-30)

u

=> =>

+3=..!....

~ F=-15cm The negative sign indicates that the combination behaves as a concave mirror.

-8 V=3x(-8)=-24 cm

Using lens formula, we get

1 11

-24

-8

5.

According to the Lens Maker's Formula, we have

1 ( -1- ~ 1 ) ,where f=10cm -=(µ-1) f R1 R2

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Optics & Modem Plzysics

Advanced JEE Plzysics When placed in medium 1, then

f=(:,-1)(~ -;J

..• (2)

=>

-in the second case, let µ be the refractive index of the liquid,

When placed in medium 2; then

.!.t = (_!:_-1)(_!__ _ _!__) µ R,, R

then

.•. (3)

2

2

From equations (1) and (2), we get 1= {µ-1) 1.6-1 9

I

48

(:,-1) G:~-1) 10.

The system behaves like a mirror of focal length

given by, 1 _ 2(µ,/µ,)

F--R-,-

1.6-1 --102

(_!:_-1)

(~-1) 1.7

µ,

20

µ=1.37

From equations (1) and (3), we get 1= (µ-1)

-(5+ ~)

Solving, we get

=> \ =91=90 cm i.e., still it behaves as a converging lens

I

1 1 36- -45=1 1=20 cm

.

2(µ,/µ,-1)"

n,

R,

Substituting the values with appropriate signs, we

get

=> ~=-10.2f=-102cm i.e., it now behaves like a diverging lens. 6.

{·: R, ->co)

Here R1 =+25 cm, R2 =-25 cm, µ 1 =1 and µ 2

=>

=23

Image coincides with object, hence, u =v =-x (say) -----+ +ve

1 _ 2{3/2) 2(3/2-1) -x - . - -25 25 2 3 1 4 -=-+-=x252525

=>

0

X=12.5cm

F =-7.5 cm

So, the system behaves as a concave mirror· of focal length 7.5 cm. 11.

For a convex lens, the distance between an object and its real image is minimum when· u = 2~ and v = 2~ When concave tens is placed in contact, then we have

1 1

V

v' = 2\t

~-2\

Given,\ =+5 cm and t =-10 cm

Shilt of image Is

The combined focal length F is given by

11111 1 ---+------+F I t 5 10 10 Since,

F=+10cm i.e., the combination behaves as a converging lens of focal

According to Lens Maker's Formula, we have

4\'

t

12.

_!__=(1.5-1)(_!__ __ 10 R, -10

=>

R1 =+10cm

.!. + 1

The path of the beam parallel to the optical axis of the

l_)

=>

Now, using,

R,

1

1

=> 9.

system and the image of object AB are· shown in figure. Image A'B' (direct and real) is obtainE!d to fu_ll scale with the object in any position. 2

2 (µ,/µ,)

2 (µ,/µ, - l)

vu R2 Substituting the values, we get

v- -15 =

The following two cases are possible.

CASE-I: The mirror is at a distance of d = f + R ,;, 2 m from the lens.

!f = (µ -11(_!__ - _!__) R R 2

~

/J.V~-

length 1o cm.

1

2 \t - 2 \ = ~ t-2\ t-2\ » \ , so the shilt of image is

tJ.v=v'-v=

=>

8.

fcomb

U

1+1=1_1_t-l v'2\\~lt

Hence, the object should be placed at a distance 12.5 cm in front of the silvered lens.

7.

1

---=-

l+-R-+<

R1

2(1.5) 2(1.5-1) -10 - +10

v=-2.14 cm

Using lens formula,

F

B'

.

_! _ _! =! , we get V u

,--'2F

.\-,>l+-1-f-----.. .... -,, ______ ...... ., --

I

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Ray Optics CASE-II:

f,=4a

The mirror is at a distance of d = f =R =1 m from the lens. The image of object A'B', also full scale, will be inverted

B

and virtual with the object in any position. A

Lens 1

Lens 2

1+- x--,---4a

1

V = V1 = ~ u -x 4a-x Similarly for Lens 2, we get

1 + - - - d - - -..

m1 =

1 v2 1

For a convex lens, distance between an object and its real image is minimum when u =2~ . Hence,

V

2

-

v2 =x+12a

(1-¾}=1

= i""=\

I

I

(b)

Using the lens formula,

~

(i.e., very

V

U

f

for the first lens

17.

_.!..._...!_=_.!...

15.

10

... (5)

v3 =-(x+Sa)

::::,

m3 = ~

x+12a

...(6).

According to the lens formula, we have,

.!V _.!U = !f

Since u = -mf , so we get

1

1

1

v+ mf = (-fl

2(~:J. 2(~-1J

R,

1

v2 -4a = 4a

=>

v 2 =7.5 cm

Since, }

1 -

same size but inverted.

For the second lens

=>

3

... (4)

the same place and m1.m 2m3 = -1 means final image is of the

v1 = 60 cm

30

x-4a

Also, v3 = -(x + Sa) means that the object and its image lie at

1 1 1 -+-=v, 30 20

V2

16 1a2' )

So, we observe from (2), (4) and (6) that m1m2 m3 =-1

1 1 1 we get ---=-, V

=>

m

1

If a concave lens of very large focal length

small power) is placed in contact with the convex lens, then its power and hence the focal length are almost unaltered. Therefore, there will be no shifting of the image.

14.

(

x+ a ,-(16a2 ) x+12a x-4a Similarly for lens 3, we get

=>

.1.=1-.\_= 1-\ µ

... (3)

16a2 =a 16a2

=>

I

... (2)

1 1 v1 -4a a x-4a 1

3\-(1-¾}=2\

µ

... (1)

4ax v1 = x-4a =>

(a)

1

-+-=V1 X 4a

B

13.

1

Lens 3

_ __.,.__ 4a - - . i

R,

=>

2µ _ 2(µ-1)

V=·-(m:Jf

So, lateral magnification is given by

-r

V

1

ml =u= m+1

¾=-( 4µr-2)

Hence, the image will be (m + 1) times smaller than the object.

F=-(4/-2] i.e., the lens is equivalent to a concave mirror of focal length r

4µ-2 16.

(a)

(b)

Let x be the distance between first lens and the object AB .

Applying the lens formula,

18.

Since a concave lens always forms an erect image whereas the given image I is on the other size·of the optic axis, so the lens is convex. Join O with I . Line 01 cuts the optic axis AB at pole (P) of the lens. The dotted line shows the position of lens. ·

.1.V - .!U = -f1 thrice, we get for lens 1,

r:::::=================================1.191

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Optics & Modern Physics

Adva11ced JEE Physics

For first reflection, let us use the mirror formula, i.e.,

1 1 1 2 -+-=-=v u f R

I

:, +(µµ;1)=:

=>

2-=-(3µ-1)

µR

V2

=>

v,=\=7cm

R

v: =-(5~;1)

Similarly after nth reflections, we get

For the second lens, we have

1 1 1 v, -(7-3)=6 4

µR

_!_;3µ-1=-2

Let parallel rays be incident on first'lens, then

V2

v1

For second reflection, similarly we get

From point O , draw a line parallel to AB which after refraction must pass through the focus F of the lens. 19.

=>

_!_ = -[(2n + 1)µ-1] v0

µR

Finally again u·sing the refraction formula,

6

~ - ~1 = µ 2 ; µ 1

,

applied at the second curved surface, we get

y 2 =2.4 cm

_!__{(2n+1)µ-1}= 1-r, 20.

(a)

!f = (µ -1)(_!_ - _!_) R R 1

R

V1

From Lens Maker's Formula, we get ~

-R

R v,=~--~

2(µn + µ-1)

2

\ =µ,-1= 1.7-1=7 =1.4 t µ,-1 1.5-1 5 (b)

21.

In this liquid the first lens will be a diverging (as refractive index of liquid > 1.5) and the second a converging one (as refractive index of liquid < 1.7 ).

24.

1

\ 1

Since the incident beam is parallel, so we have

For the second lens; we have

1

1 -5

So, the system behaves as a concave mirror of focal length

1 -20

-R() . The object will coincide with image when the object is

2 µ-1

v2 =-4cm For the third lens, we have :::::>

1 ;::;:>

V

3

1

2(µ-1)

-R \=2(µ-1)

v, =\ =10 cm

V2

2µ

\=~--R-

placed at centre of curvature. So, we get R

1

x, =21\I=µ-1

-)-00

In the second case,

... ,1)

-2µ f1 =R

i.e., rays will become parallel to the optic axis. 22.

Since the focal length is equal to two times the radius of curvature, so f =-2r

(-~) =(µ, +a1-1J(±-n

t=-2µ

=>

x,= 2ltl=~ X

µ=--'x, -x 2

2a. The rays will first get refracted, then n ·times reflected and finally again refracted. So, using

µ2 V

-

µ1

u

= µ2 - µ1 R

for first

and R = ~

x, -x 2

Now, according to Lens Maker's Formula, we have

! = (µ -1) (..!) = µ - 1 = 2-

refraction, we get .!:_ _ _!_=

v1 oo

... (2)

µ

Solving equations (1) and (2), we get

I= 3-2µo

23.

R

=>

f

µ-1 R

~

R

R

{from equation (1)}

X1

f=X 1

(_.!:._)R µ-1

v, =

= c:::===================================== 1.192

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1.

[CJ

I,

4.

I,

f f-(-2/)

=>

m=---

=>

m=3

1

[DJ Only one image will be fom,ed by this lens system, because the optic axis of both the parts coincide. Two images would have been formed if their optic axis would had been different.

i.---

8 cm ---+1

5.

Distance of image from the plane surface is

4

X1 =-=2.5cm 1.6,

µ {.. d =d"ci""} •

Maximum students are gripped by the misconception that when object lies at focus then image is formed at infinity. This is true but only for a Concave Mirror (Convex Lens) and is absolutely wrong when applied to Convex Mirror (Concave Lens) How To Proceed Then? If object placed at focus then just check out the MIRROR. If CONCAVE then image is formed at_ infinity and if Convex then

app

For the curved surface, we have 1.6 1 . 1-1.6 --+-=-4 X2 -8

=>

[BJ MISCONCEPTION

x2 :::::--3 cm

1 1 1

The minus sign means the image is on the side where the object lies. So,

apply-+-=. V U f

1,1, =(8-2.5-3) cm=2.5 cm

~

---=-

~

v=10cm

V

2.

[CJ Area of object = 9 cm 2 Also, we know that

6.

=>

A

t= A,

[

Shift t.x=25-20=(1-t}

=>

1 5=(1-1.5

x9

)1

=>

=>

L

[CJ

-10 ]' -25-(-10)

=rnJ

+20

Image will be formed at infinity, when the object is placed at focus of the lens i.e., at 20 cm from the lens. So, we have

. = mar = -A,v'(f)' AIM rea agm.1. 1cat1on =2 = -A0 u f-u =>

20

W

~

The similar thing is extended and applied here too. Here the answer fabricated by the MISCONCEPTION is 1 (but we must know this is the answer only for a Concave Mirror (or Convex Lens). For Convex Mirror we have

f m=-f-u

[CJ Critical angle between glass and liquid interface is sinC = ~ = Zµ

3/2

3

Angle of incidence at face AC is 60° For TIR to take place, we have

i>C

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Optics & Modern Physics

Advanced JEE Physics => => => 11.

2 sin(60') > ;

=>

µ<-4

3-13

12.

8.

[DJ Since no parallax exists between the images formed by two mirrors (convex and plane) hence the images for both coincide. But for a plane mirror an image is as far behind the mirror as the

object is in front of .it. Hence the image for plane mirror should be 30 cm behind it or 1o cm behind the convex mirror. So for convex mirror.

=>

13.

m=-~=3 u

=>

Since object always lies on negative side. So, u=-20 cm

'

3 I => f-(-20) => 31+60=1 => f=-30cm Negative sign indicates the mirror is concave. An Advice

J+-30 cm---30 cm~

I= 50 4 1=12.5 cm => Since R=2f => R=25 cm

=>

9.

Also Jul+lvl=D lul+3lul=B0 lul=20

~

0

1 + - - - 50

cm-----+1

I would always advice you to write _lul+lvl =D wherever you are given the distance between object and image as no error will creep in these because the MOD signs prevent the errors.

[DJ Concave lens forms the virtual image of a real object. So, we

15.

have V 1 NOW,I"f m=-u=4·

. 10 x=-cm

i'=(2'+0)

3

U=--cm

and

10 v=-- cm

14-3x---+x

3

Substituting in

Since, 6to1a1 = 180°

0

40 3

=>

=>

0+1so -2i' =180'

=>

6=2i'

=>

2'=2(2+0)

----4x------

Here, negative sign implies that i gets decreased or i' =0 . i.e., light should fall normally on mirror.

U

16.

9

10. · [CJ · Jm.,,J=2Jm..,,,,J

[BJ The minimum length of the mirror required for the purpose is-half the height of ttie person.

I=- 40 f=-4.4 cm

0

0= -2°

! =_!__.!., we get

f V 1 -3 3 -=-+f 10 40

=>

[BJ Since, o=(µ-1)A=(1.5-1)(4)=2'

=> 1=6=2° Let the mirror be rotated by an angle 0, then

u=-4x,then v=-x then 3x=10cm

=>

[AJ Since image formed is erect, hence it must be virtual. So,

..!.+.:!.=! V U f 1 1 1 -+-=10 -50 f 5-1 f 50

[BJ Focal length of mirror is independent of the refractive index of medium in which it is placed.

U=-50cm, V=+10cm

Since

[BJ 1 m=+2 1 20 => - = - 2 20-u => 40 =20-u => U=-20 cm

B

=>

I 2 I 1-(-20) 1-(-15) 1+20=2f+30 l=-10cm

17.

[BJ A divergent beam appears to converge behind the mirror thus giving a virtual image. So, a convergent beam will give a real

image.

~L Convergent Beam

- ~

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Ray Optics 19.

[CJ

10 _ 900 0 _ 180 cm 55 11 Negative sign with focal length implies that the mirror is a concave mirror.

.111 t Al ppymg v-u=1,wege 111

=>

b+a=1

=>

25.

... (1)

f=~ a+b

Further in right triangle ACB , we have 2

AC + BC

2

=AB

S" 1 mce,

1 1 v+u=t

2

=>

(a'+c')+(b'+c')=(a+b)'

=>

a2 +b 2 +2c2 =a2 +b2 +2ab

=> =>

ab=C2 Substituting this in equation (1 ), we get :::>

c'

21.

1 10 20

1 20

V=-Cffi

3

[A] Covering the lower half will just make the image less bright (not blurred) as less number of rays will be reflected as compared to the previous case.

27.

[BJ

[CJ If the mirror approaches the object or the object approaches the stationary mirror with speed v then image approaches object with speed 2v.

1 V

26.

f=a+b

20.

[BJ f=+20 cm u=-10 cm

[CJ

I mrea1=-n=f-u

22.

20"..• -···

=>

-nf+nu=f

=>

U=(n:1}

-··20° 0

:

'b

g:ui

40"

[DJ The incident and the second reflected ray make the same angle 8 with vertical. Hence, they are parallel for any value of 8 .

28.

40"

[CJ

A'

23.

[D] 1

n=

B

I f-u

f-u=nf => u=(1-n)f -(n-1)f According to the sign convention used u must always be negative.

height of wall.

=>

0

24.

29.

[DJ

0

=-4.5

30.

f m=f-u -4.5 =

=> =>

i=!:!3

360 n=-->oo

[DJ mreal

B'

From symmetry we observe that length of mirror is one third of

[A] 3

:o =

f f-(-20)

n= 360 -1=5 0

-4.51-90 =f 90 =-5.51

!=-~ 5.5

6 which is Even

31.

[BJ Ray's after reflections from two perpendicular mirrors are always parallel to incident ray irrespective of angle of incidence.

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Optics & Modern Physics

Advanced JEE Physics 32.

[CJ Power of concave lens must be Jess than that of the convex lens to form.a real image. So, net power will decrease or fqcal length will increase. Far real image v is- positive, u is negative and f 1 1 1 is positive. Applying lens formula, ---=- (substituting all V

U

38. [BJ Both concave and convex mirror give virtual image but a concave mirror gives a magnified virtual image (when object placed between F and P). Since the boy sees his image of diminished size, so the mirror must be convex.

39.

values with-sign), we get 1 , ,

Let object be placed at a distance x from mirror.

u=-X, v=-(x+10), f=-12cm

-+-=v u I

Since,

u is constant, f is iilcreasing. So v will also increase.

34.

[A]

1

1

v u I 1

[D] G C

==>

'D H ···j·······

t

:s ------i------

d

!

B

So,

1

-+-=-

1

1

+-=-

-(x+10) -x x =20 cm U=-20cm, V=-30 cm

-12.

m=~:!.= -(-ao) =-1.5 u -20

Negative sign with magnifi_catiori indicates image is real.

:E F

40.

,._ L.., J 14--2L---+1

[BJ At face AB , the ray of light suffers no devicltio11, so applying

Snell's Law at face AC, we get 35.

[BJ

1

Since an elongated- image is formed and it touches one end of the rod, so the rod must lie with one end at 2F and other end

µ = sinC = sin(45°)

=>

between 2F and F (shown in figure).

Forend A, U=-

1

3

A

3

5

.=51-51

1

µ,,,. = -,J2

51

. , , 1 S mce -+-=v u I 1 3 1 v-51= -I

2

.=-51 51 V=-2

·

1

-

/

f/3

A'

0

A

--1--+-4-~~~-',~/

,

ROD

F

-211+-f~ fil ----+I 3

•

PA'= 51 2

B

41.

[CJ 1 1

C

1

-+-=v u I ==> v-1 + u-1 =r-1 = constant

Take derivative w.r.t. time on both sides

0/>/=~-21 2

!/v·')+!/u·')=O dv ( ) _, du ( - 1) V_, -+ - 1 U -= 0 dt dt

36.

:~ =<'.(:~)

[BJ First _consider two adjacent walls (not the ceiling). If n1 is the

v'

Image speed= - u' ( object speed)

number of images formed due to these perpendicular walls, then

n '

==>

= 360 _ 1

when object moves towards mirror u decreases with passage of

90

time and hence du

n1 =3

dt

Now, when we consider the mirror on the ceiling then it will make a total of 4 images (one of the original object and th_ree images of the previous arrangement) so, total images formed equals 7. However, if the object were the observer himself then total number of images is 7 -1 = 6

37.

=-9 cms-1

dv =-(-1-)' du dt f-u dt dv ( -24 )' du di=- -24+60 di

[CJ Velocity of light is always normal to the wavefront.

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Ray Optics

dv =-i(-9) di 9 dv , -=+4 ems-

{·: 2sin(½)cos(½) =sini} =>

di

Positive value of ~: indicates that v increases with the passage

=>

of time i.e. image must be going away from mirror.

=> 42.

43.

[DJ A_:::: 60° for equilateral prism.

47.

[Bl The source cannot be seen if angle of incidence in the denser medium is greater than the critical angle.

Since, i+e=A+D => 2i=A+D

Since,

=>

2(¾A )=A+D

=>

D=~=30° 2

d h r=--=-~µ'-1 ~µ'-1

h=(1)J(%)'-1 4

h=- cm 3

[BJ Since,

uab

> U00

48.

[A] sinC=..!. µ

Vab
TIR will take place at AC if i > C i.e. 45° > C for Red Colour

Since, m=-"!... u

[m,,[<[m.,[

=>

µ,

1.39 1.41

µ=-V =>

sin CR > sin 45 Z= 3x10

=>

8

V

Green Colour

=>

V=1.5x108 ms-1

=>

v=1.5x10 10 cms-1

µG=1.44 1

=>

v,

45° 0

X

-

VlO

,

, ,,

, ,,

v,

. C =--<-=-Sin 1 1 . 45 Sin 8

,

1.47 1.41

''

sinC8 < sin45

45° 45°

,

''

'' ' -Vo

=>

49.

sini

2sin(½)cos(½)

sin(½)

sin(½)

(TIR

will take place for Blue Colour)

[C]

=;)A,

=>

A =(1.54-1) 4 ' 1.72-1

:,.

A2 =3o

sinr

n

c, < 45'

So, Red is separated from Green and Blue.

A,=(~;

[CJ i =2r . sini Since, n=-.-

=>

CG <45°

( TIR will take place for Green Colour) Blue Colour µ 8 =1.47

y

I

·45

sin CG< sin45

Relative velocity of image with respect to object is in negative xdirection as shown in figure.

~

1

·cG= .4 < .4 =sIn sm 1 4 1 1

[CJ

v,

CR> 45° (No TIR will take place)

[v0 [ =[v,[ =~(2)' +(2)' =2-12 ms·'

46.

=1.39

. CR =-->-=Sin 1 1 . 45 Sin

[D] C

45.

1

i=e=~A 4

=>

44.

½) cos(½)=% i=2cos- (%)

n =2cos(

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Optics & Modern Physics

Advanced JEE Physics 51.

59.

[AJ

sine=

= vdllnsar

µrarer

µdenser

vrare,

=>

sin8 = vder,,;er

=>

sine=.!.

asa.

= vA

Vrarer

60.

Ve

53.

54.

[DJ

Applying Snell's Law µ = s!ni , we get s1nr

v,

=>

=> 52.

[BJ Since rays after passing through the glass slab just suffer lateral displacement hence we have angle between the emergent rays

sinr = sin I µ

[CJ Since, µv = constant

=>

µ 9V9 =~Vw

=>

~(2x10')=iv 2· 3 w

:::::>

vw =2.25x106 ms-

s R 1

=>

[DJ After two reflections from two mirrors placed at right angles, the emergent ray will always be parallel to the incident ray for any value of i.

.

sInr=

sin60°

-/3

=

1

2

=> r=30° Since, PC= QC => LCPQ = LPQC =Lr= 30°

[CJ

. Angle between reflected ray QR and refracted ray

Since light has to travel from denser to raser medium so, it must

other face Is 180° - r -60' = 90°

be made incident in the dens8r medium at an angle less than

as

at the

{·.· r = 30')

critical angle. 61. 55.

Since

==> 57.

[AJ According to Snell's Law

[DJ C oc 1. A.Red

> AVlolat

CRed

> CVlolet

sini sinr

[BJ According to the problem, we have, u = -(1-1)

sinr=....!_sini

=>

sinr = ) sin60 43

I I

''

µ.

r = sin-{

....: l

µ8

=>

V=-O+1) f =+f

: : :

=µw

62.

............... ..

"'" ....

' , -.._ ._

,.._

.......... ; .........

,.,_,_,___

;3)

3

[BJ LDBM=45' => DM = htan45° = h = 32 cm

A

sini smr

µ=-.F

_,_

I+---

.

smr =

=>

3 sin45° µsini = 4i'J = ,/2 4

tanr = sinr _ sinr cosr .J1-sin2 r 3

f + 1--+t

=>

. 1 1 1 ---=-, we get Appymg I V u I 1 1 1 --+--=-(1+1) 0-1) f

=>

=>

12 -21-1=0

=>

=>

1=("2+1) cm

63.

4.J2 tanr= J1- 9 = 32

3

m

CM=htanr=32x

_______ !! --- ---

':i; ~~,_ili ~; Stone

= 3

=20 cm v23 CD =DM-CM = 32-20 =12 cm

(CJ sinC=_!_=~ µ. 4

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Ray Optics =>

µA =constant

C=49°

Apex angle of cone is 2C

= 98°

=>

.

''

A

: -.;:ec,_c - -•:•c'~ :-'·-~~---: ____________'{9: - -49;-:-~~==== ____________ =====~==-------- __ -------:-:-:-:-:-:-, :-:-:-:-:-:- - - -- - - -F: - - - - - - - --

75.

In 6OAN'

ON cosr=OA 1 OA=--

••..•(i-r)

••·••••.••.. N'

67.

second image is formed due to reflection at X2 Y2

i l ··

N

ti ;{::]''

...

cosr

[CJ For grazing incidence and emergence, we have

76.

[CJ In both A and B, the refracted ray is parallel to the base of prism.

79.

[AJ . (A+D.) sm -

r, =r2 =~=30°

s1nr, 68.

sin(~)

=2

4 sine=.!\:.= / 3 µ, 3/2

_

C

.

=Sin

-1(8) Q

and light must go from denser to rarer medium.

69.

[BJ

71.

A.Red

1 (A+Dm) v1z=sIn - 2 60+D. = 45 2 D. =15 2 Dm = 30°

=>

> A.Vlolel

80.

[CJ Critical angle C =sin-

=>

2

Further we know that at minimum deviation i = A ~Dm

According to Snell's Law µA = constant Since

.J2sin30 = sin( A+ Dm)

"' "' "' "' "'

[AJ

-

2

µ

2 According to Snell's Law, we get µ ·= ~inl

and is

brightest all other images formed further are faint.

t

tsin(i-r)

i=e=90°,

[BJ

'---'-----"-!_.._.·~--'.....··'-'

cosr

6x

µ,

This happens due to multiple refractions and reflections. The first image is formed due to reflection at X 1Y1 and is fainter. The

[DJ sin(i-r)="" OA => 6x=OAsin(i-r) In 6OAN

= µaA.a

•

-----------------

64.

µaAa =·µgA.g

''

C = sin-1

1

(¾)

i=45°

[CJ Normal Emergence implies e=O Since i+e=A+D For prism with small A,

D=(µ-1)A

(i) =30°

=>

i=A+(µ-1)A

"'

i~µA

When A > 2C , the ray does not emerge from the prism. So, maximum refracting angle can be 60° .

81.

[CJ Since, 1 µ 2 x 2 µ 3 x 3µ 4 x 4 µ 1 =1

73.

[AJ According to Snell's Law µv = constant

74.

4µ3 x3µ2 x2µ1 x1µ,1 =1 4

3

2

1 µ,

µ3x µ2x µ1=-,-

[AJ According to Snell's Law

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Advanced JEE Physics 82.

Optics & Modern Physics

[CJ Let thickness of slab be t and real depth from first side be x. Then µ

=

i

88.

Apparent Depth = !!. n

(when viewed from first side)

90.

µ = t~x (when viewed from second side)

(·.-µ=¾) => => 83.

4

••. (1)

3X

From mirror formula, we get 1 1 1 ---+--=2(6+x) 6+x f

[DJ

=>

sin(~)

Since, µ

[CJ From mirror formula, we get 1 1 1 -+-=3x X f

=>

X=9cm 3 t-9 2 4 t=15 cm

=:>

[AJ

3 1 2(6+x) =1

... (2)

sin(~) cot(~)- sin(~) 2 sin(~)

=>

6cm ,....__ X __..,.__ _ 3 X

•

14-2(6 +X)+i

=>

COS

Equations (1) and (2) give 4 3 3x = 2(6+x)

(2A) . (A+D ) =Stn

~

=>

=> =>

=>

Shift of screen = 3x - 2 (6 + x)

9X =48+8X

x=48cm

Shift of screen = 3x48-2(6+ 48) = 36 cm

84.

85.

[AJ i+e=A+D => 60+e=30+30 =:> e=0° [BJ µ

µ=-12

=>

[DJ

Apparent Depth =

92.

sin(45)=.!

86.

91.

..i.. + ..i.. 2µ, 2µ2

[A] In this case, one of the image will be real and the other will be virtual. Let us assume that image of S1 is real and that of S2 is

virtual. Then, applyin~ .!+.! = ! V U f

[BJ

=tr f =9cm

sinC1 =E:r_=.&_=..!._ µd µg µg sinC2

= µw µ,

S, I

> sinC1

....,__ X

I

I

S, I

y -+I 24-x~

87.

for 8 1 , we get

[BJ

. S mce r= =>

4 r-~

r=4x3=3m 4 Diameter = 2r = 6 m

=>

1 1 1 -+-=y X 9

d

~

-yµ2 -1

.•• (1)

for S2 , we get 1 1 1 --+--=y 24-x 9 Solving equations (1) and (2), we get x=6cm

... (2)

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Ray Optics

a 4 -=-

CONCEPTUAL NOTE(S)

c 3

This questi~m may· have following ahsWer· 6 cm from S1 a~

18cm fr9m .52 and 18cm from.S; or6cm.fram S2 • 93.

98.

sin C

[CJ Optical path length = nt So time taken

µdenser

= nt

~

To a fish. the outer world is seen in a circle of radius

-

"µ2-1

=> ,d

99.

is the depth at which the fish swims.

:::::)

=> 95.

12 r=---

[CJ

~~-1

=>

µgVg =µtVt

36 r = ..ff cm

=>

( 1.5 )x ( 2 x10') = µ, (2.5x1 o')

=>

3 30 µ,=~=-=1.2 2.5 . 25

[AJ

1 sinC=_!_=-·

(Rememberthatµwatar

~

(4/3)

µ.

=±3.)

100. [DJ

. C

sm

1 = F2

:::::-

C = 49° with vertical

=>

C=45°

=>

0=90-49°=41° with the horizon.

For i

=C

[CJ

= µw

·.

101. [CJ

µg

µdenser

=>

in denser medium angle of refraction {n rarer fJ.1edium

is 90°.

sinC = µmm,

97.

. C =5 sin 6

According to Snell's Law µv =con_stant

For the diver

96.

4/3 5/3

Real Depth µ

Apparent Depth

4 5

sinC=-=-

=>

[BJ Applying Snell's Law at the interface sep8rating two media, we

get

=>

24 413 Apparent Depth = 18 cm Apparent Depth =

102. [CJ

Given that A

-y

__________ J I 1

_i;.

I

I I

I).

8t+bj

' :

i=(A~~m )=60°

µ,=1.5=§. 2

b

' i

x-z plane (i)'

103. [BJ

"' " j' r-~-·"-, : I

c

,, ••4L"

,I

I

r+i=90 i=90-r For ray not to emerge frOm curved

•

surface

'-f I

=·~

~ - - - - - y ( o r j) From the figure, we get

(¾)(~ )= (-/c'c+~') 2

a_7 + b}

~nd

ci + d}

= Om = 60°

At minimum deviation, we have

dl

Since

µA

=> · sinC= VA=_!_ V8 2.4

[CJ

.

= µrarer =,&_

Since, µ,._v,._ = µ 8 V8

C

94.

[BJ

i>C sini>sinC

=> => =>

sin(90-r)>sinC cosr > sinC

=>

~ >1

=>

1 _ sin i >_!_

=>

1 >-\-( 1 +sin' i) n

=>

n2 > 1+sin2 i

n

... (1)

2

~

are unit vectors, so we get·

-/a' +b' =-ic'+d' =1 Substituting in equation (1 ), ~e get

,_

{·: sinC=~}

n2

n2

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Optics & Modern Physics

Advanced JEE Physics {sini--->1}

n>-./2 least value = -./2

104. [BJ

I

CONCEPTUAL NOTE(S) Critical angle incre~ses as the relative refractive index • is decreased.

110. [BJ

Dispersive Power

dµ dl.

= :~

2B

= -)!

So, as B increases, dispersive power i~creas~s. 105. [CJ

For the near end of the rod applying

!f = ..:!.U + ..!.V

Since, here u and f are negative, so

lvl=~ u-f The farther end of the rod is at infinity, so its image will be formed at focus. Hence, Length of the image is £ = lvl- f

111. [BJ u=-20cm, f=+20cm, V=?

Since.!+.!=! V u I 1 1 1 => - + ~ = v -20 20 => V=10 cm Students generally give (A) as the answer to this problem because they have in their mind that if .an object is pl8.ced at centre of curvature C, then image is also formed at C, but this is true only for the case of a convex lens or a concave mirror. So correct answer is (B).

£=~-f=__f_ u-f u-f · 106. [CJ For A

(1.5)t Total number of wa:ves = -1.(

... (1)

Total number)= ( Optical path length) of waves wavelength

113. [DJ ·

For Band C Total number of waves n,(½) (1.s>(¥) =---+--~~ l. 1. Equating (1) and (2) :::::) n8 =1.3

... (2)

Since no parallax exists between the images formed by two mirrors (convex and plane) hence the images for both coincide. But for a plane mirror an image is as far behind the mirror as the object is in front of it. Hence the image for plane mirror should be 30 cm behind it or 10 cm behind the convex mirror. So for convex mirror. U=-50cm, v=+10cm Since.!+..!=! V U f

107. [CJ Focal length of curved mirrors is independent of the refractive index of the medium in which the mirror is.placed.

=> =>

108. [BJ C

n=v

1 1 1 -+--=10 -50 I 1 5-1 i=so

=>

f = 50 4

=>

f=12.5cm

0

·

i+-30 ,m_..:__30 cm~ 14---50cm------------+1

Since R=2f => R=25 cm 114. [CJ

Since, 109. [DJ

6,~. =(µ-1)A=(1.5-1)4°=2°

1 ·01 =1- an d sin ·02 = sin µg µ..,. Since, µg > µ .... , 01 < 02 The critical angle 8 between glass·water interface is given by

2'

sine=& µ,

2'

=1.202

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Ray Optics =>

6,,.,, =(µ-1)A+(180-2i)

=>

6,,.,, =(1.5-1)4°+(180-2x2°)

=>

()total

=2°+176°=178°

121. [CJ

sinC = 4000 6000

115. [CJ

==> 0 (I)

µ

(~)

123. [CJ

Ax=3[1-(i)f

C

f).alr

3

vmad

fA.me.d

2

µ=-=--=-

AX=1Cm

1 V Now, v=+Sm, m=--=3 u => U=-24 m

Object distance = 1O+ 1 = 11 cm

116. [CJ

Since,

Apparent depth =

1

122. [BJ A virtual, erect image is obtained by using the mirror. The mirror can be both concave and convex. But a virtual image obtained by a concave mirror is always magnified and hence the mirror must be convex as we are getting diminished image.

i.3 cm+1 / ><-a-::-"'<+---10 cm--><

:::::) ==>

C=sin-

.'! n

Hence apparent separation =

=>

2 d

n

=>

! = .! __!_

f V U 1 1 1 4 -=-+-=f 8 24 24 f=6 m

For a piano-convex lens 117. [BJ According to Snell's Law for anisotropic medium n,=1

n( 0 ) sin ( 90-0 ) = constant or

f =_I'!_ µ-1

Interface

6 =_I!_

n1 sini=n(8)cos8

=>

(1)sln90 = n(0)ccs0

=>

n(0)cos0 =1

n(e)

118. [DJ Velocity of approach of man towards the bicycle =(u-v) Hence velocity of approach of image towards bicycle is 2(u-v). 119. [CJ Optical path length = µt

=>

0.5 R=3m

124. [BJ Since rays after passing through the glass slab just suffer lateral displacement hence we have angle between the emergent rays as a. 126. [AJ As shown in figure the distance between the lenses should be 30 cm.

Time taken = µt C

120. [CJ

Optic Axis 11 =20 cm fz= 10 cm

0

127. [CJ Two piano convex lens of focal length f, when combined using

optical glue will give rise to a convex lens of focal length (

L Optic Axis

i.e. size of image = size of object. => Object at 2F i.e. at a distance f from optical centre. 128. [BJ

0

~Hon

:,.--.1

:Ax: L

-+ OO

1 Since - --=.!+..!-~ fcomblriatlon ~ i ~1z

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Optics & Modern Physics

Advanced JEE Physics 134. [CJ

·0 =-1-+I __1_0_ -10 t (-10)t

4=Iµ

o--1- + I + I -10

t t

1

2 io = ~ t=20cm

129. [DJ Since, LABO= LOAB = C Further,. by definition, we have

µ

3

Applying Snell's Law at A, we get sini 3 sinC =2

=>

sini = (¾)sinC = (¾)(~) = 1

=>

i =90°

and

6=d-t µ

=> => =>

6µ=d-4µ

' ,..,, /

,

0

PA=('')'.= 4 =0.44 Ps rA - 9 136. [CJ

An 'achromatic combination' is made from a concave tens and a convex lens with greater power of convex lens so as to make the rays converge at a point. 137. [CJ ·-. p

A.2 =(~)A n2-1 1

=>

A =(1.54-1) 4• 1.72-1 '

=>

A, =¾(4•)

A

·· ..r1

130. [CJ For combination to produce dispersion without deviation. (n,-1)A,=(n,-1)A,

=>

Real Depth } Apparent Depth

D

d=10µ d=15cm

135. [AJ

', A ~,.,,, i Cfi,.c._-+-,c~

',

sine==.!= _g_

.. { · µ

0

0_••.

··C

µ

sini -_-=µ (atP) s1nr1 and r1 +C.=A

131. [CJ

~+_!_= µ-1 co -u R

=>

sini=µsin(A-C)

=>

sini=µ(sinAcosC-cosAsinC)

=>

sini=µ(sinA~1- µ~ -cosA¾)

=>

i = sin-• { ()µ' -1 )sin A - cos A}

138. [DJ At U=f

u = ___!!_ from the surface µ-1

... (1)

... (2)

{·: sinC=¾}

(focal length}

V -too

Total distance frOm centre Al

{i.e .., object is at pole} (image is also at pole} Satisfying these two conditions, only OPTION (D) is correct.

x=lul+R=R+~ µ-1

U=O

v= O

X=R+2R

X=3R

139. [CJ

132. [DJ

An air bubble in water always behaves as a concave lens and hence is always incapable to form a real image.

133. [BJ

f+(-12)

=> => => =>

·

f+(-20)

f I --=--f-12 1-20 f-20=-f+12 21 =32 f = 16 cm

Considering pole at P, we have µ 1 µ-1 -+-_=-2R 00 --R

=>

!'.= µ-1 2

=>

!'. = 1 2 µ=2

=>

1.204 i;::::::::::::::::::::::::::::::::::::::::;:::;::::::::::::::::::::::::::::::::::::::::::::::;::::::::============::;:::::::::::;;:::;:;:::::::::::::::::::::::::::::::::::::::::::=

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Ray Optics 140. [CJ

143. [DJ

The first image is formed due to the reflection from concave mirror M2• 1 1 2 => -+--=v, (-2R) -R ::::::,

1

1 2R

6x=2(1~t} 1 6x = 2(1---)c1.5) = 1 cm 1.5 Therefore, final image will be 1 cm above point P .

4 2A

-=--.V1

The two slabs will shift the image by a distance

-3

144. [CJ

p

Since

PRIIN, =:-

m,=-[ !J=-¾

=> => =>

::::> Radius of image circle= ¾a, (in magnitude).

Now the second image is formed by convex mirror M1 • The second image will be formed because the image formed ~y

the first acts as object for it. ::::>

Ob'Jeet d'rsance= t 2R - 2R

3

1

= 4R .

(-:R)-(~)

1 2 3 -=-+-

=>

V2=11

=>

R

V2

B T

u

5i=180 i::::36°

=> =>

A= 36°

S'mce tanus =t h

-+-----v, =>

A+2(2i)=180 i+4i::::180

145. [CJ

3

1

1

a=2i

Also A+2a=180°

f

f=-h-

=>

2h-----f---)''-5),-;=F-

tanO

! ~----+ef,1

Further

4R

4R 4R m =-v2= -11 =~ 2 u2 4R 11

=>

h f=--

(µ-1)A

If no lens had been there the rays would have met at A. On inserting the lens the rays meet at B. Thus A acts as an object for the lens.

=>

Since.]_ _ _!=! V U f

3 a a · =

Sorad'1usof secon d'image= a2 = 11

ima9e is

o=(µ-1)A

146. [A]

3

Similarly radius of third

=>

a3

;..-- 24 cm-----+1

3 11

=~ A

141. [BJ r+i=90 i=90-r For ray not to emerge from curved surface

=> => =>

i>C sini >sinC sin(90-r)>sinC

i- v -Iii ~-~.;..----,_ ::::>

cosr > sinC

n

=>

~ >1 n

=>

1 _ sin i >_!_

·: sinC=¾)

n2

=>

1> 1;(1+sin2 i)

=> => =>

n2 >1+sin2 i

n

-----1===1 00 .fi

I,

=>

1 1 v-c-o.3) = 0.2

=>

V=0.6m

A's OC1C2 and 01112 are _similar

n>,/2 least value=

147. [DJ

v-u=t

2

n2

::::::,

u=+24cm, v=?, f=+24cm V=12 cm Distance AB=OA-0B =24-12=12cm

{sin i---> 1)

J2

0

Hence 0.001

l1l2

0.3

0.3+0.6

I, l+0.3m+!

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11 acts as object for mirror M2 and gives an image 1; and similarly 12 ' acts as object for mirror M1 and gives image 1;.

148. [DJ

Since 11 is located at a distance 3b from M2 •

A convex mirror can never form a real image.

So, 1; must be at a distance 3b from M2 •

149. [AJ

Similarly

Given, µ=-,/2, A=60°, 6=30° For the minimum deviation, we have

153. [CJ

Substituting the values of A and µ we get

1 (4 1 )' 8=2(2,)

=30°

6m

3b from ~.

from O. Second image is at a distance 4b from O and so on. So, nth image is at a distance 2nb from O .

sin(~) sin(i)

µ

1; must be at a distance

From above we observe that the first image is at a distance 2b

i.e., the given deviation 6 = 30° is actually the minimum deviation Sm • At minimum deviation the ray inside the prism is parallel to the base of the prism (in case of an equilateral prism).

(·.-e=Jut')

a= .2;_ 16

Reflected ray_ turns through an angle 20 =22.5°

154. [CJ Distance of first image

151. [BJ CASEI

(I,)

formed due to refraction from the

plane surface of water is

::::: dreal = ..!Q_ =7.5 cm "'' µ 4/3 Now distance of this image from the plane mirror is x =5 + 7 :5 = 12.5 cm . Therefore, distance of second image d

0

n=

Real Depth Apparent Depth

(1 2 ) from the mirror will also be equal to 12.5 cm from it.

4 3

155. [DJ

CASE II

a

p n,

j'·· B A

~./r2

·.f1

V

~

P=90-r2

T y=90-r3

R sini = n, sinr,

1-(~)

1

-(-4)+(-2:r_R_

=-

~=(n-1)(~

!=(i-1)(..!. __ 1) f 3 ro-25

:::::,

f=75cm

152. [CJ

I,

=>

2

... (1)

=llz sinr2

2

2

n! sin r2 = n~ cos r,

... (2)

At!l. n2 sin( 90 - r2 ) = n3 sinr3 :;,.

tt O

M2

11 is the direct image of O by mirror M1

n! cos 2 r2 = n~ sin 2 r3

... (3)

n, sin( 90 - r,) = ( 1)sin( 90-i)

I,

I', 0

·---:-2b___,:._b b.:b b..:_2b-+< M1

2

n1 sin( 90-r1 )

-~J

=>

sin i =nf sin r1

AIC

:::::, R=-25cm According to Lens Maker's Formula

l',

r/ .·

s o.=90-r,

~+ "2 = n2-n1 -u

n,

R

Consider origin to be at the pole P.

E _.-0,\ .. 9 F

o· -{:r

"·.·

n,

. 0

p"·. •.

2 => cos 2 i = n~ cos r3 Adding (1 ), (2), (3) & (4) => 1+n!=nf+n~

... (4)

156. [AJ

P=(µ-1)(..!..-..!..J R, R 2

12 is the direct image of O by mirror M2

= c::::=================================== 1.206

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Ray Optics 158. [DJ (90-r )+(90-C)+45 = 180 :::::,. =>

3 5 v"= 51-51

r+C=45° r=45-C

1 2 v"=-51

51

Since sini = n sinr sini = n sinr

V=--

2

PA'= 51

sini=n sin(45-C) sini

2

=n(sin45cosC- cos45sinC)

OA'=~-21 2

sini= h"(cosC-sinC)

DA'=!_ 2

f

DA' 2 3 m=-=-=OA .!. 2 3

162. [DJ For a plane refracting surface, the lateral magnification is 1. So, the image of the coin will be of the same size as the coin itself.

Further, sinC = ~

163. [DJ

n

:::::,.

sini=

~[~1-~!-~

6=~=

60

2

1 ]

:::::,.

. . = ../2 1 [ vn ~ srn1 -n 1 -n 1]

2

=30

The ray passes symmetrically through the prism parallel to its base.

Hence 0 = 0°

. .-,[.fnCnf-n,J .J2

164. [CJ

l=Sln

RAY2

159. [DJ

f, +f, =36 cm

... (1)

.l,_=5

... (2)

f, :::::,.

PA2 E

0:

PA1 'A ~ f2 -..l

RAY1

fo=30cm and fe=6cm.

~ d ----+! ----x---+<

160. [BJ

------f,------+I

For drawing the above ray diagram, we must consider two rays RAY 1 and RAY 2 in such a manner that RAY 1 passes through optical centre of first convex lens and is parallel to the Principal Axis for the second convex lens (PA2).

1 2 1-2 v-(-15) = -10 1

2 15

1 30

-=---=-V

:::::,.

10

So wherever the RAY 1 cuts PA2 is the focal length

V=-30 cm

161. [CJ Since an elongated image is formed and it touches one end of the rod, so the rod must lie with one end at 2F and other end between

2F and F (shown in figure). For end A, u = Since..!+.'!.=~ V

:::::,.

U

I

1

3

1

V

51

-I

---=-

51

3

t

of

second convex lens. Similarly we have RAY 2 parallel to PA1 and let this pass through O' so as to cut PA1 at a distance equal to focal length ( i) for the first convex lens. 6.IDE and LiO'AE are similar

f/3 ,._..

A'

0

A

... (1)

ROD

p

F

LilCB and AO'B are similar· /1-y

(t +d)-x

/1

t

-1-

14-2f-J>t 1+----

fil ----f,I 3

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From (1) and (2)

\-x

d

X

i;-i; = \-d

171. [A]

f

)(-¾)

=U.:s- 1

=>

1 0.25 2 -=:--Xf' 1.75 R

=>

f'=~(175) . 2 25 f' = 3.5R

=:172. [OJ

For refraction at plane Surface. 1 1.5 1.5-1 ---+-=--(-mR) V oo =:v=-1.5 mR For refraction at the curved surface u'=-(1.5mR+R), v'->oo

Similarly we can calculate value of y . OBJECTIVE TRICK If t:. = 0 , y must be zero (satisfied by A, 8, C) · If d-> \+~, then x->"' (satisfied by B, C, D)

=:,

If d-> \, then x-> \ (satisfied by C, D) So, we observe that OPTION (C) is .the only option satisfying both the ,x and y in special situations mentioned above.

Let d' be the diameter of refracted beam. Then d = PQcos(60°) and d' = PQcosr d' cosr :::::,. -=---=2cosr d cos60° :::::,. d' = 2dcosr

1.5 -(1.5 mR+R)

"' =:,

-3 (1.5 m+1)

=:,

3=1.5m+1 1.5m=2

=:,

m=-

=:,

166. [CJ

1

1-1.5

-R

1

4 3

174. [CJ

Deviation by a sphere is 20-r) Here, deviation s = 60° = 20- r) :::::,. i-r=30° => r=i-30°=60°-30°=30°

AccOrding to Snell's Law, we have µ

= s!ni

s1nr

sin60° ,::;3 µ=--=,~ sin30° 175. [AJ

Fa

. sini 2 1 Since, sm r = = _3_ = Fa

µ

2

Also,. cosr = .J1-sin2 r

=> . cosr-=-J¾ cc:,·· d'

:::::,.

= (2)(2)JI

sinu sinr

,

=4JI

ct··~ 3.26 cm

=.!l n2 sina = .!lsinr

cm

n,

... (1)

For TIR at other end

sinC= " 2

167. [AJ

n,

1

1

1

X

f

I

~

I~

-=-+---

Also C=90-r

sinu =.!lsin(90-C)

n,

170. [AJ

1 1.5 1.5-1 --+-=--(-x) x R 2.5 0.5

. =:,

{Put r=90-C in(1))

sina.=~cosC

n,

x=R X=5R

1.208

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.'

Ray.Optics

176. [DJ

181 .. [BJ

CD is parallel to AB, so both m$dia--must .have equal refractive i · indices.

177. [CJ Since all the prisms P , Q arid R are made of same material, so the deviation suffered· by the combination remains the same as the combination forms a part of the bigger sphere. ·

·

__
- ,/3 10,/3

Total number of reflections

178. [BJ Let .R be the radius of cmvature of each surface. Then applying

=>

Lens Maker's Formula, we get

1f =(1.5-1)(.!.R +.!.) R =>

R=f

For the water lens, again applying ttiB Lens Maker's Formula, We

2

I'

3f

Since, we know that for lenses placed in contact, we have

1

1

1

1

2

Ca~)

n=3O

'

A== 2(36')

.

f

(90)' -(20)' 4(90) f=21.4 cm

184. [CJ .

4

F=T- 3t = st

sin·i

Smeeµ·=-.smr

=>

179. [BJ sini sinr

d

183. [CJ f= 02-x2 4D

=>

1 1 1 1 -=-+-+F f f I'

2

=>

=>

·-=-+-+' F. I t t,.

1

2,/3

182. [CJ Amu. =2C

.

get

=>

I

= n=-=---

i=45°

185. [CJ

n

Since

tanr= 2 h_=1 2h

=>

r=45° . • h srnr = h.JS

' ' 1 srnr = .JS 1

1 .JS ;; =-1-

.

-./2

/

·1?

,!"

,'; 0 0 o.,g,;~==~~=

0 r<--2h---'l

a

LOPCl = LOQP =0 => 30 =180' => 8=60°

186. [CJ

n=J%

180. [CJ

2=(µ-1)·(-1.._...:!...) f R1 R2 For no dispersion

d(D=O

-

0 C' C

1 0cm

AB'= Apparent Depth= x AB=1O cm

The face BC appear's to be shifted to B'C' . If x be the apparent

depth, then AB'=x and B'B=(1O-x).

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Optics & Modern Physics

Advanced JEE Physics Also, according to Laws of Reflection an image is formed as far behind the mirror as the object is in front of it => 20+x=23.2+(10-x) => 2x=13.2

=>

x=6.6cm

. Smce µ=

=>

Real Depth Apparent Depth

f

-=-3 f+u {Since we get a real image)

=>

10 µ=-=1.51 6.6

=> => =>

187. [AJ Since µ2t = µl

=>

. ( - f )' m -9ar f+u

_f_=-3 f-40 -3f+120=f 4f=120 f=30 cm

194. [DJ Axial Magnification is given by

v'

m = I along PA aK O along PA

dX=t'-t

188. [BJ

m., =¾=( f~J'

=>

i = b ( - )' f-u

1

195. [BJ From the figure it -is clear that the angle between incident ray and the emergent ray is 90° . Emergent Incident ray ray

By Laws of Reflection i=r

Also i+r=90 => i=45 According to Snell's Law sini µ=-.smr

=>

u'

Air Glass

1_5 =!!_=sin45 2

=>

sinr

189. [AJ Shift=i(1-;) away

196. [BJ easel n = Real Depth Apparent Depth

=± 3

190. [BJ According to Newton's Formula, we have X1X2

=f

=>

(10)(40)=f

=>

f2 =400

=>

f=20cm

191. [BJ A=60° i=55°' 8=46° Since i+e=A+D => 55+46=60+D => D =41° So, Dm
Case II

0 Consider origin to be at the pole P.

..!:!!.+ "2 = n2-"1 -u

R

V

±

1-(±)

_3_+_1_= _ _3_

192. [AJ At D =D. i=e

-(--4)

(-

2:)

R

R=-25cm According to Lens Maker's Formula

193. [AJ Areal Magnification = 9

~=(n-1i(~-~)

= ================================== 1.210

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Ray Optics =>

1) !=(±-1)(1-f 3 oo-25

=>

f=75cm

magnification unity the objeci must be at 2F(or C) and so image is also formed at 2F ( or C) . H€nce u =-2f = -40 cm 203. [DJ The lens will converge the rays at its focus i.e., 30 cm from the lens or 20 cm from the refracting surface, so we have PI 1 =20 cm

197. [DJ According to Lens Maker's Formula

!f =(µ -1i(J.. __!_) R R 1

=>

=>

Now, PI,= µ(PI,) =

(µ-1)

Since

=>

2

1 foc- µRed

3

2

x20 = 30 cm ~

30 cm -----+1

< µ~"lolel

ted > (.,.,let ted > t1ue

Always keep in mind that whenever you are asked to compare {greater than or less than) u, v or f· you must not apply sign conventions for comparison.

10cm

198. [DJ Since, the refractive index is increasing linearly from top to the bottom, so the light cannot travel in a straight line in the liquid as

shown in options (A) and (B). Initially it will bend toward_s normal and after reflecting from the bottom it will bend away from the normal as shown below.

!

i Rarer

Rarer

! Denser

! i

!=(µ-1)(1-J...) f R -R ... (1)

~-(µ-1)(¾)

=>

II Denser

On cutting

1=(µ-1)(1-1) r R "' =>

f=2f

205. [CJ When a lens is cut parallel to principal axis its focal length remains the same. Hence each part will have a focal length 2f.

f m=-f+u

=>

204. [CJ

j

199. [AJ U=-(f+X)

=>

Hence, the rays will converge at a distance of 40 cm from the lens.

f m=---

f-f-x f m=--

x

Negative sign confirms that real magnification is negative. 200. [DJ For a convex mirror, both (A) and (B) are incorrect. 206. [BJ According to Lens Maker's Formula

201. [DJ

U=-{f+x,)

.1__.!_ =(µ-1)(1._J...)

v=(f+x,)

v

=>

1 1 1 --+--=-

=>

1(2f+x,+x,)=f+(x,+x,)f+x,x,

~

~

f

f+X 1

R1

R2

The lens is made of two materials, so for a single object distance, two different image distances are obtained i.e. two images are formed.

Since..:!._..:!_=~ V U f f+X2

u

f

208. [BJ Let ~ and

t

be the focal lengths of the lenses of refractive

indices µ 1 and µ 2 in water, then

=X1X2

·f=~X1X2

202. [CJ Two piano-convex lenses of focal length f on combining give a

convex

lens of focal length

-½.

... (1)

To obtain a real image of

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Optics & Modern Physics

Advanced JEE Physics ... (2) Agding equations (1) and (2), we get

1 1 2(µ,-µ,) -+-= \ t µ.R => =>

::::::,

=>

...!.. = 2(µ, -µ,) 30

µ.R

(µ1-µ2 )

µ.R =so

ill<= 1 cm Object distance = 1O+ 1= 11 cm

_.:....... -··

.. -···

Substituting the .values, we get 4x15 1 (µ,-µ,)= 3x60 =3

0

C

µ

(I)

14-3 cm+i

209. [AJ Since image is formed on the screen, hence it must be real. So, Lens used must be convex {and canhot be concave) u=-45cm, V=+90cm Since, for a lens

1

1

<--><<--1 Oc m ~ t,x

214. [CJ

1

v-u=t

1 1 1 90 45 f => f=+30 cm Further again the positive sign with focal length indicates the lens is convex. ·

=>

-+-=-

215. [CJ

A

210. [BJ

a

o·

Size of Image =.":(Size of Obje.ct)

u

ih B

=>

l=~(5)=-"10cm -45· . Here again the negative sign with below origin i.e. inverted i.e. real.

indicates that it is f0rmed

u=40cm

v=40cm

f=20 cm, u=--40 cm 211. [CJ

~+~= µ2-µ1 -µ

V

=>

1 1.5 1.5-1 -(-50) + ~ 20

=> =>

R

v

...!..+12= 0.5

50 V 20 1.5 1 · v=4o-5o 1.5

10

v = 15x200 =+300 cm 10 Positive side indicates that image is real.

=>

212. [CJ

r2 =Oo =>

r1 =A=30°

and

i1 =60°

1 1 1 ,=v-u

1 1 1 1 1 vfu2040 => v=40cm The situation is shown in figure. From similar triangles PAI and IQB 5 40 h 20

-=-+-=---

=> 216. [CJ Since 0 2 = 1112

=>

4 = (0.5)!,

=>

l 2 =8cm

217. [CJ According to Newton's Formula

x1x2 =f

So, from Snell's Law, we get

µ = sini1 = sin60° = .Js sin~ sin30° .

. gives.

=>

f=.J16x25 =20 cm

= ==================================== 1.212

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Ray Optics 218. [AJ

222. [DJ For real image U=-U1, V=-2U1, f=-20cm

(~)-(~)

cos(1B0'-20)

1i+~11i-;13]1

!,

Substituting in ..:!. + .:!. = we get V U f

11

----=--2U U 20

'' '

1

''' ''' ' 0 ''0

=>

For virtual image u = -u 2

''

=>

(1-3) -cos20=-41 -1 -cos20=2

=>

cos20=..!

=> =>

20 =60'

''

''

=>

'

=>

2

=> =>

9= 30°

220. [AJ Focal Length of Lens is f1 =20 cm . Focal length of combination is fc =20+5=25 cm

1

1

U2

20

u2 =1Dcm u1 -u2 =30 cm-10 cm=20 cm

f= -40x25 40-25 f =-66.67 cm 10 p f(in cm)

1.5 D

=>

P=10D 1 f=p=0.1 m

=>

f=10cm

225. [BJ

when the space between lens and mirror is filled with a liquid then

Object is placed at a distance of 2f from the lens of focal length f i.e., the image formed by the lens will be at a distance of 2f or 20 cm from the lens. So, if the concave mirror is placed in this position, the first image will be formed at its pole and it will reflect all the rays symmetrically to other side as shown below

MO••~,''°~., oo,_, ~, •• OOj)

1 1 - =-+f' I, f 1 1 1 -=-+25 f, 20

1 25

1 2U2

224. [AJ P=P,+P2

f=20 cm As object distance = 40 cm Image distance = 40 cm

1 f,

=2u2 , f =-20 cm

223. (AJ 1 1 1 -=-+f 40 -25

1 !f =(1.5 -1)(..!. _.!.) + (1.2-1)(-oo1--14 --) 14 oo

=>

v

So, the distance between two positions of the object is u1 - u2

219. [BJ

=> => =>

,

. I . 1 1 1 Aga1nappy1ng -+-=-,we get V U f

... ' 180'--20.

=>

1

u, =30 cm

1 20

o

ft=-100cm For this liquid lens

I

I)

14--- 20 cm ---+I+- 20 cm ---....i

f,=(µ-1)(~ -~)

.!.)

=>

__1_={µ-1)(-1__ 100 -33 oo µ-1=0.33

=>

µ=1.33=3

=>

4

226. [DJ Shift=d=t(1-¾) =>

l=~

n-1

227. [AJ Focal length of piano convex lens using Lens Maker's Formula is

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Optics & Modern Physics

Advanced JEE Physics

7=(¾- 1)C~ -~)

235. [BJ

:::::> f=20 cm If point object O is placed at a distance of 20 cm from the piano convex lens rays become parallel and final image is formed at second focus or 20 cm from concave lens which is independent

of y.

0

228. [DJ Focal length remains unchanged. I oc Area of Aperture

Since angles opposite to equal sides are equal. So, in MOB We have LABO =p . In .6.CAB , external angle equals the sum

I =kD'

of internal opposite angles. So,

D=(a-P)+(a-p)

I'= 31

=>

D=2(a-p)

4

236. [CJ

229. [BJ

When magnification is m = +2 , then we have U=-X

1'=~ f In moving from air to glass, f remains unchanged while· v

V=-2X

f =+20

decreases. Hence, A should decrease.

Applying Mirror Formula,

230. [BJ f=--xm1 -m2

(See 'Displacement Method' in SYNOPSIS)

231. [BJ

02 -x2

f=-4D

=>

234. [BJ 1 2 1 -=-+F f, I,,

:::::>

1

1

237. [DJ When water is filled in the mirror, a piano convex lens is formed, so now combination contains a piano-convex lens and a mirror. The effective focal length of combination is less than the focal length of above mirror, so image is shifted downwards.

239. [DJ

1 2 1 -=.-+ft

1

-+-=2y y 20 => y=30cm So, the object has lo be moved by y-x=20cm

When 8 < C partial transmission and reflection will occur. When 8 > C , only ieflection takes place.

~

V=+2y f =+20

=>

232. [CJ

;:::}

1 1 1 -+-=-2x X 20 ::::,. x=10cm To have a magnification of m' = -2, we have· U=-y and

f= (0.9)' -(0.2)' 4(0.9) f=0.214 m 1=21.4 cm

.!.V - .!.U =!1 , we get

The given lens is a convex lens. Let the magnification be m, then for real image, we have

00

1

F=!t_

1

1

-+-=mx x f and for virtual image, we have

2

Also

.!. =(µ-1)(J_ _ _!_) t --f 00

-r

1 1 1 --+-=-my y f From equation (1) and (2), we get

ft=--

.. ,(2)

I= x+y 2

µ-1

F=---r2(µ-1) Negative sign indicates that mirror is concave.

... (1)

=>

241. [BJ Let 11

(i) (ii)

(iii)

,

12 and 13 be the images formed due to

refraction from ABC reflection from DEF and again refraction from ABC

-...1.214::=:===================================

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Ray Optics A

D

I,, =180 cm

=>

1 2 2 1 -=-+-+F 60 180 10 1 13 => -=-cm F 90 90 cm => F= 13 So, required answer to get asked condition is 2F. Hence 180 u=-cm 13 =>

0

B E

14-5 cm

C

•u

F

•1

2.5cm

Then BI,= (5)µ 9 = (5)(1.5) = 7.5 cm Now EI, =(7.5+2.5)=10 cm =>

EI2

=1O cm

246. [CJ For equiconvex lens, we have IR,l=IR,1=1=10 cm

behind the mirror

Now BI, =(10+2.5)=12.5 cm =>

Now P=2P, +P•

BI = 12.5 = 12.5 = 25 cm 3 µg 1.5 3

=>

_.! = 2(µ-1)(J__...!_)-~

=>

1 _.! = 2(1.5-1)(J_-F 10 -10

F

242. [CJ

02 -x2 f=-4D

(By Displacement Method)

=>

P=~

=>

P=~ 1-0.16

=>

p =__±__ 0.84 P=4.76D

=>

02 -x2

1

=>

~J

=> =>

=>

2

1

1

1 1 -;;- (-40)

V 15 40 V=24 cm

Applying µ 2

_

V

and

2

1

-=-+F 60 10

=> F=7.frcm For image to be formed at the same place where object is situated we have u = 2F = 15 cm 245. [DJ 1 2 2 1 -=-+-+Ff, I,, i,, (Because here we have two refractions at the concave surface of tens and two refraction for water lens). Since

1

_& = µ 2

u

-

µ, , we get

R

1 1.5 1-(1.5) -;;- (-u) = ~

20 t,=-=10cm

t=(t- X6~ -±J .

1

=15

248. [DJ

(1 '1J

1 · :r,=(1. 5 -l) 20-60

2

F=i=15 cm

v-u=F

Shift= 2 cm (downwards)

1

t, ~ co

Further 1

.!= (µ-1)(J__...!..) t R1 Rz

=>

2 1 f, i,,

Since .

t =60 cm

-J-~

-10 => F=-2.5cm , Therefore, the system will behave like a concave mirror of focal length 2.5 cm.

F

244. [AJ 1 2 1 -=-+F f, i,,

=>

R2

247. [BJ

Shift=d(1-¾J=1(1- 1

=>

R2

-=-+-

243. [AJ

=>

R1

obje~t

1 3 1 => -+-=v 2u 2R For v to be positive, we have 1 3 ->2R 2u

=>

U>3R

249; [BJ ~inaUon -), OO

1

Since - --=.:!.+.!-~ fcomblna1ion ~ ~ ~~

=>

0=-1-+.! __1_0_ -10 ~ (-10)~

==================================== = 1.215

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Optics & Modern Physics

Advanced JEE Physics 254. [DJ f . Smce m=-f+u

0=-1-+.!.+.!. -10

=>

t t

1

2

=>

10=1

=>

t=20cm

=>

-1=

,:u

==> ==>

-f-U=f U=-2f When the lens is cut, the focal length of the remaining portion is

250. [BJ f f f+(-12) f+(-20) f-12=-(f-20) f-12=-1+20 21=32 f=16cm

21. Applying Lens formula, we get 1 1 1

.-(-21) =21

==>

V--",-oo

255. [BJ 251. [CJ

1 ( R, 1 -R, 1) 10=(1.5-1)

5=(o.5>(~ -~) ... (1)

=:-

_!_ _ _!_=10

=>

R1

On immersing it in a liquid of refractive index 3

R2

-1)( ~ -:,)

-1=('µ,-1)10

~=( 'µ,

=>

(::-1]=-0.1

~=(\5-1)(¾)

=>

~=0.9 µ,

=>

f 10 => f=-10 cm Negative sign indicates diverging nature of lens in the liquid.

=>

256. [CJ Since, we have, at time t d = y+ µ(h-y) = µh-(µ-1jy

252. [DJ lul+lvl=D V

m=u => v=mu lul+mlul=D D => l u l = m+1

=>

f

U=-(m~1)

r

Since image formed is real, so it must be on positive side. Hence lvl=+v= mD m+1 For a lens

1

=>

m+1 m+1 1 --+--=mD D f

=>

f=~ (m+1)'

253. [AJ 1 (1-1) -=(1.5-1) f R eo

=>

If A is the area of the tank, then we have

at y =,;:_

{·: Ay = at}

1 1

v-u=t

=>

f !

h-y

_!_ = 0.5 .!_ 16 R R=8 cm

d,;µh-(µ-1)at A i.e., d- t graph is a straight line with negative slope and positive intercept. But d becomes constant once y =H.

=>

257. [BJ On immersing in water f increases and hence P decreases. 258. [DJ 1 1 !=(1.5-1)(- - -- ) f 0.5 -0.5

=>

!f = o.s(..3..) 0.5

=>

P=2 D

= ================================== 1.216

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Ray Optics 259. [AJ

266. [BJ Rays from O must fall normally on the mirror, only then the lens forms a virtual image at C (the centre of curvature). C

0=$i; 0='14x16 =8 cm

f

260. [A] For a lens making real image I -m=-

36cm

l+u

=> =>

-mf-mu=f

=>

U=-{1+¾)

=>

lul={1+¾)

--mu=l(1+m)

For lul to be MINIMUM m must be MAXIMUM i.e. m-, oo

=>

lulmin = f

=> =>

1 -+

1

-(36-12) x=15cm X

1 40

For lul to be MAXIMUM m must be MINIMUM i.e. m = 1

=>

267. [DJ 1

lul.,. = 21

1

1

F= (1/2) + (1/3) =5

261. [A] The ray diagram is as shown in figure A

=> F=0.2 m=20 cm Since

1 V

C

0

u

F

=>

1 1 ,-(-30) = 20

=>

V=60 cm

- - - 21 _,...;.___ 21 ____,..

268. [BJ Pcomb =P1 +P2 -xPf2

Since triangles CAI and NEI are similar, so we have

h I d/2 = 21

0=.!.+..!-~ I ~ I~

2h

=>

ct= 2

=>

h=~ 4

=>

-X=I+~

270. [DJ Apparent Separation = 2(Apparent Depth)

265. [AJ If mirror would have been absent then image is formed on the other side of lens ( at I'). So V

=>

-20

=>

Apparent Separation

= 2h µ

271. [AJ For distant vision u = --00

15

v=60cm

=>

-2-:;-=7

=> l=-2 m => P=-0.5 D For near vision u=-D=-25 cm

1

=> Since the mirror reflects the ray back, so O' serves as a virtual object and forms a real image I in front of mirror.

1 1 1 -+-=v -60 15 =>

v=+12cm

1

1

-1- -o.25=f

=> P=+3 D Hence he must use bifocal lenses with P = -0.5 D and additional+ 3.5 D (to give +3 D net)

273. [CJ In the first case. Let x be the distance of object from the mirror. Then

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Optics & Modern Physics

Advanced JEE Physics 277. [CJ

u=-x, v=+2x and f=-f

.

. 1 1 1 Usrng v+u=f,weget 1 2x

1

vv1025

m=m0 xm =-9..x---2.=-x8 U0U82,55

=>

1

-x=-1

f 2 In the second case, let y be the distance of object from the mirror. Then

=>

X=-

=>

u=-Y, v=-2y and f=-f 1 1 1 -2y

m=20

278. [BJ

1 (Shift),= 1(1-~J =9(1---) =3 cm µ, 3/2 (Shift),= 1(1- ~:) = 9(1-%) = 1 cm So, distance between two images is given by

-y=-,

(Shift), -(Shift), =2 cm

3 v=-t 2

279. [AJ

So, object will have to be moved by a distance of y - x = f .

m = m0 x ffi 8 =m 0 (

1+{'J

274. [AJ

1=-5

t

Separation =t + t 36 =--4t =>

fa=-9cm

=>

t=45cm

280. [CJ

m=1+~ f

275. [DJ For lenses placed in contact, we have 1 1 1 -=-+F \ t

J=(µ,-1l(±+i)+(µ,

-1)(~ -±)

.!_=µ,-µ2 F R

281. [CJ f

m=...!!..=10

t,

=>

fo=50cm

Separation =

t +f

= 55 cm

8

F=_R_ µ, -µ2

282. [AJ

m=1=5

276. [BJ V8 =-D=-25 cm,

1 1 1 -·=-=--+-

t

1 u,

Ve

t

t =6.25 cm.

Since Ue

1 1 =6.25- (-25)

U8 =5Cm

=>

,

V0 =15-5=10 Cl!1

283. [A] Separation= t + t Separation= 0.3 + 0.05 = 0.35 m 285. [D] For an equilateral prism, we have A=60° Since, the ray inside the prism is parallel to its base, so we have the condition of minimum deviation. So, i = e = 60° and 6m = i+ e-A = (60°+ 60°)-60° = 60°

Also, t =2 cm and 1

1

1

t

Vo

Ua

2

10

U0

-=---

=>

sin(A+6m) 2

Since, µ =

sinrn) =>

µ =sin(60') sin(30')

=-J3

u0 =-2.5 cm

= ==================================== 1.218

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Ray Optics 286. [AJ .

f

.!_ = (µ,

\

must be maximum ::::,.

{·: R,=R,}

\-too

Similarly,

289. [BJ -2 cm . .. . Lmear magn111cat1on m = - - = -2 1 cm So, the image is real and inverted lvl=2lul Let lul = x then lvl = 2x Now lul+lvl=S0-(-40)=90 ::::,. x+2x=90· ::::,. x=30cm

-1)(_!__...!.) =0 R, R,

t -> oo

So, a hollDw, convex lens of any material will behave like a glass plate.

295. [Dj The ray diagram is as shown below

.Y

So, the distance of object from the lens is 30 cm and of object is 60 cm, i.e., the lens mt.ist be located at x = -10 cm as shown in figure.

0

T

x=+50cm

x=-40cm

I

X=O

a . S ,nee, x =../2

•r 30cm

-

and

1+---c+<+--60 cm-Jo-1

P~c;, 1)

1 0cm

292. [DJ Hollow,convex lens is as shown-in figure. Applying Lens Maker's Formula, we get

0

a

Y= ../2

297. [CJ 1-16 1-6 -1+16=1-6 21=22 f=11 cm

-

=> => ::::)

+

=>

~

f1

. f2

Hollow glass lens

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1.

[B, CJ

µsini·= sinr

The.normal to the interface is along

. A-(-1<) COSI=

IAlikl

k.

+10 . 1 =20=+2

=>

r = sin·' (µsini)

=>

3=sin·'(µsini)-i

••. (1)

This is a non-linear variation of 6 with i • Also,. we have 6 to be

i= 60° => (i is the -Z-axi~}

angle Z
which Z=O

incident

ray

makes

maximum when i =·c and hence r =~.So . 2

with

6-

Z>O

=6,=~-C 2

... (2)

FIJrther When- i > C ,-then TIA takes place and·thEI incident ray is reflected back in the denser medium as shown in the tiQure . •• i .......... .

:' , / 'L,' ~6=it-2i· ,/

n"'k

µ But according to Snell's Law

,/2 sin60 =,/3 sinr =>

Sp, 6=rc-·2i • i.e., 6 decreases linearly with i. So,

sinr = v'2 ,/3

,/3

.

2

6ma.; =02 ·=rc-2C

1

s,nr = v'2 => 3.

62

r =45°

4.

[A, DJ Combl~ed Solution to 2 & 3

For i < C, ·no TIA will take place, so we have deviation (8) given-by 6=r-i

sin C = µram, µdenier

=..&_ µd

i+r' =90° According to Snelrs Law µd sini = µr sin(r')

=> µ

=261

[A, BJ

=>

Now, according to Snell's Law, we have

•.. (3)

From.(2) arid (3), we get

=> .=> =>

~=..&.=sine sin(r') µ, sini sinC sin(90-i) sin~= tani

Reflected Ray ·o(mser Riirer Refracted Ray

C = sin·' ( tani) C = sin·' ( tanr)

{Be~use i=r}

= t:::================================== 1.220

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Ray Optics 6.

[A, CJ According to Snell's Law, we have ~A,= µ2A2

=>

(1)(6000) = (1.5)).,

=>

i., = 4000

Also at m=O, v=a => O=f-a => f=a

14. [A, B, C, DJ

A

All are the consequences of the

Since frequency does not change when light goes frOm one medium to another, so·

"DISPLACEMENT METHOD TO FIND FOCAL LENGTH OF A CONVEX LENS"

C

V=-

15.

A, 3x10

5x1014 Hz

V = 6QQQx1Q-l0

[A, B, C, DJ Objective and eye piece are separated by a distance (t+t)=16.02m {OPTION {A)}

8

Angular Magnification=

_1_ = - ~ = -800

{OPTION (B)}

9.

[B, C, DJ See Displacement Method.

f, 0,02 A telescope produces an image which is always inverted.

12.

[A, DJ

In a telescope an objective is larger than the eye piece.

(OPTION (C)} Since images are formed at the same place, So one image must

{OPTION (D)}

be real and other must be virtual.

t

[B, DJ 1 1 1 -=-+f -15 30 => f = -30 cm (Diverging in nature)

,i,'

Since red deviates the least and violet deviates the maximum.

16.

lxl+IYl=32 p

Q'

So a coloured pattern with red on the outerside is observed.

P' X----+t+-Y ~ i+-

(32-x) --.i

17.

ForP

1 y -x 15 1 1 1 -+-=y X 15

=>

-y

on,

,,,(1)

1 -(32-x)

So total optical path length= I;11s,

32x-x' =240 x' -32x +240 =0

=>

x2 -20x-12x +240 = 0

=> => => => 13.

x(x-20)-12(x-20) = 0 (x-12)(x-20)=0 x=12cm,20cm

[B, CJ

{OPTION (C)}

_.,

~

1 15

1 1 1 => - + - - = -y 32-x 15 Adding (1) and (2), we get 1 1 2 -+--=x 32-x 15 32 2 x(32-x) =:is =>

[A, C, DJ Optical path length in passing from 1st medium is r1is1 • Optical path length in passing from 2nd medium is n2:52 and so

ForQ

1

{OPTION (D)}

.

1

m

Total time of flight t = - Ln;s1

{OPTION (A)}

C 1~1

For

inhomogeneous

media

optical

path

length

is

B

f

.,,(2)

OPL= n(s)ds and the ray must travel along a path in which A

time taken to go from A to B is minimum. Such paths are called stationary pathways and this is the statement of Fermat's Least

Action Principle or Fermat's Principle of Least Time. 18.

[A, DJ The final image is formed at infinity when the combined focal

length of the two lenses (in contact) is 30 cm i.e., 1 1 1 -=-+30 20 f

f=-60cm So, when another concave lens of focal length 60 cm is kept in contact with the first lens. Similarly, if µ be the refractive index of a liquid in which focal length of the given convex tens becomes 30 cm. Then from Lens Maker's Formula, we have

f-v

m=--

f

m=1-~ f

1 b

=>

Slope=--=-

=>

f

!

="-b

C

2~=(¾-1)(~,-;J

... (1)

3~=(3~2-1)(;, - ; J

... (2)

From equations (1) and (2), we get 9

µ=a

c::::=================================== = 1.221

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Optics & Modern Physics

Advanced JEE Physics 19.

[B, DJ

=>

When upper half of the lens is covered, image is formed by the rays coming from lower half of the lens i.e., the image will be formed by lesser number of rays. Therefore, intensity of image

=>

will decrease. However, complete image will be formed.

20.

When the Image is real, then u=-y,so v=-2y Again applying the mirror formula, we get

[A, CJ m = +2 , means image is virtual, erect and magnified. A virtual and magnified image can be formed only by a concave mirror

1

and that too when object lies between pole and focus.

21.

-2y

[B, C, DJ

1

1

-y=--,

=>

When passing from vacuum to a medium, frequency remains

unchanged while speed and wavelength decreases µ times.

=> 22.

[A, B, CJ For convex mirror (having positive focal length) the image is always smaller in size. For concave mirror (having negative focal length) the image is smaller when the object Iles beyond 2f .

23.

[A, BJ

30.

31.

[A, CJ When the object moves from infinity to the pole of the mirror, the virtual image moves from focus to the pole.

[B, DJ Since, for refraction at a plane surface, we have

.&=µ2 U

... (1)

V

If x be the height of the bird above the water surface, then for the light travelling from the bird to the fish, we have µ1 =1, µ 2 =µ and u=-x

:'

So, from (1 ), we get 1 µ

l+- f1 - - - -

:::::,

(-x)=v V=-µx

=>

lvl = µx

''' ''

- - - - 21, - - - - -

24.

Now speed of the bird is dx dt So, apparent speed of the bird is

[B, CJ Th~ tube length of an astronomical telescope, in normal adjustment, is (f0 + ~) and that of Galilean telescope, in normal adjustment is

(t-t)

where

t

and

t

l~:I=µ ~~

: : :, 1~;1> ~;

are focal lengths of

objective and eye piece respectively. In this case, f0 = f, so difference in tube lengths is

32.

1,-c, =(t +t)-(t-t,)=2t =21 27.

28.

[B, DJ A concave mirror can give both real and virtual magnified images. Since nothing is specified,,so m=±3 ±3 = ~

-15-u

+u=-10cm -u= -20 cm

29.

=

[C, DJ When the image is virtual, then U=-X, SO V=+2X From the mirror formula, we get 1 1 1 -+-=-

v u I

1.222

[A, C, DJ TIR takes place when ray of light traveis from denser to rarer medium.

[B, CJ A concave mirror and a convex lens_give virtual magnified.image for a particular object position (i.e. when object lies between F and P (or Cl)

=>

{·: µ>1}

Further, sinC12 Since,

µ2

µ,

>

=&.. µ1

and sinC13

:12 µ1

µ3

µ1

c,2 > C13 Smaller the value of critical angle, more are the chances of TIR. 33.

[A, CJ Since, µ

sin(~)

.A

sm2 For µ

=,/2 and om= 30°

A = 60' , we get

Further, at minimum deviation, we have

r, =r2 =

A

2

=30°

Applying Snell's Law. we get

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Ray Optics .,,, sini1 = µsinr1

34.

=>

sini, =(v'2)sin(30°)= ~

=>

i1 =45°

l+t=-1.Q=~ 16 8 Now, from (2) we get 1(~-1)=_1_ 8 16 16\'-10\+1=0

=>

[B, CJ I

1.5

=>

1.5-1

1

16\'-8\-2\+1=0 8\(2\-1)-1(2\-1) = 0 (8\-1)(2\-1)=0

=>

Using µ 2 -&= µ 2 -µ, we get V u R ' ---=-V oo 20 ;::::, v=+60cm Since v is positive, the rays actually meet.

=>

1

\=.!. or \ =8 2 P, =8D or P1 =2D

=> =>

35.

[BJ

In air, !=(µ-1)(_1_ _ _1_) R,

f

When

immersed

f=C~2=>

... (1)

R2

in

a

liquid

of

refractive

... (2)

f immersed

in

a

liquid

of

i=-¾(~.-:J

refractive

index

2µ ,

[A, DJ Real image is smaller in size if object lies beyond 2f and it is larger if object lies between f and 2f .

39.

[A, CJ A ray can pass undeviated when µ1 = µ 2 or the ray is incident normally i.e., angle of incidence is 0°.

..

(3)

'=-2(µ-1) •

40. [C, DJ For TIA, we have i>C => sini > sinC =:i, sin(45°) > sinC Since, sine=..:!.

f

n

1

[BJ When in contact, we have 1 1 1

I

n > F2

n>1.414

~

(·.- P=i=10l

10=.!+.!.

I t

1+~=10\t

·1

-~J

1)( RI1- R21)

From these two equations, we get

\yatar = 4falr = 4f

I~

6=10-0.25

It It = 16

-(3/2

\.atar - 4/3 -

6=.!+.!.- 0.25

1

Applying Lens Maker's Formula, we get

f~, =(¾-1)(~,

1 1 1 X -=-+--F' I t It ~

41. [A,C]

... (1)

When at a separation of 0.25 m, we have

I

1

v'2 >,;

-=-+F

R

38.

l=µ-1

~=(µ-1l(-f)

36.

V

V=R

From (1) and (3), we get

=>

R where µ 1 =µ, µ 2 =1, u=-R, R=-A

2'

X~. -:J

= µ 2 -µ,

V

1

i=(:µ -1)(~, -~J =>

-u

µ

index

¾=(~. -:J

When

[A, DJ

Since, .&. + µ 2

From (1) and (2), we get 1 µ-1 1=-1=>

37.

... (2)

In air, the image was inverted, real and magnified i.e., the object must be lying between f and 2f . Now the focal length has changed to 4f . Therefore, the object now lies between pole and focus and so the new image formed will be virtual and magnified .

From (1), we get

==================================== = 1.223

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Advanced JEE Physics 43.

Optics & Modem Physics 45.

[CJ Let t be the thickness of the watch glass, R be the radi1:1s of

and P) but the image is always magnified and a convex mirror

curvature at inner surface, so (R + t) is radius of curvature of outer surface. If µ be the refractive index of this thin lens (watch

glass), from Lens Maker's Formula, we get

[B, CJ , A concave mirror can give a virtual image (object lies between F can never give a real image.

46. [A, B, CJ Since, the lights used are of different colours, so they have different frequencies and hence (8) is also correct.

R

47.

[B, CJ

!f = (µ -1)(-1- ...!.) -2R --R =>

-=

' R1

.f= 2R

R2

µ-1

The focal length of lens has nothing to do with the direction frOm which the light is incident" on it.

.!V = !f = (µ-1)[...!. - _1_] --R -(R+t) l=(µ-1)(-1__ .!)=- (µ-1)t <0 f R+t R R(R+t)

49.

[C, DJ For all object positions a convex mirror (or a concave lens) forms a virtual and erect image.

So, the watch glass will have a diverging nature.

44.

µ-1 2R

.! _.! = ! = (µ -1J(J_ _...!.) vuf

!f = (µ -1)(.!R - ...!.) 2R

53,

[A, BJ

[B, CJ For a concave mirror, when object is placed at 2F (or C) real image of the same size as that object is formed at 2F (or C).

-f -f-{-2f) For convex mirror f 1 ::::) m =---=="' 1-(-21) 3 ::::)

____ ....................... i '' ' '''

mamcava

!+- d-i---2f2----+:' f, -----+<

= ================================== 1.224

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1.

[DJ

11.

1-t) < t .

Lateral displace~ent, Ax= tsin{

[AJ According to Lens Maker's Formula,

!= (µ-1)(J_ _ _!_) R1 R2

f

2.

Since, R1 = R2

[BJ

Both Statements are true but Statement-2 is not the correct explanation to Statement-1.

R, 6.

[CJ

1

Y,

16.

=>

[BJ

i

19.

[A] Both the Statements are true, and Statement-2 is the correct explanation to Statement-1.

20.

[CJ Statement-1 is true & Statement-2 is false.

21.

[BJ

When light goes from one medium to another, its frequency remains unchanged.

L 3

R,

So, power =t=O

3

4

Y1/M =Y, + (Y, -y,) ~ Y,

3

3 But Y,ia =2[ :• +(y,-y1)] => 7.

dy,/8 =2dY2 =2V di di O

A=60°

µ=-12

[DJ

When the object is virtual, a real image can be formed by a plane or convex mirror.

8.

10.

[BJ Both Statements are correct but Statement-2 is not correct explanation of Statement-1. [DJ

For a mirror, m = _f_ f-u

i = A+ Dm 2

Since, r.

= 60 + 30 = 450 2

A

= - = 30° 2

So, According to Snell's Law, we have 1 sini = sin(45°) = -12 = -12 sinr sin(30°) .!. 2 Both Statement-1 and Statement-2 are true but Statement-2 is not the correct explanation of Statement-1.

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Optics & Modern Physics

Advanced JEE Physics· 22.

23.

[AJ Both Statements.~re t~ue and Statement-2 ·1s correct expIanat·ion · to Statement-1.

[Bl

'

Both Statements are true and StatemE!nt-2 is correct explanation. of Statement· 1. 24.

25.

26.

28.

[DJ _ -· Statement-1 false and Statement-2'is true. [DJ · ·11 . I When the object is virtual, the convex mirror w1 give a "rea image.

29. '[CJ h ·1· I Angle of incidence at any location should be greater t an en 1ca angle.

[BJ . . After refraction at two parallel faces of a glass slab, a ray of light emerges in a direction parallel to the direction of incidence of white light an the slab. As rays of all colours emerge in the _same direction (of incidence of whit~ light), hence there 1s no dispersion, but only lateral !'.Jisplacement.

30.

[CJ

[CJ In search light, we need intense parallel beam of light. When source of light is placed at focus of concave mirror, only paraxial rays are rendered parallel due to ·large. aperture of mirro_r. Marginal rays give a divergent beam but in case of p~rabohc mirror, when source is at focus, beam of light produced over th.e eritire cross-section of mirror is a parallel beam.

=1.226

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1.

For convex lens :

[BJ Different angles by geometry and the given conditions are shown in figure.

1

---=-v -x 20 =>

1

1

1

---=x V 20

.•. (1)

For concave lens : 1

1

-(20-v)

=>

-(20-x) -10 1 1 ' ----=20-x 20-v -10

Salving this equation, we get

x =>

45•

µ

4.

= sin45" =-./2

3.

v=20(v'3-1) cm

[DJ

m = v, = -v = :!_ ' LI, -X X

[CJ

20(,13 -1) 20( ,/3 -1)/,/3

v'3

Magnification by concave tens is given by

STotal

= Sp + Sa + SR

SToS,

= (45° -30°) + (180° - 2 X 30°) + (45° -30°) = 150°

v,

-(20-v)

20-v

m--,-u,--(20-x) 20-x

[BJ

m,

Image formed by concave lens is virtual for all positions of object i.e., image by concave lens lies between the two lenses. For both the images to coincide, image by convex lens should also lie in between the two lenses the two lens or image by convex

lens should also be virtual.

••• (2)

Magnification by convex lens is given by

sin30°

2.

20(,13-1) v'3 cm

40-20,/3 20- 20( ,/3 -1)

( 2,13 _ 3 )

,J3 8.

[AJ Since refractive index decreases with increase of A and velocity decreases with increase of refractive index, so we have V red

9.

>

Vorang9

> Vyellow

[DJ Colour and frequency remains unchanged as it is property of source.

l+--X--+>4---20 - X ----11-1

10.

[BJ Dispersion depends on wavelength

====================================· = 1.227

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Advanced JEE Physics . 11.

Optics & Modern Physics

[DJ

21.

[AJ

Since light takes shortest path, so

60°

t = nxo C

12.

[CJ

''

Speed of light changes with refractive index.

'' '' '

/60°

------~'' ------------6=180-120=60° Laws of reflection is universally true 22.

[DJ

A=)(6-/3)' +(a-13)' +(-10)'

Ao,

A= 20 units 23.

=> => => =>

1 1 1 -+-=V 20 15 1 1 1 -=--V 15 20 1 , -=V 60

[CJ

. id-ii) (6-/3i+a-/3]-1oi<).(-1<)

COSl=-ixr--

. 10 20

1 . 60 2'

COSI=-=-· I=

24.

V=60 cm

20 0

[BJ

sini

-/3

.

F2 ..

sinr = J2., sinr = ./3s1n1

14.

[AJ Observer 0 1 cannot see the image because light will be

sini = J2 sin60° = F2 x .fa=_!_ sinr .fa -/3 ./2 ./2

absorbed by the blackened portion of the lens.

=> 16.

[DJ According to Snell's Law, we have

25

sini =.&.=Ki=60o sinr µ 2

=>

r=45°

[DJ Let the vector representing the refracted ray be

A' = 6,/31 + a..Jaj + Ck

.

ii' .(-1<) Smee, cosr=[M

sin(60°)=K, r=.90° K, = .fa 2

17.

18.

19.

=>

( 6,/31 + 8-/3] +Ck) ,(-k) )(6-/3)' +(a.fa)' +c'

[CJ

[AJ Since, e =·0° => i=r :::::,. K2 =1

=>

-C cosr=-;=~~~-=e-

=>

cos45°

,/108+192+C' -C

,/aoo + c' ·;

1

-C

F2 - ,/aoo + c'

=> C =±10-/3 Since the refracted ray travels downwards, => C=-10-/3 =>

[CJ

Since, f=R-~seca

26.

R So,for 0>0, f<2

A'= 6,/31 + a..Ja]-1 O-J3k

[CJ

IA'i=)(s..Ja)' +(a-13)' +(-10-/3)' =10v'6 27.

20.

cosr

[AJ

.A' 3, 4,1n=rA'r= 5.J2 I+ 5.J2J- .J2k

[DJ

R

I,, =R--sec(60°)=0 2

. 28.

[AJ

J

Optical path length (OPL)= ndx

= ================================== 1.228

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Ray Optics =>

f'

f

=>

OPL= (1+x')dx

i4

_B_

-2µ .-·

0 .

=>

x' OPL= ( x~g

)I' =am 4

35.

[DJ Focal length of /eris

O

Li

and L2 is given by using the Lens

Maker's Formula, so

31.

[CJ peq

.! = (µ, -1)(..!. - ..!.) I R "'

=2P,ene + Pmirror

Simila~y _1_ = (µ -1)(..!. - - 1-) ' f, ' "' (--R) => f=40 cm

and F', =-1=-~ mrror f , R

So, equivalent focal length is given by 1 1 1

¾=2(µ-1)(~-(..!i))-(-~) => => =>

1 4(µ-1)

\=50cm

=>

where P00,=i=(µ-1)(~ -~,)

-=-+-

t., I f,

2

-=--+-

I R R .!= 4µ -±+~ I R R R 1

4µ

36.

2

t,=R-R ¾=¾(2µ-1)

=>

l-2(2µ-1)

=200 cm 9

[CJ Image formed by L; lies 50 cm behind it and on principal axis

of

=>

f ..

=>

Li . This will act as an object for

L2 . 86, for L2

,

R

......................... I,

PA of L 1

32.

[CJ

4.5mm

For a plano-conv~x lens, we have _1_ = (µ

f,

.

PA of L,

-11(..!. - ..!.) R "'

=>

33.

=>

we have u=+SO cm,

[BJ When plane surface is silvered, then

Applying Len's Formula, we get 1 1 1 V 50 40

1

2

1

I=,,--~

{·.- ~--> «>}

=>

1 2(µ-1)

I=-R=>

34.

=>

[DJ

-=--+f, R H

=>

,:=R

1

PA of L, is

y = m, (4.5 mm) =>

4

y= (4.5mm)=2mm

9

Hence 12 is at a distance of (4.5-2) mm =2.5 mm from PA of

2(µ-1). 2 2µ

9

its principal axis i.e., (PA of L2 ). So, distance ofimage 12 from

,.-=,,--r

=>

1

.±

When curved surface is silvered, then 1 2 1

1 (1 1) -(-R) 2 ,.-=2(µ-1) ~- --R

1

-=-+v 50 40 200 V=-cm

Magnification produced by L, is m, =~ = · u 9 However, for ,L2 , the image 11 is at a distance of 4.5 mm above

R

~ = 2(µ-1)

=>

1

t =+40 cm

L,. 37.

[CJ Applying refraction at cuived surface formula, i.e.,

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Optics & Modern Physics

Advanced JEE Physics

EL+h=µ 2 -µ, ,we get -u V R 1

2

1

-+-=x V1 R =>

2xR

= x -R

{from pole of curved·surface}

O,

This image formed will be at a distance v'=(v,-R) from the

I

V1

C

I

2

I I I I I I

plane surface. Now again applying the above formula, but at the

plane surface, we get

I

_!_3_=0 V

V'

-R---R-

{·: forplanesurlace R-->oo}

Since, refractive index of this is 2, so th~ critical ang!B C is given by

2 -=--

=>

v

V1 -R

C =sin-•(½)= 30°

2 · Substituting v 1 = xR , we get x-R 2(x-R) = R(x +R)

Since, i =C-, so the ray grazes the plane surface.

v

40.

Since yellow is the mean colour, so µb +µ, 1.51+1.49 50 1. µ, 2 2

For the image to be virtual, we have

..!.

2(x-R)
=>

X
This condition is only sati~fied by ;

41.

[DJ Again yellow is the mean colour, so

µ ',

=~ . 42.

·38.

[CJ

[DJ . (2x)R Since, v 1 = - x-R So,'for x=2R ,we get

µ~+µ; 2

Also, U=-2R

(b)

A' =-[µ,-1] A ·µ~ -1

=>

A'= -( 1.50-1 )so= -4o 1.7?-1

43.

m, =(½)(!)

[DJ Net dispersion = ( a,

= (1.51-1.49)6°-( 1.77-1.73)4°

m2 =1

=0.02x 6°-0.04x4°

mnat = m1m2 = -1

= 0.12°-0.16° =-0.04°

Since, ml'III, =0 , so the final image formed is real, inverted and

44. [CJ

of same size.

Since,

[CJ sin(90°) =

=>

=>

sinr

=¾

0A=.J3R LACP=60° Using geometry, we get i=3QO

=>

z 60°

-1 fa .

-1 tan0= fa

=>

60° ~-----~--+x

0=-30° So, the ray is incident normally on the face -AB . Thus angle of incidence on face AC is 60° . For grazing the face AC, we have µsin(60°) =ix sin(90°)

=> 1.230

-x 10 Z=.J3+.J3 Slope=

r=30° In triangle OAC, we have OC = 2R , AC= R , so by Pythagora's Theorem, we get

=

fa z + x = 10

2

=>

=>

- a, ) +(a; - a; )

=(µ.-µ,)A+(µ;-µ; )A'

=> m1=-1 However for plane mirror, we have

sinr

.

[DJ For no deviation to take place, deviation (given by a=(µ ,-1 )A

Since, m, =(~)(~)

39.

175

by one must be cancelling the deviation due to the other.· For this (a) the prisms must be arranged upside down

v,=4R

=>

1.77+1.73 2

µ=

2

Ja

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Ray Optics 45.

[BJ

Equating equations (1) and _(2), we get

For µ =

3

2

dy

the ray will be internally reflected, so, the ray is

= y3/4

dx

normal at face BC . Hence finally, the i-efracted ray is parallel to z-axis.

::::::, :::::i-

=>

y-:314dy=dx

f.

f'

y-314dy= dx

'

=X

4y1l4

' ... (3)

The required equation of trajectory is 4y114

'

=> 46.

[BJ If BC is silvered, the ray will retrace its path. Hence equation of ray coming out Ot the prism is .Jaz + x

47.

52.

= 10 .

[AJ The image formed by the objective must lie within the· focus of the eyepiece.

49.

50.

[AJ To obtain best magnification, the object must be placed just beyond the focus of the objective lens. In this case, the first image distance from the objective is very large.

X=(256)''=4m

So the co-ordinates are ( 4 m, 1 m)

53. 48.

[DJ . At the point of intersection on the upper surface, y=1m

=>

[CJ The objective lens r11ust form a ·real image for eyepiece to magnify it.

[AJ As nA siniA = np sinip and as nA = np Therefore, ip

54.

[CJ i+9=90°, 9=90°-i,

... (1)

55.

1

2

F

f,

~)-(-!o)

=>

~=2(1.5-1)( 2~- 6

=>

----+---F 20 60 20 60

=>

F=-

= 1 because iA =90° ( Grazing incidence) = ~Ky312 + 1 =~ys12 + 1 08 1

. .

Slnl

2

8

60 .=-7.5cm 8

[AJ -7.5 22.5

3

57.

[CJ

58.

[BJ

59.

[BJ Combine solution of 57, 58, & 59 Each part will work as a separate lens and will form its own image. For any part, we have u = -0.3 m, f = +0.2 m .

V

=> ::::::,

+1

coti =~or y314

1

[BJ

'

1 ~y3/2

1

Therefore, from lens formul~,

(1)( 1) = ~( y' 1' + 1) sini

=>

1

I -7.5 m=-=--1-u -7.5+30

nA =1because y=O

because K=1.0(m)-"

I,

Since F =0 , so combination behaves like a·concave mirror.

56.

siniA

i.e., the ray Will emerge ·parallel .to the

[CJ

[DJ but tan8 = dy dx dy = coli dx Applying Snell's Law at A and B nA siniA = n8sini8

= iA = 90°

=1

boundary at P i.e., at grazing emergence.

Slope of tangent = tan8 = tan ( 90° - i) = coti

51,

=x

... (2)

1.

U

f

1

-;;-o.3

1

= 0.2

V=-0.6 m

So, each part forms a real image of the point object O at 0.6 m from the lens, as sho'Wn in figure.

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Advanced JEE Physics

Optics & Modem Physics 64.

[DJ Using Lens-Maker's Formula for the li(luid concave lens,

_1_ =(µ-1)(....!..._..!.) = _µ-1 _ 50 · -R oo R

=

_µ-1 10

3

~=µ-1 50 10 µ=1+0.6=1.6

=> =>

,->--v=0.6m---+1 Since the triangles OL,½ and OI,12 _?re similar. So, we have

65.

r1 = o Further, r, + r2 =A= 60°

1,1 2 _OB_u+v

1,L, - OA - -u=> => 60

I,I, _ 0.3 + 0.6

1,L,

0.3

[BJ Since the ray is incident nonnally, so

0.9 = 3 0.3

==>

r2 =A=60°

A

111, =3(L,L,) =3(2x0.0005) =0.003 m

[AJ

f=(µ-1)(~ -~,)

s~-----~

f=(¾-1)(¾- ~)=ix¾=¾ => 61.

[CJ

X-~

=>

-~)=-3~

f,=-3R

=>

=>

. 4 . µ 1 sme= .Jaxstn60°

=>

sine=

.

F=3R

So,

P=e-r, =0

=>

e=r2 =60°

=>

i

63

R=10cm

=>

c:::l

=> 66.

p=O

2

../3

=../3 4

k, =../3

[CJ When µ 1
[CJ If f is the focal length of the liquid concave lens, then

1 1 1 1 -3 -=-+- or-=25 10 f f 50 50 f=cm

,

4

= distance of object =15 cm

=>

1

2=µ1 µ,

F

3 :=15cm

... (1)

x µ, = µ

sin(60°)=~ µ,

0

=>

../3

So, equation (1) becomes

2 Since, image coincides with the object so, clearly rays of light must have retraced their path after reflection. This is possible only when rays of light must have fallen normally on the plane mirror. For this, the object is at the focus of the lens system.

F

=

From the graph, we observe that when µ 1 =k2

_1_=.!.+_1_=_1_ __!_=~ 'Fif,R3R3R

~

4,/32,1312

..f3µ 1 x 2

Further, deviation is given by P=e-r2

[CJ

Now

= µ,sine

µsinr2

i =A

i=(¾- 1

62.

Applying Snell's Law at AC • we get

=>

sinr2 =sinC=~

=>

sin(60°)=~

=>

2= .,/3xµ,

=>

µ1=3

3

µ,

µ,

,/3

4

{·: r, =60°)

1

8

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1/.ay Optics Therefore, second image will be formed at a distance of m6. or

k =~ ' 3

67.

__ t_)(A) below its optic axis. ( t+\-d

[BJ From the graph, we observe that

J3,

Therefore, y-coordinate of the focus of system is given by is the maximum deviation

Y=A-(t+tt-d)

and j32 is the minimum deviation

Y=(\-d)A t +\-d

=>

So, P2 = z-60° = 30° and P, "

70.

[AJ Wavefronts are parallel in both media. Therefore, light which propagates perpendicular to wavefront travels as a parallel beam in each medium.

71.

[CJ All points on a wavefront are at the same phase

=60°

$,=$,and$,=$, $, -$, = $, -$, 72.

68.

[A] From the first lens parallel beam of light is focussed at its focus i.e., at a distance \ from it. This image 11 acts as virtual object

[BJ In medium-2 wavefront bends away from the normal after refraction. Therefore, ray of light which is perpendicular to wavefront bends towards the normal in medium-2 during refraction. So, medium-2 is denser or its speed in medium-1 is more.

medium-1

for second lens L2 • Therefore, for L2

U=+(\-d). f=+t 1 1 1 1 1 -=-+-=-+-v

I u t t (\-d)

\-d

V=--t +\-d

y

•'

medium-2 75.

[DJ For both the halves, position of object and image is same, however the only difference is of magnification. Magnification for one of the halves is given as 2(> 1). This can be for the first one, because for this, lvl > [ul . Therefore, magnification, 1ml =]ti> 1. So, for the first half, we have

l~l= L.i I + - - - d - - - - · f,---d +I - - - - - f, - - - - Therefore, x-coordinate of its focal point will be

X=d+V=d+ t(\-d) t+\-d => 69.

X

It+ d(\-d) \+t-d

[DJ Linear magnification for L2 , is given by

m

~ -(~~\--d~)(\~d)-t+~-d

2

=> . lvl = 2lul Let LI= -x , then v = +2x

and lul+lvl=1.8 m :::::, 3X=1.8m :::::, X=0.6 m Hence, u=-0.6m and V=+1.2m .1111 1 1 Using-=---=----=! V U 1.2 -0.6 0.4 :::::, f=0.4m For the second half, we have 1 1 1 -~--+ f 1.2-d -(0.6+d) 1 1 1 => - = - - +

0.4

1.2-d (0.6+d)

Solving this, we get d =0.6 m

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Optics & Modem Physics

Advanced.JEE Physics Magnification for the second half will be V 0.6 1 m, =;;= -(1.2) =-2 12 and magnification for the first half is m1 = "!.... =-·() =-2 u - 0.6

76.

79.

The distance of the mirror as observed by the eye is

[AJ For direct observation, the eye E appears to be farther that it actually is. So,

X=H+!:!+..!::!_ µ

=>

[BJ The distance of eye from mirror is

81.

µH+H

So, distance of eye as seen by the fish in the mirror is H

X=(µH+H)+ 2

78.

2µ

X=H(1+}µ)

[CJ 1sin8, =-nsin02 . 9 sm 2

1 . = --s1n0 1 n

3H

=:-

X=2+µH

=>

X

9,

=H(µ+¾)

9,

[CJ For direct observation by the eye E , the fish appears to be closer than it actually is. So, we have H X=H+-

.

·

H+ !:!µ . So,

the distance of the fish as seen by the eye in the mirror is

X=µH+i=H(µ+¾) 77.

[DJ

· 1.234

2µ

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1.

A-> (r, s) B-> (p, q, r, s, t) C-> (q, r, s, t) D-> (p, r, s) (A) m < O , means real image, possible for concave mirror and convex lens. So, (A) -> (r, s). (B) m > O, means virtual image, possible for all i.e., plane mirror, convex mirror (always), concave mirror (when

(8)

object lies between focus and pole), concave lens (always) and convex lens (when object lies between optical centre and focus). So, (B)-> (p, q, r, s, t). ·

(C)

Velocity of bird as seen from water = 6 x

(D)

Velocity of bird in water after reflection from mirror

{C)

(D)

4

6 6=12

1,

1)

A-> (s, t) B-> (p, t) C-> (s, t) D-> (q, t)

4

w.r.t..bird =-6+6=0

Velocity of bird w.r.t. fish

vA. = -si +2}

=>

"A'A =VA' -VA

Similarly,

1)

5.

A-> (p, q) B -> (r) C-> (s) D-> (p, q)

6.

A-> (p, s) B-> (q) C-> (p, q, s) D-> (r)

7.

A-> B-> C-> D->

=>

8.

A-> (p, q, s)

=-10i

v, =(-1 +3]}

.±3 = 8 t

=8 + 8 =16 t = 8 J.

w.r.t. fish = 8-8= O

vA =1+at= f +{2f +j)<2l=si+2] =>

Velocity of fish w.r.t. bird = + t Velocity of image of fish after reflection from mirror in air

=8x,:l=6t

1ml < 1 , means diminished image, possible for concave mirror, convex lens, .convex mirror and concave lens. So, (C) -> (q, r, s, t). . 1ml ~ means magnified (> and same sized (= image, possible for concave mirror/convex lens (both 1ml > 1) and plane mirror (1ml = 1) . So, (D) -> (p, r, s).

2.

(A) Velocity of fish in air = 8, ,cl= 6 t

v,,=(i+aj)

(q, r) (r) (p, r, s) (p, r)

B-; (p, q)

C-> (r) D-> (p, q, s) 3.

A-> (s) B-> (p,"q, r) C-> (q, r,.s) D-; (t)

9.

A-> (p, B-> (p, C-> (p, D-> (p,

4.

A-> (r) B-> (q) C-> (p) D-> (p)

10.

A-; (p, q, r, s) B-> (q) C-> (p, q, r, s) D-> (p, q, r, s)

s) q, r, s) q, r, s) s)

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Optics & Modem Physics

Advanced JEE Physics 11.

14.

A-->(p) B--> (p) C--> (r. s) D-> (q. p)

A-> (s) B--> (q) C--> (p) D--> (q)

-~x-

· · 12.

~

1

1

V

f

~y

U

V=(u~f) 15.

~ 7

) 1

Again 1ml can be greater than or less than 1. For concave mirror : 1 1 1

(8)

~

+u

..'!.(x+y)= dx + dy =5+2=7 ms·•

(B)

dx =3+2=5 ms-1

(C)

d(2x) = 2 dx = 10 ms·• dt dt

(D)

dx =5 ms_,

dt

dt

dt

dt

µ, = µ, (no change in path) Li-0·

V=-(f~u)

=>

So, v is always negative i.e., image is always real. .

dt

A-> (q) B--> (r) C--> (s) D-> (p) A µ 2 > µ 1 (towards normal) B

-f

Further, !ml=[~=

B

µ 2 > µ 3 (away from normal)

-+-=v

C

(A)

Now, v may be positive and negative, depending on values .of u and f • Since, !ml=fu[=(u~ 1)=(

A

Ref.on ground

A-> (p. s) B--> (p, q, r, s) C---+ (p, q, r, s) D--> (q, s) (A) For convex mirror : 1 1 1 -+-=v u f

i-

C

Lr·= o on the block

µ 1 > µ 2 (away from the normal)

µ 2 > µ3 (away from the normal)

-+1

I

µ1 x

i.e., m is always less than 1 or image is always diminished.

13. A--> (s) B--> (p) C-> (q) D--> (r)

=>

1 . J2=µ 2 s1nr sin r =

}:;1

v2µ 2•

Since sinr < 1

=>

A-+ when object lies between pole and focus image is virtual, magnified and erect. B---+ when object lies between focus and centre of curvature,

D·

·image is real, inverted and magnified. C---+ when object lies at centre of curvature, image is real, inverted and of equal size. p ---+ when object lies beyond centre of curvature, image is real,

µ, < ../2µ2

ForTIR: 45°>C => sin45° > sinC

=> =>

_1_>&

-1?. µ, µ, > -!?.µ,

inverted and smaller in size.

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1.

Forward shift of point of incidence due to a single reflection

The position of the image of end b can be obtained using mirror formula, according to which we have 1 1 1 -+-=v u f where u = -40 cm , f = -30 cm

X = 0.2tan(30°) = ~ . Hence required number of reflections required is

N

Mirror Length Forward Shift

2.

2-./3 _

30

1 1 1 -+-=-v -40 --30 => V=-120cm Length of the image (rod) is given by e, =120-60=60 cm

( ~)

=>

Since, for a concave mirror, we can have both virtual and real image, so for real image, m =-5 and for Virtual image, m = +5

So.magnification rn=-dv =-(v2 -v1 ) du U2-U1

Since we know that m =_f_ f-u

=>

where f = -30 cm and m = ±5 because the image can be real or virtual. . For real image, m = -5

=>

-5=~ -30-u => U=-36cm For virtual image, m = +5

4.

lml=3

Applying mirror formula,

..!. + ..!. = ! V

f

U

for concave mirror, we get

1

v- 60 = (-40)

~30 5=---30-u => u=-24 cm Hence, the object must be placed at 24 cm or 36 cm in front of the concave mirror

=>

3.

=> =>

m=-(120-60) 60-40 m=-3

=>

v=-120cm

s

The image of the end a of the rod which lies at the centre of curvature C is formed at C .

>+-- x --+1+--60 cm ...........i Now, for the rays to again converge at S , (after reflection from the plane mirror) b'

a C

b

F

''

I

,1

1+-4Qcm~ 1+-------

Distance of S ) ( from Plane Mirror

p

1 + - - - 60 crn------.i 120 cm _ _ _ _ _ _.,.

'

= (Distance of Image formed) . by Concave Mirror

=> =>

X=120-(X+60) x =·ao cm So, the desired distance is 90 cm From reversibility principle, it hardly matters whether the ray of light is first reflected from the concave mirror or plane mirror.·

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Optics & Modem Pliysics

Adva11ced JEE Physics 5.

Since the image formed is real, so for the first case, we have m=-3

Bird(B)r

i

=> =>

' '

Since.!.+.!=! u

V

1 1 1 --+--=(-3u,) (-u,) -I

=>

f = 3u, 4

Differentiating w.r.t. time, we get

(-::)=M-:;)+(-:)

m=-2 - V2 =-2

=>

16=~(v)+u,where u=4cms-1

=>

v=9 cms-1

u,

7.

Since..!.+.!.=! V U f 1 1 1 => - - + - - = -(2u,) (-u,) -I f = 2U2

should get internally reflected i.e., should sufferTIR. From the figure, it becomes Obvious that the ray with least angle of incidence is the one which is incident almost grazingly with the Inner wall.

... (2)

3

The necessary and sufficient condition for all the rays to pass around the arc is that the ray with least angle of incidence

From (1) and (2), we get 3U1

'

''

2u2

J.

••. (3)

=>

that sin a= R; d where d is the diameter of tube

u2 -~=6cm

=>

!!.u -u =6

=>

u1 =48 cm

For TIA, n ;,.C

'

Since, v1 = 3u1

v1 =144cm

So, I= 3u, = ~(48) = 36 cm 4

Now u, =

4

sin a~ sinC

=>

--,'2:::-

=>

A~~

8.

Let at some instant bird is at a height of x · from the water

distance between fish and image of bird at this instance will be,

Since, 9 = 90° - i => lane= coli dy = coti • dx Ac_cording to Snell's Law at O _and P ~ we have µ 0 sin i0 = µP sin ip

... (1)

Since µ=~1+ay => At y=O, µ=1 =>

=

µ-1

=>

surface and it is diving downwards with v cms"-1 At this instant fish is at a depth y· below water surface. Then the

=>

1 µ

Hence, the least radius required is 12 cm.

(v,-v,}=144-108=36cm

S=µ><+y

A-d A

A=12cm

i = ¾(48) = 54 cm

9

So, shift of the screen is (v1 -v2 )

6.

=>

Since, µ=J.5 and d=4cm,soweget

and v, = 2u, = 2(54) =108 cm =>

"a"

0 For this ray, let a be the angle of lncidenC:e, then we observe

But according to the problem, we are. given that the shift of the object is 6 cm , so we get

8 '

,,

Ai-di /'R ', ~-------

7 =3

=>

t

Fish (F) ... (1)

Similarly, for the second case, we have

=>

y

u=4cms·'~,

I

=>

=>

X

'' ''

sln(so•) = (~1 + ay )sini

s=ix+y.

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Ray Optics

-' '

y A

' ,B 'I

'

'

' P(x, y)

D

i': y _._.""'::'.'.:=-----'':.le!L_;__ _ _ x

0

C

'}I.,.

0 I + - - - X------+< . .

Slnl=

i=r+r=2r Since, we know that sini µ=-.-

1 ,1+ay

~

coli= .Jay= dy

s1nr

dx

µ= sin2r ~ 2r = 2 s1nr r

d • J'ovay l.=Jdx o

=!>

11. After reflecting twice from two plane mirrors at right angles, a ray

X=2l

of light gets deviated by 180° , irrespective the angle of incidence, so the emergent ray is hence angle of deviation is 180° .

Substituting y=2m and a=2x10--a m-1 ,weget xmax

9.

aniiparaliel to incident ray and

=2000 m = 2 km

The ray diagram for the situation,is shown in figure.

R

\.

4~

I-.· .... --~--~·-~·"·

12.

From Archimedes Principle, we know that Yimmersed

= pbody

V101a1

Since, LPCQ=,-2r, LPRO=s-20-r) From the property of a circle, we get 2(LPRQ) + LPCQ = 2s => 2s-40-r)+s-2r=2•

P1;quid

"1mmarsed

r=2i-!:

= Psphera = J:...

V

p~""'

V

2

2p

i.e., half the sphere is inside the liquid. For the image to coincide

2

with the object, light should fall normally on the sphere. Using,

According to Snell's Law, we have sini

µ2

s1n1

µ= sinr

_

.&. = µ 2 - µ 1 twice, we have

V

sin ( 2i-.!:)

R

u

2

../3 = =>

sini -cos2i

sini 2sin2 i-1

::::)

2;/3sin'i-sini-,/3 =0

. ' 1±.,11+24

sm,

r,;

4v3

4

1±5

4,3

. 2

13.

i.= 60°

in~idence, then

3

Forthelens, U=-2m, f=+1.5m

From the figure, we observe that

BO=OC So, i( LOBC = LBCO = r

4

Solving this equation, we get h=1·s cm

.. Fa Slnl=-

10.

3

Further - 3- - _g_ = .L._g_ ' h-10 8 -2

r,;

Rejecting the negative value, we get

=>

v1 =12cm

::::)

(say)

and

be the angle of

1

1

1

V

-2

1.5

V=6·m

!

Since, m = (

) = -3 2

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Optics & Modern Pl,ysics

Advanced JEE Pliysics I1P = I2P and I1P12 is .L to mirror.

2~ =(1.5-1>(¾-

,,

==>

R=20cm

..!:l. _ _& = µ2 - µ 1 twice with the condition that rays

Applying

0.3m

-!i)

V u R must fall normally on the concave mirror, we get 1.5 1.2 1.5-1.2 v, --40 +20

0 1,

>+----- d _ _ _ ' _..,

2 1.5 2-1.5 ----=-d-80 v, -20

>+------6m------~

... (1) ... (2)

Solving equations (1) and (2), we get d =30 cm and v1 =-100 cm

Therefore, y co-ordinate of image formed by the lens is given by Y=m(0.1)=--0.3 m

The ray diagram is as shown in figure.

In triangle PNI2 , we have tanB = 12N NP o.3 tan 0 = NP =0.3

=> => and

1,

{·.- lane= 0.3}

NP=MP=1m d=6-1=5 m

l,

....,.___ 100 cm _______,.. 40 l+-' ~c~m.... _ _ 30 cm_____,..

x co-ordinate of final image 12 is,

1+-----B0cm------+1

X=d-1=4m

14.

Applying lens formula

.J-V _.!U =!f

17.

twice we get

f=-6cm

Using the lens formula.:!._.!.=!, we get V U f

!=12cm

,/\ ,'

V

0

1+-

.,...__ 15 cm _____..

12 cm ->oi+--- d - -

1

1

1

v,

-12

-6

---=-

1 1 1 1 ,=15-10=-30

••. (1) =:>

1 1 1 oo v,-d 12 Solving equations (1) and (2), we get v, =--4 cm

and

'

••. (2) 18.

d=B cm

Using lens formula, _!_ _ ..:!. = ! , we get V U f 1 1 1 v, - (-40) = 30 =:>

15.

Using lens formula, ..!_.!=!,we get V U f 1 1 1 -+-=v, 40 20

==>

1

1

V2

10

U

=:>

f

1 -10

=

From Lens Maker's Formula, we get

1.240

=4 cm

Va= 180 cm

v2 =-5 cm

so, the final image is formed at a distance of 5 cm from the mirror towards lens. 16.

8

Therefore, the screen has to be shifted away from-the lens by a distance X=Va-V1 =60cm

-+-===>

=(1- / )9

So. u' =-(40-Ax) =-36 cm 1 1 1 => v, - (-36) = 30

.:!.+.!=!, we get V

V1 =120cm

Shift due to the slab A>c =(1-¾)d

V1 =40cm

Using mirror formula,

f=-30cm

19.

Lateral magnification in first case is -3, so if U=-X then V=+3X Since.!._..:!.=! V U f

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Ray Optics 1 +3x

1 X

1 f

22.

-+-=4

1

3x

=1

02 -x2 f=-4D Substituting f =16 cm and x = 60 cm , we get

4f 3

X=-

D' -(60)'

.•. (1)

16

In the second case magnification is -2 , so now we have

U=-(x+1,5), v=2(x+1.5) Since.!_.!=! V U f 1 1 1 => ~--'-~ + - - - =2(x+1.5) (x+1.5) f

3 =>

D'-64D-3600=0

=>

D' -100D+36D-3600 = 0 D(D-100)+36(D-100) =0 (D-100)(O+36) =0 D=100cm

f=3.(x+1.5) 3 Solving equations (1) and (2), we get f=9 cm

... (2)

23.

For the first flare spot, the lens acts as if its right face is silvered . The equivalent focal length is given by

Using Lens Maker's Formula for both the cases, we get

_1_ =(µ, -1)(..!..-

ti,

and

R1

D' - 3600 =64D

=> =>

f

=>

20.

..!..) R2

... (1)

2(~-1)

2x1.5

2(1.5-1) +30

... (2)

R,

F=-12 cm

Using mirror formula,

1

\..,., = (µ, -1) [~:

-60

F ::::::,.

~.. =[~: -1)(~' -:,)

.!= 2(~:) F R,

Dividing equation (1) by (2), we get

f,,.

4D

But D ~ 36 cm , because it happens to be less than 60 cm . So,

1

2(x+1.5)

=

Since we know that

1

..! + ..!. =..! with v u F

u ---t co , we get

1

-+-=v 00 -12

-1)

::::::,.

Substituting the values, we get

24.

v=-12cm

Method 1 : Ray diagram is as shown

-rn-1)

14---d--+<

r..... -[¾ Jf., 4-1

=> 21.

-···-·-..

3 ~ate,=4'8;r=4x10=40cm

-----··

!,

Using the lens formula, ..:!_ _ ..:!_ = we get V U f u=+15cm, f=+30cm 1 1 1.

=>

15 cm ,...____ 40cm ----;~ d=25cm

---=-

15 30 => v =+10 cm Therefore, the focus of the rays will move 5 cm closer to the screen. The ray diagram is as shown in figure. V

Method 2: Since this combination just behaves as a plane glass plate, so the power of the combination is zero. Since

1

poomS

::::::,.

O=_.:!_+ 40

8------ ....... -----

1

1

X

=l.,m, =-i:+t- \t 1

X

(-15)

(40)(-15)

X

F

(40)(15)

--- F

X

(40)(15)

15 40 40-15 (40)(15)

X=25cm

25.

Applying Lens Maker's Formula, we get

1 ( -1+ 1) -=(1.5-1) 40 120 R,

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26.

R1 =24 cm

o:

f

The ray diagram is as shown in figure for first two steps. If the rays reflected from the mirror are parallel after passing through the lens for the second time, then 12 must lie at first focus of lens. So, the desired distance is given by

"4>l

f/2

,6! 1

10cm

_x___., , _ _ 21--+l
Applying lens formula, for L2 _, we get

1

1

V1

X

20

X=(2f+20-.!. 2

... (1)

-+-=Using, µ 2

-

V

1.5

I,

0

R,

1

f/2

x= 31

2

= 3(30) =45 cm 2

.&. =µ 2 - µ1 for unsilvered side of ~ , we get u

R

.

1

1.5-1

(-120)- (v, -10) = ~

... (2)

Solving equations (1) and (2), we get X=10Cm

= ==================================== 1.242

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Wave Optics

[co~ten~s------------------------~---~ WAVE OPTICS .............................................................................................,...................... 2.1 Solved Practice Problems ..................................................................................................2.41 Practice Exercise Sets :,

Single Correct Choice Type Questions ..................................................................2.52

:> Multiple Correct Choice Type,Questions ...............................................................2.66 :> Reasoning Based Questions (Assertion Reason Type) ........................................2.69 :,

Linked Comprehension Type Questions (Paragraph Type) •...•••..•••..•............•••..•••2.71

:,

Matrix Match Type Questions (Column Matching Type) ........................................2.TT

:,

Integer Answer Type Questions ............................................................................2.80

Answers to lri Chapter Exercise's.(ICE) & Practice Exercise Sets .....................................2.83 Solutions to In Chapter Exercises (ICE) .........................................................................., •• 2.86 Solutions to Practice Exercise Sets ...................................................................................2.92

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INTRODUCTION

he phenomenon of interference, diffraction and polarisation exhibited by light could not be explained on the basis of Newton's Corpuscular Theory. In 1678, Huygen suggested that light propagates in the form of waves. The first historic experiment in favour of wave theory was done by Focault, who in 1850 found experimentally that velocity of light in denser medium is less than that in the rarer medium which was contrary to Newton's Corpuscular Theory.

T

phase velocity or wave velocity. The energy travels outwards along straight lines emerging from the source, normally to the wavefront, that is, along the radii of the spherical wavefront. These lines are called the rays. For a point source in a homogeneous medium the wavefront is spherical.

NEWTON'S CORPUSCULAR THEORY

Newton proposed that light is made up of tiny, light and elastic particles called corpuscles which are emitted by a luminous body. These corpuscles travel with speed equal to the speed of light in all directions in straight lines and carry energy with them. When the corpuscles strike the retina of the eye, they produce the sensation of vision. The corpuscles of different colour are of different sizes (red corpuscles larger than blue corpuscles). The corpuscular theory explains that light carry energy and momentum, light travels in a straight line, Propagation of light in vacuum, Laws of reflection and refraction. However, it fails to explain the phenomenon of interference, diffraction and polarization. WAVE OPTICS

Wave optics is the study of the wave nature of light. Interference and diffraction are two main phenomena giving convincing evidence that light is a wave.

Spherical

wavefront

For a linear source of light the wavefront is cylindrical.

-

,"

-

s

.. -

Cylindrical wavefront

A small part of a spherical or cylindrical wavefront from a distant source will appear plane and is, therefore, called a plane wavefront.

WAVEFRONTS AND RAYS

The locus of all the points vibrating in same phase of oscillation is called a wavefront (WF) i.e. a wavefront is defined as a surface joining the points vibrating in the same phase. The direction of propagation of light (ray of light) is along the normal to the Wavefront. The speed with which the wavefront moves onwards from the source is called the

/ Plane wavefront

2.1

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----------- CONCEPTUAL --------·---·--·NOTE(S) Different types of wavefront Type of wavefront

Intensity

Amplitude

Spherical

Point source

LAWS OF REFLECTION ON THE BASIS OF HUYGEN'S THEORY

Cylindrical Cylindrical

Let AB be the plane wave front incident on a plane mirror M 1M 2 at LBAA' = i, where 1, 2 are th_e cqrrespondin~ incident rays .perpendicular to AB .

Light ray

Cylindrical WF

2

Line :::: -:· ...

source

~1:ne

Plane

WF

''.JJ'J:rJ ••'ft.'' J • ·..J··l 'a

I cc r

0

Aocr

0

~

rays -;-+ Light

T

According to Huygen's principle every point on AB is a source of secondary wavelets, so BA'= ct, where c is speed of light The secondary wavelets from . A will travel the same distance ct in the same time. So, AB'=ct

~

Now, LAA'B=90-i,sothat LA'AB=i, (0<90°)

HUYGEN'S PRINCIPLE

This principle is useful for detennining the position of a given wavefront at any further time if its present position is known. The principle may be staied in three parts. a) Every point on the given wavefront may be regarded as the source of the new disturbance. b) The new disturbances from each pon,t spread out in all directions with the velocity of light in the same manner as the original source of light does and ~ese new disturbances are called secondary wavelets. c) The surface of tangency to the secondary wavelets !n forward 9,irection at any time gives the position of the new wavefront at that time. This new wavefront is called the Secondary Wavefront. Titis principle explained successfully, the reflection, refraction, total internal reflection, interference and diffraction but failed to explain the rectilinear propagation light.

Also, LA'AB'=90°-r ,so that LAA'B'=r, (0
A'B ct AA' AA' From equation (1) and (2), we get Slfll=-.-=--

... (2)

sini = sinr

Li= Lr which is the law of Reflection LAW OF REFRACTION ON THE BASIS OF HUYGEN'S THEORY XY is a plane surface that separates a denser medium .of refractiVe index µ from a rarer medium. If v 1 is velocity of light in rarer medium and v, is velocity of light in denser medium, then by definition

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µ =---1.

... (1)

. v,

AB is a plane wave front incident on XY at LBAA' = Li , where 1, 2 are the corresponding incident rays normal to AB According to Huygen's principle ... (2)

The secondary wavelets from A travel in the denser medium with a velocity v2 and would cover a distance AB'= v 2t in the same time. 2 Incident

wave front

be coherent. The emission of light from any source is from a very large number of atoms and the emission from each atom is randoffi and independent of each other. Therefore, there is no stable phase relationship between radiations from two independent sources. So, for two sources to be coherent, they must be deriv~_q. fr9II1 the same parent source. In practice, coherent sources are obtained either by dividing the wavefront (~~,,in the case of Young's Double Slit Experiment, Fresnel's biprism, Lloyd mirror, etc.) or by dividing the amplitude (as in the case of thin films, Newton rings, etc.) of the incoming waves from a single source. A laser discovered in 1960, is different from common light sources. Its atoms act in a cooperative manner so as to produce intense, monochromatic, unidirectional and coherent light. Thus, two independent laser beams can produce observable interference on a screen. METHODS OF PRODUCING COHERENT SOURCES A,

DIVISION OF WAVEFRONT

In this method the wavefront is divided into two parts by the use of mirrors, or lenses or prisms. Well known methods are Young's double slit arrangement, Fresnel's biprism and Lloyd's single mirror.

2' 1'

So, from MBA' and MB'A' B.

. . BA' d. AB'. smz=-- an smr=--

AA'

=>

sini

AA'

BA AA' AA' AB' 1

BA' AB'

vt v 2t

--=--X--=-= 1

sin r

= v1 V2

So, from equation (1), we get sini =.:'!_ = µ which is the Snell's Law of Refraction. v2

smr

INTERFERENCE

When two waVes of same frequency, nearly same amplitude and constant initial phase difference travel in the same direction along same straight line, they superimpose in such a way that- in the region of superposition, the intensity is maximum at some points and minimum at some other points. This modification in intensity in the region of superposition is called Interference. The sources having the same frequency and .co~tant initial phase difference are called coherent sources. The phenomenon of interference is based on the Law of Conservation of Energy.

DIVISION OF AMPLITUDE

In this method the amplitude of the incoming beam is divided into two .parts by means of partial reflection of refraction. These divided parts travel different paths and are finally brought together to produce interference. This class of interference requires broad sources of light. The common examples of such interference of light are the brilliant colours seen when a thin film of tr~parent material like soap bubble or thin film of kerosene oil spread on the surface of water is exposed to an extended source of light. This kind of interference exists in tvvo types. a) Interference due to waves reflected from both the front and back surfaces of the film. b) Interference due to transmitted _waves. INTERFERENCE: MATHEMATICAL TREATMENT

Two waves (whether sound or light) of equal frequencies travelling almost in the same direction show interference. Consider two waves coming from sources S1 and S2 • These reach point P with a path difference t,:,c, having amplitude A 1 and A 2 •

COHERENT SOURCES

Two sources which emit light of the same wavelength with zero or a constant phase difference are .called coherent sources. Unlike sound waves, two independent sources of light cannot be coherent. Sound is a bulk property of matter. So, two independent sources of sound can produce coherent waves. However, two independent sources of light cannot 2.3

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Advanced JEE Physics ... (1)

y, =A,sin(cot-kx) and y, =A,sin[cot-k(x+LU)] ~

By Principle of Superposition, the resultant wave at P is

y = y, + y, = A, sin( cot-kx)+A,sin( cot-kx-,j,)

... (3) Substituting A1 +A,cos,j,=Acos0 and A,sin,j,=Asin0, we get ... (4) A'= A;+ Ai+ 2A,A, cos,j,

=>

ll.x=(2n+1) ,where n=O, 1,2,3, .....

,.

2

So, intensity will be maximum when phase difference ,j, is an odd multiple of ,r or path difference LU is an odd

~.

~ Imin=I1 +I,-2M=(Jf:-.,/f;)' =K(A,-A,)' The ratios

1

rnu

/min

(JI,+.,µ,)' (A,+ A,)' (JI, - .,µ,)' (A, -A,)'

I= = 4I, and Imin = 0

... (5) A, sin,j, A, +A,cos,j,

The intensity of the resultant wave is

I =-pvco'A' =KA' =K[ A; +Ai +2A1A,cos,j,] 2

I=I1 +I2 +2,Jf;i;cos,j,

Thus, when interference of two waves of equal intensities occur, the intensity of maxima becomes four times that of single wave and that of minima becomes zero.

In Young's Double Slit Experiment popularly known as YDSE, usually the intensities 11 and 12 are equal, so I1 =I,=I0

. 1

~

2 <1>=( :}=(2n+1)1t, where n=O, 1,2,3, .....

If 11 = 12 = 10 (i.e., A, =A,), we have

Equation (3), becomes y = Asin(cot-kx-0)

JA; +Ai+ 2A,A, cos,j, and tan0

~

multiple of

y=(A, +A,cos,j,)sin(rot-kx)-(A,sin,j,)cos(rot-kx)

where, A=

,j,=(2n+l),r

... (2)

y, =A,sin(cot-kx-,j,)

2 where ,j, = kLU = ( ; )LU and LU is the path difference.

~

~

... (6)

Interference term

Thus, when interference of two waves of equal intensities occur, the intensity of maxima becomes 4 times that of single wave and that of minima becomes zero.

CONDITION FOR MAXIMA: Constructive Interference From equation (6), I is maximum, when cos,j,=+1

Since, IR= 11 + 12 + 2,Jf;i; cos~, so we get

I=2I,(l+cos,j,)

~

I=4I0 cos'(¾)

PHASE DIFFERENCE AND PATH DIFFERENCE

If two waves travel different lengths of path to reach a point, they may not be in phase with each other. The phase 2 difference depends on the path difference as ,j, = " LU,

=>

4'=0, 2rc, 41t, ....

~

,j,=2n,r

~

<1>=(~}=2n1t,where n=O, 1,2,3, .....

waves.

M=(2n) ,where n=O, 1,2,3, .....

The phase difference between two light waves can change if the waves travel through different materials having different

=>

,.

where LU is the difference in length of path traversed by the

,.

2

So, intensity will be maximum when phase difference ,j, is an even multiple of 1t or path difference 8X is an even

multiple of

~

?: . 2

I= =11 +I,+2,Jf;i; =(Jf: +.,ff;)' =(A, +A,)'

CONDmON FOR MINIMA: Destructive Interference Intensity I_ will be minimum, when

cos,j,=-1 =>

if,=rc, 31t, 51t, ....

refractive indices: Suppose, we have two waves having identical wavelengths A., initially in phase, in air. One of the waves travel through medium 1 of refractive index µ 1 and length L and other wave travels through same length L in another medium of refractive index µ 2 • As wavelength differs in a medium, the two waves may not remain in phase.

The path difference after crossing through the medium is given by

LU=(n,-n,)L where n1 is number of wavelengths in medium 1 and n2 is number of wavelengths in medium 2

_.;..,, 2.4

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Wave Optics two sources was first demonstrated by Thomas Young in 1801.

Illustration 1

Two light rays, initially in phase and having wavelength 6 x 10-' m , go through different plastic layers of the same thickness, 7 x 10~ m . The indices of refractiqn are 1.65 for one layer and 1.49 for the other. (a) What is the equivalent phase difference between the rays when they emerge? (b) If those two rays then reach a common point, does the interference result in complete darkness, maximum brightness, intermediate illumination but closer to complete darkness, or intermediate illumination but closer to maximum brightness?

Light is incident on screen A , which is provided with a narrow slit, S0 • The cylindrical waves emerging from this slit arrive at screen B, which contains two narrow, parallel slits, S1 and S2 . Light emerges from these two slits as cylindrical waves. In effect, slits S1 and S1 act as individual light sources that are in phase as they originate from the same cylindrical wavefront. The light from the two slits produces a visible pattern on screen C . The pattern consists of a series of bright and dark parallel bands called fringes. The overall light amplitude at a given point on the screen is the result of the superposition of the two wave amplitudes from S1 and S2 • Two waves that add constructively give a bright fringe, and any two waves that add destructively produce a dark fringe.

Solution

(a)

ru:=(µ 1 -µ 2 )!=(1.65-1.49)(7xlO-') lll:=1.12x10_. m Since, Phase difference

=>

~ =(~" )
Max

$=( } ~_7 )t1.12x10-<) 6 1 ~ = 11.72 radian

=> (b) To discuss this case, two options arise Option 1 : Waves are in phase, then using

I =I-cos'({), we get

Schematic diagram of Young's double·slit experiment.

I= Im~cos '(11.72) - =0.81_ 2 This value is intermediate illumination closer to maximum brightness.

The narrow slits act as sources of cylindrical waves. Slits S 1 and S2 behave as coherent sources which

produce an interference pattern on screen C.

YDSE (QUANTITATIVE TREATMENT): METHOD 1

Option 2 : Waves are out of phase, then ~~, =11.72±1t

=>

~~,

=>

14 86 I=I_cos'( ; )

=>

I =0.1710

= 14.86 rad

Viewing Screen

1bis value is intermediate illumination closer to darkness.

Geometric construction fordescribingYoung's double-slit experiment. Note that the path difference between the

two rays is r2 - r1 =dsin8

THEORY OF INTERFERENCE : MAXIMA & MINIMA THEORY OF DIVISION OF WAVEFRONT: YOUNG'S DOUBLE SLIT EXPERIMENT

The phenomenon of interference of light waves arising from

Consider a point P on the viewing screen located a perpendicular distance D from the two identical slits S1 and S2 , which are separated by a distance d . Let us assume that the source is equidistant from the hvo slits and is 2.5

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monochromatic, that is, emitting light of a single wavelength A.. Under .these assumptions, the waves emerging from slits

S, and S2 have the same freque11cy and amplitude and are in phase. The light intensity on the screen at P is the resultant of light .coming from both slits. Note that a wave from the lower slit travels fartl!er_ than a wave from the upper slit by an amount equal to dsin0. This distance is called the path difference, x , where ·

fringe~ is.also equal to p = "~ . Since the quantities ;

and

d are both measurable, we see that- the double-slit interference pattern, together with equation (6), provides a direct determination of the wavelength ;\. . Youpg used this technique to make the first measurement of the wavelength oflight.

... (1)

x=r,-r,=dsin9

The value of this path difference will ,determine whether or not the two :waves-are in phase when they arrive at .p-. If the path difference is either zero or some integral multiple of the wavelength, the two waves are in phase at P and constructive interference results~ Therefore, the condition for bright fr4,ges, or constructive inte~ei-ence, at -P is given by

x=dsin9=n;\.

This result shows that the separation between adjacent dark

(n=O,±1,±2,±3,..)

· ... (2)

The index number n is called the order number of the fringe. The central bright fringe at 9 = O(n = 0) is called the

YDSE (QUANTITATIVE TREATMENT): METHOD 2

s;

Conside; that two coherent sources•of light and S2 are placed at" ; dist~ce ·d · ap:.rt and. a screen is placed· at a distance D from the plarie of th~ two sources._ ~t P be a point on the screen at a distance y frorri a point O exactly opppsite to the_ centre of the two sources S1 and S2 • If x .is path difference between the light waves reaching point P from the sourCes S1 and S2 , then

x;s,P-s,"r

zeroth order maximum. The first maximwn on either side, when n = ±1, is called the first order maximum, etc. -

Similarly, when the path difference is

pl

an odd multiple of ~,

the two waves arriving at P will. be opposite in phase and will give 'rise. to. destructive interference. Therefore, the

-~

~ is,

condition for dark fringes, or destructive·interference, at Pis given by x = dsin9 = (2n + 1)2': 2

(n=O,±1,±2...)

s

It is useful to obtain expressions for the positio~ of. the bright and dark fringes measured vertically from O to P . We shall assume that D > d and consider only points .P that are close to O . In this case, 8 is small, and so we can use the approximation .sin9 = tan9. From the large triangle ·oPQ in Figure, We see ~t

~-----~-- ---------- : l

d Q~---- ---------------

1n right angled t.S,BP,

-B

we have

S,P' =52 B' +BP' =D' +(y+f)'

... (4)

Using this result .together with equation -(2), we see that the positions of the bright fringes measured from O are given by

y

LJi...--'-- D ----<• 1Screen

' Also, in right angled sin9ztan9=';

.

~ i s,n.-----------------~~-

... (3)

,_

LiS,AP, we have ·

S,P' =S;A' +AP' =D' +(y-f)' 2

2

=>

S2P -S1P =(D' ;(y+fJ)~(D' ,{y-JJ)

=>

(S2 P+S,P){S2 P-S1P)=(y+%J-(y-%J

... (5) From this expression, we find that the separation between

any two adjacent bright fringes called Fringe Width is equal to

;\.D

d't

h

. atts,

light waves),

-1..D ;\.D ;\.D "=y -yn =-(n+l)--. P n~1 d d n=d

..... (6)

Similarly, using equation (3) and (4), we find that the dark fringes are located at . ;\.D Ya.uk =(Zn+ 1)-

2d

=

Since S2 P-S1P=x (the path difference between the two

... (7)

th~ abOve equ~tion becomes:

(S,P+; 1P)x =4y(%)~ 2yd X

2yd S2 P+S1P

In practice, the·point .P lies very close to the centre of screen, so we have

2.6

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2yd

(b) Phase difference corresponding to the given path

rue is given by

difference

2yd

X=--=D+D 2D

$=(:")rue

X=yd

... (1)

D

=>

For Maxima, we know that path difference x must be an

$=(2t)(~)

even multiple of ~, so

x = (2n)

).

2

, where

n = 0 , 1, 2, .....

=>

~ =n').., where n =0, 1, 2, .....

=>

y.=n().~),where n=0,1,2, .....

Illustration 3

At y = 0 (i.e., for n = 0 ) we get a Central Bright Fringe. For Minima, we know that path difference x must be an odd multiple of ~, so 2

x = (2n + 1)

).

2

yd

D=(2n+1)

, where 11 = 0, 1, 2, .....

).

2

, where

In YDSE, the interference pattern is found to have an intensity ratio between the bright and dark fringes as 9. Find the ratio of (a) intensities. (b) amplitudes of the two interfering waves. Solution

In case of interference, we have n=O, 1, 2, .....

=>

l.D Y. =(2n+l) d, where n=O, 1, 2, .... . 2

=>

Y.=(n+½)).~ ,where n=0,1,2, .... .

I =11 +I, +2N, cos$ (a)

r_ =I, +I, +2N, =(JI: +F,)' and Imm = 11 + 12 -2N, = (F, - F,)'

(F, +F,_)' (F, -F,_)'

Since, 1max Illustration 2

[min

If the maximum intensity in YDSE is I 0 , find the

F, +.fl, F,-F,_

intensity at a point on the screen where

(a)

the phase difference between the two interfering

).

(b) Since, I oc A2 ,

1S - .

4

Solution

=>

Imax

sources is

(~:J

=>

~=2 A,

is 10 i.e., intensity due to independent

~ . Therefore, at

~

I.i..=( I2 A2

=>

Since, l=lm,,cos'(t) where

3 1

I.i..=±=4 12 1

(b) the path difference between the two interfering beams

(a)

1

Solving, we get

beams is .'.::, 3

.

9

J

=4

Illustration 4

I =10 cos '(") =io 3

6

The intensity of the light coming from one of the slits in YDSE is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. 2.7

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Advanced ]EE Physics (c)

Solution Since, we know that I/min

Position of second minima is obtained by putting n = 2 in the equation

=(K..p; -.Jr, +.Jr,)'

11 = 210 and 12 = I0

Im~ =

1_

(,/2 ,/2 +-11)'

34

=

Illustration 5 In a Young's double-slit experiment the distance between the slits is 1 mm and the distance of the screen from the slits is 1 m. If light.of wavelength 6000 A is used, find the distance between the second dark fringe and the fourth bright fringe. Solution The position of the second dark fringe is given by

y 2 (dark)=(2n-ll"D =(4-ll"D =~(,.D) 2d

2d

y 2 =(2-.!.)1,.D =~P=~(0.2)=0.3mm 2 d 2 2

According to the problem, we have

=>

y, = (2n-1) AD, so we get

2d

2

Illustration 7 Young's double slit experiment is canied out using microwaves of wavelength A = 3 cm . Distance between the slits is d = 5 cm and the distance between the plane of slits and the screen is D = 100 cm . Find the number of maximas and their positions on the screen. Solution The maximum path] [Distance between the] difference that = coherent sources [ can be produced i.e., 5 cm Thus, in this case we can have only three maximas, one central maxima and two on its either side (for a path difference of 1,. = 3 cm)

d

The position of the 4th bright fringe is given by . nJ,.D 41,.D y,(bnght)=-a-=-d-

f 81:

Therefore, the separation is given by

1s, ...r-_-_--_--:--------1

fly= y4 (bright)-y, (dark)= (

=>

fl

10

5 6000 X 10y=zx 10"'

X

4-¾Y~

For maximum intensity at P, we have S,P-S1P=A

1 l.S X lQ°" = l.5 mm

(y;d)' +D' - (y;d)' +D' =A

Illustration 6

=>

In a YDSE , the separation between the coherent sources is 6 mm, the separation between coherent sources

Substituting d = 5 cm, D = 100 cm solving the equation, we get y=±75 cm

and the screen is 2 m . If light of wavelength 6000 A is used, then (a) find the fringe width. (b) find the position of the third maxima. (c) find the position of the second minima. Solution (a) Since fringe width is given by p= 1,.~ , so we have

and

1,. = 3 cm

and

Thus, the three maximas will be at

y=0 and y=±75cm FRINGE WIDTH & ANGULAR FRINGE WIDTH

The separation between two consecutive bright (or dark) fingers is called the fringe width (p), given by

10

(6000x10· )(2) =O.Z mm 6xl0'3 d (b) Position of third maxima is obtained by substituting P=i,.D

n=3 in the equation

2

y, =n(A~),soweget

31,.D

y, =d=3P=3(0.2)=0.6 mm D--~ r:::::l 2.8

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.Wave Optics

:w

f3=Y11+1-Yn·=7 Note that fringe width p is independent of n. That is, the interference fringes have same width throughout. The angular fringe width is given by

CONCEPTUAL NOTE(S) In VOSE alternate bright and dark bands obtained on the screen. These bands are called Fringes. ·

s LJ ·

. 'Hf -~---u1·---s,

~

·.

Screen 1 3_ Bright liiail!liil4 Dark . ht 1 3 Dark 2 Bng : 1 Bright lliil/illfiilll2 Dark

lillllliill

d)

e)

f)

14mm

----1-~~i~~;

=

21 (1000)( 400 X 10_.) 2x0.1 So, required separation is Lly=42-14=28 mm

Yn

::-:-~:~:---bri~~tn;;i~ge

2 Bright ~ 2 Dark .' 3Dark . ht_.. j, 3 Brig D . ~4.0ark

' at Central position Central fringe is always bright, -because the path difference, x =.0 a,nd hence the phase ,illSo called the zeroth maxima. The nth ·minima_ comes before the'nth maxima'. The fringe pattern obtained due to a,slit is more bright than that due to a point. If the slit widths are unequal, the minima will not be complete dark. For very large width uniform illumination occurs. If one slit is illuminated with red· light and the other slit is illuminated with blue light, n9 interf~rence pattern is observed on the screen. If the two coherent sources consist of object ?nd ws· reflected im_age, the central fringe is dark instead of bright

.__ _:o::.n•:c·c____ _- " - - - - - - - - - - - - - - - --~-

Illustration 8

In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used.' Th~ distance between the slits is 0.1 mm and the distance 'between the plane of the slits and the scree_n is 1 m . Find the minimum distance between two Successive regions·of complete darkness. Solution . . . Let 71i minima of 400 nm coincides with n,_ minima of 560 nm, then

(2n 1 -1)400 = (2n, -1)560 2n 1 -1 ='!...= 14 =21 2n,_-1 5 10 15

42mm

Light from a source consiStS of two wavelength A-1 = 6500 A and A-1 = 5200 A . If the separation between the sources from each.' other is 6.5 mm and that from the screen is 2 m, find the minimum value of y(;t:O) where the maxima of both the wavelengths coincide. Solution Let n 1 maxima of A-1 coincides with n2 maxima of A. 2 •

Then, Yn1 ~ Yn:z 71iXp d

difference, $ =0° and hence tjle Central Bright Fringe is

b) c)

7(1000)(400x10_.) · 2x0.1

Illustration 9

d = Distance between slits D = Distance between slits.and.screen A.-'=' Wavelength of monochromatic llght emitted from source a)

Y,

Next , 11th minima of 400 nm will coincide .with 8th minima of 560 nm Location of this minima is

d~=I=~ D d

n

i.e., 4th minima of 400 nm coincides. with 3rd minima of 560nm The IoCation of this minima is.

= n,,.,D d

5.= ,., = 5200 =± n,_ '-1 6500 5 Thus, fourth maxima of A- 1 coincides with fifth maxima of

'-,. The minimum value of y(;, 0) is given by

41.,D y=-d-=

4(0.65x10.. )(2) 6.5x10..,

0.8mm

Illustration 10

Two coherent sources are ·o.3 mm apart. They are 0.9 m away from the screen. The second dark fringe is at a distance of 0.3 cm from the centre. Find the distance of fourth bright fringe from the centre. Also find the wavelength of light used. Solution Second dark fringe n = 2 will be obtained at = 3'-D Y=(2n-1)"D •'2d 2d

i..D

...

2

=>

-=-y d 3

=>

A-D 2 P=a= (o.3)=0.2 m

3

Fourth bright fringe from the centre will be obtained at y, =4P=0.8 cm 2.9

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Optics & Modern Physics

Advanced JEE Pl1ysics From equation (1), we get 1,_ 2yd 2 X 0,3 X 10-,3 X 0.3 X 10-' 3D 3x0.9

=>

=>

n+l n

=>

n=4

=>

n1,.D 4x6500xl0-10 xl.2 Y, =-d-= 2xl0_,,

=>

y, =l.56xl0_,, m=l.56 mm

7

1,.=6.67xl0- m

Illustration 11 In a Young's double slit experiment, two narrow slits 0.8 mm apart are illuminated by a source of light of

I

source of monochromatic light (,. = 5000 A) is placed in front of the lens at a distance·of a= 15 cm from it. Solution

10

P=""D d

5 4

Illustration 13 A convergent lens with a focal length of f = 10 cm is cut into two halves that are then moved apart to a distance of d = 0.5 mm (a double lens). Find the fringe width on screen at a distance of 60 cm behind the lens if a point

wavelength 4000 A. How far apart are the adjacent bright bands in the interference pattern observed on a screen 2 m away? Solution Since, d=0.BxlO_,, m, 1,.=-4000 A=4000xl0-10 m, D=2m The distance between the adjacent bright bands or the fringe width is given by

6500 5200

Applying lens formula,

4000xl0- x2 m=lmm O.BxlO_,,

,! _,! = .! , we get V

f

U

1

1 1 15 10 => v=30cm V 30· , Smce, m=-=-=-2 u -15

-+-=-

Illustration 12 A beam of light consisting of two wavelengths, 6500 A and 5200 A, is used to obtain interference fringes in a Young's double slit experiment. The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm . (a) Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 65ooA (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? Solution According to the problem, we have "-, = 6500 A = 6500 x 10-10 m

v

~ ~ s

________ _)'

~ 15 cm

Fringe width p "-~

d=2 mm=2xl0_,, m

YJ

=

=>

(n + 1)1,.,D

d

d

""-, = (n + 1)1,.,

D--+1

60cm----+1

(5xrn-')(0.3) (l.5x10_,,)

In water (liquid), of refractive index µ the wavelength 1.17 x 10_,, m = 1.17 mm

the wavelength coincide

corresponds to that value of n for which

n1,.1D

0.5mm

INTERFERENCE EXPERIMENT IN WATER

(b) The least distance from the central maximum where the bright fringes due to both

30cm

0.25mm 0.25mm

P=O.lmm

D=l20 cm=l.2 m n1,.D Y, =-d-

=>

' ''

s2!'

0.5mm

Distance between two slits is d = 1.5 mm Distance between slits and screen is D = 30 cm

I

3x6500x10-10 xl.2

''

-----1

1+----

"-, = 5200 A =5200x10-10 m

(a)

'

S,t'

'

"-

to A.' =Therefore, if interference µ experiment is performed in water the fringe width decreases decreases from A.

from

p to P' , such that P=""D and P'=""'D =""D d d µd

=>

P'=~ µ

= c================================== 2.10

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Wave Optics Illustration 14

A Young's double slit arrangement produces interference fringes for sodium light (,. = 5890 A) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is immersed in. water? Refractive index . 4 of water1s -.

41,

'

;-:,

·:'' '

3

Solution

The wavelength of light in water is Angular fringe-width in air,

So,

d

d

µd

µ

d

b)

B --+-y

±

passing through slits, then

11 _a~_ W 1 -:--7:-l2

c)

Im,. Imaic

d)

2N, 11 + 12

=> e)

and if I_ =lmin, then V =0 f)

INTENSITY DISTRIBUTION

When two coherent light waves of intensity 11 and 12 with a constant phase difference $ superimpose, then the resultant intensity is given by

W2

(F.-Jw,)' =(,µ,-,µ,)' .jI;

(a,-a,)' a1 +a2 = .{vi; +,Fi;

=

l1

best. Also if I_ =0 then V =-1

a2

+.JI;

If point source is used to illuminate the two slits, the intensity emerging from the slit is ·proportional to area of exposed part of slit. In case of identical slits.

If [min = 0, V = 1 (maximum) i.e., fringe visibility will be the

= l2

a1 =a2

When white light is used to illuminate the slit, we obtain ,an interference pattern consisting of a central white fringe having on both sides symmetrically a few coloured fringes and then a uniform illumination. If qi

IR

I gl

=11 +12

h)

+2M cosq,

If x is the path difference, then

IR=I1

I 1 =I, =I0

!

+12 +2Mcos(~1txJ

In YDSE , if n1 fringes are visible in a field of view with light r of wavelength A1 , while n, witti light of wavelength A2 in \

!'

Since, IR= 11 + 12 + 2.jf;i;·cosq>, so we get (a)

Between_ nth bright and m 111 bright fringes (n > m)

I= 4!0 cos'(¼)

ilx =(n-m)~

For maxima, . q> = 2mt

=>

In:un

=0

'

Separation (file) between fringes

I =2I0 (1+cos$)

For minima, $ = (2n + 1lit

I

is the phase difference .between two waves of /

I

In YDSE, usually the intensities 11 and 12 are equal, so

I= =410

I

intensities 11 and 12 , then

I =I, +I, +zN, cos$

=>

I

the intensity of light (with respective amplitudes a1 and a2 )

With the help of the concept of visibility, the knowledge about coherence, fringe contrast and interference pattern is obtained. Fringe visibility V is defined as

~

l

If w1 and w2 are the widths of the slits and 11 and 12 is

FRINGE VISIBILITY (V)

I= + Imm

D

i

3

v r_-r_

B

PROBLEM SOLVING TRICK(S) a) lntt::rference occurs due to L_aw of Conservation of Energy. / Actually, redistribution of energy takes place. ,

ew = '-w

e - '-w - ,. - 0, - 0.200 - 0,150 w

B

Intensity distribution on the screen as a function of yin YDSE Ima,= 410 for bright fringe and Im1n =0 for dark fringe.

µ

e, = !:

Angular fringe-width in water,

D

D

'-w = !:

(b)

I i [

111

Between n bright and

(i)

mth dark fringe

If n>m then fil< =(n-m+1)~ 2

(ii)

;

I

II n<m then fil<=(m-n-¾)~

j) Identification of central bright fringe : To identify central j ___ Qrigh!.f!inge, m9n.9-chroro__c!!i.9Jight.isJepJaced_by_whiJ1q_ llgt]_t! :

===================================== = 2.11

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Optics & Modem Physics

Advanced JEE Plzysics

I

Due to overlapping central maxima will be white with red) edges. On the other side of it we shall get a few coloured 1 band and then un_iform illumination. j

j 1

' Illustration 15 In YDSE , light of wavelength 60 run is used. The separation between the sources is 6 mm and between the sources and the screen is 2 m . Find the positions of a point lying between third maxima and third minima where the intensity is three-fourth of the maximum intensity on the screen.

~

Now, NS,= dsin8 and

Solution

MS 2 = 52 P -S,P = yd (as done earlier) D

Since, ·1 = 410 cos'(¾) where l = ~

3

4(41

cos(¾)=

0)

1=mt±2:.

~

~=2nn±-"

2

~

= 310

FOR MAXIMA, Llx =(2n)

),_

2

, n =0, 1, 2, 3, .....

FOR MINIMA, Llx = (2n + 1)

6

2ny 11 d

).. D

),_

2

, n = 0 , 1, 2, 3, .....

Illustration 16

3

A coherent parallel beam of microwaves of wavelength ).. = 0.5 mm falls on a Young'~ double slit apparatos. 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 ai a distance of 1.0 m from it as shown in the figure.

2 Since, ~ = " Llx where Llx = y,d ).. D

~

Llx=ldsin0-~I

Once the path difference Llx is known, then

'7

~

~

s,

= 2mt±2:. 3

y.=(n±¼Y~

y

For the point lying between third minima and third maxima, we have

-"'--'3,,,o,_'_,-1-4--d_-_-_1_m_m_ _ _--l--+x

n=3 ~

Y, =(3-.!_))..D

~

Y,- 6

·

6

d

l--D=1m

_ 17()..D)

d

Y,

17 (0.6x 10.. )(2) 6x10"" 6

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 30° with the x-axis (as in the dotted arrow shown in figure), find the y-coordinates of the first minima on either side of the central maximum. (a)

Substitoting ~=0.6x10_. m, D=2 m, d=6 mm, we get 5.67mm

YDSE WHEN INCIDENT RAYS ARE NOT PARALLEL TO CENTRAL LINE

In this case, the r~ys reaching S1 and S2 already have an initial path difference. So, net path difference between the rays reaching the point P is given by

Solution

(a)

Given A=0.5 mm, d=1 mm, D=l m When the incident beam falls normally Path differenc~ between the two rays S2 P and S1P is Llx = S,P-S,P ~ dsin8

Llx=(NS, +S,P)-S2 P

~ . Llx=JNs, -(s,P-S,P)J=INs, -MS,I

For minimum intensity, dsin0=(2n-1).!:, n=l, 2, 3, .... 2

= ================================:::J 2.12

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Wave Optics ~

~inS=(2n-1);\. 2d

(2n~1)0.5 2x1

Path difference between the rays after passing through

2n~1

the slits S1 .and S2 is

4

S,P-S1P=l1x2 =dsin8 So, net path difference is given by

l1x = INS, +S,P-S,PI = [NS, -(S,P-S,P)[

~ S;S, =d(«D)

l1x = 111x, -11x, I= 1~·- d sin el

For first minima, we have

l1x =!: 2

Since, sin 0 :5: 1 ~

(2n-1) ,, 1 4 n:,2,5

~

~

l~-dsin81 =~

~

.'!_-dsin8=±!: 2 2

So, n can·be either·1 or 2

When n=l,wehave sin91

tans,=

~

=..!.

4

sin8=!+_.?:_ 2 2d Since d=1 mm, ;\.=0.5 mm _ . 1 0.5 => Stn 8 =-+-

=

~

1 ·v15

·.. . . When, n ~ 2 , we have sin 02 -

3

=-4

2

3'

~

~

tan8, = r.; - v7 -

Since, y=Dtan8=tan8(D=1 m)

=

1 m v15 3Y, = Dtan8, = .ff m

sin8=!+!

2 4

central maxima is

~

Since, minima can be on either side of centre O , so

.

±.ff m on the scr~e_n.

~

(b) Path aiffe~~nce between the rays before· entering the slits S1 and S, is . d NS1 =l1x1 =dsin(30°)=... (1) 2

tan8 =Jl D y=Dtan8

For sin8=!,wehave tan8= ,;.. 4 v15

there will be four minimas at positions ± _,;.. m and v15 3

2

1 3 ·0 .=' 8 =sm andsm 4 4 Since the position of first minima on either side of

So, the position of minima will be

y, =Dtan81 =

1 - 8 =±,. --sm 2 2d

1

y= ../ts m.

For sin8=~, we have tan8= ~ 4 · v7

-~

3

y= .ff m

y

YDSE FOR SOURCE NOT PLACED AT THE CENTRAL LINE. . .

If the source of light is not placed at the central line but a little beyond (slightly up or down) the central line, then the waves reaching S1 and S, will already have an initial path difference.

Path difference of waves meeting at the point P is l1x=(SS, +S,P)-(SS1 +S1 P)

================================== = 2.13

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Optics & Modern Physics

Advanced JEE Physics

=>

tu=(Ss,-ss,)+(S,P-S,P)

=>

Lu=y'd+yd D' D

To calculate the order of interference, we shall calculate

,.

L,x

n=-

---------::?lpfy

0.035x10-a m 5000x10-10 m

=>

n

=>

n=70

=> tu =701So, 7Qth order maxima is obtained at P .

_J__ 0

(b)

Screen

(c)

central line and D' is the distance of S from S1 and S2 •

2

A,

2

, n=0, 1, 2,3 .....

Illustration 18

In PROBLEM SOLVING TRICK(S)

the

YDSE

the

monochromatic

axis (as shown in the figure), where d is the separation between the two slits S1 and S2 •

n

sJ1_----:::::7f\1Pi

In the Young's Double Slit experiment a point source of ),, = 5000 A is placed slightly off the central axis as _shown in the figure.

t t

of

2

0

Illustration 17

1 mm

source

wavelength ),, is placed at a distance '!_ from the central

At the position of central maxima, we-have ill(.=

t = O.Ol = 0.02 mm = 20 µm

1.5-1 Since the pattern has to be shifted upwards, therefore, the film must be placed infront of S1 .

, n =0, 1, 2, 3 .....

FOR MINIMA, Lu=(2n+1)

(µ-l)t=0.01 mm

=>

Similarly, here once we calculated the path difference, then

FOR MAXIMA, Lu= (2n)

D,

Now, we observe that Lu= 201So, 20th order maxima is obtained at O .

where, y' is the distance of the source S above or below the

A,

At O, Lu= y,d = 10-2 mm= 0.01 mm

-----------· O

i

~ 1+ £

(a)

=1.5 m+r<---D =2 m---+'0

Find the position of the central maxima.

(b) Find the order of interference formed at O .

(c) . (a)

Find the nature and order of the interference at the point P. (b) Find the nature and order of the interference at O . (c) Where should we place a film of refractive index µ =1.5 and what should be its thickness so that a maxima of zero ·order is placed at O .

The optical path difference between the two waves arriving at P is Lu= y,d + y,d = (1)(10) + (5)(10) D1 D, 103 2xl0 3

=

intensity at O becomes !th 4

of the maximum

intensity. Given A.=6000

A and

d=6 mm.

Solution

Solution

(a)

Find the minimum thickness of the film of refractive index µ =1.5 to be placed infront of S2 so that

(a)

(Lu)".= 0

=>

y,d = y,d D1 D,

=>

=>

tu= 3.5 x 10-2 mm= 0.035 mm

2.14

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Wave Optics

d 6 y=-=-=4mm 1.5 1.5 (b) At O , net path difference is given by ~

(f)
y,d llX=-=-D, D, ~

~

(6xl0")

"-missing

(

d2

d' ,d' A.=D, 3D' 5D""

2

2xl.5

,ix =12x10-' m

Illustration 19

7

lix=120x10- m

Since, 1,, = 6000 A = 6 x 10- m ~ lll:=201,, 1

So at O , the bright fringe of order 20 will be obtained. (c)

d'

) , where n = 0 , 1, 2, 3, ... 2n+l D · By putting n = 0, 1, 2, 3 .... Missing wavelengths are

= rel="nofollow">

I=Im.,cos'(¾)

White light is used in a YDSE with separation between the sources to be 0.9 m and separation between the sources and the screen to be l_ m. Light reaching the screen at position y =1 mm is passed through a prism and its spectrum is obtained. Find the missing lines in the visible region of this spectrum. Solution

Path difference is given by

¾rm., =I-cos'(¾)

,ix= yd= (9xl04 )(1x10-') = 900 nm D

1 = 2'. 2

6

For minima Ill:= (2n + 1)~, where n = 0 , 1, 2, 3, ....

~=~=(2;)tµ-l)t

t=

A, 6(µ-1)

6000 6(1.5-1)

A,=~= 1800 (2n+l) (2n+l)

2000 A

,_

_ 1800 1

missing-

MISSING WAVELENGTH IN FRONT OF ONE SLIT IN VOSE

Suppose P is a point of observation in front of slit 51 as shown. The wavelengths missing are the ones obtained by using the condition of destructive interference, i.e., A, iix=(2n+l)-, where n=O, 1, 2, 3, .... .

2

1

1

1800 7

Of these, 600 nm and 360 nm lie in the visible range, so these will be missing lines in the visible spectrum. SUSTAlNEDINTERFERENCE

The interference pattern in which the positions of maxima and minima remain fixed is called a sustained interference.

a)

... (1)

iix=D[(1+~:)½-1]

l

Sel+ ~, 2

So, (1) becomes d' A, iix = - = (2n + 1)2D 2

l J----- -:.. .

1800 1800 3 5

CONDITIONS FOR SUSTAINED INTERFERENCE

Now iix=.JD'+d' -D

Since, D»d, so (1+ ~:

1

'~ ,____ D _ _ _..

The fundamental condition for sustained interference is that the two sources should be coherent i.e., the initial phase difference between the two interfering waves must remain constant with time. b) The amplitudes of the two waves should be equal or nearly equal. This will give good contrast between bright and dark fringes. c) The two sources should be very closely spaced, otherwise the fringes will be too close for the eye to resolve. d) The sources should be monochromatic, otherwise there will be overlapping of interference patterns due to different wavelengths, which will reduce are contrast. e) The frequencies of the two interfering waves must equal. f) The sources should be nanow. Since hvo independent sources cannot be coherent, a sustained interference pattern can be obtained only if the two sources simultaneously and, therefore, the phase difference behveen them remains constant.

==================================== = 2.15

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Optics & Modern Physics

Advanced JEE Physics ORDER OF FRINGES

If the slits are verticat as shown in figure, path difference is, M=dsiri0

n=8

If d =101. (say)

n=9 ---+---+-----------\

n = 1O

n= dcose ).

n=1Oat0=O'

s,~

CONCEPTUAL NOTE{S)

Screen

To calculate ,the .number of maximas or minimas that can be

obtained on the screen, we use- the fact that value of

This path difference increases as 0 increases. The order of fringe H is given by·

sine(or case) can never be greater than 1'. For example in the first case when the slits are vertical.

dsin0 = HA. =>

sin9=~ d

,.

dsin0

n=--

{for maximum intensity}

$ince, sine'/> 1.

The order of fringe increases as we move away from point 0 on the screen.

ni. ~ 1 d

=>

d i.

n~-

Suppose in some ,question

n=2

t s, ~

=>

d ).

comes out say 3.6 then total

number of maxihlas on the screen wm be 7. Corresponding to, n = O, ±1, ±2, ±3.

n=1

d _ _,.,__ _ _ _ _ _ _ ___, n = o

So, highest order of interference maxima is

:t .s,~

nm~

9 n - dsin ).

=[P.] ).

where [ ] represerts the greatest integer function. So, total ~ number ofmaxlmas obtained are

-

n=Oate;,0°

N=2nmax +1

When the slits are horizontat as shown in figure, then the path difference is ax= dcose

.

nP

Similarly, highest order Qf interferenc~ minima is

nm,,

=H:+f]

where [ ] represent~ the greatest integer function. So, total number of·minirTlas obtained are N =.2n~;n

Illustration 20

~ c s,

II u

s,

M-d-+I

·

Two coherent narrow slits emitting light of wavelength A. in the same phase are placed par.Jlel to each other at a small separation of 2,. . The light is collected on a screen S which is placed at a distance D(» ,.) from the slit S1 as shown in figure. Find the finite distance x such

Screen

This path difference decreases as 0 increases The order of fringe H is given by

that the intensity at P is equal to intensity at O .

dcos0 = HA. :::::::>

ep f

,.

dcos0 n=---

X

The order of fringe decreases as we move away from point

0.

~1-. - - - ~ 2_ _ _ _ _ _ _ _ _ _ _ _ _ _ _

0

f+-2).-+l

>+---- D - - - - < " i i

=

s 2.16

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,, Solution

Wave Optics

pf

Path differen_ce for w9ves reaching.at O is

s,o - s,o = 21,.

y

-----e<"'------+------{J C S2 S1

i.e. maximum interisity is obtain_ed at O . 1

Next maxima will be obtained at point P where, S1P-S2 P=1,,

i.

i+-nA-tot+--D

')1/2 "D+L2

(

Now, S1P=(D 2 +y'iv'=D l+JL,

D

... (2)

2D

2

Similarly, S2 P = (D + d) + ( y ) 2 D+d

~\

S1 -,_

S2

Substituting (2) and (3) in (1) i.e., S2P-S1P =(n-1)1., we get

t
=>

dcose = A.

Y'd 2D(D+d) Since, d =n'J...

=>

(21,,)cos8=~

=>

n1.

=>

ny' 2D(D+n1.)

=>

y=

i+----D----+IIJ

PO =tane

s,o

~=tan60'=v'3 D

=>

x=v'3D

maxima. At point P , where path difference is l. (so, x = ../3D ) we get first order maxima. The next, i.e., zero order maxima will be obtained where'path difference, ~cos ff= O 9=90° X4CO

so; our answer, i.e., finite distance of x should be x=../3D., Corresponding to first order maxima.

Illustration 21

Two point sources are d =nA apart. A screen is held at right angles to the line joining the two sources at a.distance D from the nearest source. Calculate the distance of the point on the screen, where the first bright fringe (excluding the centre one) is obser_ved. Assume D » d. Solution

At point C path difference is n1,, . Therefore, nth bright fringe will be observed. Next bright fringe is observed where path difference is (n-1)1,,, so S,P-S,P=(n-1)1,,

y' (nA.) 2D(D+n1,,)

n1.-1,,

1

2D(D+n1.) n

OPTICAL PATH

CONCEPTUAL NOTE(S) At point O ·, pa,th difference is 2A. i.e. we obtain seco~d order

=> =>

(n--1)1,, · ·

d

s

1 => cos0=2 => 8=60' Now in AS1 PO

=>

... (3)

It is defined as distance travelled by light in vacuum in the same time in which it travels a. given path length in a medium. If light travels a path length d in a medium. at d . speed v , the time taken by it will lie t = - . So, the optical V

path length is OPL=cl=c(;)=(;}=µd Since, for all media µ > 1, optical path length greater than geometrical path length. When two light waves arrive at a point by different distances in different media, the phase between the two is related by their optical path instead of simple path difference. So, 21t . Phase Difference = T( OPL).

I·--µ=;) is always

travelling difference difference

DISPLACEMENT OR SHIFTING OF FRINGE PATTERN IN YDSE

When a transparent film of thiCkness t and refractive index µ is introduced in front of one of the slits, the fringe pattern shifts in the direction where the fiim is placed. How much is the fringe shift?, . Consider the YDSE arrangem~nt shown in the figure.

... (1) 2.17

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(µ-i)t . If shi~ is ·equivalent to n" fringes then n =- - or ),,

t=~· (µ-1)

e)

µ, t

~s,

Illustration 22 Interference fringes are produced by a double slit arrangement and a piece 1>f plane parallel glass of refractive index 1.5 is interposed in one of the interfering beam. If the fringes are displaced through 30 fringe widths for light of wavelength 6 x 10_. cm, find the thickness of the plate.

l+-----D - - - ~ . ,

A film of thickness t and refractive indexµ is placed in front of the lower slit.

·The optical path difference is given by x = [(S,P-t) + µt]-S,P

=>

Solution Path difference due to the introduction of glass slab is

x=(S,P-S,P)+t(µ-1)

Since S2P-S1P=dsin0

=>

ru:=(µ-l)t

x=dsin0+t(µ-1)

Since sin 0 ~ tan 0 =

The Sh!ft, tlx is i_ndependent of the order of frihge n., i.e. Shift Of.Zero order maxima = Shift of nth order maxima: Shift is independent of wavelength.

Thirty fringes

are displaced due to the introduction of slab.

Y:D

So,

d' x=-1&+t(µ-1)

=>

(µ-l)t=3OA.

The maxima will be obtained when the path difference-is an

=>

t=--=----

even multiple of ,?:. i.e., 2

=>

t=3.6x10__, cm

=>

Llr=3OA.

D

,. x=(2n)

2

=>

,. d ' (2n)-=-1&+t(µ-1) 2 D

=>

y;= n~D -(µ-l)t~

In the absence of film, the position of the nth maxima is given by equation nAD Y, =-d-

Therefore, the fringe shift (FS) is given by FS=y,-y;c'~(µ-l)f=t(µ-l)t

\-:~="'~)

Note that the shift is in the direction where the film is introduced.

a)

CONCEPTUAL NOTE(S) 'The "enlife pattern shifts t6wards the side where the Pl.ate

is:intrOduced and there is no other change in the pattern~ TO aj~asJ:Jrei this, shift .:,Vhit8 li9ht•·.must be used: bec1;1use , With, monochrom8.tic light,all the fringes will exactly~ be slinilar.and hehce- no,Shift.Can"b8 observed. " · ' . (. . ,, c) .The·8ffective pat~ in air'is inCf~ased by an a,rnount'' (J..L-:.1)t' .~due~tojntroduction oMhe ,PlateJe .. jhe :additionai' 'path b)'.

~

=

2.18

~

3OA. µ-1

3Ox6x10·5 1.5-1

Illustration 23 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. 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. Solution Shifting of fringes due to introduction of slab in the path of one of the slits is given by

,

uy

(µ-l)tD d

... (1)

Now, the distance between the screen and slits is doubled. Hence, the new fringe width will become

P' = ,_(2D) ·d

... (2)

Given, !:!ly = J3'

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(µ-l)tD

1.(2D)

d

d

(µ-1)1 2

(1.6-1)(1.964x10-<) 2

1,.

=>

Illustration 2S

1,. = 0.5892 x 10-< m = 5892 A

Illustration 24

In a YDSE, the two coherent sources are separated from each other by 6 mm and from the screen by 2 m . A light of wavelength 6000 A is used. A film of refractive index 1.5 is introduced in front of the lower slit such that the third maxima shifts to the origin. (a) Find the thickness of the film. (b) Find the positions of the fourth maxima. Solution (a) Since third minima shifts to the origin, therefore, the fringe shift (FS) is equal to three fringe widths i.e., 3P,

A Young double slit apparatus is immersed in a liquid of refractive index µ 1 • The slit plane touches the liquid surface, A parallel beam of monochromatic light of wavelength 1,. (in air) is incident normally on the slits. (a) Find the fringe width (b) If one of the slits (say S2 ) is covered by a transparent slab of refractive index µ2 and thickness t as shown, find the new position of central maxima. (c) Now the other slit S1 is also covered by a slab of same thickness and refractive index µ3 as shown in figure due to which the central maxillla recovers its position find the value of µ 3 •

-sl

!-=:-~'i--------" ---- ()--

*=: s. s

so we have

FS=y, =3(1,.~)

1 '-1+------,D----

Since we know that the fringe shift (FS) is given by FS=(µ-1/~

(d) Find the ratio of intensities at O in the three conditions (a), (b) and (c). Solution (a) Fringe ·width is given by

(µ-l)tD = 3 i..D d d

P=i..'D =i..D d µ1d

t=~

µ-1

(b)

Position of central maximum is shifted upwards by a distance

Since, 1,. = 0.6 x 10-< m , µ = 1.5

t (3)(0.6x10-<) = 3 _6 µm

(c)

( µ'-1J10 ~ d

••

i

... (2)

Since the central maxima recovers its position, so

n=4 O=n= 3 ===n=2

t

dY=dy'

So, from (1) _and (2), we get

n=O

y',

l

(µ1-l)tD

i.---D---.. c:::::::::::::J

... (1)

Downward shift is now given by

~

f ~ -------------- Yf, 1 ~

(µ,-l)tD d

dY

1.5-1 (b) There are two positions of fourth maxima, one above and the other below the origin. So, we have

~

{-: i..·=~}

==:::,n=4

d

=>

Bright Fringe

1'D y,=1P=d=0.2mm and

y; =-7P=-7(,.~)=-1.4 mm

( µµ'1

=> (d)

-1Jm d

µ3 =µ2

µ, µ3

=µ1µ2

Since, I=I=,cos'(t) where

~ =(~" ),u: 2.19

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Optics & Modern Physics 1 tan(20°)=-

0

¼=(f)t1x

o

I oc cos'(¼) ..

2.J2

Solution

(a) The situation is in .the problem 'is shown in the figure. So, we observe that separation behveen the source_s is

In the first and third case, ·/:u = 0 while in second case, /:u = (µ2 -l)t • Therefore, the desire•ct"ratio

'

f 1 :f2 :f3 =1:cos

µ=3/2

~I

.

'("(µ,-l)t)· ,. . .1.

I

I

=~=7:-'=~-++.:---'·-~·P.

y

Lx

PROBLEM SOLVING TECHNIQUE(S)

I.

If. two thin plates are also inserted ju,~t after S1 an~ S2 , then_ our first task is to find the path difference. 1n th!3: figure shown, path of ray 1 is more.than path of ray 2 by'a distance,

,;x, = d sin a

and

X=O

1+- D

,;x, = (µ, -1) t,

X=1.03rn

= f m --+I '

d=2x0.9=1.8mm , :\. 900 Since, we !mow that :\. = - = = 600 nm

and path of ray 2 is greater than path of ray 1 by a distance. ill<,=dsin~ and ill<,=(µ,-1)t,

µ

Therefore, net path difference is, ill(=

I

X=0.03 m

3

2

(ill<,+ ,;x,)-(,;x, +,;x,) •

Applying Snell's Law at the x = 0 interface, we get 3

sin(30°) sina . 1 sma=2

3

0

Central Line

l1x =dsina=(1.si(½)=0.6 mm

Once, we know the path diff~fence illC , then

FOR MINIMA,

l.

2

,;x =(2n +1) l. , n = 0, 1,,2, .....

2

(b)

at x = 0.03 m . The region x ;e 0 is filled with a transparent liquid of refractive index

~2 . A wide monochromatic beam .

of light of wavelength 900 nm falls on the yz -plane at x = 0 making an angle of 30° with the x-axis. The sheet has two slits parallel to z-axis at y = ±0.9 mm. The intensity of the wave is measured on a screen placed at x = 1.03 m parallel to the sheet. (a) Finci the intensity at a point P on the screen where y=z=0. (b) The lower slit is covered by a transparent strip of refractiv~ index 1.4 and thickness 4.2 mm. Now find the intensity at point P. 2.20

0

2 $= " l1x )..'

0

f=f-cos'(½)=r=

, n = 0, 1, 2, .....

Illustration 26 A large opaque sheet placed parallel to the yz plane

.=

{Using Trigonometry}

Initial path difference is

2

FOR MAXIMA, ill<= (2n)

a= 20'

2 " .. x0.6xl0" =-2000n 600xl0

Net path difference at P is now l1x=(0.6 mm)+(

o

$=10001t

:::::)

[=[max

15 -1)(4.~ mm)=0.3 mm 1.4

Illustration 27 In a Young's Double slit Experiment, the light source is at distance 11 = 20 µm and 12 = 40 µm from the slits. The

light of wavelength ;\. = 500 nm is incident on slits separated at a distance 10 µm . A screen is placed at a distance D = 2 m away from the slits as shown in figure. (a) Find the values of 0 relative to the central line where maxima appear on the screen? (b) How many maxima will appear on the screen? (c) What should be minimum thickness of a slab of refractive index 1.5 be placed on the path of one of the ray so that minima occurs at C ?

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t -

s

incident at an angle = 30° on a diaphragm having narrow slits at a· separation d = 2 nim . The screen is placed at a distance D = 40 cm from slits. A mica slab of thickness t=5 mm is placed in front.of one of theslits and whole the apparatus is submerged in water. If "the central ·bright fringe is observed at C, calculate

I

I,

s,

e,

s,I

~~f 1", + A, ld _______,, C

Solution

s,J!

(a) The optical path difference between the beams arriving at P is given by

-/1.,----

ru: = (1, ;-1,) + dsin8

The condition for-maxnr{um intensity is, d.l'=nA.,wher~ n=O, ±1, ±2, ...... .

(a) the refractive index of the slab. (b) the distance of the first black line· from C. Both

sin8 =¾(ru:-(1, -11 )) =¾{n1.-(l2 -4))

=>

D ---->'-"- Screen

wavelengths are in air. Take µw = _!. 3

1

=>

sin8

=>

sin8=2(; -1) 0

=>

lOxlO

_. (nx500x10-,-20x10~)

Solution

(a)

C is,

sine =,,':.-2 20

ru: = dsin-(µ, - l)t

=>

8=sin·'(; --'2) 0 (b) Since we know that

=>

For central bright at C we have ru: = o

dsin4>=(µ,-1)t

=>

-1,;(; -2},10

=>

( µ,

=>

µ, =1.2

=>

-l) = dsin

2 =( " _,)(20x10-<)=801t 500x10

I

µslab

• µw

2 2 At C,phasediffere~ce, <1>= "~=( ")(1,-4) 1. · 1. _

=>

0

lsin0kl

=> -20:,(n-40),;20 => 20,;n,;60 Heri.ce, number of maxima obtained is N=60-20=40 (c)

To observe bright fringe at C, the mica slab should be placed in front of S2 • In that case, net path difference at

=>

µ, .. =1.2µ. =1.2(¾)=1.6

(2n-1)1.D

y

2d

.

I

2

2(µ-1)

(2n,-1)1.;D -(2n,-1)1.;D 2d. 2d · .and t..;_ are wavelengths in water.

For blacklme,

(µ-l)t=!:'. =>

0.2

=1.2

For minimum intensity at C , we have

,..

0

(b) A black line is formed at the position where both the wavelengths interfere destructively. Distance of nth dark fringe from C is given by

Hence, maximum intensity will appear at C .

-

(2xl0·3 )sin(30°) 5x10-'

-where_

sooxio... = 500 ~2xo.s

·

t..;

,.,

A; · µw

A1

4000

=>·-=-=-=-,.; ~ ,., 5600

Illustration 28

In a Young's double slit experiment a parallel beam containing wavelengths 1.., = 4000 A and 1. 2 = 5_600 A

µw Substituting these values in equation (1), we get 2.21

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y 2 =asin(rot+$)

2n,:...1=z: 2n2 -1 S

y, = asin( rot+ 2$)

.. For minimum value n, = "4 and n, = 3 •

,

r

y, =asin(rot+3$)

•

Henc!', d,istance of first blacJ< line is given by

y

(2x4-1)(4000x10-1~)40x10-2 x3 · 2x2x10-' x4

=>

y=2.lx10-1 m

=>

y=0.21mm

y. = asin[rot+(N-1)$] At angle 0, the path difference b~tween any -two successive slits is tu = d sin 8 . So, the corresponding· phase difference $ is given by 2 2 $=( ; }~= :(dsin8)

MULTIPLE SLIT- INTERFERENCE PATTERN Let us now look at the case where we have a general number N of equally spaced slits, instead of two equally spaced slits. AB an assumption, we have. shown in figure. a se.t-up of six equally spaced slits.

RESULTANT WAVE AMPLITUDE (USING PHASORS) The· above set of equations can be represented by phasor · diagram shown in figure (for a set Of six sources generalised to N sources).

p .,/

r/

./

4>\

R

-----

.,'

i

.. F

:~~-~:::______ t , I

1 ' \ \

: Set up fur 6 Equally Spaced Slits

Similar to the N = 2 case discussed already, we will make the far-field ...... assumption that the distance of the sources from the screen is much larger than the total span of the slits, which is (N -1)d. We can then say, as we did in · the N = 2 case, that all the paths to a given point P on the screen· have approximately the same length in a -~ multiplicative (but not additive) sense, which implies that the amplitudes of the interfering waves are all essentially equal and we can also say that all the paths are essentially parallel (because of far-field assumption). A close-up version near the. slits is shown in figure. Also, each path length is _dsin8 longer than· the one just above it. So the lengths take the form of r. =r1 +(n-l)dsin0.

· y1 =asin(rot)

=

2.22

r/

\

,''

'\

' 'A'

a

........

',

',

'' '

4'

.........

,\

.

...... D

'\

ii

'

If R be the amplitude of the resultant of N interfering waves, then from above phasor diagram, on extracting triangles DAG and· DAB, we get following figures to be used for evaluation of R . 0

G

,1\

~------ I o,-i'.&!

-~

To find the total wave at a given point at·an angle 8 on the screen,. we need to add up the N individual waves. The Procedure is the same as in the N = 2 case, except that now we simply have more terms in the sum. If a_ be the amplitude due to an individual source, then the equations of waves interfering at the point P' are given by

\

y

B.2

L.)' .. 2

:t11' ' 2

r ''I

.

'

:

~I

'" ~

/i:1\

• .

A

.

2

/ 2 I2 \

/ I

'' ' A'

''

1

: '' I

' '''

\

''

\

'' ''

'B.

M

I+- .2. ---+t+- .2. --+I

2

2

DAG and DAB are isosceles triangles so for triangle OAG , we have

¾=rsin(~$)

... (1)

Triangle OA~, we have ½=rsin(½)

- . . .. (2)

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Dividing (1) and (2), we get R

sm(~)

a

sm(½)

/ [ (1:~.,m:Hl ClC'lJ]1

~

y=a1~ e'·'

~

y=alm (e )

~

y-alm e

So, the resultant amplitude R is given by ... (3)

R=-[s:ntT]] If IR be the resultant mtensity, then IR =R2

~

[

,., [,{~',.m)J[,{-',.m_,-{½) ;!) _e-{1¥)]]

_ [ {•••
2

{•«(N-1)!)) 2 =sm ( rol+(N-1)

But Im [ e

1• =•'[sm(~ )]'

sm(½)

$)

2

If 10 be the intensity due to an individual source, then

... (4)

Agam here, we have directly come across the amplitude of the resultant wave given by

CHECKPOINT

For N=2,weget la =I,[

. sm -

2

s:(iJJ'

~

Since sin~= 2sin(½)cos(½) =>

-[sm(~)l ($) sm(~)]' 1. 1[ R-a

=

0

sm(½)

, where

I'.= a'

We observe that at the centre of screen, IR is indeterminate.

Ia =4I,cos'(½)=4a'cos'(½)

So, the·maximum intensity at the midpoint of the screen i.e., at 0 = 0' is obtamed by taking the limit when $--+ 0'.

RESULTANT WAVE EQUATION

2 (Please note that smce $= i_"(dsm0), so $--+0° when

To find the resultant wave equation, we shall be using the concept of complex numbers. From our knowledge of complex numbers, we know that

0 --+ 0° ). So,

e'•' = cos(rot) + ism(rot) ~

Im(e""')=sm(rol) and Re(e""'),=-cos(rot) Smee the resultant wave equation is obtamed by adding individual Waves, so we get

y = a[sm(rot)+sm(rot+ $)+ ..... + sm{rot+(N -1)$)]

-

=>

y = aim[e;"''.+ ei(mt+t) + ..... + ei(m1+(N-tlt)]

~

y = a!m[e'"'(l+ e'1 + ..... + ,,cN-1l+ )]

. l +r+r , + ..... +r N-1 =1(1-r") Smce, -.1-r

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LOCATION OF SECONDARY MINIMA(S)

IR has zero values, when

. '(N$)-o

Slil

=>

-

0.8

-

2

N$ = Integral Multiple of 2

-

N$

0 1t

4

-

=>

T=m1t, where m=0, 1,2,.3, ....

=>

$ = (2m).':.

=>

.2 -5,

N

-2,

2

However, one exception to this is when

!2

0

2:_3,c7t5'1t31t

2

4

42

2"

'"

2

"

N=4

... (5)

$=Even Multiple of .':. N

·31t -Sn -n -3n -n

2 4 42

!(a) l(O)

is also an integral

multiple of 1t, because the denominator in equation (4) is also zero.

So, f=m'rr.,where m'=0,1,2,3, .....

=>

$=(2m')1t

=>

$ = Even multiple of 1t

... (6)

So, from (5) and (6), we conclude that f, = 0, when $ = (2m); excluding $ = 0, 21t, 41t, 6,i, .....

---------------------·

Positions of Primary Maxima

i.e., $=(Even Multiple); , excluding Positions of Primary

a}

CONCEPTUAL NOTE(S) It is customary· 'not to deal With the resultant intensity

alone, buJ rather fo deal with the resultant inte!]sity relative to the maximuni intensity i.e.,· _!a_

Maxima (located at 0, 21t, 41t, 61t , ....)

•,

-

LOCATION OF SECONDARY MAXIMA(S)

derivative I, w.r.t. $(i.e., ~;) and then equating it to zero ..

(J,,__) =+--.o lim (..!e...) =;

c)

?-

~

has"a periodicity of 21t in

cl>

i.e., repeats itself for

"""

d)

This equation has to be solved numerically. However, for large N , the solutions of $ are generally very close to odd

e)

region. Just to make you understand, we are plotting

~

·

For N=2 (for two slits)~· ~e gBt,ZERO Secondary Maxima.. For N = 3 (for 3 slits), we get One Secondary Maxima. .

I, I,

The nUmber of .zeros between the ~fnain peaks· is, (N-1) , where N is the number of slits used. The number of secondary maxima (little bumps) between. the main peaks is (N-2), where N is the number of slits

used.

multiples of _2:_ excluding the values of q> = 21t±~, because N N these values will be lying well within the primary maxima Q

For N =4 (for 4 slits), we getTwo Secondary M~ima a~d so on. A point worth noting here is that the height of the se~ondary maxima (~ittle bui'nps).i.e., the bump sizes are symmetric around cj> = tt (o( in general. any, multiple of n:·).

_ _ Also,_ we know that since

=

... (1)

integratmultiples of 21t .

Ntan(½)=tan(~$)

(with$ for N=4 and N=8).

. (N$)]'

sin2

[

lim o--.o- Imax

d$ =>

I.

1, : = 1: = Nsin(½)

b)

So, dl, =0

·I

~

To find the locations of the secondary maxima (i.e., small bunips) we have to find the local maxima of I, by taking the

,lmax

2.24

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,, =,, [

s~~n~r]r

=1 t =1

d

Hence, the bump size is shortest at

4' =1t ,

because then

4, = re ,

due to which IR becomes the

least at 4' = n . Furthermore, the bump size grows as they get closer to the main peaks, as shown for various slits taken.

d

-

the denominator in the equation (1 ), will be having a maximum value at

t

-I

i

D

(a) the distances from P where intensity reduces to zero. (b) the distances from P where next bright fringe are observed. (c) the ratio of intensities of bright fringes observed on the screen.

Single slit

./.__

Solution N=2

(a)

_ _._"-~>-<--'-_,.,_"--i---'"-'--+-'-..,_~-"-<-- $ 1' Primary: Secondary

--------

------

: ______ :maximum: 1 I

maximum

In case of three slits, intensity becomes zero, when phase difference between any two waves is, 21t 4'=2mt+ , where n =0, 1, 2, ....

3

...

The corresponding pa~ difference,

N=3

~

-- ---

.. ----- ------ ------ -- ........

-- .... --

................

------ ---- .....

:J$

2 dsin0=U'J( t+zim)=n1.+½ sine=~( n+½)

''

$-2• -3

$'

-- .......- ------

~=(

A,

A,

................

A,

N=5

For small angles, sin 0 "' tan 0 = 1l D

~

y=,.~(n+½)

Y

~

500x10_,_; 0.5xl0

2(n+.!:.)= 2x10-a(n+.!:.) m 3 3

y=2(n+½) mm,wh~re n=0,1,2, .....

2 8 y=-mmforn=0,y=-mmforn=letc. 3 3 Bright fringes are obtained on the screen where (i) n=l,2,3, .... =>

(b)

A light wave of wavelength 500 nm falls upon three slits a distance 0.5 mm each from one another. A screen is placed at a distance 2 m from slits. Find

;=~(n+½)

Substituting the values, we have

As N , the number of slits, is increased, the primary maxima (the tallest peaks in each graph) become narrower but remain fixed in position and the number of secondary maxima increases. For any value of N , the decrease in intensity in maxima to the left and right of the central maximum, indicated by the blue dashed arcs, is due to diffraction patterns from the indfvidual slits, which can be ,__ _n-"e"'gl~cted he~ ________ g)

Illustration 29

~

$=2n1t,

~=(:")($)=n1. ~

dsin0=n1. 2.25

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Optics & Modem Physics camera cannot be obtained. To obtain a fixed interference pattern we must have two sources Which either have no phase difference or have a constant difference of phase. These sources are called coherent sources. It has been generally observed that coherent sources are obtained when they are derived from the same parent source. The methods for obtaining coherent sources (derived from the same parent source) are given below.

. 0 =n1.. sm d For small angles,

sin8" tan8 =JL D }!_ = n1..

D

y

d n1..D

n(soo x 10-,)(2)

d

(0.5x10"')

A.

y=2nx10"' m y=(2n) mm

{where n =1, 2, 3, .... etc.}

There are called priniary maximas.

(ii)

~=(2n+l)it, n=l,2, .... Proc_eeding in the similar manner, we get

y=2(n+½) mm

{where n=l,2,3, ....}

DOUBLE SLIT METHOD

Light from a source S is limited to a narrow beam with the help of a slit. The emergent light is made to fall upon a screen containing two slits S1 and S2 placed symmetrically with respect to the slit. Here, both S1 and S2 are illuminated by the same wavefront. Therefore the beams of light coming out from S1 and S2 have no phase difference. Thus S1 and S2 can· be treated as the coherent sources. Young used this technique in his famous Young's double slit experiment.

D

These are called secondary maxim.as. Note that y = 0 is also a secondary maxima

\?-------~-·--

because at P, ~= 1t. (c)

At principal maxirnas, we have

~

=21t, 41t, .... , etc.

s, D

----.A, ----.A,

=> - - - - - ~ A= 3A,

Coherent sources by Double Slit Method

----.A,

Resultant amplitude R = 3A,

=>

B.

{·: I oc A')

I, =910

While at secondary maximas, ( ~ = 1t, 3n, 51t..... ) A , - ----.A, ----. A,

-----A=A,

Resultant amplitude, R' = A,

=>

I~ =10

So, the desired ratio is therefore, 9 : 1

A SOURCE AND ITS OWN VIRTUAL IMAGE

Light from a source S is made to fall on a plane mirror M . Point of observation P on a screen AB receives direct light as well as light reflected from M. To an observer, reflected light appears to come from a source S1 (virtual image of S ). So interference at P takes place between waves coming from S and S' . Since S' is not an independent source, being the virtual image of S , it will have the same phase as S . Hence the two are taken to be coherent sources. Lloyd made use of this arrangement in Lloyd single mirror experiment. A

COHERENT SOURCES BY DIVISION OF WAVEFRONT

When two or more waves travel through a medium simultaneously, the resultant intensity at any point, in the medium depends on whether they interfere constructively or destructively which, in tum, depends upon the phase difference between them. Resultant intensity, at any point, remains constant with time if the phase difference between them does not change. Two inP,ependent sources can never have same phase or a constant phase difference, because if we try to have interference with two independent sources, then net intensity at any point undergoes a continuous change due to a change in the phase difference between them. As a result of this no fixed interference pattern can be observed. The interference pattern of such sources is so short-Jived that its photograph with the fastest available

=

p

s

(/)

~

2

rn

M

C.

"

BIPRISM METHOD

Light from a source S is made to fall on an assembly of two right angled prisms A and B joined base to base as shown in Figure. S1 and S2 are the virtual image of S produced by refractiqn through prisms A and B respectively. Being

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a)

phase and henc~ can be treated as the coherent sources. This type of arrangement is made use of in Fresnel's biprism experiment.

Determination of D : It is the distan~e between source . and screen. 'It can be ~easured with an ordinary metre rod.

b)

Determination of p : A low power travelling microscope is used to find the total separation x

between a number of fringes, say 20 and hence

c) FRESNEL'S BIPRISM

It is one of the convenient laboratory arrangements for producing interference fringes. It consists of a combination of two right angled prisms with their bases joined together so that their faces are inclined to each other at angle of 179° 20' . Source of light is taken in the form of a narrow slit S, illuminated by the monochromatic light and is held symmetrically at a distance of about 5 cm from the biprism. Light from S gets refracted by prism P1 and P2 , thereby, producing virtual images S1 and S2 , which can be taken as two coherent sources })reducing interference. Light beams from S1 and S2 strike the screen in the regions ED and FG respectively. EF is the common region where both the beams can be found. Therefore interference pattern can be observed in the region EF .

p=.=... .

20 Determination of d : d can be calculated by using displacement method. A convex lens is placed in between the biprism and the screen. It is observed that for two positions I.., and L2 of the lens, the images of S1 and S2 can be focussed on the screen AB . Let x and y be the distances between these images when the lens is at I.., and L, respectively. Then,

d=..{xy L,

P,

L,

"

s,.

II '

A

' ''' ''

s,.

I I I

Screen

I I 'VI

P,

Substituting for calculated.

p, D

B

~d d in equation (1), 1,, can be

LLOYD'S SINGLE MIRROR

This experimental set-up for producing interference fringes, was devised by Dr. Lloyd in 1834. Light from a source S1 in the form of narrow slit is held in such a way that the light is incident, at almost grazing incidence, upon a mirror MM'' which is blackened at the ba'ck to avoid internal reflections. S2 is the virtual image of source S1 obtained after reflection B

The separation ,,between these sources may be found by using the formula for deviation caused by a thin prism. If u is the small angle of- biprism, µ refractive index of material of biprism and a the separation of source S from biprism, then deviation c~used by prism.

from MM' . Screen AB is placed to receive light coming directly from S1 as well as that reflected from the mirror. Reflected light can be supposed to be coming from source S2 • DF is the common region on the screen where both the· beams are received and hence interference is obtained in region DF. A'

o=(µ-l)u

A

D

From Figure, d = 2ao = 2a( µ-1 )u

s,

DETERMINATION OF 1,,

'

Biprism method can be used to determine the wave length of light. The fringe width p for the interference pattern obtained is given by,

P=1,,D

D

: ,,' t-4---s, I--,:"----,.__ _ _ D------,----

d

A= Pd

' ---~----d• ,,.

'

... (1)

B' B Experimental set up for Lloyd's single mirror

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The point C lies symmetrically w.r.t. S1 and S2 and also

AP,

lies outside the interference region, zero order fringe is not visible. It can be seen by moving the screen to position A'B' so that it just touches the mirror. It will be observed that the zero order fringe at M' is dark instead of being bright as

(AS')(BS') BD

(190+5+5)(0.1) 5

4 cm

demanded by the theory of interference fringes since at M' path difference is zero. Tilis indicates that the beam which suffers reflection from MM' undergoes in phase of 1t-radian.

Llyod's single mirror can be used to determine the wave length of light. H, a is the height of source S1 above MM', then

Similarly in triangles BCS' and S'P,A, we have AP, AS' BS'= BC

d=2a

If, D is the distance of source S1 from screen AB, then fringe

=>

P=i,,o d 1'.=

2

cm

• C 3x10 8 -7 Wavelength of the light 1'.=-=--1-4 =5x10 m f 6xl0

pd= p(2a) D

(AS')(BS') BC

(190+5+5)(0.1) 10 P,P, =AP, -AP1 =2 cm AP,

width p is given by

D

p can be determined, experimentally, by using a low power microscope. Knowing p, a and D value of 1,, can be calculated. CONCEPTUAL NOTE(S) Central spot, in case of Lloyd's single mirror is a dark one instead of being bright. This proves that there is a phase change of n-radian when a transverse wave (light) is reflected from a denser medium.

Fringe width p = i,,~ Since, D=S'A =(190+5+5)=200 cm=2 cm, d=SS'=2 mm=2xl0" m p (5xl0-')(2) 2x10" Number of fringes is

5xl0-4 m=.0.05 cm

N= P,P, =40

p

Illustration 30 The arrangement £or a mirror experiment is shown in the figure. S is a point source of frequency 6 x 10 14 Hz. D and C represent the two ends of a mirror placed horizontally and LOM represents the screen. Determine the position of the region where the

fringes will be visible and calculate the number of fringes.

Illustration 31

Two flat mirrors form an angle close to 180° . A source of light S is placed at equal distances b from the mirrors. Find the interval between adjacent interference bands on screen MN at a distance OA = a from the point of intersection of the mirror. The wavelength of the light wave is known and equal to 1,, • Shield C does not allow the light to pass directly from the source to the screen. N

5 cm 5 cm

cm-+:

190

A

M

Solution

Fringes will be observed in the region between P, and P2 because the reflected rays lie only in this region. From similar triangles BDS' and S'P2 A

2.28

AP, = AS' '

BS'

BD

M

Solution

Fringe width is given by P=1'.D d where D=AB~a+b and d=S,S 2

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s,r-----

i

l

, -- •

I

:

di 81 ____ 0' I

'

......

----a.,..._ s

-----'!.-t---------------

\

1

..... .._

--r;

--

A

_-".,.

x=2µtcosr

82~-----

For maxima 2µtcosr =.n;>,

I+------ D - - - - - . a

and for minima

In !1S15B, we have

2µtcosr=(2n-1J.

.'!.=2b~ 2 2 d=2ba 2ba THEORY OF DIVISION OF AMPLITUDE REFLECTED LIGHT

If µ is the refractive index of material of film of thickness t, then path difference between the waves abc and abdef is 2µtcosr

Additional path difference due to reflection at denser medium ( at b) is .'.: 2

a

,. , 2

C

Illustration 32

A thick glass slab (µ = 1.5) is to be viewed in reflected white light. It is proposed to coat the slab with a thin layer of a material having refractive index 1.3 so that the wavelength 6000 A is suppressed. Find the minimum thickness of the coating required. Solution Optical path difference for the reflected light from coating and slab is l1x = 2µ!

For minimum intensity,

,. 2µ,t =2

m

n=l,2,3, ...

Obviously, the conditions of interference in reflected and transmitted lights are opposite to each other, therefore if the film appears dark in reflected light, it will appear bright in transmitted light and vice versa. With the use of white light, the colours visible in reflected light will be complementary to those visible in transmitted light, i.e., the colours absent in one system will be present in the other system; the sum of two constituting the white light.

p ;>,(a+b)

A.

TRANSMITTED LIGHT

In transmitted light system there is no phase difference or path difference due to reflection or transmission as all reflections take place from rarer medium. So, the effective path difference is

A/----jBf Cl µ,~1.3 iDl

t=l:....= 6000 4µ 1 4xl.3

n

So, effective path difference is

,.

--------- - - - - - - . . ·-···------3

.=>

x=2µtcosr+2

For maxima or constructive interference to take place, we have

,. =(2n) ,. 2 2 ,. 2µtcosr= (2n-1) , n =1, 2, 3, ...

-

t=1154A

CONCEPTUAL NOTE(S) Both reflected rays (one from AB and the another from CD) get a phase change of 1t •

-

------·

------·-·-

2µtcosr+ =>

Illustration 33 A parallel beam of white light falls on a thin film

2

4

For minima or destructive interference to take place 1,,

2µtcosr+ =>

2

,.

= (2n + 1)

2

2µtcosr=nt.., n=l,2,3, ...

whose refractive index is equal to - . The angle of 3

incidence i = 53° . What must be the minimum film thickness if the reflected light is to be coloured yellow ( ;>, of yellow = 0.6 µm) most intensively? Given tan53° =

i, 3

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Solution According to Snell's Law, we have sini

BASED ON INTERFERENCE (Solutions on page 2.86)

µ=-.smr

=>

4 3

In a Yourig1s Double Slit Exp~riment.carried out in a liqµid Qf retr8.ctive index· µ = 1.3 , a thiri fi1in of air is formed in front

1.

sin(53°) sinr

37° •

4 5 sinr ·

5

of the lower slit as shown in the figiJre. If a maxima of third 0 order is f0rmed at the oriQin O ; find th_e -

4 0

(a) (b)

3 sinr =5 => r=37° From Figure (a) :

thickness of the air film. positions of the fourth·maxima. The wavel8ngth Of light i'n 'is

53'

=>

3

D

A. 0 _= 0:78 µrn ~ and·

d=1090.

.. sJ ~ L--t----- ·:__·. --. _. -;~o

2

µ=1.3

I

,I

i s,U

I

D (a)

Path difference between 2 and 1 is

In YDSE , if light of wavelength thickness of ~•glass slab

ru; = 2(AD)

~

D 0

2.

~

~Airfilm_ ·

E,

I

=>

air

.

:sooo A is used, find the

(µ .=1,5)

.

~

;.,hlch should be piac~d . ' . , . ,..

befor~. the uppe;' the upper, sli( S1 so that the-' c~r\tral

ru:1 = 2BDsecr = 2tsecr

maximum now lies at a point where ·sii bright fringe was 1 lying earl!er (bef_ore inserting the slab);:

Their optical path corresponding to ru:1 is 2µtsecr

From Figure (b) :

3.

A source S of wavelength A is kept directly behind the

~lit

S1 in a double slit apparatu$. Find.the phase.difference at a

E

point O which Is equidistant- from · ~ 1 and S2

2

•

If D » d ,

what will be -~he phase difference ·-at. P if a· liquid Of refractiv_e index; µ is filled (a) between the screen ar1d·the Slits? (b) , between the slits and th0 s9urce -· s?

C

fi fils,j_·· . IP· OP=d/2

-,~r ~-------io

1 D (b)

Path difference between 1 and 2 is given by

5

ru: 2 = ACsini = (Zttanr)sini =>

(ru:),,. =ru:1 -ru:2 =2µtsecr-2t(tanrHsini)

=>

.6x

=>

4 5 3 4 =2x-xtx--2xtx-x,~ 3 4 45

L,\xnet

Since reflection takes place at the surface of denser medium, so phase difference between 1 and 2 is 1t • So, for constructive interference, we have 32t=~

15

=>

2

I= 151'. = 15x0.6 =O.l 4 µm 64 64

o

---+1

'4.

_In Solar"cells, a smcdn Solar- c;li'-°{µ =:= 3.5) 18 cocite~·wiih .a

r

thin film "of- silicon- monoxide "SiO(µ =1~45) t9 ITliniITli_?:e

I

32 = 15 f

'1,__ _ _ D

I 5.

reflective lcisses from the s·urtace. 'Determine the mil1imum thickness .of· SiO that ProduC~s the le_ast reflectiOn ··a( ·a wavelength of 550' nm , near the .ceflt_re of the- visible spectrum. • · · · A parallel beam of ·gr·een iighl' ·of waVelerlgth · 546 'nm passes -'through a slit of width 0.4 mm ., The tran?l"!litted: llgh_t is .. collected on a screen 40 cm .away;· Find the· distance between the two first order· ITlinima. Calculate the ·minimum. thickness Ct a soap bubble film (µ =1.33) that results in construt;;tive interference in the

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7.

(b) (c) {d)

Monochromatic light of wavelength 5000 A is used in YDSE , with slit separation 1 mm , distance between screen and slits 1 m . If intensity at the two slits are, 11 =41 0 , 12 =1 0 , find ~

(a)

fringe width

(b)

distance of 5 minima from the central maxima on the

.

(c)

white light (e)

14.

th

screen.

intensity at y =

i

mm.

distance of the 1000 maxima.

(e)

distance of the 5000th maxima. 15.

8.

A. 1 = 400 nm and A= 700 nm . Find minimum order of A, which overlaps with A2

=-6 µm , respectively.

Determine the phase difference (in radian) at the

16.

between x =0 and x = 1.5 µm . What then is the phase difference (in radian) at the origin between the radiation from S, and the radiation from S 2 ?

thickness 8000 A is placed in front of the lower slit, it is : observed that the intensity at a point P, 0.15 mm above the 1 1 central maxima does not change. Find the value of '1 ·

•

In a 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 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.

17.

(b)

18.

In a double slit pattern

(1' = 6000 A), the first order and

a particular reference point. If A is changed to 5500 A , find the position of zero order and tenth order fringes, other arrangements remaining the same. In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelength of light used is 5000 A . Find the distance between 7~ maxima and 11"' minima on the screen. 13.

•

Find the number of bright fringes between P and the central fringe.

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

~ =103 •

Symbols have their usual

meanings.

ray falls onto the plate perpendicularly. The refractive index

tenth order maxima fall at 12.50 mm and 14.75 mm from

Compare the intensity at a point P distant 10 mm from the central fringe where the intensity is 10

In Young's experiment a thin glass plate is placed in the path of one of the interfering rays. This causes the central , light band to shift into a position which was initially occupied

of the plate is 1.5. The wavelength is 6 x 10-7 m. What is the thickness of the plate?

In a Young's double slit experiment set up, the wavelength of light used is 546 nm . The distance of screen from slits is 1 metre. The slit separation is 0.3 mm . (a)

by the fifth bright band (not counting the central one). The

12.

In Young's double slit experiment set-up with -light of :

Suppose a slab of transparent material with thickness 1.5 µm and index of refraction µ = 1.5 is placed

origin between the radiation from s, and the radiation

(b)

•

wavelength A.= 6000 A, distance between the two slits is 2 mm and distance between the plane of slits and the screen is 2 m. The slits are of equal intensity. When a , sheet of glass of refractive index 1.5 (which permits only a , fraction T\ of the incident light to pass through) and

from 8 2

11.

Bichromatic light is used in YDSE having wavelengths ,

radiating waves in phase with each other of wavelength 400 nm . The sources are located on x-axis at x = 6.5 µm and x

10.

In a Young's double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelengths A, = 750 nm and A. 2 = 900 nm . At what minimum distance :

S1 and S2 are two point sources of radiation that are

(a)

9.

The whole experiment is carried out in a medium of ' refractive index µ

from the common central bright fringe on a screen 2 m from the slits Will a bright fringe from one interterence , pattern coincide with a bright fringe from the other?

111

(d)

The (monochromaticf so\Jrce is· replaced by another (monochromatic) source of shorter wavelength The separation between the two slits is increased The monochromatic source is replaced by source of

19.

When a thin sheet of a transparent material of thickness 7.2 x 10-4 cm is introduced in the path of one of the interfering beams, the central fringe shift to a position occupied by the sixth bright fringe. If A= 6 x 10--s cm, find the refractive index of the sheet.

DIFFRACTION : INTRODUCTION & CLASSIFICATION

When light waves pass through a small aperture, an interference pattern is observed rather than a sharp spot of light cast by the aperture. This shows that light spreads in various directions beyond the aperture into regions where a shadow would be expected if light travelled in straight lines.

What is the effect on the interference fringes in a VOSE due to each of the following operations? (a)

The screen is. moved away from the plane of the slits

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r-I--I--J---r---0

-+11).~-1 - - -d---

-- -- ---

!______ -- --- --

t + - - - D ----+1

Uniform intensity distribution

Light passing through two slits does not produce two distinct bright areas on a screen. Instead, an interference pattern is observed on the screen which shows that the light has deviated from a straight-line path and has entered the otherwise shadowed region. Other waves, such as sound waves and water waves, also have this property of being able to bend around corners. This deviation of light from a straight-line path is called diffraction. Diffraction ·results from the interference of light from many coherent sources. In principle, the intensity of a diffraction pattern at a given point in space can be computed using Huygens' principle, where each point on the wavefront at the source of the pattern is taken to be a point source. The phenomenon of bending of light around the comers of an obstacle/ aperture of the size of the wave.length of light is called diffraction. The phenomenon resulting from the superposition of secondary wavelets originating from different parts of the same wave front is define as diffraction of light. Diffraction is the characteristic of all types of waves. Greater the wave length of wave higher will be it's degree of diffraction.

phase of the light at each point in the aperture is the same. This can be achieved experimentally either by placing. the observing screen at a large distance from the aperture or by using a converging lens to focus parallel rays on the screen, as in Figure. Note that a bright fringe is observed along the axis at 8 = 0, with aitemating bright and dark fringes on either side of the central bright fringe. · Common examples : Diffraction at single slit, double slit and diffraction grating.

10

11:.' Incoming wave

B.

Screen

FRESNEL'S DIFFRACTION

When the observing screen is placed at a finite distance from the slit and no lens is used to focus parallel rays, the observed pattern is called a Fresnel Diffraction Pattern. Fresnel diffraction is rather complex to treat quantitatively. Common examples : Diffraction at a straight edge, narrow wire or small opaque disc etc.

?,

I

o~ -------- ---------

!j I+---

D ----+1

Slit

Screen

A Fresnel diffraction pattern of a single slit is observed when the incident rays are not parallel and the observing screen is at a finite distance

CONCEPTUAL NOTE($) Diffraction, can be regarded as a consequence of interference

from many, coherent wave sources. In other words, t_he phenomena of diffraction and interference are basically ~jv_alc.e_nt_._ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

from the slit.

1

FRAUNHOFER DIFFRACTION AT A SINGLE SLIT

Diffraction phenomena are usually classified as being one of two types, which are named after the men who first explained them. The first type is called Fraunhofer Diffraction and the second is called Fresnel's Diffraction. A.

FRAUNHOFER DIFFRACTION

This occurs when the rays reaching a point are approximately parallel i.e. when both the source and screen are effectively at infinite distance from the diffracting device. In this case, the incident light is a plane wave so that the

Consider that a monochromatic source of light S , emitting light waves of wavelength 1,., is placed at the principal focus of the convex lens L,. A parallel beam of light i.e., a plane wavefront gets incident on a narrow slit AB of width d as shown in figure .. The diffraction pattern is obtained on a screen lying at a distance D from the slit and at the focal plane of the convex lens L2 •

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SLIT

amount equal to the path difference

L,

J'l------=,..iPt ---

y

-====::::c=~ ot l+-----D------+< According to rectilinear propagation of light, a bright image of the slit is expected at the centre O of the screen. But in practice, we get a diffraction pattern i.e., a central maximum at the centre O flanked by a number of dark and bright,; fringes called secondary maxima and minima on either sid~ of the point O . The diffraction pattern is obtained on the screen, which lies at the focal plane of the convex lens L2 • It is found that (i)

the width of the central maximum is twice as that of a secondary maximum and (ii) the intensity of the secondary maxima goes on decreasing with the order of maxima. These observations are explained on the basis of the phenomenon of diffraction using the following mathematical treatment.

Consider Frawlhofer diffraction by a single slit as shown in Figure. Important features of this problem can be deduced by examining waves coming from various portions of the slit. According to Huygens' principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the. slit can interfere with light from another portion, and the resultant intensity on the screen will depend on the direction 0 .

}ine, where d is

the width of the slit. Similarly, the path difference between waves 2 and 4 is also equal to

(f}ine. If this path

difference is exactly one half of a wavelength (corresponding to a phase difference of 180° ), the two waves canCel ·each other and destructive interference results. This is tr'ue, in fact, for any two waves that originate at points separated by half the slit width, since the phase difference between two such points is 180° . Therefore, waves from the upper half of the slit interfere destructively with waves from the lower half of the slit. when

d . ,_ -sin 8 =2

=>

2

. 8

Slll

,_ =-

d Similarly, destructive interference (minima) occurs when the 3

5

path difference (r!.)sin8 equals 1,., '-, 21,., '-, etc. These 2 2 2 points occur at progressively larger values of 8. Therefore, the general condition for destructive interference is

·e =m-,_ sm d where

EXPLANATION & MATHEMATICAL TREATMENT

(f

(m=±l,±2,±3, ... )

... (1)

1ml,; '!_,_

Equation (1) gives the values of 8 for which the diffraction pattern has zero intensity. However, it tells us nothing about the variation in intensity along the screen. The general features of the intensity distribution along the screen are shown in Figure. A broad central bright fringe is observed, flanked by much weaker alternating maxima. The central bright fringe corresponds to those points opposite the slit for which the path difference is zero, or 8 = 0 . All waves originating from the slit reach this region in phase, hence constructive interference results. The various dark fringes (points of zero intensity) occur at the values of 8 that satisfy equation (1). The positions of the weaker maxima lie approximately halfway between the dark fringes. Note that the central bright fringe is twice as wide as the weaker maxima. Angular width of central maxima is

2 '- and. width of central d

maxima is Z~D , where D is the distance of the screen from Diffraction of light by a narrow slit of width d. Each portion of the slit acts as a point source of waves. The path difference between rays 1 and 3 or between rays 2 and 4 is equal to (d/2) sin0

To analyze the resultant diffraction pattern, it is convenient to divide the slit in two halves as in Figure. All the waves that origi:rate from the slit are in phase. Consider waves 1 and 3, which originate from the bottom and center of the slit, respectively. Wave 1 travels farther than wave 3 by an

the slit. The intensity distribution of the diffraction pattern is quite different from the interference pattern produced due to superposition of light from two coherent sources. The point 0 on the central ax:s is the brightest. The angular position (8) of n"' diffraction minima is given by dsin8 = n/,. n=l,2,3,4, ...

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Optics & Modern Physics y

Y,

-- --- ----·

I,

Y, ""Rine - 'i.ld

sine - o ----...,_ I, +:d'---lil --ntralaxis ___ frol~~-~_"_""' _""'_'---:-~----:,>-ln'-t+ensity (I) ce .t lnte nsity

y.

sin0=-}Jd

Y2

sine= -2A/d

Central

Secondary maxima

sine= 2/Jd

Sec nd

First __!_,__

121

2l. -er

),_

-d

).

0

d

D-----+<

I l 11

Plane wavefronts Screen

!

Position of the various minima for the Fraunhofer diffraction pattern of a single slit of width d.

For small angle 0, we have sin 9 .::: 8 . Thus, as shown in the figure, the angular position of the 1st, 2nd, 3rd, ... minima are 2 3 · 1y on e1t · h er s1"de of the centra1 -'- , -'-, - 1.. , ..... respective d

d

d

axis. A maximum is approximately halfway between two adjacent minima. The central bright fringe is the widest, with width

..

1

31,_

d Central

2l.

d

3).

d

I

ri r, I\ '

b)

As the slit width increases (relative to wavelength) the width of the control diffraction maxima decreases, that is, the light undergoes less flaring by the slit. The secondary maxima also decreases in width (and becomes Weaker).

c)

If d » A , the secondary maxima dl.Je to the slit disappear; we then no longer have single slit diffraction. When ,the slit width is reduced by a factor of 2, the amplitude of the wave at the centre of the screen is reduced by a factor of 2, so the intensity at the centre is reduced by a factor of 4. ·

d)

P=!:+!:= 21.. d

d

d

The secondary bright fringes are narrower. Note that as the slit width d increases, the width of the central diffraction maximum decreases. That is, there is less spreading out of the light by the slit. The secondary maxima also decrease in width and become weaker. When d becomes much greater than A, the secondary maxima disappear. The intensity I of the diffraction pattern as a function of 0 is given as

[=

ro(

S~ct

r

A slit of width d is illuminated by white light. For what value of d, will the first minimum for red light of 1..=6500A beat 0=15°? (b) What is the wavelength 1..' of the light whose first side-maximum is at 8 =15° , thus coinciding with the first minimum for the red light?

(a)

Since sin 0 .=:: ~ , so we get where

Illustration 34

ndsin0 "(yd) ct=--,_-"i D

Solution

The intensity of secondary maxima is much less. Compared to the intensity of central maximum (I0 ) , the intensity of the

(a)

dsin0 = ni.. 0

first of the secondary maxima is only 4.5%, of the second is only 1.6%, of the third is merely 0.83% ...... The successive secondary maxima decrease rapidly in intensity.

a)

If the intensity of the central maxima is 10 , then the intensity of the first and second secondary maxima are found to be ~0 and ~; . Thus diffraction fringes are of 1 unequal width and unequal intensities. Hence the ratio of the intensities of secondary maxima to centraJ maxima are

-- ~;.!=.i=~----

~ 2.34

Here, n=l, A=6500xl0-10 m, 0=15° n1..

=> (b)

CONCEPTUAL NOTE(S)

The angular position 00 of nfu minimum is given by

d

sin8n

lx6500x10°10 · sinl5° ::::: 25 · µm

This maximum is approximately halfway between the first and second minima produced with light of wavelength 1..' . Thus, by putting n = 1.5 , we get d sin 0 = 1.51..'

.---·-------

,., = dsin8 1.5

2.5 X lQ-6 X Sin 15° 1.5

430 mm= 4300

A

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Wave Optics Tilis is the wavelength of violet light. Note that the first side-maximum for light of A'= 4300 A will always coincide with the first minimum for light of A= 6500 A , no matter what the slit-width is.

b earn acqmres . . a WI"dh DA t d. Geometrical optics is based on rectilinear propagation of light, which is just an approximation. We can say that

Illustration 35

Angular width of central maximum in the Fraunhofer diffraction paHern of a slit is measured. The slit is illuminated by light of wavelength 6000 A. 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 angolar width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid. Solution (a)

2-_ • Therefore, after travelling a distance D , the diffracted d

geometrical optics is valid, if the width ~A of the diffracted beam is less than the size of the slit, that is

DA
d' => D
Given ;\, = 6000 A

Illustration 36

Let b be the width of slit and D the distance between

For what distance is the ray optics a good approximation, if the slit is 3 mm wide and the wavelength of light is 5000 A ?

screen and slit. First minima

Solution

d'

D
(3x10")' =18 m S000xl0-10

Thus, upto a distance of 18 m , we can assume rectilinear

propagation of light to a good approximation. ----D----i

INTERFERENCE & DIFFRACTION: A COMPARISON

First minima is obtained at bsin0 = A.

= rel="nofollow">

b8 = Asin0 "8

=>

8

(a)

=.?: b

different wave fronts originating from two coherent sources.

2 Angular width of first maxima = 20 = bA cc A

Angular width will decrease by 30% when A is also decreased by 30% . Therefore, new wavelength

°

4200 A = 6000 A µ

µ = 1.429 "1.43

The

..

fringes

nurumum

refractive index µ, the wavelength is decreased µ times. Therefore,

=>

(b) Here, the fringes may (b) Here the fringes are or may not be of the always of varying width. (c)

(b) When the apparatus is immersed in a liquid of

6000 µ= 4200

DIFFRACTION

It results from interaction of light coming from different parts of the same wavefront.

same width.

A' ={(6000)-(: )6000}=4200 A 00

=>

INTERFERENCE It results from (a) interaction of light coming from two

VALIDITY OF GEOMETRICAL OPTICS

When a slit of width d is illuminated by a parallel beam of light, the angular spread of diffracted light is approximately

of (c)

intensity

are dark (or perfectly

The fringes of minimum

intensity are not perfectly dark.

dark when waves are

of same amplitude). (d) All bright fringes (d) The intensity of all the possess the same bright fringes is not same. intensity. It is maximum for central fringe and decreases sharply for first, second etc. brightfringes. (e) An interference (e) In diffraction pattern the pattern consists a contrast between the good contrast bright and dark fringes is between the dark and comparatively poor. bright fringes.

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Optics & Modem Physics

Advanced JEE Physics POLARIZATION OF LIGHT

PLANE POLARISED LIGHT

According to Maxwell, light possesses electromagnetic nature. An electromagnetic wave_ consists of varying ele~tric and magnetic fields, such that the two fields are mutually perpendicular to each other and to the direction of propagation of waves. The optical phenomena i.e., phenomena concerning light may primarily be attributed to the vibrations of electric field vector in a direction perpendicular to the direction of propagation of light. In ordinary or unpolarised light, the vibrations of electric field vector are regularly or symmetrically distributed in a plane perpendicular to the direction of the propagation of the light.

It may be defined as the light, in which the vibrations of the light (vibrations of the electric vector) are restricted to a particular plane. In a plane polarised light, the vibrations are restricted to a fixed plane, so that vibrations are perpendicular to direction of propagation of light. Figure (a) represents plane polarised light having vibrations in the plane of the paper and Figure (b) represents the plane polarised light having vibrations in a plane perpendicular to the plane of the paper.

111111·

lllll! !

(a)

(b)

POLARISED LIGHT (REPRESENTATION)

UNPOLARISED LIGHT (REPRESENTATION)

In an ordinary ray of light, the electric vibrations are in all

the directions but perpendicular to the direction of propagation of the light. Such a ray of light is called a ray of ordinary or unpolarised light. It is schematically represented as shown. The arrows represent vibrations in the , plane of the paper, while the dots represent vibrations in a direction perpendicular to the plane of the paper. The phenomenon, due to which the vibrations of light are restricted to· a particular plane, is called the polarisation of light. When ordinary light i.e. unpolarised light passes through a tourmaline crystal, out of all the vibrations which are symmetrical about the direction of propagation, only those pass through it, which are parallel to its crystallographic axis AB . Therefore, on emerging through the crystal, the vibrations no longer remain symmetrical about the direction of propagation but are confined to a single plane (see Figure). A

'' p"'---,---' ' ',:-- I' ,,:~• l

UNPOLARISED LIGHT

D

--------------------------, PLANE OF VIBRATION . i ________ ,S

',,

i' PLANE i POLARISED ',,:

LIGHT

I",.... ''',',', .. PLANE OF POLARISATION l '' 17:a_.::.==------------t----- R '' __________________________ J '' '' B

C

PLANE OF VIBRATION

The plane (ABCD), which contains the vibrations of plane polarised light, is called the plane of vibration.· PLANE OF POLARISATION

The plane (PQRS) perpendicular to the plane of vibrations is called the plane of polarisation.

; PROBLEM SOLVING TRICK(S)

I

I

I

The vibrations in plane polarised light are perpendicular to the plane of polarisation. :

POLARIZATION BY REFLECTION

Polarized light may also be obtained by the process of reflection. When an unpolarized light beam is reflected, light is completely polarized, partially polarized, or unpolarized, depending on the angle of incidence. If the angle of incidence is either 0 or 90° (normal or grazing angles), the reflected beam is unpolarized. However, for intermediate angles of incidence, the. reflected light is polarized to some extent. Suppose an unpolarized light beam is incident on a surface as in figure. The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel component reflects more strongly than the other component, and this results in a partially polarized beam. Furthermore, the refracted ray is also partially polarized. Now suppose the· angle of incidence, i , is varied until the angle between the reflected and refracted beams is 90° . At this particular angle of incidence, the reflected beam is completely polarized with its electric field vector parallel to the surface, while the refracted beam is partially polarized. The angle of incidence at which this occurs is called the polarizing angle, p . From figure, we see that at the polarizing angle, p +90°+ r =180°, so that r = 90°-p. Using Snell's Law, we have sinp µ=-.smr Since sinr = sin(90°-p}= cosp, the expression for µ can be written

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cosp

Incident beam

Reflected

Incident beam

Reflected beam

polarised light is passed through analyser, the intensity I of transmitted light varies directly as the square of the cosine of the angle 0 between the transmission directions of polariser and analyser. Titis statement is known as the Law of Malus. Mathematically, according to Malus Law, we have Ioccos 2 8 ~

µ,

µ,

µ,

µ,

Refracted beam

(a)

(b)

(a) When unpolarized light is incident on a reflecting surface, the reflected and refracted beams are partially polarized. (b) T.he reflected beam is completely polarized when the angle of incidence equals the polarizing angle ep,·

I =l0 cos2 8

where I0 is the maximum intensity of transmitted light. It may be noted that 10 is equal to half the intensity of unpolarised light incident on the polariser. EXPLANATION OF THE LAW

Let the planes of polariser and analyser are inclined to each other at an angle 0 as shown in figure. Let I0 be the intensity and a the amplitude of the plane polarised light transmitted by the polariser. Polariser: Axis :

'

This expression is called Brewster's Law, and the polarizing angle p is sometimes called Brewster's Angle, after its discoverer, Sir David Brewster (1781-1868). For example, the Brewster's angle for crown glass (µ = 1.52) is

p = tan-1 (1.52) =56.7°. Since µ varies with wavelength for a given substance, the Brewster's angle is also a function of the wavelength. Polarization by reflection is a common phenomenon. Sunlight reflected from water, glass, snow and metallic surfaces is partially polarized. If the surface is horizontal, the electric field vector of the reflected light will have a strong horizontal component. Sunglasses made of polarizing material reduce the glare of reflected light. The transmission axes of the lenses are oriented vertically so as to absorb the strong horizontal component of the reflected light. Illustration 37

A ray of light strikes a glass plate at an angle of 60°. 1f the reflected and refracted rays are perpendicular to each other, find the refractive index of glass. Solution

Reflected and refracted rays are mutually perpendicular· only when the angle of incidence is equal to polarising angle, hence ip

=60°

So, refractive index is

,' '

'

''

'

' ''

' ''

'

,' --............ • •

fn~~ser

.......

/

'

' ''

I

rv..15

'

a acose 0

asine

""' ' ' Law of Malus ,'

The amplitude a of the light incident on the analyser has h\To rectangular components, (i) acos0, parallel to the plane of transmission of the analyser, and (ii) asin0, perpendicular to the plane of transmission of the analyser. So only the component acos0 is transmitted by the analyser. The intensity of light transmitted by the analyser is I =k(acos0)' =>

I=ka 2 cos 2 0

=>

I =I0 cos 2 0

where 10 = ka 2 , is the maximum intensity of light transmitted by the analyser (when 0 = 0° ). The above equation is the Law of Malus or Malus Law. ------- ,,

µ = tani, = tan60° = ,/3 = 1.732

____________________ _

LAWOFMALUS

When a plane polarised light is seen through an analyser, the intensity of transmitted light varies as the analyser is rotated in its own plane about the incident direction. In 1809, E.N. Malus discovered that when a beam of completely plane

a)

CONCEPTUAL NOTE(S) When 0::: 0° or 180°, cos0 =±1

::::-

I =10

____ So when the _transmission directions _o_f p9!af'!~er

_c~nd

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analyser are parallel or antipar8Jjiil to each other, .the maximum intensity of plane polarised light is transmitted by the analyser and is equal' to the Intensity emerging from

b)

Incident sunlight (Unpolarised)

the polariser. When 9=90°, COS9=0

=>

0

t-+•-@

--·1-----•-If---+-•

Nitrogen molecule

I= 0

So when the transmission directions of polariser and

c)

analyser are perpendicular to each other, the intensity of light transmitted through the analyser is zero. When a beam of unpolarised light is incident on the polariser, then

Scattered Light (Polarised)

I= r, (cos' 0)

¾

Since, (cos' 0) =

=>

I=&2

"iJ'Eye Illustration 38

INTENSITY CURVE

As the angle 0 between the transmission directions of polariser and analyser is varied, the intensity I of the light transmitted by the analyser varies as a function of cos2 ·8, as shown in figure. Intensity I

Two .,crossed' polaroids A and B are placed in the path of a light-beam. In between these, a third polaroid C is placed whose polarisation axis makes an angle 9 with the polarisation axis of the polaroid A . If the intensity of light emerging from the polaroid A is I,, then show that the intensity of light emerging from polaroid B will be

¼I, sin' (28) ..

I,

Solution

By Malus Law, tl)e intensity of light emerging from the middle polaroid C will be L__,.,L..__ ____,,_,L..__ _ _

90°

180°

270°

8

I1 =I0 cos2 0

360°

This intensity I, falls on the polaroid B whose polarisation

POLARISATION BY SCATTERING

When we look at the blue portion of the sky through a. polaroid and rotate the polaroid, the transmitted light shows rise and fall of intensity. This shows that the light from the blue portion of the sky is plane polarised. This is because sunlight gets scattered (i.e., its direction is changed) when it encounters the molecules of the earth's atmosphere. The scattered light seen in a direction perpelldicular to the direction of incidence is found to be plane polarised. Explanation. Figure shows the unpolarised light incident on a molecule. The dots show vibrations perpendicular to the plane of paper and double arrows show vibrations in the plane of paper. The electrons in the molecule begin to vibrate in both of these directions. The electrons vibrating parallel to the double ar_rows cannot send energy towards an observer looking at 90° to the direction of the sun because their acceleration has no transverse component. The light scattered by the molecules in this direction has only dots. It is polarised perpendicular to the plane of paper. This .explains the polarisation oflight scattered from the sky.

axis makes an angle of (90° -8) with the polarisation axis of the polaroid C . Therefore, the intensity of light emerging from B will be

I, =I, cos' (90° -8) =(I, cos' 8)cos' (90° -8) ~

I, =I,cos 2 8sin2 8=ir,(2sin8cos8)

~

I2 =.!_I,sin2 (28)

.

2

4

4

Illustration 39

Two polaroids are placed 90° to each other. What happens when N -1 more polaroids are inserted between two crossed polaroids (at 90' to each other). Their axes are equally spaced. How does the transmitted intensity behave for large N? Solution Transmitted intensity through first polaroid is 11 = 10 cos 2 0 where I, is the original intensity. Similarly, the transmitted intensity through second polaroid will be

=

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Wave Optics I 2 =11 cos 2 0=10 cos4 0

~-------=-=lC.=E_::11_ __;__ _ _--'---' I BASED ON DIFFRACTION AND POLARISATION ·I

If N polaroids are used, then

I

IN =I0 (cos0)'N As the optic axes of the polaroids are equally inclined, so ~gle of rotation e is same for each polaroid. Thus

1

A slit of width 0.025 m is placed ·in front of a lens of focal :

length

50 cm . The slit is ·. illuminated with light of 1

!

wavelength 5900 A . Calculate the distance between- the centre and first dark band of diffraction pattern obtained on !

. .!i;.=(cos0) 2N

I

a screen placed at the focal plane of the lens.

I,

Since, angle between successive polaroids is given by

( COS 2~

. 2.

!

i

90° 1t • 0 = - = - radian N 2N For large N , 0 becomes small, so we get

!

(Solutions on page 2.89)

1.

Two spectral lines of sodium D1 and D2 have wavelengths j of approximately 5890

A

and 5896

A.

A sodium lamp '

sends incident plane wave onto a slit of width 2 micrometre. I A screen is located 2 m_ from the slit. Find the spacing between the first maxima of two sodium lines as measured j on the screen.

I

JN =(1- 8~2 + ... JN =(1- 28:~2 + ...)

! 3.

which approaches 1 for large N . So, fractional intensity, is

In young's double slit experiment, the distance d between

l

the slits S1 and S2 is 1 mm.. What should the width of ''. each slit be so as to obtain 10 maxima of the double slit , pattern within the central maximum of the single slit pattern?

IN =1 I,

:::}

Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.

IN= Io

Illustration 40

. 5.

A beam of plane-polarised falls normally on a polariser (cross~sectional area 3 x 10---4 m 2 ) which rotates about the axis of the ray with an angular velocity of 31.4 rads-' . Find the energy of light passing through the polariser per revolution and the intensity of the emergent beam if the flux of energy of the incident ray is 10_, W . Solution Cross-sectional area of polaroid, A =3x 10--4 rn

i

I

i

I

6. I

I

e

a wavelength of 5000

8. 3

(Energy incident/sec)= 10- W So, intensity of incident polarised beam is given by

Since, I= I, cos' 0 where (cos' 0) =

=10

=10

2

wm-'

3

,

A.

A

is incident on it normally.

9.

Two polarising sheets have their polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned if the intensity is to drop by one-half?

10.

A polariser and an analyser are oriented so that the maximum light is transmitted. What iS the fraction of i maximum _light transmitted when analyser is rotated through (a) 30° (b) 60°? i

½

I

=~=1.67Wm_, 3x2

Light energy passing through polariser per revolution is given by E=IwAT= ~(3x10-4)(0.2)=10-4

J

Determine the angular separation between central maximum and first order maximum of the diffraction pattern due to a single slit of width 0.25 mm when light of wavelength 5890

So, average intensity transmitted is

I

I

!

A screen is placed 2 m away from a single narrow slit. 1 Calculate the slit width if the firsf minimum lies 5 mm on i eittier side of central maximum. Incident plane waves have 1

T= Zit= 2x3.14 =O.Z s ro 31.4

10" 3x10-4

A) fall at

2

Angular velocity, ro =3.14 rads-'

Energy incident/sec Area

A slit of width d is illuminated·by white light. For what value of d will the first minimum for red light (l. =6500 an angle = 30° ?

Time taken to complete one revolution,

I,

I

Two towers on the top of two hills are 40 km apart. The line joining them passes 50 m above a hill half way between the towers. What is the longest wavelength of radiowaves which can be sent between the towers without \ appreciable diffraction effects? l

11.

J

Two pola.roids are crossed to each other. If one of them is rotated through 60° , then what percentage of the incident unpolarised light will be transmitted by the polaroids?

12.

Two polaroids are !)laced at 90° to each other and the I transmitted intensity is zero. What happe.ns when one more .I L __ pplaroid Js pjaced between these two bisecting_ the_ angle 2.39

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-between them? 13.

A polaroid, .examines two adjacerit plane-polarised light beams A and B whose planes of polarisation are mutually at right angles. In one position of the polaroid_, the beam B shows, zero intensity. From this position a rotation of 30° shows the two beams of equal,inlensities; Find the intensity ratio

14.

emerging light is 3 Wm_,, what is the angle between the

~ 1,

transmission axes of the first two polarisers? At what angle

will the transmitted intensity be maximum?

of the two beams.

Show that when a ray of light is inCident on the surface of a, transparent medium at the polarising angle, the reflected

and·tral7smltted'rays are perpendicular to each other.

I

15.

Unpolarised light of i~tensity 32 wm--2 pasSes through

three polarisers such that the transmission axis of the last L_ upJ)lariser_is .crossed with _the first, it the intensity of t_he.

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Problem 1

Solving, we get

An interference pattern is observed due to two coherent sources S1 placed at origin and S2 placed at

(0, 3A, 0), where 'A. is the wavelength of the sources. A detector D is moved along the positive x-axis. Find the coordinates on the x-axis (excluding· x = 0 and oo) where maximum intensity is observed. Solution

At x = 0, path difference is 3,. . Hence, third order maxima will be obtained. At x-+ oo, path difference is zero. Hence, zero order maxima is obtained. In between first and second order maximas will be obtained. y

X

=~A.= 1.25A. 4

Hence, the desired x coordinates are

x =1.25A. and x = 4,. Problem 2

In given figure, S is a monochromatic point source emitting light of wavelength A. = 500 nm . A thin lens of circular shape and focal length 0.10 m is cut into two identical halves I., and L, 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 I., and L, is 0.15 m while that from ½ and 1 2 to O is 1.30 m. The screen at O is normal to SO. L,

s

I + - - X ----+I

' ''' '' ''

For First order maxima, we have

S2 P-S,P=A2

.Jx'+9A- -x=A. =>

Squaring both sides, we get x 2 + 9A 2 Solving this, we get

x=4AFor Second order maxima, we have

S,P-S,P = z,_ .Jx'-9,.'-x=2A=> .Jx'+ n' = (x + z,.) Squaring both sides, we get xz +9/...2 = xz +4A. z + 4xA.

=x 2 + A2 + 2xA

vt

A 0.5mm 0

'

L,

'

:....-o.1sm

.Jx 2 +9A.2 =X+A.

wi

Screen

1.30 m

If the third intensity maximum excluding central maximum, occurs at the poin_t A on the screen, find the distance OA . (ii) If the gap between I., and L, is reduced from its original value of 0.5 mm, will the distance OA increase, decrease, or remain the same. (i)

Solution

(i)

For the lens, u = -0.15 m , f .111

Thi ere ore, using - - V

U

=-

f

= +0.10

m

we get

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1 1 1 1 1 -=-+-=---+-v u f (--0.15} (0.10)

=>

I

V=0.3 m

-- tJ:r-·--- ............ ----_ ~--- 0

03 · = 2 u -0.15 Hence, two images S1 and S2 of S will be formed at 0.3 m from the Jens as shown in figure. Image S1 due to part 1 will be formed at 0.5 mm above its optics

Linear magnification, m = v =

I. For central maxima, /J.x = 0

axis (m =-2). Similarly, S2 due to part 2 is formed

sin0=(µ-l)t _sin~ d

0.5 mm below the optic axis of this part as shown. Hence, distance between S1 and 52 is d = 1.5 mm

(i-1)co.1) 2 sin0==- ~ - - sin(30°) =! . 2 50x10""

Also, D = 1.30 - 0.30 = 1.0 m = 10' mm

=> 0=30° (b) At C, 0=0°,soweget

and ;\.=500nm=5x104 mm So, fringe width is given by

l!vc = dsin~-(µ-l)t

4

P= ;\.D = (5x10 )(10') mm=! mm d (1.5) 3 Now, as the point A is at the third maxima

=>

=>

Problem 3 Light of wavelength ;\. = 500 nm falls on two narrow slits placed a distance d = 50 x 10-4 cm apart, at an angle ~ = 30° relative to the slits shown in figure. ON the lower slit a transparent slab of thickness 0.1 run and refractive

~

is placed. The interference pattern is observed on 2 a screen at a distance D = 2 m from the slits.

l!vc=(50x10-'i(½)-(¾-1)co.1)

=> l!vc.= 0.025-0.05 = --0.025 mm Substituting, l!vc = n;\., we get n l!vc --0.025 -50 A. 500x10~

OA=3P=3(½)=1 mm

(ii) If the gap between I, and L, is redu,~ed, d will decrease. Hence, the fringe width p will increase or the distance. OA will increase.

i~dex

Hence, at C there will be maxima. Therefore the· order

(c)

of minima closest to the C are -49 . Number of fringes shifted upwards is N=(µ-1)1 ;\.

( i2 -1)co,1J ~~.,....,..... 100 500x10~

Problem 4 In a modified Young's double slit experiment, a monochromatic uniform and parallel beam of light of

wavelength 6000

A ~d intensity

(1;) Wm_, is incident

normally on two apertures A and B 0.002 m respectively. A perfectly thickness 2000 A and refractive wavelength of 6000 A is placed in (shown in figure). (a) Locate the position of the central maxima. (b) Find the order of minima closest to centre C of (c)

screen. How many fringes will pass over C, if we remove the transparent slab from the lower slit?

Solution (a) Path difference is given by l!vc = dsin~+ dsin0-(µ-l)t

~ 2.42

C p

--

of radii 0.001 m and transparent film of index 1.5 for the front of aperture A

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Wave Optics 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.

Solution

Applying the lens formula _!:. _ _!:. = l_, we get V

2

PA= I( rrr,;} =: (it)(0.001) = 10_, W

{: I=:}

10

~

v=-lOcm

~

V -lQ m=-=-=2 u -5

i.e. two virtual sources are formed with distance between them 1

Power received by aperture_ B is given by

P, = I( itr;) =

2

(it)(0.002) = 4x 10C5 W

d=0.Smm

"

Lens

Only 10% of PA· and P, goes to the original direction, so

s,

'' ''

Portion of PA going to _original direction is P1 = 10-' W Portion of P, going to original direction is P, = 4 x 10-' W = (1.5-1)(2000) = 1000 A

F. "dth"p nngew1

Now, resultant power at the focal point is given by

!

10 cm - - - - - - D. -

1'.(D+lO) d

Fringes are observed between the region P and Q (waves interfere in this region only), where L d -=D 10

P = P1 + P, + 2JP1 P, cos~

~

a,:

~

i+--

~=(:")1'x= 6~; x1000=i 0

•L

0

5

Corresponding phase difference is given by

P,i"

1i

s

'' s'' '

Path difference created by slab is given by 1;x =.(µ-l)t

f

1 1 1 -+-=v 5 10

Solution Power received by aperture A is given by

U

P = 10-< + 4 xlO-' + 2J(10-<)(4xlQ-' )cos(i)

P=7x10-' W

L=Dd

Problem 5

A central portion with a width of d = 0.5 mm is cut out of a convergent lens having a focal lengt:ti of f = 10 cm , as shown in figure. Both halves are tightly fitted against each other. The lens receives monochromatic

10 Number of fringes that can be observed on the screen is given by

L

d'D

p 101'.(D + 10)

N

... (l)

Substituting the values, we get

light (,_ = 5000 A) from a point source at a distance of 5 cm from it.

3

=

2

(0.05) D 10x5x10-5(D+ 10)

Solving this equation, we get D=15 cm From equation (1), we have

(\

-V

d'

N

10,.(1+~) N will be maximum when D-), oo

(a)

At what distance should a screen be fixed on the opposite side of the lens to observe three interference bands on it? (b) What is the maximum possible number of interference bands that can be observed in this installation?

~

10--> 0

~

N

D

-

=

d' 101'.

2

(0.05) 10x5x10-5

5

2.43

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Advanced JEE Physics

Optics & Modern Physics

Problem 6

fil = dsinu =_!__ mm

400

A point source is placed at a distance !!.... below ·the 2

principal axis of an equiconvex lens of refractive index

~

Since,

2

~

and radius 20 cm . The emergent light from lens having wavelength :>. = 5000 A falls on the slits S1 and S2 separated by d = 1 mm which are placed symmetrically along the principal axis. The resulting interference pattern is observed ori the screen kept at a distance D =1 m from the slit plane.

(a)

Find the position of central maxima and its width

(b) Find the intensity at point O.

Solution

Using Lens Maker's Formula, we get 7=(%-i)(z~ ~

~ = 2,_" fil 2" x_!__xlO_, 5000 X 10-lO 400

~=10n Since, I =Im,.cos'(t) =>

I=Imax

Problem 7 The interference pattern of a Young's · double slit experiment is observed in two ways by placing the screen in two possible ways as shown in figure (a) and (b). Th_e distance between two consecutive right most minima on the screen of figure (a) using light of wavelength :>., = 4000 A is observed to be 600 times the fringe width in

the screen of figure (b) using the wavelength :>., = 6000 A . If D (as shown in figure) is 1 m then find the separation between the coherent sources S1 and S2 • Given that

d>31c,_

-~o)

2

/=20cm

Since source lies in focal plane of lens. So, all the emergent rays will be parallel. So, 1 . d/2 d tana.=-=-=-:;::sma 20 40 400

SI'

S1 S2 =d

sI

S,S,=d

}-o

s,I

Screen

Initial path difference (fil)ini.., = dsinu

s,I

I

I "" ~

(a)

(b)

0 ;,• en

"

Solution

Had the screen been perpendicular to S2 P, then P and Q' would have been the positions of first and second minima Qast two). .,

'''

s)

'' ' '' ----------~o ---::,·-"=<·i d/2 _ __,___....:::,,__=:===,..:'~:!_screen

Let the central maxima is obtained at angle 8 . Then dsin8 = dsinu sin8 =sina. tan8 = tan a. }f_ =_r!__ D 40

Dd

y=40 100

y=-d=2.5d=2.5mm 40 At O , net path difference is given by

=

2.44

s '

I

----·--10· ',P '

- - - - x, - - - . i ' ' Since the angular positions of minima do not depend on the position of the screen, so the second minima is formed at Q on the screen. For right most minima at P , we have 1

dsin01 = A.1 .

2

... (1)

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Wave Optics For small angles, we have sin el

"' tan e, = d/

(b) Calculate the minimum value of t for which the intensity at point P on the screen exactly in front of the upper slit becomes maximum. Solution

2

x,

Substituting in equation (1), we get

(a)

d'

Net path difference of the waves reaching at Q, is

... (2)

X2=-

A1

l!.x = yd+ y'd D

.For next minima at Q , we have

D'

... (3)

dsin0 2 =¾J.. 1 For small angles, we have

d

sin92 ~ tan9 2 =_1_

x,

,.._ D-..l.,1,_s,_ D'

Substituting in equation (3), we get

d'

3)..1

~,

For central maximum, Ax = 0

... (4)

X2=-

2d'

3)..1

In the second case, fringe width is given by ~=J..,D d

... (6)

(b)

Since it is given that PQ = 600~ =>

D', D

y =--y

=>

y' =

=>

y'=-sin(1tl) mm

y

1

-(f)to.Ssin(1tt))

=!!... , at point

2d' = 600 J..,D d

3)..1

=>

,

=>

... (5)

PQ=X1-X2=-

_J

2

P exactly in front of 51 , so we have

l!.x=(yd)+ ( ~) D

d' = 900J.. 1J.. 2 D

=>

d 3 = 900 X 4000 X 6000 X 10-20 X 1

=>

d' =216x10-"

=>

d=6x10-4m=0.6mm

D'

For maximum intensity, we have path difference to be an even multiple of

!:: , so 2

l!.x = (2n)!: = nJ.. 2

Substituting the values, we get

Problem 8

In a Young's double slit experiment set-up source S of wavelength 5000

A

illuminates two slits S1 and S2 ,

which act as two coherent sources. The source S oscillates about its shown position according to the equation y=O.Ssin(nt), where y is in millimetres and t in seconds. y

Is, Lx ~ - ----- - --- --- -- ---- p

0.5sin(1tt)+ 0.25 = 0.5n

=>

. ( rt I) sm

0.5n - 0.25

=>

sin(1tl)=0.5

=>

nt =-

=>

1 I= - = 0.167 s

0.5 For minimum value of t, we have n =1 1t 6

6

s Problem 9

1 mm

i s i+1mm•1• (a)

2

2mm,-~.,,

Locate the position of the central maxima as a function of time.

Two parallel beams of light P an\i Q (separation d ) containing radiations of wavelengths 4000 A and 5000 A (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure. The refractive index of the prism as a function of 2.45

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Advanced JEE Physics

Optics & Modern Physics

,.

wavelength is given by the relation, µ(:\.) = 1.20 +..!', where

(b) For, 4000 A condition of TIR is just satisfied_ Hence, it will emerge from AC, just grazingly_

A is in A and b is posiHve 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.

A

,, /

c ,/

A 8

p

r

_ sin8=0.B

- S;;,90°-C

For4000 A

B

d

~

,

C

So, deviation for 4000 A is given by

go• B

~

C

6.,.,A =90-i=90-sin-1 (0.8)~37°

For 5000 A , we have

(a) Find the value of b (b) Find .the deviation of the beams transmitted through the face AC (c) A convergent lens is used to bring these transmitted beams into focus. If the intensities of the upper and

8xl05 b µ=1.2+---,=1.2+---, =1.232 '(5000) A

the lower beams immediately after transmission from

the face AC, are 4I and I respectively, find the resultant intensity at the focus.

Solution

(a)

Total internal reflection (TIR) will take place first for those wavelength for which critical angle is small or µ is large.

Applying, µ

From the given expression of µ, it is more for the wavelength for which value of :\. is less_

,,

,

,,

,,

i=C for 4000A.

~

9=C sin9=sinC

~

(c)

C

Since sin 9 = 0-8 and sinC =_!_,so we get µ 0.8 = _!_ µ

~

0.8

The intensity of the upper beam (4000 A) and lower beam (5000 A) after transmission are 4I and I respectively, then I, =I1 +I, +2.p;i; cos~

Since no phase change takes place for the waves refracted from the lens, so qi = 0° .

~

r.=4I+I+2.J(4Ilicos(0°)

~

I• =9I

(for 4000 A.) Problem 10

1

b 1.20+-(4000)2

Solving this equation, we get b = 8_0 x 10' (A)'

2.46

0.8

65000 A =i.,,,-im
Thus, condition of TlR is just satisfied for 4000 A ~

= sin iarr

So, deviation for 5000 A is given by

i-= 8,

B

C

siniair sinimedium

1.232 = sin ia1r sin9 i., = 80.26°

A 8

For 5000 A

B

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 :\. 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 A= 648 nm , obtain the least value of t for w~ch the rays interfere constructively.

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Wave Optics Solution

Problem 11

Incident ray AB is partly reflected as ray 1 from the upper surface and partly reflected as ray 2 from the lower surface of the layer of thickness t and refractive index µ 1 =1.8 as shown in figure. Path difference between the two rays would be 1'x

=2µ 1 t =2(1.8)t =3.61

Shown in the figure is a prism of refracting angle 30° and refractive index µ, (=-Ji). The face AC of the prism is covered with a thin film of refractive index µ1 ( = 2.2) . A monochromatic light of wavelength A =550 nm falls on the face AB at an angle of incidence of 60° . Calculate

Ray 1 is reflected from a denser medium, therefore, it undergoes a phase· change of n, whereas the ray 2 gets reflected from a rarer medium, therefore, there is no change in phase ofray 2.

Hence, phase difference betweei, rays 1 and 2 would be 6.~ = 7t. Therefore, condition of cOnstructive interference will be

'

1'x=(n-½)A =>

where n=l,2,3, ...

3,6t=(n-½)A

t

'

µ.= 1.8

Least values of t is corresponding to n =1 or t.

=

=>

a}

b)

c)

sin(60°) = ,J3 sinr1

=>

sinr1

=>

'1 = 300

=>

2x3.6

A

1

=2 =A

r2 =A-r1 =30°-30°=0°

Therefore, ray of light falls normally on the face AC and angle of emergence e = i2 = 0° .

648 7.2

CONCEPTUAL NOTE(S) For a wave (whether it is sou_nd or electromagnetic), a medium is denser or rarer is decided from the speed of wave in that medium. In denser medium speed of wave is less. For example, water is rarer for sound, while denser for light compared to air because speed of sound in Water is more than in air, while speed of light is less. In transmission/refraction, no·phase change takes place. In reflection, there is a change of phase of n when it is reflected by a denser medium and phase change is zero if it is reflected by a rarer medium. If two waves in_ phase interfere having a path difference of ax; then condition of maximum intensity would be

Ax=nA., n=_0,1,2, ... But if two waves, which are already out of phase (a phase difference of n ) interfere with path difference ax, thf;ln Condition of maximum intensity will be Ax·= n=1, 2, .... ~------------ -

=>

Since, r1 + r2

=-A-

t. =-nm

=

(a) the angle of emergence. (b) the minimum value of thickness t of the coated film - so that the intensity of the emergent ray is maximum. Solution (a) Applying Snell's Law, we get sini1 = µsinr1

A 1 2 B

µ, = 2.2

(n-f}·,

--------------- ------·

(b) Multiple reflection occurs between surfaces of film: Intensity will be maximum if interference takes place in the transmitted wave. For maximum thickness, we have

1'x=2µt=A where t is the thickness of coated film

=>

t =__?:_ =~ =125 nm 2µ

2x2.2

Problem 12 A vessel ABCD of 10 cm width has two small slits S1

and S2 sealed with identical glass plates of equal tnickness. 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 alongside.

==================================== = 2.47

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Advanced JEE Physics (a)

Optics & Modem Physics

Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. •

•

L

•

(b) Now, a liquid is poured into the vessel and filled upto OQ • The central bright fringe is found to ~e at Q . Calculate the refractive index of the liquid.

"1·

0

Llx, - Llx, = 0

0

Lll'.2

0

,,,

dsin0 = 0.16 (0.8)sin0 = 0.16

D

0

sin8= 0.16 =l. 0.8 5

t --------------------fo -------- a

0

1 tan8= .fiA

0

. 0 ::::1· sm

s,

40cm

S1

I.

5 . . y2 1 tan8R:"sin8==D, 5

2 m ----+t<-10 cm

s

s~~--~c

0

Solution (a)

= Axl

D,

, 0

10

y,=-=-=2cm 5 5 Therefore, ·central bright fringe is observed at 2 cm above point Q on side CD.

Given y, =40 cm, D, =2 m=200 cm, D 2 =10 cm , .

Alternate solution for (a)

AX at R Will be zero if \

Ax 1 = ruc: 2

dsiha. =dsin8 a.=8 '

'---__JC

tana.::: tan9

40 i" tana=-=-=D, 200 5

y,

ex=·tan

Y..i =Jg_ D,

-1(1)S

02

Y, = (~: } , = (;;0 )<40) cm Y2 =2 cm

.

(b) The central bright fringe will be observed at point Q, if the path difference created by the liquid slab of thickness t = 10 cm or 100 mm is equal to &t-1 , so ·that the net path difference at Q becomes zero.

"

5

Path difference between SS, and SS, is

Llx, =ss, -ss, o o

Lll:1 =dsina=(0.8 · Lll:1 = 0.16 mm

~J(½)

Now, let at point R on the. screen, central bright fringe is observed.(i.e., net path difference = 0 ). Path difference between S,R and._S,R is Lll:2 = S,R-S,R .

o

Llx;·= d sin 8

s

... (1)

... (?-i

o

(µ-l)t=Llx,

o

(µ-1)(100)=0.16

o o

µ-.1 = 0.0016

Central bright fringe will be observ"ed when net paih difference is zero.

µ=1.0016

= t:::::================================ 2.48

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Wave Optics Problem 13

The Young's double slit experiment is done· in a

medi.;m of refractive index

± .A 3

=>

y=(~:-1}~

Substituting the values, we get y

y

SI

l------------o s, D

=>

y=4.33x10·'m

=>

y=4.33 mm

So, net path difference is 6x=6x,

(a)

Find the location of central maximum (bright fringe with zero path difference) on the y-axis. (b) Find the light intensity of point O relative to the

Corresponding phase difference, M = ( ~ )6x Substituting the values, we get

maximum fringe intensity.

15 ~=M=~( -1)(10.4x10--) 6 x10·7 4/3

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 the problem are for the given medium of refractive index

=(~-l 10.4x10.,(1.5) 4/3 0.45 X 10"3

(b) At 0, ~x, =0 and 6x2 =(~: -1}

I

(c)

yd =(.&.-1J1 D µ.

light of 600 nm

wavelength is falling on the slits having 0.45 mm separation. 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 in the figure.

s()

=>

=>

~=(~)n

_!. Ignore dispersion] 3

Now, I(~)=I-cos'(t)

Solution

Given A=600 nm=6x10·' m,

=>

l=Imaxcos '(13n)

=>

I=~I

d=0.45 mm=0.45x10°" m and D=l.5 m

SI1 (c)

s,--8----------

--- ---

For maximum intensity at O , we have

M=nA.,where n=1,2,3, ..... .

Thickness of glass sheet, I= 10.4 µm = 10.4 x 10-- m

=¾

And refractive index of glass sheet, µg

(a)

=>

6x 6x 6x , , ...... and soon A.=l,

=>

6x=(~-1)(10.4x10-'m) 4/3

=>

6x =(~-1)(10.4xl0 nm)=1300 nm

=1.5

Let central maximum is obtained at a distance y below point 0.

=>

4=

At O ,path difference is 6x=6x2 =(~:-1}

1

Refractive index of the medium, µm

6

6x1 =S1 P-S2 P= ~

Path difference due to glass sheet is given by 6x,=(~:-1}

2 3

3

4/3

So, maximum intensity will be corresponding to

=>

A=l 300 nm, 1300 nm, 1300 nm, 1300 nm, ... 2 3 4 ,- = 1300 nm , 650 mm, 433.33 nm, 325 nm, ....

The wavelength in the range 400 nm to 700 nm are 650 nm and 433.33 nm .

Net path difference will be zero, when we have Lll'.1

= M2 2.49 r-1

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Optics & Modern Physics

Advanced JEE Physics Problem 14 In Young's experiment, the source is red light of wavelength 7 x 10-' 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" m. to the position previously occupied by the 51h bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength 5 x 10-' m , the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light. Find the refractive index of glass for green light. Also estimate the change in fringe width due to the change in wavelength. Solution Path difference due to the glass slab,

t.x = (µ-1)1 = (1.5-l)t = 0.51

Problem 15 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 slits and screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 A. . (i) Calculate the fringe width. (ii) 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 minimum as the axis. • Solution Given

(i)

Due to this slab, 5 red fringes have been shifted upwards. So, we have l!u=5Ared

0.51=(5)(7x10-'m)

~

I = thickness of glass slab = 7 x 10-' m

Let µ' be the refractive index for green light, then t.x'=(µ'-1)1 Now the shifting is of 6 fringes of red light. So, we have

d=lnun,

D=l.33 m

and

i,,=6300 A Wavelength of light in the given liquid is 6300 1,,' = 2: = A "' 4737 A = 4737 x 10-10 m µ 1.33 ~

~

µ=1.33,

'dth . Frmgewi ,

1,,'D P=d

~

p (4737x10-10 m)(1.33 m) (lxlO-'m)

~

Pc"6,3x10 4 m=0.63 mm

(ii) Let t be the thickness of the glass slab.

t.x' = 61,,red ~

(µ'-1)1=61,,..,

-

( '-l)= (6)(7x10-') µ 7x10-'

~

µ'=1.6

-t--------+o 06 ·

Since the shifting of 5 bright fringes was equal to 10" m ~

sp.., = 10°" m, where p is the Fringe width

~

10" Pre, =-- m=0.2x10" m 5

A P'green

A.red

=A

i,,,,..,. =(0.2x10")(5xl0-') 7 X 10-7

1-'red }..red

~

P,.... =0.143x10"m

~

t.p = P""" -p.., =(0.143-0.2)x10" m

~

t,p =~5.71 X 10--' ID

=

2.50

't,.x

= 0.151

Now, for the intensity to be minimum at 0, this path 1,,' difference should be equal to 2

P,,..,. = ,,,,..,. • Pred

~

153 t.x = ( µglas, -1)1 = ( -1)1 µliq•ld 1.33 ~

Nowsince P=i,,o d ~ poci,, ~

Path difference due to glass slab at centre O is given by

1,,'

~

b.x=2

~

o.1st =

~

t = 15790 A = 1.579 µm

4737 2

A

Problem 16 A point source S emitting light of wavelength 600 nm is placed at a very small height h ;;_bove a flat reflecting surface AB (shown in figure). The intensity of the reflected light is 36% of the incident intensity. Interference

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Wave Optics p

fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it.

n~:

''

l' h A

(a)

L~--

•S •S

,:-:-:!

D

•S'

'

ts'

•S'

Initial

Final

Since the ray is reflected from the surface of a denser

'

What is the shape of the interference fringes on the

medium, so it suffers an additional path change of .?:_ or 2

screen?

a phase change of

(b) Calculate the ratio of the minimum to the maximum

(c)

!;. .

p

For maximum at P, path difference equals nA .

intensities in the interference fringes formed near the point P (shown in the figure).

If

H the intensity at point P corresponds to a maximum,

path difference of

calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum.

Solution (a) Since there is symmetry about the line SP, so the shape

it .

AB is shifted by x, then this will cause an additional

2(x-~) (for object and its image

taken as coherent sources). Since reflection takes place at surface of denser medium, so this will produce an ·

additional phase change of

it

,.

or a path change of - . 2

So, we get

of the interference fringes will be circular.

2(x-1)=ni..

(b) Intensity of light reaching on the screen directly from the source 11 =I, (say) and intensity of light reaching on the screen after reflecting from the mirror is I,= 36% of I,= 0.3610 •

!i_=_I_,-=-112

0.3610

0.36

=>

x=(n+1)

,. where n=0, 1, 2, 3,....

2

Now, to get minimum value of x, n must be minimum i.e., n = 0

1 JF,=0 6

(.2..-1)' 0.6 1 )' ( -+1 0.6 . (c)

=>

2x-i..=ni.. 2x =(n + l)i..

1 16

,.

=>

X=-

=>

X=

2

600 2

=300 nm

Initially path difference at P between two waves reaching from S and S' is as shown.

2.51

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This section conti.'ms Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is c~rrect. 1.

In ail interference pattern produced by two identical slits, the intensity at the site of the central maximum is I . Th'7 intensity at the same spot when either of two

slits is closed is I (A) 2 (C)

I

2,/2

(B) (D)

(B)

wave-front at a later or an earlier instant

(C) (D)

3.

6.

Longitudinal waves do not exhibit (A) refraction (B). reflection (C) _diffraction · (D) polarization

7.

The idea of the quantum nature of light has emerged in an attempt to explain (A) interference (B) diffraction (C) polarization _ (D) radiation spectrum of a black body

8.

In the spectrum of light of a luminous heavenly body the wavelength of a spectral line is measured to be 4747 A while actual wavelength of th_e line is 4700 A. The relative velocity of the heavenly body with respect to earth will be (velocity of light is 3x10 8 ms·') (A) 3 x 105 ms·1 mov;,.;g towards the earth (B) 3x10 5 ms·1 moving away from the earth (C) 3x10 6 ms·' moving towards the earth (D) . 3 x 10 6 ms·' moving away from the earth

9.

A grating has 5000 lines cm·'. The maximum order

I

,/2

In YDSE bi-chromatic light of wavelengths 400 run and 560 run are used. The distance between the slits is 0.1 mm arid the distance between the plane of the slits and the screen is 1 m . The minimum tjistance bet_ween two·successive regions of complete darkness is (A) 4 mm (B) 5.6 mm (C) 14 mm (D) 28 mm_ · In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness I is introduced in the path of one of the interfering beams (wavelength 1'. ), the intensity at the position where the central maximum ocqrrred previously remains unchanged. The minimum thickness of the glass-plate is

(A) (C) 4.

21'.

,.

(B)

21'. 3

(D) ,.

3

Young's double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe

widths recorded are (A) Pa >P, >P, (C) P,>P,>Pa 5.

Pa, P,

and P, respectively. Then, (B) P, >P, >P,

(D)

is used to'determine the velocity of light is used to explain polarization of light

I 4

I

2.

is a geometrical method to find" the position of a

P,>Pa>P,

Huygens' conception of seco11dary Waves

visible with·wavelength 6000 A (A) 2 (B) 3 (C) 4 (D) 0 10. A beam of monochromatic light enters from vacuum into a medium of refractive index n . The ratio of the wavelengths of the incident and refracted waves is (A) n:1 (B) 1:n 2 (C)· n :1 (D) 1:n'

(A)

helps us to find the focal length of a thick lens

2.52

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Wave Optics 11. In Young's double slit experiment, 62fringes are seen in visible region for sodium light of wavelength 5893 A. If violet light of wavelength 4358 A is used in place of sodium light, then number of fringes seen will be (A) 54 (B) 64 (C) 74 (D) 84 12. Monochromatic

light

of

wavelength

0.01 mm

20.

21.

(D) 1.0mm

13. Air has refractive index 1.0003. The thickness of an air

column, which will have one -more wavelength of yellow light ( 6000 A) than in the same thickness of vacuum is (A) 2mm (C) 2m

Ray optics is valid when characteristic dimensions are (A) of the same order as the wavelength of light (B) much smaller than the wavelength of light (C) much larger than the wavelength of light (D) of the order of 1 mm

of the oil film should be of the order of (A) 1 cm (B) 10 A

(C)

24. The deflection of light in a gravitational field was

predicted first by (A) Einstein (C) Max Planck

16. Though quantum theory of light can explain a number of phenomena observed with light, it is necessary to

retain the wave nature of light to explain the phenomenon of

(A)

(B) (C) (D)

25.

photo-electric effect diffraction compton effect black body radiation

17. Which of the following cannot be polarized?

(A) (C)

Radio wave Infrared radiation

(D) 10000 A

white light is observed through a red filter. The pattern seen is (A) a red cross on a black background (B) a blue cross on a red background (C) a red cross on a blue background (D) a black cross on a blue background

How many.colours are possible, then (A) 3 (B) 1

(D) None of these

5000 A

23. The blue cross on a white background illuminated with

15. There is a wavelength corresponding to each colour.

7

In Young's double slit experiment, carried out with light

22. In order that a thin film of oil floating on the surface of water shows colours due to interference, the thickness

(B) 2 cm (D) 2km

14. Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are (A) 41 and I (B) SI and 31 (C) 91 and I (D) 91 and 3I

(C)

scattering is reduced at noon

of wavelength 1,. = 5000 A, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum (taking the central maximum as zeroth maximum) will be at x equal to (A) 1.67 cm (B) 1.5 cm (C) 0.5 cm (D) 5.0 cm

5000 A

illuminates a pair of slits 1 mm apart. The separation of bright fringes in the inteiference pattern formed on a screen 2 m away is (A) 0.25 mm (B) 0.1 mm

(C)

(D)

(B) X-rays (D) Sound waves in air

18. A blue object on a white background when seen through a blue filter will appear (A) blue on a white background (B) black on a blue background (C) blue on red background (D) invisible 19. illumination of the sun at noon is maximum because (A) the sun is nearer to the earth at noon (B) rays are incident almost normally (C) refraction of light is minimum at noon

(B) Newton (D) Maxwell

Both the particle and wave aspects of the wave aspects of light appear to be used in (A) photoelectric effect (B) gamma emission (C) interference (D) classical mechanics

26. At sunset, the sun seems to be (A) higher than it really is (B) lower than it really is (C) exactly where it really is (D) lower than it would be at sunrise 27.

In Huygens' wave theory, the locus of all the points in the same state of vibration .is called a (A) half period zone (B) vibrator (C) wavefront (D) ray

28. In Young's experiment, monochromatic light is used to illuminate the two slits A and B. Interference fringes are observed on a screen placed in front of the slits. 2.53

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Now if a thin glass plate is placed normally in the path of the beam coming from the slit

33. Two coherent point sources s1 and s2 vibrating in

phase emit light of wavelength ). . The separation between the sources is 2A . The smallest distance from s2 on a line passing through s2 and perpendicular to s1s2 , where a minimum of intensity occurs is (A)

-(CJ (A) The fringes will disappear (BJ The fringe width will increase (CJ The fringe width will increase (D) There will be no change in the fringe width but· the pattern shifts ·

(CJ

.!1 2

(DJ

0

¼I,

30. In the diagram, CP represent a wavefront and AO and BP, the corresponding two rays. Find the condition on 0 for constructive interference at P between the ray BP and reflected ray OP

0

(BJ

12

4

).

(DJ

,2

(A)

(CJ

1.75 1.25

(BJ (D)

(A) B

(CJ

sec8-cos8=a

(BJ ).

).

cos8=4d

(CJ

41. (D) sec9-cos8=d

31. In Young's double-slit experiment the separation between the slits is doubled and the distance between the slit _and the screen is halved. The fringe-width becomes (A) one-fourth (B) half (CJ double (DJ quadruple 32. In Young's double slit experiment, the separation

between the slits is halved and the distance between the slits arid the screen is doubled. The fringe width is (A) unchanged (B) halved (CJ doubled (D) quadrupled

1.50 1.00

s, d

3).

(¾}, The

35. Consider a usual set-up of Young's double slit experiment with slits of equal intensity as shown in the figure. Take O as origin and the Y axis as indicated. If . . . .b 1.D d 1.D verage mtens1ty etween y1 =- - an y 2 = + 4d 4d equals n times the intensity of maxima, then n equals (take average over phase difference)

t

cos8=2d

4

refractive index of second plate is

f

(A)

3).

of same thickness, the shift of fringes is

R

A

151.

34. In a Young's double slit experiment, the fringes are displaced by a distance x when: a glass plate of refractive index 1.5 is introduced in· the path of one of the beaII1S. When this plate is replaced by another plate

29. When an unpolarized light of intensity ! 0 is incident on

a polarizing sheet, the intensity of the light which does • not get transmitted is (A) ZERO (B) ! 0

n

½(1+;) (1+;)

y

---------------s,

0

D

(B) (DJ

2(1+;) ½(1-;)

36. A plate of thickness ! made of a material of refractive index µ is placed in front of one of the slits in a double slit experiment. What should be the minimum thickness ! which will make the intensity at the centre of the fringe pattern zero? (A)

(CJ

37.

(µ-1)~

(B)

(µ-1)1.

(DJ

(µ-1)

).

2(µ-1) Two polaroids are placed in the path of unpolarized beam of intensity ! 0 such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle 0 with the polarization axis of first polaroid, is placed between

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Wave Optics these polaroids then the intensity of light emerging from the last polaroid will be (A)

(~}in' 20

(B)

(~}in' 20

(C)

(; )cos' e

(D) I, cos' e

(¾) of the maximum intensity. Angular position of this point is

sin-

(C)

sin-'G'a)

(~)

sin- (:a)

(D)

sm

(A) (B) (C)

. -1(-1,.) 4d

(A)

(2n + 1)?:

(B)

(C)

(2n +1).1:

(D) (2n + 1)_!':_

2

8

(2n + 1)?:

16

If two slightly different wavelengths are present in the the sharpness of fringes will be more than the case when only one wavelength is present

the sharpness of fringes will decrease as we move away from the central fringe the central fringe will be white the central fringe will be dark

41. Two identical coherent sources placed on a diameter of a circle of radius R at separation x( « R) symmetrically about the centre of the circle. The sources

emit identical wavelength A each. The number of points on the circle with maximum intensity is (x =SA) (A) 20 (B) 22

(C)

24

(D)

(D) 26

Laser is (A) intense, coherent and monochromatic (B) only intense and coherent (C) only coherent and monochromatic (D) only intense and monochromatic 43. Imagine a hypothetical convex lens material which can transmit all the following radiation. This lens will have minimum focal length for (A) ultraviolet rays (B) infrared rays (C) radio waves (D) X-rays 42.

44. A star emitting yellow light starts accelerating towards earth, its colour as seen from the earth will

many wavelengths uncoordinated wavelengths coordinated waves of exactly wavelength divergentbeams

the

same

47. The wavelength. of light observed on the earth, from a moving star is found to decrease by 0.05%. Relative to the earth the star is

(A) (B) (C) (D)

4

light used in Young's double-slit experiment, then

(C) (D)

refraction oflight polarization oflight dispersion oflight

consists of

1

(B)

39. In the Young's double slit experiment using a monochromatic light of wavelength 1,. , the path difference (in terms of an integer n ) corresponding to any point having half the peak intensity is

(B)

(B) (C) (D)

46. Laser light is considered to be coherent, because it 1

(A)

(A)

tum gradually red tum suddenly red remains·same tum gradually blue

45. The transverse nature of light is shown by (A) interference of light

38. In Young's double slit experiment intensity at a point is

40.

(A) (B) (C) (D)

Moving away with a velocity of 1.5x105 ms-1 Coming closer with a velocity of 1.5 x 10 5 ms-1 Moving away with a velocity of 1.5 x 10 4 ms-1 Coming closer with a velocity of 1.5 x 10 4 ms-1

48. A beam of electron is used in an YDSE experiment. The slit width is d . When the velocity of electron is increased, then (A) No interference is observed (B) Fringe width increases (C) Fringe width decreases . (D) Fringe width remains same 49. 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 fringe from the centre is 1 (A) 2 (B) 2

(C)

(Q) 16

4

50. The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

(A) 1 : 4: 9

(C)

1-~--4_ . 91t2 . 251t2

(B) 1 : 2: 3 1 9 (D) 1 : ' : ' n

n

51. A beam of natural light falls on a system of 6 polaroids, which are arranged in succession such that each polaroid is turned through 30° with respect to the preceding one. The percentage of incident intensity that passes through the system will be (A) 100% (B) 50% (C) 30% (D) 12%

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52. The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is · (A) Infinite (B) Five , (C) Three (D) Zero 53. A rocket is going towards moon with a speed v , The astronaut in the rocket sends signals of frequency v towards the moon and receives them back. on reflection from the moon. What will be the frequency of the signal received by the astronaut (Take v « c)

(A) (C)

C

C

(B) - - v

--v

c-v 2v

58. 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 1,, ), the intensity ai the position where the central maximum ,occurred previously remains unchanged. The minimum thickness of the glass-plate is 21,,

(A)

c-2v

2c

(D) ~v

-v

intensity of light at the mid-point of the screen in the first case to that in the second case is (A) 1:2 (B) 2:1 (C) 4 : 1 (D) 1 : 1

(C)

V

C

54. In Young's double slit experiment the.y-coordihates of central maxima and lQth maxima aie 2 · cm and 5 cm respectively. When the YDSE,apparatus is immersed in a liquid of refractive index ,1.5 the corresponding ycoordihates will be · · · '· (B) 3 cm, 6 cm (A), 2 cm, 7.5 cm 4 10 (C) 2cm,4cm. (D) cm, cm

3

-.

3

,55. Jn·Yoting1s double slit exp°eriment how many maxim.as

can be obtained on a screen (including the central maximum) on, both • sides of" the ·central fringe ·if 1..=ioooA',,;,d d=7oooA -

(B)

1,,

3.

(D) 1,,

3

59. In a Young's double-slit experiment, the intensity ratio of maxima and minima is infinite. The ratio of the

amplitudes of two sources (B) is unity (A) is infinity (C) is two , (D) cannot be predicted 60. Figure represents a glass plate placed vertically on a horizontal table with a beam of unpolarised light falling on its surface at: the polarising angle of 57° with the normal. The electric vector in the reflected light on , Screen .. S . vib:ate Vfith respect. t~ the plan:e .of incidenc~. in- a.

r-ill

(B) 7 (D) 4

(A) 12 (C) 18

56. A monochromatic beam ,of· light falls on YDSE apparatus at some angle (say 8) as shown in figure. A thin .sheet of glass is inserted in front ofthe. lower slit S2 , The central bright fringe (path _difference = O_) will be obtained (A)

Vertical plane

(B) Horizontal plane , (C) Plane makihg an angle of 45° with the vertical (D) Plane making an angle of 57° with the horizontal (A)

61. A clear sheet of pol!'}'oid is placed on the top of similar

At 0

(B) Above 0 (C) Below 0 (D) Anywhere depending on angle 8, thickness of plat_e 't and refractive index of glass_ µ ' 57. In Young's double slit experiment, the two slits act as coherent sources 'of equal amplitude A and wavelength A . In another experiment with the same set up the two slits are of equal amplitude A .and wavelength i,.. but incoherent. The ratio of the

are

1

sheet so that their axes make,.an ru:ig~e ~~~

(¾) with

each other. The ratio of intensity of the emergent light · to that of unpolarised incident light is ·(A) 16:25 (B)· 9:25 (C) 4:5 (D) 8:25 62. Optically active substances are those substances which

(A)

(B) (C)

produce polarized light rotate the plane of polarization of polarized light produce double refraction ·

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convert a plane · polarized light into circularly polarized light

63. If I, is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled

(A) . I, (C)

2I,

(B)

~

(D) 4I,

64. The critical angle of a certain medium is sin-1

(¾). The

polarising angle of the medium is (A)

sin-'(¼)

(B)

(C)

tan''(¾)

(D) tan-'(½)

tan-'(¾)

65. The ray of light is incident on glass of refractive index 1.5 at polarising angle. The angle of deviation of the incident ray in glass is (A) 57° (B) 33° (C) 24° (D) 114° 66. Double refraction of light is shown by (A) quartz and calcite only (B) calcite only (C) calcite and ice only (D) calcite, ice and quartz

1.3 µm

(D) 2.6 x 10-4 cm

68. The resolution of the human eye is 1'. The resolving power of the human eye is nearly

(A) (C)

360 36000

(B) 3600 (D) 360000

69. Colours of thin films are due to (A) dispersion of light (B) interference of light (C) absorption oflight (D) scattering of light 70. A person standing at a distance of 3.6 km can just resolve two poles. The distance behveen the poles is

(A) (C)

0.1 m 1m

(B) 100 m (D) 10m

·

71. A heavenly body is receding from earth such that the fractional change in A. is 1, then its velocity is

(A)

C

(B)

3c 5

(D)

5

2c 5

72. A star is moving towards the earth with a speed of 4.5 x 106 ms-' . If the true wavelength of a certain line in the spectrum received from the star is 5890 A, its apparent wavelength will be about (c = 3 x 10 8 ms-') (A) 5890 A (BJ 5978 A (CJ 5802 A (DJ 5896 A

73. Lights of wavelength 1,,1 = 4500 A, 1,,2 = 6000 A are sent through a double-slit arrangement simultaneously. Then (A) no interference pattern will be formed (B) the third bright fringe of A. 1 will coincide with the fourth bright fringe of 1,, 2 the third bright fringe of fourth bright fringe of 1,,1

(D)

1,, 2

will coincide with

the fringes of wavelength 1,, 1 will be wider than the fringes of wavelength 1,,2

74. Two slits separated by a distance of 1 mm are illuminated with light of wavelength 6 x 10-' m. The interference fringes are observed on a screen placed 1 m

from the slits. The distance between the second dark fringe and the fourth bright fringe is equal to (B) 1.0 mm (A) 0.5 mm (C) 1.5 mm (D) 2.0 mm

67. A slit of width a is illuminated by red light of wavelength 6500 A . The first minimum will fall at 9=30° if a is (A) 3250 A (B) 6.5 X 10-4 mm

(C)

C

(C)

75. Interference fringes were produced in Young's doubleslit experiment using light of wavelength 5000 A .

When a film of thickness 2.5 x 10" cm was placed in front of 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) 1.35 (C) 1.4 (D) 1.5 76. In Young's double-slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 A, coming from the coherent sources S1 and S2 •

At certain point P on the screen third dark fringe is formed. Then the path difference S1 P - S2P in microns is (A) 0.75 (B) 1.5 (C) 3.0 (D) 4.5 77. Young's double slit experiment is made in a liquid. The 10th bright fringe in liquid lies where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately (A) 1.8 (B) 1.54 (C) 1.67 (D) 1.2

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A point source emits light equally in all directions. Two points P and Q are at distances 9 m and 25 m respectively from the source.. The ratio of the amplitudes of the waves P and Q is (Al 9:25 (Bl 25:9 (C). ' 9' : 252 (Dl 252 : 9'

79. In Young's double slit experiment the y-co-ordinates of

central maxima and 10th maxima are 2 cm and 5 cm respectively. When the YDSE apparatus is inunersed in a liquid of refractive index 1.5 the corresponding · y-co-ordinates will be (Bl 3 cm, 6 cr_n (A) 2 cm, 7.5 cm 4 10 (C) 2cm,4cm (Dl cm, ~cm

3

observation be taken from :point PB - PA = !:. . Then the phase 4 waves from A and- B reaching (Al 156° (B) (C) 136° (Dl

P , such . that

difference between the . P is 140° 126°

86. Two coherent sources S1 and S2 are separated by a

distance four.times the wavelength A of the source.'The sources lie along y-axis whereas a detector moves along +x-axis. Leaving the origin and far off points the number Of points where ni.axima are observed is (Al 2 (B) 3 (C) 4 (D) 5 87. The first muumum due to a Fraunhofer diffraction

80. In Young's double ~slit experiment

~ = 10""

(d =

distance between slits, D·= distance of,screen from the slits). At a point P on the screen resulting intensity is equal to the intensity due to individual ~lit I0 • Then the distance. of point P from the central maximum is (A =6000f) (A) 2mm (Bl 1 mm (C)_ .0.5mm (Dl 4mm. 81. Th~·yci~g's double-slit experiment is carried out with

light of wavelength 5000 A . The distance b~~een the slits 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 (Bl 1.5 cm (Al 1.67cm (C) 0.5 cm (Dl 5.0 cm 82.

using light 'of ·wavelength 500 nm and a slit of )Vidth 0.5 mm will be formed at an angle (in minutes) (Al 2.42 (B) 3.43 (C) 4.84 (Dl 1.7188. Aperture of the human eye is 2 min. Assuming the

mean wavelength of light to be 5000 A , the ..,;gular resolution limit of the eye is nearly

(A)

(C)

89. In the Young's double slit experiment apparatus shown in figure, the ratio of maximum to minimum intensity on the screen is 9. The wavelength of light used is A,

then the v~lue of y is Screen

83. The blue colour of the sky is explained by

(Al

(C)

refraction polarisation

(B) reflection (D) scatteriog

84. A Young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is_

(A)

(C)

Straight line Hyperbola

(B) Parabola (D) Circle

--ho d/2

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 obseived in the same segment of the screen is given by (A) 12 (Bl 18 (C)' 24 (D) 30 .

(Bl 1 min'!}te . (D) 1.5 ~ute

2 minute 0.5 minute

y

d/2

(A)

(C)

AD d AD 3d

-<-1+--D

(B)

AD

(D)

).D

2d

4d

90. In Young's experiment, using red and blue lights of wavelengths 7800 A and 5200 A respectively, the value of n for which nth red fringe coincides, with (n+l)th bluefringeis (A) 2 (B) 3 (C) 4 (D) 5 91. In Young's experiment for interference of light with two

85. Among the two interferiog monochromatic sources A and B ; A is ahead of B in phase by 66° . If the

slits, maxima Occur at angles £Or which sin 8 = m),_ .• Here d d is

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distance of slits from the screen distance.between dark and bright fringes distance between slits width of mth fringe

92. Interference·is observed in a chamber with air present inside the chamber. The chamber is then evacuated and the same light is again used to produce interference. A careful observer will see (A) no change in the pattern (B) that the fringewidth slightly increases (C) that the fringewidth slightly decreases (D) no interference pattern

polaroid is given one complete rotation about the direction of light (A) the intensity of light gradually decreases to zero and remains at zero (B) the intensity of light gradually increases to a maximum and remains maximum (C) there is no change in the intensity of light · (D) the intensity of light varies such that it is twice maximum and twice zero 98.

93. Finger prints of a piece of paper may be detected by sprinkling fluorescent powder on the paper and than looking into it under (A) yellow light (B) brightness (C) infrared light (D) ultraviolet light

Figure here shows P and Q as two equally intense coherent sources emitting radiations of wavelength 20 m . The separation PQ is 5 m and phase of P is ahead of the phase of Q by 90° . A , B and C are three distant points of observation equidistant from the mid-point of PQ . The intensity of radiations at A, B, C will bear the ratio

•B

''

94. Two nicol prisms (polariser and analyser) have their axes at angles of 30° in between. If I is the intensity of

• light falling on first nicol, then that of emerging light is {A) 0.125! (B) 0.25! (C) 0.3751 (D) 0.5[ 95. The Young's double-slit experiment is performed with blue light and green light of wavelengths 4360 A and 5460 A respectively. If X is the distance of 4th maximum from the central one, then (A) X(Blue)=X(Green) (B) _ X(Blue)<X(Green) (C) X(Blue)>(Green)

(D)

X(Blue) X( Green)

'' '' '' ''

p : Q

•-----------
I

()-----------•

A

(B) 4 : 1 : 0 (D) 2 : 1 : 0

0 :1 :4

0:1 :2

99. In the figure is shown Young's double slit experiment. Q is the position of the first bright fringe on the right side of O . P is the 11th fringe on the other side, as measured from Q . If the wavelength of the light used

is 6000xl0-10 m, then 51B will be equal to

5460 4360

Q

s,

D____

96. A ray of light of intensity I is incident on a parallel glass,slab at a point A as shown in figure. It undergoes

partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A'B' undergo interference. The ratio Imax / Imm is I

~

s1110

B

---------. 0

s,D p

--

(A) (C)

6x10_. m 3.138 X 10-7 m

(B) 6.6x10_. m (D) 3.144xl0-7 m

100. The maximum intensity in Young's double slit experiment is ! 0 • Distance between the slits is d = 51,., where 1,. is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of · one of the slits on a screen at a distance D = 10d

(A) (C) 97.

4: 1 7: 1

(B) 8:1 (D) 49:1

Plane polarised light is passed through a polaroid. On viewing through the polaroid we find that when the

¾r,

(A)

~

(B)

(C)

I,

(D) I, 4

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Advanced JEE Physics 101. The polarising angle of diamond is 67° . The critical angle of diamond is nearest to (A) 22' (B) 34° (C) 45° (D) 60°

103. A beam of light AO is incident on a glass slab (µ =1.54) in a direction as shown in figure. The reflected ray OB is passed through a Nicol prism on viewing through a Nicol prism, we find on rotating the prism that

A~£5B 0

... ····· .

106. Two waves originating from sources S1 and 5 2 having

zero phase difference and common wavelength A will show completely destructive interference at a point P if S1P-S2P is

102. No longitudinal wave will show (A) interference (B) diffraction (C) TI R (D) polarisation

I --- -.~-

(D) ,/3)..D

(A)

(C)

5)..

2)..

(B)

(D)

3).. 4 11)..

2

107. A thin air film between a plane glass plate and a convex · lens is irradiated with parallel beam of monochromatic light and is observed under a microscope. We see

(A) (B) (C)

I

uniform brightness complete darkness field crossed over by concentric bright and dark rings

(A) The intensity is reduced down to zero and

(D) field crossed over by straight bright and dark fringes

remains zero

(B)

The intensity reduces down some what and rises again

(C) (D)

There is no change in intensity The intensity gradually reduces to zero and then again increases

intensity at a point where the. path difference is

.

.

104. When one of the slits of Young's experiment is covered

with a transparent sheet of thickness 4.8 mm, the central fringe shifts to a position originally occupied by the 30th bright fringe. What should be the thickness of the sheet if the central fringe has to shift to the position occupied by 20th bright fringe (A) 3.8 mm (B) 1.6 mm (C) 7.6 mm (D) 3.2 mm 105. Two ideal slits S1 and S2 are at a distance d apart, and

illuminated by light of wavelength ).. passing through an ideal source slit S placed on the line through S2 as shown. The distance between the planes of slits and the source slit is D . A screen is held at a distance D from the plane of the slits. The minimum value of d for which there is darkness at O is

~

s,~ 0

s ~-

(A)

l~D

108. In Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is A. is I, ).. being the wavelength of light used. The

s,~

1+- D - - - D - + 1

(B)

..fijj

~

will

be (A)

(C)

I 4 I

(B)

I 2

(D) ZERO

109. In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the sits. If the screen is moved by 5 x 10-2 m

towards the slits, the change in fringe width is 3 x 10-s m . If separation between the slits is 10-3 m , the wavelength of light used is (A) 6000 A (B) 5000 A (CJ 3000 A (D) 4500 A 110. In Young's 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 and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima now is the same as the observed fringe shift upon the introduction of the mica sheet. The wavelength of light is

(A)

5762 A

(CJ

5892 A·

(B) 5825 A (D) 6500 A

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Wave Optics 111. The phenomenon of interference is shown by

(A) (B) (C) (D)

longitudinal mechanical waves only transverse mechanical waves only non-mechanical transverse waves only all the above types of waves

112. In Young's double-slit experiment, if L is the distance

between the slits and the screen upon which the interference pattern is ob~erved, x is the average distance between the adjacent fringes and d is the slit separation, then the wavelength of light is

(A)

xd L

(B)

(C)

Ld x

(D)

xL d

1 Ldx

(A)

I=JA'+B'cos 2 ~

(B)

(C)

I=A+Bcosi

(D) l=A+Bcos~

2

r,;;;cos~

Where A and B depend upon the amplitudes of the ,two waves. 118. The time period of rotation of the sun is 25 days and its radius is 7 x 108 m . The Doppler shift for the light of

.wavelength 6000 A emitted from the surface df the sun will be (A) 0.04 A (B) 0.40 A (C) 4.ooA (DJ 40.oA 119. A flake of glass (refractive index 1.5) is placed over one

113. Interference can take place between . (A) .transverse waves only,. ]:mt not in longitudinal waves (B) lon~tudinal waves only, but not in transverse waves (C) both longi~dinal and transverse waves (D) light waves only, but not sound waves

of the openings of a double slit apparatus. The ' interference pattern displaces itself through seven successive maxima towards the side Where the flake is placed, if wavelength of the diffracted light is 1,, = 600 nm , then the thickness of the flake is (A) 2100 nm (B) 4200 nm (C) 8400 nm (D) None of these

114. Young's double-slit experiment is performed with light

120. Two coherent sources separated by distance d are radiating in phase having wavelength 1,, • A detector moves in a big circle around the two sources in the plane of the two sources. The angular position of n :a 4

of wavelength ·,_ = 6000 A . A glass plate of thickness 0.01 mm and µ =1.5 is introduced. The number of fringes shifting in the system is (A) 2000 (B) 8 (C) 120 (D) 4910

interference maxima is given as

115. The contrast in the fringes in an interference pattern

depends on (A) fringe width (B) wavelength (C) intensity ratio of the sources (D) · distance between the slits 116. In a Young's double slit experiment the source S and the two slits A and B are vertical with slit A above slit B . The fringes are observed on a vertical screen K .

. _,·(n") -

{A)

sm

(C)

tan·'(4d,,)

d

cos·1 (

4,;')

(D) cos·'(

:d)

(B)

The optical p~th length from S to B is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path l~gth from s to A is not changed, as a result. the fringe system on K moves {A) Vertically downwards slightly (B) Vertically upwards slightly (C) Horizontally, slightly to the left (D) Horizontally, slightly to the right , . . . . '

121. White light may be considered to be a mixture of waves

·117_ In the Young's double slit experiment, if the phase difference between the two waves interfering at a point

122.._The k line of sing!:¥, io,;;sed calcium has a wavelength

is ~, the intensity at that point can be expressed by the

one of'the observed galaxies,. this spectral line is located

~

expression··

with 1.. · ranging between 3900 A and 7800 A . An oil film of thickness 10,000 A is examin~d normally by reflected light. If µ =1.4, then the film appears bright for (A) 4308 A, 5091 A, 6222 A (B) 4000 A, 5091 A, 5600 A (C) 7000 A (D)

:~~~ 1: :: 1: ;~~~ 1_

of 393.3 nm as measured on earth. In the spectrum of at 401.8 nm. The speed with which the galaxy is nioving away from us, will be

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Advanced JEE Physics

(A) (C)

6480 kms-1 4240 kms-1

Optics & Modern Physics 3240 kms-1 (D) None of these (B)

123. In a YDSE bi-chromatic light of wavelengths 400 run and 560 run are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m . The minimum distance behveen two successive regions of complete darkness is (A) 4 mm (B) 5.6 mm (C) 14 mm (D) 28 mm 124. In a double slit arrangement fringes are produced using

light of wavelength 4800 A . One slit is covered by a thin plate of glass of refractive index 1.4 and the other with another glass plate of same thickness but of refractive index 1.7. By doing so the central bright shifts to original fifth bright fringe from centre. Thickness of glass plate is (A) 8 µm (B) 6 µm (C) 4 µm (D) 10 µm

sin-1 (µcos$)

(A)

90°+$

(B)

(C)

90°

(D) 90°-sin-'( s:$)

130. A beam of plane polarized light falls normally on a polarizer of cross sectional area 3 x 10-4 rn2 . Flux of energy of incident ray in 10-3 W. The polarizer rotates

with an angular frequency of 31.4 rads-1 • The energy of light passing through the polarizer per revolution will be (A) 10-< Joule (B) 10-3 Joule 2 (C) 10- Joule (D) 10-1 Joule 131. A beam with wavelength 1. falls on a stack of partially reflecting planes with separation d. The angle 9 that the beam should make with the planes so that the beams reflected from successive planes may interfere constructively is (where n =1, 2, ...... )

125. Polarisation of light establishes (A) corpuscular theory of light (B) quantum nature oflight (C) transverse nature of light (D) all the three 126. A ray of unpolarised light is incident on glass plate at the polarising angle, then (A) the reflected and transmitted rays will be completely plane polarised (B) the reflected ray is completely polarised and the transmitted ray is partially polarised (C) the reflected ray is partially polarised and the transmitted ray is completely polarised (D) the reflected ray and the transmitted ray will be partially polarised 127. To observe diffraction, the size of the obstacle

(A) (B) (C) (D)

should be of the same order as the wavelength should be much larger than the wavelength has no relation to wavelength should be exactly half the wavelength

128. A slit is illuminated by red light of wavelength 6500 A . The first minimum is obtained at 9 = 30°. The width of the slit is

(A) (C)

3200 A 6.5x10-< mm

(B) 1.24 micron 2.6x10-< cm·

2.62

Sill

(C)

• -I sm 2d

-

d

(HA)

(D) cos

-,(n:>.) -

2d.

132. A diffraction pattern is obtained using a beam of re'd light. If the red light is replaced by blue light, then (A) the diffraction pattern remains unchanged (B) diffraction bands become narrower and crowded together (C) bands become broader and farther apart (D) bands disappear 133. At sunrise or at sunset the sun appears to be reddish while at mid-day the sun looks white. The reason is that (A) the sun is less hot at sunrise or at sunset than at noon (B) diffraction sends red rays to the earth at these time (C) refraction is responsible for this effect (D) scattering due to dust particles and air molecules

(D)

129. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster's angle $. If µ represents the refractive index of glass with respect to air, then the angle between reflected and refracted rays is

=

. -1(n:>.)

(A)

134. Light of wavelength 6328 A is incident normally on a slit having a width of 0.2 mm . The distance of the screen from the slit is 0.9 m. The angular width of the central maximum is (A) 0.09 degree (B) 0.72 degree (C) 0.18 degree (D) 0.36 degree

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'

Wave Optics 135. Two point sources X and Y emit waves of same

frequency and speed but Y lags in phase behind X by 2nl radian. If there is a maximum in direction D the distance XO using ri as an integer is given by

X~D

\~

143. An optically active substance

y

(A)

A. -(n-1) 2

(B)

!:en+ l)

(D) )..(n-0

2

(A) .produces polarized light (B) rotates the plane of polarization of polarized light (C) converts a plane polarized light into circularly polarized light (D) converts a circularly polarized light into plane polarized light

,.(n+I)

136. In a Young's double slit experiment, .the slits are 2 mm

apart and are illuminated wii:h a mixture of two wavelength A.0 = 750 nm and A. = 900 nm . The minimum distance from the common central bright fringe on a screen 2m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other is (A) 1.5 mm (B) 3 mm (C) 4.5 mm (D) 6mm 137. If sound waves can be assumed fo be diffracted, which of the following objects will diffract sound waves in air from a 384 Hz tuning fork (A) A sphere of radius 1 cm (B) A sphere of radius 1 mm (C) A sphere of radius 1 m (D) A sphere of radius 10 m

138. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is (A) 1.2 cm (B) 1.2 mm (C) 2.4 cm (D) 2.4 mm 139. A nicol prism is based on the action of (A) refraction (B) . double refraction (C) dichroism (D) both (B) and (C) 140. A thin sheet of glass (refractive index 1.5) of thickness 6 micron,_ introduced in the path of one of the interfering beams in a double-slit experiment, shifts the central fringe to a position earlier occupied by the fifth bright fringe. The wavelength of light used is (A) 3000 A (B) 6000 A

cq

4500

A

(DJ 7000 A

141. Which of the following cannot be polarised?

(A) (C)

Radio waves Infrared rays

(B) p rays (D) y rays

142. When light is incident on a transparent surface at the polarizing angle, which of the following is completely polarized? (A) Reflected light (B) Refracted light (C) Both reflected as well as refracted light (D) Neither reflected nor refracted light

144. Diffraction pattern of a single slit consists of a central

bright band which is (A) wide, and is flanked by alternate dark and bright bands of decreasing intensity (B) narrow, and is flanked by alternate dark and bright bands of equal intensity (C) wide, and is flanked by alternate dark and bright bands of equal intensity (D) narrow, and is flanked by alternate dark and bright bands of decreasing intensity 145. In Young's experiment with one source and two slits, one of the slits is covered with black paper. Then (A) the fringes will be darker

(B) (C) (D)

the fringes will be narrower the fringes will be broader no fringes will be obtained and the screen will have uniform illumination

146. The distance between two coherent sources is 0.1 mm. The fringewidth on a screen 1.2 m away from the sources is 6.0 mm. The wavelength of light .used is (A) 4000 A (B) 5000 A

(CJ

6000 A

(D)

noo A

147. If three slits are used in Young's experiment instead of two,weget

(A) (B) · (C)

(D)

no fringe pattern the same fringe pattern as that wit!) two slits a pattern with fringe width reduced to half of that in the two slit pattern alternate bright and dim fringes

148. When a transparent parallel plate of uniform thickness t and refractive index n is interposed normally in the path of a beam of light, the optical path is (A) increased by nt (B) decreased by nt (C) decreased by(n-1)1 (D) increased by (n-1)1

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Advanced JEE Pl1ysics 149. In Young's experiment, monochromatic light is used to

illuminate the two slits and interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed normally in the path of the

(A)

(C)

I 2

I

(B)

(D)

I

..fi. I 4

beam coming from one of the slits, then

(A) (B)

(q (D)

the fringes will disappear the fringe-width will decrease the fringe-width will increase there will be no change in the fringe-width

150. If the Young's double slit-experiment is performed with white light, then the colour which will have maximum fringe width is · (A) Blue (B) Green (C) Yellow (D) Violet , 151. In the interference pattern, energy is

(A) (B) (C) (D)

created at the positions of maxima destroyed at the positions of minima conserved but is redistributed not conserved

152. Fluorescent tubes give more light than a filament bulb of same power because

(A) (B)

the tube contains gas at low temperature ultraviolet light is converted into visible light by fluorescence

(C) (D)

light is diffused through the walls of the tube it produces more heat than bulb

153. Energies of photons of four different electromagnetic radiations are given below. The energy value 1 eV 5 eV

(B) 2 eV

(D) 1000 eV

(C) (D)

intensity in the two situations are related as

(A)

I =I0

(q

I =3I0

15B. The phase difference between two wave trains giving rise to a dark fringe in Young's double-slit experiment is ( where n is an integer ) ZERO

(B)

(C)

2nn + 1t

(D) 21tn +.!:

2

4

159. A Young1s double-slit set-up for interference is shifted from air to within water. Then the

(A)

fringe pattern disappears

(B) fringewidth decreases (C) fringewidth increases (D) fringewidth remains unchanged 160. Two interfering beams have intensities in the ratio of 9 : 4 . Then the ratio of maximum to minimum intensity

162. The fringe pattern observed in Young's double-slit experiment is (A) a diffraction pattern (B) an interference pattern (C) a combination of diffraction and interference patterns

(D) 156. A beam of unpolarized light of intensity I is passed first through a tourmaline crystal A and then through another tourmaline crystal B oriented so that its principal plane is parallel to that of A. If A is now rotated by 45° in a plane perpendicular to the direction of the incident ray, the intensity of the emergent light will be 2.64

2,m+2:

(A)

fringe width is proportional (A) to wavelength (B) to inverse wavelength (C) to square of wavelength (D) to inverse square of wavelength

pure emission line spectrum emission band spectrum absorption line spectrum absorption band spectrum

155. If a torch is used in place of monochromatic light in Young's experiment (A) fringes will appear as for monochromatic light (B) fringes will appear for a moment and then they will disappear (C) no fringes will appear (D) only bright fringes will appear

=

(B) I= 2I0 (D) I =4I0

161. In interference with two coherent beams of light, the

154. Atomic spectrum should be

(A) (B)

intensity of the central bright fringe is I . When one of the sources is blocked, the intensity become I0 • The

iri the interference pattern is (A) 25:1 (B) 13:5 (C) 5:1 (D) 3:2

corresponds to a visible photon is equal to

(A) (C)

157. Interference pattern is obtained ol1 a screen due to two identical coherent sources of monochromatic light. The

neither a diffraction nor an interference pattern

163. In Young 1s interference experiment with one source and two slits, on~ slit is covered with a cellophane sheet

which absorbs half the intensity. Then (A) no fringes are obtained (B) bright fringes will be brighter and dark fringes will be darker (C) all fringes will be dark

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Wave Optics (D)

bright fringes will be less bright and dark fringes will be less dark

164. The distance between sources in a biprism _of angle a and refractive index µ , if the source is placed at a distance a from it is

(A)

2(µ-l)cx

{B)

(C)

(µ-l)cx

(D) (µ-l)cxa

170. In Young's double-slit experiment, we get 60 fringes in

2(µ-l)aa

165. To obtain a sustained interference pattern, we require two sources which emit radiation of (A) the same frequency.

(B) (C) (D)

~69. In the double-slit experiment, the distance of the second dark fringe from the central line is 3 mm. The distance of the fourth bright fringe from the central line is (A) 6mm (B) 8mm (C) 12mm · (D) 16mm

nearly the same frequency. the same frequency having a definite phase relationship. different wavelengths.

166. A thin mica sheet of thickness 2 x 10-< m and refractive index (µ =1.5) is introduced in the path of the first

wave. The wavelength of the wave used is 5000 A . The central bright maximum will shift (A) 2 fringes upward (B) 2 fringes downward (C) 10 fringes upward (D) None of these 167. In Young's double slit experiment, the slits are 0.5 mm apart and interference pattern is observed on a

screen placed at a distance of 1 m from the plane containing the slits. If wavelength of the incident light is 6000 A ,.then the separation between the third bright fringe and the central maxima is (A) 4 mm (B) 3.5 mm (C) 3 mm (D) 2.5 mm 168. Interference fringes are obtained due to the interference

of waves from hvo coherent sources of light with amplitudes a1 and a,_ ( a,. = 2a,). The ratio of the maximum and minimum intensities of light in the interference pattern is (A) 2 (B) 4 (C) 9 (D) 00

the field of view if we use light of wavelength 4000 A . The number of fringes we will get in the same field of view if we use light of wavelength 6000 A is (A) 40 (B) 90 (C) 60 (D) 50 171. With a monochromatic light, the fringe-width obtained in a double-slit experiment is 1.33 mm. If the whole setup is immersed in water of refractive index 1.33, the new fringe-width will be {A) 1.33 mm (B) 1mm 1.33 . (C) 1.33x1.33 mm (D) --mm 2 172. Two waves having amplitudes in the ratio 5: 1 produce interference. The ratio of the maximum to the minimum

intensity is {A) 25:1 (C) 9: 4

(B) 6:4 (D) 3:2

173. If the intensities of the two interfering beams in Young's double-slit experiment are I1 and 12 , then the contrast between the maximum and minimum intensities is good when (A) [I, - I, [ is large

(B) (C) (D)

[11 -I,[ is small either J1 or 12 is zero I,=I,

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This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/ are correct. 1.

< i,_,

(B) If ).. d < at least on~ more maximum (besides the central maximum) will be observed on the

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 106 Hz . The sources are synchronized to have zero phase difference. The slits are separated by a distance d = 150.0 m. The intensity I(8) is measured as a function of 8, where 0 is defined as shown. If ! 0 is

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

the maximum intensity, then I(8) for O,; 8,; 90° is given by ·1

f d/2

f

will.increase ;

Os,

3.

In a modified YDSE ·. experiment if point source of monochromatic light O is placed in such a manner that QS1 -OS, = ~, where ).._ is the wavelength of light and

-- 8 ---------

S1 , S2 are the slits separated by .a distance 2).. . Then

d/2

value(s) of 8 for which a maxima is obtained is/ are

* (A) I(8)= I, for 8=30°

- (!_ _____

2

(B)

I(8) =.!._. for 8 = 90°

(C)

I(8)=I0 for 8=0°

s,

4

(D) I(8) is constant for all values of 8 2.

In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is )._ . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). (A) If d =).., the screen will contain only one

4.

. -1(1) 8 . -1(5) sm 6

. -1(--1) 4 . -1 ( 7)8

(A) sm

(B)

sm

(C)

(D)

SID

--

Two coherent waves represented by 2

-rot+i) and . (2nx, y, = Asm T - rot +6") = 2.66 ================================::i maximum

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y1 =Asin( ';:'

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Wave Optics are superposed. The two waves will produce

(D) the amplitude ratio is 2

. _inter . ference a t x - x = M ll:\. (A) constr.uctive 1 2

9.

. mter1.erence . ' (B) constructive at x1 - x2 = -23,. 24

(ii) y 2 =a,sin(rot+~) (iii) y, = a1 sin(2rot)

(q destructive interference at x1 - x 2 = ~: . mter . £erence a t x - x =~ ll:\. (D) d estructive 1 2

5.

6.

To obse~e a stationary interference pattern formed by two light waves, it is not necessary that they must have (A) the same frequency (B) the same amplitude (q a constant phase difference (D) the same intensity Two point monochromatic and coherent sources of light

of wavelength A. are placed on the dotted line in front of a large screen. The source emit waVes in phase with each other. The distance between S1 and S2 is d while

their distance from the screen is much larger. Then for

S1

Four coherent light waves are represented by (i) y, = a, sin(rot)

(iv) y, =a2 sin(2rot+~) Interference fringes may be observed superposition of (A) (i) and (ii) (B) (i) and (iii) (C) (ii) and (iv) (D) (iii) and (iv)

due

to

10. If one of the slits of a standard Young's double slit experiment is covered by a thin parallel slit glass 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 (C) The bright fringes will become less bright and the dark ones will become more bright (D) The fringe width will remain unchanged 11. If the first minima in a Young's slit experiment occurs directly in front of one of_ the slits, (distance between slits and screen d = 5 cm ) then the wavelength of the radiation used can be (A) 2cm (B) 4cm

S2

•----•------------- 0 1+-d-+t

Screen

2

4 3

(q -cm 3 2 d = 3,. , there will be a total of 6 minima on· screen d = A. , there will be one maxima on the screen d = 2A , there will be two maxima on the screen

(D) - cm

3

(A) d = ,. , 0 will be a minima (B)

(q (D)

12. Two coherent sources A and B emitting light of wavelength A. are placed at positions (-D, 0) and

(-D, 3A.) respectively D»1,, y

7.

8.

If white light is used in a Young's double-slit experime~t, then (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 (q the first-order violet fringes are closer to the centre of the screen than the first order red fringes (D) the first-order red fringes are closer to the centre of the screen than the first order violet fringes

In the Young's double slit experiment, the interference \ pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that (A) the intensities at the screen due to the two slits are 5 units and 4 units respectively (B) the intensities at the screen due to the two slits can be 4 units and 1 unit respectively (C) the amplitude ratio is 3

o~-----•x (A) number of minima on y-axis is 6 (B) number of minima is more than nurnbef of maxima on y-axis (C) number offfiaxima on x-axis is 3 (D) number of maxima on x-axis is more than number of minima on x-axis 13. In the figure A , B and C are three slits each of them individually producing the same intensity 10 at P

when they are illuminated by parallel beam of light of

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Optics & Modern Physics

Advanced JEE Physics wavelength 1,. • it is given that BP '

AP =!:.2 . Also given

that d « D, then wavelength 1,. and resultant intensity I at P will be

14. White light is used to illuminate the two slits in a 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 one of the slits, certain wavelengths are missing.. Some of these missing wavelengths are b' 2b' (A) 1..=~ (B) 1..=-

(C)

d b' 1..=3d

d 2b'

(D) 1..=3d

'

----------------- p t + - - - D ~---<~

(A)

2d' D

1,.=-

(C) I= 2I,"

• 4d' D (D) •I=I,

(B)

1..=-

15. The fringe width in Young's double-slit experiment can . be increased by decreasing " (A) · separation of the slits (B) frequency of the source of light (C) distance between slit and screen (D) wavelength of the source of light

,I

.,

...

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This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STA,TEMENT 1. Bubble (C) IfSTATEMENTl is TRUE andSTATEMENT2is FALSE. Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.

Statement-I:

Statement-2:

If the phase difference between the light waves emerging from the slits of the Young's experiment is 1t -radian, the central fringe will be dark.

The fringe width is inversely proportional to the distance between the two slits in simple YDSE :

5.

Statement-2: Phase difference is equal to

2n 1,.

Statement-1: The minimllffi slit separation d for interference to produce at least one maxima other than central maxima is

times the path

difference.

31,..

Statement-1: When a thin transparent sheet is placed in front of both the slits of Young's experiment, the fringe width will

For a maxima, path difference equals n'J... . The maximum value of path difference is d .

Statement-2: 2.

increase.

6.

In Young's experiment the fringe width is proportional to wavelength of the source used.

Statement-I: In Young's double slit experiment, we observe an

To obtain interference pattern, source must be coherent. Two different light sources can never be coherent.

Statement-2:

3.

Statement-I: Two slits in YDSE are illuminated by two different sodium lamps emitting light of saffie wavelength. No interference pattern is observed.

interference pattern on the screen if both the slits are illuminated by two bulbs of same power.

Statement-2:

7.

Statement-I: In Young's double slit experiment interference pattern disappears when one of the slits is covered by transparent slab.

Statement-2: The interference pattern is observed when source is monqchromatic and coherent.

Statement-2: 4.

Statement-I:

.

Interference occurs due to superimpoSition of light wave from two coherent sources.

No interference pattern is detected when tw'o coherent sources are infinitely close to each other in sim_ple YDSE.

8.

Statement-I: When a thin transparent sheet is placed in front of both the slits of Young's experiment, the fringe width will increase. 2.69

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Optics & Modern Physics

Advanced JEE Physics

Statement-2: In Young's double slit experiment, fringe width is given by re1a· " :w . !ion e =d .

Statement-2: In Young's experiment the fringe width is proportional to waveiength of the source used.

9.

Statement-I: Total number of maxima obtained over screen ren:iains

same whether Young's Double slit experiment is performed in air or in water with same setup.

Statement-2: f3watcr

=

16. Statement-I: Interference pattern is obtained on a screen due to two identical coherent sources of monochromatic light. The intensity at the central part of the screen becOmes one-

fourth if one of the source is blocked. f3air

(in Young's double slit experiment).

Statement-2: The resultant intensity is the sum of the intensities due

µwater

to two sources.

10. Statement-I:

Interference obeys the Law of Conservation bf Energy.

17. Statement-1: Thin films such a soap bubble or a thin layer of oil on water show beautiful colours when illuminated by monochromatic light.

Statement-2: The energy is redistributed in case of interference. 11. Statement-I:

Statement-2: Colour in film are obtained due to interference between reflected light from the upper & lower layer of film.

Geometrical optics can be regarded as the limiting case of wave optics.

Statement-2: When size of obstacle Or opening is very large compared to the wavelength of light then wave nature can be ignored and light can be assumed to be travelling in straight line.

18.

as zeroth order fringe, or the central fringe.

Statement-2: Path difference between the wave from S1 and S2 ,

12. Statement-I: Light from two coherent sources is reaching the screen. If the path difference at a point on the screen for yellow

light is

3 : , then the fringe at the point will be

coloured.

reaching the central fringe (or zero order fringe) is zero. 19. Statement-1: The phase difference between any two points on a wavefront is zero.

Statement-2: Light from the source reaches every ·point of the

Statement-2: Two coherent sources always have constant phase relationship. 13. Statement-1: The maximwn intensity in interference pattern is four times the intensity due to each slit. '

Statement-2: Intensity is directly amplitude.

Statement-1: The fringe obtained at the centre of the screen is known

wavefront at the same time.

20.

Statement-1: In Young's double slit experiment interference pattern disappears when one ofthe slits is closed. Statement-2:

proportional

to

square

Interference occurs due to superimposition of light wave from two coherent sources.

of

14. Statement-1: Interference can be obtained by using two different

lamps.

21.

Statement-1: If a clean glass slide is observed under white light, one does not observe any colours. However, if this slide is

. touches with oily hands, coloU)"ed fringes appear on the slide.

Statement-2: Two different lamps are incoherent sources as constant phase difference cannot be maintained between them.

15. Statement-1:

Interference pattern is made by using blue light instead

Statement-2: These fringes are due to interference of reflected light, reflected from the upper and lower surfaces of the thin oil film.

of red light, the fringes becomes narrower.

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I

This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (DJ, out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options) Comprehension 1 In a Young's experiment the upper slit is covered by a thin glass plate of refractive index µ =1.4. The interference pattern is observed using light of interference 5000 A. Based on above information, answer the follOwing questions. 1.

It is observed that (A) the central maxima shifts upwards. (B) the central maxima shifts downwards. (C) fringe pattern will change after introduction of thin plate. (D) · none of the above phenomenon is observed.

2.

Now a thin plate of refractive index 1.7 is placed in front of lower slit then (A) central maxima will be obtained above centre. (B) central maxima will be obtain at centre. (C) there will .not be any change in central maxima after introduction of thin plate. (D) no conclusion can be withdrawn without knowing the thickness of the plate.

3.

If intensity at center is

(D) l.7 t

0.8!

(A)

t = 2777.7 A

(C) · t = 4188.8 A

4

of maximum intensity then

(B) t=3188 A (D) t = 2122.9. A

Comprehension 2 A thin film of a specific material can be used to decrease the intensify of reflected light. There is destructive interference of waves reflected from upper and lower surfaces of the film. These films are called non-reflecting or antireflection coatings. The process of coating the lens or surface with non-reflecting film is called blooming as shown in the figure. The refracting index Of coating ( n1 ) is less than that of the glass

(n2 ). 'Based

on above information, answer

the following questions. 2

Air

R.1. =n,

6.

1.5

3

minimum value of 1 is

Assume the thickness of both thin glass plate is t , the ·path difference between waves incident at center is (A) 0.5! (B) 0.3!

(C) 4.

5.

Glass

If the light of wavelength 1,. is incident normally and the thickness of film is t then optical path difference

If minima is to be obtained at center then minimum value of t is (source wavelength is 1,.)

between waves reflected from upper and lower surface of film is · ·

(A) ! = 0.6

(B)

(A) 2n1t

(C)

(D) t = __!,:__ 0.15

,.

,.

t=-

0.3

. (C)

(B)

,.

2n t-1 2

(D) 2!

2.71.

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Advanced JEE Physics 7.

Magnesium fluoride (MgF,) is generally used as antireflection coating. If refractive index of MgF, is 1.38 then minimum thickness of film required for 1-=550 nm is (A) 112.4nm (B) 78.2nm (C) 99.64 run (D) 225nm.

8.

Assuming that the thickness of film in above problem is not technically possible to manufacture, then next thickness of film required is (approximately) (A) 298.9 nm (B) 271.7 nm (C) 304.7 run (D) 550 nm

Comprehension 3 In a modified Young's double slit experiment, a

monochromatic uniform and parallel beam of light of 10 wavelength 6000 A and intensity wm-2 is incident

.

"

14. The phase difference introduced by the film is (A)

1t

radian

-A

II

-

(B) 5µW (D) 7µW

(A) 2 µW (C) 6 µW Comprehension 4

A monochromatic beam of light of wavelength 1- = 600 nm falls on Young's double slit experiment apparatus as shown in figure. A thin sheet of glass is inserted in front of lower slit S, . Based on above information, answer the following questions.

s, ~

---------- 0

s, D µ, t

S,S, = d(«D)

D

(B) at O or below 0 (C) at O or above 0 (D) anywhere on the screen

F

17. If central bright fringe is obtained on screen at O, then we have

(A) (µ-l)t=d sine

The power received at A is (A) 10.., W (B) 4x10_, W (C) 10-< W (D) 4x10-< W

(C)

µt=de

(B)

(µ-l)t = dcos8 d (D) µ-1 sine

18. The phase difference between central maxima and fifth minima is

11. The power transmitted through A is (A) 10-5 W (B) 4 x 10-s W (C) 10-< W (D) 4x10-< W

(A)

"6

(B)

(C)

31t

(D)

2

19.

12. The power transmitted through B is (A) 10-5 W (B) 4x10-5 W (C) 10-< W (D) 4 X 10-< W

4

16. The central bright fringe can be obtained (A) at 0

Based on the a~ove facts, answer the following questions.

10. The power received at B is (A) 10-5 W (B) 4 x 10-5 W (C) 10-< W (D) 4x10_. W

radian

15. The power in watt received at the focal point F of the lens is

_s

9.

2"

(D) 2: radian

(C) .': radian 3·

normally on two circular apertures A and B of radii 0.001 m and 0.002 m respectively. A perfect transparent film of thickness 2000 A and refractive index 1.5 for the wavelength of 6000 A is placed in front of aperture A . 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.

---

(B)

91t

Fringe width for the pattern obtained on screen, if 1-=600run, µ=1.5, d=3mm, D=2m and 8=30° is (A) 2x10 5 run (B) 4x10 5 nm 4 (D) 3x106 nm (C) 10 run

13. The path difference introduced by the film is (A) 10"" m (B) 10.., m (C) 10-' m (D) 10_, m

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Wave Optics 20.

Assume if

a(<~) is increased then for a given value of

µ

{A) central maxima will move downwards. (B) central maxima will move upward. (q fringe width will increase. (D) fringe width will decrease.

25. Minimum thickness of film for. destructive interference in transmitted light system is (A) 150 nm (B) 200 nm (q 250 nm (D) 100 nm Comprehension 7 A narrow tube is bent in the form of circle of radius R

Comprehension 5 In the arrangement shown in the figure, the distance D is large compared to the separation d between the slits. Monochromatic light of wavelength 1.. is incident on the slit,

as shown. Two small holes S and D are made in the tube at the positions right angles to each other. A source placed at S generates a wave of intensity 10 which is equally divided into two parts. oi,e part travels along the longer path; while the other travels along the shorter path. Both the part waves

based on the information provided answer the following questions.

information, answer the following questions.

I .

1+-

~ o----

I I

(D»d)

du

=:

0

~Screen 2D ---+i

21. The minimum value of d for which there is a dark fringe at the point O is

(A)" (B)

\

(D) not possible t~ be calculated 22. The position of first bright fringe for the minlmum value of d is

2

(q

2

below

3d

below

23. The fringe width is (A) 3D1.. · 4d

(q D1.. d

(B)

3d (D) - above 2

(B) (D)

(D) I,

(D) 31tR

2

1tR

(C)

2

(D)

2

31tR 2

Comprehension 8 When Fi'esnel 1s biprism experiment is performed in air then distance between coherent sources is 0.5 mm and distance between source and screen is 1 m . Fringe width obtained in air is 1 mm . Refractive index of biprism is 1.5. Now the experiment is performed in water having refractive

3D1..

2d 2D1.. d

If light incident ·on a thin film has wavelength as 900 nm and refractive index of film is 1.5. Based on above

information, answer the following_questions. 24. Minimum thickness of film needed for constructive interference in reflected light system is

(B) 150 nm (D) 250 nm

1tR

(C)

28. The maximum value of wavelength 1,. to produce a minimum at D is given by (A) 1tR (B) 27tR

d above

Comprehension 6

(A) 100 nm (C) 200 nm

26. Maximum intensity produced at. D is given by {A) 4I0 (B) 3I0

27. The maximum value of wavelength 1,. to produce a maximum at D is given by (A) 1tR (B) 21tR

(q ,ffJ5

d

:o (q 2I,

ff

(A)

meet at point D where a detector is placed. Based on above

index µ.

=±.If the refractive index of the biprism is µ = ~ .

3 2 Based on above information, answer the following questions.

29. The distance between coherent sources in water is

{A) 1mm (C)

1 -mm 4

,(B)

1 -mm 2

1 (D) -mm 8

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Optics & Modern Physics

Advanced JEE Physics 30. The fringe width in. water is (A) 1 mm (B) 2mm (C) 3mm (D) 4mm

¾is plac~d. The interference pattern i~ observed on a screen at a distance D = 2 m from the slits. Based on above information, answer the following questions.

31. The wavelength of light in air is (A) 4000 A (B) 4500 A (CJ 5000 A (DJ 6000 A

I

~-f -·d--------------------

Comprehension 9 . In a YDSE experiment, the two slits are covered with transparent membranes of negligible thickness which allow light to pass through it but does not allow water. A glass slab of thickness t = 0.41 mm· and refractive index µ, = 1.5 is

placed in front of one of the slits as shown in figure. The separation between the slits is d = 0.30 mm. The entire space to the left of the slits is filled with water of refractive index

µ."= ½-

A coherent light of intensity I and absolute

wavelength ;. = 5000 A is being incident on the slits making an angie of 30° with horizontal. Screen is placed at a distance D = 1 m from the slits. Based on above information, answer the following questions.

ES:._,---;;;"'IP -------- 0

~J

,.__ _ _ D - - - - . , I I

35. The angular position of the central maxima w.r.t. central line is

(B)

(A) 60° (C) 30°

36. The order of minima closest to centre C of screen is (A) 50 (B) 49 (C) 48 (D) 47 37. The number of fringes that will pass over C , when the transparent slab from the lower slit is removed is (A) 100 (B) 98 (C) 96 (D) 94

+ y t

Screen

S,S, = d( «D)

wavelength 5000 A illuminates two slits S1 and S2 , which

t+----- D ___.,

act as two cOherent sources. The source S oscillates about its

32. At point O, equidistant from slits we get (A) 9th dark fringe (B) 10th dark fringe (C) 11th bright fringe (D) 10th bright fringe

shown position according to the equation y = 0.5sin(1tt), where y is in millimetres and t in seconds. Based on above information, answer.the following questions. y

33. Central maxima is located at

Is.

5 y=--cm 6 5 (D) y=--cm 3 (B)

..fi.

----------------- p

(D) 1

2.74

38. The position of the central maxima as a function of. time is best represented by the equation

(A) -cos(1tt) (C) -sin[(1t+l)t]

2

Comprehension 10 Light of wavelength ;. = 500 nm falls on two narrow slits placed a distance d =50 x 104 cm apart, at an angle cj, =30° relative ~o the slits shown in figure. On the lower slit a transparent slab of thickness 0.1 nm and refractive index

=

Lx

s

34. The ratio of intensity at point P at y =! cm on screen 8 and maximum intensity is (A) 0 (B) 1

(C) _!_

45°

(D) 15°

Comprehension 11 In a Young's double slit experiment set-up source S of

water

5 (A) y=+- cm 6 5 (C) y=+-cm 3

C

(B) cos(1tt) (D) -sin[1t(l+t)]

39. The minimum time t at which the intensity at point P on the screen exactly in front of the upp~ slit becomes maximum is

1 (A) 2

s

(B)

1 -

3

s

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Wave Optics 1 {C) -

6

s

y

(D) 1 s

SI

40. The minimum time f at which the intensity at point P on the screen exactly in front of the upper slit becomes

s() }------------- o s,, D

minimum is

(A) 1 s 1 3

(C) - s

1

(B)

-

2

s

1 6

43. The central maximum formed on the y-axis is located at

(D) -s

Comprehension 12 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. Based on above information, answer the following questions.

mm above O

3

mm below 0

y=

11

14

3

mm above 0

44. The ratio of light intensity of point O to the maximum fringe intensity is

1

t...., --- 2 m ---+t<- 10cm

s

B~------'C

The position of the central bright fringe on the other wall CD with respect to the line OQ is {A) 2 cm below Q (B) 2 cm above Q (C) 4 cm belo~ Q (D) 4 cm above Q

42. It is observed that when a liquid is poured into the

vessel and filled upto OQ , the central bright fringe is found to be at Q . The refractive index of the liquid is (A) 1.0008 (B) 1.0004 (C) 1.0016 (D) 1.0012 Comprehension 13 The Young's double slit experiment is done in a

4

3

(D) 1

Comprehension 14 In the Young's double slit experiment a point source of 1c = 5000 A is placed slightly off the central axis as shown in the figure. Based on the information provided, answer the following questions.

1+--

slit S2 is covered by a thin glass sheet of thickness 10.4 µm

ignore

Is

ttp

,s,

~

t 1· . :t5f mm 1'-7m"--~ -'r' _10-+-1m+-----:li10

±.A light of 600 nm wavelength

½ and

2

45. Assuming the 600 nm light to be replaced by white light of range 400 to 700 nm, the wavelengths of the light that form maxima exactly at point O are (A) l300 nm, 500 nm (B) 1400 nm, 600 nm 3 3 13 0 (C) (D) l:OO nm, 650 nm ; nm, 650 nm

. 3 is fallirig on the slits having 0.45 mm separation. The lower

and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown in the figure. Assume that all wavelengths in the problem are for

1

(B)

4

S1

the given medium of refractive index

.

(D)

(C)

f---------------------f o -------- 0

medium of refractive index

3

(C) y =

{A)

1

41.

(B)

D

s, 40 cm

11

{A) y=

1 m--1----- 2

m---...

46. The order and nature of the interference at the point P

is (A) SO" order, maxima (C) 70" order, maxima

dispersion. Based on above information, answer the following questions.

(B) SO" order, minima (D) 70" order, minima

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Optics & Modern Physics

Advanced JEE Physics 47. The order and the nature of the interference at O is (A) 50., order, minima (B) 50., order, maxima (CJ 20., order, minima (D) 20., order, maxima

from each other by a distance 6 mm and 1.5 mm in the other. Now lens is removed and interference pattern is obtained on the screen. Based on the information provided, answer the following questions.

48. If the zero order maxima is formed at O , then

(A) we should place a film of refractive index thickness 10 µm in front of 52 • (B) we should place a film of refractive index thickness 20 µm in front of 52 • (CJ we should place a film of refractive index thickness 10 µm in front. of 51 . (D) we should place a film of refractive index thickness 20 µm in front of 51 • Comprehension 15 A narrow slit S

µ =1.5, µ =1.5,

s

rcreen

µ =1.5, t+-

µ =1.5,

allows monochromatic light of wavelength ). = 6000 A to fall on a prism of very small angle as shown in figure. A screen is placed at a distance l = 100 cm from the source to obtain an interference pattern. To determine the distance between the virtual images formed by the prism an experiment is done. The prism and screen are kept fixed and a convex lens is moved between the prism and the screen. For two positions of the lens (between the prism and the screen) we get two sharp point images on the screen in each case. 'fl1;e images are separated

1 oo cm ------+1

49.

Focal length of the lens is (B) 20cm (A) 16cm (D) 40cm (CJ 36 cm

so.

Fringe width of the_pattem on the screen is (A) 0.1 mm (B) 0.2 mm

(CJ 0.3 mm

(D) 0.4 mm

51. If screen is displaced slightly away from prism, then (A) No interference pattern is observed (B) fringe width remains same. (CJ fringe width decreases. (D) fringe width increases.

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Each question in this section contains statements given in two Columns which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: 1

If the correct matches are A --> p, s and t; B --> q and r; C --> p and q; and D --> s and t; then the correct darkening of bubbles will look like the following : P q r s t

A@@)(D@@

B@@@®CD c.® ®©®CD D@@)(D@@)

1.

COLUMN-I shows the effect on the fringe pattern in YDSE corresponding to the changes mentioned in COLUMN-II. Match the effects in COLUMN-I with the corresponding causes in COLUMN-II.

i

COLUMN-I

COLUMN-II

(A) Angular fringe width remains same (B) Angular.fringe width changes (C) Fringe width (linear separation consecutive fringes) changes (ID) The fringe pattern may disappear 2.

between

two

(p) Screen is moved away from the plane of the slits (q) Wavelength of light used is decreased (r) The separation between the slits is increased (s) The width of the source slit is increased

'

Figure shows a set-up to perform Young's double slit experiment. A monochromatic source of light is placed at S. S1 and 52 act as coherent sources and interference pattern is obtained 011 the screen. Match COLUMN-I with COLUMN-II keeping in mind the Young's double slit experiment. Screen

J~ ____·i------

0

s,~

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Advanced fEE Physics

CQLUMN-Il,

C01'UMN-I

(A) S is removed and two real sources emitting light of (p) Interference fringes disapp~ar. same wavelen~ are placed at S1 and S2 • (q) There is uniform iUumination on a large part of the (B) Width of S1 is two times the width of

screen.

s2·•

(q S1 is closed. (D) A thin _transparent plate is placed in front of S1 • Assuming negligible absorption by the plate. 3,

(r) (s)

The zero order fringe will not form at O . Intensity of a dark fringe will ~e non-zero, but less than the intensity of bright fringe.

Match the contents of COLUMN-I with the respective phenomenon in COLUMN-II. COLUMN-I .

COLUMN-II

(A) Shining of diamonds. (B) Light waves projected on oil surface shows seven I colours. (q Huygen's wave theory of light cannot explain. (D) Phenomena which is not Explained by Huygen's

(p) (q) (r) (s)

./

interference total internal reflection origin of spectra photoelectric effect

construction of wavelength. '.J•

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. i.. • Then match COLUMN-I with COLUMN-II. . [ .

COLUMN-I

COLlJMN:-Il

(A) For bright fringe, path difference. · (B) For dark fringe, path difference. (q Displacement of fringe when, glass plate of thickness I is placed. · (D) Distance between central maxima and first dark fringe when glass plate of thickness I is used. ·

Di.. (p)

(q) 'D(µ-1)1 d (r) ni.. (s)

5;

2d

i.. (2n-1)2

Match the contents of COLUMN-I with the respective contents of COLUMN-II.

I ,

COLUMN-I

·COLUMN-II

(p) Incoherent sources

(A) Sources of variable phase difference,-

(B) Point on a wavefront behay~s as a light source. (q) Coherent sources (q Net displacement is the vector sum of individual (r) Superposition principle . displacement. (D) Young's double slit experiment uses. ·

(s)

Huygen's principle

- ' . ,. In the YDSE appratus shown in figure, rue is the path difference between S2 P and S1P. H, now a glass slab is introduced in°front of S~, then match the contents of COLUMN-I with the respective match"': in COLUMN-II. ~

6.

p

-++-------ii' 0

s,~

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Wave Optics COLUMN-I (A) (B) (C) (D)

7.

Fringe width will Fringe pattern will Number of fringes between O and P will t;x at P will

(p) (q) (r) (s) (t)

increase decrease remain same

shift upward shift downward'

Match the quantities in COLUMN-I with their respective matches in COLUMN-II, COLUMN-I (A) Point source of light (B) Limit of resolution of telescope (C) Interference

(D) Coherent sources 8.

COLUMN-II

COLUMN,11

(p) Spherical wavefront (q) Amplitude division (r) Superposition of waves (s) Radius of lens

In the light of possibility of occurrence of phenomena listed in COLUMN-I match the ·listings in COLUMN-I to the corresponding waves in COLUMN-II.

[

COLUMN-I (A) Reflection (B) Interference (C) Diffraction (D) Polarisation

COLUMN-II (p) Non-mechanical waves (q) Electromagnetic waves (r) Visible light waves (s) Sound waves

c:::::=================================== = 2.79

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J

In this section the answer to each of the question is a four digit integer, ranging from Oto 9999. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question number X (say) is 6092, then the correct darkening of bubbles will look like the following :

X.@e@@

CDCDCDCD

®®®· @@@@ @@@@ @@ @@

~~~~

@®®®

®@e@

1.

2.

Interference pattern with Young's double slit 1.5 mm apart are formed on a screen at a distance 1.5 m from the plane of slits. In the path of the beam from one of the slits, a transparent film of 10 micron thickness and the refractive index 1.6 is interposed while in the path of the beam from the other slit a transparent film of 15 micron thickness and a refractive index 1.2 is interposed. Find the displacement of the fringe pattern, inmm. Two Coherent radio point sources that are separated by

2 m are radiating in phase with a wavelength of 0.25 m . If a detector moves in a large circle arowtd their midpoint, at how many points will the detector show a maximwn signal?

3.

central maxima (where path difference is zero) is obtained. (b) third order maxima is obtained. (a)

4.

An equiconvex lens of focal length 10 cm (in air) and refractive index

½. A concave mirror of radius of curvature 20 cm

.

is

cut into two halves m, and m, and placed at the end of the tube. m1 and m, are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principle axes of the lens.

s•-e-----

I

'~

is put at a small opening on a tube of

length 1 m fully filled with liquid of refractive index

In the figure shown the distance between the slits is d = 201., where ). is the wavelength of light used. Find the angle 0, in degree, where

--------

~2

t+20c

A slit S placed in air illuminates the lens with lighi of frequency 7.5 x 1014 Hz . The light reflected from m,

= ================================== 2.80

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Wave Optics and m, forms interference pattern on the left and EF 1

------------ p

of the tube. 0 is an opaque substance to cover the hole

created by the placement of m, and m,_ • Find : (a) The position of the image, in cm, formed by lenswater combination.

(b) The distance, in mm, between the images formed by m. and m,. (c) Width of the fringes on EF, in µm . 5.

A ray of light is incident on the left vertical face of the glass slab. If the incident light has an.intensity I and on each reflection the intensity decreases by 90% and on each refraction the intensity decreases by 10%, find the ratio of the intensities of maximum to minimum in the reflected pattern.

Screen

D>>d

10. In a YDSE (young double slit experiment) screen is

placed 1 m from the slits wavelength of light used is 6000 A . The fringes formed on the screen are observed by a student sittingdose to the slits. The student's eye can distinguish two rieighboring fringes, if they subtend an angle more than 1 minute of the arc. Calculate the maximum distance between the slits, in nun,. so that fringes are clearly visible. Give your answer to the nearest integer.

6.

7.

8.

Two slits are separated by 0.32 mm. A beam of 500 nm light strikes the slits producing an interference pattern. Determine the number of maxima observed in the angular range -30° < 8 < 30° . Interference fringes were produced by Young's double slit method, the wavelength of light used being 6000 A . The separation between the two slits is 2 mm . The distance between the slits and screen is 10 cm . When a transparent plate of thickness 0.5 mm is placed over one of the slits, the fringe pattern is displaced by 5 mm . If µ be the refractive index of the material of the plate, then find 5µ .

11. A parallel beam of white light falls from air on a thin film in air whose refractive index is ,J3 .The angle of incidence is i = 60° . Find the minimum film thickness (in nanometer), if the reflected light is most intense for i..=6oooA. 12. In a modified YDSE the region between screen and slits is immersed in a liquid whose refractive ind~x 5 · w1'th' vanes time as µ1 = - t , un til"1t reaches a stead y

2 4

state value

¾.A glass plate of thickness T = 36 µm

and

refractive index µ = ~ is introduced in front of one of 2 the slits. If the separation between the sources and the screen is 1 m and the separation between the sources is

2 mm, then calculate the speed ·of the central maxima, in mms-1 , when it is at O,

In Young's double slit experiment mixture of two light wave having wav~engths 1,. 1 = 500 nm and i.., = 700 nm are being. used. Find the position next to

y

central maximal where maximas due to both waves

coincides. ( Given 9.

~ = 1000)

0

Consider the interference at P between waves emanating from three coherent sources in same phase located at 51 , S2 and S3 • If intensity due to each source is ! 0 =12Wm-' at P and ~=!::, then 2D 3 resultant intensity at P , in wm-2 •

find the

X

µ,

13. 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 now changed to 400 nm , what will be the new number of fringes observed in the same segment of screen.

'

==================================== = 2.81

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Optics & Modern Physics

Advanced JEE Physics 14. A glass wedge of angle 0.01 radian is illuminated by monochromatic light of wavelength 6000 A falling normally on it. Find the distance from the edge of the wedge, in mm where the lQth fringe will be observed due to the reflected light.

15. In Young's double slit experiment the two slits act as coherent sources of eq"i'1 amplitude A and wavelength :>.. • In another experiment with the same set up the two slits are source of equal amplitude A and wavelength :>.. but are incoherent. Find the ratio of intensity of light at the mid-point of the screen in the first case to that in second cas.e.

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•

ICE! 1.

(a)

7.8 µm

(b)

0.6mm

8.

2.

5 µm

3.

(a)

2"[(,/d' + !' -d) + µd']

(b)

2 "(µ(Jd'

A

A

2D

+I' -1)+~) 2D

5n

(a)

2

(b)

25n 4

9.

5890

A

10.

1.2 µm

11.

Zero order maxima will remain unchanged. Tenth order will now be at 4.55 mm.

4.

94.8 nm

5.

0.546 mm

12.

8.75 mm

6.

112.78 nm

14.

4.5mm

7.

(a)

0.5 mm

15.

7

(b)

2.25 mm

16.

0.21

(c)

31,

17.

(a)

3x10-4l 0

(b)

5.49

(d)

,/3 meter

(e)

•

1

n = 5000 is not possible

18.

3.5

19.

1.5

mm

ICE II

.BAS_l=D ON Dli=FRACTJ.0N &!f'()~_filS~IOl;fj

1.

11.8mm

9,

±45° I ±135°

2.

9x10-4 m

10.

(a)

0.75

3.

0.2mm

(b)

0.25

4.

40 m

11.

37.5%

5.

12.5 cm

12.

!..

6.

1.3µm 0.2mm

13.

1:3

7.

16.

30°' 45°

8.

3.534 x 10.., rad

4

2.83

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Advanced JEE Physics

1.

B

Optics & Modern Physics

2.

D

6. . D

7.

D

8.

D

L11. 16.

0

12.

D

13.

A

B

17.

D

18.

D

19.

21.

C

22.

C

23.

D

24.

26.

A

27.

C

28.

D

29.

33.

A

34.

38.

C

39.

31.

A

32.

O·

36.

C

37.

A

41.

A

46.

C

51.

3.

---- - - -

D

5.

B

9.

B

10.

A

14.

C·

15.

D

B

20.

B

B

25.

D

C

30.

B

A

35.

A

B

40.

B

4.

43.

0

44.

D

45.

C

47.

B

48.

B

49.

A

so.

C

D

52.

B

53.

B

54.

C

55.

B

56.

D

57.

B

58.

A

59.

B

60.

A

61.

D

62.

B

63.

D

64.

B

65.

C

66.

r--n.

D

67.

C

68.

B

69.

B

70.

C

A

72.

C

73.

C

74.

C

75.

C

76.

B

77.

A

78.

B

79. 'C

80.

A

81.

B

82.

B

83.

D

84.

C

85.

A

86.

B

87.

B

88.

B

89.

C

90.

A

91.

C

92.

B

93.

D

94.

C

95.

B

96.

D

97.

D

98.

D

99.

A

100.

A

A

102.

0

103.

D

104.

D

105.

C

106.

D

107.

C

108.

B

109.

A

110.

C

111.

D

114.

B

115.

C

116.

A

117.

D

118.

A

119.

C

120.

B

121.

A

126.

B

[-

~-

[ !

42. A ----

A

-----112. A 113. C ··-------------------122. A ----------

123. . D

124•

A

125.

C

127.

A

128.

B

129.

C

130.

A

D

134.

D

135.

A

L 131.

C

132.

B

133.

136.

C

137. • C

139.

D

140.

B

B

142.

----143. ----

D

141.

B

144.

A

145.

D

A

138.

146.

B

147.

D

148.

D

149.

D

150.

C

151.

C

152.

B

153.

B

154.

A

155.

C

156.

D

157.

D

158.

C

159.

B

160.

A

161.

A

162.

163.

D

164.

B

165.

C

166.

A

167.

B

168.

C

169.

B

170.

A

171.

B

172.

C

173.

D

, __1:.:..__:_A:.:C:..... ~

6.

ABC

- - - - - -2.

7.

-·-----C

I

AB

3. AD 4. BD 5. BO ----------------------~=-__:c=----....J

ABC

8.

BD

9.

AB

10.

AGO

L__1c...1:.:.·__:_A:.:C:.__ _ _ ____::12~-~-c_______1_3_._B_D_ _ _ _ _ _ 14_._A_c_ _ _ _ _ _1_5_._·,_A_B_ _ _. - . J

= c:=================================== 2.84

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Wave Optics

- - - - - - - - - - - - - - - - ----- - - - -

I

L _ ~_ _ _ _ _ _ _ _2.:___D_____ _ _ _ --~'.-~-- _________4:_ _B____ ---------~- D [

I

6.

A

11.

A

16.

C

21.

A

7.

D

17.

D

8.D 18.

______

9.D

D

19.

10.A

A

20.

A

_ ________________________________________________

-

1.

J

---------~-- - - - - - - - - - - - - - - - -

-

j

--1

11.

A _ _ _ _ _ _ _ 2_._D_________ 3·--~---------- -- 4. -- A ----- ------- 5.__ A ___________ I _ A 7. C 8. A ____ 9. A 10. B D_______- _ C - - - - - - 1 - 2 . D _________ 13. __ C ________ 14. __ C _ _ _ ______ 15. __

16.

D

~

6.

I

17.

A

22.

C

---------

01._~_ 26. C 31.

----

____

18.

B

23.

D

19. B 20. B . ---- - - - - ---- ------------. __________24. ___~-- ____________ 25:_!-________ )

-- --------

~B ___

~A

aD

me

-------:n:--D~_-_-_--_-___-_-_-_~.:__--~ ____ - ____ 34._ C _____________ 35. C________; C

36.

B

OT

B

46.

C

51.

D

~

~

A

D

__ 42_. _ ~------ _____ :13_._ __[?__~ 47. D 48. D

=---_-_-=:: _--_4_4._-_c_-_-__-_--_--_----~--= ~s-=-_--c~===~--~-J ~

C

-

A

49.

A

50.

B

__-_: ----~--~-----~--=------_-_ ·=-:==-:-:-:=-----]

IANSWER---r

1.

I

I

6.

A-; (p, s) B--> (q, r) C--> (p, q, r) D--> (q, r, s) A-; (r) B--> (t) C--> (r) D-; (p)

A--> (p, q) B-; (s) C--> (p, q) D--> (r)

2.

A--> (p) B--> (s) C--> (r) D--> (q)

7.

3.

A--> (q) B--> (p) C--> (s) D-; (r)

4.

----------·---A--> (p, q, r, s) 8.

- 5.--: =: l~!--·-1

A--> (r) B--> (s) C--> (q) D--> (p)

C--> (r) ____ D-->(q,~

-----

B--> (p, q, r, s) C--> (p, q, r, s) D--> (p, q, r)

--·--- --- -- --------------·-- -- ------------, _ _ _ _ _ _ _ _ _3_. (a)45,(b)59 _ 4._(a)B0,(b)4,(c}60 _ _ j 2. _32 I_ 1. _3_ _ _ _ _ _ _ _ _ 5.

361

6.

739

7.

6

8.

3

~~ - - - - - - - - - - - - - - - - - - - - - --------- ----'=--=----- ____ 13.

18

10.

2

11.

100

14.

3

15.

2

12.

3

...J

2.85

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Therefore, net path difference t:.x. = (..Jd2 + t 2

-t) +

i;:;

So, phase difference Is given by

$= 2•[(../d' +l' -d)+ µd'] (b)

,_

Upwards (b)

yd -(1-.!)1= 4,_ D µ µ y=4.2mm

t,x

=> Downwards

=>

. 1(1-.!)+yd='4,. µ

2.

(µ-1)tD =>

d

4.

$=~(µ(../d'+t'-t)+i~)

The reflected light will be minimum when rays 1 and 2 (shown in figure) meet the conditi6n of destructive interference.

Rays 1 and 2 'both suffer an ad~itional phase change of 1t (or 180°) after being reflected at surface of a denser medium. The net change In phase due to 1 reflection is therefore zero and the 2

SW d

condition for a reflection minimum

s,- 2sooo A =5 µm l=-=--=50,000 µ-1

=µ(t../d' + I' -t)+ ~)

µ

Since.we are given that Shift= 5 (fringe width)

=>

3.

D

y=0.6 mm

=>

requires a path difference of

1.5-1

Hence,

=>

=>

=>

5.

,550 I= 4µ = 4 (1 _45 ) = 94.B nm

The desired distance is

=>

546x10-9 x0.4 0.4x10-3

y = 0.546 x 10"' m = 0.546 mm

·First order minima

f

y =0

y'

First order minima

i

1

=~

2D

Their optical path difference

AD =d

, ~,I --s-------il 1 ~

Geometric path difference between S2P and Sf is

D

Air

2

,.

y= I'nngew1'dh t

YJ~)d

,.

2µ1=2

f>x=SS,0-SS,O=..Jd'+l'-l

(a)

2D

When liquid is filled between slits and source S , then

= ,:;;

6.

For constructive interference In case of soap film,

= ================================== 2.86

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Wave Optics 1' = (µ-1)t = (1.6-1)(1.964 x 10"') 2 2

n=1,2,3, ....

2µt=(n-i}

For minimum thickness t, n =1 1' => 2µ1=2

10.

1=~=~=112.78 nm 4µ 4x1.33

=>

A= 0.589x10-e m

=>

1'=5890A

Shift = 5 (fringe width)

=> 7. · )a) . P= 1'D = (5ooox10·")(1) d

(b)

1x10-:i

0.5 mm

t-

t=1.2µm

y=(2n-1)w

d

At y= =>

1

3

=> d A

11,

P= (14.75-12.50) mm=0.25 mm 10-1 , ),.' 5500 p = -i:-P = x 0.25 mm~ 0.23 mm 6000 Zero order maxima will remain unchanged. Earl/er it was at 12.5 mm-12.25 mm. Tenth order will now be at 12.25 mm+ 10P' = 14.55 mm.

12.

Given,

mm, y«D

,Ix= yd D

Now resultant intensity, is given by 1=11 +I2 +2Mcos.6.4, ·

(d)

I= 41 0 + 10 + 2 g cos.6.4' =510 +4l 0 cos( 10"' 0.5x10-v

~1t) =31

t.y=(Y,,).,. -(y,),,..,

0

l:i.y= 7AD 2d

2000

13.

=> 0=30° Since, y =Dtane

As d increases, fringe width

(d)

The interference pattern due to different component colours of white light overlap (in-coherently). The central bright fringes of different colours are at the same position. Therefore, the central fringe is white. Since blue colour has the lower A, the fringe closest on either side of the central white fringe is blue; the farthest Is red.

(e)

Since in a medium the wavelength of light is 1'' =

4

9.

Since, shift =2(Fringe Width)

=>

(µ-1)1D 1'(20) d =-d-

(Pa::{)

fringe

width

decreases.

is

given

!:µ , by

Thus, fringe width decreases by µ .

14.

For bright fringes to coincide, we have

n1"-P n2A2D -d-=-d-

ITT

2 $'=( " )(1.25x10"') 400x10_,

$' = 25n

the

of the fringes remains constant.

P=1''D=W. d µd

$= 5n 2 t.X'=(µ-1)1=(1.5-1)(1.5)=0.75 µm

=1.25 µITT= 1.25 X 1Q-6

=~)

8.75x10-3 m=S.75 mm

(c)

therefore

=>

(.6.)c)ne! =_rue+ flx'

7x5x10-~xO.S 2x10

?AD d

(b)

Highest order maxima

.1>c =0.5 µm =5 x 10-1 ·m So, phase difference is given by

(2x11-1)1'D 2d

But the linear separation or fringe width increases in proportion to the distance (D) from the screen. As 1' decreases, fringe width (P oc 1') decreases.

$ = (~)(,Ix)= ( 400~·10_, )<5 X 10-7)

(b)

and

Angular separation (

n- =[f]=2000

(a)

D=0.5m

(a)

Hence, n = 5000 is not possible.

8.

d=0.1mm=10""'m,

1'=5000A=5x10·' m

n = 1000 is not very much less than 2000 Hence now .ruc =dsin0 must be used, so we get dsine = n1' = 10001'

(e)

1' - 6x10·' -12 X 10"' m µ-1 (1.5-1) .

________

For n=5 y=2.25 mm

(c)

(µ-l)t~=1'~

=)

~:

= ~:

=

~~~ =~

i.e., 51t> order maxima of larger wavelength A. 2 will overlap with

6°' order maxima of smaller wavelength A. 1 • So, Yrrin

=

SAD d

6x750x10-9x2 2x10--3

c::::=================================== = 2.87

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Advanced JEE Physics

Optics & Modern Physics

=>

Ym1n=4.5x10-3m

=>

Ymrn =4.5 mm

2 " $'=.!:+ (1.5-1)(8000+10-") 2 6QQQ X 1o-lO

=>

15.

Let n1 bright fringe of A1 overlaps with n2 bright fringe of A. 2

Then,

I= 11 +1 2 +2~ cos~

n1A1D = n2 A2 D =>

d d n1A1 =n2 A2

=>

.!1='-2 = 700 =!... n,

l.,

=>

400

$' =11.

6 The intensity at P is now given by

•

1 I'=l0 +11I0 +2~cos( ;1t)=2l0

{·: I'= 21,}

Solving this equation, we get, ~

4

4

7

17.

(a)

= 0.21

Since, file

~ (o. 3 xl0-,;(10xl0_,) _3x10-'

m

6 3 2

5

A,<~

4

P,
1

2

y=O

Y=O

l.,

.!:!!. = !.... n,

4

=>

$ = 34.52 radian= 1978°

=>

1=989'

l.,

2

Since, Ip = I0 cos

=>

implies that 7th ' bright fringe of A. 1 will overlap

(b)

2

(!)

1p=3x10-"'I0

Fringe width

p

l.D

d

(546x10-,)(1) (0. x _,) 3 10

1.82x10--3 m

with 4" bright fringe of A. 2 • Similarly 140, of A1 will overlap with 811'1

=>

of A2 and so on.

Therefore, number of fringes between point P and the central fringe are given by

So the minimum order of A.1 which overlaps with A. 2 is 7. 16.

$=(~")(file)=(546~"10-• )(3x10-<)

3 1

The ratio

=>

N=~=5.49 1.82

Without inserting the slab, path difference at P is given by

.6.x

=>

yd D

0.15x10-a x2x10-a 2

P=1.82mm

18.

Let n1 bright fringe co"!"esponding to wavelength i..1 = 500 nm coincides with n2 bright fringe corresponding to wavelength

~X=1.5'x10-7 m

i..2 =700 nm. => =>n1A.27 n2 =i;=s This implies that 7"' maxima of i..1 coincides with 5th maxima of

i..2 • Similarly 14~ maxima of i..1 wilf coincides with 10~ maxima of A.2 and so.on. So, minimum distance is

Corresponding phase difference at P is

$=(

2:

Ym1n

)

=> 19.

=>

Yrrin =3.5x10--3 m=3.5 mm

Here i..=6x10--s cm, t=7.2x10-4 cm

Displacement of the central bright fringe, file = sp 0

Since, intensity at P is given by I= 410 cos

=>

= n1~p = 7x5x10-7 x10 3

1=4l COS (¾)=21

2

(%)

2

0

0

Since file= f(µ -1)t l.

=>

f(µ-1)t=6P l.

=>

-l=6l. 6x6x10_. =0, 5 µ t 7.2x10-4 µ=0.5+1=1.5

Phase difference after placing the glass sheet is

$'=$+~(µ-1)t

=>

= ================================== 2.88

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1.

For first dark band, sine = ~

For 10 bright fringes, we get, 0, 0 =

As the diffraction pattern is obtained in the focal plane of lens, therefore

The angular width of the central maximum in the diffraction pattern due to slit of width a is

X

A

f

d

=>

10!:<2?:. d a

=>

a::;~ 5

For small e, tane = sine

x=(~} Since, l.=5900A=59x10-' m, f=50cm=0.50m d=0.025 m=2.5x10-s m X

59 x10-axO.SO 2.5x10-5

a=..!. mm=0.2 mm 5

11.8x10-a m=11.8 mm

4. Given, A, = 5890 x 10-10 m , J..2 = 5896 x 10-10 m ,

.

Smee, DF

Distance of first secondary maximum from the centre of the screen is

=)

2 d

For the two wavelengths, we have

5.

X1=~DA, and X2 =~DJ.. 2

40

m

1

40 2

km=20,000 m

Since size of Fresnel's zone, DF

=>

2x2x10--6 The linear separation between n bright interference pattern on the screen is given by

Here size of Fresnel zone DF at the middle hill must be less that

D=

3x2x(5896-5890)x10-' -SxlO~ m

fringes

in

= .Jw

..fw « 50 AD«2500 2500 2500 = =0.125 m=12.5 cm i.« D 20,000

an

Thus wavelengths longer than 12.5 cm will undergo serious diffraction effects.

nW

Xn=d Since, xn « D , the angular separation between n bright fringes

8 =x"=nA " D d

4x1Q-J

Distan\:e of either of the two hills from the middle hill is

3D t.x =X, -x, = 2d(l., -J.,)

should be

_(4x10"')'

50 m.

2 d

Spacing betweeQ the first two maximum of sodium lines is given by


Df

Thus ray optics is valid upto a distance of 40 m from the aperture.

3 DJ.

X=--

2 d

d' =T

_where d = 4 mm =4x 10-a m, A= 400 nm =4x10-1 m

d=2µm=10--6m, D=2m

3.

gJ..

201 = 2A a Since, 10 maxima of the double slit pattern within the central maximum of the single slit pattern

tane =f

2.

1

6.

Given, A = 6500 A = 6500 x 10-10 m , 0 =30° Forfirstminimum, dsin0=A

d= l. sine

6500 x10-10 sin30°

6500x 10-10

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Advanced JEE Physics => 7.

a

If is the angle between the transmission directions of the two polaroids, then the intensity of light emerging from second polaroid is

d=1.3x10-a m=1.3 µm

Given, D=2m,

X1

=5mm=5x10-3 m,

I2

l.=5000A=5x10-' m Width of central maximum is

Po

=2X 1 =2x5x10-3 m=10-2

8.

2x2x5x10-7 10-2

2D)..

Po

Io 29 = I1COS29 =2COS

Initially the two polaroids are crossed to each other i.e., 8 = 90° . When polaroid is rotated through 60° , the angle between their polarising directions will become 8=90°-60°=30°

m

So, slit width is d=

Optics & Modern Physics

'

2x10-4 m=0.2mm

2

From figure, it is clear that the angular separation between central maximum and first order minimum is Central Maximum

=>

12 =1-cos2(30°)=1-x(J3) =~1 0 2 2 2 8

=>

.!,_ =0.375

,,

So, transmitted percentage is 1-x100 = 0.375x100 = 37.5%

I,

12.

First order Maximum

Obviously, the axis of P3 is inclined at 45° to the axes of P1 and P2 Let amplitude of linearly polarised light emerging from P, be E0

).

The amplitude of light emerging from P3 is E0 cos45° = ]

8=0

-2

Next the amplitude of light emerging tro'm P2 is

0=3)._0=3). 2d 2d

0

1 ~cos45°=~x--= Eo ,/2 ,/2 2

J2

3x589Ox10-10 2x0.25x10-13

Let intensity transmitted by P1 = 10 oc E~

e = 3.534 x 1o-3 rad 9.

So, intensity after passing through P2 and P3 is

Since, I = ~ Using Malus Law, I= 10 cos 2 8

=>

I=~

4

I

; =loCOS 2 8

Thus the intensity becomes one~fourth of the maximum transmitted intensity.

1 C0S8=±,n_

=>

a= ±45°,

13.

±135°

The same effect occurs no matter which sheet is rotated or in which direction it is rotated.

10.

e

According to Mal us Law, I= 10 cos 2 8

(a)

Here e =30°

=>

(b)

l=I0

cos (30°)=1 x( 2

=>

__r__=0.75 I,

Here

e =60°

0

:J

(

=¾l 0

Intensity of emerging)= (Intensity of emerging) beam A beam B

=>

IA cos 2 (30°) = le cos 2 (60°)

1_ cos 2 (60°)

2 l=l0 COS (60°)=1 0 ( 1)'

z'

le

=-t I

__r__=0.25. 1, 11.

The planes of polarisation of light beams A and B are mutually at right angles. Initially, the beam B shows zero intensity. Therefore, 8 = 90° for beam B and 8 = 0° for beam A . When the polaroid is rotated through 30° , we have 0 = 60° for beam B and = 30° for beam A In this position, we have

14.

Let 10 be intensity of incident unpolarised light. Then the

cos 2 (30°)

G)'

(~J

=.!=,,3 3

sini P = µ From Snell's Law, ___ s1nrP sini From Brewster Law, taniP =--~ = µ

intensity of light emerging from the first polaroid will be

COSIP

I _1_ ' - 2

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Wave Optics ::::>

sinrP = cosiP

=>

sinr,=sin(90°-~)

::::>

rP =90°-iP

::::>

iP+rp=90°

The intensity of light emerging from P3 is given by 2

13 = 12 cos (90° -0)

. '( 2e)-BI,_8x3_3 ------10 32 4

Sin

Hence the reflected and transmitted rays are perpendicular to each other.

15.

Let P1 , P2

,

=¾1 0 cos2 0sin2 0 = ¾1 0 sin2 (20)

P3 be the three polarisers and 0 be the angle

between the transmission axes of P, and P2 • As P1 and P3 are

sin(20) = ../3 2 20 = 60° 0 =30°

=> => =>

crossed, the angle between P2 and P3 is (90° -0) .

So, 13

Let 10 be the intensity of the unpolarised light falling on P1 •

13

Then the intensity of light emerging from P1 will be

= ~sin2 (20)

will be maximum when

sin 2 (20) = 1 (maximum), i.e.,

sin(28) = 1 = sin(90°) ::::> 0=45°

I, l1=2 By Malus Law, the intensity of light emerging from P2 is given by

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1.

[BJ

lmax=I=(a+a) :::::>

2

1=4a2 =4l 0

When either of the two slits is covered then

I'=(a+o)' =•' =¼ 2.

~

Let nlh m1mma 560 nm, then

Ot

~

400 nm coincides with mth

minima of

5 0 4 0 (2n- 1i( ~ ) =(2m-1)( : )

10.

of

Location of this minima is,

Required distance

3.

=Y2 -

42mm

Y1 =28 mm

11.

,Ix= ni. . (µ-1)t=n1.

=>

t=~

[BJ The relation among angle of diffraction 8 , order n and number of lines per cm of the grating N is sine= NnA.. The maximum

4.

maximum

value for

n = N~ .

µ1)..1

=>

(1)1., =ni.,

=>

~=n

i.,

[DJ

=n2A2

=)

62 x5893 = n2 x4358

=)

n2 =84

[DJ

p = ).D

µ-1

d

t = -2:._ =_i._ = 2).

=>

P= (5ooox10-")(2) (10"')

=)

P=10--3 m=1 mm

1.5-1

[DJ As AR>AG>'-e :::::>

Hence

=µ2)..2

=>

For minimum value of t, n =1 µ-1

sine= 1.

n=1,2,3•..._.. 12.

=>

of

[AJ Since µ).. =·constant

n,A1

[AJ Path difference due to slab should be integral multiple of A or

=>

v& =3x106 ms~1 away from earth

order of spectrum is an integer. Thus we cannot see the fourth order, but can see the third order.

560 nm y _(2x11-1)(1000)(400x10_.) 2 2x0.1

=>

1.=6000x10_,, cm, N=5000 lines/cm, we get n=3.33. The

--=-=-

Next 11 th minima of 400 nm will coincide with B" minima

47x3x10 9 4700

vs =-i.-=

value

2n-1 7 14 2m-1 5 10 i.e., 41h minima of 400 nm coincides with 3rd minima of 560 nm. Location of this minima is, y (2x4-1)(1000)(400x10_.) 14 mm 1 2x0.4 =>

lU·C

=>

PR >J3o >Pa as j3ocA

13.

[AJ Let the required thickness be t A. So, number of wavelengths,in vacuum is _t_.

6000

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Wave Optics Number of wavelengths in air is 1.0031 6000 According to the problem

=:,

Einstein's photoelectric equation appears to be interlinking of the two aspects.

26.

_t_= 1.0031 +1

6000 =:,

14.

lm; 0

17.

6000

t=2mm

[CJ

I-,

15.

N

=(A +F,)' =91 =(F,-F,)' =[

[DJ Theoretically infinite colours wavelengths.

are

possible,

hence

....

Sun

infinite

[DJ

N' 28.

[DJ In the presence of thin glass plate, the fringe pattern shifts, but no change in fringe width.

29.

[CJ If an unpolarised light is converted into plane polarised light by passing through a polaroid, it's intensity becomes half.

30.

[BJ

[DJ The blue filter will allow only blue light to pass through. So light

from the object passes through filter. Similarly, the whi_te background will also look blue through the filter. Thus we have a

blue object under a blue background, which makes it indistinguishable.

19.

,

Image

1./

Because sound waves are longitudinal in nature. 18.

[AJ This is the effect of refraction. Rays from the sun are passing from vacuum to air when they enter·into the earth's atmosphere. They bend towards the normal making the sun to appear at a higher altitude as shown in figure.

[BJ Intensity of illumination at a point is decided by three factors; (i) power of the source P (ii) distance from the source r and (iii) angle of incidence of rays cos

e

=:,

0

I= Pcos 2

r At noon the sun's rays are normally incident making

e= 0 .

Hence I is maXiJ'l'.IUm at noon.

A

20.

[BJ Distance of third maxima from central maxima is 3l.D 3 x 5000 x 10-" x (200 x 10-') 1.5 cm X d 0.2x10-s

22.

[CJ For observation of colours, the thickness of the film should be of the order of the wavelength of visible light.

23.

[DJ The red filter will pass only red light. Hence a blue cross will appear as black. The white background will appear as red when seen through a red filter. Thus the observer will see a black cross under a red background.

24.

B PR=d PO~dsecB and CO= POcot20 = dsec0cos20 path difference between the two rays is, 6x = CO+ PO= (dsec0+ dsec0cos20) phase difference between the two rays is Li~= 1t (one is reflected, while another is direct) Therefore, condition for constructive interference should be

i'.x-!: 3l. - 2' 2

[BJ Newton's concept of light is that it is made of corpuscles or particles. All particles are deflected by earth's gravitational field. Hence they are also deflected by gravitational field. In the general" theory of relativity. Einstein predicts deflection of light by the gravitational field. The angle of deflection predicted by both are however not the same. Einstein's prediction agrees with experimental results.

31.

25.

[DJ In photoelectric effect, the photon is treated as a particle having a quantum of energy hv. It emits one electron. Here the particle and wave aspect of light appear in two sides of equation.

l.

=>

dsec0(1+ cos20) =

=>

B) =!: ( _d_)(2cos' case 2

=>

case=_!:_ 4d

2

[AJ

l.D

~ = - and

d

P' ~ l.{D/2) =.!!_ 2d

4

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Optics & Modern Physics

Advanced JEE Physics 32.

36.

[DJ

~='°

[CJ

Intensity at the centre will be zero if path difference

d d is halved and D is doubled

So, Fringe width 33.

=~

(µ-1)t=!:

p will become four times.

2

1=-'-

[AJ Path difference at s 2 is 2/,,. . Therefore for minimum intensity at

p

2(µ-1)

37.

[AJ No light,.is emitted from the second polaroid, so P, and P2 are

perpendicular to each other

P, 0

90°-0 P,e-------':.L='-=-"----P,

2l.

Let the initial intensity of light is 10

Let x be the minimum distance from s2 • Then

31. 2

l. 2

~-

X

transmission from first polaroid

11 = 1-cos2 0

2

2 Intensity of light transmitted from last polaroid i.e., from

Solving this equation we get 7l. X=12

P2 = 11 cos2 (90° -0) =~cos2 0sin 2 0 l.

2

in equation (1) we get

2

P2 =1..(2sin0cos0) =1..sin 2 20

8

x =15A which is greater than TA . 4 12 34.

38.

[CJ 0

X=

= (µ - 1) ID

~= 2n 3

d (1.5-1)tD d

and !l.x_

2

(µ- 1)t0 d

..• (1)

.6.xx(2,.,7')= 231t =,i

••. (2)

sin0=

Dividing equation (1) by (2)

3d

39.

[BJ

I= 1_ cos'(½) ¾=cos'(½)

[AJ Phase difference corresponding to

y,

=- " 2

=>

COS$=0

and that for

=+2"

=> =>

Average intensity between y1 and y2

I,. =.!n_~1' I= cos' (±)d~ =I= (n2n+ 2) 2 2

So, the required ratio is ,½(1+;)

41.

[AJ Path difference at P is .6.X

=

~

sine=..?:.

2 0.5 3 µ-1 2µ-2=1.5 2µ =3.5 µ = 1.75

Y2

8

I= 41 cos'(½)

[AJ Shift

35.

2

Intensity of light emitted from P3

= 3l.

NOTE: If we substitute s,P-s2P=

So Intensity of light after

=1-.

... (1)

s 1P-s 2P=-'F-

•

2.94

= 2(icos8) = xcos0

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Wave Optics For intensity to be maximum . ax= nl. => XCOS9=nA

=>

{n=0,1,2,3, ...) p

cos a= nA X

Since, case 'fo 1

=>

nl.t,1

.,/

X

=>

,/

,/

Using (1) and (2),

, , ''

a·

n;t,~

50.

-x-

l=lo[si:a.r ,where a.=! 1 2 For nth secondary maxima dsin8 = ( n + )A 2

nt,5 n=1,2,3,4,5, .....

=>

[AJ Laser is intense, coherent (all photons in phase) monochromatic (only single wavelength).

43.

and

2

44.

47.

51.

]'

I,

[DJ If I is the final intensity and 10 is the initial intensity then 1=1-(cos2 30°)5 2

=> 52.

_I_=.!.x(./3)" =0.12 10

2

2

[BJ For maxima 11 = d sin 8 = nA ::::) 2Asin8=n1 ::::)

sine=%

since value of sine cannot be greater 1. n=0,1,2 Therefore only five maximas can be obtained on both side of the screen.

V C

53.

::::) V=1.5x10 5 (Since wavelength is decreasing, so star coming closer)

48.

2

[ (2n/1)n

[DJ As the star accelerates towards the earth, its velocity will increase. By Doppler effect in light, a source approaching an observer shows violet shift. So the colour of the star will shift toward~ the shorter wavelength region, that is blue.

[BJ 6). l.

=.P.2 =.'.:.[dsinS] =(2n+l)• l. 2 Sin -- • . (2n+1)

.where A is wavelength. In the given

set of radiations. X-rays have the minimum wavelength. Hence focal length is minimum for X-rays. ·

a

I =I 0

[DJ The focal length increases with increases of wavelength. ioc (n-1) nearly cc :

2Io

[CJ

Therefore, in all four quadrants there can be 20 maximas. There are more maximas at 8 =0° and e = 180° . But n =s· corresponds to e=90" and 8 =270° which are coming only twice .while we have multiplied it follr times. Therefore, total number of maximas are still 20 i.e., n =1 to 4 in four quadrants (total 16) plus four more at a= 0°, 90°, 180° and 270°.

42.

.!t = 410 = 2 l2

''' ' :

s,o--~-.. s,

l. Substituting x = SA , we get

=>

••. (2)

[BJ In this case, we can assume as if both the source and the observer are moving towards each other with speed v. Hence , c-u c-(-v) c+v v = - -0 v=---v=--v c-us c-v c-v

, (c+v)(c-v) c'-v' v ~-,-~~~v=~-~~-v 2 2 2

[BJ Momentum of the electron will increase. So the wavelength

(c-v)

(A=~) of electrons will decrease· and frinQe width decreases

Since v

0

+v -2vc

c' C « c , therefore v' = - , - - =- - v c -2vc

c-2v

as ~a:l..

49.

54.

[CJ Fringe width p oc A. • Therefore, A and hence p decreases 1.5 times when immersed in liquid. The distance between central 111 maxima and 10 maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm . Position of central maxima will not 111 change while 10 maxima will be obtained at y = 4 cm .

55.

[BJ For maximum intensity on the screen dsin8 =nA

[AJ

1=41,cos'(!) At central position 11 = 410

.:.(1)

Since the phase difference between two successive fringes is 2x , the phase difference between two points separated by a distance equal to one quarter of the distance between the two, successive fringes is equal to 6 = (2n>(¾) =% radian

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Advanced JEE Physics

56.

sine= n1. = n(2000) =_I!_ d 7000 3.5 Since maximum value of sin 8 is 1 So n = O , 1, 2, 3 only. Thus only seven maximas can be obtained on both sides of the screen.

the central maximum in the diffraction pattern you add N phasers, all in the same direction and each with the same

[DJ If d sin 8 = (µ -1)t , central fringe is obtained at 0

proportional to (2N>2" and is, therefore, four times the Intensity for the narrow slit.

If d sin 8 > (µ -1)t , central fringe is obtained above O and

amplitude. The intensity is therefore N2 • If you double the slit width, you need 2N phasers, if they are each to have the amplitude of the each to have the amplitude of the phasers you used for the narrow slit. The intensity at the central maximum is

64.

If d sin 8 < (µ -1) t , central fringe is obtained below 0 57.

sin C =

[BJ

Resultant intensity I

¾. Hence

i=tan-1 (n)=tan-1

= 11 + 12 + 2Jrj; cos qi

At central position with coherent source and 11 = 12

[BJ

= 10

-

... (1) Icon = 410 In case of incoherent at a given point, lj, varies randomly with

65.

58.

= l 1 + 12 =210

Hence 1coh

= _g_

llncoh

1

%. If i is the polarising angle

tan i = n .

(%).

[CJ Reflected ray

Incident ray

time so (cos$)~=0 I,nco11

n=

... (2)

[AJ

According to given condition (µ-1)t = nA. for minimum t, n =1 So, (µ~1)tm;, =1.

Refracted

1. 1. t, =--=--=21. ma µ-1 1.5-1

ray For glass of refractive index 1.5, polarising angle is 57°, tan i = 1.5 . At this stage the angle between the refracted ray and the reflected ray is 90° . The required angle from figure is

8=57-33=24°.

66. 60.

[AJ When unpolarised light is made incident at polarising angle, the reflected light is plane polarised in a direction perpendicular to

67.

Therefore E in reflected light will vibrate in vertical plane with respect to plane of incidence.

=>

6.5 x 10-1

[DJ

=> =>

a =13x10-1 m a=1.3µm

If I is the intensity of the incident unpolarised light, the intensity transmitted by the firs~ is

½. This is the intensity of incident light

68.

2

[BJ

subtended

by

two

objects

dB

just

is the smallest

polaroid is ( ½)cos 8, where 8 is the angle between the axes.

angle

Here sine=¾, case is therefore¾.

d8 = 1' = ~0 .,Converting this into radian and taking 1 radian as 6

distinguishable,

nearly 60° (Actually 57° 18') we get the resolving power as 3600.

2 2

..!.cas e=..!.x(i) =~I

2 2 5 25 8 is the required ratio. 25

=

= a sin 30

The resolving power of the human eye is

on the second polaroid. Intensity transmitted by the second

63.

[CJ For first minimum A.=asine

the plane of incidence.

61.

[DJ The property of double refraction is shown by quartz, calcite and ice also.

70.

[CJ

[DJ If you divide the original slit into N strips and represents the light from each strip, when it reaches the screen, by a phaser, then at

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Wave Optics If x is the distance between poles and D is the distance of the eye from the poles as shown in figure we have d0 = tand9 = ~. D

Here D = 3.6 km= 3600 m

and

d0 =

~ 36 0

=> 75.

radian. Hence

72.

[AJ M

V

).

C

=>

1=~

=>

V=C

C

[CJ ).' = 1.(1-f) = 5890(1

45 ·

xio')

3x10 8

[CJ Shift, ilx=[(µ-1)t]~

X=1m.

71.

= 5802 A

=>

tD 20~=(µ-1)d

=>

1 20().~)=(µ-1) ~

=>

201.=(µ-1)1

=>

µ= 20). +1 t

=>

µ

=> =>

73.

20 X 5QQQ X 1 2.5x10-5

o-

1 0

+1

4 µ=-+1 10 µ=1.4

[CJ

i., _ 4500 _ 3 i., - 6000 -4 ::::)

4A.1

76.

[BJ For dark fringe at P

= 3A.2

i.e. the fourth bright fringe of A.1 coincides with the third bright fringe of A. 2 •

74.

llx:=1.5 mm

[CJ From here, we must take a note that central maxima ( n = O) is

Here n = 3 and A =6000

So, 11= 77.

flanked by first minima (n = 1) and then first maxima (n = 1).

[AJ

liquid, A and hence w is reduced µ (refractive index) times.

1-D x,., =(2n+1) d 2

10oo'=(5.5)oo

So, position of second dark fringe is

=>

101.·(%) =(5.5l~

=>

'i:= 5.5 =µ

=>

µ=1.8

x, =[2(1)+1]~~

=>

""------< ____ ___,

._

~v,·

Second Maxima {n = 2) Second Minima (n First Minima (n

=~~---, ._ _,

_____

x0 =n(i.~) =>

79.

[CJ Fringe width ~ oc A. Therefore, A and hence w will decrease 1.5 times when immersed in liquid. The distance between central th maxima and 10 maxima is 3 cm in vacuum. When immersed in liquid it will reduce to 2 cm. Position of central maxima will not th change while 10 maxima will be obtained at y = 4 C1"!1 .

First Maxima (n = 1)

Second Minima (n = 1) Second Maxima (n = 2)

Similarly, pbsition of nth bright fringe is

10

[BJ The intensity of light received is inversely proportional to the square of distance. Hence the amplltude of the light wave received is inversely proportional to the distance.

= 1)

=0)

1.

78.

First Maxima (n = 1)

l"'""''-------1 Central Maxima (n = O) '===~----1 a@ftFirst Minima (n = 0) i=="--------1

"-------<

6000 . =5x--=15000A=1.5m1cron 2

Fringe width ro = A~ -~ When the apparatus is immersed in a

Position of the ( n + 1) th dark fringe is

=~----i

5).

2

80.

[AJ I=4I,cos'(!) 10 =4l0 cos

2

(¾)

4W

X4=d

::::)

cos(!)=¾

So, separation is

ilx=x,-x,=%(1.~)

::::) !=2: 2

3

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Advanced JEE Physics

Optics & Modern Physics 16A.2 -4l2

=>

~=23•=(2;)AX

x, -

=>

.!=(-!}i 3 I- D

X3-

=>

6x10-7

A

;----er =3 x-

y=

X

X4

1 Q-4

41-

3).

16A.2 -9l2 6).

7 ,._ 6

=0

So, Number of points for maxima becomes 3.

D

=> 81.

87.

y=2x10-a m=2 mm

pattern appears for asin8 = nA, where n is an integer. For the

[BJ For third maximum

first minimum n=1.Then sin8=~=10-.'l rad; 1rad=57°. a Hence 8 = 0.057° = 0.057 x 60' = 3.42' .

X=n(A~) X=( 3 )5000x10·"x2 0.2x10-3

88.

1.221- = R( say ) a

[BJ n,A.1 =nzA 2

=> 83.

7

R

1-, 600 n2 =n1 x-=12x-=18 1-, 400

1.22x5x10~ rad 2x10-3

R 1.22x5x10"' x --3 2 10

x

180 d

7

[DJ

R = 0.0175 degree

Blue of the sky is explained by scattering. Scattered light obeys

R=0.0175x60 minute R=1 minute

Rayleigh's law.

I oc A~ . Hence shorter wavelengths are

scattered with more intensity. In visible light, violet and blue have the minimum wavelength. Of the two, blue is mare intense.

84.

[CJ The interterence fringes for two slits are hyperbolic.

85.

[AJ Total phase difference = Initial phase difference+ Phase difference due to path · 360° 360' I- ::::::, ~= 66°+-A-xruc =66°+-A-x = 66°+90=156°

89.

[CJ

=>

[BJ

x+1=

=>

x-1 X=2

=>

.!i. = 4

=>

11

{·: X=~I,/1,}

3

I,

=

412

i.e., if 12 = 10 then 11 =

From .6.S1S 2O,

10 = 410 cos

(S,D)' =(S,S,)' +(S,D)' (S,P+PD)' =(S,S,)' +(S,D)' y

90. X

92. Here S 1P is the path difference

=nA.

egree

I~=( ~I,/1, +1J 9 I"" ~I,/1, -1 =1

4

86.

[BJ Angular limit of resolution of human eye is

X=1.5cm

82.

[BJ If a is the width of the slit, the minimum of Fraunhofer diffraction

2

=>

~= 2• 3

=>

(2;)(~)=

=>

Y=-

41 0

(½) ~

AD 3d

[AJ (n+1)5200 =n(7BOO) => n =2 [BJ Air has a refra"ctive index slightly greater than one. So, when the chamber is evacuated, the refractive Index decreases and hence wavelength increases as a result of which fringe width increases

for maximum intensity.

(nl-+x,)' =(41-)' +(x,)'

slightly. 93. 16A.2-A2 2).

=

2.98

7.51.

[DJ Ultraviolet radiations, when fall on a fluorescent material, are converted into visible radiations. Hence we should use ultraviolet radiation.

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Wave Optics 94.

95.

[CJ Using the same argument as in the SPLUTION 104. I, 1331 -cos 30=-x-=-=0.3751 2 2 4 8

i ~ s,

1.c,--+1 d

[BJ ARed

> /.Violet

and all other colours have wavelength value lying in this region.

Further, we have X cc A , so X(Blue) < X(Green) 96.

-------- , / / ,,

[DJ

Hence corresponding phase difference

I 91 From figure 11 = - and 12 = 4 64

Resultant intensity at P

=>

_!g_=~

I= I~ cos

16

1,

2

Screen

$ = 21t x!: = ~ ,

4

2

½= $ cos ¾J =~ 2

0

(

101. [A)

If i is the polarising angle, n = tan67°

n = tani

(Brewester's law).

is greater than 2. We also know

1

- - = n. sine

sine=.!, where e is the critical angle, sine is less than 0.5.

n

Hence C is less than 30° . The only possible answer is 22° .

I By using -""- =

lmin

l

1

VI, Vtti+ rx;_+,l = [ J¼ /9 Jf' ~-1 [ ~- 1 16

11

97.

[DJ Since, according to Malus Law I= l 0 cos 2 ~

98.

[DJ

102. [DJ Only transverse waves can be polarised, while the other properties are common for all wave motion.

49 =1

103. [DJ In the arrangement shown, the unpolarised light is incident at polarising angle of 90° -33° = 57° . The reflected light is thus plane polarised light. When plane polarised light is passed through Nicol prism (a polariser or analyser), the intensity gradually reduces to zero and finally increases.

16

Since P is ahead of Q by 90° and path difference between P

and Q is

¼. Therefore at

A , phase difference is zero, so

intensity is 41 . At C it is zero and at B , the phase difference is

90° , so intensity is 21 .

99.

104. [DJ If shift is equivalent to n fringes then (µ-1)t

,

n=--

=>

noct

=>

_!g__= n2 11 n1

[AJ P is the position of 111h bright fringe from Q. From central position O, P will be the position of 10th bright fringe. Path difference between the waves reaching at P = S 18 = 10A =10x 6000x10-10 = 6x10--o m

100. [AJ Suppose P is a point infront of one slit at which intensity is to be d calculated from figure it is clear that x = . Path difference

t2 =

20 30

x4.8=3.2mm

105. [CJ Path difference between the waves reaching at P, A = .6.1 + .6.2

where .6. 1 = Initial path difference

2

between the waves reaching at P

,\=xdJ~)ct =_i_=S,=_1: D10d20204 Screen

-0,--,--02.99

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Optics & Modem Physics

Advanced JEE Physics ~=Path difference between the waves after emerging from'slits.

6, =SS, -SS, =~D2 +d' -D

113. [CJ

and 6 2 =S,0-S,O=~D'+d' -D

Interference is a physical effect.of superposition of wave motion. This can happen for any kind of wave motion, i.e., light waves,

2

=>

6=2((o'+d )Lol=2{(0'+:~)-o}

d'

sound waves, matter waves, etc.

{From Binomial expansion}

6=D

For obtaining dark at 0, 6 must be equals to (2n -1)~ i.e.,

114. [BJ The extra path difference produced by the glass plate of thickness t and refractive index µ is ( µ -1) t . Due to this if n is

the

d' ). -=(2n-1)D 2

=>

number

of

fringes

which

t=O.O1x1O...:a m, A.=6OOOx10-

10

m,

shift,

(µ-1)t=n1..

µ=1.5. This gives a

value for n = 8 .

d=/2n-21)1.D

For minimum distance n =1 so d =

117. [DJ

~

I= a~ + a! + 2a 1a 2 cos4> Put a~ +a! =A and a 1a2 =B ~

106. [DJ

l=A+Bcos4>

For destructive interterence, path difference must be an odd multiple of

118. [AJ

~.

l!J..=A.'!.. and v=rro C

107. [CJ

These concentric bright and dark fringes are called Newton's

V=7x1O 6 x

Rings.

=> 108. [BJ I= 10 + 10 + 2l 0 cos(21t)

=>

2 rc , C=3x1O8 ms-1 25x24x3600

61. =o.o4 A

119. [CJ Shift =.!!.(µ-1)t

1=410

).

A. 21t 1t For X=-. ~=-X=-

4 ). I'=l0 +10 +0

2

=>

7p =.!!.(µ-1)t ).

=.!.2

1'=210

120. [BJ

109. [AJ

. P=w =>

d J3ocD

131

D,

= D2 P, -P2 = D, -D2

132

132

6P

D2

P,

ao=o2

1., =d2

3x10-5 =A.2 = Sx _2 x10--3 =6x10-7 m=6000A 10

6x=dcose

[(n-1)t]~= 1.(2D) d d 1.= (n-,lt =5892A 2

=> 112. [AJ

).L

X=-

... (1)

For maxima at P

6x=~ From equation (1) and (2) nA. = dcos0

~

110. [CJ

=

Here path difference at a point P on the circle is given by

·--~

1

Scos- ( n;) = cos-1( ~)

121. [AJ The film appears bright when the path difference

(2µtcosr) is equal to odd multiple of!: 2 (2n-1)). i.e., 2µtcosr ~-~ where n=1, 2, 3 ..... 2

d

2.100

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Wave Optics => => =>

A

4µtcosr (2n-1)

4x1.4x10,000x10"'10 xcoso

A,

56000 (2n-1)

(2n-1) .

i. = 56000

I=~

=>

A,

18666

A,

0000

A,

5 X 48QQ X 1o-lO

(µ, -µ,)

(1.7-1.4)

8x10.. m=8 µm

A

6222

.

126. [BJ

A,

5091

A,

4308 A, 3733 A. The wavelength which are not within specified range are to be refracted.

At the polarising angle, the reflected ray is fully polarised while the transmitted ray is partially polarised. In fact a method to produce plane polarised light is by reflection at the polarising angle.

128. [BJ 122. (AJ M

).

a=B

V

6500 x10-10

C

=>

(401.8-393.3) 393.3

=>

v

=6.48 x 106

V

ms-1 =6480 kms-1

560 nm then (2n -1)400 = (2m -1)560

a=1.24x10~ m a=1.24µm

130. [AJ As here polariser is rotating i.e., all the values of 8 are possible. 12:rr 1 2. Iav =-Jid8=-= J10 cos 2 8d8

i.e., 4th minima of 400. nm coincides with 3rd minima of 560nm The location of this minima is 7(1000)(400x10.. ) - 14 mm 2x0.1 Next, 11th minima of 400 nm will coincide with 8th minima of 560nm Location of this minima is 21(1000)(400 X 42 mm 2x0.1 Required distance =28 mm

10. )

t

:::, =>

(i)

Using Mal us Law, I= 10 cos 2 8

=>

Shift rue=

a

3x1Q8

123. [DJ Let nth minima of 400 nm coincides with mth minima of

124. [AJ

=>

21to

21t

On integration we get Iav

where I ~

I

0

av

Energy Area x Time

o

=~ p A

10"'

3x10-4

10watt

-3rfi2

=.!x.!Q=E, Watt

2

3

3

. 2n: 2x3.14 1 andTimepenod T=-=---=- sec ro 31.4 5 So, energy of light passing through the polariser per revolution

=I xAreaxT=~x3x10"""x..:!.=10--4J ~ 3 5 131. [CJ

(µ-1)1

p

w B'e

'

Path difference = 2dsin8 For constructive interference 2dsinB = ni.

11+-- o, ___.,. Screen Shift due to one plate rue, =

t

=>

(µ;-1)

'(n:>.)

. B =Sin2d

133. [DJ

Shift due to another path rue, =f(µ, -1)t Nets~ift rue=rue,-rue,=.!1.(µ,-µ,)~ ).

... (1) .

During sunset and sunrise we see the unscattered part of the sunlight. Sun's rays have to travel long distance and hence are incident obliquely. All shorter wavelengths are scattered by

Also it is given that rue = 5P

... (2)

Rayleigh's law for scattered intensity I a: h.~ . Only the red-

.

Hence 5P =

t(µ,

-µ,)t

orange rays reach us. During noon the sun's rays are normally incident on the earth. Hence a portion of both the shorter

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Advanced JEE Physics

Optics & Modern Physics

wavelengths and longer wavelengths reach us. The sun looks

white.

The action of a nicol prism is based on double refraction and dichroism. When a ray of light enters into a calcite crystal, it is split into two rays O and E. Both are plane polarised in mutually perpendicular planes. One of them suffers total reflection and absorption whil_e the other comes out as plane polarised. Selective absorption is called dichroism.

134. [DJ

P='° d So, angular width, .20 is

140. [BJ

29=2p=21 D

9

d

Shift, ,1x =[(µ-1)t]%

6328x10-10 - 0.2x10""

Here ,1x = sp

0 = 3164x10-a radian

=>

e = 3164x10-e x

=> =>

9 = 0.18 degree

180

•

degree

29 = 0.36 degree

For maxima 2,n = ~"(X0)-27tl

=>

2 ;cxo)=2a(n+I)

=>

(XO)=1"(n+l)

s(1"~)=[(µ-1)t]%

=>

51"=(µ-1)t

=>

1"=(µ-1)½ 1"= (0.5)(6x10"') 5

~

A.=6x10-7 m

=>

1"=6000 A

141. [BJ Only transverse waves can be polarised. (A) and (C) and (D) are electromagnetic radiations, which are transverse waves while

136. [CJ

(P,) (P,)

(B), that is, p rays are electrons.

From the given data, note that the fringe width ·

for

i, = 900 nm

for

=: 750

is

greater

than

fringe

width

nm . This means that .at though the central maxima of

= 900

nm will be

=750 nm

and so on.

the two coincide, but first maximum for l further away from the first maxima for A.2

1

A stage may come when this mismatch equals (32

,

=750

146. [BJ

P='° d =>

then again

maxima of A1 = ~00 nm , will coincide with a maxima of A. 2

=>

=>

135. [AJ

A. 2

139. [DJ

=>

s

M1.2)

1000 =( 0.1) 1000 1" = 5000

A

111

nm , let this correspond to n order fringe for A.1 • Then it

will correspond to (n + 1t order fringe far A. 2 •

150. [CJ Using the equation p = ~). , p ~A. Of the given colours, yellow

Therefore n~D - (n + ~J.. 2D

has the maximum wavelength and hence the maximum fringe

width.

=>

n x 900 x 10-<> = (n + 1)750 x 10·'

::::)

n=S

152. [BJ The tubelight has coating of fluorescent material which converts ultraviolet light into visible light. Hence the light is more intense.

Minimum distance from

C

t I

.

nA.p

en ra maxima = -d- =

9

5x900x10- x2 x a-a 2 1

153. [BJ

= 45 x 10-4m = 4.5mm

We take one wavelength in the visible region and calculate the energy just to know its order. Let us take 6000 A (red) for

137. [CJ One of the essential condition to observe diffraction is that the dimension of the object should be of the order of wavelength of the wave. The velocity of sound in air as nearly 350 ms-1 • Hence the wavelength of sound waves produced by a tuning fork of frequency 384 Hz is A=

f

= 1 m near1y. Hence the diameter

of the sphere should be of the same order.

138. [DJ

convenience of a rough calculation. Energy E =

~.

Taking

h=6.6x10---3-4, C=3x108 , A=6000x10-10 m we get :c in joules. Dividing this by 1.6x10-19 to convert to eV, we get

E = 2 eV. For energy 1 eV, wavelength will be 12000 A. Energy 5 eV means wavelength

f

of 6000, that is 2400

A.

Both these ar8 invisible, recalling that the visible range is

Distance between the first dark .fringes is

4000 to 1000 A .

21D

Ax=d=2.4mm

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Wave Optics 154. [A].

167. [BJ

The energy levels of atoms are well defined. Hence transition

between them should give a well defined frequency, which means a single frequency, and hence a single line.

Separation n111 bright fringe and central maxima is xn

3 X 6000 X 1o-lO X 1 3.5mm. 0.5x10-3

155. [CJ The torch produces light waves which are non-coherent in nature.

168. [CJ lmax 1nm,

156. [DJ For unpolarised light

I'= .!P..cos 2

I ' =2COS I

2

~

2=9 =(81+82) a a 1-

2

169. [BJ For second dark fringe, n = 1

'(n) 4

Since, x = (2n+1)~~

I'=.!.

=>

=n~D

3 = 3W

=>

4

... (1)

2d

For fourth bright fringe, n =4. Since

158. [CJ The path difference of two waves producing destructive interference or darkness is (2n +

1)( ½)

1, 2, 3, etc. A path difference of A corresponds to a phase

1t.

::,

½corresponds to

=>

Hence the corresponding phase

=>

difference of 2v. rad and a path difference of

a phase difference of

X=n(l.~)

where n is an integer 0,

difference is (2n + 1)x .

.•. (2)

X=4(l.~) X

4

3=(3/2) X=B mm

170. [A]

159. [BJ

'.

For same field of view

n1P1 = n2P2

P'=l

=>

160. [A] 11

a~

9

½=a~=4

n,(7)=n,(l.~D)

=:>

n1A1 = n2A2

=>

60(4000) =n,(6000)

=:>

n2 = 40

=> 171. [BJ I= S.ince = (a,+a,)' - - = (3+2)' --

, Im;n

a1 -az

P'=£=1mm

3-2

µ

=>

172. [CJ

5-=~ a,

161. (A]

Using the equation ~ = ~l. ; p « l.

=:>

164. [BJ

166. [A]

Shift = i(µ-1) t ( p ")(1.5)x2x10 .. =2P 5000x10l. i.e., 2 fringes upwards.

=(a,+a2)' 81 - 82

=>

In a biprism, the distance between sources is due to deviation produced by two small angled prisms. The deviation produced by each prism is (µ-1)o:. The total deviation is 2(µ-1)o:. Separation of sources 2(µ-1)aa.

lmax lmin

173. [DJ I=

=(Jf; +,/I;)'

I..,

=(Jf;-,/1;)'

For contrast to be good 11 = 12 and hence

Imax

=411

Im1n =0

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1.

[A, C]

For, n=0, sin8=¾

The intensity of light is !(9) = 10 cos'(%)

n=±1,·sin8=-!,+! where ii=~"(,i,c)=(~)cdsin9) (i)

n=±2_,

For 9 = 30' C

3x1Q 8

(ii)

ii

2

(~)= ~

{option (A)}

8 8

.... th en

01s a·m1mma ··

The path difference on screen is > 0 For d = 3A , we have 0;5 l.

i=.': and 1(0)=0

=>

s,

[A, B, Cl Path difference at O is d if d = A, 2~, 3A, .... ., then O is a maxima. 3" 5 '" 1,fd ="2' 2' 2'

For 9 = 90' ii= ( :• )<150)(1) =. 00

(iii)

8 5

8

6.

I(8)=1 0 cos

__:_--7lp

n=4, sin8=-~

•

=>

s1n8=-a· 8

0=(~)(150)(1)=-" 300 2 2

2=4

8 •3

ri=±3, sin8=-~.?...

l.=;,=10'=300m and d=150m,weget

=>

.

81

2 2

1.5 l.

For 9=0', ii=O or i=o 2 !(9) =10

2.5 l. 0 - - - - + - - - - - - - t , i ) ( = 3l.

{option (C)}

2.5 l. 3.

[A, DJ The phase difference corresponding to 0S1 -0S2

=

Hence, the net phase difference is given by

=>

[B,

DJ

I= (,µ, +,fi;)'

g·

I,,., =(,µ,-,[I;)'

if,=2:-41tsin8 2

2:-4nsin8=n1t 2

=>

,ff; +F, =3(,Jf; -F,) 2,µ, =4,fi;

=>

1 --n 2sln0=-

11 =41 2

Since,

For maxima, we have

=>

0.5 l.

8.

2

~=-"- "(2l.sin9) 2 l.

=>

1.51.

¼iS i

4' = n1t , where n = o, ± 1, ± 2 ...

=>

4

I =(A)'

....1.

---1.

12

A2

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Wave Optics ,/o'+4d' -O=~ 2

10.

[A, C, DJ I1 =1 2 =I

11.

=(,if,

<41

45° B +----,-~-•A

Intensity is due to C only

[A,CJ

= ..Jo

Path difference

2

2

+d

=lo

D = 1 cm

-

Also, UD 2 +d 2 -D)=(2n-1)½

=> => 12.

A=D C

+F,)' =(.JI +J½J I"'=(./J-J½J >0

1_

4d 2

. gives,

14.

[A, CJ x = Path Difference = ..Jb2 + d2

-

d

l.=~

2n-1

A=2cm,

2

3

2

cm,

5

=>

X=d(n-1)

=>

X=d(1+£-1) 2d'

cm ........ .

[A, B, CJ

{ b'

00

'' y ''

Since d2 « 1, Hence

~

3 l.

(-D, 3 l.)

b' 2d

X=-

!s, \"

•----------+------+x ''

(-0, 0)

....··

d

P

!s, _;.:,...···~

--0.5 l.

-d-

-1.5l. -2.5 l. --3 l.

.rue at co on y-axis is

2

For wavelengths to be missing we must find positions of minima.

2.5 l. 1.5 l. 0.5 l.

''''

d v1rt1 +d2 =1+ 2b'}

J

At minima path difference is an odd multiple of

'' co'' 3). and -3A. there will be 6 minima and

=>

b' l. X= d =(2n+1) ; n =0, 1, 2, 3, ..... . 2 2

=>

l.

Smaxima

y

½

b' (2n+1)d;n=0,1,2, ............ .

b2

b2

b2

l.=d. 3d' 5d'"""""·"·

(-0, --3 l.) .......... 15.

Ax= 3l. (-0, 0)

o-----+-

~-------, 2.5l. 1.5l. 0.5l.

Ax=0

No. of maxima = 3

13.

[B,

[A, BJ

P= l.D

d For increasing J3 , d must be decreased and A must be

increased (i.e. frequency must be decreased).

DJ

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2.

[DI Fringe width

p=

1 ~

shall remain the Same as the waves travel

in air only, after passing through the thin transparent sheet. Due to .introduction of thin sheet, only. path difference of the wave is changed due to .which there is shift of position of fringes only, 1 which is given as .rue D(n; )t , where n is refractive index of

[DI Ax=d=nA, for n=1, d=A and here we will have three maxima.

9.

[DI Statement-1 is false but Statement-2 is tn.ie.

14.

thin sheet arid t is its thickness. 3.

5.

(DI Statement-1 is false, Statement-2 is true.

[DI

15.

If both the slits are illuminated by two bulbs of same power, no interference pattern will be observed on the screen. This- is because waves reaching at any point on the screen do not have a constant phaSe difference. as phase difference from two non coherent sources changes randomly. Therefore, maxima and minima would also change their positions randomly and in quick succession. This will result in gen·eral illumination of .the screen.

[A] Statement-1 Is true & Statement-2 Is true and Statement-2 .is correct explanation. of the Statement-1.

1

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1.

2.

[AJ It will be shifted upward

[DJ

.

Cannot be predicted without knowing the thickness of thin plate. .

3.

10.

[BJ Power received at 8,

[BJ

P8 =.!_CJ_n(0.002)' =4x10" W n

(µ, -µ,)t =rue= o.3t 4.

11.

[AJ

l. (µ,-µ,)l=rue=2

[CJ Power transmitted through A, p~ =~xrn-'W=1o-«,i,v

100

t=-l.-=J:... 2(0.3) 0.6

=>

5.

12.

[DJ Power transmitted through B,

[A]

P~ =_:!_Q_x4x10-o W=4x10-6 W 100

3 ,-~ ,/3 -I0 =I0 COS - = 4 2 2

13.

~="'

·

6.

A=(µ-1)t=(1.5-1)x2000x10·" m rue="~,.=~ 2n 6 ;. ;_ 5000 A (µ, ~ µ,)t = 6 t = (6l(o_3 ) = ----:i:a- = 2777.7 A

.6.=10-7 m

14.

2n1tmln

a.

tmln

l.

=2

550 =- =99.64 nm 4_x1.38

{·: n, < n,)

.6.

2n: X 10-7

6000x10-10

qi=~ radian 3

[DJ We know that, P cc a2 or P =ka 2

,

where k is any arbitrary

positive constant.

Now, aA =~1~---e 'aa =~4x~o-e

~ra'""!

2

::::)

15.

t

2

Resultant aniplitude, 8., = -+-~'""+_2a_A_a._c_O_S_~

[AJ

2n,t = 3l.

9.

=>

[CJ For destructive interference, we have

=>

[CJ Phase difference, qi=

[A] . Optical Path Difference (OPD) =2n,t

7.

[CJ Path difference,

3

=>

·

t = 3tm1n =298.9 nm

[AJ Power received at A,

Substituting values and simplifying, 7x10..a 2 ar=-kResultantpower, P, =k~ =7x.10--6 W =7 µW

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Optics & Modem Physics

Advanced JEE Physics 16.

[DJ Anywhere on screen because there is no relation between 8 ,

=>

d2 nyd -+-=nA.

... (1)

Also,

~ + ny:; 1 =(n+1)1.

... (2)

µ. 17.

19.

(2)-(1) we get !P=1-

For central bright fringe to be obtained at O , we have dX=O

=>

(µ-1)t=dsine

24.

[BJ &~=(2n-1h,where n=5 [BJ

[BJ increase and hence b.x increases, so that the central maxima will shift upward.

=>

25.

[BJ Net path difference is given by dsin91 + dsin82

I

=½

Is,

I

D>>d

26.

Is,

22.

=>

d=n

Since, =>

rR

=3nR _ nR = 2nR = nA 2 2 2 =11 +12 +2$j; cos~

In=(,µ,+.µ,)' when ~=2n•, n=0,1,2,3, ....

[CJ

27.

~+dL=o D 2D => Y=-2d So, the POS:ition of first bright fringe is given by d 3 y=-2d+2=-2d

=>

So, the first brightfringe is located

=

[CJ

Since the wave generated is divided equally in two parts, so we have

• For central maxima, we have dsin01 +dsin82 =0

23.

[AJ In transmitted light system, for destructive interference, we have

AX

drn)+d(!) =½<sine~ lane)

3 d' 120=2

4x1.5 1=150nm

Path difference between the waves reaching D is

-0-----20----

=>

t = 900

2_µ1 = (2n - 1).1:. 2 Again, for minimum thickness, we have n =1 900 =150 nm => I= 4x1.5

{for minimum d }

__ J~e,,._d'--=-1-..-..=-----=io

=>

[BJ In reflected light system, for constructive interterence, we have

=>

Since, filC=dsin9-(µ-1)t, so if 9 increases then dsin9 will

21.

2D

P=-1d

2µt = (2n - 1).1: 2 For minimum thickness, we have n =1 ;,_ => 2µ1=2

Fringe width P=~ =4x105 nm 20.

2D

[AJ Total path difference is given by dX = (µ-1)1-dsine

=> 18.

D

[DJ dsln81 + dsin92

2.108

=nA.

¾d below O 28.

[AJ Since &x = xR For maxima, we have &$=(2n)n, where n=O, 1, 2,3, ....

=>

(~)
=>

,._

=>

Amax= nR

max

=~=

nlT&ll

nR 1

[BJ Similarly, for minima, we have &$=(2n+1)n,where n=0,1,2,3, ....

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WaveOpHcs

2; )
=>

(

=>

). = 2nR

=>

Arna~= 2nR

yd= -10). D

34.

29.

Y=-10().~)=-%x10 4 m

=>

2n+1

max

[CJ At P, we have

[DJ

~=

Since, d., =2a(µ-1)a

and

dwater

=2a(:w -1)a

=>

l:__1

dalr

+(µ, -1)t-µ.dsin0

d>C= 10-2x3x10--4 + 1 0A= 10).+ 3A 8x1 4

,I~= 2•(10i.+ 3l.) = 20•+ 3, l. 4 2

dwator=~

=>

i

µ.-1

1t) Since cas(201t+ ~n)=cos(3;)=o So, Ip

=>

=>

=>

= 12 + 12 + 212 cos ( 201t + 3

2

I,

=.!21

Further I™ =½+½+2~(½)(½)cos(oo) 30.

[CJ Since, and

Pair

2a(µ-1)a

lmax =1+1=21

=>

1, I™

DA"

Pwaier

2a(µ-µ.)a

35.

~-1 = ~ = _2_ µ-µ. 3 4

=>

Pwatar

=>

Pwaier=3x1=3mm

P.,

=>

./21

1

=21= ./2

[CJ Path difference is given by ~

= dsin~+dsin0-(µ-1)t

2 3

31.

-- LI~---

[CJ

//1" ------

S"mce, p=d " w

32.

1 xA. 0.5x10-3

=>

1x10-a -

=>

).=5x10·' m=5000A

=> =>

= (µ, -1)t-µ.dsin0

ffi(

4 1 = Q.5 X Q.41 X 1 Q-3 -3X 3 X 10-4 X2

=>

sin8=(µ- 1)t_sin$ d

(~-1)<0.1) =>

sin8- ;Ox10-a

=>

0=30°

~

A , so we observe that

=10).

So, 1011\ bright fringe, will be formed at 0. [DJ For central maxima, we have

~=0 where, ~=i+(µ,-1)t-µ.dsin0

36.

2

[BJ

At C, 8=0°,soweget ~=dsin~-(µ-1)1 =>

~=(50x10"")(½)-(¾-1)<0.1)

=>

~

= 0.025 - 0.05 = -0.025 mm

Substituting, AX= n). , we get

n = LIX A

=>

sin(30°) = .!

.1X=5x10..f.l m

Since, A =5000

33.

~

For central maxima, ax= o

[DJ At O , optical path difference is ~

e ----

i+(µ,-1)t-µ.dsin0=0

SubstitutinQ values, we get

-0.025

_50

500x10-6 Hence, at C there will be maxima. Therefore the order of minima closest to the C are --49.

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Advanced TEE Physics 37.

=> =>

[A] Number of fringes shifted upwards is

N= (µ-1)1 ).

38.

(¾-1)(0.1) 500x10-<

100

41.

1t1:=7t 1=1 s

[BJ Given Y1 =40 cm, D1 =2m=200 cm, D2 =:=10 cm

~

[DJ Net path difference of the waves reacting at Q , is

s,

Ax=yd+y'd

D

D'

-------------------B'----'C

==-

For central maximum, l1x =O

D'

====39.

y'=--y' D y' = -(~)(o.5sin(,rt)) y' = -sin(,rt) mm

y' = sin(n+ ,rt)= -sin(,rt)

a

[CJ . d

y' =

2

, at point

5 Path difference between S81 and SSz is

P exactly in front of S1 , so we have

AX: 1 = S81 -8S2

t;x=(yd)J ~)

D

D'

For maximum intensity, we have path difference to be an even

multiple of

½,so ).

nt=~ 6

t=.!. 6

[A]

For minimum intensity, path difference is an odd multiple of

½,

so dX=(2n+1)½

=

2.110

05 0.5sin(,rt)+0.25= ~ sin(,rt) = 0

dX: 1 =0.16 mm

... (1)

~

.1.x 2 =dsin8

========-

Ax 2 = ru( 1 dsin0 = 0.16 (0.B)sin0=0.16 0 16 sin0= · =.!. 0.8 5

•.. (2) Central bright fringe will be observed when net path difference is zero. => ID<2 - ID<, = 0

sin(,rt) = 0.5n-0.25 0.5 For mir:iimum value of t, we have n =1 sin(,t) = 0.5

==-

=>

ru<: 2 =S2R-S1R

=>

40.

Ax,= dsina = (0.8 mm)(¾)

Now, let at point R on the screen, central bright fringe is observed (i.e., net path difference =0 ). Path difference between SiA and S1R is

Ax= (2n)- = ni. 2 Substituting the values, we get 0.5sin(,rt) + 0.25 = 0.5n

===-

=-

{·: n=1}

tan0=

~ 24

sine:::,..! 5 tan8::::, sin8 =Yi.=..:!. D, 5

D2 10 Y2=-=-=2cm 5 5 Therefore, central bright fringe is observed at 2 cm above point Q on side CD.

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Wave Optics

j Alternate solution for (a)

!

=>

ax at R will be zero if Ax:1 =dX:2

=>

dsina=dsin8

=> =>

a=0

Substituting the values, we get

=(~- 1)10.4x10.. (1.5) Y 4/3 0.45x10"

tan a= tan0

11. = _lg_ D1

(.&_ - 1)1~d µ_

y=

D2

=>

y, = D, ·Y, =(_!Q_)(40) cm D,, 200

=>

y2 =2 cm

=>

y=4.33x10-am

=>

y=4.33 mm=

13

3

mm

44. , [CJ

At O,· dl<,=0 So, net path difference

42. [CJ The central bright fringe will be observed at point Q , if the path difference created by the liquid slab of thickness t =10 cm or 100 mm is·equal to Ax:1 , so that the net path differe~ce at Q

ax= .ruc:2 Corresponding phase difference

t

~4' = ( 2

)Ax

Substituting the values, we get

becomes zero.

$=A$=~(~-1)(10.4x10_.) 6x10-' 4/3 . =>

a

Now, I(~)=I=cos'(f) :::::,.

s => => => => 43.

$=(~)n

(µ-1)t=dl<, (µ-1)(100)=0.16 µ-1=0.0016 µ=1.0016

I= !max cos 2

=> 45.

[CJ At O , path difference is Ax= Ax 2 = (:: -1}

[DJ Given A=600 nm=6x10-7 m,

For maximum intensity at O , we have ~x =nA, where n=1, 2, 3, ......

d=0.45 mm=0.45x10"m and 0=1.5 m

=>

A=

dX dX dX , , , ...... and so on 1 2 3

dX

=(!i:-1)t10.4x10 .. m)

dX

=(!i:-1)(10.4x10' nm)=1300 nm

sl1 =>

s,--s----------------

So, maximum intensity will be corresponding to

1

=

And refractive index of glass sheet, µg

1300 - nm, -1300 - nm, ... - nm, -1300 4 3 2 i. = 1300 nm , 650 nm, 433.33 nm, 325 nm, .... A=1300 nm

Thickness of glass sheet, t =10.4 µm =10.4 x 1a.fl m Refractive index of the medium,. µm

(1!1t)

i

=>

-

The wavelength in the range 400 nm to 700 nm are 650 nm

=1.5

and 433.33 nm(=

Let central maximum is obtained at a distance y below point 0. :::::,

yd

Ax 1 =8,P-S 2 P=[)

Path difference due to glass sheet is given by

Ax,=(~: 6.x1

=.1.X2

yd

=(!1-1)1

D

nm)

[CJ The optical path difference between the two waves arriving at P

is y,d y,d (1)(10) (5)(10) dl<=-+-=--+--3 3

-1}

Net path difference will be zero, when we have

=>

46.

:o

13

D1

D2

10

2x10

:::::,. Ax=3.5x10- 2 mm=0.035 mm To calculate the order of interference, we shall calculate dX

n=1-

µ.

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n ~~00~5xx1~~:

Optics & Modern Physics

~

Since, m = _f_

f+u

=> n=70 => 6x =70l. So, 70111 order maxima is obtained at P .

47.

=>

X

=>

X=20 cm So,weget U=-20, v=+80

[DJ

Since.!_.!=!

At O, 6x = y,d =10_,, mm =0.01 mm

V

D,

=>

Now, we observe that fix= 20A. So, 20111 order maxima is obtained at O . 48.

50.

f

l.D _ 6000x10-" x1

'"'

51.

U

f=16 cm

[BJ 0

[DJ (µ-1)1 = 0.01 mm I =o.o, - - = 00 . 2 mm=20 µm 1.5-1 Since the pattern has to be shifted upwards, therefore, the film must be placed lnfront of S, . ·

49.

100-x = 4

d

0

3x10-a

2 · mm

[DJ

P= l.D d

When D is increased, then

p is increased.

[AJ Using the concept of Displacement Method, we get 0=$i; =3mm

=2.112

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1.

A-> (p, s) B--> (q, r) C --> (p, q, r) D--> (q, r, s)

2.

A-> (p, q) B--> (s) C--> (p, q) D--> (r)

3.

6.

s,

width remains same. 7.

A--> (p) B--> (s) C--> (r) D--> (q)

8.

A4' (p, q, r, s) B--> (p, q, r, s) C --> (p, q, r, s) D --> (p, q, r)

A--> (q) B--> (p) C--> (s) D--> (r)

4.

A--> (r) B--> (s) C--> (q) D--> (p) .

5.

A-> (p)

A--> (r) B--> (!) C --> (r) D --> (p) this effective path If glass slab is introduced across increases so central maxima will be shifted downward but fringe

B--> (s) C--> (r) D--> (q, r)

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1.

Shift dY =I(µ, -1)1, -(µ,

Substituting d = 201. , we get

-1)t,I~

sinB=sin(45°)+

=> => => 2.

dy=i(1.2-1)x15x10 .. -(1.6-1)x10x10_.1(

l.S __,) 1.5x10

=>

',

3

,.1.y=3x1ff" m

dY =3 mm

4.

(a)

Since, 1- =0.25 m and d =2 m = 81. At A ·and C, AX=d=81Since this is an even multiple of A, so maximum intensity is obtained. At Band D, AX=0 Since.this is an even multiple of A, so again maximum intensity will be obtained. · B

'' ' '''

9::::59°

1~ =(¾-1)(¾+¾) => A=10cm Now applying µ 2 V

-~

u

= µ2 -

R

µ, twice, we get

3/2 __1_=3/2-1 and v, -20 10

•.• (1)

4/3 _3/2 4/3 -3/2 V2

V1

.•. (2)

-10

Adding equation (1) and (2), we get

v2 =80cm (b)

' C ------•--•---A S1 1 Sa

..,._~-... '

''

Ray 1 has a longer path than that of ray 2 by a distance dsin(45°), before reaching the slits. Afterwards ray 2 has a path longer than ray 1 by a distance dsin8. The net path difference is therefore, dsin9- dsin45° (a) Central maximum is obtained where, net path difference Is zero, so we have dsin8-dsin(45°)= 0 => 8=45° (b) Third order maxima is obtained where net path difference is 3"- , so we have dsin8-dsin(45°) = 31. 3 => sine"= sin(45°) + di.

.

The Image formed by (lens + water) system will act as an object for the mirror. This is below the axis of m1 and at the same distance as the -centre is, therefore, its image will be fanned vertically above at 1 mm from AB . Similarly m2 will fonn an image of 11 , 1 mm below CD .

D Further, between A and B seven maximas corresponding to AX = 7A , 61. , 51. , 41. , 31. , 21. and i. are obtained. Simila~y between B and C, C and D , and D and A . So, total number of maximas is N = 4 x 7 + 4 =32

3.

::i.

----------+-------

c---------1-------- ----·D i+-20

=> (c)

d=1112 =4 mm, .

Since, 1c

=>

cm......

1112 =1+1+1+1=4 mm D =80 cm

V

=f

i.---"--( 3x10' x3) - µ.f- 7.5x10 14 4

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Wave Optics P=).D =~( 3x10'14 J(80x10""J m= 6 x 10_. m d 4 7.5x10 4x10-3 P=60 µm

=> => 5.

=> -3d] 3d2 S 3P-S,P=dsine, = d [ ~ =w=l.

The intensities of the rays due to successive reflections and

refractions are shown in figure. So, 11 =0.11 0 , 12 = 0.08110

=>

,i~

= 2,

s,

pl

s,

2

3d

--

I ._ ---

0.09!,

---u 0.11,

I

- e, --------

s, If IR be the resultant intensity of the wave at the point P , then IR=l1 +1 2 +2Mcoscji

=>

=>

6.

!

1

m~

I.,,.

=>

r, = 1, + 41 0 + 2J(1,)( 41,) cos(120')

=>

l,=31,=3(12Wm"")=36Wm""

=[Jf +ll' =(19)' =361

{I_1

1O.

vi;

Angular fringe width e = !:, __ ._ d-180x60

Therefore_, number of fringe widths in a distance y are given by

r = __E.__ P ,/al,

=~

According to the given condition

Fringe width p = AD and y = ';, d v3 n=

!

0.32 X 10_, (,/3)(500x10_,)

d < 6x10-7 x180x60

=>

369.5

dmax 11.

=2.06 X 1"Q-3

m =2.06 mm

According to Snell's Law, we have

1sin(30') = F3 sinr :) r=30° 2 air

~-__c,.""'-----JL--

film Therefore, total number of maxima obtained in the angular range -30° < 0 < 30° (including the central one) is

N=2x369+1=739 7.

air The Optical path difference is given by llX. = 2µtsecr-2ttanrsini

Here, d=2mm=2x104 m, 0=10 cm=0.10 m, t= 0.5 mm =0.5x10-a m, llX =5 mm=5x10-a m,

A=6x10-7 m

AX =2,/a(tsec30')-2ttan(30°)sin(6D')

=>

AX= 4t-2{

Since, AX= ~(µ-1)1 µ- 1 = .1Xd Dt

=>

5x10-3x2x10: =0. 2

0.10x0.5x10 µ=1+0.2=1.2 5µ=6

lJ( ~)

=31

Since the Ray 1 is reflected at the surface of the denser medium so it suffers an additional phase change of re or a path change of

~ . So, for constructive interterence, we have 3t=!: 2

9.

=>

t=!:=1000 A=100 nm 6

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AdvancedJEE Physics

Since the wavelength is decreaf?ing from 600 nm to 400 nm ,

12.

so fringe width ~Isa decreases by a factor of

number of fringes will Increase by a factor of

~ = ~ . So

the

¾. Hence the new

number of fringes formed ls N=_:3_(12)=18 2 The path difference at O Is

14.

0 = 0.01 radian n =10

AX=(S,P-T).,., + T- -(S,P).,.,

J..=6000x10~ AX=yd+(µT _T) µt 1quld D

Since, 8 =.!.

At the central maxima, we have path difference AX=O

=>

cm

AX=2t=n). X

Y=_DT(!1=11,)=-TD(4:J) d µ, d 10-t

Cen~ maxima will be at O when y = O

=> =>

l=0X 2Bx =

=>

1 x=" =3x10-1 m=3mm

n,

0 l+-X-

20

~

t=4s Now, the speed of the central maxima is given by lcl)tJ 6DT V=[dt]= (10-t)'d At t=4s,wehave v=

=>

13.

6DT OT d =6d 36

1 3 v- ( )( exrn-<>) (6)(~x10_,)

Fringe width,

3x10'° ms-' =3 mms-'

15.

When coherent, then IDC at centre is zero, so, we have the resultant intensity to be

1, =41 When·incpherent, then the sources will not interfere and it wi11 be a general illumination at the point, so we have the resultant

intensity to be, I, =21

=>

.!i.=2

p = '~ «).

=2.116

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I

Dual Nature of Radiation & Matter

IContents

---- ----- ---- ----------------- - _______ _,

DUAL NATURE OF RADIATION & MATTER-----------···············································............ 3. 1 Solved Practice Problems ..................................................................................................3.15 Practice Exercise Sets

:> Single Correct Choice Type Questions ..................................................................3.21 :> Multiple Correct Choice Type Questions ...............................................................3.33 :> Reasoning Based Questions (Assertion Reason Type) ........................................3.36 :> Linked Comprehension Type Questions (Paragraph Type) ...................................3.38 :> Malrix Match Type Questions (Column Matching Type) ........................................3.40 :i

Integer Answer Type Questions ......................••..•••...•••..••...•.................................3.43

Answers to In Chapter Exercises (ICE) & Practice Exercise Sets .....................................3.45 Solutions to In Chapter Exercises (ICE) .............................................................................3.48 Solutions to Practice Exercise Sets ...................................................................................3.52

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MATTER WAVES

Light possesses dual nature i.e. it behaves both as a wave and as a particl~. In some phe~omena e.g., interference, diffraction and polarisation, it behaves as waves because they are explained on the basis of Wave theory while in some other phenomena e.g. photoelectric effect, Compton effect, it behaves as particles (photons). Since nature demands symmetry, therefore de-Broglie thought that matter must have dual nature. The particle nature of matter is well known and hence de-Broglie thought that material particles must possess wave-nature.

According to de-Broglie a material particle in motion must have a wave like character and the wavelength associated with it is given by . p

For charged particles accelerated through a potential difference of V volts, Kinetic energy i.e. EK = qV ,.

p = momentum of the particle de-Broglie assumed this expression in analogy with photon because momentum of photon is

,.

h p=-

H m is the mass of particle and v the velocity, then momentum of particle is p =mv .

h ,/2tneV

m=9-lxl0°"1 kg,

Substituting

h = 6 •62 X lO""'Js ,

e=l-6x10"19 C,weget

J1~Q i..=J1~o A=1'% A A. =

=>

... (5)

X 10-10

This is expression for de-Broglie wavelength associated with electron accelerated through a potentjal difference of V . For neutral particles (like neutrons, atoms) at temperature T , kinetic energy of most of particles, EK= k8 T, where k8 is the Boltzmann's constant.

p

p=~2mEK.

... (4)

For electrons accelerated through a potential difference of V volts

1,,

A.=!!.

So, de-Broglie wavelength ).. =.!!.._ mv If E1: is kinetic energy of particle, then

h

~2mqV

... (1)

where, h is the Planck's constant whose value is given by h =6.63x10°"' Js and

=>

... (3)

de-Broglie wavelength )..

de-Broglie's POSTULATE

,. =!!.

1..=-h~2mEK

h ~2mk,T

... (6)

The wave nature is possessed by all particles neutral of charged. The wave nature was first verified by Davisson and Germer for slow electrons.

.•. (2)

{·:EK=;:}

Illustration 1 Obtain the de-Broglie wavelength associated with thermal neutrons at room -temperature (27 °c) . Hence explain why a fast neutron beam needs to be thermalised

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Optics & Modern Physics

with the environment before it can be used for neutron diffraction experiments.

de-BROGLIE WAVELENGTH ASSOCIATED WITH THE CHARGED PARTICLES

Solution

The energy of a charged particle accelerated through

Average kinetic energy of a neutron at absolute temperature T is

potential difference V is E =.!.mv' = qV

.!.mv' =!l_k T 2 2 ' =a, =a,

2

Hence de-Broglie wavelength

L =!3.k T 2m 2

'l.=!!.= h p .J2mE

{·: p=mv)

'

Using the above formula, we get

p=~3mk,T

,.

= 12.27

.Jv

Eectnm

de-Broglie wavelength is given by '/,.

h p

h )2mqV

,.

h ~3mk8T

= 0.286 Proton

.Jv

A

A

Given m, = 1.675x10-" kg, k, = 1.38x10-23 J mol-' K-1 T=27+273=300K, h=6.63x10-" Js '/,.

6.63x10""' .J3 X 1.675 X 10-27 X 1.38 X 10-23 X 300

=a,

'/,.

de-BROGLIE WAVELENGTH UNCHARGED PARTICLES

6.63x10-20 ----m=1.45x10-" m 4.56

'l.=1.4sA As this wavelength is comparable to interatomic spacing =a-

WITH

For Neutron de-Broglie wavelength is given as 10

,.

0.286 x 10- m ~E(in eV)

N~
(- 1 A) in a crystal, so thermal neutrons can be used for diffraction experiments. A high energy neutron beam should be first thermalised before using it for diffraction.

ASSOCIATED

0.286 A ~E(in eV)

Energy of thermal neutrons at ordinary temperature is given by E=k,T

CHARACTERISTICS OF MATTER WAVES

a) b) c)

d)

e)

£) g) h) i)

=

Matter wave represents the probability of finding a particle in space. Matter waves are not electromagnetic in nahlre. de-Brogile or matter wave is independent of the charge on the material particle. It means, matter wave of deBroglie wave is associated with every moving particle (whether charged or uncharged). Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles. Electron microscope works on the basis of de-Broglie waves. The phase velocity of the matter waves can be greater than the speed of the light. Matter waves can propa·gate in vacuum, hence they are not mechanical waves. The number of de-Broglie waves associated with nth orbital electron is n . Only those circular orbits around the nucleus are stable whose circumference is integral multiple of de-Broglie wavelength associated with the orbital electron.

'l.=-h.J2mkT

where T is the Absolute temperature, k8 is the Boltzmann's constant given by k 8 = 1.38 x 10-23 JK-1 6.62x 10=-;=--~;;;;;~~---• .J2x1.67x10- xl.38x10- T 34

So,

AThermalneutro

27

23

30.83 A

.ff

HEISENBERG's UNCERTAINTY PRINCIPLE

According to Heisenberg it is impossible to measure the position and momentum of a particle simultaneously with 100% accuracy. This is called Heisenberg's uncertainty principle. Uncertainty principle successfully explains the (i) Non-existence of electrons in the nucleus (ii) Finite size of spectral lines. If /;,x and ilp are uncertainties in determining the position and momentum of the particle simultaneously, then /;,xilp ;e ..!:._, where h = 6.63 x 10->1

41t

=a>

h m!;,xilv ?. 41t

3.2

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This principle is universal and holds for all microscopic and

6.63x10-" 3.32 X 10-27 X 1911

macroscopic particles. The principle is also unaffected by experimental techniques.

).

If ~ = 0 , then 11p ----> oo and if 11p = 0 then ~ ----> oo ,i.e. if we are able to measure the exact position of the particle (say an electron) then the uncertainty in the measurement of the linear momentum of the particle is infinite. Similarly, if we are able to measure the exact linear momentum of the particle i.e., tlp = 0 , then we cannot measure the exact position of the particle at that time.

). = 1.04; 10-rn m = 1.04

tJ

Viewertj lncide~

·

\

photo~: \

a

Original momentum of electron

ti /

~

Reflected

photon

~a,. Final momentum of electron

An electron cannot be observed without changing it's momentum This principle is also applicable to energy and time, angular momentum and angular displacement. Hence,

!'.EM-;,;!!_

ill

A

QUANTUM NATURE OF LIGHT

Some phenomena like photoelectric effect, Compton effect, Raman effect could not be explained by Wave theory of light. Therefore quantum theory of light was proposed by Einstein who extended the Planck's hypothesis to explain Black Body radiation. According to quantum theory of light or radiation, the energy of an electromagnetic wave is not continuously distributes over the wave front Gust like the energy possessed by water waves). Instead Plank proposed that an electro-magnetic wave travels in the form of discrete packets or bundles of energy called Quanta. So, according to Plank, "light is propagated in bundles of small energy, each bundle being called a photon and possessing en~rgy", given by he E=hv=,_

where, v is frequency, 1,. is wavelength of light, h is Planck's constant whose value is 6-63xl0_-"-Js and c=3x10 8 ms-1

4,r PROPERTIES OF PHOTONS

M,/10 -;,; .!!_

1.

4,r

PROBLEM SOLVING TRICK(S)

a)

For numerical problems, we shall use

b)

Ax.l.p ~ h t>EAt z h &t.8 ~ h If the radius of the nucleus is r then the probability of finding, .the. electron inside the nucleus is L\x =2r and uncertainty in momentum is t\p =..!!.... .

2

3.

4.r

In the interaction of radiatio~ with matter, the radiation behaves as if it is made up of particles called Photons. This fundamental is also called as Corpuscular Theory of Light. So, light behaves both as a particle and a wave. All photons emitted by any source travel through free space with a speed equal to the. speed of light i.e. c=3x108 ms-1 • Each photon has a definite energy depending upon the frequency v of the radiation and this energy is independent of the intensity. So,

,.

E=hv=hc (injoule)

Illustration 2

Find the de Broglie wavelength corresponding to the root-mean-square velocity of hydrogen molecules at room temperature (20 °C) • Solution . Smce,

=

v

Vrms=

5.

M

~3RT

3x8.3lx293 2x 10-3

nns

4.

=1911 ms-'

6.

h h Now, A=-=-p mvrms

7.

Mass of one hydrogen molecule is given by

8.

m

2

6.02x1026

kg = 3.32 X 10-27 kg

12375 u,.·IS m · A , then E =-,_eV .

If the intensity of the light of given wavelength is increased, then there is an incr~ase in the number of photons incident per second per unit area on a surface. However, energy of the photon remains the same as long as the frequency or the wavelength is unchanged. The speed of the photon changes as it travels through different media due to the change in its wavelength. The frequency of the photon does not change when it goes from one medium to the other. In the situations wizen a photon collides witlz a material particle, the total energiJ and momentum remains conserved. However, the number of photons may not be conserved in a collision because during the 3.3

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Advanced JEE Physics collision photon(s) may be absorbed or new photon(s) may be created. 9. A photon is an electrically neutral particle which is not deflected by electric and magnetic field. 10. Rest mass of photon is zero. Since the mass of a particle m moving with a speed v is given by

14. Intensity of light (1) Energy crossing per unit area normally per second is called intensity or energy flux ,i.e. I=

m,

p

'

g

=>

I oc ]:_ r' 15. Number of photons falling per second (n) If P is the power of radiation and E is the energy of a p photon then n = - .

2

Now, for a photon, v = c, hence m0

E

=0

So from here we can conclude tl:iat the rest mass of a photon is always zero, i.e. we cannot have a frame of reference where the photon is at rest. 11. Dynamic or kinetic mass of a photon, is determined by using the Einstien's Mass-Energy Equivalence, i.e., E=hv=mc2

=>

MOMENTUM.AND RADIATION PRESSURE

An electromagnetic wave consists of photons capable of transporting linear momentum. The linear momentum P possessed by an electromagnetic wave is related to the energy E it transports according to the relation E P=-

hv

h m=-=c2 cJ...

hv

---W

C

12 The Linear Momentum of a photon is found by using the de-Broglie relation according to which, we have h C :\.=-,where :l.=p V

=>

h

p=-=c ,.

However, this resuit is also obtained by using the fact that the total energy of a subatomic particle of rest mass m0 , moving with a velocity v , having momentum p is

given by E2 =p2c2 + ni;c4

If the wave is incident in the direction perpendicular to a surface and gets completely absorbed by the surface, then equation (1), tells us the linear momentum imparted to the surface. If surface is perfectly reflecting, the momentum change of the wave is doubled. Consequently the momentum imparted to the surface is also doubled. According to Newton's Second Law, the force exerted by an electromagnetic wave on a surface is given by the equation, F=t.p

M

l(t.E)

. t.p= - Fromequation(l), -

M c M

Now, for a photon, m0 =0, so we have from above

=>

expression that

F=!(t.E) M

Intensity (I) of a wave is the energy transported per unit area per unit time.

Since, E=hv=:

(2-) A M

he

,.

=>

I=

=>

pc=-

hv h p =~=i

=>

=>

t.E =IA M

13. The number of photons N , each of energy E , emitted from a source of monochromatic radiation of w~velength ;\. and energy W and power P

E

... (2)

C

E=pc

N=w

= P = radiation power)

I=41tr2

is the rest mass of the particle.

m,=m

(f

=:

At a distance r from a point source of power P intensity is given by

mgm,

where

!t

Substituting in equation (2), F = IA C

=>

=w =Pt hv

hv

t,.E

=>

I

F

- =pressure =A

C

I prad

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!_ is also equal to the energy density (energy per unit C

(b)

p I(4itR') n=E

volume) u. Hence, P,., = u

... (3)

The radiation pressure is thus equal to the energy density (Nm-2 = Jm"). At a perfectly reflecting surface the pressure on the surface is doubled. Thus, we can write,

I

{wave totally absorbed}

prad=-=U C

21 and P.., =-=2u

{wave totally reflected}

C

=>

n

E

(1400)(4,r)(l.49 X 1011 ) 3.3xl0-19

2

l.18xl0 45

Illustration 5

A small plate of a metal is placed at a distance of 2 m from a monochromatic light source of wavelength 4.8 x 10-7 m and power 1 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square meter per second.

Solution Illustration 3

Find the energy, the mass and the momentum of a photon of ultraviolet radiation of 280 nm wavelength. Solution

(6.6 X 10-34 )(3 X 10 8 ) (4.8x10-')

.1 X _ J 4 25 10 19

Number of photons striking the metal plate per square meter per second is

Given, ,. = 280 x 10-" m n=(f)(4:,,)

E =l!c . Smce, ,_-

=>

E (4.316x10-15 eV sec)(3x108 ms-1 ) (280x10-" m)

=> 4.6eV

Mass of photon is m = ~

Illustration 6

C

=>

4.6 X 1.6 X 10-19

m

(3x10')

Find the number of photons entering the pupil of our eye per second corresponding to the. minimum intensity of white light that we humans can perceive (-10-" wm-2 ) • Take the area of the pupil to be about 0.4 cm' and the average frequency of white light to be about 6 x 1014 Hz.

8.2 X 10-J6 kg

2

Momentum of a photon is E P=-

Solution

C

=>

p

1 1 n=( =4.82xl016 m-2s-1 19 ) . 4.125xl0- (4it)(2)2

4.6 X 1.6 X 10-19 3xl0 8

Minimum intensity, I =10-10 wm-2

2.45 X 10-27 kg ffiS-l

Area of pupil, A=0.4 cm' =0.4x10--< m Average frequency, v = 6 x 10-14 Hz

Energy of one photon

Illustration 4

The intensity of sunlight on the surface of ear\h is 1400 wm-2 • Assuming the mean wavelength of sunlight to be 6000 A, calculate

E=ltv=6.63xl0-34 x6xl014 J=4x10-19 J Let n be the nllmber of photons crossing per square metre area per second.

the photon flux arriving at 1 m' area on earth perpendicular to light radiations, and (b} the number of photons emitted from the sun per second assuming the average radius of Earth's orbit is 1.49 X 1011 ffi ,

Now Intensity second

Solution

=>

(a)

(a)

Energy of a photon 12400 E = ltc = = 2.06 eV =3.3xo-" J ,_ 6000 Photonflux

IA E

(l400)(l), 4.22x10 21 3.3x10-19

=> =>

=>

= Energy incident per square metre area per

I= Total Energy of

11

photons

I = n x Energy of one photon Intensity 11 Energy of one photon 10-10 wm-2

I =--:,',,,CC.2.5xl08 m-2 s-1 19 4xl0- J

So, the number of photons entering the pupil of our eye per second is

N = n (Area of the pupil)

=>

N

= 2.5 x 108 x 0.4 x 10-1 .-, =104 s-' 3.5

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Advanced JEE Physics Illustration 7 Find the number of photons emitted per second by a 25 W source of monochromatic light of wavelength

6600 A . What is the photoelectric current assuming 3% efficiency for photoelectric effect? Given h = 6.6 x 10"' Js . Solution

Energy of each photon is

Eh:

10 66 · ::~:;;_: '

19

3x10- J

Illustration 9

A cylindrical rod of some laser material 5 x 10-2 m long and 10-2 m in diameter contains 2 x 1025 ions per m 3 • ff on excitation all the ions are in the upper energy level and de-excite simultaneously emitting photons in the · same direction, calculate the maximum energy contained in a pulse of radiation of wavelength 6.6 x 10-' m. If the pulse lasts for 10-7 second, calculate the average power of the laser during the pulse. Solution

Total energy emitted per second by 25 W source is

Total number of ions in the rod is

E=25J Number of photons emitted per second is 25 n= 8.33 X 1019 3 x10-19

N

per unit volume

N=7.85x10 190 As all the ions de-excite simultaneously, the number of photons emitted in the same·direction is also 7.85x10 19 • So, the energy contained in a pulse of radiation of wavelength 6.6 x 10-7 m is

3 % of photons ) x (charge on) emitted per second electron

3 25 _,, I=-x xl. 6 xl 0 -=0.4A 100 3x10-19 Illustration 8 A source emits monochromatic light of frequency 5.5 x 1014 Hz at a rate of 0.1 W . Of the photons given out, ·0.15% fall on the cathode of a photocell which gives a current of -6 µA in an external circuit.

(a)

Find the energy of a photon.

(b) Find the number of photons leaving the source per (c)

second. Find the percentage of the photons falling on the cathode which produce photoelectrons.

Solution

(a)

20

E = he x7.85x1019 A,

E

E=2.27eV (b) Number of photons leaving the source per second is

n, = O.l 5 x2.75x1017 =4.125x1014 100 Number of photoelectrons emitting per second .is 6x10_. n2 - - ~ ~ 3.75x1013 l.6x10-19

So,

%age of Photons ) = n2 x 100 ( falling on Cathode n,

"

...

BASED ON PHOTON PROPERTIES / (Solutions on page 3.48)

, 1.

!

The intensity of direct sunlight before it passes through the ; earth's atmosphere is 1.4 kWm 4 • If it is completely : absorbed find the corresponding radiation pressure. \ I

! 2.

\ 3.

I

According to the maxwell theory of electrodynamics an electron going in a circle should emit radiations of frequency ] equal to its frequency of revolution. What would be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed? An electron is accelerated by a potential difference of 25 volt. Find the de-Broglie wavelength associated with it.

I

. 4. 13

= % %age of Photons ) = 3.75 x 10 x ( falling on Cathode 4.125 x 1014 100 9

23.55 J

,A: .... ,, _, .,., .. ,.,,,,,,. . --1~~~1e·1 ~--·-~v~:-: :'~-~:.-:~·,,,t1H:,'" -~,,>~~-;:a.lSEt~i5m4flN 1.-

n=!'._=

O.l 2.75x1017 E 36.3x10-20 Number of photons falling on cathode per second is

6.6x10-34 x3x10 8 6.6 x10-7

Energy Average power P=-.-Tune 2355 P= J = 23.55 x 107 W = 235.5 MW 10-' s

Since, E=hv=(6.6x10"')(5.sx10")=36.3x10- J

=>

(c)

the rod

N=(2x10 25 m--3)x(3.14x(0.005)2 x5x10-2 m 3 )

Photoelectric current (I) is

I=(

=(Number of ions)x(Volume of)

15.

;

Find the number of photons emitted per second by a MW ! transmitter of 1O kW power emitting radiowaves ofj wavelength 500 m . lf 5% of the energy supplied to an incan9escent light bulb Is radiated as visible light, how many vi.Sible _light photons are .. .!31J1itted_by 1O_O_'!V~tt b_ulb~b~sume. waY.E!l~i:i91Ji. of_~IJ _visible

,.____... 3.6

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Dual Nature of Radiation & Matter photons to be 5600 A . Given h = 6.625 x 10--" Js . 6.

Calculate the number of photons in 6.62 J of radiation

The minimum energy required for the electron emission from the metal surface can be supplied to the free electrons by any one of the following physical processes.

energy of frequency 1012 Hz. Given h = 6.62 x 10-34 Js.

THERMIONIC EMISSION 7.

Monochromatic light of frequency 6 x 1014 Hz is produced by a laser. The power emitted is 2 x 10-3 W .

8.

(a)

What is the energy of each photon in the light?

(b)

How many photons per second, on the average, are emitted by the source?

Show that a free electron at rest cannot absorb a photon and thereby acquire kinetic energy equal to the energy of the photon. Would the conclusion change if the free electron was moving with a constant velocity?

9.

An electron microscope uses electrons accelerated by a voltage of 50 kV . Determine the de-Broglie wavelength associated with the electrons. If other factors (such as numerical aperture etc.) are taken to be roughly the same, how does the resolving power of an electron inicroscope compare with that of an optical microscope which uses

yellow light? 10.

An electron and proton are possessing the same amount of kinetic energy. Which of the two have greater wavelength?

11.

An electron and a photon have same de Broglie wavelength (say 1 A ). Which one possesses more kinetic energy?

12.

An electron and a proton have same wavelength. Which one p_o~sesses more energy?

EMISSION OF ELECTRONS

As we are aware of the fact that metals have free electrons (negatively charged particles) which are responsible for their conductivity. However, the free electrons cannot normally escape out of the metal surface. If an electron attempts to come out of the metal, the metal surface acquires a positive charge and pulls the electron back to the metal. The free electron is thus held inside the metal surface by the attractive forces of the ions. Consequently, the electron can come out of the metal surface only if it is supplied some minimum energy to overcome the attractive pull of the metal. This minimum energy required by an electron to escape from the metal surface is called the work function of the metal. It is generally denoted by W or $0 and is measured in eV (electron volt). One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt, so 1 eV=1.6x10-19

J

This unit of energy is commonly used in atomic and nuclear physics. The work function, W depends on the properties of the metal and the nahrre of its surface. The work function for platinum is the highest WP, = 5.65 eV, whereas it is the lowest for caesium i.e., W0 = 2.14 eV.

By suitably heating a metal, sufficient thermal energy can be imparted to the free electrons to enable them to come out of the metal. The free electrons so emitted are called as Therm ions. FIELD EMISSION

When the metal surface is subjected to very strong electric field of the order ranging from 10 3 vm-1 to 108 vm-1 , the electrons (beyond a certain limit) start coming out of the metal surface. This method of emission is dangerous and less efficient. This method of emission is also called the Cold Cathode Emission. PHOTO-ELECTRIC EMISSION

When light of certain minimum energy (or m1rumum frequency or maximum wavelength) illuminates or falls on a_ metal surface, electrons are emitted from the metal surface. TI1e emitted electrons are called pllotoelectrons. In case of Photoelectric emission, the rate of emission of photoelectrons is very low. SECONDARY EMISSION

When fast moving electrons strike a metal surface, then some of their energy is transferred to the free electrons of the metal. Due to this, when free electrons gain energy more than the work function, then they are emitted from the metal surface. These emitted electrons are called the secondary electrons. PHOTOELECTRIC EFFECT

The phenomenon of emission of electrons from a metallic surface by the use of light (or radiant) energy of certain minimum frequency (or maximum wavelength) is called photoelectric effect. The emitted electrons are called as photoelectrons. The phenomenon was discovered by Hallwach in 1888. For photoelectric emission the metal used must have low work function e.g., alkali metals. Cesium is assumed to be the best metal for photoelectric effect. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of nucleus of the atom of the metal surface. The explanation to the photoelectric effect given by Einstein is based on the Law of Conservation of Energy. Before discussing the effect further, we must understand the following terms. A.

WORK FUNCTION (OR THRESHOLD ENERGY) (W,)

The minimum energy of incident radiation, required to eject the electrons from metallic surface is defined as work 3.7 C=:J

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Advanced JEE Physics function of that surface. It is the characteristic of a metal surface.

a)

W0 =hv0 = he (in joule), where 'I., v O = Threshold frequency and '/.

0

= Threshold wavelength

Work function in electron volt is given by W.{eV)=~ e'l.0 B.

'/.

b)

1237~ (in A)

0

THRESHOLD FREQUENCY

c)

(v0 )

The minimum frequency of incident radiations required to eject the electron from metal surface is defined as threshold frequency. If incident frequency v < v0 => No photoelectron emission.

d)

For most metals the threshold frequency is in the ultraviolet (corresponding to wavelengths between 200 and 300 nm), but for potassium and cesium oxides it is in the visible spectrum ( 'I. between 400 and 700 nm)

e)

C.

THRESHOLD WAVELENGTH

('1.0)

The maximum wavelength of incident radiations required to eject the electrons from a metallic surface is defined as threshold wavelength. If incident wavelength 'I.> 'l.0 , then No photoelectron emission will take place. Illustration 10 In an experiment on photoelectric effect light of wavelength 400 nm is incident on a caesium plate at the rate of 5 W . The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit. Solution

E = 12375 = 3.1 eV 4000 Number of photoelectrons emitted per second 5 (-!.)( 10 3.lxl.6x10. q Ne (N) t=-;=f e=ne

n=

. smce, =:,

19 )

= 1 x 1013 per second

-= 1

i=(ne)=lxl013 xl.6xl0-19 =1.6x10-' A=l.6 µA

LAWS OF PHOTOELECTRIC EMISSION

We thus have the following laws of photoelectric emission, derived from the experimental observations.

=

f)

For each emitting metal, there is _a certain minimum frequency v 0 (or maximum wavelength 'l.0 ), called the threshold frequency of the incident radiation, below (above which) which no emission of photoelectron takes place, no matter how great is the intensity. The value of v0 ( or 1..0 ) is different for different emitting surfaces. The process of emission of photoelectrons is an instantaneous process. There is no time lag ( < 10..,, s) between the incidence of radiation and the emission of photoelectrons. Photoelectric effect is a one photon-one electron phenomenon i.e. even if photon has an energy sufficient to strike off 3 electrons (say) it can only strike off one electron with the excess energy being imparted to the struck off electron as kinetic energy. The number of photoelectrons emitted per second, that is, photoelectric current is directly proportional to the intensity of the incident radiation but is independent of_ the frequency (or wavelength) of light. The velocities (or the energies) of the emitted photoelectrons vary between zero and a definite maximum (vmax). The proportion of photoelectrons having a particular velocity is independent of the light intensity. The maximum velocity, vmax, and hence the maximum kinetic energy is independent of the intensity of the incident light, but depends on its frequency, increasing linearly with the increase of the frequency of the incident light.

EINSTEIN's EFFECT

EXPLANATION

OF

PHOTO-ELECTRIC

The wave theory of light could not explain the observed characteristics of photoelectric effect. Einstein extended Planck's quantum idea for light to explain photo-electric effect. According to his idea, the energy of electromagnetic radiation is not continuously distributed over the wave front like the energy of water waves but remains concentrated in packets of energy content hv, where v is frequency of is universal Planck's constant radiations and h (=6-625x10""" Js). Each packet·of energy moves with the speed of light. The assumptions of Einstein's theory are a) The photoelectric effect is the result of collision of two particles, one of which is a photon of incident light and the other is an electron of photo-metal. b) The electron of photo-metal is bound with the nucleus by Coulomb attractive forces. The minimum energy required to free an electron from its bondage is called work function, W = $0 = hv 0 • c)

The incident photon interacts with a single electron and loses its energy in two parts

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Dual Nature of Radiation & Matter. (i)

Firstly, in getting the electron released from the bondage of the nucleus. (ti) Secondly, to impart kinetic energy to emitted electron.

d)

The efficiency of photoelectric effect is less than 1 %, i.e. less than 1 % of photons are capable of ejecting photoelectrons.

EXPERIMENTAL SETUP FOR PHOTOELECTRIC EFFECT

It consists of two conducting·electrodes, the anode (Q) and

cathode (P)· which are enclosed in an evacuated glass tube ,,.~

as shown in figure.

0,0-~o


v~'"''-~,,. -"----

--=,,_

Accordingly, if hv is the energy of incident photon, then

r-fll-p-1

hv=W+K_ ~

·-~ ··:.:~

~-~~

K==hv-W

o:

... (1)

~---'W.Mr-----1, IIIt----'

. Metal

This is Einstein's photoelectric equation, where W is ~

work function and

.

1

2

Kmax = zmv= = eva

is the maximum kinetic energy of photo-electrons emitted. Equation (1) is referred as Einstein's photo electric equation that explains all experimental results of photoelectric effect and is based on the Law of Conservation of Energy. Illustration 11

Illustration 12

Ultraviolet light of wavelength 2000 A causes photoemission from a surface. The stopping potential is 2V. (a) Find the work function in eV (b) Find the maximum speed of the photoelectrons. Solution (a) Using Einstein relation he W=-i:--eV0

~

work function 5.6 eV, 0.53% of-the incident photons eject photo electrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). Given that 1 eV =1.6xto·" J. Solution

Number of photoelectrons emitted per second

~

2

=

v

max

=>

When a beam· of 10.6 eV photons of intensity · 2 wm·2 falls on a platinum surface of area lxto-< m 2 and

W= 12400 -2=4.2 eV 2000

. ·1 ·, =eV. • (b) Smce ~mv_ O

~

The battery or . some other source of potential difference creates an electric field in the direction from anode to cathode. Light of certain wavelength or frequency falling on the surface of cathode causes a current to flow in the external circuit. This current is called the photoelectric current. When the potential difference increases, the photo electric current also increases till saturation is reached. As the polarity of battery is reversed (i.e. plate Q is at negative potential w.r.t. plate P) the electrons start moving back towards the cathode. It is observed that at a particular negative potential of plate Q , no electron reaches the plate Q i.e. the current becomes zero. This negative potential for which the photo-electric current is zero is called the stopping potential denoted by V0 • Maximum kinetic energy (in eV) of photo electrons in terms of stopping potential will therefore be K_ =IV, I eV

J2 eV = 0

m

N

(Intensity) (Area) 0.53 x-(Energy of each photon) 100

N

(2)(1x10-<) x 0.53 = 6.2Sxl0 11 s-• (10.6xl.6x10·19 ) 100

t

Minimum kinetic energy of photoelectrons is 19

2(1.6xl0- )(2) 9.lxl0-:31

vmax =8.4x10 5 ms-1

K""' =0 and maximum kinetic energy is, K_ =E-W=(l0.6-5.6) eV=5 eV

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Advanced JEE Physics Illustration 13 A monochromatic light source of frequency illuminates a metallic surface and ejects photoelectrons. The photo electrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the entire experiment is repeated with an incident radiation of

~ f,

Bap

... (1)

V=-1!

V

=>

(2,5 X10" )(1.76X lQU )(2.7 X10-3 ) 3.14

v=0.38x10 6 ms-'

the photoelectrons so emitted are able to 6 excite the hydrogen atom beam which then emits a

Since, KE=.!..mv 2 =E-W

radiation of wavelength 1215 A . (a) What is the frequency of radiation. (b) Find the work function of the metal.

=>

frequency

2

W = _ _ 1 (9,1 X10-" )(0.38 X10 6 ) 49 2 1.6x10-19

Using Einstein's equation of photoelectric effect i.e.

Km~=hf-W

=> =>

where K= =13.6 eV

=>

So, when the experiment is repeated, then

t)-

12375 h(~ W= = 10.2 eV 6 1215 Solving equations (1) and (2), we get

... (2)

6

4.92x10" Hz

=>

W=4.5eV

Illustration 15 If the wavelength of the incident radiation is

increased from 3000 A to 3010 A, find the corresponding change in the stopping potential V .

Subtracting (2) from (1), we get

e(V, -V,) = (E, -E,)

W=6(3.4)-13.6

W=6.8eV

=>

V, - V,

Illustration 14

Solution

Pitch of helical path is

p =(vcos8)T = vT 2

where, T= 2mn = 21t 1tV

p=Ba

=

3.10

= he

(2-_2-)

e A.1

A. 2

8

A photon with an energy of 4.9 eV ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength B = 2.5 mT at an angle of 60° with the field direction, the maximum pitch of the helix described by the electron is found to be 2.7 mm . Find the work function of the metal in electronvolts. Given that specific charge of electron is 1.76 x10" Ckg-'.

qB

... (2)

eV,=E,-W

(b) From equation (1), we have W=hf-13.6

=>

W =(4.9-0.4) eV

Solution According to Einstein's Photo-electric equation, we have ... (1) eV1 =E1 -W

hf =3.4 eV (6)(3.4)(1.6x10-19) (6.63 X10-34 )

2

... (1)

h/-W=13.6eV

f

2

Substituting value of v from equation (1) in equation (2), we get

Solution

(a)

... (2)

W =E-.!.mv'

Ba

{·: 8 = 60°)

I·-- !l Cl=

V,-V,

6.6x10""' x3x10 (-1___ 1_)-o 012 V 1.6 X10-19 X10-lO 3000 3010 - .

Illustration 16

A light beam of wavelength 400 nm is incident on a metal of work function 2.2 eV . A particular electron absorbs a photon and makes 2 collisions before coming out of the metal (a) Assuming that 10% of extra energy is lost to the metal in each collision find the final kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions find the maximum number of collisions the electron should suffer before it becomes unable to come out of the metal.· Solution

(a) Since, E(in eV)

12375 1.(inA)

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Dual Natllre of Radiatio11 & Matter Current~)

E= 12375 =3.1 eV

4000 Energy of electron after first collision is E, =(90% of E)=2.79 eV

[·: 10% is lost}

Energy of-electron after second collision

Vs.JVs'2 ¼,

E, =(90% ofE,)=2.51 eV

0

(-JV

Hence, KE of this electron after emitting from the metal surface = (2.51-2.2) eV = 0.31 eV (b) Energy after third collision,

(+JV

Thus, the increase of frequency increases maximum kinetic energy of photoelectrons but leaves the photoelectric current

unchanged.

E, = (90% of E,) = 2.26 eV

Similarly, E, = (90% of E,) = 2.03 eV < W

C.

So, after four collisions the electron will not be able to

When frequency and intensity of incident light are kept fixed and photo-metal is changed, we observe that stopping potential (V,) versus frequency ( v) graphs are parallel

come out of the metal.

CHARACTERISTICS OF PHOTO ELECTRIC EFFECT

The following observations were made to study the effect of changes in various factors while studying the Photo Electric Effect.

EFFECT OF PHOTO-METAL

straight lines, cutting, frequency axis at different points. This shows that threshold frequency are different for different metals, the slope ( : ) for all the metals is same and hence universal constant.

A.

EFFECT OF INTENSITY

Since we know that

Intensity of light means the energy incident per second per unit area. For a given frequency, if intensity of incident light is increased, the photoelectric current increases and with decrease of intensity, the photoelechic current decreases, but

the stopping potential remains the same. In photoelectric effect current (;) is directly proportional to intensity (I) of incident light. Current (i)

eV,=hv-W

~ v, =(~}-(~) Comparing with the line y = mx + c, where m is the slope and c is the intercept on the y axis. Then, we observe that the graph is a straight line with slope

~ e

31

(a universal constant) and negative intercept

W

e

(depending on the nature of the metal).

21 I

In figure threshold frequency and work function are greater for Metal 2 as compared to Metal 1.

o~-----+

V,

Intensity (I)

Metal 1 This means that the intensity of incident light affects the photoelectric current but leaves the maximum kinetic energy

Metal2

of photoelectrons unchanged. B.

EFFECT OF FREQUENCY

When the intensity of incident light is kept fixed and frequency is increased, the photoelectric current remains the same but the stopping potential increases. If the frequency is decreased, the stopping potential decreases and at a particular frequency of incident light, the stopping potential becomes zero. This value of frequency of incident light for which the stopping potential is zero is called threshold frequency v0 • If the frequency of incident

D.

EFFECT OF TIME

There is no time lag between incidence of light and the emission of photo-electrons.

light ( v) is less than the threshold frequency (vo) , no photo-electric emission takes place. 3.11

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I PROBLEM SOLVING TRICK(S)

I j

FORMULAE FOR WORKING THE PROBLEMS ON PHOTO· ELECTRIC EFFECT ' Maximum Kln~tic Energy, of photo-electrons

EK = eV0 =Jmv!a,. If A- 0

is the threshold wavelength and

Let P be the power of a point. source of electromagnetic radiations, then intensity I at a distance r from the source is given by

I=..!_2 wm·2

v,

the threshold

1 frequency,

I

I ' I I

i Work function of photo-metal,

I.

DETERMINATION OF PHOTOELECTRIC CURRENT

41tr If A is the area of a metal surface on which radiations are incident, then the power received by the plate is

i

.

P'=IA=(_I'_,)A(W)

he 4io =hvo = -

4itr

i.,

Threshold frequency Is minimum frequency and Threshold wavelength Is maximum wavelength of incident light to cause photoelectric e~ect. Einst~in's ~hoto-electric Equation may be expressed as or EK=hv-$0

I·

_ ,

!

-.

I I

~=hc_hc

'

· cc;mdition for photoelectric emission is hv ~ 4>0 or equivalently

I

!v

~

.

-I

v 0 or equivalent y As; A0

p•

4,,~

,

Area (A)

•

GRAPH BETWEEN K_ and

V

Let us plot a graph between the maximum kinetic energy Km~ and the frequency of the falling photon v or the incident· light. According to Einstein's Photo-Electric equation, we have

hv=W+E,

=>

$=E= [_I'_,xA] .

~

1-

If v is the frequency of radiation, then the energy of photon is given by E=hv The number of photons incident ,on the pla~e per second (called photon flux) is given by

If v > v, (threshold frequency) and photon efficiency of the metal plate is. T\% , the_n the number of photoelectrons emitted per second is given by

-

K==hv-W

n=.h-[6xA]_B_ 100-

hv

100 ·

-.-, I I \

-r-

Finally, the photocurrent i is given by

Comparing with the line y = mx + c, where m is the slope and c is the intercept on the y axis. Then, we observe that the graph is a straight line with slope h (a universal constant) and negative intercept W (depending on the nature of the metal). For Metal 2, we observe that

W2 >W1 and hence (v,),>(v,),

v. Metal 1 Metal 2

i =ne where e is the charge of an electron (e = 1.6 x 10-19 C) Illustration 17 Light of wavelength 180 nm ejects photoelectrons from a plate of a metal whose work function is 2 eV . If a uniform magnetic field of 50 µT is applied parallel to the plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy. Solution ,

1,.=180 nm=18oo A 12375 E= =6.875 eV 1800 Since, Km~ = E - W = 4.875 eV Since, r = mv = ./2mK qB qB Also, 1we observe that when v = v0 , the threshold frequency, then, K=

=

=0

Substituting the values, we get r=

3.12

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Jz

X

4,875 xl,6 xlQ-l9 X 9,1 X 10-Jl 5x10-5 x1.6x10'=19

0.15 m=l5 cm

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Duai Nature of Radiation & Matter Illustration 18

! !' 1. I

j

2.

'I l

lI

i 3. I

Number of photoelectrons emitted per second is

n=(1~' )(f)(4:;,)

l

i 4.

1 )( 3.2xl0_, )( 1 ) II= ( lo' 5 X 1.6 X 10-19 4it X 0,8 X 0.8

:

(nxSxlO_, x Sx 10-3 ) =:>

(b)

Km~

n =10 5 sec-1

=E-W=2

I

eV

· 15.

Since, for an electron, we have :>..

~

= 2

I

A

150 KE(in eV)

0 :>..,=J1~ =s.66A

Further, Wavelength of incident photon is :>.., = 12375 = 2475 5

!' 6.

A

:>..,

(c)

Photoemission will stop when potential on the sphere becomes equal to the stopping potential.

(d)

K= = 2 eV0 • Therefore, the stopping potential V, is 2 volt. Let I be the desired time. Then

.,'

•

i '7.

I

V, =-l_!/.=__!!.!_= (nt)e 4ne0 r 41te0r 41te0 r V,r _ l )
2xsx10-3 9 9xlQ X 105 Xl.6 xlQ-l9

{Solutions on page 3.50} [

What will be the maximum kinetic energy of the l photoelectrons ejected from magnesium (for which the work , function W =3.7 eV) when irradiated by ultraviolet light of ·

\ a.

L--

1

I

The hydrogen· atom iii fts ground state is excited ·by m8ans of monochromatic radiation. Its resulting spectrum has six different lines. These radiations are incident on a metal plate. It iS observed that only two of them are responsible for photoelectric effect. If the ratio of maximum kinetic energy of photoelectro_ns in the two cases is 5 then find the work function of the metal. [Take ground state energy of H -atom = -13.6 eV ]. A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of 5000 A , no photoelectrons are emitted from the surface. With an unknown wavelength, stopping potential of 3 V is necessary to ·eliminate the photocurrent. Find the unknown

wavelength. A light source, emitting three wavelengths

5000

A,

6000 A and 7000 A , has a total power of 1er' W and a beam diameter of 2 mm . The power density is distributed equally amongst the three wavelengths. The beam shines ·normally on a metallic surface of area on 10-4 m2 and having a work function of 1.9 eV . Assuming that each photon liberates an electron, calculate the charge emitted per second from the metal surface.

! : : . \

I '· ; , '

j 1

1

' 1

, , :

A beam of light consists of four wavelength 4000 A , !

6000 A and 7000 A , each of intensity i The beam falls normally on an area 1

wm-2 •

10'-4 m2 of a clean metallic surface of work function [ · 1.9 eV . Assuming no loss of light energy calculate the , number of photoelectrons liberated per second. i In . an experiment on photo electric emission, following 1 observations were made : {i) Wavelength of the incident light = 1.98 x 10-1 m

(ii)

I

i

'

Determine the number of photons emitted from the source 1 per second and the number incident on unit area of the magnesium surface per second. The photoelectric work : functlo_n_ for magnesium is 3.68 eV .. Calculate the kinetic 1 energy of the fastest electrons ejected from the surface. Determine the maximum wavelength for which the , photoelectric effect can be·observed with a Mg surface.

Stopping potential

Find the (a) (b) (c)

llls

•

A 40 W ultraviolet light source of wavelength 2480 A illuminates a magnesium (Mg) surface placed 2 m away.

4800 A , 1.5 x 1o-:3

!':!_ ~ 286

t=

_:_IC'--E=llc.>cc:·,.··~·~,.>~;='_'_. BASED ON PHOTOELECTRIC EFFECT :

frequency 1.5 x :1015 sec-1

Solution

(a)

---"---".'""'"tJ

•-~~=·'"":,i'--.>-"'x/
In a photoelectric effect set-up, a point source of light of power 3.2 x 10-3 W emits mono energetic photons of energy 5 eV . The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3 eV and of radius 8 x 10-a m. The efficiertcy of photoelectron emission is one for every 106 incident photons. Assume that the sphere is isolated and electrons are instantly swept away after emission. (a) Calculate the number of photoelectrons emitted per second. (b) Find the ratio of the wavelength of incident light to the de Broglie wavelength of the fastest photoelectrons emitted. (c) It is observed that the photoelectron emission stops at a certain time t after the light source is switched on. Why? (d) Evaluate the time t.

=2.5 V .

threshold frequency. work function and energy of photo electrons with maximum speed.

Radiation of wavelength 5461 A falls on a photo cathode / ~q_~~ct~~n~ .!"i!h -~ !!1~1'!!ury,_·~ine!i~ ~e~gy of _!U ~ ~v_ j

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Advanced JEE Physics are emitted. When radiation of wavelength 1849 A falls on the same surface a (negative),potential of 4.6 V has to be applied to the collector electrode to reduce the photoelectric current Is zero. Find the value,of h 8.nd cutoff wavelength. 9.

Illuminating the surface of a certain metal alternately with light of wavelengths l., = 0.35 µm and l. 2 = 0.54 µm , It

11.

The photoelectric work function of potassium is 2.3 ei7:7f light having a wavelength of 2800 A falls on potassium, find c(a) the. kinetic energy in electron volt of the mqst energetic electrons ejected. (b) the stopping potential in volt.

was found that the corresponding maximum velocities of photo electrons differ by a factor ~ =2. Find the work function of that meta1.

1O.

=

When a sunace is irradiated with light of l. =4950 A a photocurrent appears.which vanishes if a retarding potential 0.6 V is applied. When a different source of light is used It is found that critical retarding potential Is changed to 1.1 V . Find the work function of emitting surface and wavelength of second source. If photoelectrons after emission from surface are subjected to a magnetic field Ot 1OT , what · changes will be observed in the above two retarding potentials?

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Problem 1

3.552x10-19

A mercury arc lamp provides 0.2 watt of ultraviolet radiation at a wavelength of A. = 2537 A . Assume no other wavelength to be present. The cathode of photoelectric device consists of potassium and has an effective area of 4 cm2 • The arc lamp is at a distance of 1 m from the cathode. Given that work function for potassium is W = 2.22 eV. (a) According to classical theory, what time of exposure to the radiation should be required for a potassium atom (radius 2 A) to accumulate sufficient energy to eject a · photoelectron? (b) What is the energy of a single photon from the source? (c) What is the flux of photons (number per second) at the cathode? To what saturation current does this flux correspond if the photo conversion efficiency is 10%. Photo conversion efficiency is the probability of a photon being successful is knocking out an electron. (d) Find the cut-off potential.

20

( ~-: )(4it X 10. E 12375 V (b) Smce, ,_ (in A) e

=> (c)

As per the classical theory, energy flow is a continuous process and a photoelectron will be ejected if a potassium atom receives this amount of energy over a length of time. The potassium atom is at a distance of 1 m from the source. Hence, intensity of ultraviolet radiation on the potassium surface is, 02 I=_J:_=~= · wattm-' 2 2 4itr 4it(1) 4it

Cross-sectional area of the potassium atom is, 2

A=it{2x10-10 ) =4itx10-20 m 2 Hence, the exposure time is given by,

12375 = 4.87 eV 2537

Let N be the number of photons reaching the cathode per second. Then intensity at cathode is, But energy falling per second on the cathode is (~: )(4x10_,), so

N =>

0.2x4x10_, 4itE

0.2x4x10_, 4itx4.87xl.6x10-19

N = 8.12 x 1012 photon sec-1

Since, 10% of these photons are able to eject electrons.

=>

Energy required to just eject a photoelectron from the potassium surface is equal to the work function, i.e., 2.22 eV = 3.552 x 10-19 J

E=

I'=NxE

Solution

(a)

)

(d)

i=0.lxNxl.6xl0-19 A=65nA

K= =E-W=(4.87-2.22) eV=2.65 eV

=> Stopping potential = 2.65 V Problem 2

In a photocell the plates P and Q have a separation of 10 cm , which are connected through a galvanometer without any cell. Bi-chromatic light of wavelengths 4000 A and 6000 A are incident on plate Q whose work function is 2.39 eV . If a uniform magnetic field B exists parallel to the plates, find the minimum value of B for which the galvanometer shows zero deflection. Solution

Energy of photons corresponding to light of wavelength ,., = 4000 A is

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Solution (a) Intensity of light at a distance 0.8 m from the source

£i =·12375 =3.1 eV 4000 and that corresponding to A2 = 6000 A is,

= (3.2 x 10-" Js·1 ) = x 104 watt 4 4it(0.8)2 m 2 m'

1

E = 12375 =2.06 eV 2 "6000

Energy incident on the metallic sphere in wtlt time is 2 E =1t(8xl0·') (4x104 )=8.04x10-' W

Given that the work function for the metal is W = 2.39 eV,

1

so we observe that

Energy of one single photon is E2 = 5xl.6 x10·19 J = 8x10·19

E,<Wand£i>W Hence photoelectric emission is possible with A1 only.

J

Therefore, total number of photons incident on the sphere per second is

Photoelectrons experience magnetic force and move along a

circular path.

n = 5_ 1 E2

8.04 X lQ-' = 1011 8x10-19

Since, the efficiency of photoelectric emission is one for every 10 6 • Hence, total number of photoelectrons per second is n

2

The galvanometer will indicate zero deflection when the photoelectrons complete semi-circular path before reaching the plate P. Thus, d=2r =10 cm

=>

10 10 (b) Maxi":um kinetic energy of photoelectrons K==E-W=2eV So, stopping potential V0 = 2 V

· Total positive charge on·the sphere after time t is,

r=5cm=0.05m mv

.J2Km

Bq

Bq

q = (n2 e) t = (10 5 )(1.6 x 10·")t

Further r = - = - =>

B .

-

=>

= .J2Km

4ni:; 0r

Photoemission will stop when this potential becomes the stopping potential, so

Substituting the values, we have

1 {1.6x 10·14 ) t

B. =·.J2x0.71xl.6xl0-19 x9.109x10"1 (0.05)(1.6x10-19 )

2

(9 X 10 9 Hl.6 X 10·14 )t

B""' = 5.68 x 10·5 Tesla

3.16

r

41tE 0

(BxlO")

Problem 3 In a photoelectric effect setup, a point source of light of power 3.2 x 10-3 W emits mono energetic photons of energy 5 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3 eV and of radius 8 x 10-3 • The efficiency of photoelectron ·emission is one for every 10 6 incident photons. Sphere is initially neutral and that the photoelectrons are instantly swept away after emission. (a) Calculate the number of photoelectrons emitted per second. (b) It is observed that the photoelectrons emission stops at a certain time t after the light source is switched on. Evaluate time t.

=

q=(l.6x10·")t

Potential on the sphere V = _q_

rq

where K=E1 -W=(3.1-2.39)=0.71 eV

=>

= 1 6 = 10116 = 10s

=>

2

1=111 s

Problem 4 When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, TA expressed in eV and de-Broglie wavelength A. A • The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is T8 =(T. -1.50 eV). H the de-Broglie wavelength of these photoelectrons is A. 8 = 2AA, then find

(a)

the work function w. of metal A and the work function W8 of metal B .

(b) the maximum kinetic energy TA of the electrons ejected from metal A.

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Solution

3.6x10.., Wm-' equally distributed amongst the three

K==E-W Therefore, TA ~ 4.25- WA

T,

... (1)

=(TA -1.50)=4.70-W,

Equations _(1) and (2) gives, W,-W, =1.95 eV

... (2)

that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons.liberated in two seconds.

... (3)

Solution . E('meV) = 12375 Smce, ( •) 1- inA

de-Broglie wavelength is given by h 1-=~..J2Km

=>

1 1-cc-

=>

~: =

=>

2=~. TA TA -1.5

wavelengths. The beam falls normally on an area 1 cm2 of a clean metallic surface of work function 2.3 eV . Assume

K = KE of electron

.fi.

ffe,.

=> TA =2 eV From equation (1 ), we get W, =4.25-TA =2.25 eV

Energy of photon having wavelength 4144 A 12375 E1 = eV = 2.99 eV 4144 Similarly, 12375 . E, = eV = 2.49 eV and 4972 12375 E, = eV =1.99 eV 6216 Since, only E, and E, are greater than the work function W = 2.3 eV, only first two wavelengths are capable for ejecting photoelectrons. · Given intensity is equally distributed in all wavelengths. Therefore, intensity

From equation (3), we get

co'rresponding to each wavelength~

W, = WA + 1.95 eV = (2.25 + 1.95)

=>

W, =4.20 eV

=>

T,=4.70-W,=4.70-4.20=0.50 eV

3.6x10."'

1.2x10"'

w~-2

3 or energy incident, per second in the given area

(A=l cm' =104 m') is

I =1.2x10"' xl0 4 =,1.2xl0-7 Js-1

Problem 5

Ultraviolet light of wavelengths 800 A and 700 A when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy 1.8 eV and 4.0 eV respectively. Find the value of Planck's constant.

Let '1t be the number of photons incident per unit time in the given area corresponding to first wavelength. Then,

I E1

1.2xl0-7 2.99xl.6xl0-

n1 = - = - - - - - 2.5x10 11 19

Similarly,

Solution When 800 A wavelength falls on hydrogen atom (in ground state) 13.6 eV energy is used in liberating the electron. The rest is given as kinetic energy to the electron.

Hence, K=E-13.6(ineV)

n,

I

E,

' 1.2x10-' 2.49 X 1.6 X 10-!9

3x1011

Since, each energetically capable photon ejects electron, hence total number of photoelectrons liberated in 2 s is N=t(11t+n,)

(1.8xl.6x10-")

he

10

13.6xl.6x10-"

B00xl0Similarly for the second wavelength, we. have (4xl.6x10-19 ) .

he

700x10-10

13.6xl.6xl0-19

... (1)

... (2)

Solving these two eqtiations, we get

=>

N =2(n1 +n,)

=> =>

N =2(2.5+3)x10 11 N=l.lxl0 11

Problem 7 A monochromatic

h=6.6x10""' Js

point

source

S

radiating

wavelength 6000 A , wlth power 2 watt, an aperture A of Problem 6

diameter 0.1 m and ·a large screen SC are placed as shown

A beam of light has three wavelengths 4144 A, 4972 A and 6216 A with a total intensity of

in figure. A photoemissive detector D of surface area 0.5 cm2 is placed at the centre of the screen. The efficiency

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of the detector for the photoelectron generation per incident photon is 0.9. SC

=>

This aperture ~ become new source of light. Now these photons are further distributed in all directions. Hence, at the location of the detector, photons incident per unit area per unit time is

A.

L ,~----------'- "

n

4

=>

0.6m

n3 41t(6-0.6)

Calculate the photon flux at the centre of the screen and the photocurrent in the detector. (b) If the concave lens L of focal length 0._6 m is inserted in the aperture as shown, find the new values of photon flux and photocurrent. Assume a uniform

1.052x 1016 2 4,r(S.4)

n4~=2.87x10 13 m-2s-1

total number of photons incident on the detector per unit time is

n, = (0.5x10-<)(2.87xl013 d) = l.435x10' s-1 The efficiency of photoelectron generation is 0.9. Hence, total photoelectrons generated per unit time is

average transmission of 80% from the lens. (c) If the work function of the photoemissive surface is

n, = 0.9n5 = 1.2915 x 10' s-1

1 eV , calculate the values of the stopping potential in the two cases (without and with the lens in the aperture). Solution

Hence, photocurrent in the detector is

i = (e)n, = {l.6xl0-19 ){1.2915x109 ) = 2.07x10-10 A (b) Using the lens formula, we get

Energy of one photon,

1

A.

E=3.3x10-1'

1 --0.6

1 --0.6

----=--

(6.6x10-34 )(3x10') 6000 X 10-tO

E he

v

=> V =--0.3 m i.e., image of source (say S', is formed at 0.3 rn) from the lens.

J

s

s· '' ''' ·''

Screen

nl

2 = 6.06 X 1018 3.3 X lQ-19

transmits to second medium, therefore, at a distance of

5.7 m from S' number of photons incident per unit

S-I

area per unit time will be

At distance 0.6 m, number of photons incident per unit

( ~)(1.os2x10 100

n, = 4>,(0.6)

2

=>

Area of aperture is, 2

2

S1 =~d =~(0.1) =7.85x10°" m

0.3m

Total number of photons incident per unit time on the lens are still n3 or 1.052xl016 s-1 • Since, 80% of it

area per unit time is

n,

L

----5.7 m----+1

Power of the source is 2 W = 2 Js-1 . Therefore, number of photons emitting per second is

(4,r)(S.7)2

n, =2.06x1013 m-2 s-1

i' = (2.06 x10 1')(0.s x 10-< )(0.9)(1.6 x 10-1' ) =>

the aperture,

n, =n2 S1 =(1.34x10 18 )(7.85x10°") s-1

1 ')

This is the photon flux at the detector. New value of photocurrent is given by

2

4 4 So, total number of photons incident per unit time on

~ 3.18

2

Area of detector is 0.5 cm2 or 0.5 x 10-4 m 2 . Therefore,

(a)

=>

=

This is the photon flux at the centre of the screen.

6m _ _ _.,

(a)

n3 = 1.052 x 1016 s-1

(c)

i' =1.483 X 10-10 A

Energy of incident photons (in both the cases) is

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Dual Nature of Radiation & Matter E(in eV)

Problem 9

12375 ,.(in A)

1237 => E = ~ = 2.06 eV 6000A Work function W = 1 eV Maximum kinetic energy of photoelectrons in both cases,

When a beam of 10.6 eV photons of intensity 2 wm-1 falls on a platinum surface of area lxl0-1 m 2 and work function 5.6 eV. 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies

(in eV). Take 1 eV =1.6 x10·19 J. Solution

K_ =E-W=l.06 eV

Energy of incident photons, E; =10:6 eV

or the stopping potential will be 1.06 V . Problem 8

=>

E; = 10.6 X 1.6 X 10-19 J

Assume that the de·Broglie wave associated with an electron can form a standing wave between the atoms

=>

E;

=16.96 x 10-19 J

arranged in a one dimensional array with nodes at each of

Energy incident per unit area per unit time (intensity)= 2 J

the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 A . A similar standing wave is again formed if d is increased to 2.5 A but not for any intermediate value of d. Find the energy of the electron in e V and the least value of d for which the standing wave of the type described above

So, number of photons incident on unit area in unit time is n 2 1.18xl018 16.96 X 10-!9 A Therefore, number of photons incident per unit time on given area (1 x 10-4 m 2 ) is

can form.

Solution

From the figure it is clear that 1+---2A--+

So, p{~)=2A and

=> n =1.18x1014 But only 0.53 % of incident photons emit photoelectrons, so number of photoelectrons emitted per second (n) is

N~ N

,. . (p+l)·-=2.5A

p-loops

2

=>

n = (1.18x lO")(lxl0-1)

"'

=>

n = 6.25 X 1011 K.,.,

!:'.=(2.5-2) A=o.sA 2

n = (0.53)(1.18xl0") 100

(p+1) loops

,__ _ 2.s A - - -

=> '-=1A=10-" m de-Broglie wavelength is given by

,_='!_=_h_ p ../2mK

=0

and Kinax = Ei - W =>

K_ =(10.6-5.6) eV =5 eV

=>

K-=5eV

and Kmm =0

,,,

where K is the kinetic energy of electron

=>

K=--

=>

K

=>

K = 2.415 x 10-17

=>

K =(2.415x10·") eV 1.6x10·"

Problem 10 A small plate of a metal having work function of 1.17 eV is placed at a distance of 2 m from a monochromatic light source of wave length 4.8 x 10-7 m and power 1 W. The light falls normally on the plate. Find the number of photons striking the metal plate per m 2 per sec . If a constant · uniform magnetic field of

2mA2

2

(6.63 X 10·" )

2(9.1x10-31 )(10-rn)'

J

=> K=150.8 eV The least value of d will be, when only one loop is formed. So, we have

,.

d. == 2

=>

dmin

=o.5 A

strength 10-1 T is applied parallel to the metal surface, find the radius of the largest circular path followed by the emitted photo electrons. Given /1 = 6.6 x 10""' Js, 6 1 19 c=3x10 ms- , e=l.6x10- C and electron mass m=9.lx10"31 kg. Solution

Energy of photons of wavelength 4.8 x 10-19

J is 3.19

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6.6xl0-" x3xl0' 4.8x10-7

E he

;,

4.125xl0-19 J

Power of source = 1 W = 1 Js-1 So1 rate of emission of photons from the source is 1 J5 -1 n 2.424x1018 s-1 19

4.125 X 10-

J

These photons move in all directions randomly. At a distance r from the source, the photons fall normally over a

spherical surface of area 41tr 2 • The plate is at a distance r = 2 m . Hence the number of photons striking the surface per m 2 per second is

2.424xl0"' 4.82xl0" 4x3.14x(2) The maximum KE of a photoelectron emitted from the plate n

(b) the magnitude of the electric field between the plates A and Bat t=l0s and (c) The kinetic energy of the most energetic photoelectrons emitted at t = 10 s when it reaches plate B. Neglect the time taken by the photoelectron to reach plate B. (Take e0 =8.85x10-u C'N-m-2 )

Solution Area of plates A= 5 x 10 4 m 2 Distance between the plates d = 1 cm = 10-2 m (a) Number of photoelectrons emitted upto t = 10 s are . ) number of photons falling) ( ( in unit area in unit time x area x time n=

10'

is

~ n =~[(10) x(Sx10 4 )x(10)]

he

16

Kmax=T-Wo

~

10

K- = 4.125 X 10-19 -1.17 X 1.6 X 10-19 = 2.253 X 10:,,

J

Hence the maximum velocity of the photoelectron is V

=

max

-

~

n=5x107 (b) At time I= 10 s , charge on plate A

q, =+ne=(5x10')(1.6x10-19 )

/2IC: V~.

o 19

and charge on plate B,

2 X 2.153 X 10- = 7_03 X lQS ffiS-t 9.lxl0-31 Radius of the largest circular path of the photoelectrons in the magnetic field is ~

V

q, =8x10-12 C

=

q8 = (33.7x10-12 -8x10-12 ) ~

q8 =25.7x10-12 C

Since, electric field between the plates is

mVm~ 9.1 X 10-31 X 7.036 X 105 r = - - = - - - - ~19- ~ -4 ' - 4xl0-' m = 4 cm eB 1.6 x 10- x 10-

E

(q,-q,) 2Ae,

Problem 11 T~o metallic plates A and B each of area 5 x 10-4 m 2 , are placed parallel to each other at separation of 1 cm. Plate B carries a positive charge of 33.7 x 10-12 C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on place A at t = 0 so that 106 photons fall on it per square meter per second. Assume that the photoelectrort is emitted for every 10 6 incident photons. Also assume that all the· emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV .

(a)

~

(c)

E

(25.7 -8) X 10-t2 2 X (5 X 10-4 )(8.85 X 10-t2)

---,--C----c-c---= 2x103 NC-1

Energy of photoelectrons at plate A is K=E-W=(S-2) eV=3eV Increase in energy of photoelectrons is (eEd) joule= (Ed) eV ~

(!ncreaseinEnergy)=(2x10')(10-2 ) eV=20 eV

Energy of photoelectrons at plate B is (20+3) eV =23 eV

the number of photoelectrons emitted up to t = 10 s

3.20

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This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1.

Consider a source emitting 100 W of green light at a

5.

wavelength of 500 nm . The number of photons emerging from source per second is (A) 2.5 x 1019 photon per second (B) 25 x 10 20 photon per second (C) 25 x 10 19 photon per second (D) 25 x 1017 photon per second

Which of the following graphs represents the variation of particle momentum and the associated de Broglie wavelength? p

(A) 2.

The distance d of a 100 W lamp is continuously

increased from a photocell. The photoelectric current I varies with distance d as (A) I a: d2 (B) I a: d 1 1 (D) I a:(C) I a: d' d 3.

When a metallic surface is illuminated with monochromatic· light of wavelength 1,,, the stopping potential is 5V0 • When the same surface is illuminated

(Q

6.

(C) 4.

he

61,, he

41,,

(B) (D)

~ 51,,

(C)

1..(:) 1..( ~)

(B)

p

If Planck's constant is denoted by /z and the electronic charge by e, experiments on photoelectric effect allow the determination of (A) only h (B) only e h (C) both h and e (D) only -

e

2hc

41,,

7.

The maximum energy of the electrons released in a photocell is independent of (A) frequency of incident light (B) intensity of incident light (C) nature of cathode rays (D) None of these

8.

If the energy and wavelength of electron are E and A, then the graph between logE and log),, will be

An electron of mass m , when accelerated through a potential difference V has de-Broglie wavelength A.. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference will be (A)

0) ~ >

L. ~,LL. p

with light of wavelength 31,, , the stopping potential is V0 . Then the work function of the metallic surface is

(A)

IL_,

p

(D) 3.21

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log l.

log l.

(A).·~

2x10·' T.

11

Ckg·1 , then

the diameter of the circle is nearly (A) 1.1 m (B) 1.1 mm (C) 1.1 cm (D) 11 cm

e,

14. The radiation emitted, when an electron jumps from n = 3 to n = 2 orbit is a hydrogen atom, falls on a metal to produce photoelectron. 'IJ1e electrons from the metal surface with maximum kinetic energy are made to

log E

log E log l.

If(;.) of electron is 1.76xl0

log l.

move perpendicular to a magnetic field of _!_ T in a · 320 radius of 10-3 m. The work function of metal is

(D)

(A) 1.03 eV (C) 0.86 eV log E

log E 9.

15. The radiation force experienced by a body exposed to radiation of intensity I assuming surface of body to be perfectly absorbing is

-

The total energy E of a sub-atomic particle of rest mass mo moving at non-relativistic speed v is

(A) E=m0c2 (C)

(B)

1 0V 2 E = m0 c' +-m 2

(B) 1.89 eV (D) 2.03 eV

·=~ -

E= 1 m0 v'

2

1 (D) E = 1"0C2 -2mov2 (A)

10. An electron is 2000 times lighter than a proton. Both are moving such that their matter waves have a length of 1 A . The ratio of their kinetic energy in approximation is (B) 1: 2000 (A) 1 :1 (D) 1:200 (C) 2000 :1

1tR'I .

(B)

27tR 2I C

C

(C)

41tR'I

(D) None of these

C

16. A radiation of energy E falls normally on a perfectly reflecting surface. The change in momentum of radiation is

11. The de Broglie wavelength of a particle is approximately the same as that of a photon with the

(A)

E

(B)

same energy.

(C)

(A) The energy of the particle is much greater than its

Ee

2E C

C

(D)

~

rest energy.

(B) The energy of the particle is much less than its rest

17. An electron, accelerated by a potential difference V, has de Broglie wavelength ).. . the electron is accelerated by a potential difference 4V, its de Broglie wavelength will be (B) 4).. (A) 2)..

If

energy.

(C) The energy of the particle equals its rest energy. (D) Data insufficient to arrive at a conclusion.

12. Two lumps of clay each of rest mass m0 , collide with a

(C)

4

speed of -c head on and stick together. The mass of 5

the composite lump thus formed is 10 5 (A) m, (B) m,

3

3

5

13. An electron moving with velocity 2x107 ms-1 describes·. a circle in a magnetic field of strength

3.22

2

(D)

A. 4

18. The photoelectric work function of a.metal is 1 eV. Light of wavelength ).. = 3000 A falls on it. The

(D) -m 12 °

=

2:.

·

photoelectrons will come out with approximate speed equal to (A) 10 ms·1 (B) 10 2 ms·1 4 1 (C) 10 ms· (D) · 106 ms·1

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19. A proton and an a -particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths is 1 (A) .fi. (B) .fi.

(C)

2.fi.

(D) 2

20. In an electron microscope if the potential is increased from 20 kV to 80 kV, the resolving power R of the microscope will become (A) R (B) 2R

(D)

(C) 4R

5

15

(D)

10

B10 to

5

(Neglect any thermal effect and friction is absent)

(A) E_nR2

me (C) ,. -> infinity

(B)

10 11

(C) N = f...t 0 E!

.

2h

N

(D) Data Insufficient

a

(A)

8nhce0d

(B)

8nhce0 d

27. The de-Broglie wave present in the fifth Bohr orbit is

(A) I \

.f\

v

I\

I\

(B)~

\ v/ \v/ \~0 v/v

.(D) / \

25. A small metal plate (work-function W0 ) is kept at distance d from a singly ionised fixed ion. A monochromatic light beam is incident on the metal° plate and photoelectrons emitted. The maximum wavelength of light so that the photo-electrons may go round the ions along a circle is 81te0 W0 d+e2

(D) None of these

C

\ - ~/ \v/~

= 2J.-eh0E!

are

_I_nR'

(C) / \

A=!!_

p (D) Data Insufficient

(B).

h

(B)

KC

faR'

(C)

24. A desklamp illuminates a desk top with light of wavelength ").. . The amplitude of this electromagnetic wave is E0 • Assuming illumination to be noi-mally o~ the surface, the number of photons striking the desk per second per unit area N is

(A) N =J.-e,E!

•

KC

23. For a photon, the de Broglie relation is given by

,.=.!!__

""'~~~---"-------"'

B11 in

22. In photoelectric emission the number of electrons ejected per second is proportional to the (A) intensity of light (B) wavelength oflight (C) frequency of light .(D) work function of the material

(A)

hceW,

=. r-~

B10 and 5B11 . If the atomic 5

8,rn0d

26. Aperfectly reflecting body sphere in shape of radius R is placed on a path of parallel light beam of intensity I shown in figure. One end of a spring is attached to centre of sphere and the other end to a rigid wall as shown in figure. Assuming the sphere to be in equilibrium, the spring constant of spring to be K , then compression in spring is

R

weight of boron is 10.81, the ratio of nature is 20 (A) 19 (B) 81 53

(D)

n

2

21. Boron has two- isotopes

(C)

2(.!!E..-•') w,

(C)

28. The interatomic distance between atoms in a crystal is 2.8-A, Then if such a crystal is used in DavissonGermer experiment, the maximum order of diffraction that can be observed for a beam of electrons accelerated by 100 V shall be (A) n=l (B) n=2 (C) n=10 '(D) n->oo

29. The energy of a photon of wavelength J.- is he (B) (A) he").. J.J.(D) hJ.(C) he c 30. The work function of a metal in 4 eV. Foi- the emission of photoelectrons of zero velocity from the metal surface, the wavelength of the incident radiation should be 3.23

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(A) 1700A

(B). 2700 A

(CJ 3100A

(DJ 5900 A

(C) less than vl//4

3

31. Photoelectric effect is the phenomenon in which

(A) photons come out of a metal when it is hit by· a beam of electrons ·(B) photons come out of the nucleus of an atom·under the action of an electric field. (C) electrons come out of a metal with a constant velocity which depends on the frequency and intellSity of incident radiation . (D) electrons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light and not On its inte_nsity.

l

38. The de-Broglie wavelength of a molecule of thermal energy k,T ( k, =Boltzmann constant and T = absolute temperature), is

.(A) l ~ = 2mk8T

(C) h~2mk8T

(B) (D)

h ~2mk8T It

4m 2 kiT 2

39. A proton is accelerated through a potential V . The de Broglie wavelength associated with it is (A) 12.27 A (B) 0.287 A

.Jv

32. The photoelectric effect is the ejection of electrons from the surface of a metal when (A) it is heated to a high temperature.

(B) electrons of suitable velocity strike it. (C) radiation of suitable wavelength falls on it. (D) it is placed in a strong electric field.

(D) greater than v

(C)

12.27 frn

.Jv

.Jv

(D)

0.287 frn

.Jv

40. In Davisson-Germer experiment Ni crystal acts as

(A) (B) (C) · (D)

33. Matter waves are

(A) electromagnetic waves. (B) transverse mechanical waves. (C) longitudinal mechanical waves.

an ideal reflector three dimensional diffraction grating an ideal absorber two dimensional diffraction grating

41. The angle between the incident and the diffracted electron in the Davisson-:Germer,experiment is called as

(D) neither electromagnetic nor mechanical waves.

(A) (B) (C) (D)

34. A photon of wavelength 1000 A has energy 12.3 eV. If light of wavelength 5000 A , having intensity I., falls

angle of incidence angle of diffraction angle of scattering none of the above

on a metal surface, the saturation current is 0.40 µA

and· the stopping potential is 1.36 V. The work function of the metal is (A) 2.47eV (B) 1.36 eV (C) 1.10 eV (D) 0.43 eV 35. In PROBLEM 34, if the intensity of light is made 4I, the stopping potential will become (A) 1.36 V (B) 2.72 V (C) 5.44 V . (D) 21.76 V 36. In PROBLEM 34, if the intensity of light is made 4I , the saturation current will become (A) 0.4 µA (B) 0.8 µA (C) 1.6 µA (D) 6.4 µA 37. In a photoemissive cell with exciting wavelength l, the fastest electron has a speed v . If the exciting 3 wavelength is changed to l , the speed of the fastest 4

emitted electrons will be

.

/3

(A) v\{4

=

.

.

· · 42.

In Davisson-Germer experiment maximum intens_ity is observed at "-, (A) ·50° and 54 volt (B) 54° and 50 volt (C) 50° and 50 volt (D) 65° and 50 volt

43. In Davisson-Germer experiment, an electron beam of

60 eV energy falls normally to the surface of the crystal and maximum intensity is obtained at an angle of 60' to the direction of incident beam. The inter-atomic distance in the lattice plane of the crystal is (A) 18 A (B) 3.6 A (C) 1:8 A (D) 0.18 A 44. The incorrect statement in connection with Davisson and Germer experiment is · . (A) The inter-atomic distance in nickel crystal is of the order of the de-Broglie wavelength. (B) Electrons of constant energy are obtained by the electron gun. (C) Nickel crystal acts · as a three dimensional diffracting grating. (D) Davisson-Germei- experiment is an interference experiment.

3.24

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Dual Nature of Radiation & Matter 45. In Davisson-Germer experiment the relation between Bragg's angle ~ and diffraction angle 0 is

(A) 0=90'-$

(B)

90°-$ 0=--

(C)

(D)

~~ --(1802'-0.)

0=180°-$

2

46. The ionization chamber used is Davisson-Germer experiment, acts as (A) emitter (B) collector (C) source (D) radiator 47. The distance between two consecutive atoms of the crystal lattice is 1.227 A. The maximum order of diffraction of electrons accelerated through 10 4 V will be 1 (A) 10 (B) 10

(C) 100

(D)

1 100

48. The human eye can barely detect a yellow light

(6000 A) that delivers 1.7 x 10-" watt to the retina. Nearly how many photons per second does the retina receive? (A) 50 (B) 5 (C) 500 (D) More than 5 million 49. The ratio of the specific charge of a proton to that of an CI-particle is

(A) 1:4 (C) 4 :1

(B) 1:2 (D) 2:1

SO. A photosensitive surface is receiving light of wavelength 5000 A at the rate of 10-7 Js-1 • The number 2.Sxl011 (D) 2.Sx.109 (B)

(A)

(C)

~

(B)

..J:mE .Jij

(D)

h 2mE

2mE

54. Two· photons of energy 2.5 eV each are incident on a metal plate whose work function is 4.0 eV, then the number of electrons emitted from the metal surface will be (A) one (B) two (C) None of these (D) more than two 55. Of the following moving with same momentum, the

one which has largest wavelength is an electron. a proton. an CI-particle. all have same de-Broglie wavelength.

(A) (B) (C) (D)

56. The maximum velocity of an electron emitted by light

of wavelength ). incident on the surface of a metal of ._ work-function$ is

(A)

2(hc+1'.$) m).

(B)

(C)

2(h:~).$)

(D) J2(h~-$)

2{hc-1'.$) m

57. An CI-particle is accelerated through a potential

51. Photons of frequency v fall on a metal surface for which the threshold frequency is v O • Then,

(A) all ejected electrons have the same kinetic energy h(v-v,).

(B) the ejected electrons have a distribution of kinetic energy from zero to h ( v - v0 )

53. The de Broglie wavelength of a particle of mass m moving with a kinetic energy E is

where h =Planck's constant, m = mass of electron and· c = speed of light ·

of photons received pe,r second is

(A) 2.5x10 12 (C) 2.5x10 10

2.5 eV . The stopping potential for radiation of wavelength 1500 A will be (A) 2.5V (B) 5.0 V (C) less than 5.0 V but more than 2.5 V (D) more than 5.0 V .

•

(C) the most energetic electrons have kinetic energy hv (D) the average kinetic energy of ejected electrons is hv0 •

difference of 200 V. The increase in its kinetic energy in electron volt will be · (A) 100 eV (B) 200 eV (C) 400 eV (D) 800 eV 58. Figure represents the graph of photo current i versus applied voltage ( V) . The maximum energy of the emitted photoelectrons is

52. When radiation of wavelength 3000 A is incident on a photosensitive ~urface, the kinetic energy of electrons is

-4 -3

-2 -1

0

1

2

3

4

~

==================================== = 3.25

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Advanced JEE Physics (A) 2eV (C) 0eV

(B) 4 eV

(D)

(A)

4J

Jt_ '"' iL Intensity (!)

(C)

~lL_ ~ ' ~

65. Two electrons are moving with the same speed v . One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field, then after sometime if the de-Broglie wavelengths of the two are 11. 1 and A. 2 , then

66. An electromagnetic radiation of wavelength i. has the same moffientum as an electron moving with a speed

~

2x10 5 ms-1 .

()

1. = 2.64 nm (C) i. = 3.64 nm

(A)

Intensity (I)

(C) 6.6 X 10--U J

6.6 x 10·"

J

26

J

(D) _!_xl0 6.6

(C)

and average

(C) 1.38xl017 kg

(A) Rest mass of a photon is zero

(B) Momentum of a photon of frequency v is hv

(D) 13.8xl0 20 kg

is scattered

straight backward. if the speed of electron after collision is ac , where a«: 1 then, (A) electron's kinetic energy is a fraction a of photon's initial energy. 1 of (B) electron's kinetic energy is a fraction a photon's initial energy. , (C) electron s kinetic energy is a fraction a' of

(C) Energy of a photon of frequency v is hv (D) Photons exert no pressure 63. A parallel beam of uniform, monochromatic light of 2640 A has an intensity of wavelength 100 wm·2 • The number of photons in 1 mm' of this

1

(B). 335 (D) 555

Consider a sphere of radius R exposed to radiation of intensity I as shown in

figure. If surface of sphere is partially reflection and reflection coefficient is 0.3, then radiation force experienced is 3.26

earth. distance is 1.5 X 10 8 km , The

69. A photon strikes a free electron at rest and

C

=

Sun

yearly loss in the mass of the sun is -(A) 13.8xl017 kg (B) 1.38x1019 kg

incorrect?

64.

(D) K; > 2K2

K1 < :,

68. Solar constant of the sun is cr=8.106xl0 4 Jmm·1m·2

62. Which one of the following statements about photons is

radiation are (A) 222 (C) 442

(B) 1.=1.64 nm (D) 1. = 4.64 nm

67. Let K1 be the maximum kinetic energy of photoelectrons emitted by light of wavelength 1.1 and K2 corresponding to wavelength 1.2 • If 1.1 = 21.2 then (A) 2K1 = K, (B) K1 = 2K2

61. The duration of a laser pulse is 10_. s. The uncertainty in its energy will be

(B)

(B) 1.1 > 1. 2 (D) 1.1 > i.2 or 1.1 < i.2

1.1 = 1. 2 (C) 1.1 < 1.2

60. The potential difference between the cathode and anode in a cathode ray tube is V . Then the speed acquired by the electrons is proportional to· (A) V (B) .Jv (C) V2 (D) V¥2

(A) 6.6 x 10·" J

(D) None of these

C

(A)

(D) ~r=_-················ ~

Intensity (!)

l.7itR 2I C

0.3itR2I (C) - - -

Intensity (I)

()

(B)

C

59. Which one of the following graphs represents correctly the variation of photoelectric current (i) with intensity ( I) of incident radiations

(A)

nR2 I

--T

photon's initial energy. 1

(D) electron s kinetic energy is a fraction

1 a'

of

photon's initial energy. 70. Which one of the following does not fit into the group? (A) Photon (B) Graviton (C) Proton (D) Meson 71..

Two photons approach each other. The relative velocity of approach is

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Plate

(A) .:.

(B)

C

(C)

(D)

ic

2

2c

----+ l--'mKirJl'_ __j ----+

=:

72. A particle of mass 3m at rest decays into two particles of masses m_ and ·2m having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles

---+-

(A)

(B)

(C) 1

(D)

(A)

1 2 2

(C)

73. A small ball is projected with initial speed u and at an angle 8 with horizontal from ground. The de-Broglie wavelength of ball at the moment its velocity vector becomes perpendicular to initial velocity vector is

.

(B)

h musin8

(D)

•

~

h

(B)

(C) ZERO·

(D)

KC

(C)

00

77. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV successively illuminate a metal of work :function 0.5 eV. The ratio of the maximum speeds of the emitted electrons will be "(A) 1:5 (B), 1:4 (C) 1:2 (D) 1:1 78. Light of intensity I is incident normally on a perfectly reflecting plate of area A kept in a gravity free space. If the photons strike the plate synunetrically and initially the springs are in their natural lengths, then the maximum compression in the springs is

(B) 2.4 V (D) 4.8V

V

4

(D) V 8

81.

Ultraviolet light wavelength 300 nm and intensity 1.0 wm-' falls on the surface of a photoelectric material. If one percent of the incident photons produce photo electrons, then the number of photoelectrons emitted per. second from an area of 1.0 cm2 of the surface is nearly (A) 9.61 x 1014 (B) 4.12 X 1013 (C) 1.51 X 1012 (D) 2.13x1011

82.

A proton and an a-particle are injected into a uniform electric field at right angles to the direction of field with equal kinetic energy. Then {A) the proton trajectory will be less curved than aparticle trajectory. (B) the a-particle trajectory will be less curved than proton trajectory. · (C) both the trajectories will be equally curved. (D) both trajectories will be straight.

~ch

m0c h

(D)

2IA 3KC 4IA 3KC

accelerated through a potential difference (B) 8V ·(A) 4V

speed of light c . The de-Broglie wavelength associated is given by

m,c

KC 3IA

(B)

80. A proton, accelerated through a potential difference V has a certain de Broglie wavelength. In order to have the same de Broglie wav:elength, an a -particle must be

76. A material particle with a rest :mass m0 is moving with

(A)

IA

(A) 1,2 V (C) 3.6 V

75. Photoelectrons are emitted with maximum kinetic energy E from a metal surface when light of frequency v falls on it when light of frequency v' falls on the same metal, the maximum kinetic energy of emitted photoelectrons is found to be 2E , then v' is (A) v'=v (B) v'=2v

~>h

-,.._'--.... ..,

79. The work function of a metallic surface is 5.01 eV. Photoelectrons are emitted when light of wavelength 2000 A falls on it. The potential difference required to stop the fastest photoelectrons is (h=4.14x10-15 eVs}

h mucosa

74. The force exerted by a photon of intensity 1.4 kWm-2 if it falls on a perfect absorber of radius 2 m is (A) 2.35 x 10""' N (B) 10 8 N (C) 8.35xl0 4 N (D) 8.8x10 .. N

~

,'·

Smooth surface

1

h mu

f

K

- - - - - ~ ~ -*'-~~· _.._

lS

4

(A)

_J;f Fixed

----+ L__r==

(,.,,.,) . -

l--'imrK,oo._

· 83. de Broglie waves are associated wj.th (A) moving charged particles only. (B) moving neutral particles only. (C) all moving particles. (D) all particles whether in motion or at rest.

=============================== = 3.27

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Optics & Modern Physics

Advanced JEE Physics 84.

The maximum kinetic energy (E.) of photoelectrons varies with the frequency ( v) of the incident radiation as

(A)~•

~

~) ~ •

~•

(C)

1

2

h(v-v0 )

(D)

1

2h(v+v

0)

C

(D)~•

(D) til'=2Ic

(A) .does not depend on the frequency of the incident light. (B) does not depend on the nature of cathode material. (C) depends on both the frequency of the incident light and the nature of the cathode material. (D) depends on the intensity of the incident light. 86. A particle of mass 10:.,,. kg is moving with a speed of 10 5 ms-1 • The de Broglie wavelength of the particle is· (A) 6.63 x 10-s m (B) 6.63 A

(D) 6.63 X 10-7

66.3A

ffi

87. A moving i,article is associated with a wave packet' or group of waves. The group velocity is equal to (A) velocity oflight (B) velocity of sound (C) velocity of particle

(D)

89. A photon of frequency v is incident on a metal surface whose threshold frequency is vO • The maximum kinetic energy of the emitted electron :will be (A) h{v-v 0 ) (B) h(v+v,)

90. A stream of photons impinging normally on a completely absorbing screen in yacuum exerts. a pressure Pl'. If I is the irradiance then, 2 (A) til'= I (B) til'=Ic

85. Stopping potential for photoelectrons

(C)

(D) Planck's constant

1 particle velocity

88. The maximum kinetic energy ( E= ) of photoelectrons emitted in a photoelectric cell varies with frequency ( v) as shown in the graph. The slope of the graph is equal to ·

91. An electron and a photon have same wavelength. It p is the momentum of electron and E the energy of photon. The magnitude of

f in SI

(A)

(B)

(C)

1 2c 2

unit is

1 C

(D) None of these

C

92. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is (A) 2 (B) 4 (C) 6 (D), 10

93. A sensor is exposed for time I to a lamp o( power P placed at a distance l . The sensor has an opening that is 4d in diameter. Assuming all energy of the lamp is given off as light, the number of photons entering the sensor if the wavelength of light is 1,. is P1..d 2I 4P1..d2I {A) N = hcl' (B) N =~ (C)

N = P1..d't

(D) N = PM't

4hcl 2

16hcl2

94. When

lL. v,

(A) charge of the electron

~ of the electron m (C) work function of ~e emitter (B)

a metallic surface is illuminated by a monochromatic light of wavelength 1,. , tl1e stopping potential for photoelectric current is 3V0 • When the same surface is illuminated by light of wavelength 21,., the stopping potential is V, . The threshold wavelength for this surface for photoelectric effect is (A) 61.. (B) 41.. 3 (C) 41,. (D) 81..

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DuafNature of Radiation & Matter 95. The momentum of a photon of an electromagnetic

radiation is 3.3 x 10-" kgms-1 . The frequency of the associated waves is ( h = 6.6 x 10-34 Js , c = 3 x-108 ms-1 ) (A) 3:0 x 103 Hz

(B)

(C) 7.5 X 1012 Hz

(D) 1.5 X 1013 Hz

(A)

6.0 x 10 3 Hz

96. If the wavelength of incident radiation photoelectric experiment is decreased theri (A) the photoelectric current will decrease. (B) the photoelectric current will increase. (C) the stopping potential will decrease.

in

a

(D) the stopping potential will increase.

97. The threshold wavelength for a photosensitive surface is 6000 A and the wavelength of incident light is 5000 A . Then the maximum energy of emitted

electrons would be (A) 0.041 eV (C) 4.1 eV

(B) 0.41 eV (D) 41 eV

gets accelerated by a constant electric field E . The rate of change of de-Broglie wavelength of this electron at time t , ignoring relativistic effects

(B)

eht E

(C) - .;;,

(D)

!

100. A small photocell is placed at a distance of 4 m from a photosensitive surface. When light falls. on the surface the current is 5 mA. If the distance of cell is decreased to 1 m , the current will become

(C) 20mA

(D) 80 mA

(C) 3.3x10 4 kgms-1

(D) 6.6x10..., kgms-1

substance is shown in figure The threshold wavelength is V0 volt 2 --------------------

i

I

(:6)mA

101. Einstein's photoelectric equation is E,

4

= hv- ~ . In

this

equation EK refers to (A) kinetic energy of all the emitted electrons. (B) mean kinetic energy of emitted electrons. (C) maximum kinetic energy of emitted electrons. (D) minimum kinetic energy of emitted electrons. 102. In photoelectric effect, the graph showing the variation

of cut-off voltage (V0 ) with the frequency of incident radiation ( v) is

1.4xl0 20 (D) 2x1020

(B)

106. The stopping potential ( V, ) versus frequency plot of a

(B) proton

(B)

(A) 1.4xl019 (C) 2x1019

minimum momentum of the electron may be (A) 6.6x10-20 kgms-1 (B) 3.3x10-"' kgms-1

(D) oxygen atom

(A) 1.25mA

visible light, the number of visible quanta emitted per second by a 100 W bulb, assuming the wavelength of visible lightto be 5.6 x 10-s cm , is

105. A neutron is confined to a nucleus of size 10-14 m . The

99. Of the following, the one which has the largest de Broglie wavelength for the sarn~ speed is

(A) electron (C) a -particle_

of 100 V . Its kinetic energy will be (A) 100 J (B) 100 erg (C) 100 eV (D) 100 MeV 104. If 5% of the energy supplied to a bulb is radiated as

98. An electron of mass m and charge e initially at rest

h (A) - eEt'

103. An electron is accelerated through a potential difference

5

6 7 8 14 vx10 Hz

(A) 5xl014 m (B)

6000 A

_(CJ 5000 A (D) cannot be estimated from given data 107. The energy of incident photon is 12.375 eV while the energy of scattered photon is 9.375 eV . Then the kinetic energy of recoil electron is

(A) 3 eV (C) more than 3 eV

(B) less than 3 eV (D) 21.75 eV

c::::================================= = 3.29

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Optics & Modem Physics

Advat1ced JEE Physics 108. When electrons are accelerated through potential difference of V volt, the de-Broglie wavelength associated is given by

(A)

).=t~o

A

(CJ ).=~A

(B)

).=t~o

(D)

).=ffso A

m

same potential. If their masses are me and m, respectively, then the ratio of their de Broglie

wavelength is

(CJ

(B)

m, m,

(D)

M 0c [1-(~J]

(CJ M 0 c'(1- ~)

(D)

M 0 c'[1-(~J]

strike a. totally reflecting screen per second at normal incidence so that a force of 1 N is exerted on the screen

is (A) n=5xl0 23 (C) n=5xl0 25

n=5x10 24 (D) t1=5x10 26

(B) •

~

117. A 20 arnu atom emits photon of 6.6 A while making a

~

transition from excited state to ground state. The recoil energy of the atom will be (B) 3.5 x 10-23 J (A) 1.5 x 10·23 J

110. A particle of mass 1 g is located in a box of size 2 cm. The uncertainty in the momentum of thE:_ electron will be (A) 3.3 x 10""2 kgms' (B) 6.6 x 10-02 kgms'

(CJ 3.3 X 10-33 kgms·t

(B)

116. The number of red photons {). = 663 nm) that must

V

109. An electron and a proton are accelerated through the

(A) 1

(D) 6.6 X 10->1 kgms·l

(CJ 5.lxl0-23

J -

(D) 7.5xl0-23

111. If E1 , E2 , E3 are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then

(A)

E, >E, >E,

(B)

E, >E3 >E,

~

E,>E,>E,

~

E,=E,=E,

112. The kinetic energy of electron is E, when the incident

light has wavelength ). . To increase the K.E. to 2E, the incident light must have wavelength

(C)

he £).-he

(B)

he). &+he

h).

(D)

he). &-he

E). +he

J

118. How many red photon (wavelength ). ) must strike a totally reflecting screen per second at normal incidence, if the exerted force is to be 1 N ?

(A)

!:

(B)

h

(A)

2

(A) Moc2

).

(CJ

2h

2).

h (D) infinity

119. The surface of a metal is illuminated with the light of 400 nm . The kinetic energy of the ejected photoelectrons was found to be, 1.68 eV. The work function of the metal for he= 1240 eV-nm is (A) 3.09 eV (B) 1.41 eV (C) 1.51 eV (D) 1.68 eV

120. A cathode of a photoelectric cell is changed such that the work function changes from W1 to W, ( 1'½ < W2 ) • If the current before and after changes are 11 and 12 all other conditions remaining unchanged, then (assuming

113. We wish to observe an object which is 2.5 A in size. The minimum energy photon that can be used is (A) 5 keV (B) 8 keV (CJ 10 keV (D) 12 keV 114. A certain mass of ice at 0° C melts into water at 0° C

and there by gains 1 kg mass. If initial mass of ice is m0 then,

= 2.69 x 10 kg (CJ '"o = 2.69 X 1010 kg

(A) m 0

11

(B)

Ro

(B) (D)

121. The momentum of a photon of frequency v is

(A)

~~

(B)

(C)

ltvc

(D) hve2

hev

=1 kg

(D) m0 =9xl0 16 kg

115. A star of mass M 0 , radius

=

m0

hv>W,) (A) I,= I 2 (C) I,> I,

contracts to radius R.

Energy radiated by the star assuming uniform density in each case while temperature remains unchanged is -

122. The work functions for three different metals A, B and

C are $A, $, and $c respectively with $A>$,> $cThe graphs beiween stopping potential (V0 ) and frequency v of incident radiation for them would look

like

3.30

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Dual Nature of Radiation & Matter field needed so that the rays describe the same approximate circular path ( voltmetr~-1 ) is

{A) 2x (C) 4x

(A)

(B) 3x (D) 6x

129. Which of the following graphs gives the variation of photoelectric current (I,) with the voltage (V) applied

to the electrodes of a photo cell?

(C) (A) 123. In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in (A) intensity of radiation. · ,(B) wavelength of radiation. ·(CJ frequency of radiation. (D) both wavelength and intensity of radiation.

(C)

124. Which of the following arrangements corresponds to

decreasing order of specific charge? (A) Electron, proton, a -particle (B) Proton, a -particle, electron (C) a-particle, electron, proton (D) Electron, a -particle, proton

130. An electron with speed v and a photon with a speed c have the same de-Broglie wavelength. If the K.E. and momentum of electrons is E, and P, and that of photon

125. A photoelectric cell is illuminated by a small bright

source of light placed at 1 m. If the same source of light is placed 2 m away, the electrons emitted by the cathode (A) each carries one quarter of its previous momentum. (B) each carries one quarter of its previous energy. (C) are half the previous number. (D) are one quarter of the previous number. 126. Light of frequency 1.5 times the threshold frequency is incident on a photo-sensitive material. H the frequency is halved and the intensity is doubled, the photoelectric current becomes

(A) four times (C) half

LL, .,r=_, Jc_,~, lL_,

(B) double (D) ZERO

127. The ratio of the de Broglie wavelengths of a proton and an a-particle will be 1: Z- if their (A) kinetic energies are in the ratio 1: 8 (B) kinetic energies are in the ratio 8 : 1 (C) velocities are in the ratio 1: 8 (D) velocities are in the ratio 8 : 1

is EP11 and Pp1, respectively, then the correct statement

is

~=~

(A) s_=2c E,, V

{B)

Pe 2c (C) - = -

(D) None of these

Pp1,

2c

Eph

V

131. The energy of a photon corresponding to the visible light of maximum wavelength is approximately (A) 1 eV (B) 1.6 eV (C) 3.2eV (D) 7eV 132. Planck's work was connected with (A) wave nature of matter (B) photoelectric effect (C) structure of atom (D) quantum nature of radiation. 133. The

graph shows stopping 1 - - - - for three metals,. then wavelength Stopping potential

potential

2

versus

3

128. Cathode rays moving with same velocity v describe an apprqximate circular path of radius r metre in an electric field of strength x volt metre-' . If the speed of the cathode rays is doubled to 2v, the value of electric

wavelength

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Advanced JEE Physics

(A) (B) (C) (D)

Optics & Modern Physics

Planck's constant for metal (1) is greatest Work function for metal (3) is greatest Threshold frequency for metal (1) is greatest Threshold wavelength is maximum for metal (3)

134. The work function of a substance is 4 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (A) 540 nm (B) 400 nm (C) 310 nm (D) 220 nm

135. A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. The number of photons emitted per second is

(A) l.7lxl0 31 (C) 13.27 X 10"

(C) 6,6 X 10-31 kgrns-l

(B) 6.6xl0_,. kgrns-1 (D) 6,6 X lQ-"" kgrns-l

137. An electron is moving with a velocity of

C

10

(B) 0.24 X 10-10 A (D) l.24xl0-10 m

138. The kinetic energy of the body is twice the rest mass energy. The ratio of the relativistic mass of the body to its rest mass is (A) 1 (B) 2 (C) 3 (D) infinite

139. The number of complete de-Broglie wavelengths associated with the electron in nth orbit of hydrogen atom is

(A) n

(C)

(B) 1327 x 10" (D) 13.27 X 1044

136. The largest momentum we can expect for a microwave photon is

(A) 6.6xl0-27 kgrns-1

(A) 0.48xl0-10 m (C) 0.24 X 10-!0 m

.! n

(B)

n2

(D) n'

140. In a photoelectric cell, the current "stops when the collecting plate is one volt negative with respect to the

emitting metal. The maximum kinetic energy of the photoelectrons is

1J

(A) 1 erg

(B)

(C) 1,6 X 10-1' J

(D) 1.6 x 10-" eV

. The de-

Broglie wavelength associated with it is

=

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1his section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which

ONE OR MORE is/ are correct. 1.

When photons of energy 4.25 eV strike the surface of a metal, the ejected photoelectrons have maximum kinetic energy TA eV and de Broglie wavelength :\.A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is T,=(TA-1.50)eV. If the de-Broglie wavelength of these photoelectrons is :\., = 2:\.A, then (A) the work function of A is 2.25 eV. (B) the work function of B is 4.20 eV . (C) TA= 2.00 eV. (D) T, =2.75 eV.

2.

Irnax when averaged over many pulses is 80 µA . Then which of the following statements is/ are true. (A) The charge carried by one pulse is 8 x 10-13 C

(B) Number of photoelectrons emitted per light pulse is 5

(C) Number of photons in one light pulse is 50 (D) Number of electrons carried by one pulse is 5 x 105 4.

Light from a monochromatic source is incident normally on a small photo sensitive surface S having work function ~. If power of the source is W afld a is the distance between the source and S , then

(A) the number of photons striking the surface per unit

A collimated beam of light of flux density 30 kWm-' is

. illb(WASJ timew e ---, 41chca

incident normally on a 100 mm' completely absorbing screen. If 'fYJ is the pressure exerted on the screen and Ap is the momentum transferred to the screen during a

(B) the maximum energy of the emitted electrons will be(~-~)

1000 s interval then,

3.

(A)

9" = 10-3 Nm-2

(B)

(C)

~p = 10.., kgms-1

(D) ~p = 10-5 kgms-1

(C) the stopping potential needed to stop the most

9" = 10.., Nm_,

energetic photons will be !:..(l,c-:\.~) •

A.

· · occurs onl y 1"f O,;; :\. ,;; ~ lzc (D) p h oto erruss1on

A photomultiplier tube is to be used to detect light pulses each of which consists of a small but fixed number of photons. The average photoelectric efficiency

is 10%. That is photon has 10% probability of causing

5.

the emission of a detectable photoelectron. Assume the photomultiplier gain is 10 6 and that the output current as a function of time can be approximated as shown in

figure.

When ultraviolet radiation is photoelectrons are emitted. photoelectrons to be emitted consist of (A) radio waves (B)

(C) X-rays

----------------~--·

6.

incident on a surface, no

If another beam causes from the surface, it may

infrared rays (D) gamma rays

When the intensity of a light source is increased,

(A) the number of photons emitted by the source in unit time increases

1-

_J _________ _

(B) the total energy of the photons emitted per unit time increases 3.33 i::::J

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Optics & Modern Physics

Advanced JEE Physics (D) faster photons are emitted

(A) Mirror will start to oscillate. (B) Wavelength of reflected rays will be greater than

Light rays are incident on an opaque sheet. The correct

(C) Wavelength of reflected rays may be less than that

statement(s) is/ are {A) Light rays exert a force on the sheet (B) Light rays transfer an energy to the sheet (C) Light rays transfer momentum to the sheet (D) Light rays transfer impulse to the sheet

of incident rays (D) None of these

(C) more energetic photons are emitted

that of incident rays.

7.

8.

According to Heisenberg's Uncertainty Principle, h h (A) lu!J.p ?. (B) ti.EM ;, 4rr 4rr h h (C) ti.OM.;,(D) lu!J.v ? . 4rr 4mn where the symbols bear the usual meaning.

9.

In which of the following situations, the heavier of the two particles will have a smaller de-Broglie wavelength.

(A) The particles move with the same speed (B) The particles move with the same

linear

momentum

(C) The particles move with the same kinetic energy (D) The particles have fallen.through the same height 10. The threshold wavelength for photoelectric emission

from a material is 5200 A . This material when illuminated with .................... monochromatic radiation

emits photoelectrons. (A) 1 WUV (C) 1 WIR 11.

(B) 50WUV (D) 50WIR

Light rays are incident on a metallic sheet. Then,

(A) the force exerted is Independent of frequency of light Incident (B) the force depends on the direction of light Incident (C) the pressure is Independent of frequency of light incident (D) the pressure is proportional to the area of the plate

12. The momentum of a single photon of red light of frequency 400 x lOu Hz moving through free space is (A) ZERO (C) l.65x10 .. MeV/c

(B)

14. For a 75 W point light source assuming all the electric power consumed goes Into emitted light of wavelength 600 nm, then (A) frequency of the emitted light is 5 x 1014 Hz (B) number of photons emitted per second is 2.3 x 10 20 (C) this emitted light on falling on a metal surface of work function 1.07 eV, will emit photoelectrons having kinetic energy between Oand 1 eV (D) on doubling the distance of this metal surface from the point source maximum kinetic energy of photoelectrons emitted becomes 0.25 eV 15. Photoelectric effect supports the quantum nature of light because (A) there is a minimum frequency of light below which no photoelectrons are emitted. · (B) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity. (C) even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediately. (D) electric charge of the photoelectrons is quantized. 16. If the wavelength of light In an experiment on photoelectric effect is doubled, (A) the photoelectric emission will not take place (B) the photoelectric emission may or may not take place (C) the stopping potential will Increase (D) the stopping potential will decrease

17. A metallic surface ejects electrons when exposed to green light of Intensity I but no photoelectrons are emitted when exposed to yellow light of Intensity I . It is possible to eject electrons from the same surface by (A) yellow light of some Intensity which is more than

I.

8.8 x 10-" kgms-2

(D) Data Insufficient

13. A small mirror is suspended by a thread as shown in figure. A short pulse of monochromatic light rays is Incident normally on the mirror and gets refiected. Which of the foliowlng statements is/ are_ correct?

(B) green light of any Intensity. (C) red light of any Intensity. (D) violet light of any Intensity. 18. In an experiment of photoelectric effect, light from a point source of monochromatic light of wavelength

3000 A is Incident on a metal surface. The kinetic energies of photoelectrons range from zero to 4x10-19 J,

then · (A) stopping potential for this light is 2.5 V (B) threshold wavelength for the material is 7590 A

=

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(C) stopping potential will be doubled on reducing the distance and the wavelength of light source to half (D) saturation current will be doubled on reducing the

(C) if the wavelength of source of radiation is doubled then KE of photoelectrons is also doubled (D) photoelectrons will be emitted from the surface of all the three metals if the wavelength of incident radiations is less than 200 nm

distance of source to half 19. The threshold wavelength for photoelectric emission

from a material is 5200 A . Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a (A) 50 watt infrared lamp

(B) 1 watt infrared lamp (C) 50 watt ultraviolet lamp (D) 1 watt ultraviolet lamp 20. Light of wavelength 496 nm is incident on a metal surface causing ejection of photoelectrons for which stopping potential is 1.5 V, then (A) the work function of the surface is 1 eV

(B) de-Broglie wavelength of fastest photoelectron is 100nm (C) to move the fastest electron in a circle of radius 1 m, perpendicular magnetic field B required is 4µT (D) this fastest electron when strikes zinc target can

23.

When a rnonocliromatic point source of light is at a distance of 0.2 m from a photo-electric cell, the cut-off voltage and the sahµ'ation current are respectively 0.6 V and 18.0 mA . If the same source is placed 0.6 m away from the photoelectric cell, then (A) the stopping potential will be 0.2 V . (B) the stopping potential will be 0.6 V. (C) the saturation current will be 6.0 mA . (D) the saturation current will be 2.0 mA .

24. Figure represents a graph of kinetic energy ( K ) of photoelectrons (in eV) and frequency ( v) for a metal used as cathode in photoelectric experiment. The work function of metal is 3K

2

produce X-rays 21.

0+---,,---+v

Which of the following statements about photoelectric effect is/ are false? (A) It exhibits the particle nature of radiation (B) Electrons are emitted only if the radiation has a frequency above a certain value

(C) All the electrons emitted by radiation of a particular frequency have the same energy (D) Changing the intensity of radiation changes the maximum energy with which the electrons can be emitted

22. Radiations of monochromatic waves of wavelength 400 run are made incident on the surface of metals Zn , Fe and Ni of work functions 3.4 eV, 4.8 eV and 5.9 eV respectively (take he= 12400 eVA) (A) maximum KE associated with photoelectrons from the surface of any metal is 0.3 eV (B) no photoelectrons are emitted from the surface of

-~

...............

-3

(A) 1 eV (C) 2eV

{B) 1.5 eV (D) 3 eV

25. In PROBLEM 24, the threshold frequency is nearly (A) 5xl0 14 Hz {B) 10xl014 Hz

(C) 2.5 x 1014 Hz

(D) cannot be estimated

26. It is necessary to consider light as a stream of photons to explain (A) Photoelectric effect (B) Compton effect (C) Polarization of light (D) Diffraction oflight

Ni

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'I

This section contains Reasoning ,type questions, each having four choices (A), (B), (q and (D) out of ,which ONLY ONE is correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statem~ts are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FAISE.

,. ,, .

.

. -~ ,

. Bubble (D) IfSTATEMENTl isFAISEbutSTATEMENT2isTRUE.

,. 1.

Statement-I:

Statement-2:

Threshold wavelength of certain metal is 1..0 • Light of wavelength ~lightly less than 1.. 0, is incident on the plate. It is found that after some time the emission of electrons stops.

Higher th~ ,accelerating potential, s~aller is the de-Broglie wavelength. 5.

Statement-2:

.

.

\

!

. . 1.

' ••

Statement-I: In case of .an, electron 'and a "photon having same

Electron.cannot move with a· speed of photon. 7.

Statement-I:

'.

Statement-2:

Momentum of photon is due to its energy and heilce its ' ·' ' equivalent mass.

3.

,

momentum, wavelength associated with electron is smaller.

momentum. .

'

The frequency is inversely. proportional to wavelength.

A photon ,has no rest mass, yet it, carries definite

Statement-2:

_ _

Statement-2:

6.

Statement-I:

Statement:I:

Work function of copper is greater than that of sodium. But both will have saine value of the threshold ·frJquency and threshold wavelength. :

The ejected electro~ experience force of attracti~n due to development of positive c],arges OI) plate which after certain time is adequate enough to hold .them to plate · itself. 2.

' I - ••

Statem~t-1:

r

.

_

,

The photoelectrons produced by a monochromatic light beam incident on a metal surface, have a spread in

In a photoelectric effect, the current increase~_ when positive potential of . q:>llector is increased, before · saturation of current.

their kinetic energies.

Statement-2:

Statement-2:

Ute number of emitted photoelectrons increases.

The work ·function of the metal varies as a function of

depth from the surface. 4.

Statement-I: The de Broglie wavelength of an electron accelerated through 941 volt is 0.4 A.

B.

Statement-I: A proton, a deutron and an a. ~})article are accelerated

by the same potential difference. Their ;elocities will be in the ratio of 1 :1 :-./2..

=

3.36

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Dual Nature of Radiatio11 & Matter

Stntement-2: Kinetic energy, E =qV =!:.mv 2 9.

14. Statement-1: A photon and an electron, both of energy 1 MeV has

2

same wavelength.

Statement-2: E = 22mec 2 =10 6 eV

Statement-1: Photoelectric effect demonstrates the wave nature ·of

light.

Statement-2: The number of photoelectrons is proportional to the frequency of light.

15. Statement-1:

The velocity of body of rest mass m0 is

~ c (where c

is the velocity of light in vacuum) then mass of the body is 2m0 •

10. Stntement-1: A photon has no rest mass, yet it carries momentum.

Statement-2:

Statement-2: Momentum depends more on velocity than that of

Moving mass is given as m

mass. 11. Stntement-1:

The threshold frequency of photoelectric effect supports the particle nature of sunlight.

possess the same kinetic energy.

Statement-2:

frequency, electro~ are not emitied from metal surface.

In photoelectric effect a single photon interacts with a

single electron and electron is emitted only if energy of

12. Statement-1 : Effective mass of photon varies with wavelength.

13. Statement-1: Though light of a single frequency (monochromatic light) is incident on a metal, the. energies of emitted

g'

16. Statement-1: In the process of photoelectric erruss1on by monochromatic light, all the emitted photo-electrons

Statement-2: If frequency of incident light is less than the threshold

Statement-2 : E = mc2 is the relation between mass and energy.

m,

each of incident photon is greater than _the w9rk function. 17.

Statement-1: de-Broglie wavelength of an electron accelerated 1 2 through a potential difference of V volt is 1,. = :/(; A .

photoelectrons are different.

Statement-2:

Statement-2: The energy of electrons just after they absorb photons incident on metal surface may be lost in collision with other atoms in the metal before tne electron is ejected out of metal. ·

de-Brogle wavelength of an electron is given by 1,. =..!!____ mv

3.37.=

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This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a feW questions that may have more than one correct options)

Comprehension 1 A photocell is operating in saturation mode ·with a photo current 20 µA when a monochromatic radiation of

4.

The kinetic eriergy of the electrons is (A) 0.56 eV (B) 1.32 eV (C) 0,86 eV (D) 1.76 eV

wave length 3000 A and power 1-MW is incident." When

5.

The work function of the metal is (A) 1.89 eV (B) 1.03 eV (C) 1.58 eV (D) 2.32 eV

6.

Wavelength of radiation is nearly (A) 6565 A (B) 5555 A (DJ 3535A. (CJ 4545 A

another p1onochromatic radiation of wave length 1500 A and power 5 MW is incident, it is observed that the maximum velocity of photo electron is doubled. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases. Based qn above information, answer the following questions. 1.

The threshold wavelength for the cell is (A) 3500 A (B) 4000 A·

(CJ 4500 i:,.

Comprehension 3 de-Broglie suggested that every moving particle has a

(D) 5000 A

wavelengih associated with it, which is given by 1,, =!: or , p h - . . " =..JZmK . With the help of these formulae we know that

2.

The saturation current in second case is (A) 50 µA (B) 40 µA (C) 60 µA (D) 45 µA

3.

The efficiency of photoelectron generation per incident photon is (A) 8.5%. (B) 8.25% (C) 8% (D) 8.75%

wavelength for a charged particle accelerated through a

Comprehension 2 The radiations emitted when an electron jumps from n = 3 to n = 2 orbit of hydrogen atom falls on a metal to produce photoelectrons. The electrons emitted from the metal surface with maximum kinetic energy are made to

mov_e p~endicular to a magnetic field of strength

~ T 3 0 in a raciills of 10-3 m . Based on above information, answer the following questions.

potential V is 1,,

~. V

Based on above information,

answer the following questions.

7.

Proton and alpha particle are accelerated through same . potential difference. Then ratio of their wavelength is (A) 1: ./2 (B) 2./2: 1 (C) 2:1 (D) 1:2./2

8.

When an electron is accelerated through 150 V potential difference, then the wavelength associated with it is approximately

(A) 1 A

(CJ 3 A

(B) 2 A (DJ 4 A

= ================================== 3.38

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Dual Nature of Radiation & Matter 9.

If electron and alpha particle have same momentum the ratio of their wavelength is (A) 1840: 1 (BJ

(CJ 1: 1

occur is nearly

1: 1840 (D) No!'e of these

(A) 1250nm (CJ 2250nm

Comprehension 4 A large number of identical balls each having a mass of 66.3 g are thrown with speed of 5 ms-• into a house

through two tall, narrow, parallel windows spaced 0.6 m apart, the choice of window as target being random at each throw. Fringes are formed on a wall 12 m behind the windows. Assuming the Plank's constant to have a hypothetical value of h = 6.63 x 10" Js . Based on above information, answer the following questions.

10. de-Broglie wavelength of the balls is (A) 0.2m (B) 0.02m (CJ 2m (D) 20m

(B) 2000nm (DJ 3000nm

Comprehension 6 A parallel beam of monochromatic light (,. = 663 nm) of intensity 30 kWm- 2 is incident normally on a 100 mm' completely absorbing screen for 10 s. Based on above information, answer the following questions. 16. Pressure exerted by beam on the surface is

(A) 10-5 Pa (CJ 2x10-< Pa

(BJ 10-< Pa (DJ Sxl0-5 Pa

17. Momentum transferred to the screen· during the interval is (A) 10_. kgms-• (B) 5 x 10_. kgms-1

(CJ 10-7 kgms-•

11. Fringe width obtained on the wall is (A) 0.01 m (BJ 0.2 m (CJ 0.5 m (DJ 0.4 m 12. If electrons are used in place of balls, moving with speed 107 ms-1 in the experiment then (A) Fringe width will decrease (B) No interference pattern is obtained (CJ Fringe width will not increase (DJ Interference pattern is obtained but it is not possible to observe it Comprehension 5 A surface has light of wavelength ,. = 496 nm incident on it, causing the ejection of photoelectrons for which the stopping potential is found to be 1.5 V . Based on above information, answer the following questions. 13. The de-Broglie wavelength of the fastest photoelectron emitted is

(A) 75nm (CJ 135nm

15. The threshold wavelength for photoelectric emission to ·

(B) 100nm (D) 235nm

14. To move the fastest electron in a circle of radius 1 m, the value of perpendicular magnetic field B is (A) 4.lxl0_. T . (B) 2.1x10_, T (CJ 3.2x10_, T (D) 5.2x10_. T

(D) 2 X 10"7 kgms-l

18. Number of photons striking the screen during the interval is (A) 5x1019 (BJ 10'° (CJ 2x10 20 (D) 5xl0 20 Comprehension 7 A metallic surface, when illuminated by light of frequency 8x10 14 Hz and 12xl014 Hz emits photoelectrons of maximum kinetic energy 0.5 eV and

2.0 eV . Based on the information provided answer the following questions. 19. The value of Planck's constant is (A) 6.0 x 10"' Js (B) 6.2 x 10--" Js 34 (CJ 6.4x10- Js (DJ 6.6x10"' Js 20. The work function of metal is (A) 0.5 eV (BJ 1.5 eV (CJ 2.5 eV (DJ 3.5 eV 21. de-Broglie wavelength of electron when its energy is

0.5eV (A) 8.68A (CJ 21.2SA

(BJ 17.35 A (D) 24.54A

3.39

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Each question in this section contains statements given in fwo columns, which ·have to be matched. The ~~tements in . COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles - corresponding to the answers to these questions have to be darkened as illustrated in the following examples:

If the correct matches are look like the following :

A-> p, sand t; B-> q·and r; C-> pand q; and D-> sand t; then the correct darkening of bubbles will ·

P

qr

s

I·

A@@)(v@@)

B@@@®CD c@@©®CD D@@)(v@@)

1.

In_ a photoelectric effect experiment, if f is the frequency of radiations incident on the metal surface and I is the inte~ity of incident radiations, then match the quantities in COLUMN-I with.their matches in COLUMN-II.

f

COLUMN-I

COLUMN-II ·

(A) If distance between cathode and anode is increased.

(B) If I is increas':d keeping f and work . function Constant. (q Work function is decreased keeping f and I constant.

(D) If f

is increased keeping I and work function

(p) . Stopping potential increases. (q) Saturation current increases. (r) Maximum kinetic energy of photo electron increases. (s) Stopping potential remain same.

constant. 2.

In a photoelectric effect experiment, if the following changes are made, then match the COLUMN-I with COLUMN-II.

I

COLUMN-I

COLUMN-II

(A) If intensity of incident light is increased keeping its frequency constant. (B) If work function of photo sensitive electrode is increased. (q If frequency of incident light is increased and its intensity is deCTeased. · (D) If frequency of incident light is increased keeping its intensity constant. ·

3.40

(p) Stopping potential will increase. (q) Current through circuit may stop. (r) Maximum kinetic energy photoelectrons will increase. (s) Saturation current will increase.

of

ejected

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Dual Nature o/Radiatio11 & Matter 3.

Match the properties in COLUMN-I with their respective phenomenon in COLUMN-II ..

I

COLUMN-I

COLUMN-II

(A) Photon character of radiation.

(p) Photoelectric effect. (B) Wave character of radiation. (q) Compton effect. (C) Interaction of a photon with an electron, such that photon. (r) Diffraction. energy is much greater than the binding energy of (s) Interference. electron, is more likely to result in.

'

(D) Interaction of a photon with an electron, such that photon energy is equal to or slightly greater than the binding energy of electron, is more likely to result in. 4.

Match the wavelengths in COLUMN-I to the respective matches in COLUMN-II.

I

COLUMN-I

COLUMN-II

0.1A (B) 1A (C) 10A (D) soooA

(p) (q) (r) (s)

(A)

5.

Some quant;ties related to photoelectric effect are mentioned under COLUMN-I and COLUMN-II. Match each quantity on COLUMN-I with the corresponding quantity.in COLUMN-II on which it depends.

I

COLUMN-I

COLUMN-II

(A) d.e-Broglie wavelength of photoelectron.

(B) Force due to radiation falling on metal plate. (C) Stopping potential. (D) Saturation current. 6.

de-Broglie wavelength of electron in X-ray tube. Photoelectric threshold wavelength. X-ray wavelength. de-Broglie wavelength of most energetic photoelectron emitted from metal surface in photoelectric effect.

(p) (q) (r) (s)

Frequency oflight. Work function. Area of photo sensitive plate. Intensity of light (at constant v ).

In a photoelectric effect experiment, if / is the frequency of radiations incident on the metal surface and I is the intensity of incident radiations, then match the following.

I

'COLUMN-I

COLUMN-II

(A) Work function is decreased keeping f and I constant. (B) If I

is increased keeping /

and work function

constant.

(C) If distance between cathode and anode is increased. (D) If f is increased keeping I and work function

(p) Stopping potential increa_ses. (q) Saturation current increases. (r) · Maximum kinetic ene~gy

of

photoelectron

increases.

(s)

Stopping potential remain same.

constant.

7.

With respect to photoelectric effect experiment, match the entries of COLUMN-I with the entries of COLUMN-II.

I

COLUMN-I

(A) If I is increased keeping / and

COLUMN-II ~

constant.

(B) If f (frequency) is increased keeping I (intensity) and ~ (work function) constant. (C) If

~

is.decreased keeping f and I constant.

(p) (q) (r) (s)

Stopping potential increases. Saturation photocurrent increases. Maximum K.E. of the photoelectrons increases. Stopping potential remains the same.

(D) If the distance between anode and cathode increases.

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Optics & Modern Physics

Advanced JEE Physics

8.

If radiation of energy E, intensity I falls on different kinds of surfaces mentioned, then match the quantities in COLUMN-I with their respective answers in COLUMN-IT. COLUMN-I COLUMN-IT t (A) Radiation pressure for a perfectly absorbing surface. (B) Radiation pressure for a perfectly reflecting surface. (q Radiation pressure for a surface of reflection coefficient

(p).

(p)

I

(l+p)C

(q)

2E C

(D) Impulse on a perfectly absorbing surface. (E) Impulse on a perfectly reflecting surface. (F) Impulse on a surface of reflection coefficient (p) .

(r)

-I C

(s) (t)

21 •C

E (l+p)C •

(u)

E C

= ==================================== 3.42

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In this section the answer to each of the question is a four digit integer, ranging from 0 to 9999. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question number X (say) is 6092, then the correct darkening of bubbles will look like the following:

X.@e@@ (]) G) G) (])

®®®·

®®®®

@@@@ ®

@

® ®

~~~~

@®®@

®®·®

1.

A metal plate is placed 5 metre from a monochromatic light source whose power output is 10-3 W . Consider that a given ejected photoelectrons may collect its energy from a circular area of the plate as large as ten atomic diameters (10'° m) in radius. The energy required to remove an electron through the metal surface is about 5 eV. Assuming light to be a wave, how long, in hour, would it take for such a 'target' to

followed by electrons ejected normally from the plate with maximum energy. 5.

The maximum kinetic energy of photoelectrons emitted from a certain metallic surface is 30 eV when monochromatic radiation of wavelength A, falls on it. When the same surface is illuminate_d with light of wavelength 2A. , the maximum, kinetic energy of photoelectrons is observed to be 10 eV. Calculate thewavelength A. and determine the maximum wavelength of incident radiation (both in A) for which photoelectrons can be emitted by this surface. Given h=4x10-15 eVs and c=3xl08 ms-1 •

6.

On a certain metal light of frequency v = 5v 0 falls then

soak up this much energy from such a light source. 2.

3.

Energy from the sun is received on the earth at the rate of 2 cal cm-' min-1 • If average wavelength of solar light be taken as 6600 A, then xx 10 18 photons are received on earth per cm2 per minute, find x Take 1 cal= 4.2 J, c=3xl08 ms-1 .

maximum velocity of electrons emitted is 8x106 ms-1 , where v0 is threshold frequency of metal. If v = 2v 0 then the maximum velocity of photoelectron is xx10 6 ms-1 .Find x.

Compute the typical de-Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10-10 m: 7.

4.

Light of wavelength 180 nm ejects photoelectrons from a plate of metal whose work-function is 2 eV. If a uniform magnetic field of 5 x 10-s T be applied parallel to the plate, what would be radius of the path, in mm,

A metallic sphere of radius 21 cm is kept in the path of a parallel beam of light of intensity _!_ Wm_,. The 110 force exerted by beam on the sphere is xx 10-13 N . Find

x.

================================:::::; = 3.43

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Optics & Modern Physics

Advanced JEE Physics 8.

9.

The wavelength of light incident on a metal surface is reduced from 300 nm to 200 nm (both are less than threshold wavelength). Find the change in the stopping potential, in volt, for photoelectrons emitted from the surface. Take h = 6.6 x 10"' Js.

In the photoelectric experiment; if we use a monochromatic light, the 1-Vcurve is as shown. If work function of the metal is 2 eV, estimate the power of light used (in W ). Assume efficiency of photo emission

10. If photons of ultraviolet light of energy 12 eV are incident on a metal surface of work function of 4 eV, then find the stopping potential (in eVJ.

11. A proton and an a -particle are fired through the same magnetic fields which is perpendicular to their velocity vectors. Both move such that radius of curvature of their path is the same. Find the ratio of their de-Broglie wavelengths.

to be 10-3%, i.e., number of photoelectrons emitted are 10-3% of number of photons incident on metal. I

10 µA -------~--

_ _-:::___J__________ v

-5 volt

=

3.44

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"- _·-t"f4s

,

4

• _Cif.i·

·1~-.·:

.i 'F:·"

1.

4.7x10-e Nm---2

2.

453A

3.

2.5A

9.

5.5x10-12 m

4.

2.51x1031

10.

AG> Ap

5.

1.43x1019

11.

E, >E,

6.

10"

12.

EP >Ee

7.

(a)

3.9Bx10·" J

(b)

5x1Q15

E!_ASElfQN P.HOfit:fNjP,ROP.ER'TIES

•1Jfeg!1(~ii~t;;:-,;:~( \ii/._~-:~zt?~:i:?~tt1:~~F~:-.. ;:-' ;>;':./ -~_-:... ; ~ ·/?.:.~~:tt~~~:~~-B~S.~DfQ~-!ift(C:?:h9tUECiR1¢lf:FF.Ee]i 1.

2.BeV

a.

6.6 x 1O'" Js , 5990.25 A

2.

aasaA

9.

1.9eV

3.

11.93eV

10.

1.9 eV, 4125 A, No change is observed

4.

2260A

11.

(a)

2.1 eV

5.

9.28

(b)

2.1V

me 12

6.

1.12x10

7.

(a)

1015 Hz

(b)

6.25 eV

(c)

2 eV

c:::================================== = 3.45

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Optics & Modem Physics

Advanced JEE Plzysics •

SINGLE CORRECT CHOICE TYPE QUESTIONS

ANSWERS - - -

---

C

1.

- -

C

3.

4.

A

-

B

5.

C C

6.

D

7.

B

8.

A

9.

C

10.

11.

A

12.

A

13.

C

14.

A

15.

A

16.

B

17.

C

18.

D

19.

C

20.

B

23.

B

24.

C

25.

B

28.

B

29.

B

30.

C

35.

A

21.

A

22.

A

26.

B

27.

D

- -- -

31.

D

32.

C

33.

D

34.

C

36.

C

37.

D

38.

B

39.

B

40.

B

41.

B

42.

A

43.

C

44.

D

45.

D

46.

B

47.

A

48.

B

49.

D

50.

B

51.

B

52.

D

53.

B

54.

C

55.

D

56.

C

57.

C

58.

B

59.

C

60.

B

61.

A

62.

D

63.

C

64.

A

65.

D

-

66.

C

67.

C

68.

C

69.

A

70.

C

71.

B

C

73.

C

74.

A

75.

C

76.

--

72.

C

n.

C

78.

D

79.

A

BO.

D

81.

C

82.

A

83.

C

84.

D

85.

C

86.

A

87.

C

BB.

D

89.

A

90.

C

91.

B

92.

B

93.

A

94.

C

95.

D

96.

D

97.

B

98.

A

99.

A

100.

D

101.

C

102.

B

103.

106.

B

107.

A

108.

111.

A

112.

116.

D

117.

121.

B

126.

D

131. 136.

--- - - --

•

2.

C

104.

A

105.

B

109.

D

110.

A

B

A - 113. A

114.

A

115.

D

A

118.

C

119.

B

120.

A

122.

B

123.

C

124.

A

125.

D

127.

D

128.

C

129.

C

130.

B

B

132.

D

133.

B

134.

C

135.

A

C

137.

C

138.

C

139.

A

140.

C

ANSWERS

1· L

-

MULTIPLE CORRECT CHOICE TYPE QUESTIONS

1.

ABC

2.

BD

3.

ABC

4.

ABD

5.

CD

6.

AB

7.

ABCD

8.

ABCD

9.

ACD

10.

AB

11.

ABC

12.

BC

13.

AB

14.

ABC

15.

ABC

BD

17.

BD

18.

AB

19.

CD

20.

AC

21.

CD

22.

BD

23.

BD

24.

C

25.

A

26.

AB

- --

-- -

16.

I

I

= ================================== 3.46

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Dual Nahtre of Radiation & Matter •

ANSWERS

j

REASONING BASED QUESTIONS

:- ---- ------- - - --------------· -- ------------ ---· ,_ -~- -~---·-· · - - - · -~-~- ____________ 3.__ C _________ 4.:_ __ ~---. _ _ _ _ _5_._D_ _ 6. D 7. C 8. D 9. D 10. C r- ------·-·--------------- -- --·-- -·--- ---· --· ···- .. - ----- --- -·- ---------13. A 14. A 15. A 1_ ~~--~ _ ·- - · . - - __1_2._~ _ ... -------- - -------·--··----- -------------·--------------~ 16. D 17. D

-------

j

LINKED COMPREHENSION TYPE QUESTiC>NS

·-------------- ----------------1.

1 •.

-

C

-----

6.

2.

A

- - - - ---------- -------

A

7.

-

-

-

B

16.

B

21.

B

17.

C

B

--- .

8.

[_·11._0 ____________ 12._o _____________ ,

3.

----

4.

C

5.

--- ----------- ----- -- -

A

9.

-------

---- - -

10.

B

13._B _______________ 1
15.

A

18.

20.

C

B

19.

C A

ANSWERS

---- -

--- - ---- ---

4.

5.

..

---

1200

1-·------·-- - - - - -

, 9.

,_

2.

28

6.

4

-------------

10. 8 ------------ ---------------------7

5.

I

A --> (p, q) B--> (p, r, s) c .... (p, q)

- __D- ~ ~ -

INTEGER ANSWER TYPE QUESTIONS

-----··----- - - --- -- - ------- - - . -i 1. 20

___

A-> (p) B--> (r) C--> (s)

____ o ...._(q) --····-

ANSWERS

,

I

__

MATRIX MATCH TYPE QUESTIONS

' ---- ----1. A-> (s) 2. A-> (s) 3. A --> (p, q) B--> (q, s) B --> (q) B--> (r, s) C--> (p, r) C--> (p, r) c, .... (q) D--> (p, r) .• ··- ···- ___ D--> (p, r) _ -- ·- _____ 0--> (Pl - ·--- -· 6. A --> (p, r) 7. A--> (q, s) 8. A--> (r) B--> (q, s) B--> (p, r) B--> (s) C--> (s) C--> (p, r) C--> (p) D-->(p.r) D--> (s) D--> (u) E--> (q) F--> (t)

•

- - - - - • .J

_________ J'

·---------- --------·-- - ··--··- - --------- ·-----------•

---7

B -------·

-- ---- -- ,. --

3.

7. - -·--

--

31

--

-

-- -- -

-

·-···

42

-

11. 2 ---------- -----------

·-· ---· -

-

. --·

4. 148 ------------- -------8. 2

- - ----~-- - ·-- -----

-

-- - - - -

--- -

·j /

_,

---i

________ j

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1.

For completely absorbing surface,

I Pra<1=c

1.4x103 3x108

6.

4.7x10~ Nm--2

Energy of each photon is E = hv = 6.62x 10--:w x1012 = 6.62x1042 J Number of photons present in 6.62 J of radiation energy is

calculated by using

2.

E=N(hv)

f~~=£ 21tr1 A· ).

2ncr1

·v,

N=_§_ (2,)(3 x 10 )(0.529 x10-")(1 o") (2.2" 10 8 ) 8

hv

A 7.

(a)

6.62. 6.62x10""'

10""

Energy of each photon,

•

E =hv = 6.63x10--34 x6 x1014 =3.98x10-19 J (b) 3.

For an electron, de-Broglie wavelength is given by,

l.=J150 =J150 =,/6 V 25 => 4.

N=!'_- 2x10" W E 3.98 x 10-" J

he

Energy of one photon, E = hv =-

l.

8.

6.63x10--34 x3x10 8

------- J

The total energy of an electron with rest mass m0

E = ~,..m"":-c•""+-p~'""c=• · where c is the speed of light in free space. Law of'Conservation of Energy requires, Rest mass)+ (Energy of) = ( Total energy of) ( of electron photon moving electron

104 x500 -2.51x1031 6.63x10--34x3x108 •

=> · Energy of one photon is

E

2

4

m0 c :+ hv = ~m~c + p2 c2

••• (1)

Law of Conservation of Momentum requires.

E=hc l. =>

and

momentum p is

500

Number of photons emitted per second is n = Total energy emitted per second Energy of one photon

n=

N = 5 x 1015 phoions per second.

=>

=10 kW = 1o• W

So, total energy emitted per s_econd is- 10• J

5.

P =N (energy of each photon) P=NE

=>

l.~2.5A

Power of transmitter

=>

If N is the number of photons emitted per second by the source, then Power transmitted in the beam is

( 8

6.625x10--34x3x10 -S.Sx 10_19 J 10

5600x10" Since a 100 Wbulb supplies 100 J of energy per second. So, energy released per second as visible photons is

E'=100x5= 5 J 100 Number- of photons emitted per second as visible light is E' 5 1.43x1019 E 3.5 x10"18

=>

Initial momentum)+ (Momentum) = (Final momentum) ~ of electron of electron of ~hoton

h

O+i;:=P hv c=P

=> hv=pC Squaring both sides of equation (1), we get 4 2 ·m~c + h v 2 + 2m0 c 2hv =m~c4 +p2 c2

=>

2(m,c')(hv) = 0

1--- l.=~l ... (2)

{using (2)}

This is impossible. In a frame, where the initial electron is moving with uniform velocity, the same conclusion must hold

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Dual Nature of Radiation & Matter because if a process is forbidden in one inertial frame, it is also forbidden in another inertial frame.

9.

11.

Kinetic energy of electron is E - 1mv2- 1 _(mv)2 - p2 = h2 e 2 2 m 2m 2mA2 Energy of a photon is totally kinetic and is given by

KE of an electron is EK= 50 ev·=1.6x10-19 x5x 104 J = Bx 10-15 J

he Ep =hv=-

de-Broglie wavelength of electrons is

l.

l.=-h~2mEK

=> =>

=>

----p----ee=--=m '12 9.11 x1 8 1 6.63x10--34

l.

A

X

6.63x10-11

12.07

Q--31 X

X

=-

Q-lS

-55 10-12 m- · x m

1 2 1 (mv)2 p 2 KE=-mv = - · - - = 2 2 m 2m

12.

Ep

mp

2mcA

Ee ::: T. h2 = -h-

The total relativistic energy of a particle is E = ~moc4 + p2c2

5

The kinetic energy of a particle of mass m can be expressed in terms of its momentum p as follows

As wavelength A. is same for both electron and proton,

£

Since, momentum, p =

is same for both particles and hence

p 2c2 is same for both. But rest mass m0 of a proton is greater than that of an electron, therefore, the energy of a proton is more than that of an electron of same wavelength.

As de-Broglie wavelength, A=~

p

~=p' ~J2m,-KE = Ap Pe 2m0 • KE

he 2mA2

Thus the KE of a photon is greater than that of an electron of same wavelength.

the resolving power of an electron microscope is about 10 times greater than that of an optical microscope.

=>

Ep

Ee= me

For yellow light, wavelength A =5.9 x 10-1 m As resolving power is inversely proportional to wavelength, so

-10.

(·.- P=£)

/m,

vm::

As m0 < mp , therefore, i..0 > AP Thus the electron has greater de-Broglie wavelength.

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.1.

2.

Since, E=hv=(4.316x10-" eV'sec)(1.5x10" sec-•)_=6.5 eV => K= =E-W=(6.5-3.7) eV=2.8 eV

5.

12375 Sl·nce. E("1n eV) - i.(in A) =>

12375 ;, 5 eV =8x10-" J 2480-' Number of photons emitted per second from the source is given E=

E _,12375_206 V d 2-·6000 :- . e an

by

E ·= 12375=1.77eV

p 40 19 n, =-=---=5x10 19

3

7000 Since, W is 1.9 eV, photons of energy

E Bx.10Number of photons incident ·per unit area per second on magnesium surface is 19

n = n1 = 5x10 2

•

4nr

2

~

E, = 12375 = 2.475 eV 5000

~ and E2 can only emit photoelectrons. Charge emitted per second is

.'!=(16x10-")(.!) (1D"")(lO_.) 2 t · 31tx(10-3) x2.475x1.6x10-19

1016

4rt(2)2

From Einstein's Photoelectric equation, we have K= =E-W=5-3.68 =1.32 eV

s·ince,_ =>

3.

i. (· A) om

+(1.6x10-")(.!) <;o"")(rn-•) 3 nx(10"") x2.06x1.6x10-"

12375 W(in eV)

1. = 12375 = 12375 =3363 ' W 3.68

=>

A 6.

.!S_ =5 K, =>

AE,-W = 5 AE,-W

..• (1)

.1

• •

7=9.28x10"" C=9.28 mC

E,=-0.85 eV E,=-1.51 av E,=-3.4 eV

E, = 12375 =3.1 eV 4000 12375 E, = = 2.57 eV 4800 12375 E, = = 2.06 eV and 6000

E, = 12375 = 1.7? eV 7000 Therefore, light of wavelengths 4000 A , 4800 A and 6000 can only emit photoelectrons. So, number of photoelectrons emitted per second is

A

n = l 1A1 + I2 A2 + l 3A3 E, E2 E3 E, =-13.6 eV

;;;::),

n=IA(E1E2+E2E3+E,EaJ

E,E,E,

Here, .6.E1 =E,-E1 =12.75eV and .6.E2 =E3 -E1 =12.09eV

Substituting in equation (1) and solving we get W=11.93 eV

4.

=>

3 eV=12375_12375 1. 5000 1-=226oA

(a)

~v 0 ~W=3.75x1.6x10- 18 J

;;;::),

K= =E--W =>

=>

_ (1.5x10->)(10~) 3.1x2.57+2.57x2.06+3.1x2.06] n,s . 1.6x103.1x2.57x2.06 1 n=1.12x1Q 2-

7.

=>

v 0

= 3.75x1.6x10-19 6.6x10-34

:::,'

10,s Hz

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Dual Nature of Radiation & Matter (b)

E=

=> (c)

12375 eV = 6.25 eV 1980

Kmax

hc _w 4=-1.,-

= 2.5 eV

hc_w

i.,

Kmax =2 eV

4x12375 5400

=> 8.

K mrut

=E-W=hc_hc A Ao

_ x _ x _,, 0 18 1 6 10

=>

h(3x10') 5461x10-10

he

h(3x10') A0

... (1)

10.

he 0

4.6x1.6x10-"

h(3 x 1o') 1849x10-10

h(3x10') A. 0

9.

A

11.

2

i.,

... (1)

(a)

1.l= 12375 -W

... (2)

E{in eV) = 12375 = 12375 = 4.4 eV i. (in Al 2800 From Einstein's Photo-Electric Equation, we have

Kmax =E-W

.!mv/ = he -W

... (2) 2 i., Dividing equation (1) with equation (2), and using the given fact

that v, = 2v2 , we get

•.. (1)

Since, W=2.3 eV and 1.=2800 A

From Einstein's Photo-Electric Equation, we have

.!mv/ = he - W

0. 6 = 12375 -W 4950

Solving above two equations, we get W=1.9 eV and 1.=4125 A No change is observed.

h=6.6x10.,.. Js 1.0 = 5990.25

W=1.9eV

).

•.. (2)

Solving equations (1) and (2), for h and 1.0 , we get and

_ eV 5 63

From Einstein's Photo-~lectriC Equation, we have

Further, eV,=E-W=T-"i:"""

=>

12375 3500

(b)

=>

K= =(4.4-2.3) eV

=>

Kmax=?:.1eV

Since, K= = ev,

=>

2.1eV=eV0

=>

V0 =2.1V

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1.

[CJ E=nhv _§_ = P = (-"-)hv

=>

3.

t

.

t

[AJ

10.

he s ev, =T-w he

ev, = 3;. -W

=>

4.

. 1 2 E=m 0 c2 +-m 2 0v

[CJ

... (2)

Since, E=_E__ 2m

have equal value of momentum.

'

=>

3).

W= he 6).

[BJ

11.

) nearest possible ( approximation to answer

Ee= mp =1840=2000

me

EP

3).

4W=2he

=>

=>

Since both have same de Broglie wavelength, hence both must.

he -W= 5he _ 5W =>

E=m,e'(1+

... (1)

Solving equation (1) & (2), we get

).

:0,)

=>

[A]

E2 = p2c2 + m~c"

=>

E2 =p2 Cf+E!

=>

P=~J1-(~

J

=~ (tor E»E,)

l.=-h../2meV

=>

"A.'= __ h_

For a photon

../'2MeV

=>

).' =

E=hv=he l.,

;.Jfj =>

8.

l. =he=l. ' E

12. [AJ

,

9.

=>

A.= kE2

=>

log, l. = logk -½logE

[CJ E=mc2

E=mc2

Since ~«1

=

m0 c2

5

mc2

- FITT3'

The energy of composite lump after collision will be Mc2

By energy conservation principle we get Total initial energy= Total final energy

=>

=>

=

Energy of each lump before collision is

=>

e

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_§_m .c2 +_§_m c2 =Mc2 3

0

3

0

•

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Dual Nature of Radiation & Matter 13.

[CJ

18.

mv r=qB

E=hc (4.14x10·")(3x10') A 3x10-7 => E=4.14eV So, electrons are emittec;t with a kinetic energy= 4.14-1 = 3.14 eV

2x107 r = 71.-=7:::6-x71o"''"'x--=-2-x710,-_.,., r=0.0055 m D=2r=0.011m D=1.1cm

=> 14.

[DJ Energy of falling light

1 zmv

=>

[AJ

2

=3.14x1.6x10-19

2 X 3.14 X 1.6 x1Q-19 V=

E, -E, =13.6[;- ~] 3

9.1x10-31

v=1.2x10 6 ms-1

13 6 5 AE= · x =1.89 eV 36 Photoelectrons with Kma)[ are moving on circular path, so

19.

[CJ 1.=~ mv

mv r=qB

=>

1.=--h-

mv =qBr

mJ2~V

1 p = qBr =1.6x10- x --x10-3 3200 19

=>

1.=--h. ~2rnqV

=>

i., i..

p=..:!.x10-24 =5x10-25 kgms- 1 2

,

Energy of photoelectron is K

=

=

K

=L

25x10-so

eV

20.

Now using Einstein equation, we get hv = cj> + Kmax

15.

mpQp

{·:m. =4m,. q

0

=2q,)

1.89=0.56+$ $=1.03eV

[AJ

[BJ Roc.! and 1.= ~ h oc ;., 1. 2mqV -.,V

K= =0.86 eV

=> =>

Jm·q• = 2-,/2

2m

2x9.1x10-3 1 x1.6x10-19

max

=

0

21.

F = .!.(Effective Area) C

=>

Roc.Jv

=>

~ =J~~~~ =2

=>

R,=2R=2R

[AJ A NiA, +N2A2 N1+N2 _ _ N,(10)+N,(11). 10 81

16.

N1+N2 Solving·, we get N, 0.19 19 N, = 0.81 = 81

[BJ

23.

[BJ

h he P E For a particle of mass m moving with a velocity v de Broglie relationship is given by

Fora photon A=-=-

Let

=>

p=-E

1.=~ mv

c

Ap=_E,_-(__E,_)=2E C

17.

C

24.

C

E=n(~)

[CJ

Also

1.=-h~ -hmeV So, when V becomes 4 V, A. becomes

[CJ

..!1: 2

!: 2

~

Ea = Energy 0 Volume

°

Energy=

(ie

0

E~ )(Volume of cavity)

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Optics & Modern Physics

Advanced JEE Physics 38.

Energy= ( i,,E; )(Area( ct.t))

[BJ

1 ' = k8 T -mv 2

=>

v=l~T

n~ =i<,E;(Act.t) n _ (Number of photons striking the desk) Mt (Area)( time)

=> 26.

l.

=>

~ = N =~ e0E~ Mt

2 h

39.

[BJ

..!.m v2 =eV

[BJ

2

Kx, = !(sR') =>

h - ~2mk,T

p

=>

x0 = K~ (nR')

V=

teV m,

Since, '),.. =~

m,v

30.

[CJ

According to Einstein's Photoelectric ~quation

=>

EK =hv 0 -4> =>

0=hv0 -«j>

A=

48.

Stopping potential of 1.36 V implies EK = 1.36 eV . Since hv = $0 +EK

=>

de-Broglie wavelength) for a proton accelerated t~rough a potential V

vV

J

[BJ

P= nhc l.t

[CJ

=>

(is

o 287 · r.; A

12 27 ( is de-Broglie wavelength A.= ---+.,;A for an electron accelerated vV through a potential V

(6.626x 10_,. )(3x10') l.,

=> .;4 X 1.6 X 10-19

34.

l.=--h~2mpev

( 4.14 X 10-" )( 3 :10•) 5000 x10-10 2.46=~, +1.36

=>

1.7x10-18

=>

n=S.15=5

n X 6,6 X 10-34 X 3 X 108 6000 X 10-lO X 1

~, +1.36 49.

[DJ

(e/m,) = 2 (2e/4m,)

~,=1.1eV

35.

[AJ Stopping potential is independent of intensity and depends upon frequency.

36.

[CJ

50.

[BJ According to Plank's Quantisation Law.

E=nhv=n(h:) Saturation current a: (Intensity) => Saturation current= 4( 0.4 µA)

=> 37.

=>

Saturation current=1.6 µA

10

[DJ

..!mv2 = he -W 2

..!:mv'2 =~-W 2

... (2)

3l./4 ,

Dividing,(~)

4hc_w

2

=

~~ -W l.

_, = (!l_

(6.626

t

X 10_,. )( 3 X 10') 5000 X 1o-10

11 ~=2 t .5x10

..• (1)

l.

f=(~)(~)

52.

[DJ

he -=~,+2.5 l. he (l./2)

•

=,o +

E' K

E~-2.5=hc, l.

E; =2-5+4-14 E~ = 6.64 eV

So, stopping potential is 6.64 V > 5 V

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Dual Nature of Radiation & Matter 53.

[BJ

64.

•=~

[AJ

F = .!.(Effective Area)

mv

C

If Eis kinetic energy of electron, then

=>

F=.!.(,R')="R'I

E=.!mv2

C

C

2

65. V=~

Speed of first electron may increase or decrease, depending on

the direction of electric field. However in the case of electron

,=_h_ ,l2mE 54.

[DJ

entering the magnetic field speed remains constant. Since, from de·Broglie relation

•=~

[CJ

mv

Energy of falling photon< Threshold energy. So, no photoelectric effect takes place.

=>

A1 > "-2 OR '-2 > A1

are both the possibilities.

55.

[DJ

66.

Since A=!!

p

[CJ h

i= mev

For identical p's, A are identical

=> 56.

67.

hv=EK +~

[CJ

,, ,,

hv-~=.:!.mv2

K1=hc_w

... (1)

K2=hc_W

... (2)

2

V=~

Since, A1 = 2A 2 , so (1) becomes

m,

V= • /-2~(_hc~--·~~) 58.

0

K,=~-w=.!(h )-w 2A.2

(BJ

60.

1

2

63.

K

.

= K, _ W

'

2

2

K < K2 ' 2 68.

[CJ Total energy radiated per minute from sun is

2

mv =eV

=> 61.

1

[CJ Photoelectric current oc Intensity => iccl => i=kl

[BJ

2 A2

K, = (K, +W)-W 2

=>

Stopping potential is 4 V. So, maximum K.E. is 4 eV.

59.

,=3.64nm

[CJ

Eradiated

Energy radiated annually is given by

v=~2eV , m

Etotal

~

[AJ Since .6.E.6.t 2:: h

=>

t.E( 10.,,) = 6.6 x 10.,.,

=>

.6.E=6.6x10-2fi J

[CJ

= cr( 41tR!e)

69.

=24 X 60 X 365 X Erndia!ed

Annual loss of mass= Am= E10; 11 = 1.38 x 1017 kg C

[AJ Since V=uc the electron is moving at non-relativistic speed. By Law of Conservation of Momentum h h

{where a« 1)

-+-=mv

1mmx1 mm2 C

N

=>

100x10-9 x2640x10-10

(3 X 10')' X 6.63 X 10....

N = 442 photonsmm-3

' ,'

he he mv2 -+-=mCV=--

~

A

u

A.'

... (1)

By Law of Conse,vation of Energy ..!mv2 2

= he_ he A

A'

... (2)

Adding (1) & (2), we get 2hc=.!mv'+.!(mv') , 2 a

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Advanced JEE Physics =>

mv 2 4

he

74.

mv2 2a

-=E=-+-

A. T Since, a« 1

::::)

Optics & Modern Physics

F =_f'_= IA

71.

=>

mv 2 mv 2 -«-

=>

E,=-Hm;')

=>

!mv2 =aE

2

=>

A= 41tr2 =4x3.14x(2) m2

=>

F0 • =2.35x10-4 N

2a

2

75.

[CJ

KE=hv+~ 2KE=hv'+~ => 2{hv+$)=hv'+~

'

[CJ Photons are exchange particles for electromagnetic interactions. Gravitons are exchange particles for gravitational interactions. Mesons are exchange particles for nuclear interactions. Whereas protons are not any sort of exchange particles.

76.

=>

v'=2v+f

=>

v'>2v

h

[CJ mv

Relativistic relative velocity of approach is given by

m=g-

c-(--c) 1-(c)(--c) c'

v,

77.

V-> C

=>

1--> 0

[CJ EK =hv-$0

=> =>

[CJ By Law of Conservation of Momentum, we have 0 =mv, + 2m{-v,)

=> mv1 =2mv2 Now, according to de-Broglie relation, we have

=>

(E,),=_! {E, ), 4

=>

v' _!_=v' 4

=>

~=.! v, 2

'

... (1)-

1-=!1. p

78.

[DJ

21A

Force=~

=>

C

=;,

Fx=(iKx')xa

=>

(2t)x=¾Kx'

=>

X=i(~)

[CJ The vector is perpendicular to initial velocity vector at

t = _u_ gsm8

and at this instant its speed is

v =ucot0 Since, A=

=>

=>

=

(E, ), = 1-0.5 = 0.5 eV

Similarly (E, ), = 2._5-0.5 = 2 eV

Never expect the relative velocity to grow beyond the speed of light.

73.

'

As

m-;oo

=>

72.

... (1) ..• (2)

h 1-=- where,

[BJ

=>

C

where I= 1.4 x10 3 Wm 4

(J.

70.

C

ox

..!.»1 4

[A]

J!... mv

l.=--hmucot0

A=(!u}an9

79.

(AJ KE.=EK =hv-hv0 =hv-lj,0

he 1( 4.14 X 10"" )( 3 X 10')

=>

EK =--lj,o

=>

E, -

=>

E, = 6.21-5.01

=>

E,=1.2eV

2x10-7

5.01

So, stopping potential required is 1.2 V

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Dual Nah,re ofRadiatio11 & Matter BO.

[DJ For ti.yo partiCles to have same de-Broglie wavelength ~mqV =constant

90.

mpqpV=m.. q V' 0

Force per unit area is pressure

V'=mp Qp V m.. Qo

9" =.! 6p

=>

A 61 But for a photon E =pc , where p = momentum of photon E = energy of photon

V'='!_ 8

81.

f=6p 6t

=>

~

::::;,

[CJ The time rate of change of momentum equals force.

[CJ

6E= e6p

Energy of each photon,

E

he l.

Because irradiance I is defined as energy per unit area per unit time.

E=6.6x10-19 J Power of source P=IA=1.0x1.0x10-' =10-' watt

91.

10"' 6.6x10-19

P

E

h

.e.=E e

=>.

1 10-' N'x--'--'----100 6.6x10-19

82.

he

mv=E

Number of electrons emitted is

::::)

[BJ

A., =AP

So, number of photons per sec is

N

go=_l_6E=_:t_(~)=.!{I) Ae6t eMt e

=>

6.6x10-34x3x10 8 300x10·'

92.

N'=1.51x1012 per second

[BJ Stopping Potential V5

[AJ

= Kmax e

Transverse deflection is

1 E '

_ 1at'- 1(qE)(x)' y2 2mv

y

_2'1" mv

93.

2

[AJ

E= he l.

!qEx2

Number of photons emitted is

2

2-(Jm:1')

Pt

(htrn,

For same electric field E, kinetic energy EK and length x, y a:: q ,

q is smaller for proton, so y is smaller for proton. So proton's trajectory will be less curved than a-particle's trajectory.

84.

=>

Since the radiation is spherically symmetric, so total number of photons entering the sensor is n0 times the ratio of aperture

[DJ EK= hv-hv0

area to the area of a sphere of radius l .

EK ::::::::y, v::::::::x, hv0 =constant(k)

=> 86.

y =hx-k (Equation of a straightline)

[AJ

mv

88.

=>

94.

A=~

P,t

=t,c

no

N

=n n(2d)' =Pl.Id' o 41t£2

(CJ he

"i"=$,+3V,

... (1) ... (2)

=>

l.- 6.626x10_,_, ( 10---11 )( 105 )

he 21' =$,+V,

::::;,

A.=6.63x10-a m

Subtracting 2V0

[DJ =hv-hv0

he

=n

is the equation of a straight line with slope h.

V _ he 0 - 4/o.

[AJ

$, = 41'

EK

he £2

he

89.

EK=hv-hv 0

he

he

l-0

= 41 3.57

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Optics & Modern Physics

Advanced JEE Physics

which is again equation of a straight line with slo_pe (;) .

95,

[DJ 104. [AJ

p = hv (for a photon)

5% of 100 Wis 5 Js-1

C

3 _3 x 1 0-29 = 6.6x10-34v 8

:::::,

3x10

v=·1.5x10 96.

13

Hz

[DJ Decreasing the A. of incident photon means energy of incident

=>

5=~(~)

=>

1= he

=>

t = 6.626 ;10--" x 3 x 10

=>

!!=1.4x1019

n 5l.

8

light is increased, so E1t increases and hence stopping potential

increases. 97.

[BJ

A

E

Ao

= 4.14x10-15 x3x10 8 (_!_.:!_) 10-7

K

=>

E,

5

6

= 0.414 eV 15

h=4.14x10-

=>

(2x10-")ap=6.6x10"'

=>

~P =3.3 x 10-20 kgms-1

•

106. [BJ

v 0 = 5x10 14 Hz

Wherever necessary we take h = 6.626 x10-34 Js

:::::,

t

105. [BJ Since .1x.1.p - h

E-- he_ he K -

5(5.6x10-7 )

n

Since, c=va').

eVs

6

98.

(AJ

=>

A.= 3x10 5x1014

l.=~ mv

=>

l. = 6000 A

=>

dl.=~~(v-')=~(-1)~-,dv di mdt m dt dl. h dt=- mv2 -a

=>

107. [A] Kinetic energy of recoil is EK =hv-hv'

Also V=U+at

=>

V=at

:::::,

dl. dt -

{·: U=O}

h ma2 t2 a -

h mat2

E,=12.375-9.375 Et<=3eV

108. [AJ

l.=-h../2meV Putting values of h, m, e, we get

Also ma =eE dl. h => dt=- eEt2

99.

=> =>

=>

l.=~A

=>

l. = 12.21

[AJ

l.=~ mv Since electron is the lightest particle of these four. So A. 8 is maximum.

.Jv

A

{·:q,=q,) 100. [DJ

I a:: _!_

d'

when d becomes

f,

I becomes 161

110. [AJ l>x

=-1..._ m 100

Since

t,xt,p = h

102. [BJ

hv=hv 0 +eV0

= hv-hv0

=>

eV0

=>

h h V0 =-v--v0

e

=> =>

6.6x10--34

(2/100) .6.p=3.3x10-32 kgms-1

e

= ==================================== 3.58

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Dual Nature of Radiation & Matter 111. [A]

.

=>

-h

N ;= _Total number of photons striking the totally reflecting

l. = ,/2mE

F

=>

mE = constant

=>

me <mp <mca E1 >E 3 >E2

F)..

(~hr 2h

screen =

N=5x1026

::::)

117. [AJ 112. [BJ

,/2mE =~ ).

E=hc_~

... (1)

).

2E=hc_,_ )!

=>

h' E=-2mA2

.. ,(2)

'

2E - E = hc[1- _ ).'

).

).'

E>. + he

he

).

).'

118. [CJ

2nh =1

---=-

).

l.'=(~) E>.+hc

).

n=-

2h

113. [AJ

119. [BJ

In order for scattering to occur, the wavelength of the waves must be of the same order of magnitude or smaller than the size of the object being observed. Hence the largest possible wavelength we can use in the present problem is Amax =2.5 A . Hence minimum energy is

he

Em1n=,:--

=

E

_ 12.40x10' mill2.SA

Emln

= 4.96x103

vA

eV

2

he

1

=>

~=T-2mv

=>

~=

2

1240 -1.68=1.41 eV 400

E=pc=hv hv

=>

mc2 =m 0L

(1)(3x1o')' =m0 (80x4200)

P=c

124. [AJ

m0 =.2.67 x 1011 kg

s

=~

s '

=~

• m,

115. [DJ Density of star of mass M0 , radfus R0 is M

p=--·i1tR3

m,

S=2e=e=1s a4mp2mp2p

0

Se is maximum, then comes SP and then S"

On contracting the new mass becomes

(_!!._)'

M = .'!_nR'p = M0 3 R0 Loss in mass due to contra"ction => Loss =M0 -M

=>

~ =-½mv +cl>

121. [BJ For a photon

114. [A]

3

Since,

120. [AJ Both are independent of each other.

Emin = 5 keV

=>

2x20x1.66x10--21x(6;6x10-10 )2

1 10 --48+27 E--'-'-x10-"J 40x1.66 4x1.66 E=1.5x10""" J

_!] = E

E+hc=hc ).

(6.6 x 1O""' )'

Energy radiated

125. [DJ

I oc _!_ d'

= -AM

On doubling the distance the intensity becomes one fourth i.e. only one fourth of photons now strike the target in comparison to the previous number. Since photoelectric effect is a one photonone electron phenomena, so only one-fourth photoelectrons are emitted out of the target hence reducing the current to one fourth the previous value.

= c2 AM

116. [DJ -

F=~ = ~t[i-(-i)]=N(~) { ·.- Photons rebound with the same initial value of momentum}

3.59

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Optics & Modern Physics

Advanced JEE Physics 126. [DJ Let threshold frequency be v0 • v=1.5v 0 When the new frequency is halved v' =0.75v 0 < v0

V =he_ W s eA. e

134. [CJ i.(in A)

12400 E(in eV)

So, no photoelectric effect takes place. 1.=

127. [DJ Ap=m,.va

mp Vp

=>

_!=4v" 2 v,

=>

Vp

E=nhv

=8

v.

128. [CJ 1 eE l

V

Here E=x

f

=>

10000 =

=>

~=1.71x1031

x'=4x

T( 6.6 x 10-" )( 880 x 1000)

136. [CJ Maximum microwave frequency= 3 x 1011 Hz . E hv Smce, p=-=c C ( 6.626 X 10°"' )( 3 X 1011 ) p_

.

(3x10')

Pma,c = 6.626 x 1o-31 kgms-1

=>

130. [BJ

=Unhv

t

=>

=>

137. [CJ

Given that

=>

=>

2

2 =2m v2 For same. (near1y) circular path y is same E => 2 =constant

=>

12400 =3100A=310 nm 4

135. [AJ

A,.

=..:!.at2 Y

{ ·: W = work function}

!1. = he = ...!1... p

E

mv

E=mvc

mvc

mv

=>

1 mv2 Ee =i__=_!_ EP

A.=·~

2c

Amax= 7500

A

A=0.24x10-10 m

138. [CJ

(m-mo)c2 =2moc2 =>

131. [BJ

6.62x10-34 9.1x10--31 x3x10 7

m=3m0

140. [CJ

So, E = he 1.

Stopping potential= 1 V

=>

EK=1eV

=>

i

E= (4.14x10-" eVs)(3x10' ms-1 )

7500x10-10 m

mv 2 =1.6x10-19 J

E=1.6eV 133. [BJ

eVs =KEmax =hv-W

=

3.60

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1.

[A, B, CJ hv =KE.(T) + Work function (W) => hv=T+W

=>

4.25 eV = TA + WA (for Metal A)

=>

4.70 eV = T, + W, (for Metal B)

Q

n=e ---2ons--._

----------------~---. r=

Since T, =(TA -1.5)eV

J_________ _

h Also A=-

p

8 x10-13

=>

l.=-h-J2mT

=>

).A = f t Aa TA

s·ince

AA

1 =2A

n 1.6x10_,g

:::::i,

Sx10s

Then the number of photoelectrons emitted per light pulse is n ' = ~ =5

10'

and hence the number of photons in one light pulse is 8

N=~=SO

=>

2.

0.1

TA =4T8

=>

T, = TA -1.50 gives

=>

T8 =4T8 -1.5

=>

T, = 0.5 eV

=>

TA =2 eV

=>

WA =2.25 eV

=>

W, =4.20 eV

8.

[A, 8, C, D] According to Heisenberg's Uncertainty Principle the product of (a) uncertainty in position and uncertainty in momentum h cannot be greater than - . 4s (b) uncertainty in energy and uncertainty in time cannot be

h

(c)

B,D Since 'PP=.!.=104 Nm-2 C

(d)

greater than - . 4s uncertainty in angular position and uncertainty in angular h momentum cannot be greater than - . 4s uncertainty in generalised coordinate and generalised h momentum cannot be greater than - .

4,

10. 3.

[A, B, CJ The total quantity of charge carried by one pulse of current is

f

Q = Idt

which is the area of the triangle in figure. Thus Q

=!(2ox1o"')(sox10-<>) = Bx10-" C 2

and the number of electrons carried by one pulse is

12.

[A, BJ Wavelength of UV radiation is less than 5200 A where as wavelength for IR radiation is greater than 5200 A. Hence photoelectric effect will be shown by UV radiation irrespective of its intensity.

B, C For a photon

pc-= hv

3.61

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OpHcs & Modern Physics

Adva11ced JEE Physics

14.

hv

mv' Since, qvB = - r

=>

p=-

::::>

p=8.8x10..-28 kgms-2

=>

p=1.65x10.. MeV/c

c

::::>

=>

[A, B, CJ

f=~ l.

(A)

3x108

sx10-14

600x10..

=>

N=£_

(B)

=>

hf

B = mv qr 8

=~2mKmax qr

= .J2 x 9.1 x 10-a1 x ex 1.5 ex1

1 3x91x10-3 B= .

1.6x10-19

B = 4.13x10-' T ~ 4 µT

X-rays are of order of 0.1 nm= 12.4 KeV

(C)

KE

(D)

KEmax depends upon frequency of incident photons and

= 1240 -1.07=1

600

max

22.

not distance of source.

17.

[B, DJ Energy of photon incident hv = 12400 eV 4000 => hv=3.1 eV Since, hv < work function of all metals Hence no electron will come out If 1.. = 200 nm

[B, DJ A.red > Avio1et VIBGYOR pattern shows that VIBG all have l. less than that of

yellow colour and hence can initiate photoelectric effect irrespective of intensity.

12400 = 6.2 eV 2000 Since, 6.2 eV > work function of all metals, hence photoelectron will be emitte:)d.

then hv =he=

18.

1..

[A, BJ =4x10-19 J= 4x10-19

K

1.6 X 10-19

max

=>

2.5 eV

Stopping potential = 2.5 eV

23.

Since, Km,. =hc(l_..!..)

l.

19.

l.,

the s~me. Since distance becomes 3 times, so I becomes

½.

=>

2.5 = 1240 _ 1240 300 A. 0

Hence photocurrent also decreases by this factor i.e. becomes

:::::>

A0 =759nm

~=2mA

9

[C, DJ

24.

[CJ Work function is the intercept on K-axis i.e. 2 eV

25.

[AJ 2 = 4.14 x 10-15 v

Infrared light has wavelength greater than 5200 A and UV has A< 5200

A.

So UV lamp will be able to get the photoelectrons

emitted by a surface irrespective of intensity.

20.

[B, DJ Cut off voltage is independent of intensity and hence remains

=>

[A, CJ Kmax =1.5 eV 26. l..=-h,J2mK ::::::>

A=1nm

E of incident photon

=>

=

~~!

1

v=4.8x1014 Hz

[A, BJ Photoelectric effect and Compton effect are explained on particle nature of light i.e. light is considered to be made up of a stream of photons.

::: 2.5 eV

~=2.5-1.5=1 eV

= ==================================== 3.62

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6.

[DJ h

l. =-

p

8.

,

Same

15.

[AJ

m=Hm,

for both

---,-

[DJ Kinetic energy; E =.:!.mv2 2 v ex:

~

m,

,=H

2m

=qV

1--

c'

because V is constant

V=,/3C

2

16.

[DJ K= =hV=~ KE of emitted photoelectrons varies from ze_ro to

=>

v,:v,:v.=1:~:~="2:1:1

17.

~

[DJ

,. =12};1 A

13. [AJ Energy of photoelectron emitted is different because after absorbing the photon electrons within metals collide with other atom before being ejected out of metal. 14.

[AJ

~=J E 2 =J1o'eV =1 A.ph

2rn8 c

1MeV

==================================== = 3.63

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1.

[CJ . · Smee, Kmax

he

Efficiency of photoelectron generation per incident photon is ~ =(~x100%) = 1s xhc x100 n ex PA

he

=T- 1

•

..!mv2

=he_ he

2-max.

).1

,..Ill

..!mv2 =he_ he 2 maxz A2 A111 Dividing (2) by (1 ). we get he he

... (1)

2Q X 1 a-e·X 6.6 X 10-34 X 3 X 1 Q6 X 1QQ f\= 1.6x10-19 x1x10-3x3000x10-10

••• (2)

~=(~)(66)= 66 =8.25% · 160 10 8

4.

[::J ~~:~~

mv'

-=qvB r

=

).1

[CJ

=>

).!h

~2mEK = qBr E =(qBr)'

1500 1

(2)'

3000 '• 4 4 1 1 3000-,. =1500- ••

E. =0.86 eV

5.

[BJ

E3 -E, =13.6( ; -

4

1x2 3 ------=3000 1500x2 '•

2.

=>

3

--=3000

=>

'• =4500

'•

1,)

2 3 ,1.E=1.B9 eV

Since, if,=E-Kl!IIUI

=>

A

=> 6.

~=1.89-0.86=1.03eV

[AJ

1.=~ . ,1.E

[AJ l 8 oc PA

=>

3.

2

2m

K

'• 1

=>

I~

= P2 A2

r,

P1A1

=>

~= 5x1500 20 1x 3000

=>

I~ =50 µA

7.

P,

Since, n '

=1..e

A

[BJ 1 1.oc,= ;1mq

8.

[BJ For case I, we have

n=-· he

, = 6565

[AJ i. ·= 12.21

•.. ..Jv

9.

A

[CJ Since, A=~

p

= ================================= 3.64

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Dual Nature of Radiation & Matter Further, Pe1ee1rcn = pat>ha . :::::,. 10.

17.

[CJ L\p=PA.6.t=10--4x10--4x10=10-7 kgms· 1

18.

[BJ

Wavelength will be same

[BJ

Nh -i:-=t.p

J. = ..!!_ = 6.63 X 10" mv (66.3x10")(5) => l. = 0.02 m 11.

[DJ Fringe width

=> 12.

N 10-7 x663x10..g 6.63 X 1Q--:W

p is given by

19.

P=l.D 12xo.02 d 0.6 p = 0.4 m

[AJ Since,~ =hv-W If KITIII)( and W both are in eV , but hv is in joule, then we have

= = hve -W

K

[DJ de-Broglie wavelength of electron is

So for both the frequency values, we have

_ 05

l.-

:::::i-

6.63x10" 9.1x10--31 x10 7 A. ::::,-1021 m

2 21

22

h

[BJ

l.= ~ = /150"=10A=100nm v2mE

Vii

14.

15.

B= mv =4.1x10-a T qr

[AJ

Photon energy = ~ = :::: = 2.5 eV 1

Since, Va = 1.5 V So, work function is given by

=>

:::::,. 16.

~=2.5-1.5=1 eV 12375 l.'""=--=12375A=1237nrn 1 "-min 1250 nm

h(12x10") W

... (2)

e

1:sx1.ex10·19 4x1014

[CJ

From equation (1 ), we get _ = (6x10"')(Bx10") 05 1.6x10·19 "=> 0.5=3-W => W=3-0.5"=2.5 eV

mv' qvB=r =>

... (1)

e

h = 6x10-34 Js

20.

[AJ

h(Bx10") W

Subtracting equation (1) from (2), we get h(4x10") 1.5 e

Fringe width p = AD :::: (~) x 10 z 10 m d 0.6 Fringe width is so large that it is not possible to observ~ it.

13.

1020

21.

W

[BJ

l.=-h~= 12.21 A ,/2rnK ,/v 1 A=12.27../2 A => l.= ~ -.,0.5 =>

l.=17.35A

=

[BJ _I_ 3x104

P-c-3x108

_ -

10

...i

p

a

3.65

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1.

2.

3.

4.

A--> (s) B--> (q. s) C--> (p, r) D--> (p, r)

(C)

de-Broglie wavelength of most energetic photoelectrqn A 1·227 x 10-9 1.227x10-s =12.2x1~-io m

A--> (s) B--> (q) C--> (p, r) D--> (p, r)

(D) . E=h:

-Fi

).=10A

=>

A --> (p, q) B--> (r, s) C :-> (q) D--> (p)

E=

A--> (p) B--> (r) C--> (s) D--> (q) de-Broglie wavelength of electron in X-Ray tube

5.

A--> (p, q) B --> (p, r, s) C--> (p, q) D--> (s)

6.

A--> (p, r) B--> (q, s) C--> (s) D--> (p, r)

7.

A-> (q, s) B-->(p,r) .C-->(p,r) D--> (s)

).=-h-=_h_

,/2mKE => (A)

,J2mev

). = 1.221x10·'

-Fi

12.21 A

-Fi

Accelerating potential V =1o• eV in X-ray tube

12 7 A= - ~ x10-a =1.227x10·11 m

10

(B)

).=hc=hc E eV V=2.5 V

).=0.1A Wavelength associated with X-rays

)..

12.4x:o-7 -12.4x10·11 m 10

A=1.2x10·10 m

8.

12.4x10·1 m V .

12.4x10·'=5000A 2.5

A--> (r) B--> (s) C--> (p) D--> (u) E--> (q) F--> (t)

).=1A

=

3.66

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1.

The target area is S,=n(10-9Y2"=1ex10-18 m 2

~-

The area of a 5 metre sphere centred on the light source is,

s, =4,(s)' =100 ,m'.

4.

=.!.mv 2

K max

>("' 10-") = 10-zi Js-•

=>

Assuming that all power is absorbed, the required time is,

=>

S2

100xn

2.

SeV )(1.6x10-"J)~ 2ohr

10-23 Js-1

V

=~Kmax = m

6.6x10--34x3x10 10

9

=>

x _, J 3 10 9

5.

=>

3.

(3x10-")n = 8.4

he -i:-=W+K,

de-Broglie wavelength of electrons is

1-~h~

- ./3mkT

--,.----------=== J3 ./3x9.11x1.38x3

he

m

2A

,

2

he 2(K,-K,)

(4x10-")(3x10') 2(30-10)

=>

l.

=>

A=300x10·10 m=300

"-max

A

is the maximum wavelength of the photons with which

photoelectrons can be emitted, then

~=W =hc_K

6.63x10-s

A/1\alC

10.64

=> A=6.2x10-9 m Mean separation between two electrons in a metal is r=2x10· 10 m

... (2)

Subtracting (2) from (1), we get hc_hc=K -K

If

6.63x10-"'

X 9.11X10--3I X 1.38 X 10·23 X 300 6.63x10-s

... (1)

2A =W+K2

A

m=9.11x10-" kg and T=27+273=300K

1

0.148m

When illuminated with light of wavelength 2A. ,

8 18 n = x ·~_19 = 28 x 10 photons per minute per cm 2 3 1 X=28

Mass of an electron is

=>

6

r= mvmaK = 9.1x1_0...a x1.3x10 Be 5x10-sx1.6x10·19 r=148mm

The Einstein's photoelectric equation for the first case can be written as

minute, then their total energy will be (3 x 10-19 n) joule.

=>

2x7.8x10·19 =1.3x106 ms-1 9.1 X10--31

1

=>

If n is the number of photons reaching the earth per cm 2 per

=>

2x1.6x1Q·19 = 7.8x10·19 J

= = mv2"""' r

Q = 8.4 J cm-2 min-1 Energy of a photon received from the sun is

6600x10-

A

Bev

=>

A

= he -W,

so

Q = 2 cal cm-2 min-1 = 2 x 4.2

he

"""'

The magnetic field provides the centripetal force to the electron,

1 eV

Rate of energy received from the sun

E

2

6.62x10-34 x3x10 8 180x10·9

max

t=(

31

The maximum kinetic energy of the emitted electron is

Thus, if the light source radiates uniformly in all directions the rate P at which energy falls on the target is given by,

P = (10-" watt)(§.)= (10-"

6.2x10-9

r- 2x10·10

•

0

A

1

=>

E=40-30=10eV

=>

1

=>

"= =1200 A

=~ rnax

10 eV

(4x10-")(3x10') 3 0eV 10 300x10·

(4 x10-")( 3 x10') 10

12oox10-" m

3.67 i:::::J

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Optics & Modern Physics

Advanced JEE Physics 6.

Since Kmax

=>

=> hv=7eV So, total number of photons incident is

= hv-hv 0

p

... (1)

¾ m(8x10 6 ) ' =h(Sv,-v,)

n=hv Since, Tl= 1O...'.!%

For the second case, we have ... (2)

KE= ¾mv' =h(2v, -v,)

nemltted = n X

=>

i=7=("~ )e

(8x10')' -4

v' V=4x10 6 ms-1 7.

= nR2

10x10-e =~x 10-3 x1.6x10-19 hv 100

C

=>

F = (nR').!.

=>

22 )' x 1- x -1- , F=-x(21x10--2 7 110 3x10

C

=>

=>

8.

x

(22){21x10")(21x10") 7x110x3x108

F

F =42x10X =42

13

= Px10-3 x1.6x10-19 7x1.6x10"19 x100

P=7W

10.

According to Einstein's Photo-Ele~tric Equation, we have

Kmax =eV, =hv-W

Since, eV, = he - ~ and

=>

eV0 =12 eV-4 eV

=>

eVs = 8 eV

=>

V, =8 V

i.,

11.

Magnetic force experienced by a charged particle in a magnetic

field is given by

i.,

Since, r= mv qB The de Broglie wavelength is given by

Subtracting, we get

e(V, -V,) =hc(I., -1.,) i..,)..2

I.=_!!_=~

V,-V,=hc(l.,-1.,) e

mv

A.1A2

qBr

~ = qaro.

=>

V. -V =6.6x10-" x3x10 x ~ 1 2 1.6x10-19 6x10-o

=>

=>

v - v = es x 1o-34•8+-2+19+s

Given, _!'g__=1 and q0 =2

34

=>

2

1

8

).o.

rP

32

· 33 V2 -V1 =-;::;2V 16

Since, eVa =hv- W

=>

=

-e

10 10

N

he eV,=--~

9.

1QQ

where, from the graph 1=10x10"' A

F =(Area).!. here effective Area

=>

10"

=>

5eV=hv-2eV

3.68

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Atomic & Nuclear Physics

(a) ATOMIC PHYSICS ........................................................................................................4.1 (b) NUCLEAR PHYSICS ..................................................................................................4.21

Solved Practice Problems ..................................................................................................4.48 Practice Exercise Sets

:> Single Correct Choice Type Questions ..................................................................4.58 :> Multiple Correct Choice Type Questions ...............................................................4.73 :> Reasoning Based Questions (Assertion Reason Type) ........................................4.78 :>

Linked Comprehension Type Questions (Paragraph Type) .................................. .4.81

:> Matrix Match Type Questions (Column Matching Type) ........................................4.89

:> Integer Answer Type Questions ............................................................................4.92 Answers to In Chapter Exercises (ICE) & Practice Exercise Sets .....................................4.94 Solutions to In Chapter Exercises (ICE) .............................................................................4.99 Solutions to Practice Exercise Sets .................................................................................4.106

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STRUCTURE OF AN ATOM : AN INTRODUCTION

All matter is made up of tiny particles known as atoms. There are about 105 different kinds of atoms, and they combine with each other in different ways to form groups called molecules. All matter has been found to be composed of atoms or molecules, and the basic knowledge of atoms and their constitution gives us valilable information about the behaviour of matter. THOMSON EMPIRICAL MODEL / THOMSON PUDDING MODEL

PLUM

J.J. Thomson gave the, first idea regarding structure of atom. The model is known after him as Thomson's atom model. According to this, entire positive charge is distributed uniformly in the form of a sphere. Negatively charged electrons are arranged within this sphere lying here and there. The model is popularly known as plum-pudding model. Every electron is attracted towards the centre of uniformly charged sphere while they exert a force of repulsion upon each other. The electrons get themselves arranged in such a way that the force of attraction and that of repulsion balance each other. When disturbed, electrons vibrate to and fro within the atom and cause emission of visible, infra-red and ultra-violet light. Positively Charged Matter

Thomson's atom model

spectral line. Experimentally, hydrogen is found to give several series, each series consisting of several lines. This indicated that Thomson's atom model needed modifications which was modified by Rutherford. RUTHERFORD'S ATOMIC MODEL

The correct description of the distribution of positive and negative charges within an.atom was made in 1911 by a New Zealander while working at Manchester University in England. This was Ernest Rlftherford, who was later called as Lord Rutherford for his many scientific achievements. He entered into physics during that crucial period of its development when the phenomenon of natural radioactivity had just been discovered, and he was first to realize that radioactivity represents a spontaneous disintegration of heavy unstable atoms. Rutherford realized that important information about the inner structure of atoms can be obtained by the study of coll!sions between the rushing a particles incident on the atoms of various materials that forni. the target on which the a. particle beam is incident. EXPERIMENTAL ARRANGEMENT

The basic idea of the experimental arrangement used by Rutherford in his studies was explained as follows: A piece or speck of a-emitting radioactive material is placed in a lead shield with a hole that allows a narrow beam of the a-particles to pass through it. In front of this arrangement is placed a gold thin metal foil to deflect or scatter the a particles. After passing through the gold foil the deflected particles are incident on a pivoted fluorescent screen with a: magnifier through which the tiny flashes of light were observed whenever an ex-particle struck the screen. The whole apparatus placed in an evacuated chamber, so that the particles would not collide with air molecules.

Thomson's atom model satisfied the requirements of the atom and the demands of electro-magnetic theory. According to this model, hydrogen can give rise to a single 4.1

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Advanced JEE Physics

/_____/:~:---7-_~

Zinc Sulphate~ Screen

\

Microscope

C••----

~::

~-----

a-Particle Source

Rutherford performed experiments on the scattering of alpha particles by extremely thin metals foils and made the following observations. a) Most of the a-particles, either passed straight through the metal foil or suffered only small deflections. Titls could not be explained by Thomson's atom model. b) A few particles were deflected through angles which were less than or equal to 90° . c) Very few particles were deflected through angles greater than 90° . It was observed that about only 1 in 8000 particles was found to be deflected greater than 90° . Sometimes a particle was found to be deflected through 180° . In other words, it was sent back in the same direction from where it came. The large angle of scattering came as a greater surprise. It could not be explained by Thomson's atom model. It was one of the main reasons for rejecting Thomson's atom model. CONCLUSIONS

b)

c)

=

:::::::;-------

\~

Different deviations for different a particles

Gold Foil

Also during the experiment following conclusions were made (i) If $ is the.angle made by a scattered particle with its original direction of motion and N is the number of particles available in that direction, it was found that, 1 . - - ~ oclog,N sin 4 .!. 2(ii) If t is the thickness of the foil and N is the number of a-particles scattered in a particular direction ($= constant) it was observed that ·

OBSERVATIONS

a)

:=J©

The fact that most of the a-particles passed undeviated led to the conclusion that an atom has a lot of empty space in it. a-particles are heavy particles having high initial speeds. These could be deflected through large angles only by a strong electrical force. Titls led Rutherford to the conclusion that entire positive charge and nearly the entire mass of the atom were concentrated in a tiny central core. Rutherford named this core as Nucleus. The difference in deflection of various particles can be explained as follows: a-particles which pass at greater distances away from the nucleus, shown as A and A' in figure, suffer a small deflection due to smaller repulsion exerted by the nucleus upon them. The particles like B and B' which pass close to the nucleus experience a comparatively greater force and hence get deflected through greater angles. A particle C which travels directly towards the nucleus is first slowed down by the repulsive force. Such a particle finally stops and then, is repelled along the direction of its approach. Thus, it gets repelled back after suffering a deviation of 180° .

N -=constant

t

=>

N1 =!i_ N2 t2

RUTHERFORD'S ATOM-MODEL POSTULATES

On the basis of the conclusions drawn from Rutherford's experiment, a new atom model was proposed. Titis atom model, known as Rutherford's atom model, had the following characteristics. a) An atom consists of equal amounts of positive and negative charge so, the atom, as a whole is electrically neutral. b) The entire positiv_e charge of the atom and practically its entire mass is concentrated in a small region which forms the core of the atom, called the nucleus. c) The negative charge, which is contained in the atom in the form of electrons, is distributed all around the nucleus, but separated from it. d) In order to explain the stability of electron at a certain · distance from the nucleus, it was proposed by Rutherford that the electrons revolve round the nucleus in circular orbits. The electrostatic force of attraction between the nucleus and the electron provides the centripetal force to the electron to revolve in the orbit. e)

The nuclear diameter is of the order of 10·" m . Titls can be calculated as follows. Let an a-particle having velocity v approach a nucleus (head-on) having a charge +Ze . The velocity of a-particle decreases till it comes to rest at a distance r0

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Atomic Pl,ysics from the nucleus. It is, then, repelled back along the direction of approach r0 gives the radius of nucleus. 1

Initial K.E. of a-particle

=.!.mv' 2

Final K.E. of a-particle

=0 =0

Final P.E. of a-particle

=-l_q,q,

Initial P.E. of a-particle

1

b)

4ru:o To

By Law of Conservation of Energy (U +K).,;._ =(U+K)....,, =>

.!.mv' =-l_q,q, 2

41te0 r0

r,

= _3.__(q,q,) 2

0

Illustration 1

An isolated hydrogen atom emits a photon of 10.2 eV.

4ne 0 mv

(a)

For a-particle, % = 2e => =>

So it must radiate energy continuously. Bohr calculated that this emission of radiation would cause the electrons in an atom to lose all their energy and fall into the nucleus within a hundred - millionth of a second following a spiral path. Thus the whole atomic structure should collapse. Since matter composed of atoms exists permanently as far as we know, there was obviously something wrong here. According to Rutherford model, electron can revolve in any orbit. So, it must emit continuous radiations of all frequencies. But elements emit spectral lines of only definite frequencies.

(b)

(c)

2 (2Ze')

To

= 41tEo

mv2

Solution

r,0 =-4-(Ze' )=4x9x109 ze2

41te0

mv 2

Determine the momentum of photon emitted Calculate the recoil momentum of the atom Find the kinetic energy of the recoil atom. [Mass of proton, ,nP =1.67 x 10-21 kg ]

mv2

(a)

In one of the experiments, a-particles of velocity 2 x 10 7 ms_, were bombarded upon gold foil with Z=79

Momentum of the photon is E P, =-;;

(b) Applying

the Law Momentum, we get

So, for Z = 79

P,-0

e=l.59x10-" C m=4xl.67x10-27 kg

(c)

r0

= 2.69 x 10-14

79 X (1.59 X 10-,0 )'

Conservation

~

of

Linear

P,

photon

4xl.67xl0'27 x(2x10')

K = ..!..mv2 ( v = recoil speed of atom, m = mass of 2

/

hydrogen atom)

2

2

=>

m

This gives the order of the radius of nucleus.

K =.!m(.E..) 2 m

=]!_ 2m

Substituting the value of the momentum of atom, we get

FAILURE OF RUTHERFORD MODEL

a)

5.44 x 10-21 kgms-'

P, = p, = 5.44 x 10-27 kgms-'

we get,

=>

of

atm

v=2x107 ms-1

r0 =4x9x109 x

10.2 x 1.6 x 10-" 3x10'

According to laws of electro-magnetic theory, a charged particle in accelerated motion must radiate energy in the form of electro-magnetic radiation. As the electron revolves in a circular orb!t, it is constantly subjected to

K

2 ·(5.44x10-21 )

= 2 X 1.67 X 10-27

8.86 X 10-27

J

BOHR'S ATOMIC MODEL

2

centripetal acceleration ~.

r

BOHR'S THEORY OF THE HYDROGEN LIKE ATOMS

To rectify the drawbacks of Rutherford Model, Bohr proposed a theory which applies to hydrogen atom and species like He+, Lr-+ etc. Here a single electron revolves around a stationary nucleus of positive charge Ze where Z =1 for hydrogen atom, Z = 2 for He+ etc. Bohr in defiance of the well-established laws of classical mechanics Electron spiralling inwards

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Advanced JEE Physics and electrodynamics, proposed the following postulates to support his atomic model. A.

The atom consists of central nucleus, containing the entire positive charge and almost all mass of the atom. The electrons revolve around the nucleus in certain discrete circular orbits. The necessary centripetal force for circular orbit is provided by Coulomb's attraction between the electron and nucleus. So,

where,

From(2),

4ne0

r

2

2 2

r II

= atomic number, e = charge on electron =-1 · 6 x 10~19

H.

C

z

VELOCITY OF ELECTRON IN NTH ORBIT

nh . Smce, v = - 2nmr

STATIONARY ORBITS

QUANTUM CONDITION (BOHR'S QUANTISATION RULE)

n 2 h2 s and r=--' . 1tme2 Z =>

=>

_!:_( = h)

=>

2n

i.e., mvr = n(;1t), n being integer or the principle quantum

number

( e' v= 2hs

)z

CONSTANCY OF MASS

The mass of the electron in motion is assumed to be constant.

--;;

0

v=( 2::,c )( c:) v=a(c:)

e' where a= - - is the fine structure constant (a pure 21Js 0c . 1 number) whose value is . 137

STATIONARY NUCLEUS

The nucleus is so heavy, that its motion may be neglected. E.

n' • A

2

n'

The stationary orbits are those in which angular momentum

D.

=(0.53)

'

The allowed orbits for electron are those in which the electron does not radiate energy. These orbits are also called stationary orbits.

of electron is an integral multiple of

r,

r oc-

Z

C.

nhe =--' 1tme2 Z

So, for H-like atoms

Ze = charge on nucleus,

B.

nh v=-2nmr

Put in (1), we get

=>

m = mass of electron, r = radius of circular orbit, v = speed of electron in circular Orbit,

=>

v=(1~7

)c:

i.e. velocity of electron in Bohr's First Orbit is _c_ in 137' Second Orbit is _c_ and so on.

F.

274

BOHR'S TRANSITION RULE

When an electron jumps from one stationary orbit to another, a photon is emitted or absorbed having energy equal to the difference of energies between initial and final states and being given by E;-E1

I.

E;-E1 v=--h

G. RADIUS OF ORBIT Since, we have

KINETIC ENERGY OF ELECTRON (EJ

Since, we have

1 (Ze)(e)

mv'

=hv

=>

... (1)

... (2)

2n

1 (Ze){e)

r

,2

41t&o

nh and mvr=-

CIRCULAR ORBITS

mv2

1 (Ze)(e)

mv' r

r 1 2

,2

41tso 2

Ze2 81ts 0 r

=>

-mv = - -

=>

EK=-mv = - -

,--, 4.4

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Atomic Physics

J.

POTENTIAL ENERGY (U) OF ELECTRON IN nth ORBIT

u =:,

1 (Ze)(e) 41te0

r

Ze 2

U=--41te0r

Illustration 2

K. _ TOTAL ENERGY (E) OF ELECTRON IN nth ORBIT

Total Energy= K.E. + P.E. =:,

=:,

level (n = 1) is known as the ground state. The energy level corresponding to n = 2 is called the first excited state and so · on . In this diagram zero energy level corresponds to n = oo which is the ionized state of the atom.

E= Ze' _ Ze' 81te0 r 4m;0 r

Using Bohr's theory show that when 11 is very large the frequency of radiation emitted by hydrogen atom due to transition of electron from n to (n-1) is equal to frequency of revolution of electron in its orbit. Solution FrequenCj' of revolution electron in nth orbit is given by

Ze2

l

E=--8m:0r

So, we conclude that

(2::,c)~

V

=r= 21tr = 21t(n'h_'•o)

lre,ol,tioo

1

1tme2z

Total Energy= -K.E. = (P.E.)

2

2

me' ) z f revolution = ( 4e~h3 n3

Further, frequency of transition from n to (n -1) is

E=-( 87,::i) !:

me'z'[ 1 1]

lif=-- - - -2 - Bh'e: (n -1) n'

Z'

4 2

E=-(13.6), eV n ~so

E=-( "!e4 )ch 2:

=:,

E = -(Reh)-;;,

8e0 dr 3

me z [ 2n-1 ] /if= Bh'•i. n 2 (n-1)2 When n is large then 2n-lee2n and n-leen

. n

Z'

=:,

,~e

Be,ch3

=1.097xl07 m-1 and is the

An energy level diagram for the hydrogen atom (Z = 1) is shown in the figure. E

Free Electron

l

==~r-;:;:;-;:;:;;:rt n s n

4

Bound Electron

I

-1.51 eVt-----n=3

av

11

4

Rch=Rydberg's Energy =2.17xl0-18 J=l3.6eV electron energy in first orbit of H atom.

--3.40

me 4 z2 2n

/if-=.,, 2 4 Bh e~

where R=Rydberg'sconstant =

oevl

... (1)

=:,

f transition

=(

E~h3

)z'~

... (2)

So, from (1) and (2), we observe that for large n, lrevolutJ.on

=!transition (between adjacent levels)

This Principle is also called "BOHR'S CORRESPONDENCE PRINCIPLE". . Illustration 3

Assume a hypothetical hydrogen atom in which the potential energy between electron and proton at separation

r is given.. by

n= 2

me' 4

~ =k( log,.

r-½), where k is a constant. For

such a hypothetical hydrogen atom, calculate the radius of nth Bohr's orbit and energy levels. Solution Force of interaction between electron and proton is given by

-13.6 av 1 - - - - - n = 1

The vertical axis represents energy. The (arbitrary) zero of energy is taken as the energy of a stationary electron, infinitely far from the positive nucleus. The lowest energy

F=- dU dr

= -k r

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Advanced ]EE Physics Force is negative. It means there is attraction between the particles and they are bound to e~ch other. Titls force provides the necessary centripetal force for the electron. So, we have mv 2

k

r

r

M =~) n/1 )nhBe 2 2nBe 2nm' Since, U = -MBcos180° ~

... (1)

... (2)

2mn h (e)- [$[ = Bnr 2 = " 2e

k

Illustration 5 Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV . The work function for sodium is 1.82 eV .

Since, E = U + ½mv'

So, r. =

2 + 2 =klog,r

=

=

nh and E. =klog, ( nh ) 21tvmk 2xvmk

Illustration 4 An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B • Assuming that Bohr's postulate regarding the quantisation of angular momentum holds good for this electron, find (a) the allowed values of the radius r of the orbit. (b) the kinetic energy of the electron in orbit. (c) the potential energy of interaction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B . (d) the total energy of the allowed energy levels. (e) the total magnetic flux due to the magnetic field B passing through the nth orbit. · (Assume that the charge on the electron is -e and the mass of the electron is ,n ).

Find (a) the energy of the photons causing the photoelectrons emission.

(b) the quantum numbers of the two levels involved in the emission of these photons

(c) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and (d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (Ionization potential of

hydrogen is 13.6 eV). Solution (a) From Einstein's equation of photoelectric effect, Energy of : photons causing the photoelectric [ emission ~

Solution (a)

E=U +K= nheB

v=Jf

and

k

. Smce,

.

mv r=Be

... (1)

From Bohr's Quantisation Rule, we have nh mvr=2x

(c)

K=

E4 = -0.85 eV Since, E, - E2 = 2.55 eV

2nm

. (e)( ') =-( e ( ') evr ) "' =21tr

E=K-+W=(0.73+1.82) eV

E, =-1.5 eV

4xm

M=IA= .T "'

( work ) + function

E, =-3.4 eV

... (2)

~ niv' = nhBe 2

of emitted photons

E=2.55 eV (b) In case of a hydrogen atom, E1 =-13.6 eV

r=) nh and v=)nhBe2 21tBe

l

Maximum =[.J
~

Solving these two equations, we get

(b)

M and ii will be 180° because

instead of taking electronic current, we have to take conventional current which moves opposite to electronic current. (d)

E = klog, r-

U=nheB

The angle between

Solving equations (1) and (2), we get

r= 2" ~

4nm

4nm

According to Bohr's ass~ption, we have

mvr=n(::)

= nhe

2

(c)

Therefore, quantum numbers of the tw'o levels involved in the emission of these photons are 4 and 2 i.e., from 4->2. Change in angular momentum transition from 4 to 2 · will be

V

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Atomic Physics L>.L=L2 -L4

=>

·

l>.L =

v=i=Z'R( A

z(!!__)4(_1!__) 2n · 2n

1bis relation ·holds for radiations emitted by hydrogen-like atoms i.e. -

AL=_]!.

=>

_!_ __!_)

ln/ n:

n

H(Z=l),He'(Z=2),Li.. (Z=3) and Be... (Z=4).

(d) By Law of Conservation of Linear Momentum, we have HYDROGEN SPECTRUM

Momentum of) ( Momentum of) ( hydrogen atom = emitted photon

Sinc_e, wave number (v) is given by

E mv = - ( m = mass of hydrogen atom)

=>

v=~=Z'i

C

A

E

=>

v=-,

=>

V

=>

v=0.814 ms-1

where R is th~ Rydberg constant. The various transitions for the hydrogen atom are shown in the following figure. All transitions starting from n = 2 onwards and ending at n = 1 belong to the Lyman Series. Likewise, all transitions starting from n = 3 onwards and ending at n = 2 belong to the Balmer Series. The other spectral series' names are mentioned in the figure.

me

(2.55i(1.6 X 10-19 ) (1.67 x10-27 )(3 X 108 )

FREQUENCY OF EMITTED RADIATION

If electron jumps from initial state ni, to a final state n1 , then frequency of emitted or absorbed radiation v is given by E1 -E1 =hv. or

V=

E,-Er =Z'RJ h

_!_~_!_)

l n} n:

0eV '

II

-1.51 eV -3.40 eV

HJ Balmer

_!_ __!_)

•••

0 =CO

'

11

n=5

n=4 n=3 n=2

Bracket

Pasch"eil

lnJ n:

If c is the speed of light and '.l. the wavelength of emitted or absorbed radiation, then

v=E.= ZRc(_!_ __!_) 2

A

n} n:

-13.6 e V - ~ ~ - - - - _ : _ _ - - - - - - - - n = 1 Lyman

So, Wave number (v) is given by .

INITIAL STATE

FINAL STATE

WAVELENGTH FORMULA

z

< ::;; :'.i a: w ::;;

.J

< Ill

n; =2, 3, 4, 5, 6, ...

n; =3', 1,5, 6,,7, ...

"1 =1

n, =2

_!=R(_!_-2_) 2 A

1

n~

_!=R( _!2 _ _2_) A

2

n~

FIRST MEMBERSECOND MEMBER

n;=2 to n,=1 n;=3ton1 =1

n1 =3 to n1 ;::2 n1 =4 to n1 =2·

w :c

0

U)


n1 , - oo TO n, From oo to 1

From2to 1

).=_! R

).=..±_ 3R

i.=911 A From co to 2

From 3 to 2

I.=± R

). = 36 SR

i.=3646A

z

From n1 = 4; 5, 6, 7, 8, ...

n, =3

.-

_!=R(_1_2 _ _1_) A

_3

n~

ni=4ton,=3 n;=Ston,=3

MAXIMUM WAVElENGTH (n,+1) TO n,

SERIES LIMIT,

oo

to 3

LINES FOUND , IN

UV Region

A=1216A

.-

Visible Region

i.=6563A From4h3

). =~ R

).= 144 7R

i. = 8204 A

i.=18753A

IR Region

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~

~

0

< a:

ni

=5 , 6, 7, 8, 9, ...

n,=4

_l __l)

.i.=R( 2 A. 4

n~

n,=5

n; = 6, 7, 8, 9, 10, :··

"-

_l __l)

.i.=R( ). 52

"i2

n=3 to n=4 Find the wavelength of radiation required to remove electron from first Bohr's Orbit to infinity. (d) Find the kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.

· ~ ~

(J,--J,-) n; n,

47.2=13.6Z '( 1, - 1, 2 3

Solution Wavelength of the first line of Lymen series for hydrogen atom will be given by the equation

:, =R(1l2-::{,)=3:

X

~

is

1,-J,-)

2 ... (2)

Given that A1 = A2

,,,1 ,,,1

J

3R

3Rz2 16

~

-=--

~

z=2

4

Hellce the ion x is actually He+.

The erte:r:gies of first four levels. of ion x are, E, =-(13.6)z2 =-54.4 eV

eV

n; n,

... (1)

The wavelength of second Balmer line for hydrogen like ion

eV

Z=5

i'.E=i3.6Z'(

FarlR Region

1-=74583 A

A

3 2 :,=Rz (;,-;, )= :

(b) To find i<.E,., n1 =3, n,=4.

i'.E=l3.6x52(;, - ;, ]=16.53 eV

~ E1 ..., = ( ), = -13.6 eV 2

Ionization energy is the energy required to excite the electron from n = 1 to fl

Thus, i'.E=;3,6x5

2

(;, -

= co ~

2]=340 eV

The respective wavelen~ is

,, =~ = 12400 = 12400 =36 _47 A i'.E

i'.E

340

K=-E=+340 eV U=2E=-680 eV L=~= 6·63 xl0""' 21t 21t

=

1- = 900 11R

Illustration 7 The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen lik~ ion x . Calculate energies of 0 the first four levels, of x , ·

Solution (a) Given i'.E23 = 47.2 eV

(d)

From 6 to 5

n1 =7 jo n1._=5

IR Regiori

i. ~ 40515 A

1-= 25 R

n;=6ton,=5

(c)

Since, i'.E=l3.6Z 2

I

From oo to 5

1- = 22790

Illustration 6 A single electron orbits around a stationary nucleus of charge +Ze, where Z is a ·constant aµd e is th~ magnitude of electronic 'charge. It requires 47.2 eV to excite the electron from secol\d Bohr orbit to the third Bohr orbit. (a) Find the value of Z (b) Find the energy required to excite the electron fro~

(c)

1- = 400 9R

1-=14585 A

0

IL

From 5to4

1-= 16 R

n,=6ton,=4

Ill

z ::,

From oo to 4

I\ =5 ton, =4'

l.056xl0""' Js

E,.= E), 2 =-6.04 eV and (3

E1 E, = ( )2 =-3.4 eV 4 Illustration 8 A doubly ionised lithium atom is hydrogen-like with atomic number 3. (a) Find the wavelength of the radiation required to excite the electron in Lf+ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6 eV ).

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Atomic Physics (b) How many spectral lines are observed in the emission spectrum of the above excited system? ·

= rel="nofollow">

~;, 10.2

Solution

=>

K;,20.4 eV

Given, Z=3

Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV.

.

Z' Smee, En ex: 2 n (a)

To excite the atom from n = 1 to n = 3 , energy of photon required is ,

£,43 =E,-E,

(-13.6)(3)2 (-13.6)(3)'] ( (3)' (1)'

E,43 = 108.8 eV

=>

Corresponding wavelength will be, · ,. (in Al '

12375 E(in eV)

12375 108.8

ns-_74 A

Illustration 10

A moving H-atom makes a head on perfectly inelastic collision with a stationary Li++ ion. Before collision Hatom and Li++ ion are both in their first excited states. What is the velocity of the moving H atom if after collision H is found in its ground state and Lr-+ ion in its second excited state. Take mass of hydrogen atom, '"u =1.66 x 10-27 kg and mass of Li++ to be '"zr =7,nu . Solution

(b) From nth orbit total number of emission lines can be n(n-l) . So, mnnb er of erruss10n · · 1·mes · 1s · 2

Illustration 9

A moving hydrogen atom- makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic en~rgy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state. Solution.

Let K be the kinetic energy of the moving hydrogen atom and K' , the kinetic energy of combined mass after collision.

For u++, we have Z=3 2

=>

E, =

-13.6(3) (2)'

=>

E,

-13.6(3)' (3)'

-30.6 eV -13.6 eV

Energy required for Li++, ion to go from first excited state (n = 2) to.second excited state (n = 3) is

bE=-13.6-(-30.6)=17 eV Energy released by hydrogen atom to go from first excited state to ground state is M.' =-3.4-(-13.6)=10.2 eV

So, bE-bE'=17-10.2=6.8eV is the energy that should come from loss in KE in collision. From Law of Conservation of Linear Momentum, velocity of combined mass is

(m,=7m,)

--~-----n=2

o----!S--. 0

m

m

K'

~

=!>

1 u, - 1(m, +m ) v , AKE=-m 2 2 1 1 2 Since, Af< = 6.8 eV Solving these equations, we get u, =3.9x10 4 ms-1 =>

dE = 10.2 eV

2m

--..;;-----n=1 By Law of Conservation of Linear Momentum, we have

p=p'

=> ..f2Km = ,J2K'(2m) =>

K=2K' From Conservation of Energy, K = K' + bE Solving equations (1) and (2), we get

Illustration 11

... (1) ... (2)

Find the largest and shortest wavelengths in the Lyman series for hydrogen. In what region of the electromagnetic spectrum does each series lie? Solution

bE=K 2

The transition equation for Lyman series is given by,

Now minimum value of bE for hydrogen atom is 10.2 eV, so we have bE2'.10.2 eV

¾=R(;,-:,), n=2,3, ..... The largest wavelength is corresponding to n = 2 , so 4.9

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(.!.-.!.)4

Illustration 12 The nucleus of a deuterium has a mass of 3.34 x 10-27 kg as compared to 1.67 x 10-" kg for the hydrogen. Calculate the wavelength difference between the first Balmer line emitted by hydrogen and the first Bw,;er line emitted by deuterium. Given that the mass of electron is m, = 9.109 x 10-31 kg .

_l_ = 1.097 X 107 ,._ 1

,._

=>

_l_ = 0.823 x10'

=>

,__ = 1.2154 X 10-7

=>

,._ =1215 A

ill

The-shortest wavelength corresponds to n ~ oo, so ' 7 1 (1 -=1.097xl0 . - -1) Amin 1 oo

=>

Solution The first Balmer line corresponds to the transition from n = 3 to n = 2 . In case of hydrogen atom, we have

t.E=E,-E,

'-mm =0.911xl0;'. m=911 A

Both of these wavelengths lie in ultraviolet (UV) region of electromagnetic spectrum.

=>

t.E={-

!;6-(-~~·6 )}eV=l.89eV

1

ABOUT RYDBERG CONSTANT

12375 . ( ) So, wavelength 1,. = . t.E ineV

Rydberg Constant is given by

=>

,. ;,, 12375 = 6547.6 A 1.89

me'

R=-8e~ch3 where m is mass of electron. It app"ears that Rydberg constant is same for all elements, but it is not so. Actually Rydberg constant is different for different elements. The reasC?n is that in Bohr's ~eory nucleus is assu_med infinitely heavy as compared to electron. But if the mass of nucleus is taken-into account then electron mass m must be replaced by reduced mass (µ), where

---~----n=3 --~~----·n=2

--------n=1

For ordinary hydrogen reduced mass of.proton and electron is,

mM µ= m+M

· (1.67x10°")(9.109x10"'31 ) µ, (1.67 X 10-2' + 9.109 X lo'"')

where M is mass of nucleus.

=>

=>

mM m µ=--=-m+M l+."!.

(3.34x 10-27 )(9.109 X 10·31 ) µ, = (3.34 X 10-2' + 9.109 X 10""1 )

If M-->oo(infinity), µ---->m(Mass of electron) ·

=>

Therefore µe' 8s~ch3

µ 1 =9.10408xl0-31 kg

For deuteriulll atom reduced mass of nucleus and electron is,·

M·

R

'

µ, =9.10654~10""1 kg

All energies are proportional to µ , whereas the wavelengths are inversely proportional to µ . The wavelength of photon emitted in case of hydrogen_ is given by

me'

H for a heavy nucleus, Rydberg Constant R is represented by

me'

'-,

(6547.6)(9.109x10""1 ) (9.10408xl0""')

.

6551

A

Similarly in case of deuterium, wavelength of photon emitted is,

R =-.8E~Ch3
,. _ (6547.6)(9.109xlo-") 2 (9.10654x10-31)

Then Rydberg constant for an element is given by R=__&_ 1 +."!. M Oearly this depends on mass of nucleus M and so Rydberg constant is different for different -elements. Greater is M, larger is the value of Rydberg constant R. Thus, Rydberg constant increases with increase of mass of nucleus.

=>

6549

A

,'.1,.=A,-A,=2A

.Illustration 13 Calculate the separation between the particles of a system in the ground state, the corresponding binding energy and wavelength of first line in Lyman series if such

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Atomic Physics a system is positronium consisting of an electron and positron revolving round their common-centre of mass.

Solution (m)(m)

m

Reducedmass µ=-(--)=m+m 2

. 1 rcx::Smce,

Ritz made this discovery empirically (1908) long before Bohr proposed his theory and is known as Ritz combination principle. • Generalising, we may write, labelling the photon frequency by appropriate integers, as follows: hv6111 =Es -E111

·m

For, H atom, we have ,1

=>

=0.53 A

r=(2)(0.53)=1.06 A Since, E cc m =>

{binding energy)

For, H atom, we hav:e E1 =13.6 eV E=

--~--~ -a~-~::t:o~~~~c!i~~Aa~o:

2

E,, =

10 2 · = 5.1 eV

to a-:v:::~:~~-,

principal quantum number n , then the maximum number of photons (emitted) will be (n -1) .

For, H atom, we have

=10.2 eV

... (3)

Since all combinations predicted by (3) are not actually observed, tli.ere has been an hnposition of some rules, the socalled selection rules, to eliminate certain combinations. Bohr's theory provides, as discussed above, a proper explanation of the combination principle.

0;uE~:~

13 6 · =6.8 eV

E,~i

(m
hvsm =hvsn +hvnm

{First line of Lymen series)

b)

If in a hydrogen like sample the electrons of many atoms jump to a level having principal quant1;1m number n, then 1 the maximum number of photons emitted =nC2 = n(n - ) . 2

c)

For example if many atoms are in third excited .state (n = 4), then the number of ·photons emitted for spectral

2

linesseen =

4 3 ; =6

RITZ COMBINATION PRINCIPLE

If an electron is initially in an excited state with say n = 3 , then it may transit downward from n =3 level to n =1 level directly. Alternatively, it may first transit from n = 3-> n = 2 and subsequently from n = 2-> n =1 . In the first case if v 31

~)

be the frequency of the photon emitted

... (1)

If an atom goes to excited state by absorbing certain energy, then it may emit a number of photons in succession; the sum of the energies of all emitted photons will be equal to the amount of energy absorbed. For example if an atom absorbing energy E reaches to an excited level and it returns to ground state by emitted wavelength A1,A2 and A3 , then E=hc+hc+hc

-~--~----n=3

).,1

v~

e)

_ _,___ _...,__ _~-n=2

v,. f) _ i _ __ _..___ _ _ _

n=1

ll.3

multiple of de-Broglie wavelength.

In the second case, two different spectral lines (photons) of frequency v32 and v21 respectively would be emitted given by hv 32 =E3-E2 and hv21 =E2 _,....£1 ... (2) (1) can be rewritten as

hv31 = (E, -E2 )+(E,-E,)

i.e.

2xr =ni..

=>

2xr=n(!v)

=>

mvr =n( : ) 2

This is same as Bohr's quantum condition. g)

=>

"-2

If an energetic electron strikes an electron (target electron) of an atom in ground state then electron jl.lmps to an excited state by absorbing energy equal to the difference of ground state energy and excited state energy and the remaining energy is still carried by the incident (or striking) electron. The de-Broglie quantum condition: According to deBroglie, only those orbits are allowed as stationary orbits in which circumference of orbit is equal to the Integral

"An atom possesses discrete levels" was verified by __ en_·m_e_n_t._ _ _ _ _ _ __ Franck Hertz Exp

Y31=Y32+Y21

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Advanced JEE Physics MERIT AND DEMERITS OF BOHR'S THEORY

CRITICAL POTENTIAL

The merits of Bohr's theory can hardly be overestimated. It saved physics at a time when it was in the grip of severe

crisis. But, like every physical theory, this theory also has

The resonance potential, the excitation potentials and the ionisation potentials are all included in the wider term critical potentials.

drawbacks. We enumerate below some of its merits and demerits.

RESONANCE POTENTIAL

A minimum potential V is required to accelerate the bombarding electron to an energy V (in electron-volt) in

MERITS

a)

b)

The determination of the ratio of the mass of an electron to that of a proton in terms of the reduced mass concept of Bohr's theory agrees excellently with the value obtained by other methods. The general principle used by Bohr has also been successfully applied lo a great number of phenomena such as the excitation and ionisation of atoms, X-ray spectra etc.

c)

lines (spectral) which have later been actually observed. d) The theory has been instrumental to the discovery of · heavy hydrogen (deuterium) by H.C. Urey. e) It gives a convincing explanation and a very simple and elegant picture of the origin of spectral lines. f) The agreement between the empirically determined value of the Rydberg constant and that evaluated by Bohr in terms of fundaqiental constants offers an excellent proof of the truth of Bohr's theory. DEMERITS

a)

There is an ad hoc nature in the assumptions of Bohr in that the quantum idea of the stationary orbits is mixed up with the classical idea of coulomb force.

b)

The assumption of only circular orbits is utterly unjustified. In fact, Bohr in his original paper suggested that the orbit might be an ellipse instead of a circle. The spectral series, though agree excellently in case of

d)

e)

potential.

·

EXCITATION POTENTIAL

The various values of the potential required to impart the necessary energy to excite an atom to different higher states are known as excitation potentials.

The validity of the theory is further confirmed by the

fact that the theory predicts new undiscovered series

c)

order that an atom may be excited from its ground state to the next higher state. This potential is called the resonance

hydrogen, are at variance with the theory for multielectron atomic systems, e.g. the helium, singly ionised lithium etc. In these cases, it becomes necessary to introduce a magnetic quanhlm number. It cannot suggest any explanation whatsoever for the origin of the fine structure of the spectral lines. Bohr's theory is also unable to account for the multiple structure of spectral lines. For example, the doublet of sodium, triplets of magnesium etc. cannot be explained

IONISATION POTENTIAL

The minimum potential necessary to supply the required energy to ionise an atom is called the ionisation potential or the first ionisation potential.

I

' PROBLEM SOLVING TRICK(S)

: Let us illustrate the above definitions by taking the case of : hydrogen atom. For H atom we have : 1

! 1

E = 11

18

me4.. 8c~n2h2

2.17x10- J=-13.6 8 V n2

So the energy of the 1st, 2nd, 3rd,. ......, oo-th orbits are respectively -13.6 eV, --3.4.eV, -1.15 eV ,.... , o eV. Hence, ; Resonance potential = --3.4-(-13.6) = 10.2 eV '. First excitation potential

=resonance potential

I Ionisation potential =0-(-13.6)=13:6 eV

Illustration 14 Find the quantum n.umber n corresponding to excited state of He+ ion if on transition to the ground state, the ion emits two photons in successi9n with "'?avelengths 108.5 nm and 30.4 nm . The ionization energy of H atom is 13.6 eV.

Solution The energy transitions for the given wavelengths are

f)

It cannot make any calculation about the transitions or

t.E = 12400 = 12400 =ll.43 eV

g)

the selection rules which apply to them.. · It could not explain the splitting up of spectral lines

M,, = 12400 = 12400 = 40 _79 eV

an

atom

is

subjected

to

electric

I

field

(phenomenon called Stark Effect) or magnetic field (phenomenon called Zeeman Effect).

A1

i..,

1085 304

Total energy emitted t.E = t.E, + t.E2 = 52.22 eV =>

,--. 4.12

= 10.2 eV

Second excitation potential =-1.51-(-13.6) =12.09 eV

from Bohr's theory.

when

n2

.l.E=13.6Z'(..!__..!_) eV

Hi ni

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~ 52.34=13.6x22 ( 11, - :, ) Thus, n=5 Illustration 15

A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. Find the energy and wavelength of photon emitted.

(take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system: (i) Calculate the radius of nth Bohr orbit (ii) Find the value of 11, for which the radius of orbit is approximately the same as that of first Bohr orbit for the hydrogen atom; (iii) Find the wavelength of radiation emitted when the µ -meson jumps from the third orbit to first orbit (Rydberg's constant =1.097x107 m-' ). Solution (i)

Solution

Radius of the nth Bohr orbit for hydrogen atom is

Since the binding energy is always negative, therefore,

r, =0.53

E, =-0.85 eV

m Radius of nth orbit for µ -meson is

2'

E, =-13.6n,2

2'

'

~

(ii) 13.62 2

Binding energy is En

0.85 eV

=>

n2 =4

0.53n 2 (208)2

r=---

--0.85 = -13.6, n,

~

2

. 1 Smce, roc-

Let n, be the initial binding state of the electron, then

~

n'

n'

-13.6(1)

2

r, =(8.Sx10 ... )n 2

(8.SxlO ... )n' =0.53 ~

n2 =623

:::::)

n~25

(iii) In case of hydrogen like atom,

n'2

LI.E=E,-E, =13.62'(1-;, )=12.08 eV

Let the electron now goes to an energy level n whose

Since, .1E oc m

excitation energy is 10.2 eV . Since the excitation energy .6.E is defined with respect to ground state, therefore

So for µ -meson, we have

LI.E = (12.8)(208) = 22.6 keV

E=13.62'(J..-J..) eV n1

~

n2

~

10.2=13.6x1'( ~ -J..)

~

n1 =2

1

12400 LI.E

12400 22•6 X 103

• 0.548 A

Illustration 17

n1

So the electron makes a transition from energy level ni

= 4 to

n1 =2 Thus, the energy released is .6.E = £4 - E2

~

,.

LI.E = 13.6[;, - ;, ]= 2.55 eV

A hydrogen like atom of atomic number z 1s m an excited state of quantum number 2n . It can emit a maximum energy photon of 204 eV . If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find 11, z and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is 13.6 eV. Solution

12400 = 5511 A Since, 1. = .!!E._ = LI.E 2.25 eV

Given, E211

~

Illustration 16

A particle of charge equal to that of an electron, -e and mass 208 times the mass of electron (called a µ -meson) moves in a circular orbit around a nucleus of charge +3e

E1 = 204 eV

2

... (1)

(13.6)2 (1- 4:' )=204

Also, E211 - E11

~

-

= 40.8 e V

13.6z'(J..--.!,)=40.8 n 4n

... (2)

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Advanced.JEE Physics Solving equations (1) and (2), we get n=2 and z=4 . E Smce, n

= (-13.6)z' 2 n

n

eV ,sowe h ave

Substituting m, = 100 m where m = mass of electron and z = 4

2

E1 =(-13.6)(4) eV ~

n2/t2E

0 wegetr= ' " 400nme2

E, =-217.6 eV

During de-excitation, minimum enei-gy emitted is, Emm

= E2n -

E2n-t

2 2 hE

r=---"2 zmnp e

= E4 -

(b)

As we \<now, E~ = -13.60 eV

E3 and E11 oc (-::

E _ = -217.6 _(-217.6)=l0 58 eV mm 42 32 ·

)m

For the given particle, E = (-l3 -60)( 4)' xl00=-1360 eV 4 (4)'

Illustration 18 Calculate the energy of a H ~atom in the first excited state, if the potential energy is assumed to be zero in the ground state.

and E2 = (.:_13 ·60~( 4 )' xl00=-5440 eV (2)

t.E=E 4 -E2 =4080 eV

Solution Since, we know that, in ground state, n =1, we have

,.(in A)=

TE=-KE=PE 2 ~

E; =-3.4+27.2=23.8 eV

(now)

Solution

... (1)

The quantization of angular momentum gives, 21t Solving equations (1) and (2), we get

The stopping potential for shorter wavelength is 3.95 V i.e., maximum kinetic energy of photoelectrons corresponding to shorter wavelength will be 3.95 eV. Further energy of incident photons corresponding to shorter wavelength will be in transition from n = 4 to n = 3 . E ,~,

~

=E -E = -(13.6)(3) 4 3 (4)'

2

(-(13.6)(3) (3)'

2 )

E,~, =5.95 eV

Now, from the equation, K_=E-W

we have W = E - Kmax = E4 -+ 3 - K=

m v' (a) Wehave ' r.

nh

Electrons in hydrogen-like atom (Z = 3) make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 V. Calculate the work function of the metal, and the stopping potential for the photoelectrons ejected by the longer wavelength (Rydberg constant = 1.094x107 m·1 ) · Solution

Illustration 19 An imaginary particle has a charge equal to that of an electrori and mass 100 times the mass of the electron. It moves in a circular orbit around a nucleus of charge +4e • Take the mass of the nucleus to be infinite. Assuming that the Bohr's model is applicable to this system. (a) Derive an expression for the radius of nth. Bohr orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second orbit.

mPvr,1 = -

12375 = 3 A 4080

Illustration 20

PE=2(TE)=2(-13.6 eV)

~ PE=-27.2eV However, we have assumed this energy to be zero i.e., potential energy is increased by 27.2 eV . Since, kinetic energy in all energy states will remain unchanged whereas potential energy and hence, the total energy in all states will increase by 27.2 eV . Further, first excited state means n = 2, so E2 = -3.4 eV (previously)

~

12375 t.)o(in eV)

... (2)

~

W=(5.95-3.95) eV=2eV Longer wavelength will correspond to transition from n = 5 to n = 4 . From the relation, we get ~=

A

Rz'(-1N__ __!_) 12

N;z

The longer wavelength,

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Atomic Physics

(_!_ __!_) 25

Coolidge modified the Roentgen tube. A modem X-ray tube consists of a) An electron source, preferably a filament heated by the passage of an electric current which may be varied. b). A heavy target of high melting point inclined at 45° to the path of electron beam, kept cooled by circulating cold water internally. c) A source of high potential difference applied across the filament and the target, keeping target positive with respect to filament. When the filament is heated, a fine beam of electrons strikes the target to produce X-rays.

]:_ = (1.094 X 10 7 )(3)3 1.. 16

=>

1..=4.514x10-' m=4514.A Energy corresponding to this wavelength is E = 12375 e~ A

2.74 eV 4514A So, maximum kinetic energy of photo-electrons is K==E-W=(2.74-2) eV=0.74eV Hence, the stopping potential is 0.74 V. Illustration 21 Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 A . How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV. Solution Energy corresponding to given wavelength is

E(in eV)

12375 1..(inA)

12375 = 12.69 eV 975

CONTROL OF PENETRATING POWER

The potential difference across the filament and target determines the energy and hence the penetrating power of X-rays.

Only about 1 % of incident electron's energy is converted into X-rays and the remaining 99% is converted into heat, therefore cooling device is essential with an X-ray tube.

((~·t-(-13.6)=12.69

=>

ns,4 i.e., electron excites to 4th energy state Total number of lines in emission spectrum would be N

n(n-1)

4x3

2

6

2 Longest wavelength will correspond to the minimum energy and minimum energy is released in transition from· n = 4 to n=3. -13.6 (-13.6) E,~ 3 =E4 -E3 = ( ') - ( ), =0.66eV 4 3 Hence, longest wavelength will be, = 12375 E(in eV) 1,._ = 1.875 µm

1,.

-

=>

The intensity-of incident electrons determines the intensity of X-rays , i.e., greater is the number of electrons striking the target, more intense are the X-rays produced.

NEED OF COOLING DEVICE

Now, let the electron excites to nth energy state. Then, E, - E1 = 12.69

=>

CONTROL OF INTENSITY

12375 A = 1.875 x 10 _. m 0.66

HARD AND SOFT X-RAYS

X-rays upto 4 A have high penetrating power and are called hard X-rays while those of X-rays.

PRODUCTION

When fast moving electrons strike a target of high melting point and high atomic weight (like tungsten, platinum molybdenum), X-rays are_ produced. X-rays were discOvered by Roentgen, therefore, they are also known as Roentgen rays. Roentgen discovered X-rays in 1895.

> 4 A are called soft

X-RAY SPECTRA CLASSIFICATION

In X-ray tube, when high speed electrons strikes the target, they penetrate the target. They loses their kinetic energy and comes to rest inside the metal. The electron before finally being stopped makes several collisions with the atoms in the target. At each collision one of the following two types of X-rays may get formed. A. Continuous X-rays B. Characteristic X-rays A.

X-RAYS

1,.

CONTINUOUS SPECTRUM

It consists of radiations of all possible wavelengths within a definite wavelength range having a definite short wavelength limit. These are produced due to deceleration of electrons near heavy nucleus. These decelerated electrons emit radiations which lie in X-ray region. As the electrons suffer collisions at all angles, right from the glancing collision to the direct hit, they suffer varying decelerations

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and hence radiations of all possible wavelengths within a certain range are emitted, forming the continuous spectrum. Intensity

Solution

de Broglie wavelength (i.,,) when a charge q is accelerated by a potential difference of V volt is given by h

Characteristic X-Aays

\

I,

Continuous

X-Rays

\

... (1)

For cut-off wavelength of X-rays, we have

he

qV=-

i.,,.

··-.__ wavelength l.

i., =.!!E._

The maximum limiting frequency vmax or minimwn limiting wavelength "-mm is obtained when entire kinetic energy of bombarding electron is converted to X-ray energy. If V0 is the accelerating potential, then

m

... (2)

qV

From equations (1) and (2), we get ~=

Jtf·

eV0 =hvmax =,,

ev,

-

m

h

1.8 X 10

In terms of """" eV.

o

he =--

=

B,

This relation gives the short wavelength limit of continuous X-ray spectrum.

If f is the fraction of kinetic energy of electrons converted into X-ray, then wavelength of emitted X-ray photon is given by he f(eV0 )=-;:A=~

feV,

Illustration 22 Find the cut-off wavelength of the X-rays emitted by an X-ray tube operating at 30 kV ,

Solution For minimum wavelength, the total kinetic energy should be converted into an X-ray photon,

Thus ,_ =he =12400 = 12400 ' E E 30x103

lQ X 103

0.1

CHARACTERISTIC SPECTRUM

The minimum wavelength depends on the electron energy, but not on the target material. The line spectrum depends on the element used as target. These characteristic X-rays are produced when an electron knocks out an atomic electron from one of the inner levels. The ejected electron leaves a vacancy, which is then filled by an electron falling from a higher level. When an electron jumps from a higher energy orbit Ei to lower energy orbit E2 , it radiates energy

( E, - E,) . Thus this energy difference is radiated in the form of X-rays of very small but definite wavelength which depends upon the target material. The X-ray spectrum consists of sharp lines and is called characteristic X-ray spectrum.

If the transitions are to the n =1 level, the X-rays are labelled K. , K, .. ., If they are to the n =2 level, they are labelled Ln, L/J , .... etc. In the figure shown, the energy level

diagram for an atom is drawn. The arrows indicate the transitions that give rise to the different series of X-rays.

-----------N

=0_41 A

---.---+-1-1r-+-i

-M

Illustration 23 If au X-ray tube operates at the voltage of 10 kV, find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of electron is 1.8 x 1011 Ckg _, .

4.16

X

3x10 8

"-m1n eVo

11

2

=,,

i., . =.!!E_

=,,

!L =1.Bx10 11 Ckg-1

For electron, we have

=-

V

=ii=L·=~: K. K, K,

In 1913; Moseley noted that the characteristic lines shifted

systematically as the target material was changed. He plotted the square root of the frequency of the K. line

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Atomic Physics versus the atomic nwnber Z for many elements. The straight line he obtained is shown in the figure.

where a and b are constants for a given transition. The constants a and b do not depend on the nature of target. Different values of b are as follows

b =1 for Ku - radiations of all substances. b = 7 · 4 for L. - radiations of all substances.

b =19.2 for Mo: - radiations of all substances. 0

Atomic number Z

Moseley's plot did not pass through the origin. Let us see, why?. Once one of the two electrons in the n = 1 level is ejected, an electron in the next highest level will drop to the lower state to fill the vacancy and in the process it emits the Ku frequency. For this electron the electric field due to the nucleus is screened by the remaining electron in the n = 1 level. Moseley estimated that the effective nuclear charge for the K. transition is (Z-l)e.

If the bombarding electron knocks of an atomic electron from K shell and if the vacancy is filled in from L, M, N, ....... . shells the characteristics X-rays form K -series.

For Ka-line,

EL -EK =hvK..

For K11 -line,

EM -EK =llvKp

Both these transitions are highly probable. Similarly, if bombarding electron knocks off atomic electron from L-shell and if the vacancy is filled in from M, N, ..... shells, the characteristics X -rays form L -series and so on.

Thus Moseley's Law for the frequency of the Ku line is

,[v;;: =a(Z-1) where a=

J¾ Re

MOSELEY'S LAW: CONCLUSIONS REVISITED

in which R is the Rydberg constant c is

the speed of light. The wavelength of K - lines is given by

a)

Mosley's Law supported Bohr's theory.

b)

It experimentally determined the atomic number (Z) of elements. Titis law established the importance of ordering of elements in periodic table by atomic number and not by atomic weight. Gaps in Moseley's data for A= 43, 61, 72, 75 suggested existence of new elements which were later discovered. The atomic numbers of Cu, Ag and Pt were established to be 29, 47 and 78 respectively. When a vacancy occurs in the K-shell, there is still one electron remaining in the K-shell. An electron in the L-shell will feel an effective charge of (Z-l)e due to +Ze from the nucleus and -e from the remaining K-shell electron, because L-shell orbit is well outside the K-shell orbit. Wave length of characteristic spectrum is

c)

1 2 f=(Z-1) [1- n ,] where n=2,3,4, .....

d) e)

Illustration 24

Show that the frequency of K, X-ray of a material equals to the. sum of frequencies of Ku and Lu X-rays of the same material.

t)

Solution

g)

_! = R(Z-b)'(..!__2_)

The energy level diagram of an atom with one electron knocked out is shown above. Energy of Ku

X-ray is EKu =EL-EK

of KJJ

X-ray is

h)

Energy of X-ray radiations is given by

t;E = hv =he= Rhc(Z-b)'(..!._2.)

E,:,. =EM-EK

A

and, of Lu X-ray is Eta.= EM -EL

i)

thus, EK~ =EKu +ELu or vK~ =vK.. +vL~

n~ ni

If transition takes place from n2 = 2 to n1 = 1 we get

the Ku line, for which we have (i)

MOSELEY'S LAW: GENERAL TREATMENT

According to Moseley, for a given transition, the frequencies of characteristic lines emitted by different elements obey the relation

v=a(Z-b)

tfi ni

A

a=

J3~c

= 2.47 x 1015 Hz

(ii) Frequency is given by v

2

=>

also called Moseley's Law.

Ko:

3 2 2 =RC(Z-1) (1-..!.)= RC (Z-1)

22

4

2

vK 0

=2.47x10 15 (Z-1) Hz

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(iii) E•• =10.2(Z-1) eV (iv) In general the wavelength of all the K-lines are

Solution Energy associated with the electron in the K shell is approximately

given by 2._=R(Z-1)'(1-J,.)

AK

~

where n=2,3,4, .... While for Ka line

AK«

2

E, =-(74-1) (13.6 eV)=-72474 eV

n

1216 (Z-1)

An electron in the M shell is subjected to an effective nuclear charge that depends on the number of electrons in the n = 1 and n = 2 states because these electrons shield the M electrons from the nucleus.'

•

=---, A

Illustration 25

In Moseley's equation, we have

:J1 =a(z-b), where

a and b are constants. Find their values with the help of thfll e o owine: d ata. z Wavelength of K. X-rays. Element Mo Co

42

27

Since, there are eight electrons in the n =2 state and one remaining in the n = 1 state, hence roughly nine electrons shield M electrons from the nucleus, so we have zeH=z-9

Hence, the energy associated with an electron in the M shell

is

-o.71A 1.7ssA

-13.6z!, V 32 e

Solution

EM -EK =[--6384-(-72474)] eV = 66090 eV

,.

C

where. f=-

So, for the first element, we have

=a(z,-b)

... (1)

{t=a(z,-b)

Illustration 27 X-rays are incident on a target metal atom having 30 neutrons. The ratio of atomic radius of the target atom

and :He is (14)1/'.

and for the second element, we have

... (2)

(a) Find the atomic number of target atom. (b) Find the frequency of K. line emitted by this metal. R=l.lx10 7 m-1 , c=3x10 8 ms-1

From equations (1) and (2), we get

Jc(_},-

V e

32

9 Therefore, emitted X-ray has an energy equal to

..fl =a(z-b)

Jt

2

(13.6)(74-9)' eV=-6384eV

EM=

Since, According to Moseley's Law we have

-13.6(z-9)

Solution

kJ=a(z1 -z2 )

... (3)

1

(a)

From the relation r oc A'

Now, we know that c = 3 x 108 ms-1 and further it is given to

us that A1 = 0.71 X 10-10 m =>

(

=>

A, =56

z, =42 and

z, =27

(b)

Solving above three equations, we get

z, =56-30=26

1)

Illustration 26 Determine the energy of the characteristic X-ray (Kil)

. 3Rc( Z-1), f,.=Rc (Z-1 l'(l Smee, 21 - 22 =-4 Substituting the given values of R, c and Z, we get

emitted from a tungsten (z = 74) target when an electron drops from the M shell (11 =3) to a vacancy in the K shell (11=1).

4.18

=(14)½

z, = A, - numberof neutrons =>

a=5x107 (HzJV' and b=l.37

=

i )' 1

A2 = 1.785 X 10-lO m

!,a = 1.55 X 1018 Hz

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Illustration 28

Characteristic X-rays of frequency 4.2 x 1018 Hz are when transitions from L-shell, produced K-shell take place in a certain target material. Use Mosley's Law to determine the atomic number of the target material. Given Rydberg constant R =1.lxl0 7 m-'. Solution

For K-series, b =1

nm,. =1 and (sine)_ =1

INTENSITY OF TRANSMITTED X-RAY

-l)'(J___J__) n; ni

The intensity of monochromatic X-ray beam after penetrating a thickness x of a target material is given by I= I 0 e-µx, where µ is a constant called Absorption Coefficient. Its value depends upon nature of material. µ increases with increase of A and atomic number Z of absorbing material. µ is maximum for lead.

Substituting the values, we get

4.2x 1018 = (1.1 x 107 )(3x10')(Z-1)'G-¾)

=> => =>

n

For ma;ximum wavelength

Hence equation (1) has solution only for l. 5 2d .

ni ni

v = Rc(Z

l.= 2dsin8

:::::} "-mu =2d

AE = hv = Rhc(Z-b)'(J___J__)

=>

where n is an integer, called order of maxima. Equation (1) is called Bragg's equation

(Z-1)' = 1697 Z-1=41 Z=42

BASED ON ATOMIC STRUCTURE & PROPERTIES (Solutions on page 4.99)

PROPERTIES OF X-RAYS

a)

b) c) d) e)

f)

g) h) i)

X-rays are electromagnetic waves of very short wavelength of order of 1 A . Therefore, they can exhibit properties of reflection, refraction, interference, diffraction, polarisation like ordinary light. Due to this property they help in the study of crystal structure. They travel in vacuum with speed of light i.e. c =3x10 8 ms-1 They are electrically neutral, hence cannot be deflected by electric and magnetic fields. They do not possess magnetic moment. They have ionising power. Therefore, when they pass through a gas, the gas is ionised. They have penetrating power. They can penetrate light substances like wood, flesh, thick paper, thin sheets of metals, but cannot penetrate heavy substances like lead, calcium, barium sulphate etc. When incident on certain metals, they liberate electrons. This effect is called photoelectric effect. They cause fluorescence in many substances like barium, cadmium, zinc sulphide etc. They have destructive effect on living tissues. Therefore the persons working with X-rays· often wear lead clothes.

1.

A doubly ionised lithium atom is hydrogen like with atomic number 3. (a) Find the wavelength of the radiation require to excite the electron in Li++ from the first to the third Bohr orbit (Ionisation energy of the hydrogen atom equals 13.6 eV). (b) How many spectral lines are observed in the emission spectrum of the above excited system.

2.

Find the Ionization energy of a doubly ionized lithium atom.

3.

An electron and a proton are separated by a large distance and the electron approaches the proton with a kinetic energy of 2 eV . If the electron is captured by the proton to form a hydrogen atom in the ground state, what wavelength photon would be given off?

4.

Electrons in hydrogen like atoms (Z = 3) make transitions from the fifth of the fourth orbit and from the fourth to the third orbit. The resulting radiation are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work function of the metal and the stopping potential for the photoelectrons ejected by the longe~ wavelength.

5.

Use Moseley's Law with b = 1 to find the frequency of the

K. X-rays of La(Z = 57) if the frequency

Ot

the K" X-rays of

Cu(Z=29) is known to be 1.88x1018 Hz.

BRAGG'S LAW

When an X-ray beam of wavelength A is incident on a crystal of inter planar spacing d at grazing angle 8, then the directions of diffraction maxima are given by 2dsin8 = nl. ... (1)

6.

When the voltage applied to an X-ray tube is increased TJ = 1.5 times, the short wave limit of an X-ray continuous spectrum shifts by M = 26 pm . Find the initial voltage applied to the tube.

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Optics & Modern Physics

Find the .cutoff wavelength

for the continuous X-rays

~(d)

coming from an X-ray tube operating at 40 kV . (0)

8.

Certain gas of Identical hydrogen like atoms has all .its

atoms in a particular upper .energy level. The atoms make transition to a higher energy level when a monochromatic radiation, having wavelength 1654 A, is incident upon It. Subsequently, the atoms emit radiation of only three:

different photon energies. (a) Identify the atom . (b) Obtain the ionization energy for the gas atoms•. (c) If the atoms of the gas are to be excited to such a level which gives radiation ·of only six different photon energies, what should be energy of incident radiation.

9.

1 0.

Electrons are emitted from an electron gun at alniOst zero velocity and are accelerated by an-electric field E through a distance of 1 m . The electrons are now_ scattered by an_ atomic hydrogeri: sample in grounds state. What shoulg be the minimum value of E so that red light of wavelength 6563 A may be emitted In the hydrogen? A hot gas emits radiation of wavelengths 460

A,

831

A

and 1035 A only. Assume that the.atoms have only two excited states and the difference between consecutive

energy levElls decreases

14.

Using the known values for hydrogen atom, calculate (a) . radius of third orbitfor Li"' (b) speed of electron in fourth orbitfor He· .

15.

Find an 8xpression for the magnetic dipole ·moment and magnetic field Induction at the center of a Boh(s hypothetical hydrogen atqm in the nth orbit of the electron in 0 terms of universal constants.

16.

A small particle.of mass m moves In such a w_ay that the potential energy U = ar2 where ~ iS a, ·constant and r is the distance of the particle from the origin. Assuming Boh(s model of quantization of angular momentum and circular orbits, find the.radius of nth allow~d orbit. .

17.

(a)

of K0 line of target atom B at 40 kV , then determine

aS energy IS increased. Taking the

energies of the ground state and the first e~cited state.

(b)

A mixture of hydrogen atoms (in 'their ground state) and .hydrogen like Ions (in their first excited state) are being

excited by electro!'ls which have· been accelerated by same potential difference V volts. After excitation when they

come directly into ground state, the wavelengths of emitted light are fotind in the ratio f5 : 1. Calculate the minimum value of V for which both the ,atoms get excitea after collision with electrons. Also find the atomic number of 0,ther ion and the energy of emitted llghtby hydrogen atoms and ions. 12.

A stationruy .He• emitted a photon corresponding to the first line of Lyman series. This photon liberated a photo elecJron from a stationary· hydrogen atom in the gl'ound state. Find the velocity of the photoelectron.

If two times the 1-,,,. of continuous X-ray spectra 61 target atom A at 34.3 kV, is same as the wavelength

energy of the highest energy state to be zero. Find the

11.

ihe

Find wave number of the p~oton emitted for transition n =3 to 11 =1. What Is the minimum energy that an electron will have after interacting with this atom- in the .ground state if the initial kinetic energy of the electron is (i) 6 eV (ii) 11 eV

the atomic number of the atom B . _ Stopping potentials of 24, 100, 11 O and 115 kV are measured for photoelectrons emitt~d from a certain ele_ment when it is radiated with mo~nochromatlc X-ray. If ·Uils element is used as a target in an X-ray tube,· what will be the wavelength of K0 line?

18.

If potential energy In first orbit is taken to be zero, then find the kinetic energy, potential energy and total energy in first and second orbit of hydrogen atom.·

19.

Wavelengths belonging to Balmer 'Series lying in the range of 450 nm to 750 nm Were used to eject photoelectrons from a metal surface whose work- function is 2 eV . Firid (il1 eV) the maximum kinetic energy of the emitted photoelectrons. Take h9 = 1242 eV nm.

20. The potential energy of a particle varies as :

130 The energy levels of a hypott,elical ·one electron atom are shown In the figure. "'-------~o ev n = 5 - - - - - - - - - - 0 . 8 0 eV n=4 -1.45eV

U(x)=[E,, o,;x,;1 0 X>1 For o,;x,;1, the de-Broglie'wavelength Is J..1 -and for

x> 1 , the de-Broglie wavelength is }.2 • Total energy of the '

particle is 2E, . Find

n =3 - - - - - - - - - 3 . 0 8 eV

21.

n=2 -----,,l.-----5.30eV 10 3

,

!:,_. J..,

An electron·in a hydrogen like atom Is In an excited state. It has a total energy of -3.4 eV , Calculate. (a) the kinetic energy, (b) the de-Broglie wavelength of the electron.

n = 1 °'---- ·_.1_e_v_ _ _ -15,6 eV (a) (b)

Cc)

Find the Ionization potential of this atom. Find the short wavelength llmit of the series terminating at n = 2 . Find the excitation ootential for the state n = 3 .

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NUCLEAR STRUCTURE

The nucleus of an atom consists of two types of particles, protons and neutrons. A proton has a positive charge equal to 1 · 6 x 10-" C and a mass equal to 1 · 6726 x 10-27 kg . A neutron has no charge and its mass is 1-6749 x 10-21 kg. Thus a neutron is slightly heavier than a proton. Protons or neutrons, being the particles present inside nucleus, together are called Nucleons. The total number of protons in the nucleus is called its atomic number (Z) . The total number of nucleons in the nucleus is called its mass number (A) . If N is the number of neutrons, then, A=Z+N No electrons are present inside the nucleus. If X is the chemical symbol for an element then its nucleus is represented as or as 2 XA

1 u = ~ ( mass of one atom of 6C12 ) = 1 · 66056 x 10-21 kg 1 NUCLEAR RADIUS

Assuming that the nuclei are spherical, their radii are well represented by the empirical formula R=R,A1/'

where R, =1-lxl0-15 m=l·lfermi(fm) NUCLEAR DENSITY

The density of a nucleus of mass M and mass number A can be written as

M

p=--

:x

¾7tR' fx(R,A1/3)' =>

ISOTOPES

Nuclides having the same charge number

{Z) but different

mass number (A) are called isotopes. All the isotopes are

Nuclides having the same mass number (A) but different atomic number ( Z) are called isobars.

(Ax1.67x10-") kg 3

±x(l-lxl0-15 m) A 3

pce2.9x1017 kgm.,,

1bis comes out to be -10 17 kgm-3 , which is extremely large

as compared to the density of ordinary matter which is -10 3 kgm-,3.

chemically similar and hence they occupy the same position in the periodic table. ISOBARS

Aamu

CONCEPTUAL NOTE(S) Since density is independent of mass number A, so all nuclei same density. So, whether two nuclei are isobars, or isotopes or isotones. They must possess same density.

THE NUCLEAR FORCE

ISOTONES

Nuclides having the same neutron number ( A-Z) but different mass number (A) are called isotones. ATOMIC MASS UNIT (u OR AMU)

Atomic and nuclear masses are generally expressed in terms of atomic mass unit (a.m.u.).

The force which binds the protons and neutrons inside the nucleus is neither electrical nor gravitational. It is an entirely different kind of force called the strong nuclear force. This force is extremely complex in nature. Some of its main characteristics are mentioned below. a) Nuclear forces are attractive in nature. Their magnitude, which depends upon inter nucleon distance is of very high order.

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c)

d)

e)

Nuclear forces are charge independent. Nature of force remains the same whether we consider force between two protons, between two neutrons or between a proton and a neutron. These are short range forces. Nuclear forces operate between two nucleons situated in close neighbourhood only. Nuclear forces decrease very quickly with distance between two nucleoI\S. Their rate of decrease is much more rapid than that of inverse square law forces (Coulombic forces). The forces become negligible when the nucleons are more than 10-14 cn:i apart. Nuclear forces are spin dependent. Nucleons having parallel spin are more strongly bound to each other than those having anti-parallel spin.

YUKAWA THEORY OR MESON THEORY OF NUCLEAR FORCES

According to this theory, a nucleon consists of a core surrounded by a cloud of mesons, which may be charged or neutral. The mesons constantly get exchanged, back and forth, between two neighbouring nucleons. In this process the two nucleons remain bound to each other. A.

PROTON-PROTON INTERACTION

It is the force between two neighbouring protons. It is due to the exchange of rr, 0 meson between them. It is represented in' the form of a reaction as folloWs.

Proton P1 emits rr,0 and gets coi:verted into a proton P/, having different co-ordinates. So, P1------>P/ +rr:o This 1t0 is absorbed by P, which also gets converted into a

P------>N'+1t+

b)

1t"· meson is emitted by the neutron whiclt is absorbed by the proton. N-->P'+1t·

P+1C------>N' DOG-BONE ANALOGY

The above interactions can be explained with the dog bone analogy according to which we consider the two interacting nucleons to be two dogs having a common bone clenched in between their teeth very firmly. Each one of these dogs wants to take the bone and hence they cannot be separated easily. They seem to be bound to each other with a strong attractive force (which is the bone) though the dogs themselves are strong enemies. The meson plays the same role of the·common bone in between two' nucleons. RADIOACTIVITY

The phenomenon 0£ Spontaneous emission of radiations (o., p, r etc.) oy certain nuclei is called radioactivity. The phenomenon of radioactivity was discovered by Bacqrierel. LAWS OF RADIOACTIVE DISINTEGRATION A.

RUTHERFORD-SODDY LAWS

a)

Radioactivity is nuclear disintegration phenomenon. It is independent of all physical and chemical conditions. The disintegration is random and spontaneous statistical process. It is a matter of chance for any atom to disintegr_ate first. The radioactive substances emit o. or P particles along with r -rays. These rays originate from the nuclei of disintegrating atoms and form fresh radioactive products with different physical and chemical propertief?.

b)

new proton P2'. ·Hence,

P2 + 7to------> P2' B.

Due to exchange of "- meson.

c)

NEUTRON-NEUTRON INTERACTION

It is the force between two neutrons. It is also due to exchange o~ 1r0 between_ them

•

N 1 -----+.N/ + n°

The rate of decay of nuclei (- dN) is directly proportional-t~

N2 + 1to-----+-N2'

the number of undecayed nuclei (N) in the sample at time

•

•

_t. C.

PROTON-NEUTRON INTERACTION

It is the force between a proton and a neutron situated close to each other. It can be take place in following two ways. a)

Due to exchange of

:r,+

=>

' dN --ocN dt

"

dN =-1.N dt where A is constant of proportionalitr called Decay_ Constant or Disintegration Constant.

=>

meson.

Proton e~ts 1r+ meson and gets c'?nverted· to a neutron. Another neutron absorbs this 1t+ meson_ to get converted to a proton. So,

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Nuclear Physics N

CONCEPTUAL NOTE(S)

N,

N So, at t =T,~ ,. , N =--.!!... 2

N

fTw

•

N~

'2Tt:2

•

No

3Tw

•

N0

nT1.,

- 0----:---:-,, 2 ~ 22 -.....,........ ?:3 - ~ ..... -

0.368 N,_

where 11= t

,

N0 2"

Time Lapsed

T,,

Half Life

·---·---------------T2T3T4T5T

~

NdN

,

f-=-Af dt

N

N

0

~

log,( ( ) = _:,.t

~

e

=>

N=Noe-At

~

AT1/, = log, 2 = 0.693

~,

1: _ 0.693

where, N is the number of un-decayed nuclei in the sample at time t and N O is the number of un-decayed nuclei in the sample at time t = 0 (initially).

B.

DISPLACEMENT LAWS

When a nuclide emits ex -particle, its mass number is reduced by four and atomic number by two, i.e., XA-----+ z-2 yA-4+ 2 He4+Energy z

b)

When a nuclide emits a p-particle, its mass number remains unchanged but- atomic number increases by one, i.e., zXA-----+:+1 yA +_1 13° + V + Energy

where V is the antineutrino

I

0

I

c)

t=0

1 (=100%)

0

t=Tv,

½(=50%)

_!, (=50%)

.!.4 (=25%)

~

1=3T1/L,

t=10(r1/,)

t=nT1/,

n ----->1 H' +_1 p + v (antineutrino) 0

When a proton is converted into a neutron, positive P-particle or positron is emitted. 1

l/2 - - , . -

t = 2T1/,

The P-particle is not present initially in the nucleus but is produced due to disintegration of neutron into a proton. i.e., 1

H 1 --->0 n1 +1 13°+v (neutrino)

When a nuclide emits a gamma photon, neither the atomic number nor the mass number ~ges.

HALF LIFE (T,J

The half life period of a radioactive substance is defined as. the time in which one-half of the radioactive substance is disintegrated. If N 0 is initial number of radioactive _atoms present, then in ~ half life time T1/,, the number of undecayed radioactive atorru; will be N, and in next iu;lf life'

(=75%) 4 7 C - (=87.5%) . 8

½(=12.5%) 2 (1)"

I l

~99.9%

=0.1%

(½J

I

2

•. . {1-(½J}

Illustration 29

At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 seconds the number of undecayed nuclei reduces to 12.5%. Calculate (a) mean life of the nuclei (b) the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number. Solution (a)

In 10 s, number of nuclei has been reduced to half (25% to 12.5%). Therefore, its half-life is 11/2 = 10 s

2

N0 andsoon. 4

=-21

Table : Fraction of active/decaved atom at different time : Re~1.aining fraction . Fraction•of atoms of active atoin.s N -N Time (t) decayed -·-0-·- ..(N/N,) probability N, •. , of survival probabilitv of decav

So, we coriclude that the number of an-decayed nuclei in the sample decays exponentially with time.

a)

-Uiµ.

Relation between half-life and mean life is 4.23

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Advanced JEE Physics I

m=

number of atoms.

=_i_ =~ S log,(2)

0.693

c)

. T 0.693 S mce, 112 = - .· :>.

tmean =14.43 5

=>

=>

(b) From initial 100% to reduction till 6.25%, it takes four half lives. 1=41v2 =4(10)s=40s

=>

1=40 s

1

°i:"' ' = 0.693 (T,,,) = l .44 (T,,,)

i.e. mean life is about 44% more than that of half life. Which gives us 't > T,t/2l So, we conclude that

100%~50%~25%~12.5%~6.25%

=>

1

Tv2 . = 0.693't

:::::,.

Tv2 <-r

MEAN LIFE(<)

The mean life or average life of a radioactive substance is equal to the average time for which the nuclei of atoms of the radioactive substance exist. The average life of a sample can be calculated by finding the total life of all the nuclei of the substance and then dividing it by the total number of nuclei present in the sample initially. Mathematically

=>

,.1

't=-

Illustration 30

The mean lives of a radio·active substances are 1620 years and 405 years for a emission and p emission respectively. Find out the time during which three-fourth of a sample will decay if it is decaying both by a emission and P emission simultaneously. Solution Let at some instant of time t, number of atoms of the radioactive substance are N . It may decay either by a emission or by f3 emission. So, we can write,

If the effective decay constant is A , then 1'.N =1'..N +1'.,N CONCEPTUAL NOTE(S)

a)

From N = N0 e-11 , we get

log,(~)= -:>.t

=>

1 1 1'.=1'.. +1'., = 1620 + 405

=>

1 1'. = - - year-1 324

=>

No =Noe-u 4

=>

-1'.t = log,(¼) =-1.386

=>

( _!_}=1.386

So, the slope of the line shown in the graph i.e. the

magnitude 91 inverse of slop~ of log,(~) vs t curve is kn~wn as mean life ('t).

=>

324 1=449 year

Illustration 31 b)

From N=N0e-u,if .

=>

t=..!.=t :>.

N = N0 e-1 = N0 (

¾) =0.3iN

0

A radioactive nucleus can decay by two different processes, the half life for the first process is t 1 and that for the second process is t 2 • Show that the effective half-

= 37% of N0 •

i.e. mean -life is the time inteival in which number of undecayed atoms (N) becomes

.!a

37% of original number of atoms. or It is the time in which number of decayed atoms (N0

~ = .!. + .!. . f

tl

t2

Solution -

N)

becomes ( 1-¾) times or 0.63 times or 63% of original

=

life t of the nucleus is given by

times or 0.37 times or

Let at any instant t, number of nuclei in radioactive sample be N . Then it may decay by either of two different pro~esses. So,

4.24

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Nuclear Physics

-d: =-(d:), -( d:),

dN d A=--=--(N0e-').t)=AN0 e-A1 =Aae->-1 where

=>

'/,.N = '/,. 1N + '/,. 2N

A, = ('/,.N0 ) is the activity at time t = 0 .

=>

'/,.='/,. 1 +'/,. 2

~

-=-+t t, t,

1

1

dt

dt

So, activity of a radioactive sample decreases exponentially with time.

1

"/ -t-0.693) ( . v,- --,.UNITS OF RADIOACTIVITY

Illustration 32 Prove mathematically that mean life or average life of

a radioactive substance is t,m

=..!. .

A.

BACQUEREL

In S.I. system the unit of radioactivity is bacquerel. 1 bacquerel = 1 disintegration/sec= 1 dps.

'/,.

Solution

B.

Let N be the number of atoms that exist at time t . Between t and t+ di let dN atoms are decayed, then Mean or average life is

It is defined as a unit of activity equal to 10 6 dps.

0

f tdN

C.

RUTHERFORD

CURIE

The traditional unit of activity is curie. It is defined as 3, 7 x 1010 dps which is also equal to the radioactivity of 1 g of pure Radium.

tw =.!:!!.__ 0

f dN

N,

. Smce,

=>

dN

--='/,.

Illustration 33

N

The half-life of Cobalt 60 is 5.25 years. How long after its activity have decreased to about one-eight of its original value? Solution The activity is proportional to the number of undecayed atoms. In each half-life, half the remaining sample decays.

dt

dN=-'/,.Ndt

-f t'/,.Ndt 0

-N,

Since, N

=N 0 e-u, so

Since,

(½)(½)(½)=½,

therefore,

three

half-Jives

or

15.75 yr are required for the sample to decay to !th its

N,

8

This integration is done by parts to get the result

original strength.

ACTIVITY OF RADIOACTIVE SUBSTANCE (A)

The activity of a radioactive substance means the rate of decay (or the number of disintegration/sec). This is denoted by A A, ='/,.N,

Illustration 34 A count rate-meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. (a) Find the decay constant (b) Also, find the half life of the sample Solution

Initial velocity A;= dNI ='/,.N0 = 4750 dt 1=0

... (1)

... (2)

Final velocity A 1 = dNI = '/,.N = 2700 dt 1=5 0

Dividing (1) by (2), we get

==================================== = 4.25

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Advanced JEE Physics

743x101•,-0) 2 xl0' (1-e-(';:°:'""}) . ( 0.693 ) 14.3x3600

4750= N0 2700 N, The decay constant is given by

Solving this equation we get t = 14.3 hr. A = 2.~03 log, ( ~:) =>

Illustration 37 There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half life of neutrons be 700 s, what fraction of neutrons will decay before they travel a distance of 10 m? Mass of neutron equal.to 1.675 x10·" kg.

A= 2.303 10 (4750) = 0_113 min-1 5 g, 2700 ·

Half life of the sample is

T = 0.693 = 0.693 = 6_14 min. A 0.113

Solution

Illustration 35 The half-life of radium is 1620 years. How many radi~ atoms decay in 1 s in a 1 g sample of radium. The 1

atomic weight of radium is 226 gmoi-

N = ( O~~l )( 6.02 x 10

= 2.66 x 10 atoms

The decay constant is

A 0.693

T1/2

0.693 (1620)(3.16x10')

=>

{2k =

v=

v-;;; s

=> 21

)

2

•

Solution Number of atoms in 1 g sample is 26

K 1 , . Smce, =-mv

Taking one year = 3.16 x 10 s Now, t.N = iN = (1.35 x 10-11 )( 2.66 x 10 21 ) = 3.6 x 1010 s-1 . l!.t

Thus, 3.6 x 1010 nuclei decay in one second.

Illustration 36

A F" radio nuclide with half life· T =14.3 days is produced in a reactor at a constant rate q = 2 x 109 nuclei per second. How soon after the beginning of production of that radio nuclide will its activity be equal to R =10 9 disintegrations per second?

10 2.5xl03

Fraction decayed is

N

N,{1-,-")

N,

N,

!:!_ =1- e-0.6930dxlo-3/700

=>

~=3.96x10-'

7.43 x1013

dN dt

Now, -=q-iN

1~=f

=>

N=i(1-,-")

o

N,

Illustration 38 A radionuclide X is produced at constant rate a . At time t = 0 , number of nuclei of X are zero. Find

(a) the maximum number of nuclei of X . (b) the number of nuclei at time t . Decay constant of x· is A.. Solution (a) Let N be the number of nuclei of X at time t . Rate= a

Rate =AN

:

{given} Rate of formation of X = CL Rate of disintegration =AN Number of nuclei of X will increase until both the rates will become equal.

dt

A

Substituting the values, we get

a=A.Nrrur,(

=>

CL Nmax=i

(b) Net rate of formation of X at time t is, dN -=CL-AN dt

=

_,,

Therefore;

=>

o q-AN

1 -e

N,

Solution

10' 0.693 14.3x3600

= 2_5 xlO' ms·'

4x10-' s

=> 1 _35 x 10-u s-'

7

V

19

2x0.0327x1.6x10· 1.675 X 10'27

4.26

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Nuclear Physics

_!!'!_=dt

=>

o.-1.N

Integrating with proper limits, we have

=0.648 =64.8%

and of U238 is 35.184%

o

N=~(1-,-")

=>

X 1QQ%

Similarly, share of U235 is 0.016%

f_!!'!_ =fdt a-J..N o

i

1.

This expression shows that number of nuclei of X are increasing exponentially from Oto ~ . 1.

Illustration 40 Uranium ores on the earth at the present time typically have a composition consisting of 99.3% of the isotope 92 U238 and 0.7% of the isotope 92 U:z.;s • The half

Jives of these isotopes are 4.47x10'y and 7.04xl0 8 y respectively. If these isotopes were equally abundant when the earth was formed, estimate the age of the earth.

N

Solution

o/1-. ---------------

Let NO be number of atoms of each isotope at the time of formation of the earth (t = 0) and N 1 and N 2 the number of atoms at present (t = t). Then

... (1)

N1 =Noe-;,11

and N 2 Illustration 39

Natural uranium is a mixture of three isotopes ~U, ~U and ~U with mass percentage 0.01%, 0.71% and 99.28% respectively. The half life of three isotopes are 2.5 x 10 5 years. 7.1 x 108 years and 4.5 x 109 years respectively. Determine the share of radioactivity of each isotope into the total activity of the natural uranium.

=>

N1 N,

(1. 2 -1. 1 )t=log,(::) t=(-l-)log,(99.3) 1.2 -1. 1 0.7

Substituting the values, we get 0.693 7.04 X lo'

0.01 0.71 d 99.28 m, m2 = m an m3 = m 100 100 100 m1 m2 m3 Now, N 1 = , N2=and N 3 = M1 M2 M3 en m1 =

=>

(;:;J;,

=

~l_ + ~_!_ + ~_!_ M 1 T1

=>

M3 T3

0.01/100 1 --~-x--234 2.5 x 10 5 years

R, R

M 2 T2

(

1

0.693 4.47 X 109

lo

(99.3) g, 0.7

t=5.97xl09 y

Illustration 41 In the chemical analysis of a rock the mass ratio of hvo radioactive isotopes is found to be 100 : 1 • The mean lives of the hvo isotopes are 4 x 109 years and 2 x 109 years respectively. H it is assumed that at the time of formation the atoms of both the isotopes were in equal proportional, calculate the age of the rock. Ratio of the atomic weights of the two isotopes is 1.02 : 1.

where M 1 , M 2 and M 3 are atomic weights.

R

... (4)

Equating (3) and (4) and taking log both sides, we get

Let m be the total mass of natural uranium.

R,

... (3)

N, = 99.3 0.7

=>

R, 1.1 N1 Shareof U"' ,-= R 1.1N 1 + 1.2 N 2 + 1.,N,

=>

= i"2-1"iJ1

N,

Let R1 , R2 and R3 be the activities of U234 , U 235 and

Th

... (2)

Further it is given that

Solution

U238 respectively. Total activity R = R1 + R2 + R,

= N 0e-;,

21

1 )+(0.71/100)( 1 ) 0.01/100)( 234 2.5xl0 5 235 7.lxl0 8 + (9928/100)( 1 ) 238 4.5 X 109

Solution At the time of observation i.e., at time t, we have

3.= 100 m2 1

{given}

Further it is given that

================================== = 4.27

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Optics & Modem Physics

A,= 1.02

masses of the final neutral atom and the neutral

A,

helium- atom.

1 b)

Number of ato'ms N = m A

N, N2

m, A, m, A,

100 1.02

(i)

... (1)

-=-x-=--

P-decay : There are different simple type of P-decay p-, P' and electron capture. of

time of formation, then

Noe-'-i' Noe-~t

N1 N2

().2-"-1)1

Jr

(p-)

is an electron. Emission

involves transformation of a neutron into a

proton, an electron and a third particle called an antineutrino ('ii) .

Let N 0 be the number of atoms of both the isotopes at the

-=---=e

A beta minus particle

(ii)

... (2)

P- decay usually occurs with nuclides for which the neutron to proton ratio (

~ ratio)

is too large

Equating (1) and (2), we get · for stability.

e('-,-A,)1 =100 --

(iii) In p- decay, N decreases by one, Z increases by one and A doesn1t change.

l.02

(1.2 -1.,)t = log, (100)-log, (1.02)

(iv) p- decay can occur whenever the neutral atomic mass of the original atom is larger than that of the final atom.

log, (100)- log, (1.02)

(2x~O' - 4x~O')

(v) Nuclides for which N is too small for stability can

Substituting the values, we have

z

t =1.834 x 1010 years

emit a positron, the electron's antiparticle, which is

NUCLEAR RADIATIONS

identical to the electron but with positive charge. The basic process called beta plus p• decay

p-> n+P' +v (v = neutrino)

According to Rutherford's experiment when a sample of radioactive substance is put in a lead box and allow the emission of radiation through a small hole only. When the radiation enters into the external electric field, they splits into three parts (a-rays, P-rays and r -rays) y-rays

a -rays

+

+ + ·+

+

(A)

a)

){ y-rays a. -rays

x X

W decay can occur whenever the neutral atomic mass of the original atom is at least two electron masses larger than that of the final atom (vii) The mass of v and 'ii is zero. The spin of both is (vi)

!2

x

p-rays ' '

'

in units of

!!.__ The charge on both is zero. The

21t spin of neutrino is anti.parallel to it1s momentum while that of antineutrino is parallel to it's momentum.

Magnetic ){ field

(viii) There are a few nuclides for which f3+ emission is not energetically possible but in which an orbital electron (usually in the k-shell) can combine with a proton in the nucleus to form a neutron and a neutrino. The neutron remains in the nucleus and the neutrino is emitted.

(B)

p+p+-+n+v

a -decay : Nearly 90% of the 2500 known nuclides are radioactive; they are not stable but decay into other nuclides (i) When unstable nuclides decay into different nuclides, they usually emit alpha (a) or beta (P) particles. (ii) Alpha emission occurs principally with nuclei that are too large to be stable. When a nucleus emits an alpha particle, its N and Z values each decrease by two and A decreases by four. (iii) Alpha decay is possible whenever the mass of the original neutral atom is greater than the sum of the

c)

y -decay : The energy of internal motioil of a nucleus is quantized. A typical nucleus has a set of allowed energy levels, including a ground state (state of lowest energy) and several excited states. Because of the great strength of nuclear interactions, excitation energies of nuclei are typically of the order of 1 MeV, compared with a few eV for atomic energy levels. In ordinary physical and chemical transformations the nucleus always remains in its ground state. When a nucleus is placed in an excited state, either by bombardment with high-energy particles or by a radioactive transformation, it can decay to the ground state by

= ================================== 4.28

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Nuclear Physics emission of one or_ more photons called gamma rays or gamma-ray photons, with typical energies of 10 keV to 5 MeV. Titls process is called gamma (y) decay. All the known conservation laws are obeyed in y-decay.

The intensity of y -decay after passing through x thickness of a material (µ=absorption co-<efficient)

Table: Properties of a, Features

!

1. Identity

' 12

is

given by

I= I,,-'"

p and y -rays

a -particles

P-particles

y-rays

Helium nucleus or doubly ionised helium atom ( 2 He4)

Fast moving electron (-P' or P-)

Photons (E.M. waves)

+2e

-e

ZERO

4mP

m,

Massless

::::: 107 ms-1

1 % to 99% of speed of light

Speed of light

4MeV to 9MeV

All possible values between a minimum certain value to l.2MeV

Between a minimum value to 2.23 MeV

'100

10,000 (100 times of p upto 30 cm

Charge

I I

3. Mass 4mP (m, =mass of

proton = 1.87 x 10-"

14.

Speed

5. Range of kinetic energy

6. Penetration power

(r, P, a)

1 (Stopped by a paper)

(100 times of a )

of iron (or Pb ) sheet

I

7. Ionisation power

I

·10,000

100

1

Deflected

Deflected

Not deflected

Line and discrete

Continuous

Line and discrete

Produc~ heat

Produces heat

Produces,. photo-electric effect, Compton effect, pair production

(a>P>r) I

18. Effect of electric or· ' magnetic field

'

9. Energy spectrum

I10. Mutual interaction with I

'

matter

I

z-2 yA-4 + 2He-1 Z

XA ~ na yA' Z'

=>

1l

"

A-A 1

b)

I

I

zXA ~Z+lyA + _1eo+v zXA ~ z.XA

zXA~zXa+"(

-

=> n,=(2n.-Z+Z')

=---

4

RADIOACTIVE SERIES

a)

'

!

a-decay

XA z

11. Equation of decay

I

If the isotop~ that results from a radioactive decay is itself radioactive then it will also decay and so on. The sequence of decays is known as radioactive decay series. Most of the radio-nuclides found in nature are members of four radioactive series. These are as follows

Inumber Mass 4,z

Table · Four radioactive series Series Stable end Parent (Nature) product Thorium (natural)

., Th'"

a2Pb200

Integer

11

I

52

Neptunium II 52 I 411+1 (Artificial) 93Npw B""' c:================================= 4.29 = 83

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Advanced JEE Physics Uranium (Natural)

4n+2

Actinium (Natural)

I 4n+3 ' c)

d)

92

U"'

s9Ac221

82

Pb 206

51

82

Pb'"'

51

-

Find the time at which NY is maximum and determine the population of X and Z at that instant.

I I

The 4n+1 series starts from 94 Pu 241 but commonly known as neptunium series because neptunium .is the longest lived member of the series. The 4n + 3 series actually starts from

SUCCESSIVE DISINTEGRATION EQUILIBRIUM

92

AND

Solution

Let at time t =t , number of nuclei of Y and Z are NY and N 2 • Then

z

U235 • RADIOACTIVE

Suppose a radioactive element A disintegrates to form another radioactive element B which intern disintegrates to still another element C ; such decays are called successive disintegration. B

Rate of disintegration of B = dN, = -1..2 N 2 dt

J(

Net rate of] (. Rate of Rate of formation = disintegration - disintegration ( ofB of A ofB

J

:

l.,N,

~ : ~ l..,N,

( d~x) = -AxN x

... (1)

( d~y) = AxNx -1..yNy

... (2) ... (3)

and ( d:z )=AyNy Given Ny(t) = N,1..x [e-'"' -e·','] Ax - Ay . . dN (t), For Ny to be maxrmum __Y _ = 0 dt i.e., AxNx = AyNy {from equation (2)}

=> =>

(N e-i.xt) = ')..

A X

EQUILIBRIUM

In radioactive equilibrium, the rate of decay of any radioactive product is just equal to it's rate of production from the previous member. i.e.~ A1/'{1 ='A. 2N 2

Rate

Rate equation of the populations of X, Y and Z are

C

Rate of disintegration of A= dN, = -1.. 1N 1 (which is also the dt rate of formation of B)

Zero N,

Zero N,

Nx= N 0 e. . .x1

~

l.,

A

N0 =10w

t=O t=t

O

A -A

No'A.x [e-Ayl _ e-Axt]

y

A, -'}.. X

y

e-Ayt

Y=---1

_X_ _

A.y

e-Axt

=>

Ax= e
= rel="nofollow">

(Ax -1..y )tlog, (e) = log,G:)

Illustration 42

=>

I

A radioactive nucleus X decays to a nucleus Y with a decay constant Ax = 0.1 s-1 • Y further decays to a stable

Substituting the values of Ax and Ay, we have

=>

!:1.- N2 -

'!2 _·(TtJ2)2

'A.2 - N1 -

'!1 -

(Tt/2\

nucleus Z with a decay constant Ay =2. s-1 Initially 30 there are only X nuclei and their number is N 0 =10 20 • Setup the rate equations for the populations of X , Y and Z . The population of the Y nucleus as a function of time is given by Ny(t) =

=

{(,.:~~Y)

}{exp(-1..vt)-exp(-1..xt)}

Av

1

(Ax-Av)

log,(,.x) 1..y

0

t ( · ~_!_log,( /)=15log,(3) O.l 30

30

=> !=16.48 s The population of X at this moment Nx

=>

= Noe-i.xi = (1020)e-Co.1JC16.48)

Nx =1.92xl019

4.30

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{From equation (4))

Since, Ny= NxAx

A.y

=>

Ny =(1.92xlO")(O/) =5.76xl019

,.

30 => => =>

'

Nz=N,-Nx-Ny 20

19

Nz=l0 -l.92xl0 -5.76xl0

II

19

Nz =2.32xl019

i

Number of nuclei of B as function of time t .

3.

The radioactivity of a uranium specimen with mass number 238 Is 2.5x1-04 s-1 , the specimen's mass is 2 g. Find the half-life.

4.

(b) Time t at which the activity of B is maximum and the value of maximum activity of B . Let the number of nuclei at any instant be shown in the table

A

B

C

At t=O

N,

0

0

At t

N,

N,

N,

: 5. I I I

... (1) Now, dN, =1,.(N, -N,) dt

I

dN, =1,.N e·" -fN dt a 2

=>

dN, +1,.N,dt=1,.N0 e·"

=>

e"dN2 + 1,.N,e"dt = 1,.N0 dt

=>

d(N,e")=1,.N0 dt

I

~

The disintegration rate of a certain radioactive sample at any instant is 4750 disintegrations per minute. Five minutes later the rate becomes 2700,per minute. Calculate (a) decay constant and (b) half-life of the sample.

'

the roots of a plant The plant was allowed 70 hours to 1 settle down and then activity was measured in its fruit. If the , activity measured was 1 µCi , what percentage of activity is f transmitted from the root to the fruit in steady state?

I

iI I

I 7.

A sample of 1 g of 1:ai with a half life of 2.7x107 year decays into a stable isotope .of thallium by emitting a particles. (a) What is the activity of the sample? {b) What will be the activity of the sample after 2 years? (c) After-what time does the activity reduces to 25% of the original activity?

'

N,

'

i

0

0

I a.

N 2 =AN0 te-u

'

Ac'l2. 7 has a half life of 21.8 years with respect to 1' radioactive decay. The decay follows two parallel paths, one I leading the Th 227 and the other leading to Fr223 • The percentage yields of these two daughters nuclides are 1.2% and 98.8% respectively. What is the rate constant in yrs-1 , for each of the separate paths?

In an agricultural experiment, a solution containing 1 mole J of a radioactive material (T\12 =14.3 days) was injected into j

Integrating, we get

f d(N,e")= J1,.N,dt

1 1

I 6.

I

=>

!

!

Solution . (a)

! Find the amount of heat generated by 1 mg of Po210 preparation during_ the mean life period of these nuclei if the 1 emitted alpha particles are known to possess kinetic energy : 5.3 MeV and practically all daughter nuclei are formed I directly in the ground state.

Illustration 43

(a)

{Solutions on page 4.102)

At time t = O, number of nuclei of a radioactive substance are 100. At t = 1 s these numbers become 90. Find the ,· number of undecayed nuclei at t = 2 s .

2.

'

Consider a radioactive disintegration according to the equation A -), B -), C . Decay constant of A and B is same and equal to A. Number of nuclei of A, B and C are N 0 , 0, 0 respectively at t =0 . Find

BASED ON NUCLEUS PROPERTIES & RADIOACTIVITY

I

I inside a closed volume. The radioactive decay constant for I A number N0 of atoms of a radioactive element are placed \

the nuclei of this element is A1 • The daughter nuclei that j

(b) Activity of Bis, R, =1,.N2 =)..'N,te·"

form as a result of the decay process are assumed to be radioactive, too, with a radioactive decay constant A2 •

For maximum activity, we have dR, =0

l \

Determine the time variation of the number of such nuclei. , Consider two limiting cases, when A.1 » A2 and A1 « A2 • j

dt

=>

1=2.),_

9.

Calculate the probability that a radioactive atom having a mean life of 10 days decays during the fifth day.

=>

R_ = 1,.N, e

10.

Old wood from an Egyptian tomb, 4500 years old has C-14 activity of 7.3 dis. min- 1 g-1 • 'Old wood known to be 2500

I f

I 1

j

years old has a C-14 activity 9.3 dis. min-1 g-1 •

================================::::J = 4.31

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Advanced JEE Physics (a) · What is half life !or C-14? (b) What is the activity of fresh wood?

I/ 11.

Determine the amount of

l

provide a source of a particles, of 5 millicurie strength. If

84 Po

210

l!.Jn = [zm, +(A-Z)m.J-m( 2 XA)

_

(polonium) necessary to

hail life ot.polOnium is 138 days, given 1 curie =3.7x.10

10

This mass defect exists in the form of binding energy of nucleus, which is responsible for binding the nucleons into a small nucleus. So,

Binding energy of nucleus = (!!.m)c' = (931.5)!!.m (in MeV)

disintegrations/sec. 12. The specific activity of a preparation consisting of l radioactive Co58 and non-radioactive Co59 is equal to

j

2.2x10 12 dissec·1gm. The half 'life of Co58 is 71:3 days. Find the ratio of the mass of radioactive cobalt in that preparation to the total mass of the preparation.

!

1 13.

The half Jives of radioisotopes P32 and P33 are 14 days

and 25 days respectively. These radio isotopes are· mixed in the ratio of 4: 1 of their atoms. If the initial activity of the mixed sample ·is 3 mCi, find the activity of the mixed ·isotopes I

- ----- -- _ _ __J

_ after60days. ------

andb"m d"~g energy per nuc1eon

If the masses are taken in atomic mass unit, the binding energy is given by B.E.=[(zm, +(A-Z)m,,)-m( 2 xA)]931.5 MeV Dividing the binding energy by the number of nucleons A , in the nucleus, We obtain the binding energy per nucleon. The stability of a nucleus is measured by the binding energy per nucleon. The more the binding energy per nucleon of the nucleus the more stable a nucleus is. Packing fraction

MASS ENERGY EQUIVALENCE

According to Einstein the mass and energy are equivalent i.e., mass can be converted into energy and vice-ve_rsa. The mass energy equivalence relation is

l!.E = c't.m Accordingly, .annihilation of 1 kg· mass is equivalent to energy given by /1E =1 X (3 X 10 5 )

=>

l!.E=9x1016

2

is the mass defect per nucleon. So,

= i'.m A

a}

CONCEPTUAL NOTE(S) It is not the binding energy which accounts for th8 stability of nuclells.

b)

The stability of nucleus is governed by binding energy per

J

l1E=(1.67x10·")(9x1016 ) 21

J

Illustration 44

Find. the binding energy of

1

;c? Also find the

binding energy per nucleon. Given mass of !H

16

(1.67 x 10- )(9 x10 ) l!.E -'-----'--'c~-~ eV 1.6x10-"

=1.0078 u,

:11 = 1.0087 u, :'C = 12.00004 u .

l!.E = 931.5 MeV

Solution

MASS DEFECT AND BINDING ENERGY

It has been observed that the mass of a nucleus is always less than the mass of its constituent nucleons (i.e., protons + neutrons). The difference between the total mass of the nucleons and the mass of the nucleus is called the mass defect (i'.m) . This is due to the fact that when nucleons

combine to form a nucleus, the binding energy of nucleons is liberated. The binding energy is equal to the work that must be done to split the nucleus into particles constituting it.

One atom of 1;c consists of 6 protons, 6 electrons and 6 neutrons. The mass of the rm-combined protons and electrons is the same as that of six ~ H atoms (if we ignore the very small binding energy of each proton-electron pair). Mass of six ) H atoms = 6 x 1.0078 = 6.0468 u Mass of six neutrons = 6x1.0087 = 6.0522 u Total mass of component particles = 12.0990 u Mass of 1;c atom = 12.00004 . Mass defect ·= 0.0990 u

m( XA) be the mass of nucleus, 2

mp

f

(f)

nucleon. The more the binding· energy per nucleon, the more stable a nucleus is.

Energy corresponding to annihilation of 1 amu of mass is

Let

(t.m)c' =-A--

Binding energy = (931)(0.099) =92 MeV

= the mass of proton and

Binding energy per nucleon = ~~ = 7.66 MeV

mn =mass of neutron then, the mass defect is given by 81n = mass of constituent nucleons - mass of nucleus

= c::================================== 4.32

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Nuclear Physics Illustration 45

peaks corresponding to nuclei 2 He4 , 6 C12 and 8 Q 16 • The peaks indicate that these nuclei are more stable than those in their neighbourhood. It confirms the reason for extraorciiuary stability of a-particle .

A neutron breaks into a proton and electron. Calculate the energy produced in this reaction in MeV . Mass of an electron = 9 x 10-31 kg, Mass of proton = 1.6725 x 10-27 kg , 1.6747 x 10-27 kg . Speed

Mass of neutron =3x10 8 ms-1 •

of light

c)

Solution

Mass defect (~m) =

binding energy per nucleon is maximum and it is equal to 8.8 MeV.

Mass ) ( mass of proton ) of + ( neutron mass of electron 21

d)

27

~m = [(1.6747x 10- )-(1.6725x 10- +9xIO-'" )] ~In=

0,0013 X 10-27 kg

Energy released Q = ~c' 2

Q=(0:0013x10-27 )x(3xl08 ) =1.17xl0-13 J 1.17x10-13 1.6xl0-

--~~ 0.73 x IO' eV = 0.73 MeV 19

Q

VARIATION OF BINDING ENERGY PER NUCLEON WITH MASS NUMBER A

The graph represents the average binding energy per nucleon in MeV against mass number A . It is observed that the binding energy for nuclei (except ,He', ,c 12 and 8 0 16 ) rises first sharply, reaches a maximum value 8.5 MeV at A=S0 and then falls slowly, decreasing to 7.6MeV for elements of higher mass number A = 240 . Following facts can be concluded from this curve. Region of Greatest Stability

,_..._______

-,•

9.0

e)

..

O" / '

~ c·;,

;'·'

.

7.0

6.0

After mass number 20, binding energy per nucleon increases gradually and for mass number between 40 and 120, the curve becomes more or less flat. The average value of binding energy per nucleon in this region is about 8.SMeV. For A=56( 26 Fe 56 ), the

..

5.0

After mass number 120, binding energy per nucleon starts decreasing and drops to 7.6 MeV for uranium. This low value of binding energy per nucleon in case of heavy nuclei is unable to ha:ve control over the repulsion between the large number of protons. 'Such nuclei are unstable and are found to disintegrate by emitting a-particles . The emission of cx-parti~e not only decreases repulsive force .inside the nucleus but also increases the value of B.E./ A of the nucleus due to its extraordinary stable structure ( a-particle has large binding energy). It is call~d a-decay. Sometimes, the heavy nuclei increase the value of their B.E.f A by emitting an electron. It is called P-decay . Inside the nucleus, an electron does not exist. It is created at the time of f3-decay due to conversion of a neutron into proton. The f3-decay leads to increase in Coulomb's repulsive force, but it increases B.E./ A and also improves the neutron-proton ratio. All such nuclei, which undergo a and P-decay are called radioactive nuclei. The binding energy per nucleon has a low value for both very light and very heavy nuclei. In order to attain higher value of binding energy per nucleon, the lighter nuclei may unite together to form a heavier nucleus (process of nuclear fusion) or a heavier nucleus may split into lighter nuclei (process of nuclear fission). In both the nuclear processes, the resulting nucleus acquires greater value of binding energy per nucleon along with the liberation of enormous amount of energy.

4.0

NUCLEAR STABILITY

3.0 I. I·

2.0 I

1.0 0

a) b)

I

I

I

I

50

I

I I I I

I 100

I

I I I I 150

200

I

I

I

250

The binding energy per nucleon for light nuclei, such as 2 1 H ,isverysl)1all (=1 MeV). The binding energy per nucleon increases rapidly for _ nuclei up to mass number 20 and the curve possesses

Among about 1500 known nuclides, less than 260 are stable. The others are unstable that decay to form other nuclides by emitting a, P-particles and y -EM waves. (This process is called radioactivity). The stability of nucleus is determined by many factors. Few such factors are given below : a)

Neutron-proton ratio

( ~ Ratio)

: The chemical

properties of an atom are governed entirely by the number of protons (Z) in the nucleus, the stability of

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Advanced JEE Physics an atom appears to depend on b9th the number of protons and the number of neutrons. (i) For lighter nuclei, the greatest stability is achieved when the number of protons and neutrons are approximately equal (N ~ Z) i.e.,

(iii) Only five stable odd-odd nuclides are known : H 2 , 3 Li 6 , 5 Be10 , 7 N 14 and 75Ta18/J 1 c)

~ =1 .

(ii) Heavy nuclei are stable only when they have more neutrons than protons. Thus heavy nuclei are neutron rich compared to lighter nuclei (for heavy nuclei, more is the number of protons in the nucleus, greater is the electrical repulsive force between them. Therefore more neutrons are added to provide the strong attractive forces necessary to keep the nucleus stable.) (iii) Figure shows a plot of N verses Z for the stable nuclei. For mass nurn0er upto about A= 40. For larger value of Z the nuclear force is unable to hold the nucleus together against the electrical. repulsion- of the protons unless the number of neutrons exceeds the number of protons.

ARTIFICIAL TRANSMUTATION - NUCLEAR REACTIONS

A radio-active substance breaks up by emitting radiations. · The daughter nucleus, left behind, has different physical and chemical properties and is assigned a new place in the periodic chart. Thus radio-activity is the phenomenon by which a substance gets converted into another one. Titls change can be brought about by artificial method, by bombard~g a given nucleus with some radiation. The particles constituting the incident radiation must possess sufficient kinetic energy so as to penetrate into the given nucleus. As they enter the given nucleus, a compound nucleus (an intermediate state) is formed which is generally unstable. The compound nucleus then breaks up to produce product nucleus by emitting radiation. The process is, schematically, represented as

1-::·o 4 < - - - - - - - - - - - - - - -

I

.~gi __:_j;_..__-+

P +

0

(~::J (~J

LAWS GOVERNING NUCLEAR REACTIONS

E

a)

"

C: C:

i

z"'

b)

c) .32 40

4815.6J6~ft;ff~80·_···aa!-96~

Atomic number (Z)

d) At Bi(Z = 83, A= 209), the neutron excess in

N -Z = 43. There are no stable nuclides with 2>83. Even or odd numbers of Z or N : The stability of a nuclide is also determined by the consideration whether it contains an even or odd number of protons and neutrons. (i) It is found that an even-even nucleus (even Z and even N) is more stable (60% of stable nuclide have even Z and even N ). (ii) An even-odd nucleus (even Z and odd N ) or odd-even nuclide (odd Z and even N) is found to be lesser sable while the odd-odd nucleus is found to be less stable. 4.34

->

electrons have no contribution to it. Such reactions are known as nuclear reactions.

.c

·

(~~J

This is a reaction in which only the nuclei take part. Orbital

~ :;;

b)

+ T - > C*

e::~J (!:~J

96it-~,---,, __ -;,_

~~i':

Binding energy P.er nucleon : The stability of a nucleus is determined by value of it's binding energy per nucleon. In general, higher the value of binding energy per nucleon, the more is the stability of the nucleus.

e)

Law of Charge: The electric charge involved in nuclear reactions must be same before and after the reaction. So, charge number is conserved in nuclear reactions. Law of Number of Nucleons: The total number of nucleons involved in a nuclear reaction must be same before and after the reaction. So, mass number is also conserved in nuclear reactions. Law of Conservation of Energy: The total energy (rest mass energy + K.E.) of the reacting particles must be equal to the total energy of the product particles. Law of Conservation of Linear Momentum: The total linear momentum of the reacting particles must be equal to the total linear momentum of the produCt particles. Law of Conservation of Angular Momentum: Total angular momentum of nuclei before and after reaction must be the same.

Q-VALUE OF A NUCLEAR REACTION

Consider a nuclear reaction, schematically represented by equation I+T->P+O

Let E1 , Ep and £0 be the kinetic energies associated with I, P and O respectively while the target Tis at rest initially. Q-value of a nuclear reaction is given by,

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Q=E,+E0 -E1

Reactants

Let m11 my., mp, m0 , respectively,·bethe masses of I, T ,P and 0 BEFORE REACTION

AFTER REACTION

Energy of I= m1c2 + E1

Energy of P = mpc 2 + Ep

Energy of T = m,c'

Energy of O = m0 c2 + E0

Total energy of the system= m1c2 + E1 + "'rc2

Total energy of the system= mpc2 + Ep + m0 c2 + E0

By Law of Conservation of Energy,

m1c + E, + 1tlrC2 = mpc 2 + Ep + m0 c2 + E0 2

o

Q = E,

+ E0 -E, =[(m1 +m, )-(m, +m0 )]c'

where t.m is the mass defect between initial and final particles. CASE I Exo-ergic reactions: A reaction is said to be exoergic if Q is positive.

o

(m,+m0 )<(m1 +m,)

;u 7.01600-3m,

; Be 7.01693 - 4m,

: H 1.0783 - lm,

!n 1.0866

Total 8,02383- 4m,

Total 8.02559-4m,

The Q -value of the reaction Q = -0.00176 u = -1.65 MeV

The energy is supplied as kinetic energy of the bombarding proton. The incident proton must have more than this energy because the system must possess some kinetic energy even after the reaction, so that momentum is conserved. With momentum conservation taken into account~ the minimum kinetic energy th~.t the inciderit particle can be found with the formula. E., =-(1+ :)Q=-(1+f)(-1.65)~1.89 MeV

Illustration 47

The part of mass which disappears gets converted into the energy in accordance with- Einstein's Mass-Energy equivalence. CASE II Endo-ergic reaction: A reaction is said to be endoergjc if Q is negative.

o

Products

(m, +m0 ) > (m, +m,)

i.e., the sum of the masses of product particles is greater than that of reactant particles. For this reaction to proceed, the incoming particle must possess an energy equivalent to the mass defect. The minimum amount of energy that a bombarding particle must have in order to initiate an endoergic reaction, is called Threshold Energy

A neutron with kine ti~ ~nergy K = 10 MeV activates a nuclear reaction n+ 12C-----.+ 9 Be+a. Find the kinetic

energy of the alpha particles outgoing at right angle to the direction of incoming neutrons. Take ll = 931.5 MeV and threshold energy of reaction (E,.)= 6.17 MeV. . Solution

C

n

Q------+ m,

E,,,.

Qm,

Since,

Q+K, =K, +K4 where,

mp is the mas~ of the projectile i.e. the

nucleus used to hit the ,target and 111r is the mass of the target nucleus

..• (1)

Applying Law of Conservation of Linear Momentum Along x-axis

.J21n,K, = .J2m,K, cos8

... (2)

Along y-axis Illustration 46 How much energy must a bombarding proton possess to cause the reaction.

iLi+iH-+:Be+~n Solution

.j2m,K3 = .J2m,K, sin8

... (3)

Squaring and adding equations (2) and (3), we get_ 111tK1 + m3 K3 = m4 K4

K4=m,K1+m3Ks m4 m4 Substituting value of K, in equation (1) and rearranging, we get =>

Since the mass of an atom include the masses of the atomic electrons, the appropriate number of electron masses must be subtracted from the given values.

r:::================================== = 4.35

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Advanced JEE Physics . -E. w h ere, Q ( ) 1+""

_m,

Suppose the parent nucleus ~X be at rest before decay. Let -va and Vy be the· velocities· of the a-particle and the daughter nucleus. Applying the Law of Conservation of Momentum, we get myVy = ma Va •.• (1)

--o:l 7 - 5 69 M V -1 -- . · e 1+12

~

(1+¼)K, =-5.69+(1-¼)(10)

~

K, =2.21 MeV

As the energy Q -released in the decay process appears in the form of kinetic energy of a- particle and the daughter nucleus, So :we have

ALPHA DECAY

'Alpha decay is a process in which an unstable nucleus transforms itself into a new nucleus by emitting an alpha particle (a helium nucleus, ~He). Since an a -particle has two Protons and two neutrons, so after an a. -decay, the parent nucleus is transformed into .a daughter nucleus with mass number smaller by 4 and atomic number sg,aller_ by 2. An alpha decay can be expressed by the equation

1 2 1 2 Q 2mttva +2myVy = Substituting the value of Vy from equation (1), we get1 2 2 1 . v2 +-m mo.Va =Q 2 a a 2 m~ 1

21.22

2mymQv(l +2mava =myQ

:x---+ ~jY + !He+Q

½(my+m(l)mav! =myQ

Here Q is the energy released in the 'pi-ocess and is determined from Einstein's mas~-energy relatioii. which gives

v'=(-

K =.!:.m Cl • 2 a_«.

my

my

+m(l

)Q

Since, my = (A - 4) amu and ma = 4 amu, theref~re,

) 2 Q= ( mx·-mr-mHt c

where mx, my ._and mHt are the masses of the parent nucle1;1s

X, daughter nucleus Y and the a -particle respectively. The energy Q is shared by the daughter nucleus X and the a -particle. As the parent nucleus is at rest before its a -decay, the a -parti~les are emitted with fixed energy and hence are mono-energetic. This energy can be determined by Applying the Laws of Conservation of Energy and Momentum. For example, uraru'!'Il (238) on emitting an u -particle changes into thorium (234) as

~u~~Th+ !He+Q Similarly, polonium (208) is transmuted into lead (204) as

'!Po---,--),~Pb+ !He+Q Generally, the nuclei with mass number 210 or more undergo a -decay. In such nuclei, the long range repulsive forces between the protons dominate Over the short range nuclear forces which bind the various nucleons together. By emitting a. -particles, these nuclei achieve greater stability. An a -particle has a high value of binding energy (= 28 MeV) . After the emission of an a -particle, the binding energy per nucleon of the emitting nucleus increases and the residual nucleus·becomes more stable. SPEED OF EMITTED a-P.ARTICLES

Consider the alpha decay process equation i.e.,

1X------+t~Y + ;He+Q

For example, in the a.- decay of a randon nucleus ~Rn, we have Q=5.587 MeV

~ ~

. K.

=( A~4)Q

( 222 - 4 ) x5.587 MeV 222

K. = 5.486 MeV = 5.486 x 1.6 x 10-19

J

Since, m. = 4 arnu = 4 x 1.66 x 10·21 kg Va

~

=

2 X 5.486 X 1.6 X 10-19 -1 ms 4 X 1.66 X 10-2'

v. =1:62xl01 ms·1

THEORY OF a-DECAY (TUNNELLING EFFECU

The a.- particles emitted by different raclioactive nuclei have kinetic energy ranging from 4 to 9 MeV. The nucleus of an a- emitter posses a barrier of height about 25 MeV. Figure shows a plot of the potential energy U of the system consisting of the a- particle and the residual nucleus. The a- particles are short of about 16 to 25 MeV of energy, needed for the emission. There~ore, classically, we cannot explain the· emission of a.- particles by radiOactive nuclei.

The speed of the emitted a- particles can be calculated by using the Laws of Conservation of Energy and Momentum.

=

'

"

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u

Given : Potential barrier

my :::: 228.03 amu

'"• = 4.003 amu

m(;n) = 1.009 amu

m(!H)=l.008 amu

1 amu=l.66x10-" kg=931.5 Mevc-2

K =KE of a-particle

Solution .(a) The given equation is,

2

2Ze r=-4ne0K

~x~~Y +!He

A=228+4=232 and 92=z+2

Plot of potential energy U of an a-particle as a function of distance r from the centre of the residual nucleus.

In 1928, Gamow, Congdon and Gurney explained the emission of a.- particles in terms of the penetration of the nuclear potential barrier on the basis of Quantum Theory. According to this theory, we have a) An ct- particle may exist as an entity (already formed) inside a nucleus before it escapes from the nucleus. b) The ct- particle is in a state of constant motion inside the nucleus with a speed of about 10' ms-1 • c)

Quantum mechanically, even an ct- particle having insufficient kinetic energy has a small but . finite probability p of its crossing the potential barrier.

As the size. of the nucleus ~ 10-14 m and speed of a- particle 7 ~ 10 ms-1 , the a- particle takes about 10-21 s to move across the nucleus. Thus a- particle .Presents itself before the

potential barrier 10 21 times in a second. The probability of escape of an ex- particle from a nucleus will be P=pv

z=90

~

V

~

v(l =4x106 ms-1

(b)

•=

l.lxl0 2 x2xl.6x10-19 x3x10 3 4.003 x1.66 x 10-"

Applying Law of Conservation of Llnear Momentum, we get

~

~

v,

m,

(4.003){4x106 ) (228.03)

vy =7x10 4 ms-1

Therefore, energy released during the process is

21

s-1 ) , so P is sufficiently large and the ex- particle can tunnel through the energy barrier which is classically impossible. Hence a.- decay occurs as a result of barrier tunnelling. As v is large (10

~

~

E

(1.66x10-") {2xl.6x10-13 ) 2

[C4.003){4x10') +(228.03){7x10')'] MeV

The barrier tunnelling explains why every ~U nuclide in a sample of ~U atoms does not decay at once, even when its

decay process has a positive Q value. Consequently, the half-lives for a.- decay of most of the alpha unstable nuclei are very long. For example, the half-life of ~U for ct- decay

~

34 E = 0.34 MeV = 0. amu = 0.000365 amu 931.5

· Therefore, mass of ~ X = m, + m. + 0.000365 = 232.033365u So, mass defect is

is 4.5 x 109 year .

l'.m = 92(1.008) + (232-92)(1.009)-232.033365 ~

l'.,n = 1.962635 amu

Illustration 48

~

Binding energy =1.962635x931.5 MeV =1828.2 MeV

A nucleus X-initially at rest, undergoes alpha-decay, according to the equation

BETA DECAY

~x~~Y:ra (a)

The process of spontaneous emission of an electron

Find the value of A and z in the above process.

(b) The ct -particle in the above process is found to move in a circular track of radius 1.1 x 10 2 m in a uniform magnetic field of 3 x 103 T. Find the energy (in MeV) released during the process and binding energy of the parent nucleus X .

(e-)

or a

positron (e•) from.a nucleus is called beta decay. Like a- decay, j3-decay is a spontaneous process, with a

definite disintegration energy and half-life. It is also a statistical process, obeying the Law of radioactive Decay.

In beta minus (p-) decay, the m~s number of the radioactive nucleus remains m1changed but its atomic 4.37

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Advanced JEE Physics number increases by one. An electron and a new particle antineutrino (V) are emitted from the nucleus, as in the decay. ~P----+

::s + e- + V

In general, the beta minus decay may be represented as

:x----+ 2 :iY + -~ e + V

However in j3- decay, the disintegration energy is shared in all proportions between the three particles i.e., daughter nucleus, electron (or positron) and antineutrino (or neutrino). As a result, the kinetic energy of the electrons (or positrons) is not fixed. Their energy varies from zero to a maximum value K=. Thus j3- rays have. a continuous energy spectrum, as shown in Figure.

OR 1X----+ 2 jY+Jr_+V The electron emitted from the nucleus is called a beta particle, denoted by p- .

~r

(P') decay, the mass number of the radioactive

E " "C;t;: e o :;; a.

In beta plus

I

.0 ,,,

nucleus remains unchanged but its atomic number deCI'eases by one. A positron (e') and a new particle neutrino ( v) are emitted from the nucleus, as in the decay.

I

"' "\,

I

,; a:

'fiNa----+~Ne+e+ +v

0

In general, the beta plus decay may be represented as

"' "'

'

/ ./

0.2 0.4 0.6 Kinetic energy (MeV)-

The distribution of the kinetic energies of positrons emitted in the decay of :cu.

1X----+ 2 ~Y++~e+v

OR ~X----+ 2 jY +_p+ +v The positron so emitted is called·a beta plus particle

(P')

The positron iS an antiparticle of electron. It has a positive charge equal in magnitude to the charge on an electron and has a mass equal to the mass of an electron. Similarly, neutrino and antineutrino are antiparticles of each other.

Both are massless, chargeless particles having spins

±.!.. 2

Although a nucleus contains no electrons, positrons and neutrinos, yet can it eject these particles. It is believed that electrons, positrons and neutrinos are qeated during the process of beta decay. If the unstable nucleus has excess neutrons than needed for stability, a neutron converts itself into a proton. So in a beta-minus decay, an electron and an antineutrino are created and emitted from the nucleus via the reaction given by

The maximum kinetic energy or end Point energy K= must

be equal to disintegration energy Q. When the electron (or positron) has maximum energy, the energy carried by the dau'ghter nucleus and neutrino is nearly zero.

-- ------·-- --·CONCEPTUAL --- . --~NOTE(S) a)

b)

Since neutrinos (or antineutrinos) ·are massless and chargeless, they interact so weakly with matter that it becomes very difficult to detect them. They can penetrate through earth without being absorbed. By ingenious experiments, neutrinos have been' detected and their mass and spin or intrinsic angular momentum have been measured. Electroa:i capture. Some proton rich nuclei capture one of the atomic electrons (usually from the K shell). A proton in the nucleus combines this electron forming a neutron. A neutrino created in the process is emitted from the nucleus.

n---+p+e-+V

p+e- ...+n+v

If the unstable nucleus has excess protons than that needed for stability, a proton converts itself into a neutron. So in a beta-plus decay, a positron and a neutrino are created and emitted from the nucleus via the reaction given by p---+n+e++v Clearly, a beta decay process involves the conversion of a neutron into a proton or vice versa. These nucleons have nearly equal masses. That is why the mass number A of a nuclide undergoing beta decay does not change.

The entire process may be represented as ~X+e-----+l,Y+v This process is called electron capture or K-capture. The vacancy created in the K shell is filled by trans"ition of electrons from the outer shells. This results in the emission ---~_l_i:~aracteristic X-rays.

Illustration 49 Neon - 23 beta decays in the following way: ~Ne----=,~Na+ _~e+V

CONTINUOUS ENERGY SPECTRUM FOR BETA RAYS In both ex- and p-decays, the disintegration energy Q

depends on the nature of the radionuclide. In the ex-decay of a particular radionuclide, every emitted a- particle has a definite amount of kinetic energy.

=

.,.-'-~

Find the minimum and maximum kinetic energy that the beta particle -~ e can have. The atomic masses of 23 Ne and

"Na are 22.994511 and 22.989811, respectively.

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Solution Reactants ~Ne

Usually, y- rays are emitted after a- or p- decay, but there are long lived radioactive nuclei that emit only y- rays.

Products

22.9945 - lOm,

~Na 22.9898-11 m, CLASSIFICATION OF NUCLEAR REACTIONS

-~e-me Total 22.9945 - lOm,

Total 22.9898- lOm,

Nuclear reactions can be classified into the following categories.

Mass defect = 22.9945 - 22.9898 = 0.0047 u A.

Q = (0.0047)(931) = 4.4 MeV The P-particle and neutrino share this energy. Hence the energy of the P-particle can range from 0 to 4.4 MeV . GAMMA DECAY

The process of emission of a y- ray photon during the .radioactive disintegration of a nucleus is called gamma decay. As the emitted r- ray photons have zero rest mass and carry no charge, so in a y- decay the mass number and atomic number of the nucleus remain unchanged and no new element is formed. A y- decay can be expressed as

:x

(Excited state)

-----+

;x

(Ground state)

+

y

A nucleus does not contain photons, yet it can emit photons. These photons are created during the emission process. We know that a nucleus can exist in different energy states. After an a or a P- decay, the daughter nucleus is usually left in the excited state. It attains the ground state by single or successive transitions by emitting one or more photons. As the nuclear states have energies of the order of MeV, therefore, the photons emitted by the nuclei have energy of the order of several MeV. The wavelength of such high energy photons is a fraction of an angstrom. The short wavelength electromagnetic waves emitted by nuclei are called y- rays.

An example of y- decay is shown through an energy level diagram shown in Figure. Here an unstable ;co nucleus is transformed via a J3- decay into an excited :Ni nucleus, which in tum reaches the stable ground state by emitting photons of energies 1.17 MeV and 1.33 MeV, in two successive y- decay processes.

ELASTIC SCATTERING

The incident particle gets deflected without any change in its energy, i.e.,

iHe+ 1~Au-------+ 1;:Au +~He The bombarding particle passes sufficiently at large distance away from the target nucleus so as to get repulsion which changes its direction of motion without any change in its energy. B.

INELASTIC SCATTERING

If the bombarding particle passes close to target it gets deflected. Due to strong repulsion, the target particle also acquires some energy. So the energy left with the scattered particle is less than that it had initially. 1 l

1 H + 73 Li-'>.7Li+ 3 1H

~Li means existence of ~Li in one of its excited states. C.

SIMPLE CAPTURE

The incoming particle is captured by the target nucleus. The product nucleus which is generally in the form of excited state decays to the ground state by emitting y-ray of energy hv.

D.

DISINTEGRATION (NUCLEAR TRANSMUTATIONS)

The intermediate compound nucleus breaks up and ·results in a product nucleus and an outgoing particle. The product nucleus has different chemical properties as compared to the target particle. Majority of nuclear reactions belong to this category. Such nuclear disintegrations are called Nuclear Transmutations. a) Disintegration by a-particles (i)

(a,

p)

reactions

:x+~He-)>::~c* -)>~:iY +~H

E,= 1.17 MeV

{C • is Compound Nucleus} E,= 1.33 MeV

EXAMPLES

The historical experiment of Rutherford is an a induced transmutation, an ( a, p) reaction.

~Ni

7N14 + ;ie4-------+ eF1s·-------+ i!O + iH1

Energy-level diagram showing the emission of r-rays by a ~Co nucleus subsequent to beta decay.

and is exoergic in nature (Q > 0). Other useful (a, p) reactions are

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Advanced JEE Physics 1

1

~B + !He-----+ ;N *-------+

:c + ~H

1

~Na+ !He-------+~AI *-----+:Mg+ ~Al+ !He-----+

:H

EXAMPLES 11 8 +1H-----+ ,2C*-----+ 11c+1n

~;p *-------+ ;!Si+ :H

6

:~sc + !He-----+ :~v *------+:Ti+ :H (ii)

I

6

6

0

t~c~ ,~B

{

(a= -1. 763 MeV) (Half-life = 2.5 min)

( u, n) reactions

"N 'H -----+ 12 "M 9 • -----+ 12 "M 9 + 'on 11 a+,. { ~Mg~;:Na

; X+ ~He-------+ 1:;c *-> ~;iY + ~n {C* is Compound Nucleus)

(Q = -4.84 MeV) (Half-life= 12.3 s)

EXAMPLES

1 54 241

54 1 Cr + H---t 55 Mn 25Mn * ----t 2 5+ 0n

~Li+ !He-----+ ~8 •------+ ~8 + 6n 1

1

~Mn~ :er

{

:se + !He-----+ :c •------+ :c + 6n 1

1

(0=-2.16 MeV) (Half-life=310 days)

1

!F + !He-----+ ~Na•------+ ~Na+ ~n

63 ~

~Na+ :He-----+ ~Al*-------+ :Al+ 6n

b)

Cu+ 11H----t 64 Zn *-----+ 63 Zn+ 10n ~ ~

· :zn~ :cu {

Disintegration by protons

(i)

(p, u)

reactions:

a-particles. The

(Q=-4.15 MeV) (Half-life=38.5 min)

When

the

reactions

yield

The

(p, a.) reactions are usually

(iii)

exoergic and have the general form ; X + ~H->

(p, n)

(p, y)

reaction is always endoergic.

reactions:

When

the

reactions

yield

y-photons . The compound nucleus formed by absorption of proton by the target nucleus does not emit any nuclear particle but goes down to the ground state emitting one or more y-photons . The

~:ic *-------+ ~~iY + ~He {C • is Compound Nucleus)

EXAMPLES

~u+:H->!Be*-------+:He+iHe (0=4 MeV)

(p, y) reaction is the radiative capture of proton.

;u+:H------+!Be*->!He+iHe (0=17.35 MeV)

The general equation of (p, y) reaction is

1

:x +~H-----+::ic * -----+t:~c + r

1

~s+:H------+ :c·------+!Be+iHe (Q-8.59 MeV)

{C • is Compound Nucleus}

The product nucleus !Be is highly ustable and } decays almost immediately as !Be-------+!He+iHe, { so that the final reaction gives three a-particles 19 20 F 11H-------+ 10 Ne 9

+

*-------+

4 16 0 2He 6

+

(a --

EXAMPLES

;u + iH-----+ :se •-----+!Be+ r ;N +~H----t 1!0 * ----t 1!0 + y

1

812 MeV)

~1Mg + iH-----+ ~Al* ----t ;';Al+ y (Q = 2.38 MeV)

gAI +iH----t ;!Sr* ----t ~Sr +y

:cu +:H-------+ :zn *-------+:Ni+ !He

(iv)

(Q = 3.76 MeV)

(ii)

(p, n)

reactions:

When

the

reactions

When

the

reactions

yield

: X + !H ----t A-£x + ~ H

neutrons. The general equation of this type of reactions is

This is an example of direct reaction without any formation of the compound nucleus.

:x+!H----t1:iC*----t 2 .'.iY +~n

EXAMPLES

{C * is Compound Nucleus}

~Li+ ;H-t

the product nucleus being isobaric with the target

:u + ~H

:se + ; H ~ :se + ~H

nucleus.

c)

Disintegration by neutrons (i) {n, a) reactions: When

: X (the same as target nucleus):

=

reactions:

deuterons. The general equation of this type of reaction is

yield

Since two isobars differing in Z by unity cann~t both be stable, the product nucleus is p+ active, decaying by p• emission (or electron capture) into

(p, d)

a-particles. The general equation for reactions is

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yield

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:=;y + ~He

followed by :H-4 ~He+ -~e + V0

The radiative caph.µ'e raises the target nucleus to an excited isomeric state and by releasing the excitation energy as y photons, the product nucleus becomes an isotope of the target nucleus. The isotopic product nuclei are generally pactive, as it has a higher neutron-proton ratio compared to the original one. In fact, this method of inducing P-activity is used extensively with copious supply of neutrons from reactors.

1

EXAMPLES

: X + ~n-------+ A+iC *-------+

{C* is Compound Nucleus) The (n, a)· reactions are usually exoergic, i.e. Q is positive, particularly for medium heavy nuclei. EXAMPLES

:u+~n-Jo;u·--t:H+:He (0·=4.785 MeV)

~B+~n-J, 1~B*->;U+!He (0=2.79 MeV)

fiCt+~n-Jo:;c1•------4-~p+:He (0_=0.935 MeV) The first two reactions are utilised in the construction of neutron detectors as they have fairly large cross-sections. The first reaction also gives a method of .producing tritium which is _useful in nuclear fusion.

(ii)

followed by ~H

d)

Disintegration by deutrons (i)

(d, a)

When

the

reactions

The general equation of

yield ( d, a)

reactions is

The general equation for ( n, p) reacti~ns is

1x+ iH-> ~::c *-> t:Y + ~He

:x +~n-------+A+;c *-------+ 2 ~Y +!H

{C * is Compound Nucleus)

{C * is Compound Nucleus)

The Q-values are usually positive and the reactions exoergic. Some example of ( d, a)

The product nucleus Y is an isobar of the target nucleus X with Z-value one unit lower and is thus p- active decaying to the target nucleus. So,

So,

reactions:

a-particles.

(~' p) reactions: When the reactions yield protons.

2

~rr , :He+-~e+V0

12

reactions are given as under. EXAMPLES

~Y ~ ~X

the (n, p) reactions are such that the initial and

(O =22.4 MeV)

final nuclides are identical. The process therefore appears to be a conversion of neutron into a proton and an electron.

(0=13.57 MeV) 1

1

1

:0 +~H--+ :F *------+ jN+;He

EXAMPLES jN+~n------+ 1~N*------+ 1:c+:H (Q = 0.627 MeV)

1

~~Na+ ~H--+:iMg *------+ ~riNe + ;He

(o =6.9 MeV) 21 Al+ 2H------+ 29Si 13 1 ~'"

~Al+~ n------+:!Al*------+~Mg+iH (a= -1.83 MeV) followed by ~Mg

10~

(n, y)

reactions:

When

12

2

(0=6.7 MeV)

1

:iAl+~ e+Ya ~;Mg+ ~H------+:AI *-----+~Na+ !He

Only the first reaction is induced by thermal neutrons.

(iii)

* ___,,.·2sMg + 4He

the

reactions

(0=1.96 MeV)

yield

y-photons . This is the most important neutroninduced transmutation, known as radiative capture of neutrons, and has the following general equation.

Since the a-particles ejected from the compound .nucleus are to cross high potential barrier, the (d, a) reactions occur at fairly high energy of deuteron and for low Z target nuclei.

(ii)

~x +~n------+ A+_ic * ------+A+1Y +r

( d,

p)

reactions: When the reactions yield protons.

The general equation of ( d,

{C * is Compound Nucleus} The product nucleus is thus the same as the compound nucleus in the ground state. The ( n, 'Y) reaction is always exoergic (Q > 0) and can be induced by almost zero energy neutrons.

p)

reactions is

AX+2H->A+2C*->A+1X+1H 1 Z+l Z 1

Z .

{C * is Compound Nucleus) The Q-values are usually positive and the reactions exoergic. For ~ome light nuclei, however, Q may be negative. Som~. examples of ( ii, p) reactions are as under.

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Advanced JEE Physics EXAMPLES

(iv) (d,

;u+:H-------+!Be*->:Li+~H (Q=-0.193 MeV)

:c + fH-------+ jN *-------+ :c +~H

1

1

1

.

Optics & Modern Physics

t)

reactions: When the reactions yield tritium.

The general equation of (d,

(Q = 2.72 MeV)

t) reactions is

~X+iH-),~:;c*-------+A-~y +~H

{C * is Compound Nucleus} (Q =4.74 MeV)

~~P+;H-------+:8*-------+fsf+~H 1

1

The product nucleus Y is an isotope of the target nu~leus X . The cross-section of such reactions is low. Some examples are as under.

(0=5.71 MeV)

1

~Ag + ;H-------+ ~Cd *-------+ ~Ag +~H

~Li +iH-----+ !Be* ---+~Li +!H

(0=4.6 MeV)

( Q = -0.996 MeV)

The products of (d, p) reactions are usually radioactive

(iii) ( d, n)

reactions:

When

the

reactions

neutrons. The general equatioil. of ( d,

is

:x +iH-------+ :::c

n)

yield

(Q = 4.59 Me\T)

reaction

(> 20 MeV) of 2p), (d, 3n) etc.

NOTE : At higher energies *

------+:!~Y + ~n

deuterons

{C * is Compound Nucleus)

With some exceptions, the ( d, n) reactions are exoergic and the Q-values are positive. The product nucleus Y is an isotope of the compound nucleus. Some examp"ies of ( d, n) reactions are given below. EXAMPLES ;u+;H-------+ !Be*-------+ !Be+~n (a= 15.024 MeV) 1

!Be+fH-------+ ~8 *-------+ 1~B+~n 1

1

1

1

1

1

(Q = 4.36 MeV)

:c+;H-------+ ;N *-------+ ~N+~n (Q =--0.283 MeV)

:0 +.:H-------+ :F *-------+ ~F + ~n

(a= -1.625 MeV)

~c1+;H-------+:Ar "-------+\1:Ar+~n (Q = 6.28 MeV)

When deuterons bombard deuterons both (d, p) and ( d, n) reactions may be observed because of two alternative decay schemes for the compound nucleus.

:;Ag + iH-----+ 11';Cd *-----+ 1:ca + 2 ~n

e)

Photo disintegration ( y-induced transmutations) This type of reactions, called photo disintegrations or photonuclear reactions, occur when sufficiently highenergy photons enter into a nucleus. The energy ol the incident photon must be greater than the binding energy of a nuclear particle (separation energy) like neutron, proton, a-particle etc. to produce

(y, n), (y, p), (y, CI) etc. reactions. The photo disintegration of deuteron, discovered by Chadwick and Go!dhaber, deserves special mention, for it was from this reaction that they evaluated neutron mass. i H +y-----+~H *-------+ ~H +~n

The product ~H is tritium; an isotope of hydrogen. Its nucleus Is called triton which is p-active .

!Be+r-----+!Be*-----+!Be+~n This reaction is used for preparing the source for photoneutrons. Other y-induced reactions are:

~c +y-----+ ~c *-----+ !Be+!He

1

(Half-life =12.4 yr)

;B + y-----+ ;n * -------io !Be+ ~H

The other product ~He is a stable isotope of helium. The tritium may be bombarded with deuterons to

produce (d, n) reaction ~H+~H-------+:He*-------+!He+~n

(0=17.6 MeV)

The reactions with beryllium and deuterium serve as sources of neutrons. A thick beryllium target if bombarded with 1 MaV deuterons (accelerated in a cyclotron or Van de Graaff generator) yields about 108 neutrons per sec per µA deuteron-current absorbed in the target.

( y, n reaction)

Another example of ( y, n) reaction is

(a= 4.03 MeV)

~H +~H-------+ :He•-------+ :He+ ~n (a= 3.26 MeV)

(d, 2n)! (d,

1

the

~H·+~H-------+ !He•-------+ ~H +~H

the

reactions in Which more than one particle (two or more) is emitted from the compound nucleus become important, e.g.

( y,

CI reaction)

1

1

( y, I reaction)

1

~B + y-------+ ~B *-----+!Be+ ~H

1

(y, d reaction)

!Be+r-----+:Be*-------+!Be+~n

( y, n reaction)

!Be+ y-----+ !Be*-----+ ~Li+ !H

(y, p reaction)

The last ( y, n) reaction serves as a convenient source of neutrons. Neutral 1t-mesons can be artificially produced by the interaction of y-rays with hydrogen and deuterium.

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1H +y~rc

0

+!H

CONCEPTUAL NOTE(S) When slow neutrons are bombarded on

0

iH +y~1t +~n+!H The first reaction can occur only if the energy of the y-photon is not less than the threshold value equal to the mass-energy of the neutral pion mass. Steinberg could produce rt0 -mesons by bombarding light targets such as hydrogen or beryllium with high energy x-radiation from an electron synchrotron.

92

a)

b}

r+!Be~!B*~!Be+~n c)

{Photon)

Illustration 50 A proton is bombarded on a stationary lithium nucleus. As a result of the collision two a particles ·are produced. If the direction of motion of the u particles with the initial direction of motion makes an angle

¼), find the kinetic energy of the striking proton.

Given binding energies per nucleon of Li7 and He4 are 5.60 and 7.06 MeV respectively. (Assume mass of proton :;::: mass of neutron). Q value of the reaction is given by

Q=(2x4x7.O6-7x5.6) MeV =17.28 MeV Li7

0

a--

a a

d) e)

0

0

where KP and Ka. are the kinetic energies of proton and a

particle respectively. Applying Law of Conservation of Linear Momentum, we get

K, = l6~0 cos 0 = (16K.)(¼)'

=

~=~

2

The fission of U238 takes place by fast neutrons.

0

Applying Law of Conservation of Energy for Collision, we get . . . (1) K, +Q=2K

=

In nuclear fission the sum of masses before reaction is greater than the sum of masses after reaction, the difference in mass being released in the form of fission energy. The phenomenon of nuclear fission was discovered by Otto Hans and F. Strassmann in 1939 and was explained by N. Bohr and J.A. Wheeler on the basis of liquid drop model of nucleus. It may be pointed out that it is not necessary that in each fission of uranium, the two fragments Ba1 ' 1 and Kr 92 are formed but they may be any stable isotopes of middle weight atoms. The most probable division is into two fragments containing about 40% and 60% of the original nucleus with the emission of 2 or 3 neutrons per fission. So, average number of neutrons produced per fission is 2.5. Most of energy released appears in the form of kinetic energy of fission fragments.

CHAIN REACTION

A.

~2m,K, = 2~2m0 K0 cos 0

U235 +0 n1 - i ,56 Ba141 +36 Kr 92 + 3( 0 n1 )+200 MeV

If on the average more than one of the neutrons produced in each fission are capable of causing further fission, the I\umber of fissions taking place at successive stages goes on increasing at a rapid rate, giving rise to self sustained sequence of fission known as chain reaction. The chjtin reaction takes place only if the size of the fissionable material is greater than a certain size called the critical size. There are two types of chain reactions.

Solution

p

U235 , the fission takes

place according to reaction

iH+y~1t0 +iH

cos_, (

92

... (2)

{-:- m. =4m,} ... ~

Solving equations (1) and (3) with Q = 17.28 MeV, we get K, = 17.28 MeV NUCLEAR FISSION

The splitting of heavy nucleus into two or more fragments of comparable masses, with an enormous release of energy is called nuclear fission.

UNCONTROLLED CHAIN REACTION

In this process the number of fissions in a given interval on the average ·goes on increasing and the system will have the explosive tendency. This forms the principle of atom bomb . If a nuclear reaction is uncontrolled then in about 1 µs, energy of order of 2 x 103 J is released. B.

CONTROLLED CHAIN REACTION (AS IN A NUCLEAR REACTOR)

In this process the nwnber of fissions in a given interval is maintained constant by absorbing a desired number of neutrons. This forms the principle of nuclear reactor, consisting of the following parts: a) Fuel: The fuel is U235 or U233 or Pu 23<J b) Moderator: A moderator is a suitable material to slow down neutrons produced in the fission. The best choice as moderators are heavy water (D20) and graphite (C). c) Controller: To maintain the steady rate of fission, the neutron absorbing material known as controller is used. The control rods are made of Cadmium or Boron-steel.

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Advanced JEE Physics d)

Coolant: To remove the considerable amoW1t of heat produced in the fission process, suitable cooling fluids known, as coolants are used. The usual coolants are water, carbon-dioxide, air etc. Reactor shield: The intense neutrons and gamma radiation produced in nuclear reactors are hannful for human body. To protect the workers from such radiations, the reactor core is surronnded by concrete wall, called the reactor shield.

e)

0.693 Also, "- = - - = 0.005 per day 11/2

~

(- dN)=m (0.005)(6x10")rn per day di 210 So, electrical energy produced per day is

E (0.005)(6x10")m x _ x _,. 8 4 10 • 210

J

Since, E=l.2xl07

(given}

CRITICAL MASS

~

If the amount of uranium is too small, then the liberated neutrons have large scope to escape from the surface and the chain reaction may stop before enough energy is released for explosion. Therefore, in order for explosion to occur, the mass uranium has to be greater than some minimum value, called the critical mass.

Activity at the end of 693 days is R

Illustration 51

· Polonium

{~Po)

emits

!He

particles and is

converted into lead {~Pb). This reaction is used for producing electric power in a space mission. Po 210 has half life of 138.6 days. Assuming an efficiency of 10% for the thermoelectric machine, how much 210 Po is required to produce 1.2 x 107 J of electric energy per day at the end of 693 days. Also find the initial activity of the material. Given: Masses of nuclei "

0

Po= 209.98264 amu, ""Pb= 205.97440 amu,

Avogadro number

=6 x 1023 moi-1

~

R, =R(2) =32x-=4.57x10 21 per day

Electrical energy produced = 8.4 x 10-

=

4.44

1021 7

Illustration 52

In a nuclear reactor, fission is produced in 1 g for U"' (235.0439 u) in 24 hours by a slow neutron (1.0087 u). Assume that 35 Kr" (91.8973 u) and 56 Ba141 (140.9139 amu) are produced in all reactions and no energy is lost. (a) Write the complete reaction (b) Calculate the total energy produced in kilowatt hour. Given 1 u=931 MeV. Solution

The nuclear fission reaction is 92 U235 + on1 ~ 56Ba141 + 36Kr92 + 3 on1 Mass defect ,:'.rn = [ ( m, + rn,) -( m,, + rnK, + 3m, )]

mn = 256.0526 - 235.8373 = 0.2153 u

~

14

J

6 02 10 · x " 235

2.56 x 1021

Q = (5.12 X 10 23 ) X (1.6 X 10-13 ) = 8.2 X 1010

J

Q = 8•2 X lQto kWh= 2.28 X 10 4 kWh

3.6x106

J

Po is required to produce the desired energy,

N=.!:!....x6x10" 210

5

Q = 200 x 2.56 x 1021 = 5.12 x 10" MeV

Energy liberated per reaction = (£\rn)931 MeV = 8.4x 10-13

then

d ~(1)" per ay=." 2

Energy released in fission of 1 g of U 215 is

£\rn = 0.00564 amu

Let m g of

21

693 n=--=5 138.6

Number of atoms in 1 g

and

Since, 2~Po-------+::Pb+iHe

210

710

~

Solution ~

0.005x6xl0 xl0 210

Energy released, a = 0.2153 x 931 = 200 MeV

~ He = 4.00260 amu .

lamu=931MeV

23

=

where, n is the number of half lives

REPRODUCTION FACTOR

It is the ratio of the rate of neutron production and the rate at which the neutrons disappear. · Whether a mass of active material will sustain a chain reaction or not is determined by the reproduction factor (K). If K ;, 1 , the chain reaction will be sustained. If K -1 , the mass is said to be critical.

m=l0g

NUCLEAR FUSION

The phenomenon of combination of two or more light nuclei to form a heavy nucleus with release of enormous amount of energy is called the nuclear fusion. The sum of masses before fusion must be greater than the sum of masses after fusion,

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Nuclear Physics the difference in mass appearing as fusion energy. The fusion of two deuterium nuclei into helium is expressed as 1

2

He3 nuclei.

Net result is

H 2 + 1H 2 ----.+ 2 He" +23-8 MeV

,H' + 1H1 +,H1 + 1H1 - > 2He4 +2 1~ 0 +

It may be pointed out that this fusion reaction does not actually occur. Due to huge quantity of energy release, the helium nucleus 2 He4 has got such a large value of excitation energy that it breaks up by the emission of a proton or a neutron as soon as it is formed, giving rise to the following reactions.

2v+2y+Energy (24-6 MeV) (ii)

Carbon-Nitrogen cycle For the main sequence stars with extremely high temperatures, Bethe suggested an alternative to proton-proton cycle called the Carbon-Nitrogen cycle. The cycle is

H + 6 C12 ~ 1 N13 • + r + 1.95 MeV

1

H 2 + 1H 2 - > 2 He3 + 0 n1 +Q(=3-26 MeV)

1

1

H 2 + 1H 2 - > 1H 3 + 1H 1 +Q(=4·04 MeV)

,N13" - ) . 6 C13 + ,~0 + v(neutrino) + 2.22 MeV 13

The fusion process occurs at extremely high temperature and high pressure just as it takes place at sun where temperature is 107 K. So, fusion reactions are also called Thermo-nuclear reactions. Nuclear fusion has the possibility of being a much better source of energy than fission due to the following reasons. a) In fusion there is no radiation hazard as no radioactive material is used. The fuel needed for fission (U-235 etc.) is not available easily whereas hydrogen needed for fusion can be obtained in huge quantity. c) The energy released per nucleon is much more in fusion than in fission. However, the very high temperature and pressure required for fusion cannot be easily created and maintained and as such it has not been possible as yet to use fusion for power generation.

8

C +,H1-> 7 N14 +r+ 7.54 MeV

14 7N 80

+ ,H1-> 8 0

150

15

- > 7N

15

,N +

1 1H - } ,

150

+tP'°'

+ y + 7.35 MeV +v

+2.7 MeV

12

sC + 2He4 + 4.96 MeV

Net result is

,H +,H1 + 1H1 +,H1-> 2He4 +2,f3° + 2v + 3y Energy (26 · 7 MeV)

b)

For sun, both the cycles occur with equal probability. Stars with masses between 0.4 to 2.5 solar mass produce energy mainly by C-N cycle rather than P-P cycle. Stars with masses 0.4 solar mass or lower (which ·constitute the bulk of stellar population in our galaxy) mainly derive their e~ergy frorJ'! P-P cycle.

Illustration 53

It is proposed to use the nuclear fusion reaction: tH1+1H2 =2He"

CONCEPTUAL NOTE(S)

a)

b)

c)

For the fusion to take place, the component nuclei must be brought to within a distance of 10·14 m . For this they must be imparted high energies to overcome the repulsive force between nuclei. This is possible when temperature is enormously high. The principle of hydrogen bomb is also based on nuclear fusion. To start a fusion bomb very high temperature is required. This is achieved by incorportating an atom bomb within the nuclear bomb. The source of ene_rgy of sun and other stars is nuclear fusion (or thermo-nuclear reactions). There are two possible cycles: (i) Proton-Proton cycle In 1938, Hans Bethe suggested that the stellar energy is produced by thermonuclear reactions in which protons are combined and transformed into helium nuclei. This is known as proton-proton cycle and is applicable for relatively low stellar temperature. The cycle is

H1 + 1H ~ 1H2 + ,~ + v (neutrino)+ 0.42 MeV 1

1

0

... (1) 2

3

,H +,H1->2He +r+S.5 MeV 3

3

4

1

1

2 He + 2 He - > 2 He +,H + 1H + 12.8 MeV

... (2) ... (3)

in a nuclear reactor of 200 MW rating. If the energy from above reaction is used with a 25% efficiency in the reactor, how many grams of deuterium will be needed per day. (The masses of 1 H 2 and ,He' are 2.0141 and 4.0026 u respectively). Solution

Energy released in the nuclear fusion is Q=tJ.mc' =ti.m(931) MeV

=>

Q =(2x2.0141-4.0026),r931 MeV = 23.834 MeV

=>

Q=23.834xlO'eV

Since efficiency of reactor is 25 % So effective energy used

=~ x 23.834 x 106 x 1.6 x 10-19 J 100

=9.534x10-13

J

Since the two deuterium nucleus are involved in a fusion reaction, therefore, energy released per deuterium is 9.534 X 10-13 2

For 200 MWpowerperday number of deuterium nuclei required

The reactions (1) aQd (2)_ occur _twice to yi~ld two

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Advanced JEE Plzysics 200xl0 6 x86400 9.534 xlO-n

c)

3.624xl0 25

d)

2

Since 2g of deuterium constitute 6 x 10

23

nuclei, therefore e)

amount of deuterium required is

2 X 3.624 X 1025 6xl023

120.83 g/ day f)

Radioactive isotopes are used to locate the position and extent of cancer. Radioisotopes are used in locating tumours within the brain. Radioactive Cr 51 is used to locate the exact position where the haemorrhage might have taken place inside the body. Water contents of the body are measured by using deuterium and tritium as tracers. Radio gold is being used for the treatment of leukaemia.

Illustration 54

g)

In the fusion reaction ~ H + ~H-)- ~He+ ~11, the masses of deuteron, helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released 1 amu = 931.5 Mevc-'.

B.

IN INDUSTRY

a)

Radioactive Carbon-14 is used to study wear and tear of the position of an engine. C-14 is mixed with the ring. After some time, the engine oil is analysed to detect the presence of any radiation in it. In case of wear and tear the radiations are found to be there. Radio Cobalt is used for testing fields and castings by taking their photographs with y-rays.

Solution i'.n! =2(2.015)-(3.017 +1.009)=0.004 amu

b)

So, energy released is 6E = (0.004x931.5) MeV =3.726 MeV 3.726 Energy released per deuteron = - - = 1.863 MeV

C.

2

. k N umb er ofd euterons m 1 g

26

6.02xl0 ===-=2

3.0lxlO"

So, energy released per kg of deuterium fusion is

E = (3.0lxlO" x 1.863) = 5.6x 1026 MeV

NUCLEAR HOLOCAUST

The estimate of after effects of the atomic (or nuclear) explosion is termed as nuclear holocaust. If a fusion bomb explodes, then a nuclear holocaust will not only destroy every form of life on earth but will also make this planet unfit for life for all times. The radioactive waste will hang like a cloud in earth's atmosphere and will absorb snn' s radiations, thus causing a long nuclear winter. One can imagine this only by the mathematical figures quoted, according to which energy liberated by fission of 50 kg of U235 is equal to 4xl015 J which is the energy available from 20,000 tons of Trinitrotoluene ('TNT). USE OF RADIOISOTOPES A.

IN MEDICINE

Radioisotopes are extensively used in medicine: a) Radio iodine is used to determine the condition of human thyroid gland. lodine-131 is administered orally to the patient. After a sufficient time, the activity is measured. From the observations it can be interpreted whether the gland is over-active, normal or underactive. b) Amounts of sodium and potassium in the body is measured by using Na-24 and K-42 as tracers . .._:_.. 4.46

·

AS TRACERS

The radioactive isotope has identical chemical properties as another stable isotope. Therefore, by mixing it With stable isotope we can trace the presence or distribution of the element in a biological or physical system by detecting the radiation emitted by radio-isotope of that element. The radioisotope in such a case is said to be a Tracer. Thus, a radioactive tracer is a radioisotope which, when mixed with a chemically similar element or artificially attached to a biological system, can be traced by radiation detecting devices. Following are the few examples of radioisotopes acting as tracers: a) Phosphorous-32 mixed with phosphorous manure has been used to study the process of extracting food from soil by various plants. b) Radio carbon is being used for research in photosynthesis in plants. c) The progress and absorption of sodium chloride in the body can be studied by feeding the person with radioisotope sodium-24 alongwith sodium chloride. d) With the aid of radioisotopes, the rate, place and sequence of formation of the organic constituents of a living body can be studied.

{Solutions on page 4.104)

. 1. I

Consider two decay reactions .

:_u----+ 2~~Pb + 10 protons+ 22 neutrons

(a)

2

(b)

2

I I I

'

::U----,, 2:Pb + 8 !He+ 6 electrons

Are both the r~actions posslble?.

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r-1

II

I 2.

Given : Average binding ;n~y of 2:U = 7.57 MeV-, that

ofl

2:Pb = 7.83 MeV and that of !He= 7 MeV per nucleon.

i

Find the minimum kinetic energy of an a -particle to cause

1

the reaction 4

1

He, H and

14

N + 4He-+ 170 + 1H. The masses of

17

14

'3.

0 are respectively 14.00307u, 4.00260u,

In a neutron induced fission of

235 92 U

energy of 185 MeV is released. If a

which is equivalent to 1 kg of U235

I ,o.

I

235 reactor is 82 U

continuously operating it at a power level of 100 MW , find the time it takes for 1 kg of uranium to be consumed in this / I

I: 4.

2

and atomic mass of 2 He4

~Bi decay as per following equation.

a-!

11.

=4.00388 amu .

8 protons and 8 neutrons are separately at rest. How much Given:

~Bi-----+ a:,Tl + !He

The kinetic energy of a particle emitted is 6.802 MeV. Calculate the kinetic energy of Tl recoil atoms.

1

:0 nucleus?

'

Mass of neutron =1.008665u Mass of hydrogen atom =1.007825u

,

energy will be released if we form 2

6.

a+ 7 N14 ------+ 8 017 + p when

of 7 N14 = 14.00752 amu, Atomic mass of 8 0 11 = 17.00453

Find the increase in mass of water when 1 kg of water absorbs 4.2 x 10 J of energy to produce a temperature riseof1K.

In a nuclear reaction

Given that atomic mass of 1H1 =1.00814 amu, atomic mass

j

3

5.

•

particles of KE 7.7 MeV were bombarded on nitrogen atom , ; protons were ejected with a kinetic energy of 5.5 MeV. (a) Find the Q -value of the reaction (b) Find the angle ~ between the direction of motion of proton and a -particle.

!

nucleus, usable

reactor.

the mass of coal with calorific value of 30 kJgm-1

l

1.00783u and 16.99913u .

!

(b)

I'

N,

l

200

/9.~ssuming. the splitting otU215 - ;~cle~~b·;~t;; MeV j energy, find i (a) the energy liberated in the fission of 1 kg of U"' and

Mass of

1

:0 atom =15.994915u

1

------ ____ __,I

Neon-23 decays in the following way ~Ne----+ ~Na+ -~e + V , Find the minimum and maximum kine1ic energy possessed by the beta particle ( _~e) . The atomic mas~es of 23 Ne and "Na are 22.9945u and 22.9898u, respectively.

7.

The 11

radionuclide

11

c

decays

according

to

11

C------+ 8 + e... + v. The maximum energy of the emitted \

·positrons is 0.961 MeV . Given that atomic mass of mass of

11

11

I C=mc:=11.011434u, atomic I

8=m~=11.009305u, and the mass of positron

=mP = 0;0005486u, calculate disintegration energy

Q and

compare it with the maximum energy of the emitted positron given above. (1u-931 MeV). 8.

It is proposed to use the nuclear fusion reaction H2 + fi 2 ------+ J-ie4 in a nuclear reactor of 200 MW rating.

1

If the energy from the above reaction is used with a 25% efficiency in the reactor, how many gram of deuterium fuel will be needed per day. Given that the masses of 1H2 and 2

He4 are 2.0141 atomic mass units and 4.0026 atomic

mass unit respectively.

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Problem 1 A gas of hydrogen like ions is prepared in such a way that the ions are only in the ground state aild .the first excited state. A monochromatic light of wavelength · 1216 A is absorbed by the ·ions. The ions are lifted t~ higher excited states and emit radiations of six wavelengths, some higher, some •ower or some greater than the incident wavelength. (a) Find the principle quantum number of the final excited state. (b) Identify the nuclear charge on the ions. (c) Calculate the value of the maximum and minimum wavelengths. Solution

(a)

Energy corresponding to ;\. = 1216 A is, E = ~ eV=l0.177 eV 1216 Now, total six wavelengths are obtained in the erniSsion spectrum, hence from n(n-l) . 2

6

We have n = 4, i.e., after exCitation the single electron junips t? 3rd excited State or n = 4 . (b). Now it may jump either from n = l or n =.2 .

-i~-----,m----n=4 --+------"'"4-l-~--n=3

- + - - - - - -~'4--''Lh--n=2

- - - - - - ___y__n_n=1 (a)

(b)

(c)

If it jumps frOm n =1, then in emission spectrum all the six photons haVe energy equal to or less than the energy of absorbed photon or the wavelength of emitted photon is either equal to O! greater than the wavelength of absorbed photon While in the question it is given

=

that the emitted wavelengths are either less than or greater than or smaller than the wavelength of absorbed photon. Which is possible only in the second case, i.e., when electron jumps from n =2 fo n = 4 . Hence, E4 -E, =i0.177 eV

=>

(c)

-13.6z'- -(-13:6z')=l0.l77 42 22 .

Solving !his, we get z = 2 Maximum wavelength corresponds to rmmmum ertergy, i.e., a trarisition from n = 4 to .n =3 . Thus, t.Emln =E,-E, ( .:13_6)(2) 2 ·[-(13.6)(2)'] (4)' . (3)'

=>

t.Emin

=>

t.Emln = 2.64 eV

=>

,.

12375 =4687 A =~= max 8Euun 2.64

Minimum wavelength Corresponds to maximum ener_gy: i.e., a transition from n 4 to n =1 . Hence, ~Emax = E4 - Ei

=

2

=>

(-13.6)(2) _ [(-13.6)(2)'] 2 (4) , Ct)'

=>

t.E...;. =51 eV

=>

,. . = 12375 = 242 A = 51

Problem 2 A gas of hydrogen like·atoms can absorb radiations of 68 eV. Consequently, the atoms emit radiations of only three different wavelengths. AU the wavelengths are equal or smaller than that of the absorbed photon. (a) Find the initial state of the gas atoms. (b) Identify the gas atoms. (c) Calculate the minimum wavelength of the emitted radiations.

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(d) Find the ionization energy and the respective wavelength for the gas atoms.

(a)

Solution (a)

s·mce,

Solution

N

n(n-l) 2

3

1 Since, E" =- ~ eV, so we have n -18 E =-=-18eV l

=>

n=3 i.e., after excitation atom jumps to second excited state. Hence n1 =3.So ni canbelor2

(1)'

-18 E =-=-4.5eV 2 (2)' -18 E,=--2 =-2eVand (3)

-18

E, =--, =-1.125 eV (4)

The energy level diagram is shown in figure. - - - - - - - - E4 =-1.125 eV - - - - - - - - E , =-2 av

n1 = 1

If ni =1 -then energy emitted is either equal to, greater than or less than the energy absorbed. Hence the emitted wavelength is either equal to, less than or greater than the absorbed·wavelength. Hence n1 ,;t:. l. If n;

= 2, then

- - - - - - - - E2 =-4.5eV

E,, ~ Ea and hence Ae :5 Ab

- - - - - - - - E , =-18 eV

(b) The excitation potential of stage n = 2 is E,-E, =18-4.5=13.5 V

So, n1 =2

(b) Since, E3 -E, = 68 eV

(c)

=>

(13.6)(z'i(¼-¼)=68

=>

Z=6

Energy of the electron accelerated by a potential difference of 16.2 V is 16.2 eV. Since we observe that

E, -E1 =-1.125-(-18) = 16.875 eV > 16.2 eV 12375

(c)

and E3 -E1 =-2-(-18)=16 eV <16.2 eV

28.43 A

With this energy the electron will be able to excite the atom from n = 1 to n = 3 , so we have the possible wavelengths corresponding to the· transitions from 3-->1, 3--;2 and 2-->1. Hence

(13.6)(6)'(1-½) (d) Ionization energy is 2

IE= (13.6)(6) = 489.6 eV

,_ _ 12375 32 -E,-E,

A= 12375 =25.3 A 489.6

,_ = 12375 31

Problem 3 The energy levels of a hypothetical one electron atom 1 are given by En =- ~ eV, where n =1, 2, 3, ..... n (a) Calculate the four lowest energy levels and construct the energy level diagram. (b) Find the excitation potential of the stage 11 = 2. Find the wavelengths (A) which can be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 V . (d) Assuming these atoms to be in the ground state, will they absorb radiation having a wavelength of 2000A? (e) Also calculate the photoelectric threshold wavelength of this atom. (c)

E3-E1

12375 -2-(-4.5)

4950

A

= 12375 = 773 A 16 12375 -4.5-(-18)

917 A

(d) The energy corresponding to 1. = 2000 A is given by 12375 E= =6.1875 eV 2000 whereas the minimum excitation energy is 13.5 eV (n=l to n=2). Hence this is not possible. (e) Threshold wavelength for photoemission to take place from such an atom is, 12375 1. mm. = 18 = 687.5 A

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Adva11ced JEE Physics

= · dE=76xl0

Problem 4

Two hydrogen like atoms A and B are of different masses and each atom contains equal number of protons and neutrons. The difference in the energies between the first Balmer lines emitted by A and B is 5.667 eV. When

eV The wavelength of K. line (transition of electron from L shell to K shell) is, i. (in A}= 12375 '· ,iE(in eV)

the atom A and B, moving with the same velocity, strike

a heavy target they rebound back with the same velocity. In the process, atoms B imparts twice inomentum to· the target than that A imparts. Identify the atoms A and B . Solution

.,.w

~-~=3

i. K,

wavelength (in A) for the spectral lines for transitions originating at n = p and terminating at 11 = 1 are given by , = 1500p' ,. , p -1

Applying Law ·of Conservation of Linear Momentum on atom A and heavy target, we get

{a)

mAu=Mv~ -mAu

=>

2m8 u=Mv2 •

m8

(c)

=2Mv1

=2mA

... (2)

.

2Za

.,

1500p'

(a) Smce, l\,=-,-p -1

neutro~, so we ha'Y·e mA =2;lA =ZA

Calculate the ionization potential of this element.

Solution

Since, both A and B contain equal number of protons and

'"s

Find the wavelength of the least energetic and the

(b) Construct an energy level diagram for this el_ement showing _the ene_rgies of the lowest three levels.

Similarly for atom B and heavy target, we get

=>

where p=2,3,4 .....

most energetic photons in this series.

2mAu=Mv1

Given Mv2

= 12375 = o.163 A 76xlQ3

. Problem 6 For ~ certain hypothetical one-electron atoni, the·

5:667=13.6(2!-Z!)(;, -;, )

=

3

=

... (3)

1. = 1500[----½-) 1--

p'

Zs

So, "-max corre:sponds to least energetic photon with p=2 . , ..

From these equations we get

z, =land z, =2 i.e., A is 1 H' and Bis ,He' (both having single electron).

i.= =1500[~)=2000 A

1-4

Problem 5

Stopping potential of 24 kV, 10 kV, 110 kV and 115 kV are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic Xray. H this element is used as a target in an X-ray tube, what will be the wavelength of K. line? Solution

Stopping potentials are 24 kV, 10 kV, 110 kV and 115kV If the electrons are emitted from conduction band then the maximum kinetic energy of photoelectrons would be 115xl03 eV.

li they are emitted from next

(b)

Amin

-corresponds to most energetic photon with p -+ oo

=

i.""" = 1500 A

i.-, =1500

A

=

12375 E.c:E1 = ·eV=8.25eV 1500

=

E1.=-8.25eV - - - - - - - E , . = -0.95 av - - - - - - - E2 =-2.05 eV

mner .shell maximum kinetic

energy of photoelectrons would be 110 x 103 eV and so on. For photoelectrons of L shell it would be 100xl03 eV and for K shell it is 24x103 eV. Therefore, difference between energy of L shell and K shell is, M= E, -E, =(100-24)x103 eV

- - - ~ ~ - ~ E 1 = -8.25 eV

Further, A. 2 ....1 = 2000 A 12375 _eV=6.2 eV 2000 E, =-2.05 eV

E,-E, =

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Similarly ). 31 =1?.°o[

1

~¼)

~

= 1687.5 A

E=6.63xl0 6 MeV

Problem 8

12375 E,,-E, = eV = 7.3 eV 16875 E, =-0.95 eV (c)

It is proposed to use nuclear reaction 5-1Po 210 -----; 82 Pb 206 + 2 He 4 to produce 2 kW electric power in a generator. The half life of polonium (Po'") is 138.6 days . Assuming efficiency of the generator be 10%, calculate

Ionization energy is E,~. =8.25 eV

how many grams of (Po 210 ) are required per day at the end of 1386 days. (b) initial activity of the material Mass of nuclei : Po'" = 209.98264 amu , (a)

So, Ionisation Potential equals 8.25 V Problem 7 A radionuclide with half life 1620 sec is produced in a .reactor at a constant rate 1000 nuclei per second. During each decay energy 200 MeV is released. If production of radio nuclides started at t = 0 , calculate

rate of release of energy at t = 3240 sec. (b) total energy released upto t = 405 sec. (a)

Pb"" = 205.97440 amu , ,He'= 4.00260 amu (1 amu = 931 MeV)

Solution (a) Llm= 0.00564 amu 25.25 MeV =8.4x10-13 J

0 693 Since, ). = · = 0.005 per day

Solution (a) Let N be the number of nuclei at time I, then net rate

Iv,

Let m g of Po210 are r~quired per day for the reactor,

of increase of nuclei at instant t is,

then

dN di

-=o.-).N

n

{where o. = rate of production of nuclei}

~

0

,.

23

0.005x6.02xl0 xm per day 210 So, energy produced per day is dN) 'N - =A ( -dt

1_.:!!'1._= Idi a-AN 0

(6.02xl023 )m 210

N=~(1-,-")

... (1)

.

E

Rate of decay at this instant ~

R=1'.N=a(1-,-") Hence, rate of release of energy at this time

released in each decay)

~ 1

=cx(l-,-")(200) MeVsecSubstituting the values, we get

Rate of release of energy = 1000 ( 1- e1: 20

(b)

(200)

E=(l2xl06 )mJ

Now, 10% of (12xl06 )m equals 2xl03 x24x3600J

= R (energy

0.693 ~3240)

23

0.005x6.02xl0 xm xS.4 x 10_13 J 210

m

2xl03 x24x3600 l.2xl0 6

R =AN= (o.oos)( ~

144

g

144 )(6.02x10") 210

R = 2.064 x 1021 per day

Rate of Release)= l.5 x 10, MeVsec_1 ( ofEnergy · (b)

Total number of nuclei decayed upto time I is ext- N

Now, R=R,Gr

/.,. n = 1386 = 10 ) \ 138.6

10

R, = (2) R = 2.11 x 10 24 per day

Total number of)= o.t~~(l- ,-") ( decayed nuclei 1'. Hence, total energy released upto this instant is

E=[ o.t-f(1-,-")]<200) MeV

Problem 9 Suppose a nucleus initially at rest undergoes a. decay ~c_cording to equation

~X----+Y+a

Substituting the values, we get 1620 E = 1000 x 405- 1000 x 200 MeV _ (1- e '-"'·"')] 0 693 [ 1620

At t = 0 , the emitted a. particle enters in a region of space where a uniform magnetic field ii = B01 and electric field

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Optics & Modem Physics

E= E0 i exist. The a particle enters· in the region with velocity V = v0 J from x = 0 . At time t = ../3 x 107

,n!J,

qaEo

(c)

sec,

When an a -particle is emitted with velocity v, from a stationary nucleus X, decay product (nucleus Y) recoils. Then by Law of Conservation of Linear Momentum, we have

the particle was observed to have speed twice the initial speed v0 , then find

myvy =mavo

the velocity of a particle at time t . (b) the initial velocity v 0 ofthe a particle (a) (c)

v = m.v, =( 4.003 )(lO') ' m, 221.03

v, =1.81xl0 5 ms-1

the binding energy per nucleon of a particle.

Given that m(Y) = 221.03r,, m(a) = 4.003u, m(n) = 1.009u,

m(p)=l.00Su mass on a particle m. =~x10-" kg charge

Total energy released during a -decay of nucleus X is E = K.E. of nucleus Y + K.E. of a -particle

on a particle q =3.2x10-19 C and 11t=931 Mevc-2

=>- E=2,myvy+zm(lv0

3

1

0

Solution

(a)

E

Magnetic force on a particle, (at t = 0)

2

1

2

1 ·66 xl0-" [(221.03)(1.81x10 5 ) 2 + 2xl.6x10-"

F. = q(vx ii)= q.[( v,j}x(B 1)]

(4.003)(10')

0

=>

fm =-qavoBok

~

Force due to electric field (at any time I) 0

]

E=2.11MeV

Hence, Mass lost during a -decay is

F, =qE =q E,i

2

!~~ u = 0.0023u

Mass of nucleus X is

Hence, the particle will move in a c!rcular path in y-z plane due to magnetic field and at the same time it will µ,.ove along x -direction. The resultant path is therefore,

m, = ( m, +m. +0.0023)u ~

a helix with increasing pitch.

m, = 225.0353u

Mass defect in nucleus

y

X, ti.m=92m, +(225-92)m. -m, =1.898u

v,

So, binding energy per nucleon is

BE

A

1.898x931.5 MeV = 7_86 MeV 225

{;)-------+-+e.-Z

Hence., velocity of particle at any time t can be written

as, ii=( q~~o

. respectively. At t = 0, probabilities of getting a and Pparticles from the radionuclide are equal. Calculate their respective probabilities at t = 1620 s . If at t = 0 , total number of nuclei in the radio nuclide are N O • Calculate the time t when total number of nuclei remained

1)1 +v cos8]-v,sin8k 0

where 0 =rot= Baq,:i. t

m. (b)

Speed of particle at any time t is V

( m.Et)'

= ~ + v!

{·.· sin2 0+cos2 9=1}

. Given v=2v0 at t=(J3X107 ) 2

(2v,) ={"3x107 )

2

+v;

'"u.

qaEo

Problem 10 A radionuclide consists of two isotopes. One of the isotopes decays by a. -em1ss1on and other by f3-emission with haH lives T1 = 405 s , T2 = 1620 s,

N undecayed becomes equal to - 0 • 2 Given, log10 2 = 0.3010, log 10 5.94 = 0.7742

and

x'+4x-2.5=0, x=0.594. Solution

,so,weget

At t = 0, probabilities of getting a and p particles are same. This implies that initial activity of both is equal, say

R,. Activity after t = 1620 s is

=

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R=R,(2)"' =16. R, 2

~

1

and R2

(n+21trl)=RZ'(n(n+2;) (n + l)

Total activity R = R, + R, = _2._ R, 16

=>

=2 and

R

I

=>

Since, r ocZ

9

=>

p particles is R, =_ll_

R, =R Not= N.,_ T, T, Not =2 N., 4 1

... (3)

n'

Probability of getting a particles is R, R

Since,

... (2)

Equating equations (1) and (2), we get

=R,(½):: =~

Probability of getting

1 (n+l) -=-).. 21tr

2 (n + 1) r=---r,

11

Substituting in equation (3), we get

9

11

02

R(ll)'(n)(n+2) (n+ 1)

=>

(n + l) = (1.09x107 )(11)(21t) x (0.529 x10-10 )(n' + 2n)

Solving this equation, we get n=24

Let N 0 be the total number of nuclei at t = 0 then, Problem 12

_N, d N _ 4N0 Nm - an "' - - -

5

A radioactive element decays by )}--emission. A detector records n beta particles in 2 seconds and in next

5

Given, that N 1 + N 2 = No 2

=>

Let

'

2-seconds it records ~n beta particles. Find mean life

'

(2)"' + 4N,5 (2)'"' = N, 2 '2

N, 5 2

4

... (1)

(½)';'° =x, then the equation (1) becomes

correct to nearest whole number. Given log, 121 = 0.6931, log, 131 =1.0986 . Solution

Let n0 be the number of radioactive nuclei at time

4

x +4x-2.5=0

I= 0 . Number of nuclei decayed in time I are given by

=>

X=0.594

n0 (l - ,-><), which is also equal to the number of beta

=>

(½)1620' = 0.594

particles emitted during the same interval of time. For the given conditio_n.

Solving, we get t=l215 s Problem 11 If the wavelength of the nth line of Lyman series is equal to the de-Broglie wavelength of electron in initial orbit of a hydrogen like element (Z =11). Find the value of n.

Solution n line of Lyman series means transition from (n + it state to first state. th

1-) 2-RZ'(1-).. (n+l)' h

hr

mv

mvr

(21t)(hr) (n + l)h

... (1)

(~+¾n )=n,(1-e-")

... (2)

Dividing equation (2) by (1), we get 1-e4 }. 1.75 =------::,,:1- e

=>

l.75-l.75e-" =l-e-4>

=>

1.75e-n-e--4,. =~

... (3)

4

Let us take e-21.

=X

Then the above equation becomes

x' - l.75x + 0.75 = 0

... (1)

2

de-Broglie wavelength in (n + 1t orbit is

,_

n = n0 h- e-")

21tr

= (n +1)

1.75± ~(1.75) -(4)(0.75)

=>

X

=>

x=l and -

2 3 4

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Adva11ced JEE Physics e-21• =1 or e-n =~ 4

But e-2l. =1 is not accepted because which means A= 0 .

the mass of the daughter nucleus is 223.610 amu and that of the a-particle is 4.002 amu. Determine the total kinetic energy in the final state. Hence obtain the mass of the parent nucleus in amu. (1 amu = 931.470 MeVe-')

Hence, e-2;.. = ~ 4

=>

-21dog.(e) = log, (3)-log, (4)·= log,(3)-21og, (2)

=>

1 A=log,(2)- 1og,(3)

Problem 14 A nucleus at rest undergoes a decay emitting an o.-particle of de-Broglie wavelength, ,. = 5.76 x 10-15 m . If

Solution

2

Given mass of a-particle, m = 4.002 amu and mass of

Substituting the given values, we get

dau"ghter nucleus,

,. = 0.6931-½x(l.0986) = 0.14395 s-1

M = 223.610 amu de-Broglie wavelength of a-particle,

So, mean life

A.= 5.76 X 10-lS

1

!w =-i;:=6.947

S

So, momentum of a-particle is

h 6.63x10°"' k _1 p=-i;:= 5.76x10-1' gms

Problem 13 Nuclei of a radioactive element A are being produced at a constant rate a. . The element has a decay constant A,. At time t = 0, there are N 0 nuclei of the element.

(a) Calculate the number N of nuclei of A at time t . (b) if a= 2N0A., calculate the number of nuclei of A after one half-life of A and also the limiting value of N as' t --4- 00,

=>

p = 1.151 X 10-l9 kgms-l

By Law of Conservation of Linear Momentum, this should also be equal to the linear momentum of the daughter nucleus (in opposite direction). Let K1 and K2 be the kinetic energies of a-particle and daughter nucleus. Then total kinetic energy in the final state

is

Solution

(a)

Let at time t, number of radioactive nuclei are N Net rate of formation of nuclei of A

dt ~=dt a.-,N

=>

K =i._(~+..!_)

=>

K=p'(M+m)· 2 Mm

2

N =.!(a-(a- ..N 0 )e-") A.

K

=10-12 J

K

... (1)

a= z,.N, and t --1/,t - log.(Z) in A.

Substituting

10-u · 1.6 xl0-13

m:;

3 N=-N, 2

a.

=>

J+(d::;~~!rJ+[=::tJ ( ;::::J=( nucleus a-particle nucleus (.6.m)

a= 2W0 and t-> oo in equation (1),

N=-=2N, A. N =2N0

6.25MeV

K=6.25MeV , 6.25 Mass d e1ect, 1'.m = - - - = 0.0067 amu 931.470

equation (1), we get,

(ii) Substituting we get

m M

Substituting the values, we get

o

Solving this equation, we get

(i)

p'

2M

Since, 1 amu = 1.67 x 10-21 kg

f ~=fdt

Noa-AN

p'

2m

K=K1 +K2 = - + -

dN =a.-A.N

(b)

ID

=>

m,.,..,, =(4.002+223.610+0.0067) amu

~

mparent

= 227.62 amu

Hence, mass of parent nucleus is 227.62 amu.

= ==================================== 4.54

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Atomic & Nuclear Physics Problem 15 A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, orginating from all possible transistions between a group of levels. These levels have energies between -0.85 eV and -0.544 eV (including both these values). (a) Find the atomic number of the atom. (b). Calculate the smallest wavelength emitted in· these

1,. =

0.693 = 0.693 h-1 t,12 15

=> A.=0.0462h-1 Let R,, be the initial activity then

R,, = 1 microcurie = 3_.7 x 10' dps Let r be the activity in 1 cm3 of blood at I= 5 hr, then

transistions. (Take /Jc=1240 eV-mn, ground state energy of hydrogen atom =-13.6 eV)

=>

Solution (a) Total 6 lines are emitted. Therefore,

R be the activity of whole blood at time I= 5 hr Total volume of blood should be

n(n-_1)

2

=>

. ' per second r = 296 d'1smtegration

60

r = 4.93 disintegration per second, and

V=!

6

r

V

n=4

= R,,e-" r .

S0 transition is taldng place between ,n th energy state 1

Substituting the values, we get

and (m + 3 )"' energy state.

4

V =(3.7 x10 )e-
E. =-0.85 eV

cm3

4.93

=>

=> ~=0.25 m Similarly, E••, =-0.544 eV =>

=> =>

-13.6(:: )=-0.85

-13.6

z' (m+3)

2

Problem 17

-0.544

The element curium ;!8Cm has a mean life of 1013 seconds. Its primary decay modes are spontaneous fission and ex-decay, the former with a probability of 8% and the latter with a probability of 92%. Each fission releases 200 MeV of energy. The masses involved in decay are as follows:

:;'Cm= 248.072220 u,

;!' Pu = 244.064100 u and : He = 4.002603 u . Calculate the power output from a sample of 10 20 <:;m atoms.

(1 u=931 mevc-1 )

t.E= =-0.544-(-0.85)=0.306 eV 1,.min

=_!!I;__=

t.E=

\1 = 5.95 It.

... (1)

z ... (2) -(-)=0.2 m+3 Solving equations (1) and (2) for z and m , we get m=12 and z=3 (b) Smallest wavelength corresponds to maximum difference of energies which is obviously (Em+J -Em). =>

V = 5.95 x 10 3 cm3

Solution The reaction involved.in a -decay is

1240 = 4052.3 nm 0.306

:!Cm ---+ 2!! Pu +; He

2

Problem 16

Mass defect ·ll.m = mass of 2;cm - mass of

A small quantity of solution containing Na" radio nuclide (half-life = 15 hour) of activity 1.0 microcurie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 hour shows an activity of 296 disintegrations per minute. Determine the total v~lume of the bond in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person.

Similarly,

(1 curie = 3.7 x 1010 disintegrations per second)

Mean life is

Solution A. is the disintegration constant, then

=> =>

2 ,:

Pu - mass of ; He

llm = (248.072220-244.064100 -4.002603)u

llm = 0.00551711 Therefore, energy released in a -decay will be

E. =(0.005517x931) MeV =5.136 MeV E&sinn

= 200 MeV

{given)

1

tmean =1013 s=A,

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So, disintegration constant A.= 10-13 s-1

~2K1m = ~2 (4m) K, sin 0

Rate of decay at the moment when number of nuclei are 1020 _dN =AN=(10-")(10 20 )

=>

dN =10 -dt

7

Squaring and adding equations (1) and (2), we get K+K1 =4K, ... (3) ~

dt

... (2)

4K2 -K1 =K=65eV

... (4)

Now, during collision, electron can be excited to any higher energy state. Applying Law· of Conservation of

d"ISmtegration · . per. second

Of these disintegrations, 8% are in fission and 92% are in a -decay.

Energy,.we get

K =K1 +K, +'1E -3.4 e V - - - - - - - - - n = 4 -6.04eV n=3

Therefore, energy released per second is

P=~=(0.08x107 x200+0.92xl0 7 x5.136) MeV t

-13.6 e V - - - - - -.......--1---n = 2

P=~=2.074x10' MeV

t

So, power output (in watt) is

-54.4 eV-------4>--+-..,__ n =1

P = energy released per second (Js-1 ) ~

He (Z = 2) 0

P=(2.074x10')(1.6xl0-13 )

~

So, power output is P = 3.32 x 10-s W

6.E

65=K1 +K, +'1E

... (5)

Can have the following values, '1E1 ={-13.6-(-54.4)} eV=40.8 eV.

Problem 18

A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90° with respect of its original

Substituting in (5), we get K, +K, =24.2 eV

direction.

Solving (4) and (6), we get K, = 6.36 eV and K, = 17.84 eV

(a)

Similarly, when we put AE = '1E,

Find the allowed values of the energy of the neutron and that of the atom after the collision. (b) If the atom gets de-excited subsequently by emitting radiation, find the _ frequencies of the emitted radiation.

[Given:Massof He atom =4x(massofneutrons) Ionization energy of H atom = 13.6 eV ] Let K, and K2 be the kinetic energies of neutron and helium atom after collision and l!..E be the excitation

'1E=48.36 eV Put in equation (5), we get K1 +K, =16.64 eV

K,

LX

4m

'1E=AE3 ={-3.4-(-54.4)} =51 eV

Applying Law of Conservation of Linear Momentum along x-direction, we get

P;=P1 ./2Km = ~2(4m)K2 cos9

... (1)

Similarly, Applying Law of Conservation of Linear

=

Put in equation (5), we get K1 +K, =14 eV

... (8)

Now, solving (4) and (8), we get K, = -1.8 eV and K, = 15.8 eV But since the kinetic energy cannot have the negative values, the electron will not jump to third excited state i.e., n=4.

K,

~

... (7)

Similarly, when we put

energy.

~ K=65eV,"

'1E={-6.04-(-54.4)} eV

~

Solving (4) and (7), we get K1 =0.312 eV and K, =16.328 eV

Solution

(a)

~

... (6)

Therefore, the allowed values of K1 (KE of neutron) are 6.36 eV and 0.312 eV and of K, (KE of the atom) are 17.84 eV and 16.328 eV and the electron can jump upto second excited state only (11 = 3). (b) Possible emission lines are only three as shown in figure. The corresponding frequencies are

momentum along y-direction, we get

4.56

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----.-,---n=3 11 2

Hence, mv' =-1__ (Ze)(e) r

--------'L-+--.-n=2

... (1)

r2

4ne0

Further, it is given that Bohr model is applicable to .this system also. Hence,

3

Angular momentum in nth orbit is L = nh 21t

-----...L..----'-n=1

v,

h mvr=n21t nh :::;, V=-2mnr Substituting in (1), we get

(E, -E,)

:::;,

h {-6.04-(-13.6)} xl.6 xl0-19 6.63x10-34

=>

v,

=>

v1 =1.82xl015 Hz

... (2)

n2h2 e nme Z Substituting Z = 3 and m = 208m,, we get

r=--' 2

E3-E1

V2=--h-

{-6.04-(-54.4)} xl.6 x10·19 6.63x10·34

=>

v,

=>

v2 =11.67xl0 15 Hz and v3 =E,-E, h

=>

v,

=>

v 3 =9.84x10" Hz

r,

n2h2eo 624mnee2

(b) The radius of the first Bohr orbit for the hydrogen atom is

{-13.6-(-54.4)} X 1.6xlQ·l9 6.63x 10·34 Equating this with the radius calculated in part (a), we get

Hence, the frequencies of emitted radiations are

1.82 X 1015 Hz, 11.67 X 1015 Hz and 9.84 X 1015 Hz Problem 19

A particle ·of charge equal to that of an electron, -e, and mass 0.208 times of the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge +3e. (Take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system. (a) Derive an expression for the radius of the nth Bohr orbit. (b) Find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom. (c) Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit to the first orbit. (Rydberg's constant

n' z624 ~

=1.097xl0' m·1 )

---

-,,

g --·

Kinetic energy of atom is 1

Ze2

2

8m:0 r

K=-mv 2 = - Ze'an dh t e potenti.al energy ·u IS =- 41t£0r

Ze'

~

Total energy E11 = - - 81tt0r

Substituting value of r, calculated in part (a), we get E n

But (-

=1872( m,e n2

4 )

8t~lt2

m{:) is the ground state energy of hydrogen ·

8e0h

atom and hence is equal to -13.6 eV

Solution

If we assume that mass of nucleus is very much mass of mumeson, then nucleus will be assumed to be at rest, only mumeson is revolving round it. (a) In nth orbit the necessary centripetal force to the mumeson will be provided by the electrostatic force between the nucleus and the mu-meson.

(c)

n::::i25

', '' '

<$J m=208m, ,' ,,

,'

= -lS,72 (13.6) eV=

254 9 2 ~ · eV n

=>

E,

=>

E3 -E, =-25459.2(¼-¼)=22630.4 eV

n

So, the corresponding wavelength, is '-

/

=>

,.(in A)= 12375 22630.4 ,. =o.546 A 4.57

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This section contains Single Correct Choice Type Questions. ,Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

(C) 126.2 days

1. . A radioactive substan~e is -being pi-o_duced at a constant rate' of 200 nuclei. The decay coilstant of the'. subst~ce is

1 s-1 • Assuming that initially there are no nuclei present, the time (in second) after which the number of nuclei will become 100 is · (A) 1 s (B) . log,(2) s

1 (C) - - - s log.{2) 2.

6.

Radioactivity is

(A) (B) (C) (D)

spontaneous process irreversible process self disintegration proc~ss all the above

There are two radio nuclei A and B out of which A is an alpha emitter and B is a beta emitter. Their disintegration constants are in the ratio of 1: 2 . The ratio of number of atoms of A and B at any time t so

In the beta decay, the inass number and atomic number of the daughter nuclei respectively (A) increase by 2 (B) decrease by 1 (C) do not change at all

Half life of a radioactive substance A is two times the half life of another radioactive substance B . Initially the number of nuclei of A and B are N, and N, respectively. After three half lives of A number of nuclei of both are eq~~- Then the ratio N" is . N,

that probabilities of getting alpha and beta particles are same at that instant is

-(A) 1:2 (C) 2: 1

(A)

(B) · 1:e (D) e: 1

The magnetic field at the centre (at nucleus) of the hydrogen like atoms (atomic number = z) due to ·the motion of electron in nth orbit is proportional to n3 , n4 (A) -;_,· (B) z

z' (C) 5.

n'

(DJ

1

(B)

3

(C) 4.

(D) specific charge

(D) mass number remains· same but atomic number n:tcreases by 1

8.

3.

a-particle emitted during various radioactive processes have same (A) speed (B) momentum

(C) kinetic energy 7.

(D) 2 s

(D) 1262 days

z'

,,,

9.

1 6

1

(D) 1

4

8

The angular momentum of an electron in an·. orbit is quantized because it is a necessarj., condition for the compatibility with (A) the wave nature of electron. (B) particle nature of electron. (C) Pauli's exclusion behaviour. (D) None of these.

The half life of radon is 3.8 days. After how many days will _!_ th of the radon sample remain behind 10 (A) 1.262 days (B) 12.62 days

10. In a sample

of a

radioactive substance, the fraction of

the initial number of nuclei will remain undecayed after

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T

a time t =-, where T is the half life of radi9active 2 substance is 1 1 (B) {A) .fi. 4 1

(C)

(D)

2.fi.

16. The total energy E of a sub-atomic particle of rest mass m0 moving at no·n-relativistic speed v is

{A)

E =m0c2

(B)

1 'E=-mov2

(C)

1 v2 E = m0c + 2m 0

(D)

E·=m0 c' - 1 m0 v '

1

2'

2

2

.fi.-1 17.

11. In a characteristic X-ray spectra of some atom superimposed on continuous X-ray spectra

An electron is lying initially in the n = 4 excited state. The electron de-excites itself to go to n =1 state directly emitting a photon of frequency v 41 • If the same electron first de-excites to n = 3 state by emitting a photon of frequency v 43 and then goes from n =3 to n =1 state by emitting a photon of frequency v 31 , then

W {C)

~=~+¾ v43 ·=v 41 +2v31

00

~=~-¾ (D) Data Insufficient

(A)

P represents K. line

18. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation.

(B)

Q represents K, line

Infrared radiation will be obtained in the transition

w

2~1 (C) 4~2

{C) Q and P represent K. and K, lines respectively (D) Position of K. and K0 depend on the particular atom

12. In X-ray tube when the accelerating voltage V is halved, the difference between the wavelengths of K0 line and minimum wavelength of continuous X-ray spectrum (A) remains constant

(B) becomes more than two times (C) becomes half (D) becomes less than two times

n3

(C)

atom. It is E, when hydrogen i?n is collided and E3 when helium ion is collided. Then

(C)

215

n2

(D)

21.

E,<E,<E,

vocn

(B)

h

(D)

n

22.

_!i_ .fi.n 2h

.fi.n

In Hydrogen spectrum the wavelength of H 0 line is 656 nm , whereas in the spectrum of a distant galaxy, H. line wavelength is 706 nm . Estimated speed of g'!laxy with respect to earth is (A) 2xl0 8 ms-1 (B) 2xl07 ms-1

(C)

1 At to decay to -th of

16

h

2n

14. The half life of ·,,, At is 100 µs . The time taken for the

radioactivity of a sample of

1 vocn

20. The spin angular momentum of an electron is equal to

(A)

(B)

3~2 5~4

19. When an electron has a transition from the state ( n + 1) to state n where n is quite large, then the frequency of the emitted radiation ( v) will vary with n as 1 1 {A) voc(B) voc-

13. The minimum kinetic energy of an electron, hydrogen ion, helium ion required for ionization of a hydrogen atom is E1 in .case el~ctron is collided with hydrogen

(A) E1 =E,=E,

00 (D)

2xl06 ms-1

(D)

2xl05 ms-1

its _initial value is (A) 400 µs

(B)

6.3 µs

If R be the Rydberg's constant, the energy of an electron in the ground state of the H atom is given by

(C)

(D)

300 µs

(A)

40 µs

Rh

15. The rate of decay of a radioactive element (A) increases with increase in time (B) decreases with increase in time (q remains constant with increase in time (D) decreases exponentially with time

(C) -Rhc

1

(B)

c

Rhc (D)

he

R

23. The activity of a radioactive substance i~ R1 at time t1

and R2 at time 12 {> ~1 ). Its decay constant is

1,..

Then

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Advanced JEE Physics

(A) R1t1 =Rit2 R - ,, (C) - 1 - " 2- = constant

(B)

R, = R, ,'(•,-•,l

respectively. A light beam containing wavelel)gths of 550 nm , 450 nm and 350 run with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is

(D) R, = R,l(', -•,)

t2 -tl

I

24. In the PROBLEM 23, number of' atoms decayed between time interval t1 and t2 are

(A)

2 Jog/ \R, -R2 )

(C) 1,.(R,-R,) 25.

(B)

R,e-"• -R,e-"'

(D)

(R, ;R,)

Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is (A)

(C)

1

(B)

4

1

(D) 1

26. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy -E0 • Its potential energy is (B) 1.5E0 (A) -E, (C)

(D) E,

2E0

27. The de-Broglie wavelength of electron in ground state

of an hydrogen atom is (A) 0.53A (C) 1.s2A

radius. The value of n is (B) 2 (A) 1 (C) 3 (D) 4 29. The angular momentum of an electron in the hydrogen 3h atoms is - , where, h is the Planck's ·constant. The .

2it kinetic energy of this electron is (A) 1.51 eV (B) 3.4 e'v (C) 4.35 eV (D) 6.8 eV

30. A y -ray photon is emitted · (A) after ionization of an atom (B) due to conversion of a neutron into a protonjn the nucleus

(D) due to conversion of a proton into a neutron in the nucleus

31. Photoelectric effect experiments are performed using three different metal plates p, q and r having work

·

4.60

(D)

~, = 2.5 eV

32. Two radioactive materials X1 and X2 have ·decay

constants 101,. and 1,. respectively. If initially they have the same number of nuclei, then the ratio of the number. of nuclei of X, to that of X, will be

(A)

1 101,. 11

101,.

(B)

(D)

:!. e

after a time

1 111,. 1 91,.

33. A potential difference of 20 kV is applied across an X-ray tube. The minimum wavelength of X-rays generated is

(A)

0.26A

(B)

(C)

0.16A

(D) o.so A

0.62A

34. White X-rays are called 'white' due to the fact that (A) they are electromagnetic radiations having nature same as that of white light. (B) they are produced most abundantly in X-ray tubes. (C) they have a continuous wavelength range. (D) they can be converted to visible light using coated screens and photographic plates are affected by them just like light. 35. The half life period of a radioactive element X is same as the mean life time of another radioactive element Y .

Initially both of them have the same number of atoms.

(C) after de-excitation of a nucleus

~, = 2 eV,

(C)

(C)

(B) 1.06A (D) 3.33A

28. The difference between nth and (n + l)th Bohr's radius of H atom is equal to it's (n-l)th Bohr's

functions

(B)

3

4

2

(A)

and

Then,

(A) X and Y have the same decay rate initially. (B) X and Y decay at the same rate always. (C) Y will decay at a faster rate than X. (D) X will decay at a faster rate than Y.

~. = 3 eV,

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36. It has been observed that the most stable nuclei have generally the neutron to proton ratio as

(A) 1:2 (C) 1:4

(B) 2:1 (D) 1:1

45. In a hydrogen atom, the binding energy of the electron in the ground state is E, , then the frequency of revolution of the electron in the nth orbit is

(A) E1n

elements

(D) Radioactive elements cannot be produced in the laboratory 38. Which of the following statements is correct? (A) Beta rays are same as cathode rays. (B) Gamma rays are high energy neutrons. (C) Alpha particles are singly ionised helium atoms.· (DJ Protons and neutrons have exactly the same mass.

(C)

.!i_ nh

2E,n

(D)

2E,

h

nh

~6. Magnetic moment due to the motion of the electron in n"' energy state of hydrogen atom is proportional to (A) n° (B) n (C) n' (DJ n' 47. A radioactive isotope is being produced at a constant rate X . Half life of the radioactive substance is Y . After some time the number of radioactive nuclei become constant. The value of this constant is

•.

(B)

(AJ XY

39. The approximate value of quantum number n for the circular orbit of hydrogen 0.0001 mm in diameter is (A) 1000 (B) 60 (C) 10000 (D) 31 40. Which of the following is its own antiparticle? (A) photon (B) electron (C) proton _ (D) it-meson

(B)

h

37. Select the wrong statement. (A) Radioactivity is a statistical process (B) Radioactivity is a spontaneous process (C) Radioactivity is the neutral characteristics of few

XY (C)

X

y

(DJ (XYJlog, (2)

log,(2)

48. Which one of the following statements is wrong in the

context of X-rays generated from an X-ray tube? (A) Wavelength of characteristic X-rays decreases when the atomic number of the target increases

41. In Bohr's Model of hydrogen atom. The ratio between the period of revolution of an electron in orbit of n = 1

(B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target (C) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube (D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube

the

to period of revolution of the electron in the orbit n =2 is

(A) 1: 2 (C) 1: 4 42

(B) 2:1 (D) 1:8

K. wavelength emitted by an atom of atomic number Z =11 is A . Find the atomic number for an atom that

emits K 0 radiation with wavelength 4;. (A) Z=6 (B) Z=4 (C) Z=ll (D) Z=44

(A) isothermal process (C) adiabatic_process

continuous X-rays has values from

(B) isobarric process (D) isochoric proces~

50. Consider the fission reaction 236u------,,xn7 + y111 +n+n

"

.

i.e. two nuclei of same mass number 117 are found plus two neutrons. The binding energy per nucleon of X and Y is 8.5 MeV whereas of U236 is 7.6 MeV. The total energy liberated is (A) 2000MeV (B) 200MeV (C) 20MeV (D) 2MeV

(A) 0 to oo (B) ,._ to oo where Amin > 0

(C) 0 to Am~ where A- < oo

44.

Assuming that all laws of thermodynamics can be applied, to a nucleus, the a. decay of a nucleus may be regarded as an

43. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the

(D)

49.

Amin to Amax where O< "-mm < Amax < oo

The half-life of the radioactive radon is 3.8 days. The time, at the end of which l/20th of the radon sample will remain undecayed, is (given Iog 10 e = 0.4343) (A) 3.8 day (B) 16.5 day (C) 33 day (D) 76 day

51. The activity of a radioactive sample is measured as 9750 counts (minute)-1 at I= 0 and 975 counts (minute)-1 at t = 5 minute. The decay constant is nearly (A) 0.922 min-' (B) 0.691 min-1

(C)

0.461 min-'

(D) 0.230 min-1 4.61

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Advanced JEE Physics 52.

(q. -(N11.1,-,/ +N11.2,-,,,) 1. N 1. ,-<>-,•>,l• (D) -N . 1 1 2 2

Magic numbers are (A) 2, 8, 18, 28, 50, 164... . (B) 2, 8, 18, 50, 82, 164... . (C) 2, 8, 20, 28, 50, 82, 126 ... . (D) 2, 20, 50, 82, 164, 224 ... .

59. If an electron has, orbital angular momentum quantum

number I= 7, then it will have an orbital angular 53.

The half life of radium is 1600 years. The fraction of the sample of radium that would remain after 6400 years is 1 1 (B) (A) 4 2 1 1 (C) (D) 16 8

54. A and Bare isotopes.Band Care isobars. ff dA, d, and

de be the densities of nuclei A, B and C respectively then (B) dAd,>dc (D) dA =d, <de (C) dA =d, =de

momentum equal to

(A) 7(;n)

J7(;J

(C)

(C)

(B)

(A)

R, - R,

=constant

t2 -fl

3/

(C) R,=R1 exp ( t -t ) (D) R, = R1 exp(..!L) Tt, 57. The electron in a hydrogen atom makes a transition

from an exited state to the ground state. Which of the following statements is true? (A) its kinetic energy increases and its potential and total energies decrease

(B) its kinetic energy decreases, potential energy increases and its total energy remains the same

(C) its kinetic and total energies decrease and its potential energy increases

(D) its kinetic, potential and total energies decrease 58. At time t=O, N 1 nuclei of decay constant 1.1 and N 2

(A) -N1N 2,-<,-,•>,J•

(B)

=

-( ~:),-(',•',)<

C

C

2

(B)

C

(D)

137

11 C

274

62. A hydrogen atom emits a photon corresponding to an electron transitio_n from n =5 to n = 1 . The recoil speed

of hydrogen atom is almost (A) 10-< ms-1 (B) 1

(C) 4 ms-

2x10-2 ms-1

(D) 8x102 ms-1

63. A beam of electron is used in an YDSE experiment. The slit width is d . When the velocity of electron is increased, then (A) no interference is observed (B) fringe width increases (C) fringe width decreases (D) fringe width remains same

and ,.Zn 61 are 63.9298 u and 63.9292 u respectively. It can be concluded from these data that : (A) both the isobars are stable (B) Zn" is radioactive, decaying to Cu" through P-decay (C) Cu" is radioactive, decaying to Zn 61 through y-decay

64. Masses of two isobars

29 Cu"

(D) Cu" is radioactive, decaying to Zn" through P-decay

nuclei of decay constant A. 2 are mixed. The decay rate of mixture is

A. The

61. The speed of the electron in the first orbit (ground state) of the hydrogen atom in te~ of velocity of light c is

mass number of element. (A) A=l (B) A=12 (C) A=16 (D) A can take up any integral value from 1 to 110

radioactive sample is T , then

(D) "56(;n)

o.on

(A)

time ! 1 and R2 at time t, ( !2 > !1 ) : H mean life of the

42(;n)

60. The radius of the first orbit of hydrogen is 0.528 radius of the second orbit of hydrogen is (A) 4.752 A (B) 2.112 A (C) A (D) 0.142 A

55. If u denotes 1 atomic mass unit. One atom of an element has mass exactly equal to Au, where A is

56. The ac~vity of a sample of radioactive material is R1 at

(B)

65. The ratio of molecular mass of two radioactive substances is 3/2 and the ratio of their decay constant

is be

4.62

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(A) 2

(B)

8 9

73. A radioactive sample S1 having an activity of 5 µCi has twice the number of nuclei as another sample S2

(C)

(D)

9

which has an activity of 10 µCi . The half lives of S1

4 3

8

and

66. The electronic transition in Li 2+ ion that emits radiation of wavelength same as the wavelength of second Balmer line of H -atom is W 4~3 ~ 6~3 (C) 12~6 (D) 12~9 67. The ratio between total acceleration of the ele~tron is singly ionized helium atom and hydrogen atom (both in ground state) is (A) 1 (B) 8 (C) 4 (D) 16

68. A radioactive substance X decays into another radioactive substance Y. Assuming that initially oniy X was present, 11._l" and AY be the disintegration

constants of X and Y, N, and N, be the number of nuclei of X and Y at any time t , then the number of nuclei NY will be maximum when

(A)

~. Nx -Ny

'-, "-x -Ay

(B)

N,

'-,

N,: -Ny

"-x -Ay

69. The ratio of the maximum wavelength of the Lyman series in hydrogen spectrum to the maximum wavelength in the Paschen series is

(B)

3

(A) 105 (C) 70.

52 7

(D)

74. In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is (A)

(B) (C) (D)

75. The potential difference across the Coolidge tube is 20 kV and 10 mA current flows through the voltage

supply. Only 0.5% of the energy carried by the electrons striking the target is converted into X-rays. The power carried by X-ray beam is P . {A) P=0.1 W (B) P=l W (C) P=2 W (D) P=lO W

76. For large principal quantum number n the frequency of revolution of electron is equal to the frequency of transition of the electron between two adjacent orbits. This frequency f is proportional to

(A) :,

(B)

:,

1~8

(C)

(D)

1 -;;o

m-1

{B)

(C)

s-1

(D) year

For a number of values of Z and v which plot gives a straight line. 1 (A) v vs Z (B) V VS

z

is

(D)

V

(C)

V V

72. Electrons with de-Broglie wavelength ),. fall on the target

in an X-ray tube. The cut-off wavelength of the emitted · X-rays is (A) ,_ _ 2mc),. 0 h

2

(B)

Jv

vs

z

(D)

V

VS

..fz

78. Solar constant of the sun is a=8.106x10 4

4

2

1 n

77. A metal of atomic number Z is used as a target in a Coolidge tube. Let v be the frequency of the K. line.

mm-1

71. The velocity of an electron in the first orbit of H atom is v . The velocity of an electron in the 2nd orbit of He'

(C)

E(';;u) > E('i,iI) + E(;Y) + 2E(n) E(';;U)<E('i_iI)+E(;Y)+2E(n) E(';;U) < E(':Ba) + E(:;Kr) + 2E(n) E(~U) < E(':Ba)+ E(:;Kr) + E(n)

6

(A)

(B)

can be

15

The SI units of decay constant is

(A) 2v

s2

(A) 20 years and 5 years, respectively (B) 20 years and 10 years, respectively (C) 10 years each (D) 5 years each

'-o = 2h·

Jmin-1m-2

and average sun earth distance is 1.5 x 10s km . The yearly loss in the mass of the sun is (A) 13.8 x 1017 kg (B) 1.38 x 1019 kg

(C) 1.38x1017 kg

(D) 13.8x10 20 kg

me

79. Which of the following processes ~epresents a y -decay? (A)

•xz+r--+AXz_, +a+b 4.63

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Advanced JEE Physics (B)

AXZ +lnO~A-3XZ-2 +c

(q

AXz------>AXz+ f

(D)

A

Xz +e_1 ~AXA-1 + g

80. Binding energy per nucleon vs mass number curve for nuclei is shown in the figure. W , X , Y and Z are four nuclei indicated on the curve. The process that would release energy is ·c

y

j

8,5

8.0 ~>7.5 E'

~

~ .5 5.0 w~

C

'6

ffi

0

30

90

87.

In the above decay, how many p particles are given out? (A} 4 (B) 5 (C) 6 (D) no comments

88.

An electron jumps from the fourth orbit to the 2nd orbit of the hydrogen atom. If R=lD' m-1 , the frequency of the emitted radiation will be (A) 1-x105 Hz 16

__,_____X w --~-----1-----, ' '' ''' '' '' ''' '' '' 60

(B) 5 (D) no comments

(A) 4

(q 6

_2_x10 15 Hz 16

(B)

l_x1015 Hz 16

(D) ~x10 15 Hz 4

89. The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state is

120

Mass number of nuclei

(A) Y------>22

(B)

W------>X+Z

(C) W------> 2¥

(D)

X------> Y + Z

7.418 ms-1

(A)

4.718 ms-1

(B)

(C)

4.178 ms-'

(D) 7.148 ms-1

90. A hydrogen atom emits a photon of energy 12.1 eV. Its 81.

Packing fraction is given by the relation A-M A (B) (A) A M-A M-A M (D) (C) - ·M-A A

82. In the following nuclear reaction,

orbital angular momentum changes by /iL . Then t,.L equals (B) 2.11 X 10-;,.i js (A) 1.05 X 10-;,.& js (D) 4.22x10-;,.i Js (C) 3.16x10-;,.i Js Y

stands for

~AI+1 He----7~P+ Y

(A)

;n

(B)

~, e

(C)

\H

(D)

'.,e

(C)

83. If 1 mg of U"' is completely annihilated, the energy liberated is (A) 9xl010 J (B) 9x1019 J

(q

9x1018

J

(D) 9x1017 J

84. The fraction of the initial number of active nuclei which

remains undecayed after half of a half-life of the radioactive sample is 1 (A) 4

(C)

1 .f5.

(B)

1

2./2

(D) ./2-1

85. The instantaneous concentration N, the initial concentration N 0 , the radioactive constant A can be written as

(A) N =N,e" (q N=N,e-'

(B)

91. The Bohr radius of the fifth electron of phosphorus atom (atomic number 15) acting as a dopant in silicon (relative dielectric constant 12) is (A) 380.9 pm (B) 390.8 pm

N =N,e_,.,

(D) N=N,e'

930.8 pm

(D) 830.9 pm

92. A photon collides with a stationary hydrogen atom in ground state inelastically. Energy of the colliding photon is 10.2 eV . After a time interval of the order of macro second another photon collides with same hydrogen atom inelastically with an energy of 15 eV . What will be observed by the detector? (A) 2 photon of energy 10.2 eV (B) 2 photon of energy 1.4 eV (q One photon of energy 10.2 eV and an electron of energy 1.4 eV (D) One photon of energy 10.2 eV and another photon of energy 1.4 eV

93. The half-life of

131

I is 8 days. Given a sample of

time t = 0 , we can assert that

131

I at

1

(A) no nucleus will decay before I= 4 days (B) no nucleus will decay before I =8 days (C) all nuclei will decay before I = 16· days (D) a given nucleus may decay at any time after I= 0

86. How many a. particles are emitted in the decay 1:X-),Yiat9 4.64

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Atomic & Nuclear Physics 94. The radius of second orbit of an electron in hydrogen atom is 2.116 A . The de-Broglie wavelength associated with this electron in this orbit would be (A) 1.058 A (B) 2.116 A (C) 6.64 A (D) 13.28 A 95. During a negative beta decay (A) an atomic electron is ejected (B) an electron which is already present within the nucleus is ejected (C) a neutron in the nucleus decays emitting an electron (D) a part of the binding energy of the nucleus is converted into an electron 96. Consider the spectral line resulting from the transition n =2 --> n =1 in the atoms and ions given below. The shortest wavelength is produced by (A) hydrogen atom (B) deuterium atom (C) singly ionized helium (D) doubly ionized lithium

97. The angular momentum (L) and radius (r) of a hydrogen atom are related to each other as

(A)

Lr= constant

(B)

Lr'= constant

(D)

Lr-, = constant

(C)

98. The energy _of a photon is equal to the kinetic energy of a proton. The energy of the photon is E . Let 1..1 be the

de-Broglie wavelength of the proton and 1,.2 be the wavelength of the photon. The ratio

~ 1..,

(A) ZERO (C) 13.6 eV

to

(A) E' (C) E-1

100. Imagine .an atom made up of a proton and a hypothetical particle of double the mass of electron but having the same charge as the electron. Apply the Bohr Atom Model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength 1,. (given in terms of Rydberg constant R for the Hydrogen atom) equal to SR

(B)

~ SR

R

(B) infinite (D) 10.2eV

atom correspond to increasing ~ ' C values of energy i.e., B EA< Ea< Ee. If A.1, "-21 A.3 are "'2 the wavelengths of radiations . A corresponding to the transitions C to B, B to A and C to A respectively, which of the following statements is correct?

(B)

1.., = 1..,1,.,

"-1 + A.2 (D) 1..,'=1..,'+1..,' IO~. If elements of quantum number greater than n were not allowed, the number of possible elements in nature would have been

1

(A)

2

(C)

6

n(n+l)

1 n ( n+1)(2n+l) .

104. The radius R of a nucleus changes with the nucleon number A of nucleus as 2

(C)

RocA' RocA0

1

(B) RocA3 (D) RocA

105. The net force between two nucleons 1 fm apart is F,_ if

(B) E'1' (D) E-'

99. The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. The kinetic energy of the electron is (A) 0 (B) 13.6 eV (C) 6.8 eV (D) -13.6 eV

(A) -2._

.!

102. Energy levels A, B, C of a certain

(A)

is proportional

(D)

101. The binding energy of the electron with n --> oo in H atom is

1

(C) Lr 4 = constant

18 SR

both are protons, F2 if both are neutro:15, and f 3 if one is a neutron and the other is a proton. (A) F,
(C)

F1
(D) F,=F,
106. The half life of the radioactive substance is 40 days. The substance will disintegrate completely in (A) 40 days (B) 400 days (C) 4000 days (D) infinite time 107. A radioactive nuclide is produced at the constant rate of

n per second (say, by bombarding a target with neutrons). The expected number N of nuclei in existence t seconds after the number is N O is given by

(A) N =N,e-"

(B)

N =f: +N0e-"

(C) N=f:+( N,-f}-" (D) N=f:+( N,+f:}-" (where 1,. is the decay constant of the sample)

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Optics & Modern Pl1ysics

Advanced JEE Physics 108. The absorption coefficient of X-rays for. a given wavelength is larger for (A) lithium (B) lead (C) aluminium (D) copper

109. For X-rays the wavelength ,. (in A) in terms of energy E (in keV) is given by (A) ,_ = 6.20 E

(B)

,_ = 12.40 E

(D) 110. As ·per Bohr Model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionised Li atom ( Z = 3) is (A) 1.51 (B) 13.6 (C) 40.8 (D) 122.4 111. The K«

X-ray emission line of tungsten occurs at

A. = 0.021 nm . The energy difference between K and L levels in this atoms is about , (A) 0.51 MeV (B) 1.2 MeV (C) 59 keV (D) 13.6 eV

112. If the atom· "" Fm"'7 follows the Bohr's model and the

radius of last orbit of

100

Fm 257 is n times the Bohr

radius, then find n (A) 100 ·

(B) 200

(C) 4

(D) 1

118. The ionisation potential of hydrogen atom is 13.6 volts. The energy of the atom in n = 2 state will be (A) -10.2 eV (B) -6.4 eV (C) -4.4 eV (D) -3.4 eV 119. A radioactive nuclide can decay simultaneously by two different processes which have individual decay constants "-, and A- 2 respectively. The effective decay constant of the nuclide is ,. given by 1 1 1 (A) "-=~"-1"- 2 (B) -=-+-

"- "-, "-2

4

113. A nucleus with mass number 220 initially at rest emits an a-particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the a-particle (A) 4.4 MeV (B) 5.4 MeV (C) 5.6 MeV (D) 6.5 MeV 114. The radioactivity ofa sample is R, at time 11 and R2 at time 12 • If the half life of the sample be T , the number of atoms that have disintegrated in time ( 12 - 11 ) is ·proportional to

(A) R1t1 -R,12

(C)

117. The energy associated with a thermal neutron is of the order of (A) 10KeV (B) 1 Kev (C) 0.1 MeV (D) 0.DlMeV

(R,-R,) T

(B)

(R,-R;f1

::cu

and ::zn are 63.9298 u and 63.9292 u respectively. It can be concluded from this data that (A) both the isobars are stable. (B) 64 Zn is radioactive, decaying to "Cu through P- decay. (C) 64 Cu is radioactive, decaying to "Zn through y- decay. (D) "Cu is radioactive, decaying to "Zn through P- decay.

121. A fraction f, of a radioactive sample decays in one half

(D) (R,-R,)T

life, and a fraction

115. In the given reaction, the radioactive radiations are emitted in the sequence as zXA----+z+t yA~z-1 TA--l->z-t TA-&

(A) a, p, y (C) y, a, P

120. Masses of two isobars

(A) (B) (C)

A decays in one mean life.

f, > /2 f,
(D) Data insufficient to arrive at a conclusion

a,

(B) p, y (D) a, y, P

116. The graph between the instantaneous concentration ( N) of a radioactive element and time (I) is

122. The half life of 131 I is 8 days. Given a sample of time t = 0 . We can assert that (A) no nucleus will decay before I= 4 days (B) no nucleus will decay before t =8 days

131

I at

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(C) all nuclei will decay before t = 16 days (D) a given nucleus may decay at any time after I= 0 123. In the process of fission, the binding energy per nucleon (A) increases (B) decreases (C) remains unchanged (D) increases for mass number A< 56 nuclei but decreases for mass number A> 56 nuclei 124. A proton, a deutron and an alpha particle are accelerated through potentials of V, 2V and 4V respectively. Their velocities will bear a ratio (A) 1:1:1 (B) 1:.Jz:1

(C)

.Jz:1:1

(D) 1:1:.Jz

series, v 2 be the frequency of the first line of the Lyman series, and v 3 be the frequency of the series limit of the Balmer series. (A) v 1 -v 2 =v3 _(B) v 2 -v 1 =v3 v3 =

1

2(v 1 +v,)

(D) v 1 +v 2 =v3

126. The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A. The wavelength of the second spectral line in the Balmer series of singly' ionized helium atom is (A) 1215 A (BJ 1640 A

(C) 2430 127.

22

A

(D) 4687 A

Ne nucleus, after absorbing energy, decays into two

a- particles and an unknown nucleus. The unknown

nucleus is (A) Nitrogen (C) Oxygen

131. In an excited state of hydrogen like atom an electron has a total energy of -3.4 eV . If the kinetic energy of the electron is E and its de Broglie wavelength is '/.., then (A) E=6.8eV,'/..-6.6x10-10 m

(B)

E=3.4eV, '/..-6.6x10-10 m

(C)

E=3.4 eV ,'/..-6.6x10-11 m

(D) E=6.8eV, '/..-6.6x10-11 m

125. Let v 1 be the frequency of the series limit of the Lyman

(C)

130. Let the potential energy of a hydrogen atom in the ground state be zero. Then its energy in the first excited state will be· (A) 10.2 eV (B) 13.6 eV (C) 23.8 eV (D) 27.2 eV

(B) Carbon (D) Boron

132. The radioactivity of a substance is measured in terms of disintegration per second. 3 x 10 10 disintegration per second is (A) 1 eV (B) 1 MeV (C) 1 rutherford (D) 1 curie 133. The shortest wavelength of the Brackett series of a hydrogen like atom having atomic number z, is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of z is (A) 2 (B) 3 (C) 4 (D) 6 134. The shortest wavelength in the Lyman series is 912 Then the longest wavelength in this series is (A) 1216 A (B) 1824 A (C) 2434 A (D) 3648 A

135. The maximum angular speed of the electron of a hydrogen atom in a stationary orbit is (A) 6.2 x 10 5 rads-1 (B) 4.1 x 1016 rads-1

(C) 128. According to Bohr's theory of hydrogen atom, the product of the binding energy of the electron in the nth orbit and its radius in the nth orbit (A) is inversely proportional to n3 (B) is proportional to n2

(C) has a constant value 7.2 eV A

(B)

(C)

I< T

(D)

I=.!_ T l>T

(D) 9.2 x 10 6 rads-1

(A)

,/')..R('/..R-1)

(B)

~

(D) ,/')..(R-1)

129. If the half life and the mean life of the radioactive element is denoted by I and T respectively then l=T

2.4 x 1010 rads-'

136. A hydrogen atom is in an excited state of principle quantum number n . It emits a photon of wavelength A when returns to the ground state. The value of n is

(D) has a constant value of 10.2 eV A

(A)

A.

137. A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is (A) 6h (B) 12h (C) 24 h (D) 128 h

c:=================================== = 4.67

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138. Consider <>-particles , P-particles and y-rays each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are (A) <>, p, y (B) <>, y, p

(C)

p,

y,

(D) y,

(l

p,

(l

(B) y-decay (D) fission

(C) fusion

140. If N 1 atoms of a radioactive element emit N 2 beta

particles per second, then the decay constant of the element (in s-1 ) is (A)

(B)

N, N,

N,

N,

(C) N,log,(2)

(D) N,log,(2)

141. The decay constant of a radioactive sample is A.. The half-life and mean-life of the sample are respectively given by

(A) ! and (log, )2 A. A.

(B)

(C)

(D) --""-and! (Iog,2) "-

A.(log,2) and

i

(log,2} and! A. A.

where r0 is a constant.

Assuming Bohr's model to be applicable, write variation of r11 with n , n being the principal quantum number.

1 n 1

(B)

(C) ,. oc n'

(D) r oc-

r oc-

"

"

n'

144. After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is (A) 6000 (B) 9000 (C) 3000 (D) 24000 145. If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements would be

=

4.68

149. Beta rays emitted by a radioactive material are

(A) (B) (C) (D)

electromagnetic radiations the electrons orbiting around the nucleus charged particles emitted by the nucleus neutral particles

Cl

143. The electric potential between a proton and an electron

(A) r, oc n

148. The shortest wavelength of X-rays emitted from an X-ray tube depends on· (A) the current in the tube (B) the voltage applied to the tube (C) the nature of the gas in tube (D) the atomic number of the target material

the number of nuclei at time t = 0 , then maximum number of nuclei possible are

(A) . the use oflead shielding (B) passing them through heavy water (C) elastic collisions with heavy nuclei (D) applying a strong electric field

f,),

147. The binding energies of nuclei X and Y are E, and E2 respectively. Two atoms of X ftlse to give one atom of Y and an energy Q is released. Then (A) Q>E2 -2E1 (B) Q<2E,-E, (C) Q=E,-2E, (D) Q=2E,-E,

150. A radioactive nucleus is being produced at a constant rate a per second. Its decay constant is A, . If N O are

142. Fast neutrons can easily be slowed down by

is given by V =V, log{

(B) 32 (D) 64

146. The binding energy of ap electron in the ground state of He is equal to 24.6 eV. The energy required to remove both the electrons is (A) 24.6 eV (B) 38.2 eV (C) 49.2 eV (D) 79 eV

139. The equation: 4;H+ ~iHe 2+ +2e- +26 MeV represents (A) p-decay

(A) 60 (C) 4

(A) "-

(B)

(C) N,

(D)

N +~ 0 A.

151. The X-ray beam coming from an X-ray tube will be (A) monochromatic (B) having all wavelengths smaller than a certain maximum wavelength (C) having all wavelengths larger than a certain minimum wavelength (D) having all wavelengths lying between a minimum and a maximum wavelength 152. A radioactive isotope A decays into another isotope B which has a half-life equal to half the half-life of A . Both isotopes emit <> -particles during their decay B decays into a stable nucleus. If a sample consists initially of atoms of A only, then the net activity of the sample initially (A) decreases with time (B) increases with time (C) remains constant (D) any of the above may be true

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Atomic & Nuclear Physics 153. The probability of survival of a radioactive nucleus for one mean life is

(A) 1-~

(B)

e

(C)

log, (2)

(D)

e

L,, =Lu and /Eu/
(C) Lu =Lu and/Eu/>/Eu/

I

Lu
(D) 1

_ log, (2) e

154. The binding energy per nucleon of 1 H 2 and

2

He 4 are

I.I eV and 7 MeV respectively. The energy released in the process 1 H 2 + 1H 2 = 2 He4 is

(B) 16.6 MeV (D) 23.6 MeV

(A) 20.8 MeV (C) 25.2 MeV

(B)

161. The potential energy associated with an electron in the orbit (A) increases with the increase in radii of the orbit (B) decreases with the increase in the radii of the orbit (C) remains the same with the change in the radii of the orbit (D) None of the above 162. Suppose

155. In a hydrogen atom, the electron is in nth excited state.

It comes down to 'first excited state by emitting ten different wavelengths. The value of n is (A) 6 (B) 7 (C) 8 (D) 9 156. There are two radioactive substances A and B such that decay constant of B is twice that of A . Initially both have equal number of nuclei. After n half lives of· A , the rate of disintegration of both are equal. The value of n is (A) 16 (B) 4 (C) 2 (D) 1

157. The

figure

shows

(B) directly proportional to each other (C) equal to each other (D) not related to each other

J,, be the frequencies for

165. If P, , P1 and P, be the penetrating powers of a , jl and

the c-qrves a, b and c respectively (A) f, = f, and I, (B) f, = J,, and I,= I,

,'1,

and I, = I,

(D)

f, = J,,

and I, = I,

158. The ratio of the energies of the hydrogen atom in the first to the second excited state

(A) 4: 1 (C) 4:9

(B) 1 : 4 (D) 9:4

159. The ratio of the wavelength of first line of Lyman series to the first line of Balmer series is ·

(A) 1 : 4 (C) 27: 20

(B) is more. than the daughter nuclei by two (C) equal to the neutron number (D) exceeds the number of neutrons in the daughter nuclei by unity by one positron ~mission 164. Half life period and mean life time of a radioactive element are (A) inversely proportional to each other

with anode potential for a photosensitive surface for three different radiations. Let I,, I, and I, be the intensities and f,, , f, and

f, = f,

163. Mirror nuclei are those in which the number of protons in the parent nuclei (A) is less than the daughter nuclei by one

the

variation of photocurrent

(C)

a radioactive substance disintegrates completely in 10 days. Each day it disintegrates at a rate which is twice the rate of the previous day. The percentage of the material left to be disintegrated after passing of 9 days is (A) 10 (B) 20 (C) 25 (D) 50

(B) 5 : 27 (D) 20 : 27

y radiations respectively then

w

~

~>~>~

167. A star initially has 10 40 deutrons . It produces energy 2 2 3 via the processes and · 1 H + 1 H -------+ 1 H + p 1H

160. A hydrogen atom and a Li... ion are both in the second excited state. If Ln and Lu are their respective electronic angular momenta and EH and Eu their respective energies then, (A) LH >Lu andiEHJ>IEul

~=~=~

~ ~<~<~ ~ ~=~<~ 166. A radioactive nuclide emits an a, j3 and a y ray in close succession, the atomic mass of the end product is reduced by (A) 8 a.m.u. (B) about 4 a.m.u. (C) 2 a.m.u. (D) 1 a.m.u.

2

+ 1H 3 -------+ 2 He4 +n. If the average power radiated

by the star is 1016 W, the deutron supply of the star is exhausted in a time of the order of (A) 10' s , .(B) 10' s (C)

10u s

(D) 1016 s

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(The

masses

of

nuclei

are:

m(H')=2.014u,

(A) 2h

(B)

h 2it

(D)

m(p)=l.007u, m(n)=l.008u, m(He')=4.001 u) (C)

168. The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA . The number of electrons striking the target per second is (B) 5xl06 (A) 2xl0 16 (C) lx1017 (D) 4xl015 169. The wavelength K. of X-rays produced by the X-ray tube is 0.76 A . The atomic number of the anode material of the tube is (A) 30 (B) 40 (C) 50 (D) 60

177. An electron in the ground state of hydrogen has an

angular momentum½_, and an electron in the first excited state of lithium has an angular momentum L2 •

(A)

(C)

L, = L, L, = 2L,

(B) L1 = 4L, (D) L, = 2L,.

178. A certain radioactive element disintegrates for an interval of time equal to its mean life. The fraction of the original amount that remains undecay~d is

scattered at an angle 60° (A) 100

(B) 85

(C) 70

(D) 55

e

(B)

(C)

e'

(D) """Ji

179. If N

171. The kinetic energy of an electron in the first orbit of H atom is 13.6 eV. The total energy of an electron in the second orbit of He+ (A) 13.6 eV

(C)

-13.6 eV

were not allowed in nature, the possible number of elements would be (A) 4 (B) 32 (C) 60 (D) 64

173. The ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is

(A) 2: 1 (C) 8:1

1

denotes the concentration of a radioactive

dN a:N dt

_ dN dt

a:

e"

(D)

dN --ocN

dt dN -N --oce dt

watt is

(A) 3.6xl026 (C) 36x10 26

(B) 0.36 X 1026 (D) 0.036x10 26

181. Order of magnitude of density of uranium nucleus is ( mp = 1.67 X 10-27 kg)

14

(C) 10 kgm""

(B) 6365 A (D) 6563 cm

(B)

180. In the sun about 4 billion kg of matter is converted to energy each second. The power output of the sun in

(A) 1020 kgm-'

(B) 4:1 (D) 16: 1

174. The wavelength of the first line of the Balmer series in the hydrogen atom spectrum is

(A) 6563 A (C) 6563 m

(A) (C)

172. If the elements with principal quantum number n > 4

e

element, then the rate of change of concentration with time ( t) can be written as

iS (B) 3.4 eV (D) -3.4 eV

1

(A) 170. In Rutherford experiment the number of a-particles

scattered at angle 90° is 25. How many particles are

h 2 h 4it

(B)

10'.' kgm""

(D) 1011 kgm-'

182. The electron emitted in beta radiation originates from

(A) (B) (C) (D)

inner orbits of atoms free electrons existing in nuclei decay of a neutron in a nucleus photon escaping from the nucleus

175. An electron with kinetic energy E eV collides with a

hydrogen atom in the ground state. The collision is observed to be elastic for (A) 0<E
176. The angular momentum of the electron in hydrogen atom in the ground state is

=

4.70

183. Electrons with energy 80 keV are incident on the

tungsten target of an X-ray tube. K shell electrons of tungsten have -72.5 keV energy. X-rays emitted by the tube contain only (A) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of - 0.155 A . (B) · a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths.

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(CJ the characteristic X-ray spectrum of tungsten. (D) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of - 0.155 A and the

' characteristic X-ray spectrum of tungsten.

184. The intensity of X-rays from a Coolidge tube is plotted against wavelength A as shown. The minimum wavelength found is 1., and the wavelength of the K. line is Ak • As the accelerating voltage is increased I

(1.,-1.J increases

(CJ Ak increases

[Mass of the helium nucleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu] (A) 7.6 MeV (B) 56.12 MeV (CJ 10.24 MeV (D) 23.4 MeV 189. Consider the following reaction 1H2 +1H2------+ 2He4. +Q.

If m{iH') =2.0141 u; m( 2 He4 ) =4.0024 u.

1.,

(A)

188. If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is

1.

i.; (B)

The energy Q released (in MeV) in this fusion reaction is (A) 12 (B) 6 (CJ 24 (D) 48

(1.,-1.,) decreases

(D) A" decreases

190. For uranium nucleus how does its mass vary with volume? 1 (A) mocV (B) mocV

e' 185. The quantity - - has a value 2he0 c

(A) _!_ ms-1 137 1 (CJ 137

(CJ m «

2 -1 -ms 137 2 (D) 137 (B)

186. A radioactive sample consists of two distinct species having equal number of atoms initially. The mean lifetime of one species is t and that of the other is St . The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a Junction of time. Which of the following figures best represents the form of this plot?

N

192. A radioactive material of half-life T was produced in a nuclear reactor at different instants, the quantity produced second time was twice of that produced first ·time. If now tlwir present activities· are A1 and A 2 respectively then their age difference equals

N (B)

N

(CJ

(D) mccV2

191. A radioactive element X converts into another stable element Y . If half life of X is 2 hr, initially only X to be present and after time t, the ratio of atoms of X and Y is found to be 1 : 4 , then time t in hr is (A) 2 (B) 4 (CJ between 4 and 6 (D) 6

(A) (A)

.Jv

lo;, 2 llog.( ~'

JI

(CJ _ log,T 2 J1og,(~)1 2A,

(B)

1:

Tl log.( )I

(D) rJ1og,(~)1 2A,

193. de-Broglie wavelength of an electron in the nth Bohr orbit is An and the angular momentum is In, then

N

(A)

In cc"'"

(B)

1 1. oc" I.

(CJ

"-n OC J~

(D)

1.

(D)

187. The largest wavelength in the ultraviolet region of the

hydrogen spectrum is 122 nm . The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is (A) 802 nm (B) 823 nm (CJ 1882 nm (D) 1648 nm

1

ocn

I!

194. When an electron in the hydrogen atom in ground state absorbs a photon of energy 12.1 eV, its angular momentum (A) decreases by 2.llxl0-34 Js (B) decreases by l.055xl0-34 Js

(CJ' increases by 2.11 x 10-34 Js

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(D) increases by 1.~ x 10-34 Js 195. A radioactive nucleus A finally transforms into a stable

nucleus B . Then A and B can be (A) isotones (B) isotopes (C) isob.;,s · (D) None of these

200. When a hydrogen atom emits a photon during the . transition n = 5 to n = 1, its recoil speed is · approximately (A) 0.1 mms-1 (B) 3 mms-1 (C) 4 ms"' (D) 800 ms-'

.

. 3

196. Number of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time t = 2 s : Then number of nuclei at time t = 4 s will be (B) 790 (A) 700 (D) 810 (C) 800

(; =1.8 x 1014 Ckg·1 for an electron)

'197. H 92 U238 changes to 85 At 210 by a series of a and , decays, the number of a and p decays undergone is (A) 5and7 (B) 7and5 (C) 7and7 (D)° 7and9 '

P

198. In hydrogen and hydrogen like atoins the ratio of di#'erence of energies E2n - E,. and E4,. - £2,. varies with atomic number z and principle quantum number 'n as

z'

z'

(A)

--;;,

(B)

~-

(C)

-nz

(D)

(;)°'

199. Consider the electronic energy level diagram of H -atom. The photons associated with shortest and longest wavelengths would be emitted from the ,atom by the.transitions labelled

---------~~n=oo ~--~------•~A~n=4 -=•D__-+"c-~-----n=3 ' B

---~~-~----n=2 -----------n= 1

(A) (B) (C) (D)

=

4.72

D and C respectively · C and A respectively ,

C and D respectively A and C respectively

201. A potential difference of 10 V is applied 'across an X-ray ·tube. The ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced is

. I

1 () A 10 4

(B)

(C)

(D) 1

2~

1 100

202 During a nuclear fusion reaction (A) a heavy· nucleus breaks into two fragments by itself (B) a light nucleus bombarded by the1111¥ neutrons · breaks up (C) a heavy nucleus bombarded by thermal neutrons breaks up (D) two light nuclei combine to give a heavier nucleus and possibly other products 203. Four physical quantities are listed in Column I. Their values ar~'listed in Column II in a raildom order Column I · Column II A Thermal energy of air (i) 0.02 eV molecules at room temperature.

·B

Binciing energy of heavy nuclei (ii) 2 eV per nucleon.

' C X-ray photon energy. (iii) 10 keV iv 7 MeV D Photon ener of visible Ii ht. The correct matching of columns I and II is given by A B . ·C D iv iii ii (A) iv. iii ii (B) iv ii iii (C) iii ii iv (D) 204. Of the various series of the hydrogen spectrum, the one which lies wholly in the ultra violet region is (A) Lyman series (B) Balmer series (C) Passhen series (D) Brackett series

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This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/ are correct. 1.

The momentum of a single photon of red light of frequency 400xl012 Hz moving through free space is

(B) 8.Bxl0-78 kgms-2 (D) Data Insufficient

(A) ZERO

(C) 1.65x10.. Mevc-1 2.

5.

The radius of the orbit ofan electron in a Hydrogen like

An electron makes transition inside a hydrogen atom. The orbital angular momentum of the electron may change by h (A) h (B)

h 2it

(C)

(D)

h 4it

atom is 4.5a0 , where a0 is the Bohr radius. Its orbital

31

apgular momentum is ' . It ;s given that h is Planck · 2it constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) (A)

9 32R

(C)

SR

"9

3.

(B) '

6.

following decrease in the excitation?

(A) potential energy (C) kinetic energy

9 16R

7.

4

(D) 3R

H+, He+ and o++ all having the same kinetic ·energy pass through a region in which there is a uniform

magnetic field perpendicular to their velocity, The masses of H+, He+ and o++ are 1 u, 4 u and 16 u respectively. (A) H+ will be deflected the most. (B) o++ will be deflected the most. (C) He+ and o++ will be deflected equally. (D) All will be deflected equally,

An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. Which of the

In Bohr's model of the hydrogen atom, let R, V, T and E represent the radius of the· orbit, speed of the electron, time period of revolution of electron and the total energy of the electron respectively. The quantities proportional to the quantum number n are (A) VR (B) RE V (C) (DJ E R

I

8.

Let mp be the mass of a proton, m,. the mass of mass of a : Ca nucleus. Then,

Which of the following statement(s) is (are) correct? (A) The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons

(B) The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons

(C) In.nuclear fission, energy is released by fusing two nuclei of medium mass (approximately 100 amu) (D) In nuclear fission, energy is released by fragmentation of a very heavy nucleus

~

neutron, M 1 the mass ~f a ~Ne nucleus and M2 the

(A) M, = 2M1 4.

(B) angular speed (D) angular mo~entum

M 2 <2M1

(C)

9.

(B)

M, > 2M1

(D) M1 <10(m0 +mr)

The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation

{A) (B) {C) (D)

the intensity increases ,the minimum wavelength increases the intensity remains unchanged the minimum wavelength decreases 4.73

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Advanced JEE Physics electron in a hydrogen atom makes a transition "1. ---+ n2 where, "1. and n2 are the principal quantum numbers of the two states. Assume the Bohr Model to be valid. The time period of the electron in initial state is eight times that in the final state. The possible values of n,_ and n2 are

10. The

(A) n, =4, n, =2 (q n1 = 8, n, = l

16. Consider an atom whose energy level diagram is shown

in figure. ---------E=O - - - - - - - - - - E5 =-8eV - - - - - - - - - - E , =-13 eV - - - - - - - - - - E , =-15 eV

(B) n,_ =8, n, =2 (D) n, = 6, n, = 3

- - - - - - - - - E , = - 1 9 eV

11. The wavelength of K. X-rays for lead isotopes Pb208 ,

Pb 206 , Pb"" are 1,.1 , (A) ,.,=,.,=,., (q ,., < ,., < ,.,

1,.2

and 1,.3 respectively. Then (B)

,_,>,.,>,.,

(D) ,., = ~,.,,.,

12. A radioactive sample has initial concentration NO of nuclei.

(A) The number of undecayed nuclei present in the sample decays exponentially with time. (B) The activity (R) of the sample at any instant is directly proportional to the number of undecayed nuclei present in the sample at that time. (q The number of decayed nuclei grows linearly with time.

(D) The number of decayed exponentially with time.

nuclei

grows

13. Suppose the potential energy between electron and .

2

proton at a distance r is given by - Ke, . If m be the 3r mass of electron, then application of Bohr's theory to hydrogen atom in this case shows that (A) energy in the nth orbit is proportional to n' (B) energy is proportional to m"' (q energy the nth orbit is proportional to n-2 (D) energy is proportional to m' 14. The mass number of a nucleus is

(A) (B) (C) (D)

always less than its atomic number always more than its atomic number sometimes equal to its atomic number sometimes more than and sometimes equal to its atomic number

----------E,=-25eV

Suppose an atom starts at level 3 then (A) Shortest wavelength photon the atom can emit is

1.24x10-7 m (B) Longest wavelength photon that it can absorb is 6.2x10-7 m (q Lowest frequency photon that can ionize the atom is 3.62x10 15 Hz (D) The number of ways of de-excitation of atom to ground state is 3 17. A positronium atom consists of a positron and electron revolving aronnd their common centre of mass. Then

compared to hydrogen atom the positronium atom has (A) ground state energy half of hydrogen atom (B) Rydberg constant half of hydrogen atom (q radius of first orbit of electron double that in case of hydrogen atom (D) velocity of electron in first orbit same as in case of hydrogen atom 18. The ground state and first excited state energies of hydrogen atom are -13.6 eV and -3.4 eV respectively. li potential energy in gronnd state is taken to be zero. Then (A) potential energy in the first excited state would be 20.4 eV (B) total energy in the first excited state would be 23.8.eV (q kinetic energy in the first excited state would be 3.4eV (D) total energy in the ground state would be 13.6 eV 19. When a hydrogen atom is excited from ground state to

15. A muonic hydrogen atom is a hydrogen atom with electron replaced by a muon whose mass is 212 times the mass of an electron, then (A) Bohr radius of the muonic atom is 250 fm

(B) Ground state energy of muonic atom is 2883 eV . . h,, groun d state ts (q angu1ar momentum m 2 d . 106h . (D) angular momentum m groun state 1s - -

"

first excited state then

(A) its kinetic energy increases by 10.2 eV . (B) its kinetic energy decreases by 10.2 eV . (q its potential energy increases by 20.4 eV. (D) its angular momentum increases by 1.05 x 10""' Js 20. An electron orbiting in a circular orbit around the nucleus of an atom

(A) has a magnetic dipole moment

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(BJ exerts an electric force on the nucleus equal to that on it by the nucleus (C) does produces a magnetic induction at the nucleus (D) has a net energy inversely proportional to its distance from the nucleus

21. In the Bohr model of the hydrogen atom (A) the radius of the nth orbit is proportional to n2 (B) the total energy of the electron in the nth orbit is inversely proportional to n (C) the angular momentum of the electron in an orbit

~·

is an integral multiple of

26. From the following equations pick out the possible nuclear fusion reactions (A) ,c"+,H'---->,C"+4.3MeV

(B)

6

C12 + 1H 1 ----> 7 N 13 +2MeV

,N"

(C) + 1 H 1 ----> 80 15 + 7.3 MeV (D) 92U235 +on1_____,. s-iXe140 +36Sr94 +ont +on1 +y + 200MeV 27. An electron in hydrogen atom first jumps from second excited state to ground state and then from first excited

7t

(D) the m_agnitude of the potential energy of the electron in any orbit is gre::i.ter than its kinetic energy

(D) Gap between energy levels will now be doubled

\

22. The electron in a hydrogen atom makes a transition from 2nd excited state to the ground state. Then (A) it's KE. increases and total energy decreases (B) both its KE. and total energy increases (C) frequency of emitted photons may be 4.6 x 1014 Hz (D) frequency of emitted photons must be 2.9x1015 Hz 23. The energy, the magnitude of linear momentum and

orbital radius of an· elect,ron in a hydrogen atom corresponding to the quantum nwnber n are E, P and r respectively. Then according to Bohr's theory of hydrogen atom

.!.

state to ground state. Let the ratio of wavelength, momentum and energy, of photons emitted in these two casesbe•a, band c respectively,then

(A)

.

1

a=

.(B)

5 b=27

(D) c=-

28. A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n . It can emit a maximum energy photon of 204 eV . It makes a

transition to quantum state n , a photon of energy 40.8 eV is emitted, then (A) Z=2 (B) Z=4 (C) n=l (D) n=2

29. Which of the following physical quantities in hydrogen atom are independent of the quantum number n . The symbols have their usual meanings.

PEr is proportional to

(B)

P.IS proportion . al to n E

(A) vr

(B)

(C) Er is constant for all orbits (D) Pr is proportional to n

(C) Er

(D) v2 r

24. If potential energy in hydrogen atom with electron in ground state is taken to be 13.6 eV, then

(A) potential energy in the first excited state would be 34eV (B.) total energy in the first excited state would be 37.4eV (C) kinetic energy in the first excited state would be 44.2eV (D) total energy in the ground state would be 27.2 eV 25. If electron of the hydrogen atom is replaced by another particle of same charge but of double the mass, then select the correct option(s). (A) Bohr radius will increase to double value (B) Ionisation energy of the atom will be doubled (C) Speed of the new particle in a given state will be one fourth of what electron will possess in the

4 5 27

(A)

n

9

c=-;;

:,

30. Select the correct statement(s). (A) X-rays travel faster than ultraviolet rays in vacuum. (B) Balmer series of H-spectrum is found in visible region. (C) The characteristic X-rays are produced due to jumping of electrons from higher to lower shell. (D) In photoelectric emission process, the maximum energy of the photoelectrons must increase with increasing intensity of incident light. 31. Total energy of electron in the first orbit of hydrogen

atom is equal to the (A) total energy of electron in 2nd orbit of (B) total energy of electron in 3rd orbit of (C) total energy of electron in 2nd orbit of (D) total energy of electron in 3rd orbit to

He' He' Li" Li"

same orbit

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32. Which of the following transitions in He• ion will give rise to a spectral ]ine having the same wavelength as

some spectral line in the hydrogen atom? (A) n=4 to n=2 (B) n=6 to n=2 (C) n=6 to n=3 (D) n=8 to n=4

(C) the kinetic energy is proportional to \ n (D) the angular momentum is proportional to n 39.

series (A) it may emit another photon in the Balmer series.

33. When atomic number Z is doubled in atoms, which of the following statements are consistent with Bohr's

(B) it must emit another photon in the Lyman series. (C) the second photon, if emitted, will have a wavelength of about 122 nm . (D) it may emit a second photon but the wavelength of this photon cannot be predicted.

theory? (A) Energy of a state is doubled. (B) Radius of an orbit is doubled. (C) Velocity of electrons in an orbit is doubled. (D) Radius of an orbit is halved. 34. Let A,, be the area enclosed by the nth orbit in a

40. The graph between

(A) will pass through the origin. (B) will have a slope of four units at all points. (C) will be a monotonically increasing nonlinear curve. (D) will be a straight line.

metal 1

(C)

E

-

2

(D) None of these Hydrogen atoms absorb radiations of wavelength 1'.0 and consequently emit radiations of 6 different wavelengths of which two wavelengths are shorter than 1'.0 • Choose the correct altemative(s).

1/1. nm-'

"dirtl ·na1 to -h e tan 8 IS ec y proportio , where 1· 1 1s e

2

·-

3

_.__~:---~

-·

The final excited state of the atoms is n = 4 The initial state of the atoms may be n = 2 The initial state of the atoms may be n = 3 There are three transitions belonging to Lyman

(A)

V1V2 V3=-V1 +V2

(B)

V3=V1+V2

series

(C)

A. 3 = ~ A1 +A.2

(D)

A3=A1+A2

38. For the electron in the nth allowed orbit is . . proportion . al to -1 (A) th! e mear momentum 1s n

(B) the radius is proportional to n ,--, 4.76

0.004

41. The wavelengths and frequencies of photons in transitions 1, 2 and 3 for hydrogen like atom are }.1 , A. 2 , A. 3 , v1 , v2 and v 3 respectively. Then

. E

(A) (B) (C) (D)

0.002

Planck's constant and c is the speed of light. (D) The violet colour light can eject photoelectrons from metals 2 and 3.

2

37.

metal 3

(A) Ratio of work functions $1 : $2 : $3 = 1: 2: 4. (B) Ratio of work functions $1 : $2 : $3 = 4: 2: 1.

(A) ionisation potential is _.§_

(C) totaI energy IS

metal 2

8

0.001

· 36. ff the potential energy of the electron in the first allowed orbit in hydrogen atom is E then its

2

and stopping potential (V) of

V

35. A particular hydrogen like atom has its ground state total energy - 54.4 eV, then (A) its atomic number is 2 (B) it can absorb a photon of 40.8 eV (C) in its ground state it cannot emit photon (D) for its ground state its potential energy is -108.8 eV and kinetic energy is +54.4 eV

-

i

three metals having work function $1 , $, and $3 in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/ are correct? [Here 1'. is the wavelength.of the incident ray]

hydrogen atom. The graph of log,(~:) against log,n

. . (B) kinetic energy IS

Whenever hydrogen atom emits a photon in the Balmer

42.

In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is 1'. . The intensity of light falling on slit 1 is four

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times the intensity of light falling on slit 2. Choose the correct choice(sJ. . (AJ If d =1-., the screen will contain only one maximum. (BJ If 1-. < d < 21-. , at least one more maximum (besides the central maximum) will be observed on the

8 6

4

screen.

(q 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 (DJ 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 43. Assume tJ\at the nuclear binding energy per nucleon

(!)

versus mass number (AJ is as shown in the

figure. Use this plot io choose the correct choice(sJ given below.

2 O

100

200

A

(AJ Fusion of two nuclei with mass numbers lying in the range of 1
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This section contains Reasoning type questions, each having four choices (A), (B), (q and (D) out of which ONLY ONE is correct. Each question contains STAUlMENT 1 and STATEMENT 2. You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1. Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble.(C) IfSTATEMENT1isTRUEandSTATEMENT2isFALSE.

,

Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.

1.

5.

Statement-I:

corresponding to transition from n = 2 to n = 1 is much greater as compared to transition from n -~ oo to n =2 .

increased, the wavelengths of the characteristic X-rays

do not change.

Statement-I: In .a hydrogen atom energy of emitted photon

If the accelerating potential in an X-ray tube is ·

Statement-2:

Statement-2:

When an electron ·beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray

Wavelength of photon is directly proportional to the energy of emitted photon.

energy. 6. 2.

Statement-I:

Statement-I:

Magnetic moment of an atom is due to both, the orbital

According to classical theory, the proposed path of an electron in Rutherford· atom model will be parabolic.

motion and spin motion of every electron.

Statement-2:

A charged particle produces a magnetic field.

According to electromagnetic theory, an accelerated particle continuously emits radiation.

Statement-2: 7.

Statement-I: Total energy in an orbit is negative in an atom.

3.

Statement-I:

Statement-2:

An electron in hydrogen atom passes from n = 4 to n =1 level. The maximum number of photons that can be emitted is 6.

B.

Statement-2: No. of photons emitted can never be more than 5. 4.

Electron is bounded by electrostatic attraction between electron and nucleus.

Statement-I: The wavelength of first Balmer line of deuterium is slightly more than that of hydrogen.

Statement-I: Between any two given energy levels, the number of absorption transition is always less than number of . emission transition.

Statement-2:

Statement-2:

Absorption transition starts from the lowest energy level only and may end at any higher level. But

In.the centre of mass of..an atom reference frame both nucleus and electron are non-stationary.

level and end at any energy level below it.

emission. transitions may starts from any higher energy

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Atomic & Nuclear Physics

9.

Statement-1: Total energy of revolving electron in any stationary

15.

orbit is negative.

Statement-2: Energy is a scalar quantity and it can take positive and

Statement-2: The mass of a- particle.

negative value. 10. Statement-1: In outermost stationary orbit, energy of electron is least negative.

Statement-1: The ionising power of J3- particle is less compared a- particle but their penetrating power is more.

16.

P- particles

is less than .the mass of

Statement-1: A certain radioactive substance has a half life period of

30 days. The disintegration constant is 0.0231 day-1

Statement-2: In outermpst orbit, electron is at maximum distance

•

Statement-2: Decay constant _varies inversely as half life.

from nucleus. 17.

11. Statement-1:

Speed of electron in the n"' orbit of hydrogen atom is 1 VH oc-, n

Statement-1: , Half life of a certain radio-active element is 100 days, After 200 days fraction left undecayed will be 50%. Statement-2:

1'~ --(-1)' , where symbols have usual meaning. N, 2

Statement-2: Speed of electron in n"' orbit of hydrogen like atom is

-N

Z'

VHlike

oc-

18.

n

fl- particles emitted from radioactive nuclei has continuous energy ranging from zero to a certain maximum value.

12. _Statement-1: Atomic hydrogen gas excites to third excited state. The number of spectral lines in emission spectrum obtained is 6.

Statement-2: In

Statement-2:

fl- decay

a neutron is converted into a proton, an

electron and an antineutrino. -The total energy emitted in P- decay is shared by P- particle and antineutrino.

Nu~ber of spectral lines in emission spectrum can never be less than 6. 13. Statement-1: In a hydrogen atom energy of emitted photon corresponding to transition from n = 2 to n =1 is much greater as compared to transition from n ---+ co to

Statement-1:

19. Statement-1: r- photons are emitted during annihilation process of electron and positron.

Statement-2: High energy photons are.emitted due to conversion of

n=2.

mass into energy.

Statement-2: Wavelength of photon is directly proportional to the · energy of emitted photon

20. Statement-1: 25% of the radioactive nucleus remains active after

200 days for an element of half life 100 days. 14. Statement-1: The difference between the wavelengths of series limit

of the Lyman series of spectral lines and that (the series limit) of the Paschen series is equal to the wavelength of a spectral line of the Lyman series (for the hydrogen atom).

Statement-2: The wave number of an atomic transition can. be

Statement-2: N 21.

=N,(½)f where symbols have usual meaning.

Statement-1: If binding energy per nucleon increases after a nuclear reaction then the reaction is exothermic.

Statement-2: calculate~ from the formula V = R( ~ 2

- : 2 ),

where V

is the wave number, R is the Rydberg's \constant, m and n' are the quantum numbers of the initial and final states.

H there is decrease in mass in nuclear reaction then the reaction is exothermic.

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Optics & Modern Physics

Advanced JEE Physics 22. Statement-1: If the current in the filament of electron gun in a X-ray tube is increased, the penetration power of X-rays is increased.

26.

Statement-1: Radioactivity is independent of the physical and chemical conditions of substance.

Statement-2: Radioactivity is a property of the nucleus.

Statement-2: Increasing current increases the number of electrons

emitted by the electron gun. 23. Statement-1: It is necessary to keep high vacuum in Coolidge tube to produce X-rays.

27. Statement-1: The Wave number. corresponding to the transition between the atomic levels n = 3 and n = 2 of a hydrogen atom i.e. V32 ,.is related to the wave-numbers

v

Statement-2: High vacuum is kept in Coolidge tube so that the electron emitting from the filament of the tube may not lose their energy in colliding with the atom of the gas in the tube. ,·

21

25.

Statement-1: The wavelength of the K. line from an element and that of the K, line satisfy the relation i,,. rel="nofollow"> ).!\, Statement-2:

for a hydrogen atom by the relation

=.!.

for a transition is given by

"'·· --¾-),

.!.A.lffll =R(·n12

where R is the m Rydberg constant and m, n are integers representing the initial and final principal quantum numbers.

the expression

f,) .

constant for an atom.

31

The' wave-number V""'

f, and /, respectively, then frequency of first line of

Statement-2: _ Difference of energy level between two orbits is

v

Statement-2:

24. Statement-1: If maximum frequency of Balmer and Paschen series is

Balmer series is (!, -

and

V31 =V21 +V32·

28.

Statement-1: When light is passed through a sample of hydrogen atoms in ground state, then wavelengths of absorption lines are same as wavelengths of lines of Lyman series in emission spectrum. Statement-2: In ground state hydrogen atom will absorb only those radiation which will excite to higher energy level.

The energy separation between the K0 levels is smaller

than th~! of the K, levels .

=

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This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of competitiveness there may be a few questions that may have more than one correct options) Comprehension 1

In a mixture of H -He· gas ( He• is singly ionized He atom), H atoms and He+ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid. Based on above information, answer the following questions. 1.

2.

3.

The quantum number in He+ ions is (A) 2 (C) 4

II

4.

The minimum potential difference through which electron may be accelerated from rest to cause the ejection of K-level electron from the atom is (A) 1.875xl0 4 V (B) 3x10-1' V 14 (C) 3.33 x 10 V (D) 10 V

5.

Energy released when L-level electron moves to fill the vacancy in the K-level is (A) 3 X 10-lS J (B) 3x10-1' J (C) 2.7 X 10-lS j (D) 2.7 x10-1' J

6.

Kinetic energy of the electron emitted from the M-level is (A) 3 x 10-1• j (B) 3 x 10-17 j (C) 2.7 X 10-lS j (D) 2.67 X lQ-15 j

of the state finally populated (B) 3 (D) 5

The wavelength of light emitted in the visible region by He+ ions after collisions with H atoms is (A) 6.5x10-7 m (B) 5.6x10-' m (C) 4.8xl0-7 m (D) 4x10-7 rn The ratio of the kinetic energy of the n = 2 electron for the H atom to that of He+ ion is 1 (A) 1 (B) 2 4

(C) 1

as X-ray or may all be used to eject an M-level electron from the atom. Based on above information, answer the following questions.

(D) 2

Comprehension 2 Simplified picture of electron energy levels in a ----M certain atom bombarded with high energy electrons is shown in the figure. The impact of -3 x 1O16 J ----L one of these electrons has caused the complete removal -3 x 10 ,.r. J - - - - - K of K-level and is filled by an electron from the L-level with a certain amount of energy being released during the transition. This energy may appear

Comprehension 3 When a particle is restricted to move along x-axis between x = 0 and x =a, where a is of nanometer dhnension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x =a. The wavelength of this standing wave is related to tlie linear momentum p of the particle according to the de-Broglie relation. The energy of the particle of mass m is related to its

'

linear momentum as E = L . Thus, the energy of the 2m particle can be denoted by a quantum number n taking values 1, 2, 3,.. .. ( 11 =1, called the ground state) corresponding to the number of loops in the standing wave.

.::======================================== = 4.81

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Advanced JEE Physics

Optics & Modem Physics

Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a . [Take h = 6.6 x 10-34 Js and e = 1.6 x 10-19 C ]. The allowed energy for the particle for a particular

7.

value of n is proportional to

(A) .-, (C) .-1

8.

(B) (D)

a

2

2

a

!fthemassoftheparticleis m=lxlO-.,"kg and a=6.6 the energy of the particle in its ground state is closest to (A) 0.8 meV (B) 8 meV (C) 80meV (D) 800meV

9.

The speed of the particle that can take discrete values is proportional to 3

(A)· n2 (C)

,,-1

(B) /" (D) n

1

n'

Comprehension 4 Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, H known as deuteron

i

and denoted by D can be thought of as a candidate for fusion reactor. The D-D reaction is iH + iH-------+~He+ n+energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of : H nuclei and electrons is known as plasma. The nucl_ei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time 10 before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 x 1014 scm" . It may be helpful to use the following : Boltzmann constant k = 8.6 x 10-5 eVIC1 ; 2

-'-=1.44x10-9 eVm. Based on above information, answer 41tE 0

the following questions. 10. In the core of nuclear fusion re'actor, the gas becomes plasma because of (A) strong nuclear force acting between the deuterons (B) Coulomb force acting between the deuterons (C) Coulomb force acting between deuteron-electron pairs (D) the high temperature maintained inside the reactor core

=

I·

11. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 x 10-15 m is in the range (A) lx109 K
masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if ( 1"1_ + m2 ) < M . Also two light nuclei of masses m3 and m4 can rmdergo complete fusion and form a heavy nucle~s of mass M' only if (1~ +m 4 ) > M'. The masses of some neutral

a toms are Enven m th e ta ble b eow 1 1.007825 u . ;H /H

;u

6.015123 u

~Gd

151.919803 u

;H

3.016050 u

:zn

69.925325 u

2:Bi

208.980388 u

2.014102 u

;u

7.016004 u

:Pb

205.974455 u

!He :se

4.002603 u

2

2

~Po

81.916709 u 209.982876 u

(1 u = 932 Mevc-2 ) Based on a~ove information, answer the following questions. 13. The correct statement is (A) The nucleus !Li can emit an alpha particle

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Atomic & Nuclear Physics (B) The nucleus ~Po can emit a proton

(C) Deuteron and alpha particle can undergo complete fusion. (D) The nuclei ;::zn and can undergo complete fusion

::s,

14. The kinetic energy (in keV) of the alpha particle, when the nucleus 2:Po at rest undergoes alpha decay, is (A) 5319 (B) 5422 (C) 5('07 (D) 5818 Comprehension 6 A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fix;ed positive ions surrounded by free electrons can be treated as neutral plasma. Let N be the number density of free electrons, each of mass m . When the

electrons are subjected to an electric field, they are displa'ced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency cop, which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency w , where a part of the energy is absorbed and a part of it is reflected. As co approaches cop, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals. Based on above information, answer the following questions. 15. Taking the electronic charge as e and the permittivity as e0 , use dimensional analy'sis to determine the correct expression for roP (A)

~ ~

(B).

fJJie

(D)

~

0

(C)

0

a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e., n ~ p + e- + V,, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (Ve) to be massless and possessing negligible energy and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 x 106 eV . The kinetic energy carried by the proton is only the r€coil energy. Based on above information, answer the following questions. 17. What is the maximum energy of the anti-neutrino? (A) ZERO

(B) Much less than 0.8xl0 6 eV (C) Nearly 0.8x10 6 eV (D) Much larger than 0.8x10 6 eV 18. If the anti-neutrino had a mass of 3 eve-' (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron? (A) 0,; K,; 0.8 xl06 eV (B) 3 eV,;K,;0,8xl06 eV (C) 3eV,;K<0.8xl0 6 eV (D) 0,;K<0.8xl0 6 eV Comprehension 8 A radionuclide with half life -T is produced in a reactor at a constant rate P nuclei per second. During each decay, energy E0 is released. H production of radionuclide is

started at t = 0 . Based on above information, answer the following questions. 19. The'number of nuclei in the radionuclide at any instant t is given by PT ( t1og,2) (A) N = - - 1-e T log,2

2

16. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N ~ 4 x 1027 m-3 • Take t 0 ~ 10-11 and m R:' 10-30 , where these quantities are in proper SI units (A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm

(B)

N=PT(l_,-f)

(C)

N =PTlog,2(1-e

(D) N=PT(l-e

2~)

20. The rate of release of energy as a function of time is

Comprehension 7 The f3- decay process, discovered around 1900, is

(A)

-""•·') PE ( log.°z 1-e T

basically the decay of a neutron (n) . In the laboratory, a proton (p) and an electron (,-) and observed as the decay

(C)

PE,Iog,2(1-e

products of the neutron. Therefore, considering the decay of a neutron as a two body decay process, it was predicted theoretically that the kinetic energy of the electron should be

n:,,,_)

(B)

''°;·') (D)

~i

( PE, 1-e

T

PE,(1_,-,',)

21. Total energy released upto time t is (A)

Hog,2) PTE ( PtE0 - - ' 1-,-Tlog,2

c:================================= = 4.83

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Advanced JEE Physics

(B) (C)

PtE,-PTE, ( 1-e

Optics & Modenz Physics llog,2)

Comprehension 11 Nuclear reactions

r

PtE0 -PTE0 log,2(1-e

"°;·')

Comprehension 9 and B, both convert into a

stable nucleus C . The nucleus A converts into C after emitting two a- particles and three P- particles, the nucleus B converts into C after emitting one ex- particl~ and five P- particles. A time t = 0 , nuclei of A are 4NO and that of B are N 0 • The half-life of A for converting to C is 1 min and that of B is 2 min. Initially number of nuclei of C

If atomic numbers and mass numbers of A , 2 2 , A1 and A2 respectively, then (A) 2,-22 =6

and B

are Z1

Given that the atomic masses are

mUH-atom) =1.007825 amu (u) m(;H-atom) = 3.016049 arnu (u) m(;H-atom)=2.014102 amu (u) Based on above information, answer the following questions. 28. Protons are incident on : H at rest. The threshold energy for the reaction is (A) 5.4MeV

(C) 2MeV

23. The number of nuclei of C, when number of nuclei A and B are equal is (A) 2N, (B) 3N, 9N, 2

(D)

of

29.

(B) lOMeV (D) 8 MeV

When ; H are incident o;, protons, then threshold energy is

(A) l0MeV (C) 16 MeV

(B) 8 MeV (D) 20MeV

SN,

24. The rate of disintegrations of · A time t0 (in minute), then (A) 10 =4 (C) t, =8

artificial

called threshold energy. Consider the reaction given by p+ :H------+iH + iH

(B) A, -A, =4 (C) Both (A) and (B) are correct (D) Both (A) and (B) are wrong

(C)

for

reactions, energy input is required in the form of kinetic energy to be converted into nuclear binding energy. The minimwn energy required for the reaction to take place is

are zero. Based on above information, answer the following questions. 22.

performed

reactions, exoergic and endoergic. In exoergic reactions energy is released. In endoergic reactions energy has to ·supplied for the reaction proceed. In exoergic reactions nuclear energy is converted into.kinetic energy. In endoergic

(D) None of these

Two radioactive nuclei A

are

transmutation of elements for there are two types of nuclear

30. Which of the following statement is correct? (A) Less energy is required for nuclear reaction if light

2

and B are equal at

part is at re~t and heavy particle is incident

(B) More energy is required for nuclear reaction when heavy particle is incident and light particle is at

(B) t, =6 (D) t0 =2

rest

(C) Threshold energy does not depend on which particle is at rest

Comprehension 10 The electric current in an X-ray tube (from the target to the filament) operating at 40 kV is 10 mA. Assume that on an average, 1 % of the total kinetic energy of the electrons hitting the target is converted into X-rays. Based on above information, answer the following questions.

The heat produced in the target every second is (A) 29.6 J (B) 396 J (C) 3600 J (D) 360 J

4.84

Comprehension 12 Consider a different atomic model in which electron

k

27. Minimum wavelength of X-ray produced is nearly (A) 0.3 A (B) 3 A (C) 30 A (DJ 300 A

=

reaction

revolves around the nucleus (proton) at a separation r under the action of force which is different from the electrostatic force of attraction. The potential energy between an electron and the proton due to this force is given

25. The total power emitted as X~rays is (A) 40 W (B) 400 W (C) 4 W (D) 0.4 W 26.

(D) Thr.eshold energy does not depend on Q-value of

by U = -"""i'" where k is a constant. Based on above r information, answer the following questions.

31. The radius of nth Bohr's orbit is

=

=

it (A) r=-vkm nh

(B)

2it r=-vkm nh

4it ,,r =-vkm

(D)

r

(C)

nh

= Bit ./km nh

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Atomic & Nuclear Physics 32. The velocity in the nth orbit is given by

(A)

(C)

V V

nh 81t'm../km

l

nh'

(B)

V

(D)

V

81t'm../km

'

(C) VocA 3

81t'in../km

33. The kinetic energy of the electron in nth orbit is given by (B)

KE

BE

A

2

128rt m k

(D) KE

E, E,

34. The potential energy of the electron in the nth orbit is

E,

given by

(A) PE (C)

PE

n4h4

(B)

64kn4 nz2 n2h2

PE

(D) PE

128krt4 m2

(C) TE

I I

I I

I I

I

I

l

I I I

I

l

I

l28k1t4m2 n4h' 256k1t4 m2

I I I

I I I

I I I

I I I

I

I

I

N,

I

N3 N2 N1

Mass number

39. The Q value of the reaction is

(A) E1N1 -(E,N, +E,N,) (B) , {E,N, +E3 N 3 -E,.N,)

(C)

E,N, +E,_N1 -E,N,

(D) E1N 1 +E,N,-E2 N2

40. If Mw is the mass of W, Mx is mass of X and My js

256krt4 m2

mass of Y nucleus, choose the correct statement.

Comprehension 13 -All nuclei consist of two type of particles i.e., protons and neutrons. Nuclear force is the ~trongest force and the stability of nucleus is determined by the neutron-proton

ratio or mass defect or Binding energy per nucleons or packing fraction. The shape of nucleus is calc;ulated by quadrupole moment. The spin of nucleus depends on even or odd mass number. The volume of nucleus depends on the ~ass number. The whole mass of atom (nearly ·99%) is centred at the nucleus and the_'magnetic moment of nucleus is measured in terms of the nuclear magnetons. Based on above information, answer the following questions.

W ~>~>~

00 ~<~<~

Mw My Mx (C) - . - < - > N, N2 N3

Mw My Mx (D) - - > - < N, N, N,

N1

41.

N,

N3

N1

~

N,

~

e>

•

•

C

w

C,

(A)

'5.5

"'s

C

~

w

(B)

"' C

'8 ' C

m

'jjj

,g

charge dependent short ranges forces non conservative force spin dependent force

,g Atomic mass number

Atomic mass number

~

37. Binding energy per nucle6n is maximum (A) for lighter order element Qow mass number) (B) for heavier order element (high mass number) (C) for middle order element (D) equal for all order elements

N,

The graph representing the relationship between atomic mass number A and the total binding energy (BE) of the nucleus, for nuclei heavier than Z is

36. The correct statement(s) about nuclear force is/ are

(A) (B) (C) (DJ

____ J___ J__ y ' w _____ J' ___ J' __ ~-~ I

128rt m k

(D) TE

z .-~-..x -------

n4h4

35. The total energy of the electron ii, the nth orbit is given by n-1h 4 n4h4 ~ (A) T E = - (B) TE 4 2 4 2

128rt m k n4h4 256krt'm'

(D) ·VocA

Comprehension 14 Consider the nuclear fission reaction W ~ X + Y . Based on the graph given showing binding energy per nucleon vs number of nucleons, answer the following.

n4h4 4

(B) ,VocA3

(A) V .,'. A'

n'h

~

•

•

C

w

(C)

C,

C

'6 C

m

C

~

s

w

(D)

C,

C

'6

ifi

/

iii ,g

,g

Atomic mass number,

Atomic mass number

38. Volume (V) of the nucleus is related with mass number (A) as

::::::================================= = 4.85

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Advanced JEE Physics 42.

Optics & Modem Physics

When a slow neutron is captured by a U 235 nucleus, a fission results which releases 200 MeV of energy. If the output of a nuclear reactor is 1.6 MW , the rate of nuclei (per sec) undergoing fission is (A) 8 x 10" (B) 1 x 1013

(q lx10"

5xW"

(D)

43. The binding energy per nucleon versus mass number is best represented by the curve MeV I

9

,,--~-

\

1---~\

,,,"\ o " " - - - ' - - ~ - ~ ~ • Mass No. (A) ' , 120 .," A

~-' C

(A) A (q C

(B) B (D) None of these

Comprehension 15 In a nuclear fusion reactor, the reaction occurs in two stages.

Stage-I: Two deuterium ( 1 D2 ) tritium ( 1 T 3 )

nuclei fuse to form a

Comprehension 16 Many unstable nuclei c~ decay spontaneously to a nucleus of lower mass but different combination of nucleons. The process of spontaneous emission of radiation is called Radioactivity in which the decay rate is actually exponentially decrease with time. Also, radioactive decay is a statistical process which is independent of all external conditions. Based on above information, answer the following questions.

the following statements is correct ? (A) TM >TH

(B) Both are equal (q TM
0.693 (C) - - s n

0.693n N

(D) - - - s

nucleus with a proton as a by-

product. The reaction 1113-y be represented as

D(D, p)T. Stage-II : A tritium nucleus fuses with another deuterium nucleus to form a helium ( 2 He 4 )

(A) R111 = R,t,

(B)

R, - R, = constant

(C)

R, = R,,-,t•,-•,I

t2 -t1

(D) R,t, = R,t,

Comprehension 17 An .electron orbits a stationary nucleus of charge +ze where z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron form the second Bohr orbit to third Bohr orbit. Based· on above information, answer the following questions.

m(,D')=2.014102u (atom)

m(. T') = 3.016049u

49. The activity of radioactive substance of decay constant A. is R1 at time t1 and R2 at time t2 ( > t1 ) , then

nucleus with a

neutron as a by product. The reaction is represented as T(D, n)a..

Given that

(D) 0.484%

47. If TH is the half life and TM is the mean life, which of

,,,.,,, ...

-~ "ffi -9 MeV

(q 0.384%

(atom)

m(,He')=4.002603u (atom)

m(. H') = 1.007825u (atom) m(,n') =1.008665u Based on above information, answer the following questions. 44. The energy released in the Stage-II of fusion reaction, is

(A) 4.033 MeV

(q 40.33 MeV

(B) 17.587 MeV (D) 1.7587 MeV

45. The energy released in the combined reaction per

deuterium, is given by (A) 4.207MeV (C) 6.207MeV

(B) 5.207MeV (D) 7.207MeV

46. The percentage of the mass energy of the initial deuterium is released is

(A) 0.184%

=

4.86

(B) 0.284%

50. The value of z is (A) 5 (C) 3

(B) 4 (D) 2

51. The radius of first Bohr orbit is (A) 0.53 A (B) 0.106 A (CJ o.318 A (D) 0.53nm Comprehension 18 In quantum mechanics, some quantities are discrete and cannot be continuous. One of these quantities is the energy. Energy can only take certain values, like E1 , E2 , E3 , E4 , .•.. , which are called energy levels. The energy cannot take any

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Atomic & Nuclear Physics values between E1 and E2 or E2 and E3 or E3 and E4 , etc. Certain transitions from one energy level to another result in the emission of a photon of radiation, whereas others can only take place if a photon is absorbed. The energy levels in a newly discovered gas are expressed as

(C)

(A)

nh(:)

(B)

(C)

n;(:)

(D)

.

moment of the orbital current due to the electron moving in its orbit and the magnetic field B is

-64

(C)

is best represented as

'•',

' ',

',_

4nnh

e

(B)

2nnh

(D)

e

(D) MN,

60. The number of decayed nuclei after time t is m -it (A) -m N ,e-J..1 (B) -e M M

' , (D)

4m

The initial number of nuclei present is m {A) (B) MN, N, m

(C) .!!!..N M A

..

(D)

Comprehension 20 A radioactive sample has mass m, decay constant A and molecular weight M . H Avogadro number is NA then answer the following questions. 59.

•,

2nhBe m

nhBe m nhBe

58. Total magnetic flux due to the magnetic field B passing through the nth orbit is 1t nh nnh (A) (B) 2 e e

(C)

53. The ground state energy is (A) -144 eV (B) +144 eV (C) -244 eV . (D) None of these

{A)

(B)

2m

52. The ionization energy of the gas is (A) 244 eV (B) 576 eV (C) 144 eV (D) +13.6 eV

54. The graph of En versus \ n

nhBe

(A)

-144

-36 Based on the information given, answer the following questions.

2nh(:) n:(:)

57. Potential energy of interaction between the magnetic

------~------, I n E,,(eV) 2 3 4

(D) J2neh 1tBe

56. The kinetic energy of the electron in nth orbit is

En=-E1z2

n' 2 where, -E1z is the ground state energy. Taking z =1 for simplicity, but not assuming that the gas is hydrogen, an experiment is designed to measure the energy as a functions of the level. The results obtained in the experiment are given in the table.

J neh 2nBe

,

'•,

(C) .!!!..N (1-,-") M

' ...

A

(D) .!!!..(1_,-'-') M

61. The activity of sample after time t is Comprehension 19 An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr's postulate regarding the quantisation of angular momentum holds good for this electron of charge e and mass m , answer the following questions.

55. The radius of nth orbit of the electron will be

(A) J2nh nBe

(B)

,nh

V27tlk

m~;'" }-"

(A)

(

(C)

MN,1',-" m

(B)

:NA1"(1-e-'-')

(D) None of these

Comprehension 21 In atom X , a single electron orbits around a stationary nucleus of charge +Ze where Z is a constant and e is the magnitude of the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to the third 4.87=

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Advanced JEE Physics

Optics & Modern Physics

Bohr orbit. Take ionization energy of the hydrogen atom 13.6 eV. . ' 62. The value of Z is. (A) 3 (C) 5

(B) 4 (D) 6

63. Kinetic energy of electron in ground state of atom X is (A) 122.4 eV (B) 170 eV (C) 340 eV (D) 680 eV 64. The wavelength of electromagnetic radiation required to remove the electron from 3'• Bohr orbit to infinity is

(A) 329 A (C) 546 A

,.

(B) 428 A (D) 636 A

Comprehension 22 The nuclei of' a radioactive element X are being produced at _a constant rate a and this element decays ti;). a stable nucleus Y with a decay constant 1,. and half-life T1/2 •

At time t = 0 , there are N O nuclei of the element X . ~ased on the information given, answer the following questions.

65. The number N x of nuclei of X at time t :a T1/2 is

a-1,.N, . 21,. " (1,.N,+f)1i _(C) (A)

(B)

(21,.N -a).! ' .o . 1,.

(D)

a+'>..N~, 21,.

'

68. The principal quantum number of initially excited level B is . (A) 1' ' 1 (B) 2 (C) 3 (D) None of these 69. The magnitude of ionisation energy of.gas atoms is (A) 13.6 eV (B) 12.7 eV (C) 54.4 eV (D) None of these

'

' 70. The maximum and minimum energies of the emitted photons · (A) 12.75 eV and 0.66 eV (B) 13.6 eV and 12.75 eV (C) :i3.6 eV and 0.66 eV (D) None of the'se Comprehension 24

. '

The figure shown an energy level diagram for the hydrogen atom. Several transitions are marked as I, II, ill, IV, V, & VI. The diagram is only indicative and not to scale. Based on the information given, answer the. following questions.

66. The number Ny of nuclei of Y at time I is

(A) at-( a-tNo}-" +(

particular upper (excited) energy level B and there are no atoms.in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing monochromatic light of photon energy 2.7 eV. Subsequently the . atoms emit radiation of only sb< different photons energies. Some of emitted photons have energy 2.7 eV. Some have·more and some have less than 2.7 eV. Based on the information given, answer the following questions.

a-tN,)

"'----------.

4

'

(B)

(D)

2

a-tN, }-" at-( a-tN, }-"

(2)+I(a-1,.N,) (A) ~lo ,.g, 2 1,.'

'.

4.88

JVI

71. A Balmer series photon is absorbed in (A) II (B) III (C) IV (D) VI

(A) 291nm (C) 487nm

i a-tN,) f Iog,(2)-2( a-tN,)

(B) 364nm (D) 652nm

73. ,The transition that occur, when a hydrogen atom is

Comprehension 23 . , A gas of .identical hydrog€n like atoms has some ato~, in lowest (ground) energy level A and some atoms in ,

V

72. The wavelength of. the radiation involved in transition fl is

flog,(2)+½( a-tN,)

(C) . log,(2l'-½( (D)

1vl

I

67. The number Ny of nuclei of Y at l=T1/2 is

(B)

II

at+(a-~,}-"-(a-tN')

(C) at+(

mT-

'

3

irradiated with radiation of wavelength 1025 A (A) I (B) II (C) IV (D) V

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Each question in this section contains statements given in two colwnns which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: ' 1

If the correct matches are A-> p, s.and t; B-> q and r; C-> p and q; and D-> sand t; then the correct darkening of bubbles will look like the following :

P

q

r

s

I

A@@)0@(D B@@)0@G) c@@)0@G) D@@©®CD

1.

Some laws/processes are given in COLUMN-I. Match these with the physical phenomena given in COLUMN-II. [

(A) (B) (C) (D)

2.

COLUMN-I Nuclear fusion Nuclear fission p-decay Exothennic nuclear reaction

COLUMN-II (p) Converts some matter into energy (q) Generally possible for nuclei with low atomic number (r) Generally possible for nuclei with higher atomic number (s) Essentially proceeds by weak nuclear forces.

Some laws/processes are given in COLUMN-I. Match these with the physical phenomena given in COLUMN-II.

I

COLUMN-I (A) Transition between two atomic energy levels (B) Electron emission from a material (C) Mosley's Law - (D) Change of photon energy into kinetic energy of electrons

COLUMN-II

(p) Characteristic X-rays (q) Photoelectric effect (r) Hydrogen spectrum (s) P-decay

4.89

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Advanced JEE Physics 3.

Match the quantities in COLUMN-I with their dependence on the principal quantum number n and the atomic number Z in COLUMN-IT. I. COLUMN-I

(A) (B) (C) (D) (E)

4.

COLUMN-II

Angular speed Time period Angular momentum

l

n' Z' (q) n Z' (p)

Magnetic moment

Magnetic Field

(r)

7

(s)

7

Z'

The energy, the magnitude of linear momentum, magnitude of angular momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number n are E, p, L and r respectively. Then according to Bohr's

theory of hydrogen atom match the expressions in COLUMN-I with statement in COLUMN-II.

I

COLUMN-I

COLUMN-II

(A) Epr

(B)

(p) (q) (r) (s)

E.

E (C) Er (D) pr 5.

l

is independent of n

is directly proportional to n is inversely proportional to n

is directly proportional to L

If f, is the maximum frequency of emitted photon of Lyman series, f, is minimum frequency of the emitted photon of Lyman series and f, is maximum frequency of the emitted photon of Balmer series, then match the relations in given in COLUMN-I with the corresponding relations in COLUMN-IL COLUMN-IT

(B)

f, f,.

(C)

(f, - f,)

(A)

(D) (f, 6.

(p) greater than

(q) is negative (r) less than / 2 (s)

f,)

(B) Inverse photoelectric effect (C) De-excitation of Be+4 atom from second excited state (D) K X-ray photons of molybdenum Z = 42 0

(p) (q) (r) (s)

COLUMN-IT

FewMeV 20KeV 54eV O.leV

Match the processes given in COLUMN-I with their characteristics in COLUMN-II.

I

COLUMN-I (A) . Alpha decay

(B) Beta decay (C) Positron emission (D) Electron capture

=

is equal to / 3

Match the quantities in COLUMN-I with their respective v~ues in COLUMN-II. ,. '·. COLUMN-I (A) Pair Production

7.

f,

4.90

COLUMN-II

(p) (q) (r) (s)

Mono energetic particles are emitted Poly energetic particles are emitted Angular momentum is conserved Can take place inside and outside nucleus

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Atomic & Nuclear Physics 8.

Match the quantities in COLUMN- I with their respective values in COLUMN-II. COLUMN-I COLUMN-II (A) 1 rutherford {p) 1 disintegration/ sec (B) 1 becquerel (q) 3.7xl010 dis/sec (C) 1 curie (r) 10 6 dis/sec (D) Activity of 1 g Ra"' (s) 1010 dis/ sec

9.

Match the processes given in COLUMN-I with their characteristics in COLUMN-II. [

COLUMN-I

COLUMN-II

(A) a- decay

{p) Atomic

of

the

product

p• - decay (C) p- -decay

(q)

of

the

product -nucleus

(B)

(D) Electron capture

(r) (s)

number decreases Atomic number increases

nucleus

Atomic number of the product nucleus not necessarily changes some mass is converted into energy

10. Match the processes given in COLUMN-I with their characteristics in COLUMN-II. [

COLUMN-I

COLUMN-II

(A) a-decay

{p) (q) (r) (s) (t)

p-- decay (C) p• - decay (B)

(D)

r- decay

Mass number decreases Atomic number decreases

Mass number does not change Chemical symbol of nucleus changes Energy is released

11. Match the reactions/processes given in COLUMN-I with their characteristics given in COLUMN-II. COLUMN-I

COLUMN-II

(A) Spontaneous radioactive decay of an uranium nucleus initially at rest as given by reaction

~u------+ ~Th + i He+ .... (B) Fusion reaction of two hydrogen nuclei as given by reaction

{p) (q) (r) (s) (t)

Number of protons is increased Momentum is conserved Mass and energy are inter convertible Charge is conserved Angular momentum is conserved

:H + :H-'J>iH + ....

(C) Fission of U"' nucleus initiated by a thermal neutron as given by reaction

6n+~U-'J> 1,:tBa+ :Kr+36n (D) p-- decay (negative beta decay) 12. Match COLUMN-I of the nuclear processes with COLUMN-II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists.

[

COLUMN-I

{A) Alpha decay (B) P'-decay

(C) Fission (D) Proton emission

COLUMN-II

{p) (q)

i:0~1;N + ....

~U~2;,Th+ ....

(r)

1

(s)

~Pu~ 1:La+ ....

:Bi~ 1~Pb+ ....

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In this section the answer to each of the question is a four digit integer, ranging from 0 to 9999. The appropri;te bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question number X (say) is 6092, then the correct darkening of bubbles wil! look like the following :

·

X.@e@@ CD (D'G) CD

®®®· @@@@

,@@©@

®@@@

-~~~~

'@@®
@®e@

1.

How many head-on, elastic collisions must a, neutron

6.

have with deuterium nucleus to reduce its energy from 1 MeV to 0.025 eV. 2.

The binding energy of an electron in the ground state of He atom is equal to E, = 24.6 eV. Find the energy required, in eV, to remove both, electrons from the

7. · A "'Cd radionuclide goes through the transformation

chain. 118Cd

atom.

3.

4.

Find the quantum number n corresponding to nth excited state of He+ ion if on transition to the ground state the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm . The ionization energy of the hydrogen atom is 13.6 eV. From what material ( Z value) is the anode of an X-ray tube made, if the K. . line wavelength of the characteristic spectrum is 0.76 A ?

5.

7

Li target is bombarded with a proton beam aµrent of 10°" A for 1 hour to produce 7 Be of activity 1.8 x 10 8 disintegrations per second. Assuming that one 7 Be radioactive nucleus is produced by bombarding 1000 protons, determine its half-life, in days, to the nearest three digit integer. A

At time t = 0, activity of a radioactive substance is 1600 Bq·, at J=8 s activity remains 100 Bq. Find the activiiy, in ilq, at t = 2 s .

30 min

118ln

4,Smin

1185n

(stable)

The half lives are written below the respective arrows. A time t = 0 only Cd was present. Find the percentage of nuclei transformed into stable ov~r 60 minutes.

8.

The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. Find the ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium.

9.

A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that log, 2 = 0.693, find the fraction of the initial number of nuclei (expressed in nearest, integer percentage) that will decay in the first 80 s after preparation of the sample.

10. A silver sphere of radius 1' cm. and work 'function 4.7 eV is suspended from an insulting thread in free-

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Atomic & Nuclear Physics space. It is under continuous illumination of 200 run

wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The

17. Emission spectrum of hydrogen atom has two lines of

maximum number of photoelectrons emitted from the sphere is Ax 10 2 (where 1
what series does a spectral line belong if its wave number is equal to the difference of wave numbers of

z. 11. The activity of a freshly prepared radioactive sample is 10 10 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10-" kg . Find the mass (in mg) of the radioactive sample.

Balmer series with wavelength 4102 A and 4861

A . To

the above two lines? What is the wavelength of this line in A ?Given, R=l.097xl07 m-1 • 18. A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n . The excited atom can make a transition to the first excited state by successively emitting two phOtons of energy 10.2 eV

and 17 eV respectively. Alternately, the atom from .the 12. A proton is fired from very far away towards a nucleus with charge Q = 120e, where e is the electronic charge.

It makes a closest approach of 10 frn to the nucleus. Find the de-Broglie wavelength (in units of fm) of the proton at its start.

Take the proton mass, m, =(¾)x10-27 kg,

!'. = 4.2 x 10-15 JsC-1 , e 1 frn =10-15 m .

-

1

same excited state can make a transition to the second excited state by successively emitting two photons of

energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z .

(Ionization energy of H-atom = 13.6 eV) 19. The ionization energy of a hydrogen like Bohr atom is 4rydberg. (a) What is the wavelength of the radiation emitted, in

- = 9 x 109 mF-1 and

41ts0

A, when the electron jumps from the first exdted state to the ground state?

13. A stream of a. -particles is incident on a sample of hydrogen gas. What should be the minimum kinetic energy of a. -particles, in e V , to ionize the hydrogen

(b) What is the atomic number Z of the atom? (c) Also the radius of the first orbit for this atom as a multiple of the Bohr's radius a0 is ao , where * is

atoms.

not readable. Find • .

14. Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic potential energy of H atom becomes -1.7 eV. Now a photoelectric plate having work function W = 2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie

wavelength of the ejected photoelectrons in

A

to the

nearest integer.

15. A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These

levels have energies between -0.85 eV and -0.544 eV (including both these values). (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions, in

A.

(Take ground state energy of hydrogen atom to be -13.6 eV). 16. A neutron with an energy of 4.6 MeV collides with protons and is retarded. Assuming that upon each collision neutron is deflected by 45° find the number of collisions which will reduce its energy to 0.23 eV.

•

20. A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation to the nearest three digit integer , in A , required to excite the electron in Li++- from the first to the third Bohr orbit. The ionization energy of the hydrogen atom is 13.6 eV. 21. The electric current in an X-ray tube operating at 40 kV is 10 mA. Assume that on an average 1 % of the

total kinetic energy of the electrons hitting the target are converted into X-rays.

(a) (b)

What is the total power emitted as X-rays, in watt? How much heat, in joule is produced in the target every second?

22. The mean lives of a radioactive substance are 1200 yr 600 yr for a. -emission and P-emission respectively. Find out the time, in year, during which three fourth of a sample will decay if it is decaying both by a. -emission and P-emission simultaneously. Given log, (4) = 1.4 and

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IB30 1.

2.

.

. . _' ·_ ..'.: .

(a)

113.74 A

(b)

3

,; -~: ': ;' . _. .

. ..... ·-~: :.··, ,:~ BASEDlONrATOMIClSJRUCliUREl&lP.ROP.ERlilES 14.

122.4 eV 15.

3.

793.3 A

4.

2eV, 0.754V

5.

7.52x1018 Hz

6.

15865 volt

7.

o.31A

8.

(a)

He

(b)

54.4 eV

(c)

10:2ev

9.

12;1 eV

10. ·-26.9 eV, -12 eV 11.

10.2 V, 2, 10.2 eV, 51 eV

12.

3.1x106 ms-1

'13.

(a)

15.6 eV

(b)

2335A

(c)

12.52 V

(rl)

1.01x107 m-1

(e)

(I) 6 eV , (ii) 0.7 eV

16. 17,

18.

(a)

1.587 A

(b)

1.095 ~ 1o• ms·'

neh µonm2e1 4nm ' 8s0 h6n5

(~r

.

Samit'

(a)

42

(b)

o.163 A

FlrstOrblt: 'KE=13.60eV, PE=0eV, TE=13.60eV Second Orbit: KE= 3.40 eV, PE= 20.40 eV, TE= 23.80 eV

19.

0.55 eV

20.

-./2

21.

(a)

3.4 eV

(bl

6.63 A

= ================================== 4.94

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Atomic & Nucleqr Physics

1.

81

2.

1,55x106 J

3.

1.4x1QH' s

4.

3.81x.104 year-1 and 3.14x10-2 year-1

5.

(a)

0.113 min-'

(b)

6.132mln

9.

0.39

10.

(a)

5724 years

(b)

12.58 dismin-19-1

6.

1.2sx10-11 %

11.

1.12x10.. g

7.

(a)

L42x10" per year

12.

1.88x104

(b)

1.41x10" per year

13.

0.205 mCi

(c)

5.4x10' years

BASED ON BINDING ENERG:V. l/1 NUCl!EAR REACTIONS 1.

No; first reaction is not p6ssible. :

2.

1.54 MeV

3.

8.78 day

4.

4.7x10-14 kg

5.

0.1308 MeV

6.

K.., =0, K= =4.4 MeV

7.

0,961 MeV

120.35 g

9.

(a)

8.09x10" J

(b)

2.?xio• kg

(a)

-1.18 MeV

(b)

43°

10.

11.

1.

B·

2.

D

3.

C

6.

I , 11.

D

7.

D

8.

C

12.

D

13.

16.

C

18.

B

17. · 22;

A

21.

A

23.

26.

C

27.

D

28.

'31.

A"

32.

D

33.

36.

D

37.

D

, 41.

D

42.

46.

B

51.

I·

8.

127.6 MeV

4.

D

5.

B

D

9.

A

10.

A

B

14.

A

15.

D

D

19.

A

20.

C

D

24.

D

25.

B

D

29.

A

30.

C

B

34.

C

35.

C

38.

A

39.

D

40.

A

A'-

43.

B

44.

B

45.

D

47.

C

48.

B

49.

C

50.

B

C

52.

C

53.

D

54.

C

55.

B

56.

C

57.

A

58.

D

59.

D

60.

B

61.

C

62•.

C

63.

C

64.

D

65.

C

66.

C

67.

B

68.

D

69.

D

70.

C

71.

B

72.

A

73.

A

74.

A

75.

B

I

4.95

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Advanced JEE Physics 76.

B

Optics & Modern Physics

n.

C

78.

C

79.

C

80.

C

81.

D

--------82. A

83.

A

84.

C

85.,

B

86.

A

87.

A

88.

C

89.

C

90.

B

91.

A

92.

C

93.

D

94.

C

95.' C

96.

D

97.

D

98.

B

99.

[iii1.

A

102.

B

103.

D

104.

106.

D

107.

C

108.

B

109.

B

L11~.- C.

112.

D

113.

B

114.

D

B

-B - -

-

100.

C

105.

C

110.

D

115.

B

--·----

116.

D

117.

D

118.

D

119.

D

120.

D

121.

B

122.

D

123.

A

124.

D

125.

A

126.

A

127.

B

128.

C

129.

C

130.

C

131.

B

132.

D

133.

A

134•. A

135.

B

136.

B

137.

B

138.

A

139.

140.

A

141.

B

142.

B

143.

A

144.

D

145.

A

146.

D

147.

C

148.

B

149.

C

150.

A

C

151.

C

152'.

B

153.

B

154.

D

155.

A

156.

D

157.

A

158.

D

159.

B

160.

B

B

165.

C

161.

A

162.

D

163.

D

164.

166.

B

167.

C

168.

A

169.

B

170.

A

171.

C

172.

C

173.

D

174.

A

175.

B

176.

C

177.

C

178.

B

179.

B

180.

A

181.

B

182.

C

"183.

D

184.

A

185.

C

186.

D

187.

B

188.

C

189.

C

190.

A

191.

C

192.

C

, 193.

A

194.

c

195.

B

200.

C

196.

D

197.

,c

198.

D

199.

C

201.

D

202.

D

203.

A

204.

A

1.

BC

2.

AC

3.

A<;,

4.

AD

5.

BC

6.

BC

7.

ACD

8.

CD

9.

AD

10.

AD

11.

AD

12.

ABD

13.

AB

14.

CD

15.

ABC

16.

ABC

17.

ABCD

18.

ABCD

19.

BCD

20.

ABCD

21.

AD

22.

AC

23.

ABCD

24.

ABD

25.

BD

26.

BC

27.

ACD

28.

BD

29.

BCD

30.

BC

31.

AD

32.

AD

33.

CD

34.

ABD

35.

BCD

36.

ABC

37.

ABD

38.

ACD

39.

BC

40.

AC

41.

BC

42.

AB

43.

BD

7 ~

:::J

7

= c:::==================================:::::i 4.96,

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Atomic & Nuclear Physics

B 2. D 3. C 4. D 5. C I -----------·----------------------------'· C 7. A 8. A 9. B 10. B

I

1. ~6.

7___,

L_21_. ~ _ _ _ _ _ _ _1_2._c _ _ _ _ _ _ _1_3._c _______ 14_._0_ _ _ _ _ _1_5_._s ____ 16.

A

17,

D

18.

A

19.

A

20.

D

21.

B

22.

D

23. ' A

24.

A

25,

B

a

A

~

A

a

1.

C

2.

C

3.

A

4.

A

5.

C

6.

D

7.

A

8.

B

C,1.

9.

D

A

12.

B,

13.

A

28.

J

A

10. D - -C - - - - - - 1~ - -A- ~ - - - - 1~ -------------- C __, 16. B 17. C 18. D 19. A 20. B ------------···------------ - - - - - - ---------_ _ _2_4. -s-----·-2'--5-_- c - - - ' - - 7 21. A 22. B 23. C

l- - - - - 26. B

r---

~ ' - - C ----- 36. B ;-- 41. C

27. ~

---~ 42,

~--------

A

29. C 30. C ---------K D a C j --------~ B • A

-~----~---·-·-· D - - - - - -~ - -B- C

~

D

D

43.

A

--------

____

---4-4-.- B - - - ~ - - - 45.

C R A • C ~ C ------------L. 51. B 52. B ------'-~~ __D_______ _!;~:.,._A_______

a

C

~

A ' - ~--- - - - - - -~-

B

L!.:: ..E _ •

A

-

B

D

----------~---~

B

~

C

--~=----_-________ _m

A

C

j

A

55,

~

---

------------------, ~ D

~ C KA -- - - - -------------

C .

~ C ---1-2-.- c - __ :_-_--~--

~

~

B

A

_:·-,_-a-.-_-_o_______-__:::_:::_-=----_-__-_-_-_

ANSWERS•-· 1.

6.

A- rel="nofollow"> (p, q) B-> (p, r) C ->-(P, s) D-> (p, q, r)

2.

A-> (p) B .... (q) C-> (r) D-> (q)

7.

A-> (p, r) 3. A-> (r) B-> (p, q, s) B ->,(p) C-> (p) C->(q) D..; (q) D-> (q) ______________ E-> (s) A-> B .... C -> D->

(p, r) (q, r, s) (q, r) (p, r)

A->(q,r,s,t) 12. A-;(q) 8-> (q, r,,s, t) 8-> (p) C-> (q, r, s, t) C-> (s) ~---D_->_(~p,_q~,_r,~··~t)_ _ _ _

11.

8.

·4,

A-> (p, s) B-> (q, s) C-> (q, s) D-> (s)

--··-·~-- - - -

A-> (r) B .... (p) C -> (q) D-> (q) '-"----

£.=.i9,__________

9.

5.

A-> (p)

----~,

B-> (p)

c ....!_Jq)1 D-> (s)

-------·--10. A-> (p, q, ~. t)

A-> (p, s) B -> (p) C-> (q, s) D -> (p)

B-> (r, s, t) C-> (q, r, s, t) D -> (r, t)

-----~

================================== = 4.97

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Advanced JEE Physics

I1. 5. I 9. 13. \11. 21.

Optics & Modern Physics

8

2.

79

3.

100

6.

800

7.

4

10. 8

68

·5

4.

·41

11.

31 1.

14. 4

15.

(a) 4, (b) 40441

16. 24

26206

18. 6

19.

(a) 300, (b) 2, (c) 2

20.

(a} 4, (b) 396

22.

8.

12.

i

.

.

114

560

/

= ================================= 4.98

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1.

(a)

E3 -E1

(-13.6)(3)' , (3)'

t.

[(-13.6)(3)']=lOB. 8 eV (1)'

=> (b)

=

=>

t.

=t.(~:=;J

=>

i. =1.88 x 10" ( 57 -l)' 29-1

=>

t. = 7.52x10"

Number of lines in emission spectrum is

N = n(n-1) 2

N . (3)(:-1)

=>

3 6.

2.

S'

mce,

E

n-

(13.6)2' n2

e

Since, A=

V,

4.

IE= 122.4 eV

Energy .of electron in ground state of hydrogen atom is -13.6 eV . Earlier it had a kinetic energy of 2 eV Therefore, energy of ·photon released during formation of ,hydrogen atom, / &E = 2-(-13.6) =15.6 eV

1. = 12375 = 12375 = 793.3 &E 15.6

A

7.

=>

0.26 = (12375)(3~)

=>

V = 15865 volt

E E

= • - E'

(-13.6)(3)' (4)'

i. . (in A)- 12375 mm V (in volt) 8.

(a)

=> (b)

=>

[(-13.6)(3)']=2.754 eV (4)'

(c)

2

So, E = E, -E, =(13.6)(4)(.!-J...) = 10.2 eV , · 4 16

Hence stopping potential is 0.754 V

5.

Using the equation,

Ji= a(2-b)

IE=2'(13.6 eV)=(2)'(13.6 eV) IE = 54.4 eV

N=n(n-1)= 6

Again using Einstein's Photo-Electric Equation, we get

Kmax =E-W=0.754 eV

2'(13.6)(.!-.!)= 12375 4 9 1654 2=2

Therefore, it fs helium atom. Ionization energy is

For longer wavelength, we have

(-13.6)(3)' (5)'

12375 _ 0_ 31 A

40 X 103

As the atoms finally emit radiation of only 3 different photon energies final excited state corresPonds to n = 3 . So, the initial excited state corresponds to n = 2

=>

[(-1~:i!(3)']=5.95 eV

Using Einstein's Photo-Electric Equation, we get W =E-K= = (5.95-3.95) eV =2 eV

&E=E -E ' '

(.!._1_) V 1.5V

Cutoff wavelength Am1n is given by,

For shorter wavelength, we have ,i

V,

where i., - i., = 26 pm = 0.26 A

(1)'

=>

12 75 ~ , so we have

i., _ i., = 12375 _ 12375 = 12375

IE- (13.6)(3)' eV

3.

Hz

V

So, ibnisation energy, n =1, for u++ (Z =3) is

=>

(2"'-1)' Zcu -1

\:u

(b =1)

9.

Energy of photon corresponding to i. = 6563 A is

Since, b-= 1, so we get

4;99

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Optics & Modeni Physics

Advanced JEE Physics AE=

12375 eV=1.88eV 6563 ~--~-~-n=3 1.88 eV n = 2

Since, K=·-½mv 2 =4.352x10-

V=

110

· 2x4.352x1'0-18 9.1x10-31

=>·

v=3.1x10 6 ms-1

(a)

From figure, we observe that

12.1 eV 13.

lonisati?n) =15 _6 eV ( Potential

~------n=1 This is the difference in energy between n = 3 and n =2 .

10.

(b)

'

(e)

AE 31 = -3.08_-(-15.6) = 12.52 eV

Given E3 =0

(d)

,, =460 A =>

E, -E, = 12375 =26.9 eV 460

Since, E3

=>

0-E1 =26.9eV E, = -26.9 eV

=> 11.

E3 -E 2 = 12375

-~--~--n=3 1'-'''---n =2 -h---'-

E2 -E1 =10.3 eV > 6 eV

(i)

(ii)

E, -E, =10.3 eV <11 eV So, the electron can excite the atom.

z'(..!._..!.) 22 nf

14.

(1-..!.) A1 1 n2 he =E z'(1-..!.) A. °1n~ o

(a)

... (1)

z = 3 for

u+2 • Further we know that

Substituting, n = 3, z = 3 and a 0

n' z

rn = -a 0

=0.529 A , we get

r3 for

u+2

... (2)

(3)' r, =w(0.529)

Wavelength of emitted light is

he= E

K•• =(11-10.3) eV=0.7eV

=>

When hydrogen atom is excited, then we have

o

A-'

,,,

1035 E2 =-12eV

eV=E

12.52 - 12375

Hence, the striking electron cannot excite the hypothetical atoms. So the electron will keep its energy with itself. => Krnrn=6eV

A =12 eV

(1-..!.)

A-'_

....:!...~1.01x10 7 m-1

=> (e)

eV = Eo 1 n2 When ion is excited, then

5.3

1 _. AE,, ,,, - 12375

- ' U . - - - - - n =1

Further, ,, = 1035 A =>

min

Therefore, excitation potential-for state n = 3 is 12.52 V

=O

=>

= 12375 = 2335

A

Hence the single electron in the hydrogen atom should excite at least upto n =3 and for this the minimum energy of the striking electron should be 12.1 eV. ·

(b)

z = 2 for He+ . Also we know that

... (3) Vn

... (4)

A =1.587 A

z

= r,V1

Substituting n=4, z=2 and v,=2.19x10 6 ms-1 ,weget

2

for He+,

Further it is given that

!:1. =~ ,, 1

v 4 =(¾)<2.19x10') ms-1

... (5)

=>

Solving the above equations, we get

v4 =1.095x106 ms-1

2=2, n=2, n,=4 and V=10.2V _Energy of emitted photon by the hydrogen atom is

15.

AE = E, -E, = 10.2 eV

Magnetic moment µ=NiA=(f)(rcr

and by the ion is

=>

AE' = E, -E, = (13.6)(2)'(1- ~) = 51 eV 1 12.

µ

=(-e_-)(.r')= evr 2rcr/v

2

We know that mvr = nh

2, Solving equations (1) and (2) neh

Energy of photon of the flrst line of Lyman series is

E = E, -E, = (13.6)(2)'(1-¾) = 40.8 eV

2

)

... (1) ... (2)

µ = 4rcm

Energy required to ionize the hydrogen atom is 13.6 eV. Therefore, kinetic energy of electron emitted from the hydrogen atom is

K=(40.8-13.6) eV=27.2eV =>

=4.100

K=4.352x10-18 J

Magnetic induction, B = µoi = µoe 2r 2rT

=>

B=

µ 0ev (2r)(2.r)

µ 0ev 4.r'

... (3)

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Atomic & Nuclear P/zysics From Newton's Second Law, we have 0

2

mv2

4m;:/! =-r-

Orbit

K(eV)

U(eV)

E(eV)

First

13.60 3.40

0

13.60 23.80

Second

e'

=>

v2=-4m:0mr Solving these equations, we get µ nm2e 7

".(4)

19.

20.40

Wavelengths corresponding to minimum w.avele_ngth {Amin) or maximum energy will emit photoelectrons having maximum kinetic energy.

B=-'-8E0h5n5

{Amln)

belonging to Balmer·series and Lying in the given range

(450 nm to 750 nm)

16. The force at a distance r is,

corresponds

to

transition

from

(n=4 to n=2). Here,

F=-dU =-2ar dr

V -E4 = -13.6 2 = -0.85 e (4)

Suppose r be the radius of nth orbit. Then the necessary centripetal force is provided by the above force. Thus,

mv' -=2ar

13 6 and E =- · =-3.4eV ' (2)'

... (1)

r Further, the quantization of angular momentum gives,

=>

IIE =E, - E,

So, Kmax

nh

mvr=-

=>

... (2)

2,

=2.55 eV

=Energy of photon-work function

K==.2.55-2=0.55eV

Solving equations (1) and (2) for r , we get

r =( n'h' Sarni' 17.

(a)

20.

)v•

=>

(Kinetic energy K,) =(Total energy)-(PE)

=>

K,=2Eo-Eo=Eo

=>

1'--h~ ' - ~2mE,

Given that

2 {12375 X 10-,0} =____1:_-,---,'34,3 x l O' 1.09x10'(Z-1)'(1-¾) (b)

For Q::::;;x::::;;1, PE=E0

For X>1, PE=O

=> Z=42 Energy of K, line =100-24=76 keV =>

1' = 12375 Ka 76x10 3

=>

1',. =0.163 A

... (1)

:::::)

Kinetic energy K 2

=>

A

= Total energy

= 2E0

h

' ~2m(E,) From equations (1) and (2), we get

A

... (2)

~=-,/2.

",

18. Since, E, =-13.60 eV

21.

Also, we know that

(a)

TE=-KE=PE =>

2 K1 =-E, =13.60 eV

=>

U1 =2E 1 =-27.20 eV

Further, E0

Kinetic energy of electron in the orbits of hydrogen and

hydrogen like atoms So, Kinetic energy

(b)

=[Total energy[

= 3.4 eV

The de-Broglie wavelength is given by 1'=_11_=_h_ p ,/2Km

13.6Z' =---, -

where K is the kinetic energy of an electron Substituting the values, we get

n

1'(6.6x10,.. Js) - ~2(3.4x1.6x10·" J)(9.1x10'" kg)

=>

E =_§_=-3.40 eV ' (2)'

=>

K,,, 3.40 eV and

::::)

A=6.63x10-10 m

=>

U2 =-6:80 eV

=>

1' =6.63 A

Now U1 = O , i.e., potential energy has been increased by 27.20 eV. So, We will increase U and E in all energy states by 27.20_eV while kinetic ·energy will remain unchanged. So, we have

c:=================================4.101

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Rahul Sardana with www.puucho.com

1.

In 1 second 90% of the nuclei have remained undecayed, so in another 1 second 90% of 90 i.e., 81 nuclei will remain

After 70 hours actiVity A=R 0e-11

undecayed.

2.

1 Since, N = 0-3 x 6.02 x 1023

210

=2.87 x 1018

During one mean life period 63.8% nuclei are decayed.

R1

R, =>

A1N =

14 3 ·

is

=2.92x1017 per sec

'

~~rcentag8 ~f ). = 3.7~10" x100= 1.26 x 10-11 % ( ActIvIty Transmitted 2.92 x 1017 .

= 5.06 x 1021 7.

1 N=--x6.02x10" 109

(a)

A=~ 2.5x104 4.94x10-1e s-1 N· 5.06x1021 => . tV2-- 0 ·693 -14 ). - • X 10" 5

4.

0 693 -( · x24)(ro)

In fruits, the activity was observed 1 µCi or 3.7x 10" per sec·.-..,. Therefore, percentage of activity transmitted from root to the fruit

E =0.638x2:87x1018 xs:3x1 .6x10-13 J= 1.55x106 J N = __g_x 6.02x1023 238 Since, R = i.N

•

A=(3.37x1017 )e

:::::,

Hence, energy released is

3.

R0 =3.37x1017 per sec

=>

R=AN

3

1 x--x6.02x1023

R=1.42x1014 peryear

=>

1.2

O.BQ

2.7x10' 109

_ .

(b)

After 2 year, R =Roe-11 = (1.42 X 1014 )eE--0.693/2.7•101](2)

(c)

R=1.41x1014 peryear After 2 half lives activity reduces to 25% of the original value, so we have

=>

i.,N = 98.8

"-2 = 82.3.3)..,

... (1)

0 693 Further, i. = · year·'= 0.0318 year·• 21.8

t = 21,,, = 5.4 x 1O' years

... (2)

Also (-dN)=(- dN,)+(-dN;) 'dt dt dt => AN = A,N + A2 N

8.

Since N = "- 1N° (e-li1 -e-l.z1 ) '

=> A=A1 +A2 Solving equations (1), (2) and.(3), we get A, =3.81x10-4 year-1 and A. 2 =3.14x10-2 year-1

... (3)

2

"-2

-A,

When _A1 » A2 we have, N2 ~ N0 e-1.z1 Physically this means that parent nuclei practically instantly transform into daughter nuclei, which then decay According-to the Law of Radioactive Decay with a certain decay constant.

N, 5.

(a)

(b)

6.

R=R0e-" :::,.

2700 = 4750e-s1

=>

i..=0.113min-1

tv2

=-i.- = 6.132 min

~N, ------------ ---i.,

0.693

R, =l.N R0

°·

693

14.3x3600x 24

x 6.02 x 1023 per sec

=4.102c::=================================

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Atomic & Nuclear Physics -N - !:I_ N -,, 2 oe

12.

'

Let x the desired ratio, then, mass of Co 58 in 1 g be x

i.,

When A, « A2 , then N2

N = ~ x 6.02 x10 23

=>

r::: !:l.N0 e-'"'1

i.,

58

Given, 2.2 x 1012 =AN=

i.e., N2 versus t graph in this case is as shown in.the figure.

=> 9.

x ~ x 6.02 x 1023

58

X=1.88x10-3

Probability of a nucleus to decay in time t is _.!:!_ = 1- ~-U

N,

10.

0 693 ·

-

71.3x24x 3600

(a)

(b)

=1- 8-(flO)(S) = 0.39

13.

Since, R = J.N - Nlog, ( 2 )

Iv,

==>

R =R,e-" =>

7.3 =s.se-1..<4500- 25001

=>

A =1.21x10-4 year-1

=>

0.693 Iv, = -,.= 5724 years

R oc _!i_

1,,

N, . G,ven, N,

=>

Ro

= Reu =(7.3)e(1.21-10.:.i)(45ool

R0

= 12.58 dismin- 91

1

d an

(tv,),

14

(Iv,), = 25

=>

R, N, (t,,), 4 25 100 -=-x--=-x-=-

=>

R,=~~~x3=2.63mCi

and

R, =3-2.63=0.37 mCi

R,

Further applying, R =R,e-"

==>

4

=1 N,

(t,,),

1 14

14 ·

After 60 days, we have 10

11.

N=!:!,_ 5x10:-3x3.7x10

A

(

0.693 ) 138x24x3600

=>

m=( 3.2x10" )(2 lO) 9 6.02x10"

=>

m:=1.12x10--a ~

_x , 3 2 10 5

R;

_0.693,.60 14

=R,e-i..1 = (2.63)e

= 0.135 mCi

0.693 60

R;

= R2 e-i..21 =(0.37)e--g-~ =0.07 mCi

So, total activity is

R' =R; + R;

= 0.205 mCi

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1.

The reaction is possible (spontaneously) if th!3 binding energy of products is larger than that of 2

:u .

(a)

4.

Total binding energy of •::u=238x7.57=1801.66 MeV

m=

E

c'

4.2x10' k (3x108 ) ' g

m =4.7x10-14 kg

::::-

2

and binding en~rgy of :Pb=206x7.83=1612:981 MeV the binding energy of products is less than that of 2:U .

(b)

2.

5.

=>

v')(M;}

6.

O=-1.20MeV

Kmtn

7.

=1.54 MeV

=>

(KE),,= 6.082x

4 MeV 208 (KE),, =0.1308 MeV

Please observe that here, atomic masses are given (not the nuclear masses), but still we can us~ them for calculating the mass defect because mass of electrons get cancelled both sides. Thus, Mass defect
o =(0.0047u)(931.5 Mevu-')

0=4.4 MeV Hence, the energy of beta particles can range from O to 4.4 MeV.

Km,, =loi(:: + 1) =>

(KE),, =(KE).(~)

=> =>

Hence, the minimum kinetic.energy of a-particle to initiate this reaction is give_n by

K =(1.20)(4.00260 +1) "" 14.00307

=>

=>

Q value is negative, so the reaction is endothermic.

=>

v = M.. v..

(KE) =.!Mv' =.!M(M;v;)=(.!M Tl2 2 M2 2aaM

(931.5M~V)

=>

=mv

M Kinetic energy of Tl atom Is

Since, the masses are given in atomic mass units, so, we shall first calculate the mass difference between reactants and products in the same units and then multiply the mass difference by 931.5 Mevu-1 • Thus, we have

=(14.00307u + 4.00260u -1.00783u -16.99913u)

Law of Conservation of Linear Momentum, we haVe M.. v..

Note that the protons and neutrons are free and do not have any binding energy. Hence, this reaction is not possi~I~ spontaneously. It can take place only if 1801.66-1612.98=188.68 MeV of energy is supplied from outside. In the second case binding energy of products is larger than the binding energy of the parent nucleus. Hence, reaction is possible spontaneously.

0

By

o =Jm; ~(m; +m,)Jc' where m~ and m~ are the nuclear mass of

11

c

and

11

8 , so

m8 = m~ +Sm 0

3.

Total energy released is 1 tlE = - -x6.02x1026 x185x 1.6x10-13 J = 7.58x 1013 J 235 Power P =100x106 =109 Js-1 Therefore, time 7 5 1013 I= · ~~9 sec=B.78 day

me =m~+6m8

=>

O=[(m,-6m,)-(m,-5m,+m,)]c'

=>

0=[mc-m8 -2mp]c

2 {-:

(~:::~)=(;

0:~::~)}

0 =(11.011434-11.009305 -2 X 0.0005486)931 0=0.961 MeV The disintegration energy is equal to the maximum energy of the emitted photon.

=> =>

= ================================== 4.104

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Atomic & Nuclear Physics 8.

.6.m = 2(mass of 1H2 )-(mass of 2He 4 )

=>

P.;' =Pi +P~ +2P0PHCOS$

=0.0256u

2mctKa


=2moKo + 2mHKH

~-~--~-

+2)(2m,K,)(2m,K,)cos$

Total energy required per day is

E = 200x10 x24x3600 J =1.728x 10 6

13

... (1)

J

Let m be the mass of deuterium required. Then energy required for reactor is

5 2 E' = ( ~ )( m~ )<6.02 X 10") (23.85 X 1.6 X 10-") 1 0 This should be equal to 1.728 x 1013 J 4 X 1QQ X 1. 728 X 1Q13

By Law of Conservation of Energy, we have

m 25x6.02x10 23 x23.85x1.6x10-13 g m=120.35g 9.

0

(a) Total number of atoms in 1 kg of U238 is 1 N =--x 6.02x1026 =2.53 x10 24 238 => Total energy released, (b)

m =>

10.

K,=O+K -K,=(-1.18+7.7-5.5) MeV =>

LlE=8.09x10 13 J

K 0 =1.02 MeV

$~43°18' 11.

Since, .6.E=(8mP+Bmn-m 0 )x931.5 MeV

=>

8.09x1013 =2.7x109 30x10' g g

;.E = 127.6 MeV

m=~.7x10 6 kg

(a)

Q-value=(
(b)

By Law of Conservation of Linear Momentum, we have

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1.

[BJ Let N be the number of nuclei at any time

t . Then

dN =200-).N dt N

dN

f200-AN

'

A= 0.693 3.8

I

'

N= 200(1-e-•) Given-thatN=100 and ).=1s-•

=>

100=200(1-e-')

=>

.-,

=>

t = log, (2) s

10

1

~{

log10= 0.6931 3.8

=(i)

2.3= 0;69\ 3.8 t =12.62 days

[CJ

8.

[DJ Three half-lives of A is equivalent to six half-lives of 8 . Hence, we have

9.

[AJ

AA=..!_ ..

N,

- = e 3.s 10 --

).

3.

_!i.=J_

Since

fdt

2

Probabilities of-getting a and p-pa.rticles are same. So, rate of disintegration are equal.

4.

= ABNB

=>

A.ANA

=>

NA =Ae =2 NB 'A.A

nh mvr=2•

[DJ

B n

= µoln

=>

•2rn

-"--= (2nr) mv n

where, -

h

mv

10.

-

=de-Broglie wavelength

[AJ Fraction of nuclei which remain undecayed is

f=_.!!=Noe-11 =8-11 No . No

~)(!) -1.,_,,, T 2=02

f=e -(

f=-1-=_1 e'og../2

../2.

5.

=4-106

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Atomic & N11clear Physics 11.

[CJ Since, EK. < EK,

17.

[AJ According to Ritz Combination Principle {form< i < n}

=> 12.

AK.. > AK,

e.g.,

v 4 ....1 =v 4 .... 3 +v 3 ....1

[DJ

Since, ,l,l.=l.,. -l...

. .. (1)

When V is made half, Amin becomes two times, however AK.

19.

remains the same. => I::,.'),.,' = AK. - 2/1.mln

[AJ

,:=R,[n;- (n~1)'] .!_=R [ (2n+1) ]

From (1 ), we have Amin

=>

or

Y4 ....1 =V4 .... 2 +v2 .... 1

= AK.. -

A.

H

/::,,.'),,.

!-R )..-

tit...'=AK. -2AK,, +2AA

=>

tiA'=2M-AK.

=>

M'<.2LU

n2(n+1)2

2 2 n {··• forlargen, 2n+1=2n n2(n+1) =n4 } Hn4

1

2

i = Ra;;, v=~=2RHc

13.

[BJ Assuming that ionization occurs as a result of a completely inelastic collision, we can write

n'

l. 21.

mv,=(m+m")V

[BJ M

_v

where m is the mass of incident particle, mH the mass of

,:-=+c

hydrogen atom, v 0 the initial velocity of incident particle and V

{-) sign to be used for approaching (+) sign to be used for receding

the final common velocity of the particle after collision. Prior to

'

E _mvo ,- 2

,l,l.

=>

collision, the KE of the incident particle was

l.

E=(m+mH)V

2

23.

2

m v~

2

[DJ

Let R0 be the initial activity. Then

2(m+m,)

R,

The decrease in kinetic energy must be equal to ionization energy, so we have

R2

E; =E, -E =(~)E,

=R0 e-At,

and R2 = R0 e-i.12

= e1,.(11-1,J

R,

m+mH

=>

V

+-

V=2x10 7 ms-1

=>

The total kinetic energy after collision

706-656 656 C

R2

=R1e1.,(1i-1,J

§..=~ E0

m+mH

24.

i.e., the greater the mass m , the smaller the fraction of initial kinetic energy that will be used for ionization.

[DJ

R,=AN1

N _R,

i-T

Time lapsed

T,o

N - R2 ' - l. So, number of atoms decayed is N=N1 -N2

4

Ti~e lapsed =400 µs

N=(R't')

=> 16.

[CJ E=mc2

25.

[BJ After two half lives ..!.th fraction of nuclei will remain undecayed. 4 Or, ~th fraction will decay. Hence, the probability that~ nucleus

Since "!... « 1

4

C

=>

2

decays in two half lives is

E=m 0 c (1+;;2 )

.·

¾·

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Optics & Modern Physics

Advanced JEE Physics 26.

[CJ The necessary centripetal force to the satellite is provided ·by the gravitational force of attraction. mV2 GMm i.e. -r-=~r,-

l

33.

= _.!_ 9).

[BJ ).=~

eV,

.!mv' = GMm = KE 2 2r

6,626 X 10-24

).

PE of the satellite is given by

X

3 X 1Q 8

( 20x10' )( 1.6x10-")

PE=-GMm r

).=0.62 A

Now KE+PE=E 0

=>

35.

GMm _ GMm =E ,2r

r

0

GMm=E 2r

or

;::::)

'

_ GMm =2E

And (-dN) =).,N dt

=>

Decay rate of Y > Decay rate of X

39. [DJ Since, rn oc n2

=> =>

rn+i -rn

[DJ

0.000 1 x 1O_, = ( 0.5 x10-" )n' 2

=r11_1 {

(n+1)' -n' =(n-1)' n=4

[A]

3h (h) S.mce, -=n.2rt n=3

42.

21t

;::::)

n2 =1000

;::::)

n = 31

Because -~dius of nth orbit is ~ual to} r0 =n r0 where r0 =0.529A

[A]

.!oc(Z-1)' ).

13 K =JS_= ·6 =1.51eV " (3)' 9 31.

y

(-:),>(-:),

=>

~ 0.53 A

). = 2n(0.53) A= 3.33 A

Given,

=).,N

di '

II.= 21tr

29.

Av >Ax

[DJ where r = Radius of first orbit

28.

Av

Since, (- dN)

r ' Which is equal to PE.

27.

[CJ 0.693 Ax

=>

~=(z,-1)' Z1 -1

A. 2

Solving this, we get, 2 2

[A]

1240 E, = = 2.25 eV 550

43.

or

1 4=

(z,;_1-1)'

=6

[BJ The continuous X-ray spectrum is shown in figure.

E, = 12400 = 2.75 ev 450 1240 E, = =3.54 eV 350

E,_

E1 cannot emit photoelectrons from q and r plates. E2 cannot emit photoelectrons from r . Further, work function of p is least and it can emit photoelectrons from all three wavelengths. Hence magnitude of its stopping potential and saturation current both will be maximum.

32.

-+---'---~---). )...

All wavelengths > "-m1n are found, where ).m;,

[DJ· N1 =Nae:1011 N2

;2375 /

V in volt

Here, V is the applied voltage.

= N0e-;.1 ~ = ..:!. = 0 c-10J.+).Jt N, e

=

44.

[BJ Using

2 N = N0 e-u where A= 1090

~"

9).t = 1

= 1090 { 2 ) 3.8

= ================================== 4.108

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Atomic & Nuclear Physics N _Q_=Ne

rog.(2)t

°

20

9750 2 3 A.=..! x 2.310910 = · = 0.461 min-1 5 · · 975 5

s.s

Solving this equation with the help of given data we find

t = 16.5days. 45.

53.

~ =(i)'

[DJ The binding energy is numerically equal to the kinetic energy of the electron, so

1 ' =E1 -mv 2

... (1)

Since, mvr = nh

... (2)

2.

N

54.

[CJ Since nuclear density is independent of mass of nucleus, hence all possess equal density.

56.

[CJ

Dividing Equation (1) by Equation (2), we get

_y_ = 2E1 2rcr

nh

If R0 be the initial activity of the sample, then R1= R0e;.li and R2 = Roe-)J2

f = 2E,

nh 46.

47.

where A.=

[BJ Magnetic moment e Angular momentum 2m => Magnetic moment cc Angular Momentum

=>

Mccn

[DJ

R

R, =R, exp 58.

[CJ Number of radionuclei become constant, when X =l.N

[DJ

=>

R =: "-1N1e-;.,1A.2N2e-'-:!'

=>

R = A.1N1A.2N2e-(i..,+l2J1

[DJ

2,

Since decay is to be regarded as a statistical spontaneous process, hence a. decay can be regarded as an Adiabatic process.

60.

=>

L=v'S6(2:)

[BJ ~ = n2Ro

61.

R,

=>

R,=2.112A

[CJ

V0

U, = 1793.6 MeV 62.

1(cZ) n

=137

[CJ Since Lip=!! ).

=>

l.=!lo (R')=2.31o (Ro) tg"R tg"R

= 4( 0.528 A)

=>

Speed of electron in the nth orbit of H like atom is

[BJ Total initial binding energy is 236x 7.6

R = R0 e-;.1

and for sample 2,

• h L=JnCn+1)-

[CJ

[CJ

Ri = A.1N1e-;.,1

When they are mixed, we have R = R1R2

=>

51.

~t,)

R2 = "-2N2e-;.,1 .

[BJ Cut-off wavelength depends on the applied voltage not on the atomic number of the target. Characteristic wavelengths depend on the atomic number of target.

Total final binding energy is 2( 117)( 8.5) => U, =1989 MeV =>
e,

Rate of decay of sample 1,

59.

=>

e-;.t,

/·.- L=n;.)

N=-X-=~ log, (2) log,(2) y

50.

).

-;.12

R1

).

49.

{·.- Mean life = T =.!}

~=!__=e;.(1,-12)

N=.".

48.

i

mv=hR(!--1...) H H 1 25 V=hR,(24) mH 25 v=4 ms-1

Decay constant

=================================== = 4.109

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Optics & Modem Physics

· Advanced JEE Physics 63.

vHa". =a(c(22 ))=ac

[CJ As velocity (or momentum) of; electron is fncreased, the wavelength

(A.=%) will decrease.

Hence, fringe width will decrease ( ro oc A). 72. 64.

In beta decay, atomic number is increased by 1 whereas the mass number remains the same. Therefore, following equation can be possible

p=-i::

Kinetic energy of striking electrons p2 h2

~Cu 4:zn+_1e0

65.

[AJ

Momentum of striking electrons h

[DJ

K=-=--2

2m 2mA. This is also, maximum energy of X-rays photons.

[CJ

Activity R = ,N Number of nuclei (N) substances, so we have R oc,

he h2 ~= 2mA.2

per mole are equal for both the ~

,._ _ 2mA.2c 0 h

=> 73. 66.

Activity of

[CJ

R(J...-J...)=R (_!l_nf __!l_) n! H

22

42

.

H

Substituting n1 = 6 , n2 67.

=12

(z)'

aoc

A., N2 "-2 = 2N1

=>

a oc z3

=>

~=(~)' =8 a,

~ 74.

75.

[DJ

Net rate of formation of Y at any time t is X

69.

AXNX

-A.YNY

[BJ Total power drawn by Coolidge tube is VI . ~ P1o1a1 = VI =200 W Since only 0.5% is carried by X-rays, so

=>

= A.YNY

=>

[DJ

!:,.

(¾-~)

., n-¾)

T, =4 T,

P=u~~)P• .,

Maximum wavelength of Lyman series will correspond' to the transition of electron from n = 2 to n = 1 and maximum wavelength of Paschen series will correspond to n·= 4 to n =3.

=>

IT =half life = 10 ~ 2 )

[AJ

. . dN NY 1s maximum when rif =0

=>

2N 1

T2 = ~

Rest mass of parent nucleus should be greater than the rest mass of daughter nuclei.

1

dN, N dt= A.

T,

s, )

Given N1 = 2N2

{torn= 1)

(l)

(activity of

=> =>

=>

2

,,N, =21 (,,N,)

[BJ

v'

s, = .:!.

=> makes both sides equal

a=-r

68.

[AJ

76.

1

P=-(200)W 200 P=1W

[BJ

e2 cz

7

108

71.

e'

where u = - 2he0c

78.

[CJ

Total energy radiated per minute from sun is E,adlate<:t

is the fine structure constant ( u = ~ ) 1 7

=4.110

= er( 41tR~)

Energy radiated annually is given by

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Atomic & Nuclear.Physics Etotal = 24 X 60 X 365 X Erad1ated

=>

Annual loss of mass =.6.m=

E1

f

1

87.

[AJ Charge number decreases by 4, but actually it must decrease by 8 (due to emission of 4a-particles ). Hence 4p particles must have also been emitted.

88.

(CJ

=1.38x1017 kg

C

80.

[CJ . If it was A

Y-;22 Reactant: R=60x8.5=510 MeV Product: P = 2 x 30 x 5 = 300 MeV 6E =-210 MeV ENDOTHERMIC If it was B W->X+Z R = 120x7.5 = 900 MeV P = 90x 8+30 x5 = 870 MeV 6E=-30MeV ENDOTHERMIC If it was C W-;2Y R = 120x7.5 = 900 MeV P = 2x 60 x8.5 = 1020 MeV 6E=120 MeV EXOTHERMIC If it was D X-> Y +Z R = 90 x 8.0 =720 MeV

P = 60 x 8.5 + 30 x 5.0 = 660 MeV 6E=-60 MeV ENDOTHERMIC 81.

.1.=R (.1.-J...) A " 4 16 Since c= v"A. C

V=-

A

V:::

89.

84.

E=9x1010 'J

)

h(24) p =.§_=R C • 25 Recoil momentum of H-atom will also be p. mv-=p

p

V=-

(1.097x10')(6.626x10-")24

m

=> 90.

(25)(1.67x10·") 1

V=4.178ms-

[BJ E =-n2e 13.6 V

=>

E1 =-13.6 eV

=>

E, =-3.4 eV

=>

E, =-1.50 eV

=>

E, =-0.85 eV

From above we can see that E3 -E1 =12.1 eV i.e. the electron must be making a transition from n = 3 to n = 1 level.

[CJ N ( -1 )" Since-=

N,

where n

=>

2

Time lapsed Half life

6L=(3-1)_11_='1_

2,

6L

Tv, According to the problem n = 2 - = ..:!. Tv2 2

86.

6

Momentum of photon emitted is

[AJ E=mc2

=>

X( /

E=R• hc(1-J...) 25

A

E=(10-")(9x10")

7

[CJ

s·mce

=>

X 10

16

f=6m=M-A

83.

8

v=J!._x10 15 Hz

[DJ

A

3 X 10

[AJ M=235-219=16 Number of a-particles emitted = ~= 4 4

=> 91.

•

6.626 x 10-" 3.14

.6.L=2.11x10-34 Js

[AJ For the fifth electron to act as a dopant it must lie in the valence shell of P.

So n = 3 . (TRICKY HINT!) Z = 15 . n2h2&0 n2 Smee rn = - - , = - r0 1tme Z Z where r0

= 0.529 A= 52.9 pm

=>

15 (52.9 pm)

r, =

9

(for Phosphorus]

... (1)

When phosphorus acts as dopant in silicon, the expression for Bohr radius is

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Advanced JEE Physics

Optics & Modern Physics 98.

, n2h2E [ =--=kr: n ·1rme2z n

h

92.

r; =12r, =12[ 1~(52.9)pm]

=>

r; = 380.88 pm

A2

he

99.

[BJ

Since K.E. = -T.E.

[CJ The first photon will excite the hydrogen atom (in ground state}

excitation a photon of 10.2 eV will be released. The second

=>

K.E.=+13.6eV

100. (CJ

En

=-me' ~,m Bn h

photon of energy 15 eV can ionise the atom.

= mass of electron

E0

mhypc!hetical e4

Hence the balance energy i.e., {15-13.6) eV=1.4 eV is

=>

E

retained by the electron. Therefore, by the second photon an electron of energy 1.4 eV will be released.

=>

E = _2(Rhc) " n'

=>

~~ ) = J.N

Therefore, decay process lasts upto

n

Amruc

94

Amruc

=>

2(_11_) 2.t

photon

~=E 3-E 2 =-2Rhc(.!_.!)

).

[CJ

2rt

8n2h2 s~

~=2Rhc(~)

t 4 co Therefore, a given

nucleus may decay at any time after t = 0.

mvr = n~ =

=

The longest wavelength (or mmImum energy) corresponds to transition between adjacent states. i.e. n=3 to n=2

[DJ Number of nuclei decreases exponentially N =N0 e-.u and

Rate of decay ( -

94.

!:l.a:E112 i.,

E

in first excited state (as E2 -E 1 = 10.2eV ). Hence, during de-

93.

or

~= ../2mE

{·: &=k<,)

r~ =12rn

=>

[BJ

=

36

=~

SR

102. [BJ According to Ritz Combination Principle Ec ... A =Ec...a + Ea...A

h mvr=n So, de-Broglie wavelength is ~

he he he -=-+A3

A. 1

A. 2

i. = _t,_ = nr = (3.14)(2.116 Al= 6.64 A mv

95. [CJ

103. [DJ For each principal quantum number n, number of electrons permitted equals the number of elements corresponding to the quantum number.

Following nuclear reaction takes place 1 0 0 n 4,H1 + _ 1e +V V is antineutrino.

96.

=>

[DJ

of Elements

Shortest wavelength will correspond to maximum energy. As value of atomic number (Z) increases, the magnitude of energy

97.

(Total Number)= '-'2n' 'n(n+1)(2n+1)

104 [BJ

in diffetent energy states gets increased. Value of Z is maximum for doubly ionised lithium atom (Z=3) among the

inR3 a:A

given elements. Hence, wavelength corresponding to this will be least.

=>

[DJ

n2 h2 Smee, r = -'-,e ,m •

E

3

'

Ra:A 3

105. [CJ The nuclear force of interaction between any pair of nucleons is identical i.e. force between two neutrons (F2 ) equals the force between neutron and proton

L=n(;J

=>

nh=2•L e, (2,L)' r- 2

=>

Lr-2 = constant

=>

=4.112

e 1tm

3

L,

( F3 )

•

However, between two

protons net force is equal to the resultant of nuclear force between them (attractive in nature) and electrostatic force between them (repulsive in nature). Hence F1 is a value lesser than F2 and F3.SO F,
'

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Atomic & Nucleilr Physics 107. [C]

113. [BJ Given that K, + K, = 55 MeV

dN=n-AN di Because the population N is simultaneously increasing at rate n and decreasing due to decay at rate AN.

=>

f~=fdt Non-AN 0

=> =>

.!.en(n-ANo)=t l. n-l.N

P1 =P2 =>

~2K,(216m)=~2K,(4m)

as =>

P=-/2Km K, = 54K,

N=~+(N -~)e-" l. ' l.

OBJECTIVE TRICK

114. [DJ ~

=AN 1 and

(

A is same for a given sample)

N1 -N2 =R,_R,=(R,-R,) A

I is intensity at a distance x and

N -N

µ is absorption coefficient; [µ] = L-1

1

µ is maximum for lead as lead has maximum ability to absorb radiations in a minimum distance.

A

A

(R,-R,)T oc(R -R )T 2

0.693

1

2

117. [DJ

1 E= k,T

109. [Bl

2

l.= he

at T=300K

E

=>

=AN2

Number of atoms disintegrated in time ( t2 -t,) is

Where 10 is intensity at x =O,

=>

R2

As N2
108. [Bl I= l 0e-µx

=>

... (2)

Solving equations (1) and (2), we get K, = KE of ex- particle =5.4 MeV.

At t = 0, N = N0 which is satisfied by (C) only.

=>

... (1)

From Conservation of Linear Momentum,

l.

E = -½< 1.38 x 10-" )(300)

( 6.626 x10_.., )( 3 x10') 1.6x10-" E{in keV)

2

12.42 A E(in keV)

1.6x10·19

E=0.01 eV

=>

E(in keV) ,.

E= 1 (1.38x10-")(300)eV

=>

,___ 1.242x10...g m

118. [DJ

E =-13.6 eV

110. [D]

n' = -3.4 eV

=>

Z'

E2

E, =,(13.6eV) n

=>

E,=9(13.6eV)

=>

En= 122.4 eV

120. [DJ Expected mass of Cu must be less than that of zinc. So it is unstable and radioactive, decaying to Zn through p decay. 121. [BJ

111. [C]

~

,."- = 0.021 nm= 0.21 A

is the fraction decayed in one half life Tv 2 and

Since, A.Ku corresponds to the transition of a'l electron from

is the

T,, = 0·693 = 0.693T l.

L- shell to .K - shell, therefore,

=>

t

fraction decayed in one mean life ( T) .

-(· V)- 12375 _ 12375 _ 58928 V E,- E,rne -l.(inA)- 021 e

=>

liE ~ 59 keV

So,

T > Tv2

Further, since the fraction decayed is ( 1-e-u)

112. [D]

(rm)=(:')( 0.53 A)= (nx 0.53)A =>

m'

-=n

z

m = 5 for ,00 Fm

_(5)'_1 n----100 4

257

(the outermost shell) and z = 100

t =1-e-i.r =1-e-1 =0.632 t >I

However, if we had· been asked about the fraction of sample undecayed, then we must give our answer in the light of formula

~ =e-lt N,

where N is number of undecayed nuclei in sample at

time t.

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Advanced JEE Physics

Optics & Modem Physics 130. [CJ

124. [DJ

Since we assume the potential energy to be zero in the ground

V=J2~V

state. So,

125. [AJ Series limit (i.e. shortest wavelength) of Lyman implies transition

Total Energy=U+K=Kinelic Energy => Total Energy= 13.6 eV (in ground state)

(·.- U=0)

First line (or longest wavelength) of Lyman series implies

If the potential energy is not assigned a zero value, then total energy is -13.6 eV. So, we conclude that making potential energy zero increases the

transition from 2---+ 1

value of total energy by 13.6-(-13.6) = 27.2 eV.

from oo---+ 1

Series limit of Balmer implies transition from oo ---+ 2 -- ---

- - - , - - - - - - - - - - - 00

---t--+--+-----n=3

_,,-~-+------n=2

Now actual energy in second orbit

131. [BJ Since, E = -

=> ~--~~----n=1 According to Ritz Combination Princip_le

=>

=-3.4 eV

Hence new value is (-3.4+27.2) eV = 23.8 eV

13 6 ·

n'

. -34-_ 13.6

=>

• -

n2

n=2

A=_!:_

E.,, ... 1 = E""...2 + E2_.1

Since,

hv1 ·= hv3 + hv2

Velocity of electron in second orbit is

mv

C

C

v, = 2(137) = 274 126. [AJ

=>

, =

6.626 x 10"' (9.1 X 10-'"l(

:, =R[¾-¾]

_!__ = 4R[_! ,,

4

2.J 16

=>

,,=121sA

127. [BJ

~Ne-----+2;He+~ X => A =14 => 2=6 i.e. ~4 C is the unknown nucleus.

X

274

=> A - 6.6 x·10-10 m Further Kinetic Energy =-(Total Energy) =->

A. 2 =~ "-1 27

3 10')

E = -(-3.4 eV) = 3.4 eV

133. [AJ Shortest wavelength of Brackett series corresponds to the transition of electron between n1 = 4 and n2 4 co and the shortest wavelength of Balmer series corresponds to the transition of electron between n1 = 2 and n2 4 co • So,

(2'>(1;~6)=(1~6)

=>

z2 =4

or Z=2

134. [AJ

128. [CJ

1 2 and r" cc n n

En oc 2

The longest waveilength in a series is obtained when a transition takes place between the lowest consecutive levels. Here transition must take place from n = 2 to n =1

So, Eln is independent of n

=>

E,r, = (13.6 eV)(0.53 A)

=>

E,r,=7.2eVA,

=>

El1 = constant

135. [BJ Maximum angular speed will be in its ground state,.so, we have v, 2.2x106 ms- 1 ro =-= max r1 0.529 X 1 o-lO m

=>

=4.1X10 ra"dS16

ffimax

1

129. [CJ

t = 0.693

'

T=_!

=> =>

'

4.114

t=0.693 T T>t

136. [BJ

.

he =Rhc(1-_!_)

,

=>

n'

n-

~

\/'i:R-1

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Atomic & Nuclear Physics 137. [BJ

145. [AJ The maximum number of electrons in an orbit are 2n 2 • lf n > 4, is not allowed, then the number of maximum electrons that can be in first four orbits are:

From R=R,(iJ Wehave. 1=64(iJ

2(1)' +2(2)' +2(3)' +2( 4 )' =2+8+18+32=60 Therefore, possible elements can be 60.

or n = 6 = number of half-lives.

t=nxt,12 =6x2=12 h

146. [DJ When one electron is removed, the remaining atom is hydrogen like atom whose energy in first orbit is

138. [AJ Penetrating power is maximum for r - rays, then of p- particles and then a - particles because basically it depends on the

velocity. However, ionization power is in reverse order.

139. [CJ j3 and y - decay take place from a radioactive nucleus. Hence, options· (A) and {B) are wrong. During fusion process two or more lighter nuclei combine to form a heavy nucleus.

1.=!1_

(24.6 + 54.4) eV = 79 eV energy is required.

147. [CJ

148. [BJ Shortest wavelength or cut-off wavelength depend only upon the voltage applied to the Coolidge tube.

N=N, A= N2

149. [CJ

N,

Beta particles are fast moving electrons which are emitted by the nucleus.

142. [BJ Heavy water is used as moderators in nuclear reactors to slow down the neutrons.

143. [AJ

U =eV =eV,log,(

AN= a.

1. 152. [BJ At time t = o, A gets converted to B which has a higher activity. Therefore, activity increases.

2

mv eV0 --=-

r

Maximum number of nuclei will be present when Rate of decay= Rate of formation

N=~.

dr r This force will provide the necessary centripetal force. Hence,

150. [AJ

=>

t)

lfl=[- dU[= eV,

=>

Therefore, to remove the second electron an additional energy of 54.4 eV is required. Thus, to remove both the electrons

N

Since, R = N2 particles per second and

(2=2)

E, = -54.4 eV

During fusion, binding energy of daughter nucleus is always greater than the total binding energy of the parent nuclei. The difference of binding energies is the energy released. Hence, 0=E2 -2E1

140. [AJ Activity of a radioactive substance R = AN

=>

E, =-(2)'(13.6 eV) =>

r

153. [BJ

v=l!F

... (1)

P(survivaO = N(t) = Noe-)J = e·~t No No

... (2)

For, t=-, we have

Moreover,

1

nh

mvr=2•

'

Dividing equation (2) by (1), we have

.1.e

P(survival) =

mr- ( -n h ) -~ - 21t eV0

154. [DJ Energy released is given by

6E= (

144. [DJ Activity reduces from 6000 dps to 3000 dps in 140 days. It implies that half-life of the radioactive sample is 140 days. In 280 days (or two half~lives) activity will remain ..:!.th of the

. 4

initial activity. Hence, the initial activity of the sample is 4 x 6000 dps = 24000 dps

=>

Total Binding ) (Total Binding) -2 Energy of 2He4 Energy of ,H2

6E=(4)(7)-2(1.1)(2)=23.6 MeV

155. [AJ n(n-1) Number of possible emission lines are - -when electron 2 jumps from nth state to ground state. In this question this value is

====================================4.115

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....,_ Optics & Modim .,

Advanced JEE Physics

162. [DJ On the last day we have 100% decay i.e. on the ninth day 50% decay must be there or 50% must be left.

N= (n-1)(n-2) 2

=>

(n-1Hn-2)

10

2 Solving this, we get

167. [CJ

,H2 + ,H2-------+ iH3 + p 2 4 3 1H + 1H -------+ 2He + n :::::,. 3 1H2 ~ 2He4 + p + n

n=6 156. [DJ Let AA

= A.

and A8

= 2A

Initially rate of disintegration of A is AN0 and that of B is .

(2l.)N0

•

=>
=>

t,m = -0.026 u

=>

l;,El=c'l;,mJ

=>

,iE = ( 9 x10")( 0.026 x1 .67x1D"")

(Half-life of B) =½(Half-life of A)

=>

t,E = (931.5)(0.026) MeV.

So, after one half-life of A or two half-lives of B , we have

.6.E = 3.87x10-12 J => As each reaction involves 3 deutrons, so total number of

After one half-life of A , rate of disintegration of A will becomes AN 0 0 and that of B would also be , because "· 2 2

t"'-;.:r:AN -, •


(-~~). =(-~~). ~

1040

reactions involved in the process

n=1

3

If each reaction

produces an energy .6.E , then 157. [AJ Saturation current is proportional to intensity while stopping potential increases with-increase in frequency. Hence, f3 = t while Ia < lb 158. [DJ First excited state is n = 2 and second excited state is n = 3 . Also,

40

.

10 28 E1o1a1 =-.6.E=1.29x10 J 3

E1ota1 =Pt Time of exhaustion of the star

t 1.29x10" 1016

:::::,.

.t=1.29x1012 s

Ea::_!_ =>

" n' E, 9

E3

168. [AJ

l=.9.=ne t t

=4

159. [BJ

_!_=Rs(.!-.!) l., 4 9

10-" =(-'/-){ 1.6 1o-")

=>

3.2 X

=>

('t:)=2x10"

X

5

169. [BJ

l., 36 5 l., = 3 = 27

_!=Z'R

4

A

160. [BJ Second excited state implies n = 3

=>

L,=n(;,)=3(;,)

=>

Lu =n(;.)=3(;,)

(_l __!_)

"'n~n~

For K"' line, n1 =1 and n2 =2

=>

_! = Z'R• (~) l. 4

=>

2=~3l.~.

=>

2=39.9=40

Z'

JE,J=--,.(13.6) eV n

=>

JE,J=

1'

9

.

(13.6) eV

=>

3' JEuJ= ,(13.6)eV 3 JEul=13.6 eV

=>

IEu I= 9IE,J

=>

IEul>JE,J

=>

170. [AJ Since in Rutherford experiment 1 Noc---

sin'(!) =>

N90 N60

sin4 ( 30) sin" ( 45)

=1.116c==================================== www.puucho.com

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Atomic & Nuclear Physics 180. [A)

N.,

=>

N50

16

.!!! = 4 X 109

=T

t

4

E=mc2

N60 =100

=>

=>

f )c'

=>

_§_=4x10 9 x9x1016

=>

_§_=3.6x10 26 Js-1

=>

_§_=3.6x10" W, t

=Uf

171. (CJ

Total Energy=-{K.E.)=-13.6 eV 172. (CJ

t t

2

For quantum number n we have 2n electrons. For For For For

n =1 n =2 n=3 n =4

kgs-l

we have 2 elements we have 8 elements we have 18 elements we have 32 elements

181. [BJ

So, total number of elements is 2+8+18+32=60

A(1.67x10·")

p

inR3

·3 173. [DJ

R., :::::,

a:

'

n2

Since, R = R0 A 3 where A0 = 1.1 fm

An ocn-4

=>

p

A(1.67x10"")

_±nA3A

Hence

A2

Ai=

3

n~ 16 n~ = -

p = 2.9 X 10 kgm-3

=> 183. [DJ

174. [AJ

A0

±=Rs(¾-¾) =>

0

17

=~c =1.55 x 10-11 m

=>

l.=~=6563A 5R,

175. [BJ For an elastic collision to take place there must be no loss in the energy of electron. The hydrogen atom will absorb energy from the colliding electron only if it can go from ground state to first

excited state i.e. from n = 1 to n =2 state. For this Hydrogen atom must absorb energy E, -E, = -3.4-(-13.6) =10.2 eV.

l.,=0.155A

is the minimum wavelength of continuous X-rays which carry energy equivalent to energy of Incident electrons. As this energy of incident radiation is more than that of K-shell electrons, the characteristic X-rays appear as peaks on the continuous spectrum.

184. (AJ Ac

decreases

with

increase

in

accelerating

voltage

in

accordance with the expression given by

l. _ he c - eV

So, if the electron possesses energy less than 10.2 eV it would never loose it and hence collision would be elastic.

Wavelength for K 11 line is not affected as it is due to the

177. [CJ

h L,=(1)2n

... (1)

{Using Bohr's Quantisation Rule)

In the first excited state of Li h L, =(2)-

2,

=>

L2 =2 L,

electronic transition between n = 2 and n = 1 in the target element. Hence ( AK - Ac ) increase with increase In the accelerating voltage.

185. [CJ

... (2)

~

and

178. (BJ

N=N0e-" 1

1

- - ;,. Fine structure constant (a) = 2he0C 137

!,d/M0L'T'

187. [BJ The series in U-V region is Lymen series. Longest wavelength corresponds to minimum energy which occurs in transition from n=2 to n=1.

Fort= Tav =I, we hav:>

=> i\l= No e

122 = (

1 /R )

... (1)

¾-d,

Th8 smallest wavelength In the Infrared region corresponds to maximum energy of Paschen series.

==================================== = 4.117

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Advanced JEE Physics

•=

1/R

... (2)

(_1__..1..) 3' a:,

193. [AJ

'-,"-,

, =(2nr) •

n

'

Solving equations (1) and (2), we get · 1"=823.5 nm

{·.- r o::·r1~. 2• I·: J• = nh)

188. [CJ ·

4(2He4) =a01a Mass defect, t,m = { 4( 4.0026 )-15.9994} amu = 0.011 amu

ground state (n =1) to second excited state (n = 3). Therefore change in angular momentum is t,L=L,-L,

189. [CJ 4 2 2 1H + ,H ------t 2He + Q

=>

t,m=m(;He')-2m(,H')

=>

t,m=4.0024-2(2.0141)

=>

.6.m·= --:-0.0258 u

t,L = 3 (__11_) - __11_ = !1_ 21t 21t 1t t,L= B.BxlO""' Js=2.11x10.,.,. Js 3.14

195. [BJ

Since, Q = c 26.m

=> =>

A and B can be isotopes if number of p-decays is two times

a= ( 0.0258 )( 931.5) MeV

the number of a-decays .

Q =24 MeV

190. [AJ Nuclear density is constant hence, mass o:: volume or m oc V.

191. [CJ Let N0 be the number of atoms of X at time

t = 0.

Then at

t = 4 h (two half lives), we have

N =No and N =3No

4

x

=>

4

V

x

ln 2s only 90% nuclei are left behind. So, in the next 2s 90% of 900 i.e., 810 nuclei will be left. 197. [CJ Let number of a-decays be x and number of p-decays be y. Then 92-2x+y=85 2x-y=7

... (1)

and

238-4x = 210 => X=7 Substituting this value in equation (1 ), we get Y=7

3

and at t = 6h (three half-lives), we have N

196. [DJ

=>

.Nx=_! N,

194. [CJ After absorbing a photon of energy 12.1 eV electron jumps from

Energy released per oxygen nuclei =(0.011)(931.48) MeV=10.24 MeV

N

and N

=___Q_

8

V

7N =-' 8

198. [DJ

NK _ 1 N, -7

4

The given ratio Therefore,

¾lies between i and f 199. [CJ

t lies between 4 h and 6 h

.

Since, t,E=K(-1,---1,-)=hc

n1

192. [CJ Since, A,

=>

= constant

= AN0e- 1t 1

. 1. 1 ('N,) A

t, = A ago

••. (1)

'

n2

A

ln the given diagram, emission from n =4-+ n = 2 would give photon of maximum energy (shortest wavelength) and n = 4-+ n = 3 transition would give photon of ITlinimum energy (longest wavelength).

Since, A, =(:\.)(2N0 )e·",

1 (2AN,)

=>

_ 1 t2 -i age ,;:-

=>

t, -t, =..!.log,(~)

200. [CJ ... (2)

' ' =>

2A,

t, -t, = log~( 2 ) log, (

~J

Energy of photon is E=13.6(1,- ~) eV 2 => E=13eV Since, momentum is conserved, so we have E -=mv C

=>

=4.118

E

V=-

mc

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-Atomic & Nuclear Physics =>

v- 13x1.6x10-19 -1.67x10--27 x3x108

_

4 m5-1

201. [DJ For incident electron, we have h h h '• - p - ,J2mK - ,J2meV Far shortest wavelength of X-rays, we have

1-2 - ~ -

=>

eV

!:i.:.! /eV 1. cV2m 2

Substituting the values, we get

,,

!:1 =1

c=========================================i = 4.119

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1.

2.

[B, CJ , For a photon pc=hv hv c

4.

=>

P=-

=>

p=8.Bx10-2a kgms-2

=>

p·= 1.65x10-6MeVc-1

5.

21t

[B, CJ

2n

(According to Bohfs Quantisation Rule)

4.5a0

Since, n1 and n2 are integers (n 1,n2 >1),so n2 -n1 is also an

zn'

integral value and hence AL must be an integral multiple of

_I!_ . This is satisfied by OPTION (C) and OP.TION (8).

n' = a0 Z

=> 2=2 Possible transitions are 3 --+ 2 , 3 -t 1 and 2 --+ 1

2n

6.

[B, CJ

1 1 Ucc-oc-

For 3--+2,wehave

r

1 -R(2)'(1 ))=4R(9::4)- 5R 1 4 9 · ~ 9

r

i =R(2)'(1-¾)=4R(¾)= 3~

n Locn

1=~

32R

7.

[A, C, DJ In Bohr's model of Hydrogen atom Roe n2

·For 2--+1,wehave

V"' .!_

i =R(2)'(1-¾) = 4R(¾) = 3R 1=-2..

3R

[A, CJ mv .J2mE r=-=-qB qB

. Jm

=>

n'

vcc~ and

For 3--+1, we have

3,

n'

Koc~cc_!_

=>

=>

~

So llL =(n, - n, )_I!_

21t

n=3

Since, rn =a0

=>

[A,DJ

I\

Let electron jump from n1 --+ n2 •

Since L= nh = 3h

=>

,!rn

In nuclear fusion two or more lighter nuclei are combined to form a relatively heavy nucleu~ and thus, releasing the energy.

[A, CJ

=>

Deflection "'

=>

n

1 To::n3 and E a: 2

n

=> => => =>

VR oc n TEocn T -ocn R V n _., E

r,ocq

=4.12oc::=================================

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Atomic & Nuclear Physics 8.

[C, DJ

14.

For ~~Ne nucleus to exist

in case of 1H2 , mass number is greater than its atomic number.

M1 <10mp+10m0

M1 <10(mP +mn)

~gca

Further since

[C, DJ In case of 1H1, mass number and atomic number are equal and

15.

has 20 more nucl~ans and thus it requires

[A, B, CJ Since Bohr's radius is given by n2h2£

more energy to hold all of them together and hence M1 :;r: 2M2

r=-' nme2Z

instead M1 > 2M2 because some additional mass defect must

occur to provide an additional B.E. to

~gca

=>

nucleus.

1

ra:m

_!_=me

9.

r0

12375

mµ

V (in volts)

With increase in V, Am will decrease. With decrease in Am

Since, Fa: m

energy of emitted photons will increase. And hence intensity will increase even if number of photons emitted per second are constant. Because intensity is basically energy per unit area per

unit time.

10.

[A, DJ

~

T,

E=E,(::)

=> =>

E = (13.6 eV)(212) E=2883 eV

Angular momentum is given as

Toc n3 =>

=>

L=mvr= nh

2,

=8

For ground state n = 1 , ~o we have

satisfied by both (A) and (D)

11.

L=__tl__

[A, DJ Characteristic X-rays depend upon the atomic number Z , so A,=A 2 =A3 OR

12.

,,

2•

16.

=~,,,,

[A, B, CJ he E S.mce, -=A 1. 1.=~

[A, B, DJ

t;E

N = N0 e-~

1

· he

Total number of undecayed nuclei equals

=>

Ll

(N0 -N)

(N,-N)=N,(1-e-")

=>

Activity R =-1.N = dN

[A, BJ

=>

dU Smee, IFI =dr

=>

Ke 2

Ke 2

=-,r

mv2

7=-,nh

2,

1.

mi,

=

12400 =1240A=1.24x10-' m 10(eV)

,

12400

= = 2(eV) =6200A=6.2x10-' m

... (2)

=>

15 X 1.6 X 10-19

= 6.63 X 1Q-3 1V

15

=> v=3.62x10 Hz The total number of ways of de-excitation of atom to ground state is 2.

... (3)

17. ... (4)

Since total energy E is half the potential energy, so we have Ke2 Ke2 Ke2n6 E=--= 6r' 6Kfm 3 So, total energy is Ea: n6 and Ea: m-3

min

Lowest frequency photon that can ionise the atom will have to remove the electron from n = 3 state, so we have E =15 eV

From (2) and (3), we get

r= (4•'h2·n2 e'K)!:!_ =K1n2 (m)

= =/1 E~

... (1)

According to Bohr's Quantisation Rule, we have mvr=-

,

Absorption will take place when transition takes place from n=3ton=4.

dt

.

max

Emission will take place when transition takes place from n = 3 to n=1

which is growing exponentially with time.

13.

and

Amin =:-E

N = Number of undecayed nuclei in the sample at time t.

[A, B, C, DJ Since reduced mass is

mm

m

µ = -1 -2- = m1 +m 2 2 Now, energy is directly proportional to mass, so E0 oc m

=>

Ea:µa:~ 2

====================================4.121

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Optics & Modern Physics

Adva11ced JEE Physics =>

E E,

e·(m)

~-- ....

2

=R0 hc

Since, E0

So, Rydberg constant is R0

E = ----2... a: m

he

=>

Recµ

=>

_I:!_=~=R0 m 2

,'

I

(KE) 1 = 14.4x10·" 10 eV =13.58 eV 2xO.53x1O-

=>

(KE)= 2

=>

KE decreases by 10.2 eV

',

' ' ' '' ':' 'tCM l '' '' ,,' ' ',. . . . 4' . .--.,."', ,

=>

Now PE=

14.4x10·"

2xO.53x10-10 x4

1 e' 4n&0 r

eV=3.39eV 10

14.4x1O....:..===---ev

=>

14.4x 1o·" eV=-27.1eV (PE)= 1

n2 h2 e

=>

(PE)2 =- 14.4x10·

nme Z

=>

PE increases by 20.4 eV

e'(m)

0.53 X 1o-lO

Radius of orbit is

10

r = - -2 '

0.53 X 10-iO X 4

-6.79 eV

So.for n=1, 2=1,wehave Now Angular momentum is L = mvr = nh

h 2e 1 r =--'-«2 0

nme

2n

m

=>

1 µ

::::)

rix-

::::)

_!_=~=2 r, µ

21.

[A, DJ

r0 «

Velocity of electron in first orbit is

e'

V=--

22.

2hs,

18.

h 2n

n' Z

and IPEl=2(KE)

[A, CJ

which is independent of mass, hence for positronium atom and hydrogen atom velocity of electron is same in both cases.

2 Since, v = CRZ (-;.- ~ ) n2 n1

[A, B, C, DJ

=>

v,~, =(3x10')(1.1x10')(1)(¾-¾)

Under normal conditions, the total energy (E) , kinetic energy (K) and potential energy (U) are

=>

V 3 _,. 2 =(3.3x10

=>

16 5 v 3 ... 2 = 36· x1O15 =4 .6x1014 Hz

E1 =-13.6 eV] K1 = 13.6 eV

for ground state

U, =-27.2eV

15

{

5 ) 36

Similarly for transition from 3 to 1, we have

v,~, =(3x10')(1.1x1o'l(1-¾)

E, = -3.4 eV] K2 = 3.4 eV for first excited state U, =-6.8 eV

=>

v 3 .,.1 =2.9x1O 15 Hz

Please note that (0) is incorrect as it says "must be" because from second excited state i.e., n = 3 , photons are emitted for electron transitions from 3 ~ 2 , 2 --t 1 , 3 --t 1 .

If PE in ground state is taken to be zero, then KE remains unchange~, however new PE and TE are increased by

27.2 eV. So, for ground state, we have

23.

E, =(-13.6+27.2) eV=13.6 eV K, = 13.6 eV (same)

[A, B, C, D] Eo::_!_oc_!_ r2 n2

... (1)

Pocvoc..!.

... (2)

r o:: n2

... (3)

U,=(-27.2+27.2) eV=OeV For first excited·state, we have

n

E,=(-3.4+27.2) eV=23.8.eV K, = 3.4 eV (same) u,~(-6.8+27.2) eV=20.4eV 19.

[B,C,DJ Ground state is n = 1 So, first excited state is n = 2

1 e' KE=-41te0 2r

(for Z=1)

24.

=>

PEr oc _!

=>

p -« n E

=> =>

Er oc n° Pr oc n

n

[A, B, DJ Actually, in the ground state, we have

=>

KE= 14.4x10-10 eV 2r Since, r = O.53n2 A

KE=+13.6 eV PE=-27.2eV and TE=-13.6 eV In the first excited state, we have

=4.122==================================

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Atomic & Nuclear Physics KE= +3.4 eV, PE= -6.8 eV , TE= -3.'l eV Now PE and TE both will be increased by 13.6-(-27.2)=40.8 eV KE remains unchanged being independent of reference. 25.

30.

Intensity of incident light may increase.by decreasing distance of source without increasing frequency.

31. [A, DJ

[B, DJ . 1 S ,nee, r cc m and· E0 ?= m

Since, E, =-(13.6>(~:) eV E,=-13.6eV for 2=1 and n=1 Similarly for Z = n , we have E=E, =-13.6 eV

For an orbit, mvr = nh is a constant, because n = constant

2.

26.

i.e., for second orbit of He· and for·third orbit' of

vccm

=> [B, CJ

32.

A

4(b;- ~' )=(q;- p;)

(1--1-) nf n!

a=2p and b=2q So, a and b are even integers greater than 1, satisfied by (A) and (D).

For a photon of energy E , mOmentum p is

33.

V

3

f- 22

4

~-3 2

36

A1

,, =-1--1- =5

ground state orbit and v 0 is velocity of electron in ground state.

34.

~=a=27 ,, 5

=>

2'

::;,·

5.. =!:!. = '- =_§_ 'E he 1 27 2

,,

5

1

27

a

2

So, b=E.t=5..=c=2. 28.

[B,

2

E2

27

DJ

a 2n4 =·1ty

=>

A= nr

=>

An= A1n4

=>

loge An = log Al+ logn4

=>

log 8 An -loge A1 = 41090 n

=>

log.(::)= 41og N

1

C=-=-

P2

[A, B, DJ

r=a 0 n2

he Further

[C, D] 22 n2 z E=E 0 2 , r=a 0 - , V=V 0 n 2 n where E0 is energy of electron in ground state, a0 is radius of

p = _§_ , where v is the speed of light.

=>

4

b2 - a2 = q2 - P2

This equation will be satisfied when

(J__J_) nf · n~

E oc

4

~

S"1nce, E =~ he =>

[A, DJ Let the transition in He· ion be from level a---+ b and that in H atom be from p ---+ q . Then A1 = 1 2 gives

[A, C, DJ From n=3 to r:,=2

.:!. oc

u++, we have

energy equal to -13.6 eV

In fusion reaction, two or more lighter nuclei combine to form a comparatively heavier nucleus.

27.

[B, CJ

log,(~)

logen

0

So, th~ graph will be a straight line passing through origin having a slope of four units at all the points.

1 t.E=204=13.6 2'(!---) 1 4n 2

35.

40.8 = 13.6 2'(-1,--\-)

=>

E _ -13.62' V ,--n-,~ e

n 4n Sati~fied for, Z = 4 and n = 2 29.

Since, E1 = -54.4 eV

[B, C, DJ

1

voc-

n

{-.- V

Eoc __!_

n' r cc n2

=>

~ , Er ,

V

v 2r are independent of n . •

[B, C, DJ

=(

2~:J¾}

~

- 13.62

•

2

(1)' .

= -54.4

~

Z 2 =4 => 2=2 Also, 40.8 eV is the difference between two energy levels n=2 and n=1. Also the electron cannot ·tall from the ground state and hence cannot emit photon. So,

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Advanced JEE Physics

Optics & Modem Physics

_1_ = 0.001 nm-' 1.,, - 1- = 0.002 nm-'

-E=K and E=~ 2

38.

[A, C, DJ Since,

1.,,

p = mv where v a: ..:!.

- 1- = 0.004 nm-' 1..,

n

1

:::::,

poc-

Since, KE oc ! and r oc n

r 1

KEoc-

41.

n'

2,

39.

1.,,

=>

1.,, =2sooA

= sooo A

[8,CJ

According to Ritz Combination Principle, we have E3 =E, +E 2

Further L = nh :::>

=>

So violet colour can eject photoelectrons from Metal -1 and Metal-2.

2

:::::,

1.,,=10000A

Violet colour has wavelength 40~0 A.

n

.

=>

Locn

=>

hv 3

=hv1 + hv2

=>

v3

+v 2

=>

-=-+-

=>

A-~ A, + "-2

[B, CJ Any transition in the Balmer series must end up at n = 2 . This must be followed by the transition from n = 2 to n = 1 by

emitting a photon of energy 10..2 eV . This 10.2 eV photon

corresponds to a wavelength of about 122 nm , which belongs

to the Lyman series.

=V1

C

C

C

A.3

A,

A2

I·- v=fl

,-

·

42. · [A, BJ 40.

[A, CJ From the relation, he· ·

eV=T-$

For d =A, there will be only one, central maxima. For A < d < 2.l , there will be three maximas on the screen corresponding to path differerice, AX= O and /!x = :U.

or

This is equation of straight line. Slope is tan 8 = he

e

=>

$, :$,: $, =~:~ :~= _1_,-1- :-1-= 1:2:4 101 A02 Aro

Ao, A~ '-ro

43.

[B, DJ In fusion two or more lighter nuclei combine to make a comparatively heavier nucleus. In fission, a heavy nucleus breaks into two or more comparatively lighter nuclei. Further, energy will be released in a nuclear process if total binding energy increases.

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1.

[BJ

10.

Cut-off wavelength depends on the accelerating voltage, not the characteristiC wavelengths. Further, approximately 2% kinetic energy of the electrons is utilised, in producing X-ray. Rest 98%

is lost in heat. 2.

4.

11.

[CJ Speed of electron in H like atom is vH u1<e

[DJ According to classical electromagnetic theory, an accelerated

charge continuously emits radiation. As electrons revolving in circular paths are constantly experiencing · centripetal acceleration, hence, they will be losing their energy continuoUsly and the orbital radius will go on decreasing and form spirals and finally the electron will fall on the nucleus.

[BJ At outermost orbit total energy is zero and electron is free from the . influence of the nucleus of the atom.

12.

[CJ

N=

µD >µH

=>

Ro>RH

=>

A. 0
13.

[CJ Lyman series - its energy is in the ultra violet region Balmer series - its energy is in visible region

In the centre of mass frame both the nucleus and the electron revolve about the common axis passing through the centre of mass.

17.

[DJ After 200 days the number of un-decayed nuclei in the sample will be

5.

[CJ Since ')..

t.o

= ~ , so the aE

-

.

[AJ

[AJ Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.

9.

initial number of un-decayed nuclei in the sample

initially.

Both Statement-1 & Statement-2 are correct & Statement-2 is the correct explanation of Statement-1.

8.

¾the

wavelength will be inversely proportional

the energy difference between the levels. The energy difference is more when the transition takes place from n --+ oo to n =2 than when the transition takes place from n =2 to n=1.

7.

4(4-1)

=S 2 and this ·depends on number of energy levels available for transition.

oc R ex: µ , where µ is the reduced mass of the system.

= rel="nofollow">

2he0 n

1 No. of lines in emission spectrum is N = n(n - ) 2

[DJ

i

= (~)~

22.

[DJ Statement-1 is false, the· penetration power depends upon accelerating potential. Statement-2 is true, increasing current increases the temperature of filament causing it to emit more electrons.

24.

[AJ

h\ = 0-E, and ht= 0-E, => h(\-t)=E,-E, So, Statement-1 and Statement-2 both Statement-2 correctly explains Statement-1.

are

Correct

and

[BJ Total energy is negative because electron is bound to the atom due to coulomb attraction and in the bound system energy is negative.

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1.

[CJ

Energy given tiy H-atoms ii, transition from n = 2 to n := 1 is equal to ·energy taken by He... atom in transition from n = 2 to n=4. n =2 -----'9

5.

· .

-3.4 eV IIE =10.2 eV n = 1 - - - - - ' - - - - - - - - -13.6 eV H-atom

t

3x10-15 1.6x10-

V --"---'-"19

~:: _____I,_____ IIE = 10.2 eV n=2-----'· ·

: ~:v

[DJ The difference of. the energy will be gained by the emitted electron as kinetic energy, so KE= 2.7x10-15 -3x10-17 KE=2.67x10-15 J

=:,,

[A] nl. a=-

n = 1 - - - - - - - - - - -54.4 eV He' 2=2 [CJ Visible light lies in the range, l., = 4000

.6.E=2.7x10-15 J

-13.6·eV

7.

2.

[CJ li.E=3x10-15 -0.3x10-15 J ~

6.

2

c::::, c::::,

A

to l., = 7000

l.=2a=_t,_=_h_

.n

... (1)

a'

8.

[AJ

K, -(1)'-1 2 -4

[BJ From equation (1), we get n2h2 E=-Ba2m In ground state n =1 h' ~ E, = Bma2

Substituting the values, we get E,=8meV

Kin~tic energy· K ~ 2 2

9.

[DJ From equation (1 ), we get Pa: n ~

[AJ

eV=3x10-15 4.126

./2Em

Ea:__!_

). = 4.68 X 10·' ffi

4.

p a

12375 A= 4687.5 A 2.64

KHe• -

c::>C>X=a

./E oc ]_

Wavelength of photon corresponding to this energy,

=>

0

•

IIE., = -3.4-(-6.04) =2.64 eV

3.

0

------a------+<

A.

From energy level diagram of He+ atom we can see that in transition from n =4 to n = 3 energy of photon released will lie

l. =

0

X=O

Energy of photons corresponding to these waveleingths (in eV) would be, 12375 12375 E = =3.09eV E = =1.77eV . ' ' 7000 ' 4000

between E1 and E2

1.875x10' V

~

mva:n

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Atomic & N11clear Physics

=103E

E 11.

"

[AJ Fror'n conservcition·of mechanical energy, we have . U1 +K1 =U,+K, =>

Now, since E"

0+2(1.5 kT)=-1-(e)(e)+O 4ru;0

=>

d

Substituting the values, we get T =1.4x10' K

12.

13.

0

15.

""

=(

m,,.

mPb +ma

)a

E" =(2ii:4)a E

[BJ As given in the paragraph, a reactcir is termed successful, if nt0 >5x1014 scm-a

(5.422) =53m MeV =103 105

[CJ

.:-~~-~ m-!l xC 2 xNm-2

{N =kg ms"'}

KgxC 2

=R

(CJ

m(~H) +m( ;He)= 2.014102 + 4.002603 =>

2

16.

[BJ A= 2nc

m(~H)+m{;He)=6.016705 u

(0

Since, m{:u) = 6.015123 u 1"= ~~6oonm

=>

m1 +m2 > M 1 So, (A) is incorrect. . m{:H) + m{':;Bi) =1.007825 + 208.980388

vrne;

m{:H)+m{':;Bi)=209.988213 u

17.

[CJ M~ximum energy of the antineutrino will be nearly 0.8 x 106 eV .

18.

(DJ Minimum kinetic energy of electron can be zero or greater than zero. But maximum ki!'letic energy will be less than 0.8x10 6 eV.

19.

[A]

Since, m( 2!~Po)=209.982876 u

=> m1 +m2 >M So, B is incorrect m{ ~H) + m(;He) =2.014102 + 4.002603 =>

m(~H)+m{;He)=6.016705 u

Since, :Li=6.015123u

Since, dN =P - ,N

=> (m,+m,)>M' So, (C) is correct and hence deuteron and alpha particle can go complete fusion. m{;:zn) + m(:!Se) =69.925325 81 .916709

dt

dN log, 2 --=dt, where A.=--

P-,N

+

=>

T

PT ( _.,,,,_,) N=--1-e T

m{;:zn)+m(:!se)=151.842034 u

loge2

Since, 1:Gd=151.919803 u => m3 +m4 <M' So, (D) is incorrect. 14.

20.

[BJ

(

21.

'""-')

[AJ Energy released upto time t is

m{;He)=4.002603 u m( 2~~Pb)+m(;He)=209.977058 u Now, ilm =209.977058-209.982876 => ilm = 0.005818 u => Q = IIE = 0.005818 x 931.5 ~ O=5.419467MeV=5419.467keV => 0=5419.5 keV

T

Rate of energy release is AE 0 =PE0 1- e - T -

[AJ 2 ~Po.= 2:Pb +!He+ .dE

m(':\Pb)=205.974455 u

( _.,,,,_,)

Since, A=AN=P 1-e

Ereleased

=(Pt-N)E0

=>

22.

[BJ

z, -(2)(2)+ (3)(1) = z, -(2)(1) + (5)(1) = z, 2 1 -22 =4 A,

-(4)(2) =A, -(1)(4) .. =A,

By Law of Conservation of Momentum, we have

O=p" -P1ead

=>

Pa

=>

~2m 0 E0 =V2m,,.E,..

=>

4E.,_

=·P1ead

23.

(CJ For A

= 206EPb

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Advanced JEE Physics

Optics & Modern Physics 30.

From above calculations we observe that less energy is required for a nuclear reaction, which a light particle is incident on a heavy target than if a heavy particle is incident on a light target.

For8 N ~ N o ~ No 0

2

4

31.

After 4 minute, we have

24.

[8] Given AA =R8

25.

= AaNa

A.ANA

=>

(lot 2)(4N,e-•••) =N,Cot 2)•-""

=>

t0 =6 min

mv

( n'h' )

=>

m 41t2m2r2

=>

32.

=400 -4 =396 W

., 16x2mk r =~

=>

r = 4n ,.iii,iZ

[DJ

Substituting (3) i~ (2), we get nh V

2nm(~~ ,.lii,iZ)

_ A 03

[A] If x+X----.+Y+y

33.

For the above nuclear reaction, threshold energy is given as

E,.=-0(1+::)

KE1

= iH

n"'h 2

Bn'm,./ii,iZ

1(n'h') =2mv = 2m 64n4m3k

=> 34.

When protons are incident on ~H , then X

V-

[8]

.

The Q~value of reaction is given by Q = (1.007825 + 3.016049-2x 2.014102) X 931.5 => Q = -4.033 MeV

2

KE-

n4h4 128n4 m2k

[D]

PE=U=-~

r'

and X = ~H

=>

1 007825 K =4.033 Mev(1+ · ) • 3.016049 K. = 5.381 MeV

kn 4h4

=>

[CJ When ~H is incident on protons

=4.128

... (3)

nh

=>

=>

4k

=7

4k

=>

[A]

=>

... (2)

=7

4x"'m

P101a1 =400W

12400 40x1000

29.

... (1)

. nh

[BJ Total power is P= Vl=40x10 3 x10x10--3

=>

=74k

=>

. n2h2

ptotal - pemitted

28.

2

-,-=7

V=~2nmr

So, heat produced per second is

27.

4k

=>

[C]

=>

r'

mv2

Now, according to Bohr's Quantisation rule, we have

p =(1 o/o) X 4Q X 103 X 1 Q x1 Q-3 => P=4 W 26.

F= 4k

=>

9 0 N0 =(4N0 +N0 )-(~+~')= ~

=>

[CJ

F=-dU dr

NA=Na=~o

=>

[Cl.

35.

[CJ

x=~H and X=!H

TE=KE+PE

K =(4.033 Mev)(1+ 3.016049) • 1.007825 K• =16.10 MeV

=> =>

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Atomic & Nuclear Physics 37.

38.

[CJ

50.

[AJ

B.E. per nucleon for intermediate nucleus is more than lighter or heavier nuclei.

Since

[DJ

=>

.!. = RZ' (J... -J...) 22 32

' ')..,

h·c =(Rhc)z'(.!.-.!.) 1c 4 9

where Rhc = 13.6 eV

Since,

39.

42.

=r0 A3'

rNu

1c

(i1trg )A

=>

VNu

=>

VNu oc A

=

Also, given that he= 47.2 eV

[BJ Q = (N,E, +N,E,)-N,E,

51.

=>

5 47.2=13.62'(3 6)

=> =>

2 2 ~2s 2=5

[BJ

Since, '" = (0.53)

[DJ

, = o.53 A

Sfnce, 1 kW= 3.1 x1013 fission

sec =>

n'

2 A

'

1.6x103 kW=a1.6x10 3 x3.1x1013

5

,, =0.106

A

= 5 x 1016 fission per second

52. 44.

[BJ The reaction for the second stage is given by

=> => => 45.

4

1Ta +p2-l-2He4 +on, +E2

=>

Lim= (3.016049 + 2.014102) -(4.002603 + 1.008665)

So, ionization energy

E,=4E,=4(-144)=-576 eV

E, =0.01888x931=17.587 MeV

53.

[DJ E1 =-576 eV

54.

[AJ

s·mce,

E = 21.62 MeV

2 62 So, energy released/ deuteron is \

E

n

E,22 = -7

= 7.207 MeV So, graph of En vs

46.

deuterium released

55.

is a straight line passing through origin

[BJ

-==evB

M-t,.an

=>

n

mv'

[AJ

,: _ .!_

48.

7.207x 100 0.384% 2.014102x931.4

~

with negative slope. The distance between successive points is non-uniform.

[CJ

( % of Rest Mass Of)

47.

= 576 eV

.ti.m=o.0188Bu

[OJ Total energy released E=E1 +E2 =4.033+17.587

=>

[BJ

s·rnce, E2 = E,

H-

•.. (1)

r

d ,: _ 0.693 'A,

V

eB

... (2)

m

TM> TH

According to Bohr Quantisation rule,

[CJ

A= dN = 1.N di

=>

nh mvr=2rr nh vr=~2rrm

... (3)

From (2) & (3), we get

T = 0.693

... (4)

r=~

n

49.

[CJ R1 = R0 e-J.1 and R2 = R0 e-1:?1 ~=e"(1,-t1)

R,

56.

[CJ Kinetic energy, K == imv2

K=.!.m(n'h') 2 2 2

mr

R = R,e-i.(i,-1,) 2

.•. (5)

K=inh(:)

:::::=================================== = 4.129

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Optics & Modern Physics

Advanced JEE Physics 57.

[AJ

::::,

. e

eB 2nm

1·=-=--

T

Since, Area= nr

2

=

N, =±(u-(u-,N,)e'(!¥))

n:(;:)

N, =.2.[ u-(u-1-N,)e·"'-']

=> =>

'loge2 we have =-,-.

At t=Tv 2

2

... (1)

N, =.2.[u-(u-1-N,)e·"]

=>

2.r S .1nce, T = v

M = iA = ( e'B )(nnh) = neh 2nm

Be

2m

/-

Sine~, e109•x = X

Since, PE=U=MBsin(90°) e-lo9o2

U = nhBe 2m

= elog.m ::: .!. 2

N,=±[u-(u-;N,)] 58.

[BJ

N =a.+ANo ' 21-

~ = 1tr 2B

~ = n(~)B = nnh Be

59.

e

66.

[CJ

[BJ Further, due to the decay of X to the stable nucleus Y, we have

No= mNA

dN, =1-N dt X

M

60.

Substituting value of N. from equation (1) and solving, we get · [CJ

The number of undecayed nuclei after time t is N = N0 e-1,,

1

==>

So, the number of decayed nuclei is

N' = N, -N = N, (1-e·")

67.

u-1-N,) _,, (u-1-N,) Nv=a t + ( -,- e - --,-

... (2)

[CJ

2 Substituting the value of Tv, = log~( ) in equation (2), we get N =~log (2)+(u-1-N,)-(u-1-N,) 2 ).,

YA,e

u

[CJ 68.

AE=13.6 z'(_!__ _ _!__)

n~ n!

63.

=>

47.2=13.6 z'(¾-¾)

=>

Z=5

[BJ

E2 -E 4 = -2.5 eV < 2.7 eV So, the electron will be making a transition to n = 4. ==> n=2 4 (-0.85 eV)

t

[CJ

'

Total energy TE= 13.6(5)' = 340 eV Since, TE = KE => KE=340V 64.

-z

( ) 1 (u-1-N,) --A-_

Nv =iloge 2

62.

/,,.

[AJ

he

13.6(Z)'

1"

n'

3 (-1.5 eV)

2 (-3.4 eV)

n = 1 (-13.6 eV) 69.

he n2

1- -

'' '' ''' ''

(13.6 eV)Z'

[AJ

E =-13.6z, 8

'

n'

A 6.63x10-34 x3x10 x9 13.6 X 25 X 1,6 X 10-19

:::>

1-=329A

and E0

=E2 =-3.4 2 2 =E4 = -0.85 2 2

E0

Now, E0 -E 8 =2.7eV 65.

[DJ

=>

dNx = a.-AN dt X

The ionisation energy is IE= 13.6 eV

Z=1

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Atomic & Nuclear Physics 70.

[AJ Emax =E 4 -E1 =12.75 eV and

72.

Emio

=>

= E4 -E3 = 0.66 eV

=>

he -0.85+3.4=2.55 eV=-l. 12400 l.= =4862A=486.2nm 2.55

[CJ E -E ' '

73.

= he l. 13.6x(1)' 16

(-13.6)x(1)' 4

he

[DJ

t.E =he= 12400 = 12 _09 l. 1025

========================================= = 4.131

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1.

2.

A-->(p,q) B--> (p, r) C--> (p, s) D --> (p, q, r)

For t,wehave "1=2 and n2 ---+co k ~=4

A--> (p, r) B--> (p, q, s) C--> (p) D--> (q)

3.

A--> (r) B--> (p) C--> (q) D--> (q) E--> (s)

4.

A--> (p, s) B--> (q, s) C--> (q, s) D--> (s)

\-t =~ 6.

A--> (p) B--> (q) C--> (r) D--> (q) y---+ e- +e+ ~ For pair production, we have E = 2m8 c 2 = 2 x 0.51 MeV = 1.02 MeV Inverse photoelectric effect is X-ray production and energy involved in it is of order of tens of KeV For de-excitation of Be+4 from first excited state, we have z2 42

E =-x 13.6 =,x13.6

EocJ_

n' Z => E • 54.4 eV = 55 eV For K X-ray photon of molybdenum, we have

n'

1

:::)

p oc n

:::>

ro::n2

0

E(K0 )=hv= hx3 cR (Z-1)'

4

E

1 1 -cc- and Epra:p n n

::::::,

E(K ) =_:l_(hcR)(Z-1)' =_:l_E, (Z-1)' 0

4

.!.. cc n3 and Er = constant p

5.

A--> (p) s:...(p) C--> (q) . D--> (s) f oc

7.

(J.._J..) n~ n!

::::::,

1

°2

For ~.we have n1 =1 and n2 ...+oo

=>

=>

E(K0 )=¾x13.6x(42-1)' =17.146x10' eV

=>

E(K

0

•17 KeV

)

A--> (p, r) B--> (q, r, s) C--> (q, r) D--> (p, r) For a- decay, we have

f = k(~---i), where k is a constant 0

4

\ =k

,

~x~A~~Y + !He Q =(KE)v + (KE)"

Since, (KElv « (KE)0 , we have

O=(KElo

For ~.we have n,=1 and n2 =2

= c::========================================::! 4.132

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Atomic & Nuclear Physics So, kinetic energy of all the emitted a- particles is equal to Q i.e., mono-energetic a- particles are emitted. Angular momentum is conserved in a- decay For ~- Decay, we have

a neutron decaying to a proton, so 1Y + V + e·

(a) (b)

8.

:x------t z:

A-; (r) 8--> (p) (q) D--> (q)

c-.

Q ~(KE),+ (KE), + E,

Since, (KE), « (KE),

9.

A--> (p, s) 8-; (p) C--> (q, s) D-; (p)

10.

A-; (p, q, s, t) 8--> (r, s, t) C--> (q, r, s, t) D--> (r, t)

11.

A--> 8--> C--> D-->

=> a~(KE),+E, Also note that, (a) E-.. is the energy of anti-neutrino and E,, takes on values from zero to maximum. Hence poly-energetic particles are emitted. i.e., poly-energetic antineutrinos are emitted. (b) Due to emission of antineutrinos spin angular momentum is conserved. (c) Neutron can decay in free space i.e., outside nucleus For Positron emission, we have a proton decaying to a neutron, so

~X------t l 1Y + v + e+ Same explanation as above in case of Beta decay can be applied here too. Proton cannot decay in free space i.e., outside nucleus because rest mass of proton is less than that of neutron.

For Electron capture, we have ~X+e- --),l1Y +v Q =(KE), +E,

Since, (KE), « E,

=>

angular momentum is conserved. it cannot take place outside nucleus i.e., in free space.

a~E.

All the neutrinos emitted are of equal energies and their energies are approximately equal to 0. That is monoenergetic neutrinos are emitted. So, for electron capture we have that

12.

(q, (q, (q, (p,

r, s, t) r, s, t)

r, s, t) q, r, s, t)

A--> (q) 8--> (p) C--> (s) D--> (r) In alpha decay, charge number decreases by 2 and mass number decreases by 4. In ~+ - decay, charge number decreases by 1 and mass number remains same. In proton emission, charge no, decreases by 1 and mass no, decreases by 1.

•

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1.

From our knowledg-e of Collision Theory, the fraction of kinetic energy lost by neutron is

AK 4m,m, K,-(m,+m,)'

4(1)(2) (1+2)'

3.

8 9

where K1 is the initial kinetic energy of neutron and .6.K is the energy loss.

After first collision 6K1 =¾K 0

After second collision AK 2

=¾K,

So, total energy loss is

.

•

AK= .6.K1 +AK2

4.

and so on

=>

(13.6)(2'>(1- n~)=52.1eV

=>

1-2=0.96

=>

nz5

Since =>

8

+ ... +AKn =9(K0 +K1 + .... +Kn-1)

=> =>

where, K1 =K0 - AK1 = ~

0

K,

E -E = 12375 + 12375 " ' 1085 304

5.

=~, =(¾)' K,

1

.

{·.- 2=2}

n

.! = R(2-1l'(..!._1-) · i2 2 2

' A

_ ~ _,, 0 76 10 2-1~40

(1.09x10')(2-1)'(¾)

Zz41

At time

t, let say there are N atoms of 7 Be (radioactive). Then

net rate of formation of 7 Be nuclei at this instant is,

dN di

10"' 1.6x10"" x1000

l.N

dN =6.25x10" -l.N di

"I

1 9 =1-8 ( - -~) 1 AK=-K 9 o

1-l 9

=>

1 K, - AK 0.025 9° =-K--=105

·O

11

_ _!_ 10 (6.25x10 -AN0 A _ge 6.25x1011

Since,

AN 0 =Activity

of

)= 7

3600

Be

at

t=1 hr=1.8x108

disintegrations/second

'

=> 9"=4x107 Taking log both sides, we get n=8 2.

11

where N0 are the number of nuclei at t = 1 hr or 3600 second.

gn

Since, K, =10' eV and AK=(10'-0.025) eV

=>

""'I dt

dN

!e.25x10 -AN

=>

_ _! 10 (6.25x10"-1.8x10')= 3600 A ge 6.25x1011

A =8 x 10~ sec-1 Therefore, half life

=>

After removing the first electron it will become He· ion. The

=>

IE=13.6(2')=54.4 eV

0.693 8x10 t,,, ~ 100 days

tV2 = --_,- = 8.66 x 106 sec = 100.26 days

ionization energy of single electron in He .. ion (Z = 2) is

Therefore, total energy required to remove both the electrons is given by

E=(24.6+54.4) eV=79 eV =4.134

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Atomic & Nuclear Physics 6.

10.

Since. A=R,(¾J

he

~-$=BV0

where, n is the number of half lives.

ne

V --o - 4nl:or

Given, R=~~

16

=>. 1240 eV-4.7 ev=( xne) eV

.(.!.)'

A,= A

200

' 2

9x19 Xnx~.6x.10-19

=:>

__ _ 62 47

1600 Bq i.e., 800 Bq.

1.5x10-2 ~ 100 n 9x1.6x10-10

n=4 Four half lives are equivalent to 8 s . Hence, 2 s is equal to one half life. So in one half life activity will remain half of

7.

4m:or

9

At time t=t

11.

N, = N,e-" and

N2 = NoA1 (e-Ai' -e-¼') -A1

A:

dN dt

N=10'°xG:)

=:>

m1o1a1=Nxm 1 m101a1 =1010 x10 9 x10--25 x10 6 mg

-<e-'' -e-"'l]

N =N [1-e-" __l._, =>

Na

= 1- e-1,1 -

No 0 Since, A.1 =-

and

·:i = 3

_l._,-(e->.,' - e-i..2')

"-2-~1

12.

120exe 4n:£0 (10x 10-15 )

0.0231 min-1

A2

2

N,

120e2

"-=!:x 41t,i:;0 x10-"" e

0.0154-0.0231

2x120xm

/r---1-0~-,~,-x-3___

A= 4.2x 10-1s x

(e-o.ll231>ro _ 8 -0.0154•60)

Ysx109 x2x12oxsx10- 21 A=7x10-15 m

N, =1-0.25+3(0.25-0.4)=0.31

N,

= .!!..x 4nso X 10-14 2m

0.693 0 0154 mrn . _, =4S=. t = 60 minutes N3 =1 _ 8 --0.02a1>60 0.0231

l.

m10ta1 =1 mg

A2-A1

o

=-l.N =-10"

=>

Ns=No-N1.-N2 3

10

.

l.=7fm

N

%age =..2.x100% =31%

13.

N,

Let m 1 and m2 be the mass of a-particle and hydrogen atom. By Law of Conservation of Momentum, we get

8.

m1u, = (m1 + m2 )v

V=~-.P, e e

where u1 is the initial vetocity of the incident a-particle and v is the final common velocity (or velocity of centre of mass),of the particles. · By Law of Conservation of Energy, we get

h Slope=e It is same for both, so

Ratio =1

1

,

1(

) ,

E

2m1v, =2 m1 +m2 v +.6. o

9.

where .6.E0 is the Ionization Energy The loss in KE of the a -particles must be gained by the atom as ionisatio_n energy, so we h'ave·

=>

~=e-i..1

=>

N -~•so -=e 1386

=>

N,

.!.( m m2 )v~ =.6.Eo 2 m +m 1

1

N, 0.693~80 N -=e~ N,

=>

!!.=e--0.04 N,

=>

.!'!. = ( .!.)"" N, e

·.. -= N= (1)'·" =

Fraction of nuclei decayed

1- -

N,

1- -

e

2

=>

1 +m2) E , = (m 1 K1 =-m1V1 -- t,. o 2 m,

=>

K 1 =(1+ ::).6.E0

=>

K, =(t+f)<13.6) ev

=>

K 1 =68eV

0.04 =4%

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Optics & Modern Physics

Advanced JEE Physics 14.

4.6x106 0.23 Taking log both sides, we get n~24

U=-1.7eV E=~=-0.85 eV = -

2

13 5 -

2 n' ::::> n=4 Ejected photcielectron will have minimum de-Broglie wavelength corresponding to transition from n =4 to n = 1 , so we have

17.

IIE = E, -E, =-0.85-(-13.6) = 12.75 eV

,. =

=>

15.

150 A J10.45

=>

A.2

m

m

Neutron

Proton

n:

22

n2 =4 Now difference of wave numbers of above two lines is

_!_ __!_=_!=A

... (1)

A.1

... (2)

=>

A.2

A,

1 _ 1 _ 1 -R( 1

"-1"-2"-

J)

n:n:

Transition n2 ~ n1 or 6 ~ 4 corresponds to second line of Brackett series, whose wavelength is given by

_!=R(..!.._..!..) 2 2 A

4

6

Substituting the values, we get

K,

Q1---, K,

n:

22

n1 =6

mass of proton = m

~

A,1

Substituting values of A. 2 and R, we get

(13.6)2' 0.544 n' (13.6)2' 0.85 => (n-4)' Solving equations (1} and (2). we get 2=4 and n=20 l. _ 12375 40441 A m•• - -0.544-(0.85)

Mass of neutron

_!_ = R(_!_ - _!_)

..!._ = R(_!_ - _!_)

=>

16.

A

Similarly, for Balmer Series, we have

Since N=n(n- 1)=6 ' 2 => n=4 i.e., if n1 =n, then n2 =n-4

(b)

l., =4861

Substituting values of A.1 and R , we get

(for an electron}

l.=3.8A~4A

(a)

A,

Given, l., =4102

For Balmer Series, we have

Using Einstein's Photo-Electric Equation, we get Kmlll( = .6.E-W =10.45 eV

=>

n

l.=26206

Lx

0

A

y 18.

From the given conditions

... (1)

E, -E, = (10.2+17) eV = 27.2 eV

... (2)

and E, -E, = (4.25 + 5.95) eV = 10.2 eV Equation (1) - (2) gives E3 -E2 =17eV

K, Applying Law of Conservation of Linear Momentum along y-direclion, we get J2mK 1 sin45° =~2mK2 sine

... (1)

along x-direction, we get

~2mK0 -~2mK, cos45° = ~2mK, case Squaring and adding equations (1) and (2), we get K 2 = K1 + K 0 - ~2K0K1

=>

2 2 (13.6)(¾-¾)=17

=>

2'(13.6i(; )=17 6

~

2 2 =9

~ 2=3 From equation (1), we get

..,(2) ... (3)

2'(13.6i(¾- ~' )=21.2

By Law of Conservation of Energy, we get

K2=Ko-K1

... (4)

=>

Solving equations (3) and (4), we get =>

K - K, ,- 2

=> i.e., after each collision energy remains half. Therefore, after n

collisions, we have

K, =K,GJ

19.

(3)'(13.6{¾- n1, )=27.2 _! - _!_ = 0.222 4 n' 1 ,=0.0278 n

=> =>

n2 =36 n=6

(a)

1 rydberg =2.2x10-" J=Rhc Ionisation energy is given as 4 rydberg, so

=>

=4.136

0.23=(4.6x10'i(¾J

IE=8.8x10-18 J .

S.Sx 1o-,s 1.sx10-19

55 eV

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Atomic & Nuclear Physics ::::;i,

Energy in first orbit is

21.

E, =-55 eV

n

10x10--3 6.25x101s 1.6x10-19 Kinetic energy of an electron is

Energy of radiation emitted when electron jumps from first

excited state (n = 2) to ground state (n = 1) is E

Total number of electrons striking the target per second

3 K =40x10 eV =40x103 x1 .6x1Q-19 =6.4x1Q-15 J Total energy of electrons striking the target (per second) is E = 6.25x1016 x6.4x10-15 =400 J

3

E,, = ( ), -E, =- E, =41.25 eV 4 2 ~o, wavelength of photon emitted in this transition is

l.= 12375 = 3 ooA 41.25 (b)

Let Z be tt~e atomic number of given element. Then

(a)

Total power emitted as X-rays is 1% of 400 W. So P=4W

(b)

Heat produced per second is

H=(400-4) Js·'=396Js·'

E, =(-13.6)(2') => => (c)

-55=(-13.6)(2') 2=2

Since, r = ~

22.

, so the radius of first orbit of this atom is

When

an

substance

(13.6)z'

m

e n' Substituting z =3 , we get

1 Tm

E =- 122.4 eV

n' 122 4 E, =; =-122.4 eV

=>

emission

1 T,

= _l_

1 1200

).~

1 600

Tm =400 yr

Since, t = .:!.109,

122.4

A

and E3 = - -2- =-13.6 eV (3)

=> llE = E, -E, = 108.8 eV The-corresponding wavelength is

=>

1 T.

(1)

l.=

J3

-=-+-=--+-

'

=>

and

= disintegration constant for p -emission only

Mean life is given by T

V

a

A" = disintegration constant for a -emission only

A.11

E,

by

simultaneously, the average disintegration constant A.av is given by A.av =A.a +All where

20.

decays

(N,) NI

= Tm log,

(N,) N,

=>

t=4001og,(~i)=400log,(4)

=>

t=400x1.4= 560year

12375 A=12375A= 11374 A 108.8 · tiE(in eV)

:\.=114A

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