Grb Optics Total

Dr. P. K. Sharma Er. Anurag Misra

Understanding Physics

Optics & odern Physics FOR (Main &Advanced) & ALL OTHER ENGINEERING ENTRANCE EXAMINATIONS

~, 9\><1.",~o11flw> S"'1" UNDERSTANDING PHYSICS '1~~3.12..0" . .

o

OPTICS & ERN P YSICS

For JEE (Main & Advanced) & All Other Eng!neering Entnince Examinations

By :

Dr. P.K. Sharma

Er. Anurag Misra

(M.Sc., Ph.D.)

(B.Tech. IIFROOrkea)

_0 ȣ1

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No part a/this book may be reproduced in any form or by any means wilhout the prior written permission oilhe authors and the publisher. The authors and the publisher have made every effort to provide allthentic. accurate a1ld up-to-date matter ill this book. However. they do 1I0t take atly legal respollsibility/or any mIsinterpretations or errors inadvertently overlooked.

PREFACE There has been a lot of change in JEE p~tl c rn from this year. It )Viii bcconducted in two phases- Main & Advanced. No:w board marks will also play role-in deciding the selection. The relevance of thi!; book is more in thi s changed scenario. We have dealt with the theory and concepts in details. it will help the student in Ihcirboard preparation as well as JEE (Main & Advanced) preparation. In this book. we have put a large number of problems in different fomlals. Single correct choice questions are segregated in Icvcl-l and 1cvcl-2 as per their toughness. Lcvcl-l problems are more relevant for JEE (M ain) while levcl-2 problems for JEE (Ad,·anccd). Miscellaneous solved examples have also given for better understanding of the concepts and their applications. We have put 3 lot of effort to make this book error free hut ifsomc still persi st then don't hesitate in informing us so that theyean be rectified.

We are thankful to our students who have given us the insight of the areas which are problematic for them in physics. Thcy have always been an inspiration to do better and something different in the subject. We also want to thanks Mr. Manoj Kumar Bathla and Mr. Vishvnath Bathla, Directors, GR. Bathla Publications Pvt. Ltd. There is always a scope for improvement in anything what we do. So we will appreciate any suggestion or criticism, from our readers, which is intended for the further improvement ofthe book.

. Er. AnuragMisra

Dr. P.K Sharma.

(B.Tech. IIT-Roorkec)

(M.Sc .• Ph.D.)

July, 201 3

[I am tbankfulto my wife Ranjana and my kids Mimansa and Siddhant for theirpaticnce and encouragement to write this book]

, Note : Stua... Lts and honourable teachers may feel free to give valuable suggestions on the ~il suggcstionsgrb@gmai l.com toimprovethequalityofbook.

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

CONTENTS Chapters

Page No.

Chapler-1 GEOMETRICAL (RAY) OPTICS 1. 1 Reflection and Refraction of Light 1.2 Mirror and its Properties 1.3 Image Fonnation due to Spherical Mirror ' 1.4 Image Formation due to a Plane Mirror I.S Properties of Refracting Surface 1.6 Image Fonnation due to Spherical Refracting Surface 1.7 Image Fonnation due to Flat Refracting Surface 1.8 Total Internal Reflection 1.9 Pri sm 1.10 Dispersion 1.11 Angle of Deviation 1.12 Angle of Dispersion 1.13 Dispersive Power 1.14 Prism and Spectrum 1.1S Combination of Two Prisms 1.16 Lenses and its Properties 1.17 Image Formation due to Lenses

1.18 Si lvering of Lenses 1.19 System of Thin Lenses 1.20 Optical Power and Focal Length 1.21 Cutting of Lens .:. M iscell a neous Examp les .:. Assignments ~

Conceptual Questions ~ Multiple Choice Questions (A) Only One Choice is Correct Level-I Level-2 (8) More Than One Choice/s is/are Correct Match the Columns Comprehensions ~ Subjective Problems Level-l ~ Subjective Problems Level-2

== ==

,

1- 202 4 5 II

16 19 21 24 25 27 28 29 29 . 29 30

32 35 38 40

42 46 51 76

78 89 100

107 110 120

128

(vi)

.:. Answers .:. Hints an d Solu tions

141 147

Chapter-2

WAVES OPTICS 2.1 Huygen's Principle 2.2 Condition oflntcrfcrcncc of Light 2.3 Optical and Geometrical Path 2.4 Phasor Addition of\Vave ~ 2.S Young's Double Slit Experiment (VOSE) 2.6 Intensity Distribution 2.7 Fringe Shift due to Thin Plates 2.8 Change in Fringe Width due to Change in Medium 2.9 Change in Phase due to Reflection 2.10 Interference in Thin Films due to Reflection .:.. M iscellaneou s E xa mpl es .:. Assi~nrll c n t s => Conceptual Questions => Multiple Choice Questions (A) Only One Choice is Correct Level-I

Lcvcl-2 (8) More Than One Choice/s is/arc Correct => Match the Columns => Comprehensions => Subjective Problems Level-I => Subjective Problems Levcl-2 .:. Answers .:. Hints a nd So lutions

203- 282 203 204 206 206 207 209 211 2 13

214 214 217

231

231 235 241 244 246 250 · 251 259 262

Chapter-3

PART ICLE AND WAVE NATURE OF RADIATIONS (LIGHT)

283- 369 3.1 Maxwell's Laws ofElectrornagnetic Induction 3.2 Electromagnetic Waves . 3.3 Black Body Radiation: Planck's Law 3.4 Photoelectric Effect 3.S X-rays 3.6 X~ray Diffraction: Bragg's Law -

283 285 286

290 297 . 300

(vii)

3.7 Mosley's L~w and Characteristic X-rays 3.8 Einstein's PhOlon Concept and Dual Nature of Light .:.

~ l i s c(' II :l II C OU S

302

304 3)0

Exa mp les

.:. Ass i", II I11(' lIt s ~ Conceptual Questions ::) Multiple Choice Questions (A) Only One Choice is Correct Level-l Lcvcl-2 (D) More Than One Choicc/s is/arc Correct ::::::) Match the Columns ::) Com preh ensions :::::) Subj ective Prob lems level-I :::::) Subjective Problems lcvcl-2 .:. Answers .:. I·lints and Solut ions

32 )

323

330 336

341 342

347 ~ .

348 35 1

354

Chapter-4 PARTICLES AND WAVE NATURE OF MATTER 4.1 4.2 4.3 4.4 4.5 4.6 4.7

,

370-489

Introduction Composi ti on of an Atom Thomson's Model (1897) Rutherford's Mode l (1911) a-particle Experiment Atomic Spectrum and Energy Levels Bohr's Model (Idea ofQuanlis31ion by Using Balmer's Formula) Mechani sm of Emission and Absorption: Atomic Hydrogcn Spectrum 4.8 Bohr's Corrcspondencc Principle 4.9 Frank·Hertz Experiment : Confirmation of Atomic Encrgy Levcls 4.]0 dc-Broglic's Maltcr Waves 4.11 ~xperimcntal Confirmation of Wave Nature of Matter by Davison·Germcr and Gcorge D. Thomson (1927) 4.12 Double Slilinterferencc of Electrons 4.13 Wave Func tion tV : Max Born's Intcrprctation ofW 4.14 Schrodinger's Wave Equation 4.15 Particlc in a Box 4.16 Heisenberg's Uncertainty Principle 4.17 Wave·Particle Duality of Matter and Radiation: Complementary Principle (1927.1928) .:. Miscella neou s Ex a m p les .:. Assign m ents ~ Conceptual Questions

370 37 1 37 1 373

374 376 382 386

387 387

390 391

392 396 398 399

402 405

420

(viii) ~

MUltiple Choice Questions (A) Only·One Choice is Correct Level-I · Level-2 (8) More Than One Choiccfs is/arc Correct ~ Match the Columns ~ Comprehensions ~ Subjcctive Problems Level-I ~ Subjective Problems Level-2 .:. An swers -:. Hi nts :lIld Solutio ns

423 427 431 433 435 440 44 2 449 453

Chapter-5

NUCLEAR PHYSICS 5.1 Nucleus 5.2 Nature of Nuclear Forces 5.3 Binding Energy and Stability of a Nucleus 5.4 Law of Radioactive Decay 5.5 Successive Disintegration and Radi~active Equilibrium 5.6 Secular and Transient Radioactive Equilibrium 5.7 a-decay

5.8 (\.decay 5.9 y-decay S.10 Nuclcar Reactions

..

5.11 Conservation Laws in Nuclear Reactions 5.12 Endoergic Reactions in Laboratory: Reaction Thresholds S.l3 Nuclear Fission 5.14 Nuclear .Fusion .:. Misc.-:ll a neous Examp les .:- Assignme nts ~ Conceptual Questions ~ Multiple Choice Questions (A) Only One Choice is Correct Level-l Level-2 (8) More Than One Choice/s is/are Correct ~ Match the Columns ~ Comprehensions ~ Subjective Problems Level-l ~ Subjective Problems Level-2 -:. Answe r s .:. Hints and Sol utions

490-567 490 492

493 495 498 499 501 503 507 509 5 11 511

512 517 520 ·529

530 538 543 545 547 549

550

552 555

1

Geometrical (Ray) Optics Study

,

~

.

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1,10 1.11 1,12 1.13 1.1,4 1.15 1.16 1.17 1.18 1.19 1.20 1,21 .

1,1

m

Poi~ts

Reflection and Refrac tion of Light Mirror and its Properties

Image Formation due to Spherical Mirror Image Fonnation due to a Plane Mirror Properties of Refracting Surface

Image Formation due to Spherical Refracting Surface Jmage Fonnation due to Flat Re fracting Surface To tal Internal Reflection ,

Prism Dispersion Angle of Dc viation Angle of ~i spersion

.

.

,

"

, r.

. '."

,

,

w.

Dispersive Power

Prism and Spectrum Combination of Two Prisms

,

.

Lenses and its Properties

Image Formation due to Lenses Silvering of Lenses System o f Thin Lenses Optical Power and Focal Length

Cutting of Lens .

l

. ,

t

, .. . .

..

Reflection and Refraction of Light

When we look into waler we can see OUf faces due to reflection ofliSht. We can sec the objects inside water as the light coming from these objects is transmitted to air. This tells liS that. Definition : A ray of light partly returns to its initial medium al the boundary of two media. This is called reflection ofl ight wave. Furthennore the ray oflight is partly

GRB Unde~tandinl! Physics Optics and Modem Physics

2

transmitted to the second medium with different velocities which is called refmction of light.

Mirror: If the boundary is silvered. total energy of the ray is reflected back without any transmission. Then the reflecting surface is called mirror (or reflector).

--

Ught wave is partly refiected and partly transmitted

~[

-

For a mirror, transmission is negligible

Refractive Index : When a ray of light moves from vacuum or air to a medium (glass, say) its speed decreases. The ratio of speed of

Medum-, (air)

r: =n

Medlum-2 (air)

light in air to that in a medium is called refractive e index of .he medium

Since,c = ~ andv= &oJ.10

The wavelength decreases when light Is refracted from air to a medium

h

vilE

putting,uJ.l 0 :::. I for transparent material and.!... = E r <0

we can write where E r = relative pennitlivity of the medium. Since frequency oflight is a property oflhe source oflighl (but not oftbe medium), it remains, the same in refraction. Then, the wave length of light changes. The wave length of light in a medium is I..=~

f

or

=(~)~ /).=l..no . 1

wher~ 1..0 = wave length of light in air.

3

Gtomttrical (Ray) Optics

Laws of Reflection Reflection oflight is the process of deflecting a beam oflight in the same medium. Experiments in reflection have yielded the following two laws: I. The incident ray, the reflected ray, and the nonnal , aJilie in the same plane, called the plane of incidence. The in cident and reflected rays are on the opposite sides of the nomra!. " n 2. The angle of incidence is always equal to the angle of reflection.

Li = L r

That is,

These angles are measured wi th nonnal to the point of incidence. _ _ _ _~_ _ _ _ 1 Let. Cj = unit vector along the incident ray, = unit vector along the reflected ray, i = unit vector along tangential direction, ,i = unit vector along omward nonnal Li=Lr=L9 Let ej = sin9 i -cose n ...(1)

e,.

cr =sin9 i + cos9 n

,

n

... (2)

" ---""---_1

From equation (I) and (2), we get,

e; = 2eas9 ,i ... (3) e,· n= (I) (I) cos (n- 9) = -cos9

Cr Also,

By putting the value ofeos9 in equation (3),

"

cr -Cj =-2(e; ·n) n

I

er =

ej - 2(e; . n) Ii I

..

e

n-e

It is the vector fonn of the law of reflection . --..: --, . ~ •• 1!.lc ~ A rayoJUghtis ;nciden/onaplanem;"~ra~onga l'ec/ora; + bj - ck.

. .

Tlte normal on incidence point is along i ; :. Find a unit vector along the reflecled ray.

a7 + bj - ck

A

A

j

+j

Solution: Here e; = 7~2="",~~ and n = - Also,

e,. = ej

er = =

Ja'+b' +c' -2 (ej ·ii) ii

ai+hj -ck Ja'+b'+ c' I

Ja'+b'+c'

,fi

-2[

]i+j

a+b Ja' +b' +c',fi,fi

..,

[-bi -aj -ck]

Ans,

4

1.2

CRB Undentanding Physics Optics and Modem Physics

Mirror and its Properties

Types of Mirrors, Principal Axis, Pole, Centre of Cun'uture and Focus: There are thrce types of mirrors. i.e., concavc, plane and convex . If we drop a perpendicular to the mirror. it is called principal axis (PA) and the point of intersection ofPA and the mirror is called IlOlc P. The centre of curvature C of the mirrors lies on the principal axis. It lies infront of concave mirror, behind convex. mirror for plane mirror C lies at infinity. Thc distance between centre of curvature and pole is called radius of curvature

PA

" ' P [ ---- PA

·'R" .

,

Focus: A thin beam of light along principal axis I' F C is used at a point on the principal axis. This is called PA focal point F. The focal point lies infront of concave mirror and behind a convex mirror and lies at infinity for plnne mirror. The distance between pole and focal point is called focal length denoted by '1". Real and "irtual object : The point of intersection of two (or more) incident rays (mys 0 coming towards the mirror) for a reflecting system (surf,lce or mirror) is c
p-ve

q+ve

q-ve

From and back space of a mirror are A and V space respectively

Geometrical (Ray) Optics

5

If the centre of curvature and focal points lie in rcal space, radius of curvature and rocallength arc positive. If the curvature and focal points li e in virtual space, the radi us of curvature and focal length arc negative. For real objects. object di stance is pos itive and for virtual object. object distance is negative. For rea l and virlual image. image distance is positi ve and negative respecti vely.

Hence. for convex mirror R is - ve, Ijs -ve; for concave mirror R is +ve and tis +\'9. for real and virtual object p is +ve and -ve respectively. for real and virtual image p is +ve and -ve respectively.

1.3

Image Formation due to Spherical Mirror Consider an extended (thin and crect)

real object AB on the principal axis. Bring (draw) three (or any two) rays 1,2 and 3 from the lap oflhe obj ect ray· ] is parallel to principa l ax is and passes through the focal point after reflection . Ray·2 is reflected at the pole. Ray·3 passes through the centre of curvature and therefore retraces its path (because of nonnal incidence). ]f we produce the three rays, they will inlerset at A' which is obviously the image of A. Since A is directly below B. the image of B. that is B t , can be found directly above A' in convcx and plane mirror 'and directly below At in case of concave mirror.

® (j)

i - - - - P --"'-..;" The image is real inverted.

(j)

, ,,,A· , , , , , ! ~--p--..,.!q ,, -

A,

,,

,

The image Is virtual and erect.

GRB Understanding Physics Oplics and Modem Ph)'1ics

6

Hence, the image is real and inverted in concave mirror; crect and virt\lal in plane and convex mirror for the real object. However. for different types of objects and its locations, the size, shape and nature orlhc image will change which will be throughly discussed in the examples. (j)and@-~--'iB=-_~~+--------;~B'

..,,"

-' ,

-'00

,

-'

00

Ao o

o o o

o

!,® o

The image is virtual and erect.

Mirror equation: Let us consider the real image A' B' oflhc object AB as described earlier. In the similartriangles A' 8'0 and ABO, we have A'B' =A'O='f. . .. (i) AB AO p In similar llS, A' B'F and 8 10F t

A'

A

F

o

o o o

-,

,

A'B' A'F p - f -- = - - = -B,O FO f A'B' p - f -- = -... (ii) AB f Using cqns. (i) and (ii), q p- f -= - p f

B,,----, Bo

o , 0

--q--#-~

° i. + .i

o o

or

:----p---+j.: o )

Since; = r,MBO and 6.A 'B'O are similar

or Discliss the positions and natllre oftile images 0/an erect real object placed on the principal axis jonJled due to a concave and convex mirror. Solution: Concave mirror: (i) Object at infinity. image is point size, real as the focus. (ii) Object between 2Fand infinity, image is real, inverted, diminished formed between F and 2F. (iii) Object at 2F. image is real inverted , same size formed at 2F. (iv) Object between Fand 2F, image is inverted, real magnified formed beyond 2F.

------

---

G~metrical

(Ray) Optics

7

, ,Ps

(v) Object at F, image as ghostly as infinity. (vi) Object between pole and focus. image is virtual, magnified, erect when the object is approaching from focus to the polc, its image size decreases 10 the size of the object. Convex mirror: For all positions of the real object, the images are virtual, erect and diminished formed between focus and pole. The image size increases from zero to size of the object.

(I) In concave mirror for real object images can be real and virtual. magnified. same size and diminished formed at all positions between +00 to~. (ii) In convex mirror, tor real object, images cannot be real; all images are erect virtual. diminished formed between -F and pole.

8

GRB Undmtll.nding Physics Optics and Modtm Physics

~%'4"!'1!:lc :I

Diagrammclrical(), silO\\' differcll1 types o/images produced by COI/ ' cave (III(/ com'cx mirror for rcal and virtllal point objects placed 011 tile prillcljml axis of lite mirrors. Solution : Concave mirror : Real

object and real image

,.., ...... /

o

C

Real object and vinuai image

c

Virtual object and real image

c

-----Convex mirror: Real object and virtual image

o

Virtual object and virtual image.

o

Virtual object and real image. , 0

C

9

Geomelrical (Ray) Oplics

In concave mirror for virtual object, virtual image is not possible; In convex mirror, for real objecl real image is nol possible.

Magnification Definition: Magnification of an image is defined as the ratio of size of the image 10 that of the object. There arc two types of magnifications, i.e., lateral or transverse magnification and linear or longitudinal magnification.

B

,,A' ,,,, ,,,, ,,' ,, ,,

So

'"

,,,F ,

0

!--, ,

,,

, ""," ,, ", ,,

s"

A,

0

p

J

",,

~

"

Lateral magnific ~~ion: It is defi ned as the ratio of height of the image to.that of the object

h· fro

A'B' AB

m=-' =- In the similar triangles A'B'e and ABO, .",.... A'B' A'O q m= -- = - - = AB AO P

or

! m= -~l

where -ve sign is given because A'B' and AB are opposite by directed. Similarly comparing the similar triangles A' B'F and B10F, A'B ' m=-- = - B,O f

q-f

or

! m=qjf!

eRB Understanding Physics Optics and Modem Physics

10 Aliter: From mirror fonnula

I I I -+=-

P

q

!

multiplying both sides by p. or

I+E = p q !

or

1+_I_=P (-m) !

~

or

111 = -

!

p- ! Similarly multiplying both sides by q.

r

'!.+l='!. P !

1m

or

=-(q

/ JI m

For concave mirror, when p --+ 00 , m -4 0 when p --+ /. m --+ <Xl when p --+ 0, m --+ +1 when p --+ 2/. m --+ - 1. For convex mirror, m is + ve given as m=If.=L=!-q P p+! !

+1

-i--ir-!2i'::::::=:=--o :F

-1

12F

p

-----t----' ,

,,, ,

Magnification versus object distance for concave mIrror

when p --+ 00, m --+ 0 when p -+ D, m -+ +1. Linear magnification: Linear magnification of an image is defined as the ratio of thickness of the

m +1

image to that of the object given as

I m'=~ I

o

Magnification versus obiect Relation between lateral and linear distance for convex mirror magnification : For thin objects, we can get thin images, then differentiating the mirror equation we have

p

11

Geometrical (Ray) Optics

d(~ + ~)=d(7 ) (.: J = constant)

__ lOp __ I Oq =0

or

p'

q'

~ =-(~r

or

I m'

or

= _m

2

1

,

1.4

Image Formation due to a Plane Mirror

Since f = 00 for a plane mirror, putting l = 0 in the mirror equation! +! = 0, we

f

p q

have

o If the object is real, pis +ve. Hence, q is negative. If the object is virtual, p is -ve. Hence, q is +ve.

Real object and virtual image

Virtual object and real image.

o

1. Real object - real image, and virtual object - virtual image are not possible in plane mirror. 2. Magnification is m = +1. 3. Object distance = image distance 4. If object is real, image is virtual; tf object is virtual, Image is real.

~':r".

Draw the images of a point object due to two plane mirrors placed perpendicular to each other. Solution: Two beam coming from the object 0 get reflected twice at two mirror surface to form the direct images II and 12 and the other indirect image 13 (image of J1 _/I.. ~ e~~ . AI

and image of / 2) as shown in the figure.

12

GRB Understanding Physics Optics and Modem Physics

o

I,

:--.....:....:..:... -..-

......

-""V

,, ,, II

...... .. .. .. \.1

r---------rL--"-'''''--'---'''''--I

....'(

:

'

... ,.... .. '>,.. "' ... ......'<. ..

lI

I

..........

, ......... I ...... ... ... - -

...... - -

I II

"

..... ..

.

\

.. .. ' ..' , ' ..'

I,

I,

The two density reflected beam reach the eye to sec these three images.

Number of Images Formed by Two Inclined Plane Mirrors Let 0 be the angle between mirror. (a) If3~OO = even number = m. Then. number of images fonned, n = m - I. (b) If3~O° = odd number = m

o

(i) Number of images fonned n = m - L of the object is placed on the angle

bisector. (ii) n = m of the object is not placed on the angle bisector.

(e) 1[360 ~ integer, then number of images " = integral part of 360°. 0

o

a

Effect of Rotation of Plane Mirror Direction of incident ray I is kept fixed while mirror M I is rotated by angle e to position M 2. Reflected ray R I changes to position R2 and nonnal changes from N I to N,. From figure it is clear that LABC = 2AI> + 5 =2(+ +9) or 5 = 28 That means when incident ray is fixed, and mirror rotates through an angle 8. (a) Thcn reflected ray rotates through the angle 28 in the same sense as the mirror rotates.

13

Geometrical (Ray) Oplies

N, N

,,, ,,

,,," ,,

10 '

R_

5........... 7

~' Q +O

-

:::::::::::::::~;;;;~ "

(1 119"

"

-

0:'-::::::::'-_~~-_-M, M,

(b) The angular velocity and angular acceleration of new reflected ray becomes twice as that of mirror. w, =2ro m : ando. , = 2a. m (00, = angular veloci ty of reflected ray OOm = angular velocity of mirror) -:: Note: If mirror is fi xed and incident ray rotates:
-. '" ,, ,

,

0'

,

,

Fixed Mirror

Note 1. The minimum length of a plane mirror to see one's full height in it is

1,

where H is the height of man. But the mirror should he placed in a fixed position which is shown in figure. A

. Man

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 two similar triangles ADF and BFC, AD = BC = x (say), similarly in triangles CDG

andDGE, CD = DE = y (say) Now. we can see that height of the man is 2(x + y) and that of mirror is (x + y). i.c., height of the mirror is halrthe height ofthe man. The mirror can be placed anywhere between the centre line DF (of AC) and DG(ofCE).

14

GRB Understanding Physics Optics and Modem PhysiC!

Note 2. A man is standing exactly at midway between a wall and a mirror and he want to see the full height of the wall (behind him) in a plane mirror (infront of him). The minimum length of mirror in this case should be

1.

where H is the height of wall . The ray diagram in this case is drawn in the figure. 2x

2y1

o

J

Wall

MirrOf

Man

d

d

In triangles HBI and IDC, HI = Ie = x (say). Now, in triangles HBI and

ABF. AF=FB HI BI AF 2d -=x d

or

or AF = 2x Similarly we can prove that DG = 2),. ifCK = K.! == y. Now, we can see that height ofthc wall is3(x + y)while that of the mirror is (x+ v). A poinl source of light S. placed at a

distance L tn/ront o/the centre ofa mirror

I\-------

of width d, hangs vertically on a wall. A man walks in/ ront of the mirror along a d

lim' parol/ello the mirror at a distance 2L from it as shown. The greatest dis tance over which he can see the image of the Ughl source in the min-or is:

(a)d / 2

(b)d

(e) 2d

(d)3d

Solution: (d) The ray diagram will be as shown in the figure.

HI=AB=d Since,

DS=CD=-d 2 AH =2AD

s

L 2L

Geom~trical (Ray)

15-

Optics

c _____ .1 _________ A,

B,

•"

,:0

G

H

,s :E ------1-------F

J

=2CD =2-d = d 2 /J = d

GH Similarly,

GJ =GH +HI +/J =d+d+d=3d

Ans.

Two plane mirrors M J and M2 are inclined at angle 9 as shown. A ray of light l which is parallel to M J strikes M2 and after two reflections, the ray 2 becomes parallel to M2. Find the angle 9. M, 2

k---+---l

Solution: Different angles are as shown in the figure. In triangle ABC,

a=900-9

A

c

9 +9 +9 = 180

0

9 =60 0

ADS.

Laws of Retl'llCtlon When a light ray is incident on a surface separating two transparent media, the ray bends at the time of changing the medium. This phenomenon is called as refraction and is governed by two laws: (1) The incident rays, refracted ray and the nonnal drawn at the point of refraction lie in the same plane. (2) Angle of incidence i and angle of refraction r is related as ~n i = n2 SIO r nl

GRB Understanding Phr-ics Optics and Modem Phys.ics

16

The above law is called Snell's law. 0= 0 1

•

,, ,I

, n2 :> n,

(i) If n2 > 11]: then sin i > sin r nnd hence i> r. The refracted ray bends towards"

the normal. (ii) If n2 = n\; the ray goes undeviated. (iii) If tl2 < II] ; sin i < sin r and hence i < r. The rcfracled ray bends away from the nonnal. (iv) If i = 0; r = O. Then the incident ray goes undeviated.

1.5

Properties of Refracting Surface'

Real and \'irtual object: An object is defined as the point of intersection of two (or more) incident rays for any refracting surface (like refrac ting surface as described earlier). The perpendicular distance between refracting surface (pole) is called object distance p.

Real and virtual image: An image formed due to refraction is defined as a point of intersection of two (or

more) refracted rays. If the refracted rays converge to a point. thai is called real image, if they diverge from a point, that is called virtual image. The perpendicular distance between image and pole (refracting surface) is called image distance

q. Real and ,"'irtual space: It is clear due to transmission of light that, infront of a refracting surface no refracted rays can exist. Hence the fron l space of a refracting surface is called virtual space of V -space. Since we can really touch the refracted rays behind the refracting v-space surface the back space is called real space or R-space. • Sign convention: For convex surface, the Incident ray centre of curvature is lying in R-space. Hence radius R of curvature is positive. For concave surface, the centre of curvature lies in V -space,

~Realimage ...... Refracting surface

I<~:('" . ~-

I Virtual image

.•

• Refracted ray

Refracting surface

Geometrical (Ray) O ptics

V-space

17 V-space

A-space

-A--<

- - A -C

R-space

Hence R is - ve. For real and virtual object, the object distance are positive and negative respecti vely. For real and virtual image the corresponding image distances are posit ive and negat ive respectively. I ft he focal points lie in R-space the corresponding focal length positivc, if it li es in V-space foca llcngth is negative .

TraCing Ray-diagram; Image F.ormation due to a PointObje,ct Relation between object distance and image distance Descartes' formula: Let two incident rays I and 2 fro m a real point object get refracted and Ihe refracted rays 3 and 4 intersect at 110 fonn as point image I. " ,

"'..

n,

....... "''''''' .. - h

o

o

a

"

4

2

; ' - - p - . \'

I----A

C

-I

ql------i

Fo r refraction at P, using Snell 's law, Itl sin i = " 2 sin r (assuming Itl < 1/2) For small angles i and r for paraxial my OP and PI, sin; -=:::; and sin r'::.r Then, 1I ,;= 1I2 r In triangle ope, i = 9+P In the triangle PCI, r = p- a Substituting j and r from eqns. (ii) and (iii) in eq (i), " I (9 + P) = " 2 (p - Cl )

II , puning 9=-

p

p = -II ,

R

and a = -II,m cqn. ('IV)

q

and simpli fying the tenns_~,_ __ __

..."."'l

-It ( +-112 -_"2- II , P q R

" ,(i) ,,,{ii) , ..(iii) " .(iv)

GRB Understanding Physics Optics and Modem Physia

18 .

.

,.,. d rays caned principal rays; ray -1

To trace the Image, we need two most sImp t Ie

.'

.

obliquely incident; ray -2 incident normally along the princIpal axIS. The meetmg

point 01 the corresponding refracted rays is called image. 11

II

For vinua\ object, put P""'_ ve and usc -1. +..1. =

"2 - 111

R

h . .. to get t e Image position.

p q Then draw the ray diagram. We have shown same possible types of images and their locations considering "2 > II, and "2 < "I as show in the diagrams.

--- --

c

...

-- _... ---

---- ..

--- -----

"'- ...

, "......... , ..... "- ... o

c

---_-------... _.... --~

--..

...

o

0

Important point .. The image position is decided by

n,,1'I2. nature of the surfaces (flat, convex or

concave). nature of the object (real or virtual), object positions. Hence by suitable adjustments of all the above parameters we can get different types of images as exhibited in the example.

Focal points: When we put q :: co and p = II in the equation nl + ~ = n2 - "I P q R' we have II = nlR "2

"I

When we put p =co andq =/2 in the eqn. nl +"2 :: n2 - "I; p q R

i

,

"

Geomt:trical (Ray) Optics

j

We have V-space

F,

A-space

V-space

A-Space

•

12

= n,R n2 - nl

Prove that (i) ~ = n2 = "2 - "I If1l 1121 R' (ii) f, + 12 = R. Solution: (i) Since the focal points FI and F2 lie in V-space and R-space, c - n,R and .Ic2 = + n,R wit. h t he Sign . . r .J I = convention of local length. Assuming n2- II 1 "2 - 111 "2 > "1, as R is + ve for convex surface, we have ~=.!!L.=n2-nl

If1l 1121

Proved

R

(ii)

or

Proved

Fr and ~ are known as object and image focal points respectively. 2. For concave surface Ii lies at A-space and ~ lies in V-space. 1.

1J!

Image Formation due to Spherical Refracting Surface

Consider the tip B of the extended object AB. Draw three principal rays; ray-1 from the tip ofthe object parallel to the principal axis; ray -2 goes through the centre C of curvature undeviated and ray - 3 gets refracted at the pole O. Assuming "2 > "" the rays -I and 3 bend towards the normal CQ. All the three rays after refraction meet at D' which is regarded as the image of the tip D of the object. Since the lowest point A of the object is located directly below B, we can find its image A' taken on the principal axis

direclly above B'.

.', 18-

GRB Undmlanding Ph),SLCS Optics and Modem Physics /

n, n,

B

"I

Ass'uming n2 > the rings -1 and 3 bend towards the normal and the normally incident ray 2 goes undevialed to meel at B' which Is the image 01 B

We can choose any two principAl rays to find the image of a point real object.

Magnification Lateral magnification: Lateral magnification is defined as the ratio of hei gh Is of image and object given as

A' C PA' - PC m = - - = ,;;",,-;;", BC BP + BC

q- R p+R

or

111= -

n1 "2 _ +- P q

Since,

... (i)

"2 - II I .

R

,

substituting R = 1/2, - "I in eqn. (i), ~+~

p

q

Aliter: In the 6S, ABO and A' B' O. for small heights of object and image (paraxial rays). "I

or,

sin; = 1/2 sin r "1; :::'n2 r

--~~----~--~~£ Geometrical (Ray) Optics

r.

II{

or

A:)=112(An

or

A~~'( =::~) = ::~~

or

III - r12 P

_ IIJq

(Proved)

magnificafIOn.. ItiS . dCfimed as the ratio ·ofthickncss of image and object givenLinear as

, oq

1/1=-

01' Taking differentials of both sides orthe equation ~ + ~

P

l/ ,

=112 -

R

111 ,

we have

, m'

or

= oq = _('1.)2~ dp

or

11/'

P

112

= '_ ~('1.)' "2

P

Relation between lateral and linear magnification: m'

IQ )' =_1l2(I/ n1 "2P

or ,112 m 2

m =---

II]

or

Image Formation due to Flat Refracting Surface , , p' R ' h' I "I "2 Real pomt obJect: uttmg = 00, In t e lonnu a - + - = , , p q

, II] EJj

112 - III

R

,

"2

q= - -P

For real object , p is +ve; then q is - ve; this tells us that the image is virtual formed in virtual space

GRB Undentanding Physics Optics and Modem Physics

. 18

A-space

",

"2

--I --, ,, , ,,

,

o '.

p--oj

If nl > n,.lq l< lpl If nJ < n2.jq l >lp1 V-space

n1 <"2

A-space

n2

____ r_______ _

-------------o

•

O- -oj

q

Virtual point object: For virtual object, put p = - ve, R = co, in the fannula ~ + n2 = n2 - "I to obtain

p

q

R

-

A-space ::~fi--------·

~v~=p=aoo~_______--4_~~~ -~--~-~--~-~--~----t> \o---q .II 10 o

_\

I

GeomdncaJ (Ray) Optics

Then, the image is real ( ': q >0) fonned in real space. Ifn2> IlI, Q > P Ifll2 < Ill . q < P

v-space

R·space

~-~--=-=-=--=-~j~:t:'~::~:~=~~O~-=--=-====:=--' ~ ~~-P---ql----.l.1 Extended object: If the object is extended. with the help of two principal rays from the tip of the object, obtain its image 8 1 (if " 2 > II ] ). B 2 (if 112 < "l)' Since A is directly below B . find Al and A2 directly below the points B] and 8 2 as show in the figure. ~

",

I,:

______________;B~_~--:BJ!2,--+-_~____ @

-----...

"

... __ _

I

,

A,

0 ---.. ___

21

I " __

\

\

: \

,

--1.-- \

A

nz < ",

where R = 00 for flat surface

then, for real object we will get virtual image for virtual object we will get real image If

If

>""

"2 Iql>/pl "2 <"" Iql
GRB Undentanding Physics Optics and Modem Ph),sia , ------------~~~~~~~~~~~----~-

Afish is situated at a depth 0/40 em i/l water ( ,,:::

j).

Filld (i) image a/the ftsll, ( ,' ,' )

,:.:'c,,-p;;;'hc..,.,o/the Imter tank (or fiIS') I ,

__,-='".::.a:..:'

apparelll depth

(iii) velocity a/the image o/the fish if,., mOl'es dOll'lIwilh a speed 0/ 5 em/so Solution: (i) Putting 111 == II, 1/2 == I. R -= to and p == +pin Ihe famm la, II , -" \ +.....:. == p q

' " - II I ~

R

;

~ ®

!! + ! ::: O p q or or

q

3

p

q ;::- ~x 40em - 30 em (virtual image)

Ans.

;:: " = 1

Ans.

(ii) The real depth . == 12 Apparent depth q

If

, '

t

4

=

(i i i)

I

9.. ::: -V". where 1/::: ~

3

a

,, ,

,

pCD

, ,

'-

tip

- = 5 cmfs, tI,

~,; \ =H'~~ \ =!-4:3\ X5 ::: +!1 cmfs downwards

Ans.

4

Important points 1. Apparent Depth ==

Real depth A.1. .

"

2. ~image 11 vobJect for refraction as flat surface.

1.8

Total Internal Reflection

Definition: Similar to a mechanical wave. an emw, that is. a my of light undergoes reflection and refraction when at the interface of two media. As discussed in last section. when a ray of light moves from denser medium (medium of higher refractive index) 10 rarer medium (medium oflower Rl) not only it bends away from the nannal upon part refraction (transmission) but also part reflection takes place obeying

Geometrical (R-ty) Optics

--~~----~----~~ the len s of reflecti on Thepart" I fi . then some critical an' .ta re ractlon tak.es place for all angles of incidence lesser 0 (tota l) energy of tl glc c· FO:I~hc angles ofmclde nce i greater than critical angle, all h . Ie wave W I be reflected hack into the original medium This p enorncna IS called total internal reflection (TIR). .

,, ,,, ,

,, ,,,

,,

O{~

, ,, Critical angle: Ifu2 < Ill. r> i. Then for i = SCI r = 90° (the refrncted ray grazes

the interface). According to Snell's Jaw, "I sin ac = "2 sin 90° or

ec

· - tI12 =sm -,

"

"I "

Partial reflection takes place for any angle of incidence, for n, < ~. However, if 11-] > "2 , partial reflection takes place for some angle of incidence 0 < Be and a > Be total inlernal refleclion takes place. At normal incidence (i = 0), the ray is refracted without any deviation. At oblique incidence, the refracted ray deviates towards the normal when the light moves from rarer fo denser medium and deviates away from the normal when the ray moves from denser to rarer medium. However, in all cases the velocity of light changes.

1.9

Prism

Definition: A prism is a medium between two flat planes inclined at an angle A called angle of prism. Let us consider a ray of light of given frequency (monochromatic light) is refracted hvice at the points P and Q with angles of devintion 8, and 8 2 respectively. Then, the net deviation of the my. that is. the angle between incident and emergent. I

•

~

18

GRB Undentanding Physics Optics and Modem Physics

.-. ---~~~.:.::::::::.==--'-M

-- -- ---

Imergent ray

Deviation: Ray is D=6 J +62 =(i-r)+(e-r') = i+e - (r+r')

... (i)

InPMQN, If + A'= 1800

In

PNQA'+r+r'=180 Then, r+r' = A Substituting r +,.' = A in eq (i), D=i+e - A LImiting angle of incidence, condition for grazing emergence and maximum deviation: If the emergent ray grazes the second plane,

,

,......

.. r=r 9O-r' " e=90° and r' =9" Then,

r=A-r'=A - 9 c

Hence, the deviation is maximum, Drrax =;-r+90-,.'

--D

'2~;. « -~====~==~~::~----------------------------------,--'~ ~ ~ Geometrical (Ray) Optics

~-

= i +90-(r+r' )

IDJTD)I. = 90 +

imin -

Atl

where, ;m = sin - l (Jn2 - l sin A - cos A) Minimum devlltion : We can prove theoretically and experimentally that, for

minimum deviation.

I i=e and

r=, I

-------

This means that the refracted ray must be

parallel to the base of the pri sm. Putting r = r' in A = r+ r',

6 (or 0)

I r=~ I Putting e = i and D = Dm in D = i+e- A,

I

i= A +2Dm

I

Substituting i and r for minimum deviation in

For minimum deviation refracted ray Is parallel to the base of the prism D

Snell's Jaw,

sin i n = - .-- • we have smr Om

,, ,, ,, ,, -i---,, ,, ,,

sin .4 Ie 1- e 2 Minimum deviation for thin prisms: Ifthe angle Deviation Is minimum when / of prism is small A« JO O ), the deviation will be small .

•

..

Then,

. A+Dm_A+Dm d' A _ A sm an sm - - -.

This yields,

or

2 2 Dm +A =n A

2

2

A

I Dm = (n-l)A For A :s; 100 .Dmin =(n -1}A

1.10 Dispersion

When a ray of light moves from airto any medium its speed in the medium is given as,

I v= t=~' 1

where n == JE; = Rl of the medium.

"

--"------

,',

; . '_________~G~RB~~U:"n"'d~,~""'ta~n~d::;n~g:._P.:h::,..:;,,::..:O::.:::Pt::.;~"a_nd::.:.M.:.od:..:.:,~m_P:..:.:h,..:.::: ;" .

. . .

d'

d

The relative penmttlV1IY Cr or a me mm epe of light given as,

nds on the frequency or wavelength

• which is know as Cauchy's law. Different freque ncy oflighl travel with different speeds in a medium. This phenomena is called dispersion and the medium is said to be dispersive.

,

.

~ - ! - - - - - ), "

n decreases with

>.

1.11 Angle of Deviation If a ray of light (of a given frequency) falls obliquely on a refracting surface, it is refracted with an angle r given as II sin r :;:: 110

sin i The ray of light dc~'iatcs from its original direction by nn angle. 0=; - r

,

.

Since 11 = a + ~ and )., red is maximum: the RI ofthe medium for red is minimum;

I.'

hence "r" for red is maximum. Then. The angle of deviation 0 is minimum for red light.

Similarly.).. violet is minimum, the Rl of the medium for violet light is maximum, r is minimum and hence 5 v is maximum.

n

,

Violet light has maximum deviation.

Geometrita1 (Ray) Optics zj 29 ------~~----------------------~~ 2 · :---

1.12 Angle of Dispersion Then, when a ray of sun (white) light containing seven colors (wavelengths) IS refracted with different index of refraction thm depends on thc frequency of the component waves (colors). Hence the beam diverges from the point of incidence having seven different colors; red with minimum deviation and violet with maximum deviation. Thus, all seven rays are dispersed with an angle between red and violet given as

I~

- B" - B"I

The red light deviates minimum and violet where 4> is known as angle of deviates maximum. The angle of dispersion is ¢ = 5v - Sf dispersion or angular dispersion. The

angular dispersion measure the quantity (amount) of dispersion:

1.13 Dispersive Power Since the ray ofyello\\' light lies at the middle or mean position of all seven rays, the deviation ofthc yellow light is called mean deviation given as is )' or 5 m:an' The angle of dispersion. When divided by the mean deviation. that is deviation of yellow light gives us a quantity ca lled dispersive power given as.

/

.

"=

~ =B, - B,/

5 mean

5y

_

COil a glass slab splitlhe colors of a SUlI figlll ?

Solution: At (he first surface thc sun ray is split into seven colors (mys). At the second surface all the rays more pardUel with the incident ray after emerg ing from the slab . In consequence, at the colors wil! again superimpose to yield white light. Hence. when the emergent beam fall s on a white wall or screen we cannot see any color. Then. we need the second surface so that the emergent ray will not be parallel and that is possible in a prism.

1st sUrlace'

"'

1.14 Prism and Spectrum In a prism. a beam (ray) of white light emerges with a diverge beam of seven colors with different angles of deviation. If the emergent beam falls on a screen !wall I photographic plate, weean see a pattern of seven colors is red ... yellow ... violet. ofa rainbow. This pattern of colors is called "Color spectrum".

GRB Undentanding Physics Optics and Modem Physics

Screen

DIspersion of light in a prism

fu .!ft..,~

For an incident sun light. find the (;) mean deviation (m angular dispersion (iii) dispersive power of the prism of small angle A for minimum deviation. Solution: (i) The mean deviation is lirrcan =liy =(ny - l)A

ADS.

(ii) The angular dispersion is

q,=li v- lir = (n. -I)A - (n, -I)A = (n v - nr)A

ADS.

(iii) The dispersive power is ",=

$

~~

= (nv - nr)A (ny-I)A

Ans.

Lll Combination of Two Prlams DI..,.,.1on without Mean Deviation When we put a prism of apex angle A) with an inverted prism of apex angle A2. with the assumption that the emergent yellow ray has zero deviation as the yellow ray will be parallel with the incident white light, The emergent yellow light is panlIMtl to the

incident sunlight causing zero average deviation

• Geometrical (Ray) Optics --------~-----------------------~

Putting,

8 =8 Y1

- on

BYI

= By>

·.31

=0,

... (i)

The net angular dispersion is

~ =~I-~' (.:~

=001°.., -rollin

q. =oo]l5 Y1

or

Using eqns. (i) and (ii), , = (00\

-

= roB,.) .. . (ii)

-rolO n

«2)0 Yl =

(WI -

(02)0)2

I ~ =(rol-ro,)("" -I)AI - (rol-ro')("n -I)A,I

or

MHn devIlllon wIthout dioper.Ion As the emergent beam consists of parallel rays, the angular dispersion is zero.

~

=0

~I - ~' =0

or

~I =~,

or or Then the mean deviation is

OlIOy] =002°)'2

... (i)

... (ii)

Using eqns. (i}.ra_n_d..:(_ii)c.,_ _ _.,.-_--:-_ _ _ _--:-_ _:-,

or =(n>l

-1)A{I -;;-)=(nn -1)A2(~-1)

It is important to note that, Since the prism 1 is joined with the -inverted" prism -2, the net deviation is or :: 0Yt - 12 and the net dispersion is , = ., - ~. The ~minus· sign is due to the fact that, prism 1 tries to bent the rays down and prism -2 tries to bend the rays up.

°

• 32

A

GRB Understanding Ph},sics Optics and Modem Physics

~~.~ ' ------~~~~~~----~ 1.16 Lenses and Its Properties

.. . transparent me d III · m bounded by two curved DefinatlOn : A lens IS (spherical/cylindrical) refracting surfaces. In some cases. we have lenses of one surface

R,

: Thin lens

Thick lens

curved and the other is plane surface. When we neglect the thickness of a lens we call it "thin lens", Types of lenses : Depending on the nature of surfaces of the tenses we categorise them shown as following: (i) Bi-convex : Two convex surfaces. (ii) Bi-cOIlcave : Two concave surfaces. (iii) Plano-convex: onc surface is convex and other is flat surface. (iv) Piano-concave one surface is concave and the other is flat. One surface (v) Convexo-concave convex and the other is concave. one surface (vi) Concavo-convex concave and the other is convex.

" "

0

TI

Bi-convex

Bi-concave

]

Mono-convex

~

Plano-concave

Convexo-concave

Concavo-convex

Incident ray

7

, ,,

- ,, -,,

f'_......

'"'"~

,

...... l...

Tangents

, ,,, , ,,,

o

-,'-'

,,

c, Normals

,

Emergent ray The incident ray and emergent ray are paraliel when the refracted light passes through the optic centre.

33

Geometrical (Ray) Optics

Optic centre (or pole) : It is a point on the optic (principal) axis inside the lens through which a ray of light will pass through it will emerge parallel to the incident ray suffering no deviation . . Then, the triangles CIOB and C 20A are similar. Hence, the position of optic centre is given as

o

C20 = C2 8 C,A

cia c,a -R, --= cia R,

or

Therefore, the position of the optic centre (or V-space pole) does not depend on thc choice of the rays. It is · a unique point. IncIdent ray Sign convention: A lens divides the space into two parts. The front of the lens where incident rays exist is called virtual space or V·space. The back space of the lens is called real space o r R-space because really the refractcd rays exist in backspace, (i) For real object, p is +ve and for virtual object. p is - ve, (ii) For real image, q is + ve and for virtual image, q is - v~. V-space

.'

A·space

• Refracted ray

R-space 1

2

- -

V·space

C,

- -

A·space

2 -R,.--++--IR,, - -C,

For convex lens, Rl is +ve, R2 is -ve, For concave lens, RI is -ve, R2' is +ve.

(iii) For the centre of curvaturcC\ of J st surface of convex lens, R\ is + ve as it lies in real s'pace similarly R2 is - ve as C2 lies in virtual space. Forconcavc lens, C I and C 2 lie in virtual and real space respectively, Then. R\ is - ve and R1. is + ve of concave lens. n (iv) If the focal length is + ve, the focal point lies in real space and vice-versa. The focal length is n' n' n n' given as

} =(;. -I)U, -

n

jcan be +ve. - ve and infinte depending on n', n, RI and R2 . Ifjis + ve, the lens is converging (but

.1

2

1

2

The focal length of a lens . depends on n, n', ~ and R2

34

GRB Understanding Physics Optics and Modem Physics

need not be convex); iff is - vet the lens is diverging (but need not be concave). Lens-equation : Tracing ray diagram, using a real object AB. we gel a real image .due to the bi - convex lens. Compari ng the similar triangles DA = AB DA' A' B'

ABDandA'B'D,

... (i)

,, ;

C

B

. ,,

Q

,, 'A'

F

A ,

,, , ,,, , ,,,.

,, ,

D ,

,, ,, ,, ,, ,

p

,

'--:

Comparing the similar triangles CDF and A'B'F, DF CD AB

.. . (ii)

FA' = A'B' :::: A'B' Using the cqns. (i) and (iO,

DA = DF DA' FA'

or

E.=L q q- f

or

pq - pf=qf

or

(p+q)f = pq

or

I ~ +~=;I

Focal points: The above lens equation tells that if we inter change the object and image, p and q will be inter changed and the equation remains unaltered . This

means that, the two focal points Fl and F2 will be syrrunetricalJy situated unlike single refracting surface

satisfying the reversibility oCthe rays.

--;~I,,=~,,=tt,\=~'~'=~·Il-F,

F2

Geometrical (Ray) Optics

35

1.17 Image Formallon due to Lenses Taking an ClTect object AB on the principal axis. Let us find its image by trucing ray-diagram. First o f all chose the lip B of the objcct . Being two rays arc parallel to the principa l axis which a fter image refracti on passes through the focus F; the other passes through th e optic centre undev iated . The two refracted rays meet at B ' in real space for bi-convex lens and in virtual Space in bi-concavc lens. H' is thc image of the lip B of the object. Since the lowest point A of the object is located at the principa l axis directly bel ow B its image A' will be formed directly above B' for bi-convex lens and below B' for bi-concave lens as shown in the figure . Hence, we can gel the real image A' B' in bi-convex and virtual image in hi-convex lens. B

1

2

o

A'

A

Real and magnified image

S

FA

S'

1

At

Virtual and dimlshed image

Real image is inverted and virtual image is erect for an erect object due 10 a lens. The rays 1 and 2 are catled principal rays.

Lem;-maker formula (equation) : Let the point object 0 on the principal axis givc a final image /2 due to the convex 0 - concave lens of radii of curvature R] and R2 Rl = n sUlTounded by two media of Rl "I aJld 112 respectively.

GRB Understanding Physics Optics and Mod~m Physics

36

1

",

"'

---------q

I'

q,

p

I,

I,

-I

For refraction at surface t, image II will be fonned fOrlhe real object O. ~ + 112 = 112 - III

P

ql

... (i)

RI

For refraction at surface 2, image 12 wi ll be formed for the virtual object J I :

_" _ ' _ + fli = ~"" I ;,--.::",,' q R,

... (ii)

(- q ,)

Eqns. (i) + (ii) yields

",(!+1)=("2 -11,)(_1 __I ) p q RI R2

!+1.("' _I)(_1 __I) 1.("'_ 1)(_1__I) f

or

p

or

q

RI ' R2

fli

"I

RI

R2

Magnification : We can sec that the linear dimensions of an image are different from that of the object. Considering the enlarged (magnified) imagc the lateral magnification m is defined as the ratio of height of the image and object.

",

A

ho

-

op

",

"

8

q

oq B'

P

I,

I I

'" '" , " , '" '" " , h, '" To "

"

0, ' 00

A'

"

• •

Geometrical (Ray) Oplics

--~~~~~~------------~----------~,

37

,, ---

Latera I (tra nsvcrse) mai'g:::n~ifi:::IC:;.~tl::o:::nC:::--'''_"7--' Im.'lgc height h; 111= = -Object height - 110 In the similllf triangles ABO and A'O'O,

m= A'B' - AD

DB' =- -DB =

I

or where Ji

-q p

m=

-~ = - P ~ill

= n2; p "" power of the lens. p

The lateral magnification is numerically equal 10 the ratio of image and object distances. For real image mis +ve (because qis + vel and for virtual image mis +ve (because q

is -vel.

•

Linear (longitudinal) magnification: The linear magnification m' is defined as the ratio of thickness of the image and object.

•

1 m' = 6q 6p

I .

Rel.tion between m' and m :

m' =

{'!.)' P

= _(

Ii )'

p-

Ij

Relation between m and m ' : According to lens eqn. :

• or

1)1 • .,"

_ _I dp _ _1 dq =0 p2

•

! +! = 1 p q I q2

or

I~I =(~)'

or

jlm'l= m2 1

"

,

38·

GRB Undtrstanding Physics Optia and Modem Physics

---4~--------~====~~~----~An object of height 2 cm is placed at a distance of 10 cm from a bi-convex lells oj RI = 20 cm, R2 = 30 em and n = J.5. Find

_.L_-+-+___ -...J

the : (i)!ocallength of the lens.

10em

I

(U) image di.{tan ce. (ii,) image size, (iv) Ii/fear magnification. Solution: The focal length of the lens is (I)

or

-I =(n - I) ( -+ - ) = (\.5 - 1) ( -+II II ) Rt R, 20 30

J

Ans,

J=24cm

(ii) PUI. p =+ lO,q = 1.

f

= +24cm in!+! = !toobtain.

p q

J

~+! = l.

24

10 q or

(iii)

q=

-liO

em (vinual image)

hi = mho = ( -;

Ans.

}Q

= (-12017)x2

(10) = +;4 (crect vinual image)

(iv)

Ans.

m' =m 2

j Ans.

1.18 Slivering 01 Lenses When we silver are refracting surface of a lens. obviously it will behave as a mirror. An incident ray fonn a point object 0 gets refracted tv.·ice by the un-silvered and reflected once at the silvered surface to form the final image I. Let nJ = RI of the surrounding medium and"2 = refractive index orlhe lens. For refraction at surface I, the image is II ... (1)

Geomdrical (Ray) Optics 39 ----~~~----------------------.--2 1

n,

n,

o ,, ,

,,, ,,

,, ,,, ,, ,,

I,

------

,, :,

I,

,, ,, ,,

, ,,, , , ~1,..~--,q2---.f-----q'-----+i·:, , '

,

~,,~---p---~ !

The image I, serves as a virtual object for the mirror (silvered surface 2) fonning the image 12 , Applying mirror formula,

I + - I = -2 ' -(-g,) g, R,

... (ii)

Then, the image 12 acts as a virtual object for the refracting surface 1 fonning the final image at I .

_"_2 + ~ = _",-I-."-'",,, - q2

q

- R2

... (Hi)

I I I -+-=wehave p g [,rr'

Substituting

Find Ihe eJJeclive focaiienglh ofa hi-convex lens ojreJraclive index n and radius R. Solution: where,

2("' ) f:ff =(:~ -1)(~1 +;2 )+ ~ . R,

= R2 = R.

nZ

= n,

n, = 1

40

GRB Und~rslanding Physics Optics and Modtm Physics .

----~~~--------------------~~--~----------

r

Then,

_

R

Ans.

J
1.19 System of Thin Lenses DC\'iation ora ray: The deviation of the ray produced by the lens is

o=a+p

B=l!+ !!

or

p q

o="(~+~)

or

I

r p-HH---QI------i'l

.,

' [gJ s =-fh

:. or

Effective focal length of two lenses of focililengths II and h placed at a separation d: Let the ray parallel to the princip~l axis suf(e~ a net deviation 5 due to the system of two co-axial lens.

h,

i--2 "'- ...

" B

..

D

E

----

F

I--(f'-d)~

41

Geometrical (Ray) Optics

where3 1

=

deviation due to the lens-l and

32 = deviation due to the lens-2

"I ' II

Putting

31= -

'"

and 3 2 = - , we have

12

8 =!!l+"2

II 12

or,

"I

h\ 112 -=-+-

I

II 12'

where I = focal length of the given system of two-lenses .

.

1 = _1 + n2/n l I II 12

or

'" DF II -d h h; = BF = II ' we ave

Since,

1

1

(f1-dY/I

II

12

-=-+

or

I !

= _I +_1 _ -.!L

I II I, 1112

The position oflhe effective lens is given as

x = ("I - "2)eo,6 =

;,

("1/",)/6

., ,

= '" - ",/(",/f)

=(I-~~ )r

.

-

.,

-

,

~

.

-

l..;...:.

x = Id , from the 2nd lens.

I,

When the lenses are in contact d = 0 ; 1 1 1 - ,.. -=-+Then, ~

, - -

..

I I, 12

Combination of lenses placed in contact: For the sake of simplicity let us assume that two convex lens of focal lengths II and 12 are placed in contact. To find the effective focal length of the combination, let us take a point object 0 on the principal axis. The lens I will produce an image II which setVes as a virtual object for the lens 2 to produce a final image I.

GRB Understanding Physics Optics and Modern Physics

42

~ - --

0

,

2

--.. -- .... --

I,

.--1

.,

p

•

Then, applying Lena fannu la for the object for the lens I,

1+ _' =1.. P q , II

... (i)

The image I) behaves as a vi rtual object for lens 2. Setting p = -q I' and q = q. f = h in the lens fannula for lens 2. .. .(ii)

_'_ +1=_' q h

- q, Eqns. (i) + (ii),

or

l+ ! =..!. +_ ' p q I, h _,_ = 1.. lorr II h

+_,

Similarly. for many lens placed in contact,

~

~ 1.20 Optical Power and Focal Length Ca) Renecting surface: The optical power of a reflecting surface is given as

p=l=l I R'

, ---R--:I----------I--IR-P Is +ve

P is-YG

where R = radius of curvature of the reflecting surface. For concave mirror, R is + ve, and fo r convex mirror, R is - ve.

-,

Geometrical (Ray) Optics

43

The optical power of the mirror does not depend on the surrounding medium. (b~ Refracting surface: The optical power of a refractmg surface is given as

p = "2 - Ill R Convex surface: If n2 > nl (and R is +ve ), P is +ve. If n2 < nl (and R is +ve), P is -yeo

n,

n,

-R-C

Pis+velfn2>n, Pis-veifn 2
-

Concave surface: If " 2 > "1 (and R is -ve). P is -yeo If "2 < III (and R is - ve), P is +vC.

n,

n,

C-A--! P Is -va If n2> n, Pis +ve If n2< ",

The optical power of a refracting surlace depends on the refractive index media at both of its sides and the radius of curvature oltha interface. If Pis + va, the surface is said to be converging and if P is -ve, the surface is diverging.

The relation between optical power and object focallengthlo (or 11) is given as

P=~,

II

/p=n 2 ;n l .

where

The relation between optical power and image focal length fi (or h) is given as

p =

n2

h'

where Find the optical pOl..'er ofa (i) convex (if) concave mirror ojradius of CUn'olllre Rr = 20 em and R2 = 30 em respectively.

Solution: (i) P (ii)P

=1.. =..L x 100 =- 100

Ans.

=1.. =_2_x 100 = +6.660

'Ans.

RI

R2

- 20

(+30)

Find the (I) optical power of the inter/ace (m corresponding focal lengths. R = 20 em.

44

GRB Undmtanding PhysiC! Optic! and Modem Physics

-,

- 1.5 x 100-_ -5 I p _ ", - n,_4/3 _ _ 0 5o-1U t'Ion.'(')

R

+20

6

(..II ) f 1 =II,- =1.5 - - =-1.8m P

h

Ans.

- 5/6

= " 2 = 4/3 = -1.6 m

P

Ans.

• I J

n, = 1.5

Ans.

- 5/6

(c:) Thin lenses: A thin lens has two refracting surface 1 and 2. The power thin lens is then scalar addition of powers of each refracting surface given as.

or the

P=P\ + P2 1

2

The focal lengths orthe lens are ~ . n, P

and

f2 '" !!g, where P "" P

n(J. - R' 1 !'2.Rn R\

2

+

2

RfJ! without sign I

conv~ntion.

Find ,he optical power alldloco/leJlgths ofa bi-com 'ex lens ofradius ofcun'oturc RI = 20 em, Rz "" 30 em. Assume that the RI a/the fells is 1/ =1.5 and the RI a/the media at both sides are II] = 4 / 3and " 2 = 1.6.

Solution:

p = n( _1 _ _I_)+~ _ ~. RI R2 R z R,

"I

= ~ "2 =1.6 3

where n = 1.5, "I = 4 /3 and "2 = 1.6 Putting RI = + 20 em, Rz = - 30 em, we have

P = {1.5(1. +1.) + 1.6 _ 4/3} x 100 0 20 30

- 30

20

= 0,50

"I

4/3

n2

1.6

Ans. 8

f '=P=1 /2 =3 m h =P=1I2=3.2m (d) Thick lens: Let the powers ofthe interfaces I and tofthe thick lens be PI and P2 . Then, the powerofthe thick lens is given as P dPIP2 P _ n-nl P _p, - 1 +2 . 1- -n R,

,. Ans.

.,

Ans.

,

45

GcomdricaJ (Ray) Optics

P2 =

and

"2

-1/

R2

where n = R.I. of the medium between the interfaces and d = distance between the interfaces .

...

-

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

~1UI'!!tU.'!.!..

Find the optical powero/thc glass ball ofR.I. = Solution: The power of each interface is

PI

II Qlld r(ldills

R = I 1tI.

=P, =11 - 1 =1.5-1 =_,

d=2R

+R

R

2R

n

and Then, the optical power of the glass ball is

-A

P=P, +P2 _ dPI P2 II

, , =2R +2R-

(2R>(~ )(~) 3/2

= (2 /3)R =?D 3 (e)'Thln lens separated by a distance d : The previous fonnula can also be used as the power of the combination of two thin lenses separated by a distance d as shown in the figure

r-- - - - - .

ADS.

",

4----d

",

" PI :::: Power of the lens I. in the given media. P2 :::: Power of the lens 2, in the given media. d = distance of separation between the lenses. n = R.I. between the intervening spacc. A bi-concaJ..'e lens ofoptical power -W;s kept at 0 distollce 0/20 cm from a bi-convex lens ofoptical power +3 D in air. Find the power of the combination. .

Solution:

Pelf

= PI + P2 -

dP,P2 =-"-" n

Urn j-2D)(+3D) = (-2D) + (-I'3D) _ ,,5,--,,--;-_ _

,

Ans.

GRB Undmtanding Physics Optics and Modem Physics

46

1.21 Culling 01 len. 1. The focal length of a lens without sign convention is given as

PA

L(!!_IX _IRI __R2I) f 11'

Hence, when we cut a piece from a lens parallel to its PA , its focal length remains unchanged. Then it behaves as the original lens. 2. When we CUI the lens perpendicular to its PA, we obtain two plano-concave or convex tenses of focal lengths II and h given as

_I =(!! _I)_IR\

II

11'

and _I = J n, _ 1)( _1 ) without sign convention. Since II :t. h in general, h \ 11 R2 these two lenses behave as different lenses. If we shift the cut ponions the combined lens will be different from the original lens. 3. Jfwe cut and remove a ponico of a lens, the PA is not disturbed. Hence the

remaining two lenses behave as the original lens and form same image of the object for paraxial beam.

L,

,,, ,,,

,

__ _ __ L ___

, __________ _

~

o

4. lfwe cut the lens along (orparaUel to) PA and shift them along the PA. their PA do not change. Then, generally two image will be formed on the PA for a single point object 0 placed on the PA . .

..

L,

""""==:::::: O~~I' I,

,

47

Geomelrical (Ray) Optics

S. Referring 00.4, jfwe shift the cut lensesLI and0 perpendicular to the original PA. their principal axes will be shifted parallel to Ihe original PA. Then the images II and 12 will be formed not on the original PA.

L,

,,

- -----~------ PAl , PA

,,

o

____ -L _____ .

PA

2

"

6. Ifwe put two cut pieces (halves) at a small angle theirprincipai axes will also be rotated through same angle. Hence, we will get two distinct images II and 12 for same object 0 placed on the original PA.

----, "-

,,

---- ----

"

7. Ifwe put halves side by side in case (i), the effective focal length is infinite as net deviation of the ray will be zero. In case (ii). the net deviation will be double giving size two the focallcngth / -I= -I+ -I or feq =-

/",

/

/

2

Case ~)

A

-

Case (Ii)

" 1\

"~

V ..

f

..

~f"--I When the principal axis is disturbed after cutting and shifted, images for same object.

r

we

can obtain

two

, GRB Undentanding Phy~ics Oplia and Modem Ph}'lics

We have dealt the geometrical (ray) optics in terms of real and

virtua., spaces ~nd

real and virtual objects and images. There is another method to deal with ray optiCS. In which we use new Cartesian sign convention. Since il is also a co~mon met~od we will just discuss its approach and write down the expressions for thiS convention which we have already derived with real-virtual space convention .

New Cartesian Sign Convention:

O/P means optical centre/pole AB Is object. y ve t+

At

Incident ray

(+ve)

•

B

Principal

O/P

•

(-ve)

axis or . Optical axis

L.

"'

Mirror or lens Is kept at OIP. (1)

The distances measured in the direction of incident ray are taken as positive and those in the direction opposite to incident ray are taken as negative.

(2)

Distances are taken as positive when above the principaVoptical axis and negative when below the axis.

(3) The distances of the objects and images are measured from the optical centre or pole (i.e., Ol?) which is regarded as origin of Cartesian coordinate system .

Mirror formula:

1u.1 + v.1 ::: 11 · f'

(For both convex and concave mirrors)

u is the distance of object from pole v is the distance of image from pole fis the focal length of mirror.

All the values should be put along with sign. Formulae Related to Mirrors :

(1)

f:::~:

(2)

Power of a mirror is defined as,

relation between focal length and radius of curvature of spherical mirror.

p

=_

1 f (in m)

=

100

d"

f (in em) loptar

(3) Transverse magnification (when object is placed perpendicular to the principal axis).

.'

49

Grometrical (Ray) Optics

1m"" 6= -: I; (4)

I = Size of image, 0 = Size of object

Longitudinal magnification (when object is placed with its length along the principal axis);

- vtl = - -d dv . (for small objects) u2 -u, u

I (V2 mL = 0 = (i)

1

1

In case of small linear objects, from - + -

v u

=-1I

We have

so

i.e., (ii)

mL=_dV du

2 = [~l2 =rn u

If a 2-D object is placed with its plane perpendicular to principal axis (see figure) its magnification, called superficial magnification, will be :

a

mba

•

ma

area of image (rna) (mb) 2 ms = area of object = axb ;:::: m (iii) In case of more than one optical component, the image formed by first component will act as an object for the second and so on. So over-all magnification,

!:L !:L ... =~xqxq ...

I It m=o=~x~x~ Refraction at Curved Surfaces:

o (i) .... 3 _Ill _ 112 - Ill +}.I3 - }.I 2

(ii)

vu -

(iii) Lens formula:

where

R,

I~-~=~I

H:m

-1)

R2

[~t -~2l

R

GRB Und~nlandin8 Physics Optics and Modem Physia

50

(iv) Magnification for refraction as single curved surface,

! m =~ (v)

I

Magnification when lens Is put with different mediums on either side,

rn th~~~~;es,

If medium is same on both

1'1

111 = )13·

61'3 \)

I ~I

m= (vi) Power of a lens kept in air, p =

1

f (in m)

=

100

t (in em)

0

>,'

r

----

-~

51

GeomdricaJ (Ray) Optics

~-"

MISCElLANEOUS EXAMPLES

~

,~~~

-

~

Example 1. Ifray of light suffers successive rcneclions at two mirrors inclined at an angle 9, aftcr falling on the first mirror at an angle of incidence a , its total deviation is independent of a. What happens when 0 = 900 ? Soludon : Total deviation of the incident ray due to reflection from mirrors M I and A12 • i,e.,

5 = 5, +5, = (l800-2a) + (1800-2~) = 360 0-2(a +~)

. ~ .(1)

Since (90°--a)+0+(90°--/l)=1800, O=(u+~) ... (2) D From equations (I) and (2),6 = (360 -20). Thus, the total deviation is independent of the M2 angle of incidence (on either mirror) and is J dctennincd only by the anglc between the mirrors. 1 If 0 = 90·, 5 = 3600-2x90· = 180° ,,, , Thus, when the mirrors are al right angles. a ray a'a aftcr successive reflections from bOlh travels parallel to itself, bUI in a d irection opposite to thai before incidence, i.e., the direction oflhe ray is reversed in this case. Example 2. A convex mirror of radius ofcurvalurc 30 cm fonns the image of an object at a distance of to cm from the pole of the mirror. Find the position of the object and the magnification.

Solution: v

f

= R = 30 = +15 cm (for convex mirror f is positive)

2

2

=JO em, (for convex mirror image is always virtual, so v is +ve)

I I I -=-I I -,u=--= I Iv 15xlO =-30cm -=-+-, I u v u I v I-v 10-15 u is negative. So the object is placed infront of the mirror. Magm·fi· Icatlon = m = -v = -10- = --.I u (-30) 3

The image is diminished. ADS. Example 3. A square wire of side 3.0 cm is placed 2S cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire? (The centre of the wire is on the axis of the mirror, with ils sides nonnal to the axis). Solution: u=-2Scm,/= - IOcm, v=?, hi =3 .0cm

GRB Und~ntanding Physics Optics and Modem Physics

52

l-!_l ~-'-+!. ~ +25-10 ~ -15 ~ -3 v-I" - 10 25 -25 xlO 250 50 II, -

=-~x!tJ It

Area ofthc image =

=_(-50 3)=_2Cm 3x - 25 X

hi = 4cm 2.

ADS.

Example 4. A thin rod of length f 13 is placed along the optical axis of a concave mirror or focal length f such that its image which is real and elongated just touches the rod. Calculate the magnification. Solution: As in question. image touches the rod, i.e., image and object coincides, hence one cnd of the rod I u' should be at the centre of curvature. It is also written that 1-------1-----image is enlarged, it indicates that the orientation rod .-~_-"-"_-,,_-,, _-,,,,,==-~-II p ..... should be towards focus then only we can get enlarged Image .. -:0:.3 ___ 11 v image along the principal axis. Let I be the length of the image.

or

10--------

J=m/ 3

Then,

m=-I-

Now,

11'=11+ /

1 13 Also, one end of the image coincides with the object, u' = 2/. 3

:::::) u=2/-/ =5/. 3

V

3'

=-(11+/ +m/ ] 3

3

Putting in mirror fonnula, we get I +l~! => 3 +l~! ,,+(j13)+(mI13) " I 51+1+"!/ 51 I I 2 3 => - - = - :::::) "' = Ans. m +6 15 2 Example 5. A small plane mirror strip is kept at a distance of 10 em infront of a convex mirror, with its plane nonnal to the principal axis. An object is placed at a distance of20 cm from the plane mirror, as shown in figure. Calculate the focal length of the convex mirror. if the image fonned by the plane mirror and the convex mirror coincide, without p,uallax. Solution: The distance between the object and plane mirror = The distance between image and plane mirror.

,, ,,

4M

o

P

... -----------t----20cm 10cm

----....j

10cm

This means, the image is formed behind the convex mirror, at a distance of 10 em. because,OP = 20 em. u ~ -(20+ 10) ~ -30 ern, v = +IOem.[ ~?

•

~mdricaJ (Ray) Optics

53

l=l+lor/ = ...!!!:.., /= - 30xIO _ -300_ 15 11 V u+v - 30+10--20-+ em

f

ADS.

Example 6. A concave mirror and convex mirror of focal length 20 em each arc placed facing each other. 80 em ap~~ (sec figu.rc). An object of height 4 em is kept midway between them. Fmd the pOSitIOn and ,height afthe image fonned by reflection, first at the convex and then at Ihc concave mmor. Solution: For reflection nt the convex mirror, f = +20cm, II = -40 em I'

= _fi_"_ = 2_0--;x-;:,(,--4-;:-0",) 11- / -40 - 20 = 13.33 em

II is the image formed by the convex mirror, which acts as the object for the concave mirror. For reflection at the concave mirror,

/ = -20 em, II = - (80 + 13.33) = - 93.33 em I' = /11 = -20 x (- 93.33) = -25.45 em 11- / -93.33 + 20 Magnification produced by convex mirror = ml = - 13.33 = -0.333

40

Magnification produced by concave mirror = m2 = 25.45 = 1.27

20

Total magnification = m = mlnJ2 = (-0.333»( 1.27 = -0.42 Size of image = -0.42 x size of object = (-0.42) x 4 = -1.68 cm The final image is inverted, real and of height 1.68 cm. Example 7. Find the velocity of image w.r.t. ground. 5 mI' The mirror is at rest and the object is moving perpendicular to the axis with 5 m1s as in figure.

t

Solution: and

m=L

O~-

=

(-20) =-2 /-11 (- 20)-(-30) v = -mu = -(-2)(-30) = -60em

ADS.

f:::20cm

mmh) 30cm

For tbe velocity component perpendicular the principal axis :

v'

(V,Mh =- - (VoMh

~

(V,M) .l ' 0 [Sinee(VOMh =0]

II'

For the velocity component along the principal

(V,M)" = m(VOM h + ho dm dl

•

:.l ~

GRB Understanding Physics Optics and Modem Physics

54

[Since. ho = 0]

(V,M h = (-2)(5) =-IOmis ~

~

~

VIG = VH•1+ VMG

=-

VIG = -10 +0 = - 1001/5, i.e.• 10 m/s (moving downwards) Ans. Enmple 8. Find the velocity of image in a situation as shown in the figure . 15m1s

1=20cm

2m1s

53'

o fo- - - - -30 - - - - - - -em Solution: ~

~

.

Vo = Velocity of object = (9f + 12j)mls andVM = Velocity of mirror = -2; m

m=~ =

f -u

- 20 = -2 -20-(-30)

For velocity component parallel to optical axis ~

~

(V,M) U = - ""(VOM) II

,

~

. = -441. mls

(V,M) U = (-2) x (9+2)1

15 mls 12 mls

530

-;o""''''--~---l 9 mls

2 mls

For velocity component perpendicular to optical axis ~

(V,Mh

~

= (VoMh = (-2)12j= - 24j

~

~

mls ~

:. VIM = Velocity of image w.r.t. mirror = (VIM) II + (VIM h ~

Also,

~

=(-447 -24j) mls

~

VIM = VIG-VMG ~

or

V, = (-44i-24j)-2i = (-46i - 24j)mis

Ans.

• 55

Geometrical (Ray) Optics •

El:8mpl~ 9. A point objcct is pla.ced at the bottom of a lake of depth h. If the refractive index efwaler is II, find the: (a) apparent depth of the lake (image distance), (b) apparent shift of the object. Solullon: (a) The refracted rays I and 2 meet at 1. That 1 is the image of O. The image distance is AI = ABltan r 2

,

:::ABI sin r (for small r, tan r==sin r) =

•

(AO tan iYsi n r

:::AOsin i sin r

(,: tan i ==sini)

" " Apparent depth

A

, ', "

, B

= -

Then

Realdepth

I

=~

where n = Ratio refractive index of 2 and I. Ans. (b) Hence, the apparent shift is 10 =AO-AI Apparent shift =

o

AO( I - ~)

Ans.

Important point When the point object Is placed below many transparent layers

of refractive index n"Il2.t1J etc. and the corresponding thicknesses I,. 12 .. . etc. the apparent depth of the object, that is, the image position is given as App. depth =

/

dn;..

The apparent shift of the object = 1:t, (1- ~)

/

'2

,, n,

o

eRB Undentanding Physics Optics and Modtm Physics

56

2

Example 10. A ray oflight strikes the face I ofa transparent cylinder of refractive index" at an anglc i. Find (a) sin j such that the refracted ray will graze the surface 2. (b) the reflection of the incident ray.

Solution: For the refraction at B at critical anglc of incidence. applying Sne1l's

law,

sin Ie sin 90

M

1

"1

n

-=~ =",-

or

. .

I =-

Stn1c

... (i)

n

8

, I, , ,,

r

-------------'c n

Applying Snel1's law at A.

sin; 11"1 -=~

or

"

= -

sin,. III I sin i =nsin r

... (ii)

r+ic = 90 0

... (iii)

In the" ABC, Solving cqns. (i). (ii) and (iii),

sini=~n2 - 1

ADS,

Since for any value of i sin i is always less than one,

.In' -I < I or

,,

Ans.

Example 11. A ray of light is incident on the 0 surface 1 of the glass slabs with an angle ;, If the 1' refractive index of the slabs is " . find the (a) lateral shift of the emergent ray (b) answer is (a) if j is small, r--""\-------Crl (c) apparent shift of the object when viewed from the side 2 of the slab ifsmalJ angle ofi. Solution: (a) In the /) ABC. the lateral shift is

,

x = ABsin(i - r)

=(cosr)Sin(iAD

2

r)

(": AD = AB eoSr)

57

Geometrical (Ray) Optics

si ~.;:.n",(i_ - .:..
(": AD =1)

= 1-

or

x = - 1 (sin icos r - cos isin r) cos r

. . . (i)

,,' A

o

B

According to SneWs law, sin i = ~ sinr . sin i Stnr = or n and

J

· 2

Jn 2 _ sin 2 n

j

cos r = I - sm r = :!.::.'---=::....:

Substituting sin r and cos:

~n eqn. ;i),

[Sin i n2 _ sin 2 i_cos ;sin ;]

J

Jn 2 _ sin 2 i n n n

x

=}n;-~:~s~::~ ;]

Ans.

(b) When i is small sin i:: ,. and cos i:: I Then,

Ans.

(c) When we keep an object 0, its image i is fonned at a distance 01 ahead of the object towards the slab.

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60

Using eqns. (i), (ii) and (iii),

e =gn - I[sin{ A -sin - t (b) The deviation is D = i+ e - A

i

= +sin - I[sin{ A _ sin

-t i)}]i :

Enmple 16. Find the image of the object 0 due to the concave mirror of radius of curvature R = 40cm.

Solution: p = -30cm,

J= R =40 = 20cm 2

2

Then

! =.L!

or

+20 - 30 q = 12cm

q

i)}]

Ans.

Ans.

A

------ --

---~------E

----------" 30em

J

p = _1___1-

1-

ADS.

•

Example 17. A point object 0 is placed at a distance of 10 cm from a silvered glass slab of" = 1.5, thickness 5 cm. Find the final image of 0. Solution: The position of image I, is given as, All = OA x 1.5

=lO x ~ = 15cm 2

Geometrical (Ray) OPba

61

The position of image 12 is BI2 = BI. = 15 + S=20cm. The position of the final image 13 is

o

81 3 = RI2 = 20 = 40 COl from 8 II 312 3

--i

10cm

-

ADs.

-

5an

I--

.-

-- --- ---------- -----:/ .. c==: --~-----

0

---'::.-:....-----==~-

8

A

1'< II .

n= 1.5

'-5cm-

Example 18. Find the distance x between the glass slab and convex lens, if the final image will fonn at the object. The radii of curvature RI = 20 cm, R2 = 30cm and R1 of the lens is n = 1.5. Convex lens

Glass slab

r' o

r-

, 7cm ->

n =1.5

x

I) V

Solution: The focal length of the lens is given as

1. = (n-IJ(_I +_1 )

[

R,

R2

-1)(2~ + 31 0)

or

; = (1.5

or

[=24cm

As the fmal image will coincide with the object. after the refraction through the lens, the ray should move parallel to the principal axis. Then. 0'/1 = [=24cm.

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GRB Undcntanding Physics Optics and Modem Physics

II ../. ......... 0

-

I, A

0' B ,

15cm

I· all

,"

V

·1

={I-D={I-IIs)= I

em

Hence,

A/t '= AO - Ol l =7-1=6cm 0'/\ = All + AD + BO' = 6 + 3 + x = 24 em x = IS em Example 19. The object 0 is located at a distance of20 em from the lens off = 5 em. Find the o value of x. if the final image after reflection from the mirror will be formed very close to the lens. Solution: Since the image will be formed as the Then or

~

lens, 2\" = q

x=~

or

2

ADS.

20em"-

... (i)

" 'u~

Where the image distance q is given by

L _l+l q

p f

=-.!.+!

20 5

63

Geometrical (Ray) Optics q =20 3 By using eqns. (i) and (ii),

or

... (ii)

x = 10 em. 3

-

Example 20.. ~ point object has velocity 110 = 2cmfs when It IS placed at a distance ofp = 30 cm from a lens moving with a speed 11m = 2 cmfs towards left. Ifthe focal length of the lens is! = 20 Vo -0em, find the velocity of the image.

r--

Solution:

or

Ans.

Vm

--H--x 30cm -to-

-

Vm

.

,

-

Vo

o

or

; ; ~-~m(~)' +~

or

-:j =-{~- ~>(~)' +~

or

... (i)

!=! _!=.L-L

q/p2030 q=60cm

or

.....

Substituting, q = 6Ocm. p = 20 cm, 110

;;;

~

=2 i.

-t Om

A

=- 2; , is eqn. (i),

~_[(2iH_2i)](~~)2 +(-21) = -38

i emfs

Example 21. A candle is placed at a distance of 30 cm from a very long glass cylinder having hemispherical end of radius of curvature R = 10

Ans. n = 1.5

em.

Find the (a) image, (b) magnification of the , -./ 30cml~~_~ image. Solution: (a) The position of the image is given as

n,+n2=n2- n, p q R I 1.5 (1.5 - I) -+-~ 30 q +10

or

q=90em

Ans.

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64

•

m= nlq =_1 x90=2:1 n,p 1.5 30 Enmplc 22. A bi·convcx lens of focal length 20 em is placed at a (b)

Ans.

distance / = 6cm from a plane mirror. If the medium between mirror and lens is water, find the equivalent focal length of the system.

Solution: The reflection of LI will give L2. Thus. we have two 'identical causes LI and L2 arc separated by a distance d = 2/in a medium of R.I. = n. Hence, the equivalent focal1cngth of the system is given as l,

______ L _____ _l _____ -l _____ _

_1_

2

----- - ~--- - -

------1-----------1----------,-----:~~~~¥~~~~ :..-_:..:::.::.1 ___-_":.::.:. ------1------------1------

or

ADS.

Geometrical (Ray) Optics

65

Example 23. A bi-convex lens and a plano-concave lens are "\ and 112 kept in contact. The plane surface is silvered, the refractive index of the lenses arc 11\ = 4/3 and "2 = 1.5 respectively. Putting R\ = 20cm, R2 = 30cm, find the: (a) focal length of the combination,

"1

(b) image distance when a thin object is kept at distance of p = 60cm from the combination, (c) lateral magnification.

Solution: (a) There arc two lenses. Forlens l(in air),

liJ,

=

(LI)(.l+.l) 3 20 30 J, =36em

For lens 2 (in air),

1 2

)2 =(1.5-+3~) or 12 = -60cm For mirror f m = 0 because R ~ 00. The ray moves twice through the lenses 1 a.nd ~ an~ reflected once by the mirror. Hence, the focal length of the combmalton IS

1=2+-'-+_1_ J J, h Jm =2+2

(b)

36 -60 or 1=45cm The image distance q is given as

ADS.

1 1 1

q=J-P

1 1 = 45- 60

or (e)

q = + 180cm the image is virtuah

Ans.

Lateral magnification is q (180) mlat

or

mlat

=-p=-@

=-3

Ans.

GRB Understanding Ph)"ia Optics and Modem Physia

66

Example 24. An equi-convex lens has R.I.= " = 1.5 and radius of curvature R = 20cm, find the. (a) power of the lens in air, (b) rocal1cngth of tile lens in air, (e) power of the lens in a medium of R.I.= n = 1.4. (d) rocallength of the lens in a medium ofR.I.= n = 1.4.

n:1 .5

(e) power of the lens in a medium ofR.I.= n ~ 1.5. ({) focal length orthe lens in a medium ofR.I.= n = 1.5, (g) power of Ihe lens in a medium ofR .I.= " = 1.6,

(h) focal length of the lens in a medium orR.I.= 1.6. Solution: (a) The power ofthe lens in air is Pair

= (n _1)I R = (1.5-1);oem

CI

F

= +5 0 Ans.

(b)

(e)

The focal length of the lens in air is n. 1 f = -"'- = --'----;-1 Pair 1I20cm-

or f =20cm The power of the lens in the medium of n = 1.4

Ans.

=(n-no)~

P"""

= (1.5 -1.4);0 em - I

(d)

= + 10 The focal length of the lens in the medium is

(e)

frrcd = ~ =.l.:L = 140cm p""" 1/ 100 Power of the lens in the medium of "0 = 1.5 is P = (n-no)£=(O)1.=O R 20 .

(f)

The focal length is f = 1.5 = 00

(g)

Power of the lens in the medium of no = 1.6 is

p

P=(n-no)£=(1.5-1.6)1.= __ 1 em-I =-ID R 20 100

Ans.

Ans.

Ans. ADS.

Ans.

,

•

Geometrical {Ray} Optics

(h)

The focal length of the lens is

1=1.6= p

Ih

= -160em

Ans.

-I/IOOcm I

A bi-convex lens can be canvassing if it is placed in a medium of R.I. greater than that of the lens. It behaves as a simple glass slab of zero power or infinite focal length if the R.I. of the lens and the medium are equal. The lens becomes diverging when the RD of the lens is lesser than that of the surrounding medium.

•

Enmple 25. Two lenses L] and L2 having focal lengths IIII = 10cm and 1121 = 20cm arc separated by a distance of5cm. Find the (a) focal effective length of the vision of two lens .

Solution: (a)

The focal length of the combination is

-- - ,'5om~' ,

'V ;.,---. 1=

III, I, + h-d (10)(-20) 10-20 - 5

(b)

•

= 40 cm 3 The position of the effective lens from L2 is given as X=

=

dh II +h-d

Ans.

.'

5(-20) =20 em 10+(-20)-5 3

Ans.

68

GRB Undenlanding Physics Optics and Modem Physics

(c)

II

For refraction at LI :

1+1=1 p q

f

I ,I

00

q

p' -

f.l

1 +1=.1

or

---

10

"

q-

or q = 10 For refraction at L], : II behaves as a virtual object for L2. Hence, p = -Scm.

q'

/' = h = -20, Thcn,_1 +1 = _1_ -5 q'

- 20

Ans.

q' = 23° em from L], towards right

Example 26. A bi-convex lens ofR.1.

/I

= 1.5, radii of curvature RI = 20cm and

R2 = 30cm is placed in air. Find its: (a) focal length (b) image distance for the object 0

o

(c) ml al

t-

(d) m lincar (e) the height of the image if the height of the object is ho = 2cm. Also draw the ray diagram. Solution: (a) The focal length of the lens is given as

n

I 4Oan

; =(::- I)(~, -~,)

:1 = (\5-1)(+~0 - (-~O»)

or or (b)

f =24cm Putting p = +40cm, f = 24cm in lens fannula

1 1 1 wehave -+-=p q f' , _1_+1 =1.

+40 or (e)

Ans.

q

24

q =60em

m,.. = -(~) = _::: = _;

An .. Ans. Ans.

69

Geometrical (Ray) Optics

Ans. (f)

Ray diagram

1-

p•

q

Example 27. A plano-convex lens of R.I. 1.5 and radius of curvature 10 cm is placed in air. A point object 0 is kept at distance oflO cm from the lens. find (he image.

Solution: (a)

The focal length ofthe lens can be given as

-,

2

LJ:_""m_\, ; =(:: -1)(;\ -;,) Putting n2 = 1.5, nl = I, RI = +lOcm and Rz = 00 (for plane surface) in the above formula, we have

; = (\5-11~) = ;0 or f =20cm. Then using the lens fonnula,

1+1:1 p q J _ 1- +! = _1_ +10

q

+20 ,

.. • GRB Undmtanding Physics Optics and Modem Physics

70

Ans.

q = -20cm The image is point and virtual. Example 28. A thin erect object is placed infrant of a concave mirror of R = 20cm at a distance of 15 em. There is a spherical (concave) refractive surface of R' = 30 cm and n = 1.5. As shown in the figure. Find the (a) position of the final image, (b) total magnification of the final

n=1.5

image.

R'::3Ocm

Solution: (a) For the mirror,

""""'"

R':3Ocm

n:l .S

0

, p

0:'-p'-'

,

,,"

,,,'2 ,

g'

q = 30cm The image J I behaves as a virtual object for the concave refracting surface. Hence, p' = -(q - 20) = -(30 - 20) = - lOem,", = = 1.5, R' = -30 for the refraction. Putting all these values in nl + n_, = "":;"-;-","",, p' q' ~ R'

I,",

or or (b)

_1_ +!l = l2..::.!

-10

-30

q'

q'= 18cm

Ans.

Then. the lateral magnification of 12 relative to 1\

q'

n,

p'

",

mlat = - - x -

18 x_ I =---

(-10) 1.5

=§ 5 The magnification of [lis mI'

at

= -q = -30 = -2 p 15

Geomdric:a1 (Ray) Optics

71

. . .IS m = mlat . m, 6 Th en Ihe Iota 1magm.rIcatlOn =.5 x - 2 = 5-12 llli

Ans.

Example 29. A thin object of height Ira = 2cm is kept as a distance of20 cm from a spherical interface of radius of curvature n,=1.5 R = 30cm separating two media of R.I. nl = 1.5 j and "2 = 4 / 3. Find the: n2''' !. hO 3 (a) image position,

+~

(b) mlat, (e) mlinear, (d) height of the image.

Solution: (a) Put "1 = 1.5,"2 =

~, p

= 20 em N' ,

R = +30cm

... ...... ... n,=1.5,._./_ _. ._ .... ... ... ~/ n2=.!.. =:, ':::':'" 3 L~[±:q::r--.c:..::=-.,_c:_:__~_::--

~ + n2=n2 - "1

Then, or q =

;§;

p q R 1.5 + 4 / 3 = 4 / 3 - 3/ 2 20

q

20cm \ _

_p_l\

+30

'--'-...---

-16.55 em. Ans.

(b)

n, pn, = _(-16.55)(1.1.) 20 4/3

q mlat = - - -

= +0.931 (e)

m];~

= - m,'

(n,)=

al "I

Ans.

_(0.931),(4 /3)

1.5

= - 0.77 (d)

Ans.

himage = Imllt l hobject

=0.93Ix2cm = 1.86cm

Ans.

Example 30. Find the images due to the refractions at the curved (spherical) surface of R = 20 cm and the flat surface. ,r----;--,

"2=~ 3

Solution: Put "I = I, "2 = 4/ 3 ,p = +20cm, R = +20cm in the formula

. GRB Und~ntanding PhysiC! Optic! and Modem Physics

72

~ + "2 = "2 - "I to obtain

(a)

p

q

R

/' '"

/

~;.f:f:......

"1'::::1

.---- - - --6 40em

~-------

.- .

I

20em

"2- 413

.....,::

---

c

_2Ocm~_ 40cm

I\,

_1_+ 4 / 3 = 4/ 3-1 +20 q +20 0'

(b)

Ans.

q = -40cm.

The image is virtual. For refraction at the flat surfacc:p' = 40 +20 + 40 = IOO cm

,

,

. .

" 1 + "2 = "2 - "I

Ii

q' R 4/3 + 1 = 1- 4 / 3 100 q' 00 q' = -75cm (virtual)

Ans.

Example 31. A white light gets dispersed when it passes through a pri sm. The average R.1. is n. If the refracti ve indexes of the emergent rays I and 2 have the difference III] - "2 1= Sn, find the angle of dispersion SD.

Solution: Snell's law for minimum deviation D;

I "

...........

............

" ,or

0'

Ans.

•

73

Geometrica1 (Ray) Optics

Example 32. A point object 0 is placed in front ofa silvered glass slab of thickness 5 em, R.t. (II) = 1.5, at a distancc ofS cm. Find the location of the final image.

Solution:

For surface 1, nl = I , n2 = I.S, P = 5cm, ~+n2

p

=n2

R = 00

-n l

R

q

! + 1.5 = 1.5 - 1 5 q 00 q = -7.5cm The virtual image 1\ is fonned 2 due to the refraction of _ surface I . This image will . ... behave as a real object for the · .... ..::,::: refracting surface 2. Then the -:::'''l ......- .. , object distance 1,__ -- ......0/

I

"I

p=7.S+S = 12.5cm. Since ~ q'·'2.Scm the mirror is plane, q=7.Scm -1,_-"9.:::.!.1,,.66<m""'' '-..J1 q'= -p= -12.5 cm. Then the .. virtual image 12is fonned. This image will same as a real object for refracting surface I .Hence for refracting surface 1• p"=12.5+S = 17.Scm,nj = 1.5,ni = I,R= oo

. .

Then, or

r

.

,

~ + _" 2 = "".. ,-:-,..:"",

R Q+_I = 1-1.5 p"

17.5

q"

q"

00

or q' = 11.66cm. Hence, I ) will be fonned at a distance of 11.66 cm to right from the first face. Ans.

GRB Undenlanding Physics Optics and Modem Physics

74

Example 33. A ray of light is incident at the origin 0 as 90 medium whose R.I. varies withy as n = ~ Ky3/2 + 1. Find

= 90°, It travels , in0 a y

I

~

the: (a) trajectory of the ray in the medium, (b) point of emergence at P, if y = ,1,Q = 2 m, (e) angle 9 of emergence.

,, , ,,

,"

,

, ,,

p

Medium

,, ~.k'~----------.' o 00

no

Solution: (a) no sineo = nl sinOI = n2 Sina2 = nsin8 for multiple layers. Putting = 1, 0 0 = 90° (approx), y p

~O~~~

_______ ,

nsinO = 1

.. .(i)

The slope of the path is dy =

dx

.'

tane = cosO

... (ii)

a = !. we have cosS = ~n 2 -1. Puttingcos9 in eqn. (ii). " dy = ';'""_-1

Putting sin

... (iii)

dx

Putting

n = ~Ky3f2 + I in eqn. (iii),

or

dy =~Kll2 +1_1 dx y - 3I4 dy =,[Kdx

or

Jy-3I4 dy = JK Jdx

or or

y

x

o

0 yl/4

= .fKx

2 4 y=K x ;PuttingK= 112../2,

,

i

• 75

Geometrical (Ray) Optics

x'

y=-

8

(b)

Putting Y= 2m, we have x =2m . Hence, the ray will emerge at a point P(2, 2).

(e)

At)'=2m

Ans.

n=~Ky]/2+1

= _1- x2,/2+1 2,/2 =,/2 Hence, or

nsinO = I

e =~n-lm=Sint~)=4S0

The ray will strike the other face of the slab of 45° with the nonnal as then came out of the slab parallcl to its face. .

CRB Undmtanding Ph)'S ics Optia and Mod~m Physia

76

_ .......... ...... .. -

........ ..........--

ASSIGNMENTS

......... . .......... .. .. ...... .----.. _-.-

rJ Conceptual Questions 1. When light move from are medium to the other, why does it bend towards or away from Ihe normal? 2. At the time of sun rise why does sun appear red? "1

3. When we look a fish moving inside watcr from above the watcrdoes its npparent position depend on the (a) real depth offish in watcr (b) angle of vic wing? 4. When hot air comes out of the oven or flaming about the hot sand afthe rivcr bcd, things like trees, stones seem to be vary, why? 5. What is mirragc ? Why is this form?

6. Can we have (produce) 1.00% reflection or refraction of a ray of light of its boundary?

7. How arc refractive index and relative pcnnitivity of a medium for a particular wavelength of light related? 8. In lotal internal reflection why do we use the word "total". 9. 10. II. 12. 13. 14.

What are the conditions of (i) maximum (ii) minimum deviation of a prism? What is basic definition of mirror? What are real and virtual space S for mirror and refracting surfaces? What is the definition of an image in reflection and refraction? What is thc definition of an object? What is the differencc between real and virtual object?

15. What is the difference between real and virtual image? 16. For a (i) real (ii) virtual object what types of images are not possible due to convex and concave mirror? 17. Drnw the graph of magnification versus object distance for a (i) concave (ii) convex mirror. 18. What will hap~n t~ the size and nature of.the image when an object (real) approach from mfinny towards the pole of a (I) concave (ii) convex mirror? 19. What is the relation between lateral and linear magnification? 20. Can we produce a real image in a plane mirror? 21. Fo~ spherical refracting surface describe the nature of images of real and virtual objects. 22. How many focal points are there for a lens and a mirror?

•

77

Geometrical (Ray) Optics

23. What is the relation between object and image focal points of a (i) lens (ii) spherical refracting surfaces? 24. What is optic center of a lens? 25. What happens to the power of a mirror when the mirror is emerged in water? 26. How can we changc the powcr of a lens? 27. What is thc meaning ofpowcr of an interface of two transparent mcdia? . 28. For a (i) real (ii) virtual object which types of image are not possible due to concave and convex thin lenses? 29. Can you make a convex lens diverging or concave lens converging? Explain. 30. Ifwe cut a lens parallcl to its principal axis does it focallcngth change? 31. If the rays are not paraxial how many focal points and images will be formed due to a lens and mirror? _ 32. What happens to the focallcngth if a lcns is silvered? 33. Does the formula optical powc.r p

=: }

valid for thin lenscs pl~ccd in a medium.

34. Why should a light pulse slow down when moves from air to a transparent medium?

,.

GRB Undmtanding Physics Optics and Modem Physics

78

rJ Multiple Choice Questions lA) Only One Choice Is Correct

Level·1

1. When a ray of light changes a medium which of the following does not change? (a) Wavelength

(b) Speed

(e) Frequency

(d) Amplitude

2. Refractive index of a material depends on the: (a) amplitude of light (b) wavelength of light (c) intensity oflight (d) initial phase of light 3. The anglc q, between the reflected & refracted rays is:

(a) 90· . . -1"1· ( C) 1[-sm -Stn-I

",

4. For minimum deviation Dnjn

(a) j j

(d) <-2;

= 30°, the refractive index of prism is:

(e) 312 (b) ./i 30° the maximum deviation is: (b) 45· (e) 30·

5. In Q4. if i = (a) 60· 6. In Q-S. the refractive index of the prism is: (a) 1.5 (b) 1.41 (e) .fif2

(d) none of these

(d) 75· (d)

Jfi3

79

Geometrical (Ray) Optia

7. Which of the following p - q graph is valid for concave mirror or converging lens? p

p

(a)

(b)

p

(e)

p

(d)

8. Which of the following p - q graph is valid for convex mirror or diverging lens? p

(a)

p

(b)

p

p

(e)

(d)

-::::::=f-t---

q

GRB Understanding Physics Optics and Modem Physics

80

9. An object 0 is placed at a distance x from the point ofthc glass slab of Ri= nand thickness I. The distance), of the image 1 is: Glass slab

--(a)! n

10. A glass slab of refractive index '" n and thickness I is placed in between the forces F and the converging lens L. The image will be fonned at a distance x = L

A

- ---

F

, (a)t(I-~) (c) /

(b)/+I(I-~)

+!

(d)/

n

-t(I-~)

11. A ray of light enters into the prism nonnally with the face I and comes Qut

normally with the face 3. The minimum R.I. of the prism is:

(a) cosec ~

(b) tan 9

(c) sec 9

(d) cosec9

81

Geometrical (Ray) Optics

12. The roy enters perpendicular to the surface and emerges at the surface 3 nonnally being reflected at the face 3. If AB "" AC. then the apex angle 0 "" A

o

B'+-'3------"'.C (a) 60· (b) 45· (c)36· (d) 72· 13. A fly has a velocity v= of + bJ + ck . Its image has a velocity:

(a)

-01 - bj - ck

+bj-ck (d) 01 -bj - ci. (b)-al

- of +bJ+ck 14. An insect moves along the dotted line I. Its image moves along the line: (c)

1-

5'~','........,

,,

/3

> ,

" " ~ 1

(a) 2

"- .... (c) 4

(b)3

"4

(d) 5

15. A ray coming from an object is reflected totally at the slant surface of the prism. The possible orientation ofthe image is:

Obi""

(a)

-->

(b) <--

(c)

t

GRB Understanding Physics Optics and Modtm Physics

82

16. A point moves perpendicular to the mirror with Vo = 27 m /s. The mirror moves with Pm =

m _v: -,

-7 m / s . The velocity orlhe image is:

(n) -4im / s

(b) -37 ml s

(c)

-27 m /s

(d) nooe orlhcsc

17. A lens made of two materials 1 and 2 is cut at the principal axis and shifted slightly as shown in the figure. The number of images is:

(b) two (d) four (c) three 18. Two thin prisms I and 2 are jointed as shown such that the incident ray A and the emergent yellow ray B arc parallcl. The dispersive powers of the prisms are rot and (02 respectively. The ratio of mean deviation produced by the prisms I and 2, is: (a) one

B

(d) I : I

19. In the above question iflhe red and violet rays are parallel, the ratio of mean deviation ofyeUow light produced by the prism 1 and 2, is: I" ~

t

(e) I : I

(d) ~rol /ro2

83

Geometrical (Ray) Optics 20. The optical power ofa mirror is P in air. Its optical power in a liquid ofR.1. =

It

is:

(a) P I" (b) nP (e) P (d) none of these 21. A convex lens offocallengthJis placed in contnct with a plane mirror. The focal length o f the combination is:

(a)1

(b)

21

(d) ""

22. An equi...convex lens of focal length combination is (R.1. = n):

J is

silvered. The focal length of the

n

I

(c)

(b)2n_l

I

(d) I(n -I)

2(11 -I)

211-1

23. The number of images due to tbe three mutually perpendicular plane mirrors, of the object 0 is: y

, (b) 5

(0) 3

(c) 9

(d) 7

24. Which of the following point P is a virtual object?

(.)

(b) p

(c)

"p

(d)

"p

,,

• GRB Understanding Physics Optics and Modem Physics 25. The dimension of dispersive power is: (a)[Mol- 'To] (b)[ MOloTo] (c)[Molr']

(d)[Ml'T -']

26. Which can split sun light? (3) Glass slab (b) Convex lens (c) Concave mirror (d) Prism 27. A light wave travels with different velocities in different media. This is called: (a) refraction (b) polarisation (e) dispersion (d) diffraction 28. Different tight waves travel with different velocities in same medium. This is called: (b) dispersion (a) refraction (e) polarisation (d) interference 29. A convex lens of focal length 4Qcm is in contact with a concave lens of focal length 25cm. The power of the combination is: (a) - 1.5 dioptcrs (b) -6.5 diopters (d) +6.67 dioptcrs . (e) +6.5 diopters 30. A ray of light from a denser medium strikes a rarer medium at an angle of incidence i (sec figure). The reflected and refracted rays make an angle of 90° with each other. The angles of reflection are rand r . The critical angle is:

, " (n)sin - 1 (tan r)

(b) sin - l (tan i)

(c)sin - l(tan/')

(d) tan - , (sin I)

31. Spherical aberration in a thin lens can be reduced by: (a) using a monochromatic light (b) using a doublet combination (c) using a circular anr.i.ilar mark over the lens (d) increasing the size ofthe lens 32. A real image of a distant object is fonned by a plano-convex lens on its principal axi~. 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

•

85

Geometrical (Ray) Optics

(d) is the same whichever side of the lens faces the object 33. 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 center of curvature. A point object is placed at Cit 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 OCI (c) virtual and located at a point between C and 0 (d) real and located at a point between C and 0 34. A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids Llor ~having refractive indices ,. qand J.l 2 respectively (J,i 2 )0 J.lI )0 I). The lens will diverge a parallel beam oflight ifit is filled with : (a) air and placed in air (b) air and immersed in Lt (c)L,and immersed in L2 (d) L2 and immersed in LI 35. A point source of light B is placed at a distance L in front of the center ofa mirror of width 'd ' hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a di stance 2L from it as shown in figure. The greatest distance over which he can see the image of the light source in the mirror is:

-I I

,, B

,, ,,

l..!o....i

I

2l

(a) dl2 (b) d (e) 2d (d) 3d 36. In a compound microscope, the intennediate image is: (b) real. erect, and magnified (a) virtual, erect, and magnified (c) real, inverted, and magnified (d) virtual , erect, and reduced 37. A ray of light passes through four transparent media with refractive indices 1l101l2,Jl3 and Jl4 as shown in figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have: (a)~1 =~2 (e)~3 =~4

(b)~2 = ~3 (d)~4 =~I

86

GRB Understanding Ph)'lics Optics and Modem Ph),ics

38. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prism Q and R of identical shape and of the same material as Pare now added as shown in the figure. The ray will suffer:

a

R

(a) greater deviation (b) no devialion (c) same deviation as before (d) total internal reflection 39. Which one of the following spherical lenses does not exhibit dispersion'? The radii ofeurvaturc of the surface of the lenses are as given in the diagrams.

(d)

:'\l RLJ'"

40. Two plane mirror A and B arc aligned parallel to each other. as shown in the figure. A light ray is incident at an angle of300al a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times Ihe ray undergoes reflection(ineluding the first one before it emerges out) is: ,8

213m

.2m

30'

A

(a) 28 (e) 32

(b) 30 (d) 34

Geometrical (Ray) Optics

87

41. A ray oflight is incident at the glass-water interface at an angle i. 1t merges finally parallel to the surface of water. Then. the value ofJ.Lg would be: Air

IlUi '" 4/3

Water

Glass

(a) (4/3),in i

(b) Il'in i (e) 4/3 (d) I 42. A beam of while light is incident on glass-air interface from glass to air such that green light just suffers tola1 internal rencction. The colors of the lighl which will come out to air are: 1 Increases

•

I II IIII

V

AI, (Green light)

•

SGVOR

G~S$

(a) violet, indigo, blue (b) all colors except green (c) yellow, orange, red (d) white light 43. An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation,

s

(a) PQ is horizontal (b) QR is horizontal (d) anyone will be horizontal (c) RS is horizontal 44. A point object is placed at the center of a glass sphere of radius 6 cm and refractive index 1.5. The distance of virtual image from the surface is: 006em (e) 12em

004em (d)gem

t

88

,

GRB Underslanding Ph),!ics Optics and Modem Physics

45. A convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is 213. Their equivalent focal length is 30 em. What arc their individual focal lengths? (b)-IO, IS (a) - IS, 10 (c)7S, SO (dl-7S, SO 46. Focnllcnglh of the plano-convex lens is 15 em. A small object is placed at A as shown in the figure. The plane surface is silvered. The image will from at:

• A

•

200m

•

(a) 60 em to the left oflcn s (b) 12 em to the left of lens (c) 60 em to the right of lens (d) 30 em to the left of lens 47. A ray of light travelling in water is incident on its surface open to air. The angle of incidence is Q, which is less than the critical angle. Then. there will be: (a) only a reflected ray and no refracted ray (b) only refracted ray and no reflected ray (c) a reflected ray and a refracted ray and the angle between them would be less than 180°-29 (d) a reflected ray and a refracted ray and the angle between them would be greater than 180°-29 48. In an experiment to delennine the focallenglh (J) ofa concave mirror by the Il-U method, a student places the object pin A on the principal axis at a distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping hislhereye in line with PA. When the student shifts hislher eyes towards left, the image appears to the right of the object pin. Then: (a) x < [ (b) [ < x < 2[ (c) x > [ (dl [ > x > 2[

,

Geometrical (Ray) Optics

89

Leve/-2

I. A glass slab of thickness / = 3cm. R.I . =

II

= ~ is placed between the objcct 0 and

,

the concave mirror of focal lengthf = 4.Sm. It is observed that the image is fonned the object. If we draw the slab, the image will be fonned at a distance:

\

"",n

x

(a) 10 em

(b)30em

2. In the previous question x = (a)gem (b)8em 3. An insect moves with a velocity velocity oflhe image is:

vi

)

(c) 80 em

(d) 90 em

(e) 10 em

(d) 9 em

towards the inclined plane mirror. The

_-lLJ=--_+x

4. A point object moves in +x-direction with v= I mlsalong the principal axis ofthe concave mirror of rocallengthf= 10 m. When the mirror moves with a velocity Vm = -7 mls and the object isat a distance of p = IS m. the speed oflhe image is: -

-

lm1s

tmls

--, 15m

(a)

-Simi,

(e) -6iml'

I

(b) -9iml' (d) none of these

-

r r

, GRB Undentanding Physia Optics and Modem Physics

90

5. When the fish moves up with a velocity V== 5 mls, the image of the fish seems to move with a velocity:

'ilf

.Y.

:=---~:=1~:=:-~

:=-

m~mfm~iW

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

(b) IS mls up (e) lS misdown (d) 2° mlsup 4 4 3 6. The image oCthe point object 0 due to the thin prism ofnpex angle A = 4°, R.I. = 1.5, object distance x :;:; 10 em wilt be: (a)Smlsdown

A

<~o<::::::.:::::::it}'.Ji»~,

-

(a) ~; em above 0

,-I

(b) ~; em left of 0 on the x-axis (c) 2.S em above 0 (d) 20 em above 0 7. A ray at a height h parallel to the principal axis is incident on a concave mirror of

radius of curvature R. If" = ~ • the value of x:

(b»

~

(e) < !! 2

(d) is indeterminate

•

91

Geometrical (Ray) Optics

8. For a ray oflight incident at the interface form rarer to denser medium nl < n2. which of the following graph of angle of deviationS versus angle of incidence i is correct:

,,

~ (b)

, (d)

9. In the above (previous) problem. if nl > n2 which of the o-i graph is correct?

• (b)

(a)

, (e)

(d)

, GRB Undenotanding Physics Optics and Modem Physics

92

10. A light ray 1 enter at A into the solid transpnrcnt rod of U-shnpe having circular cross section. The light ray 2 emerges oul of the tube at B. lrlhe R.I. orthe rod is II .

Then d : R

1

(a)

s: II

2

(e) ::;: (II-I)

(b»(n - I)

II. The equivalent RJ ora composite glass slab comprising ofthicknessx\ and Xl is:

I (a) nix, + 1l2 X2 x\ + x 2

(b) III + 112 2 12. The oplical powcrofa lens ofR.l. =

'z

I

(e) (11\ + 1l2 ):!!. 2 x2

(d) 2n\1I2 " 1+ 112

in air is P . When it is immersed in a liquid of R.t = II', its optical power will be P'. Then p ' = II

p

(a)

-I n -I

II'

(b) 11 - 11 11 -I

• (e) ", n

13. Following the deviation 8 - angle of incident i graph, the angle of prism is: (a)-x+y+z (b)x+y-z (c)x + z - )' (d)x - y + z

• (d) n-n

,

n

z - - -- __ _

,

y

14. A person moves horizontally with a speed ", and the plane mirror mO\'e5 vertically with a speed () 2. The speed of the image is:

(v, = ,[im/s.v, = ~m/s. e=45'

. .

(a) i +2j

(b)2i-j

(c) - 2i + j

(d) none of these

. .

~mttrical (Ray)

93

Optics

15. An object 0 moves with a speed ul parallcl to the mirror M which moves with a velocity u2. The velocity of lhe image is:

V't v,

Mil

r , -r v, (e)

(d)

-v, •j

16. The mirror and objcct moves along x-axis with velocities ul and L'2 respectively. The velocity of the image is:

17. A point object 0 is placed at Ihe principal axis so that the image will be fonned at by the cqui-convex lens ofmdius of curvature R. Then the R.I . of lens is:

o

1- , (a)R _ 1 ~

(b)R + 1 ~

(e) !! +1 x

(d)noneoflhese

18. The image produced by the plane-convex lens coincides with the objcct. lfradius of curvature is R. the R.1. of tile lens is:

(a)!! +1 .T

(b) R _'I

x

(e)~ + I

19, A bi-convex lensL is cut into two parts I and 2. These pans are kept side by side as shown in figure A. and B. The ratio of focal length of the systems A, Band Cis: (b) 1:2:1 (a) 1:1 :1

(e)2: 1:1

(d) 1:1 :2

(d) Rlx L

,, ,

,

2

A

¢=

1: 2

C

"*

, , B

GRB Und~nt.anding Physics Optics and

94

Mod~m Physics

20. A bi-convex lens is cut into two equal halves I and 2 symmetrical to the principal axis xx'. The two halves are arranges as shown in the diagrams BtC and D . The ratio of optical powcrs of the systems A, B. C and D is: (assume! = focal length of the lens (A).

CiJ

x

("I

(AI

CJ \Y

• 2

_I

-I

••

(01

(el

(a) 2:1:1 :2 (h) 1:1:2:1 (e) 1:2:1:2 (d) 1:2:1: 1 21. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam normally on the face AB is totally reflected to reach on the face BC if: A

•

"

-------------------------(a) sinO

~~

(h) sinO

~~

(h)~dnO ~ ~ 3 9 (d) sinO ~ ~ 9

•

95

Geometrical (Ray) Optics

22. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance II from the pole of the mirror. The size of the image is approximately equal to: (a)b

_b_) u-f b(L)' II-f

112

'"

("~f )

(b)b

(d)

(

23. A beam ~f light consisti~g of red, green, and blue colors is incident on a rightangled pnsm. The refractIve indices of the material of the prism for the above red. green, and blue wavelengths are 1.39, 1.44, and 1.47, respectively. The prism will:

• • • 45' - - _ . '--_----'L:>. (a) separate part of the red color from the green and blue colors (b) separate part of the blue color from the red and green colors (c) separate all the three colors from one another (d) not separate even partially any color from the other two color 24. A thin prism P]with angle 4 0 and made from glass of refractive index 1.54 is combined with another thin prism P2made from glass of refractive index t.72 to produce dispersion without deviation. The angle of the prism P2 is: (b) 4 0 (c)3° (d)2.6° (a) 5.33 0 25. Two thin convex lenses of focal length fland hare separated by a horizontal distance d (where d < flo d
!

+--+1-: ,

(a)x= flh ,y=d

fl

+12

(b)x = fl(h+d),y= d f,+h-d fl+h

d--:

1

,

GRB Understanding Phy1ics Optics and Modem Physics

96

(c) x

=1112 + d(/I - d) y = &(/1 II+h-d'

d) II+h- d

(d) x = /112 + d(/I-d) y= &(/I-d) II + h - d ' 11 + 12 d 26. A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placing: (a) a concave mirror of suitable focal length (b) a convex mirror of suitable focal length (e) a convex lens of focal length less than 0.25 m (d) a concave lens of suitable focal length 27. The focal lengths of the objective and the eyepiece ofa compound microscope arc 2.0 em and 3.0 em, respectively. The distance between the objective and the eyepiece is 15.0 em.The final image fonned by the eyepiece is at infinity. The two lenses arc thin. The distance, in em, of the object and the image produced by the objective measured from the objective lens. arc respectively:

•

(a) 2.4 and 12.0 (c) 2.0 and 12.0

. (b)2.4 and 15.0 (d) 2.0 and 3.0 28. A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The center of curvature is in the glass. A point object Pplaccd in air is found to have a real image Q in the glass. The line PQcuts thc surface at a point 0 , and PO = OQ. The distance PO is equal to: (a) 5R (b)3R (c) 2R (d) 1.5R 29. 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) convergent lens offocallength 3.5 R

(b) convergent lens of focal length 3.0 R (c) divergent lens offocallength 3.5 R (d) divergent lens of focal length 3.0 R

30. A diverging beam of light from a point S baving 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 s glass slab is I and the refractive index n, then the divergence angle of the emergent beam is:

,

(a) zero (b) a (c) sin -I(~)

(d)2Sin-l(~)

,

,I

,, ,

I '

t I,

•

Geometrical (Ray) Optia

97

31. A rectangular glass s lab ABCD of refractive index n,is immersed in water of refractive index n2 (n[ < n2 ). A ray of light is incident at the surface AB of the slab as shown.The maximum value of the angle ofincidenee a IIlJ.I{ such that the ray comes out from the other surface CD is given by: A ----. J.il

(a)gn-,[::: co{gn-t: ))]

(b)gn-'[n, co{gnt,~ ))] (d)gn-t: )

(C)Sin-l(:~)

32. An observer can see through a pin-hole the top end ofa thin of height II, placed as shown in the fisu.re. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 211, he can see the lower end of the rod. Then the refractive index of the liquid is:

·t~~. 2.

(b)

J5 2

(d) ~ 2 33. The size of the image of an object, which is at infinity, as fanned by a convex lens of focal length 30 cm is 2 em. If a concave lens of focal length 2Ck::m is placed between the convex lens and the image at a distance of 26cm from the convex lens, calculate the new size of the image: (a) 1.25 em (b) 2.5 em (d) 2 em (e) 1.05 em

, CRB Undtntanding Ph)'Jics Optics and Modem Physics

98

34. A container is filled with waler (JJ. = 1.33) upto a height of 33.25cm. A convex mirror is placed IS em above the water level and the image of an objec~ plae.cd at the bo«om is fonned 25 em below the water level. Focal length of the mirror IS:

-

~'''f'''''''''

10:000. ,, ,,, ,, ,, ----J-----~:::~3~~:~~~ ----:J':.-.-::-:. ::.: m ._! .3;;l-. r----__ =:. -.-_

. ----,,------

-:: -:-:.. .:(~-:-=::~

1 ?Scm

(0) IS em (h) 20 em (c) - IS.3Icm (d)IO : m .. 35. The graph shows relationship between object di ~ tance and Image dIstance for an cquiconvcx lens. Then, focal length of the lens IS:

31 yom 30

------,', ,, " , ,, " ,

fttt_+---I----'---..:.J' u cm-31

(0) O.SO ±O.OS cm (c) s.oo ± O.OS cm

-30

-20

10

0(9,+9)

- 10

(b)O.SO±O.IOcm (d) S.OO ± 0.)0 cm

36. Rays oflighl from Sun fall s on a biconvex lens of focal length/and the circular image of Sun of radius r is formed on the focal plane of the lens. Then,

(a) area of image is Jtr2 and area is directly proportional off (b) area of image is 1U 2and area is directly proportional to f2 (c) intensity of image increases iffis increased (d) if lower half of the lens is covered with black paper. area will become half

99

Geometrical (Ray) Optics

37. A light beam is travelling from region I to region IV(rcfer figure). The refractive indices in regions I,lI,llI and IV are no, no /2, "0/6 and no/8, respectively. The angle of incidence for which the beam just misses entering region IV is:

e

Region I

o

Region 1\

Region III

Region IV

"0

"0

"0

2

6

o

"0

o

0.2m

O.6m

(')'in-'(3 /4) (b) sin -'(1 /8) (c) 'in - '(1 /4) (d) sin -'(1 /3) 38. Two beams of red and violet colo~ are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be: (8) 30° for both the colors (b) greater for the violet color (e) greater for the red color (d) equal but not 300 for both the colo~ 39. A ball is dropped from a height of20 m above the surface'ofwater in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line offall of the ball. is looking at the ball. At an instant when the ball is 12.8 m above the water surface, the fish see the speed of ball as:

(a) 9 m,-' (b) 12

m,-'

(c) 16 ms-' (d) 21.33

m,-'

-, 100

GRB Understanding Physics Optics and Modem Physics

(8) More Than One Cho/ce/s Is/are Correct I. When a ray of light enters into a new medium:

(a) it gets reflected at the boundary(interface) (b) it gets refracted at the interface (e) the amplitude or lhe transmitted (refracted) wave decreases (d) its wavelength must change

2. For minimum deviation Dm : (a) i = e (b)Dm (e) r (d)

~

~2i-A

r

11 = Sin( A +2Dm }Sin ~

3. For maximum deviation Drrcu: (a) emergent must graze the surface(face) (b) incident ray must graze the surface (e) Drm.'( = 90o+e_ A (d) Dmin = 90 o+i - A 4. When a ray light is incident on a glass slab, it: (a) moves slowly in glass (b) bends towards the nonnal (e) may bend away from the nonnal (d) will not have a net deviation 5. A stationary plane mirror is lying parallel to the y • z plane. If an object moves ~

~

with a velocity Uo its image moves with velocity u2 . Then:

~

~

~,

(c)uoy=v;y

6. For a convex mirror or diverging lens, we can have: (a) real object - rca) image (b) real object - virtual image (c) virtual object - real image (d) virtual object - vinual image

z

Mirror

'"

.

f

~metrical

101

(Ray) Optics

7. For a concave milTor or converging lens, we can have: (a) real object - real image (b) real object - virtual image (e) virtual object - real image (d) virtual object - virtual image 8. Setting sum appears: (a) rcdish (b) bigger (e) higher (d) none of these 9. When a ray oflighl moves from denser to rarer medium (n2 < nl), with gradually increasing angle" of incidence: ,, (a) It deviates away from the nonnal UpIO a critical anglc (b) more light will be reflected gradually with increasing i , (e) total light will be reflected if;> ic (d) least amount or light will be reflected at i = 0° , I,' , 10. In which easels the image is fonned at the site of the object?

,

",

, ,

~,

(a)

(b)

(e) .:{-.>_c

f

c

(d)

1 J. The deviation ''0 min .. a prism versus the angle of incidence "i .. is given. Then:

•

6ma -- ----------------

6

c\m

,,

,, ,, --.,-t---',,, ,,, "'-;

--.,- ----------

, ,

o

"".

(a) 0 = il + i, - A (b) ifo.,;" = 30°. A = 30°

(c) foro max. ii

= e = 90° and sin ii = nsin(A - 9 c ). where 9(" = sin -I !

n

GRB Understanding Physi« Optics and Modem PhysiC!!

102

12. The distance "d" between two identical concave mirrors ofmdius of curvature R so as to coincide the image I on the object 0 itself, which is placed at the mid-point, can be:

•

o

-d-

(a) R (b) 2 R (e) 3 R (d) RI2 13. A man is standing inside a cubical room of side H as shown in the figure . A plane mirror is fitted at the front wall.

'=:::"'::T=--------'

-r j- : : =

I

Ho

-Hl4-

Y, Y

I

(a) the minimum size "h" oflhe mirror to see the full image ofthe man is Ho 12 (b) the minimum size "h" of the mirror so as to view the beside(behind) wall is 3Hn. (e) the position 'Y' orlhe mirror so as to view his full image is Yo

2

(d) the position "y" ofthe mirror so as to view the full image orlhe behind wall is 2at (H + HoY7. 14. A thin lens made oflhin sheet of glass has air inside it. Then: (a) it has focal length equal to infinite Ai, (b) it can deviate light rays (c) it can disperse light (d) all of the above

eRB Undcntanding Physic. OPIKs and Modem Physics

102

I

. -I 1

(d)omu = 90o +ii -A,whercil = sin - [nsin(A - Oc )} , where9 c =S1n

~

12. The distance "d" between two identical concave mirrors of radius ofcurvalure R so as to coincide the image J on the object 0 itself. which is placed at the mid.point, can be:

o• -d-

(a) R (b) 2 R (c) ) R (d) RI2 13. A man is standing inside a cubical room of side Has shown in the figure. A plane mirror is fined at the front wall. Then,

II T H

Ho

II

-Hl4-

Y,

1 -,

-,-" Y

3HI'-ll

(a) the minimum size"h" of the mirror to seethe full image of the man isHO 12

(b) the minimum size "h" of the mirror so as to view the beside(behind) wall is 3Hn.

(c) the position "y" ofthe mirror so as to view his full image is )'0 2 (d) the position 'Y' the mirror so as to view the full image of the behind wall is 2a. (H + Ho Y7.

of

14. A thin lens made of thin sheet of glass has air inside it. Then: (a) it has focal length equal to infinite (b) it can deviate light rays (e) it can disperse light

(d) all ofthe above

•,

103

Geometrical (Ray) Optics

]5. A cube of length ~x is present so that its point face is located on the center of curvature C of the concave mirror. Then :

(a) Imlincarl = lmlatl2

(b) the image will be a bigger cube (c) the front face of the image is larger than its back face (d) the upper face of the image is curved 16. Which orthc following statcment/s regarding image and object islare correct? (a) Any object placed in virtual space is called virtual image (b) We can nOI see a virtual object (c) We can not see a virtual image (d) Total inlernal refleclion causes clearer images 17. A man stands on a flat mirror M I in front of plane mirror M 2:

(a) Ihe mirror M I can fonn inverted image (b) the mirrors MJand M2fonn three virtual images (e) if M I moves along x-axis the images do not move (d) if M 2moves along y-axis the images do not move 18. If an object. the plane mirror and the image move with velocities ~

~

~

vo. Vrn

and Vi respectively. ~

~

(a)

V

omz = Vlmz ~

~

(b)

vorn)"

= u Im y

~

(e)

~

U

om: = Vim:

~

~

~

~

(d) VOy = v iy and vo: = v I:

,

104

GRB Undentanding Physics Optics and Modem Phytia

19. In the above question, if we substitute the plane mirror by a concave mirror:

p-

q ->

(a) v Om l

~-(9l~; p m, ....

->

(b) uOm, ->

(c) v Om z (d)

y

:# vi m )'

'*

->

v imz

-;Om, ~ ( ~F ; ,

->

and v Oz

~ -(~

F;,

z

20. Which of the following point is/are virtual object?

(a)

(b)

(e)

(d)

11. A plane wave becomes spherical if it is reflected or refracted by: (a) a glass slab (b) plane mirror (c) convex mirror (d) concave lens

Geom~lrical

105

(Ray) Optics

22. The ray I undergoes rpflection and refraction due to the glass slab. Then, the rays: 1 2

3

4

(a) 2 and 4 may superimpose destructively (b) 2 and 4 may super impose constructively

(c) 1,2,4 and 3 may be in phase (d) 3 and 4 may be out of phase 23. When a light wave is reflected which of the following cannot remain same?

(a) Frequency (c) Intensity 24. Refractive index can be: (a) +ve

(c) zero

(b) Phase (d) Velocity (b) - ve

(d) one

25. Two sodium lamps can not be: (a) monochromatic (b) coherent (d) ofsarne power (c) of same intensity 26. The light wave when changes the medium which of the following can change? (a) Frequency (b) Speed (c) Direction of motion and wavelength (d) Refraction index 27. A converging lens is used to fonn an image on a screen. When the upper half of the lens is covered by an opaque screen:

(a) half the image will disappear (b) complete image will be formed

(e) intensity ofthe image will increase (d) intensity of the image will decrease 28. In an astronomical telescope. the distance between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length loof the objective and the focal length feofthe eyepiece are:

(a)/o =45cmand/. =-9cm (b)/o =5Ocm and I, = lOem

(c) 10 = 7.2em and Ie = 5cm (d)/o =3Ocm and Ie =6cm

·r

106

GRB Understanding Physics Optics and Modem Physics

29. A planet is observed by an astronomical refracting telescope having an objective of focal length 16 m and an eyepicce of focal length 2 em. Then: (a) the distance between the objective and the eyepiece is 16.02m (b) the angular magnification orlhc planet is - 800 (c) the image of the planet is inverted (d) the objective is larger than the eyepiece

30. A ray oClight traveling in a transparent medium falls on a. surface separating the medium from air at an angle of incidence of 45°. The ray undergoes total internal reflection. lfn is the refractive index oCthe medium with respect to air, select the possible value(s) of n from the following: (a) l.J (b) 1.4 (c) 1.5 (d) 1.6 31. A student perfonned the experiment ofdctcrmination offocallength ofa concave mirror by 1/- V method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24cm. The maximum error in the location of the image can be 0.2 crn. Thc 5 sets of(lI. v) values recorded by the student (in em) are: (42, 56). (48, 48), (60, 40), (66, 33), and (78, 39). The data set(s) that cannot come from experiment and is (arc) incorrectly recorded, is (are): (a) (42, 56) (b)(48,48) (c) (66, 33) (d)(78,39) 32. A ray oflight ofwavelengthAo and frequency lIoenters a glass slab ofrcfractive index ~ from air. Then: (a) its wavelength increases, frequency decreases (b) its wavelength decreases, frequency remains same (c) its wavelength increases, frequency remains same (d)

6~ =~o(~ -I}nd 6v = 0

- -- - - -

,

107

Grometrical (Ray) Optics

(lMatch The Columns 1. An object (arrow) is placed in front of plane mirror in each case. Find the image. Column I

,

(a)

Column J[

,

/

(P)

.

(b)

(e)

(d)

I\

(q)

\/

I

\

\

(r)

,

(s)

/

2. A wave front approaches to a mirror or lens as shown. Find the reflected or refracted wave front. ColumD II

Column I

(a)

~....... ""

(b)

(P)

~(

(q)

~

"',,.

-ttt- .I.

-,

--<

m

v-'" (e)

h., I I

., (d)

,

GRB Undmtanding Physics Optics and Modem Phy1ics

10a

(r)

~

~

*,,'\

~

(s)

(( (.)77-

3. Match the mirror/lens with graphs (m = magnification). Column II

Column I m

., (a)

Convex mirror

(p)

-

0

1"-

-

)\

m

., (b)

Convex lens in air

(q)

p

0

f

,, ,, , ,

p

if ,

• (e)

Concave mirror

(r)

,

0

1\

~

p

109

Geometrical (Ray) Optics

q

(d)

Concave lens in air

(s)

i\ \1'

'-

,,

. 4.

Match the columns A and B.

ColumaA

Column B

(a)

Glitterin2 of diamond

I (D)

Interference

(b)

Rainbow in rainv day

I (a)

Total internal reflection

(e)

Rainbow colors in thin films of (r) water

I (d)

Glitterin2 of dust particles

I (s)

Double refraction and dispersion o fli~ht

Scatterin2

5. Two transparent media of refmctive indices III and J.I3 have a solid lens shaped transparent material of refractive index J.I2 between them as shown in figures in column II. A ray traversing these media is also shown in the figures . In column I different relationship between J.I). J.I2 and J.IJ8l'e given . Match them to the ray diagrams shown in column II.

Columa J

Columa II

(a)

J.II < J.I2

(P)

(b)

J.II > J.I2

(q)

~ ~~

110

GRB Undentanding Physics Optics and Modem Physics

(e)

~,

= ~3

,]:,

(r)

r(s) (d)

112 >"') (1)

~~2' - .,

y;

.,

t)Comprehensions Pa.sage-1 (Newton'. displacement method) This is a method to calculate the focal length ofa lens by displacing it along the

principal axis. For two different positions I and 2 of the lens, we can get two distinct images 11 and J2 of the object 0 on the screen. The distance D between the object and screen is kept constant. If d = distance between two positions 1 and 2 of the lens, give answer the following questions: 1

2

0----1 1. The focal length of the lens is:

(a)D'-d' (b)D'+d' (e)D'-d' (d) D' - d' D 4D 4d 4D 2. If hi and h2 are the heights of the images II and 12. the height of the object is: (a) hI + h, (b) Jhlh, (e) hlh, . 2 hl+h, 3. To get two sharp images 11 and J 2 on the SCreen: (a) D = 2/

(b) D > 4/

(e) D < 4/

(d)

Ih, + h'

'/1' (d) D > 2/

111

Geometrical (Ray) Optics

Passage-2 (Combination of thin lenses) When we keep two lenses LI and L2 having two optical powers in air PI and P2 respectively, at a distance d, the combination behaves as a lens of optical power PP2

P = PI +P2 -d-'-. n

_d_

...,,"'" where n = R.I. of the medium between the lenses. If d = 0 ( the lenses are placed in contact), P = PI + P2 using the above infonnations, give answer the following questions: 1. A equi-coneave lens nl = 1.5. R =20 em is placed in contact with a plano-convex: lens "2 =4/3 and R =20 cm. The optical power of the combination in Diaptore is:

(b) - 10/3

(c) ~

(d) 1013 3 2. The magnification of the.final image of the object 0 is: (a) - 519 (b) 915 (c) 514 (d) - 9110

(a) 8/3

3. Two lenses ofR.I. = 1.5 focal lengths II = 15 cm and/2 = 10cm in air are kept in water at a distance of 5 cm. Then, the effective optical power of the combination is: (a) 20/9 D

"

,

GRB Understanding Physics Optics and Modem Physics

112

Passage - 3 (Dl8perslon and Prl8m) When a ray of sun light passes through the prism, each color of light has its own speed in the glass. The seven colors come out of the prism with different angles of deviation. Red deviates least and blue deviates maximum. Thus a prism can disperse a white light into seven colors which is called color spectrum.

For a thin prism. the angle of deviation is given as 15 = (n -1)A . The angle of .......... r · · ~ ~ · d ISperslOD IS given as,*, = Uv -Ur' The dispersive power 0 a pnsm ISID = ome:m = S y ' Using the above ideas, give answer the following questions.

I. A prism bas R.I. for violet and red nv = 1.523. nr = 1.5145. If the angle of prism is A = 2°, the dispersive power oftbe prism is:

(a)

(b)

(e)

(d)

0. 1639 1.593 1.8 0.18 2. The rays (i) and (ii) are the incident and emergent tays of the achromatic combination of two thin prisms. The ratio of(nv - nr ) for the prisms I and 2 is (A and A' are the angles of the prisms 1 and 2 respectively):

(i)/)f.<:2:S;~1

(il)

1

(e) A - A' A +A'

(a) A'

A

(d)

p;

3. Ifin the combination of prisms the emergent red, violet rays arc parallel, the ratio

of mean deviations in the prisms I and 2 is:

,

2 n'

,

y

n

(a)(I-::)

1

(b)(I-::)

(e) ""

""

(d) ""

""

,

113

Geometrical (Ray) Optics

Passage - 4 (Measurement of R.I. of a Liquid) When we keep a bi-convex. lens on a plane mirror, the combination will act as a mirror, as shown in the diagram.

1. In diagram,flx. where f ::: focallcngth of the lens in airis :

,I

~~LJ (a) 1:1 2. In diagram.

,

f'

(b) 2: 1 where f

(e) 1:2

(d) ./2:1

::: focal length of the lens in air is:

0'.__+ __

I

r

~t~LI (a) 1:1

(b) 2:1 (e) 1:2 (d) 1:./2 3. When we introduce a liquid between the lens and mirror as shown in figure, assuming the radius of curvature of the lens R ::: 20/3 and x~ = 40 em :

0'._____

I I

". L

(a) 1.5

(b) 1.66

(e) 1.52

(d) 4/3

I

'r

,

GRB Undentandins PhysKs Opbcs and Modern PhYlia

114

P ....ge-5 (TJR)

When a beam of light moves from denser to rarer medium. a fraction of its energy gets reflected and rest of the light gets transmitted. For any angle of incidence i less than a critical (maximum) angle 9 this happens, if; > OCt total energy will be reflec~ed back. This is called total internal reOection(TlR). Following the above ideas, give answer the following questions.

,,, I :

",

"3 1. The incident ray cannot emerge in the mcdiwnofR.I . n] if(assuming"r slightly

greater than sin

-t( :~ ):

(a) n2 has any value

(b) n2 < n]only (c) n2 < nl only (d)nl < n3 andn2 < n,

2. If we think of a medium wbose refractive index decreases along y-axis. For a ray of light incident at 0, assuming so many thin strip of medium and applying Snell's law of refraction, which of the following path is correct for the refracted

ray. y

•

3

12

,

/:0
x

(b)

\

115

Geometrical (Ray) Optia

(e)

(d)

3. The interface xx' separates the space into I and 2 having two media. If media I has R.I . =" and media 2 is air, taking negative R.I. into account, which of the following diagram is correct for total internal reflection?

2

'---,1---

(al

A

(b)

(e)

~

(d)

P."IIIO 6 (OptIcal Pow . .) The opti~l power of an interface is defined as, p="2-"t,

R where R = radius of curvature oftbe interface. Putting p = co, q = h in the fonnula ., 1p+'2 Iq~

R

(=

- h .A-

n,

P)

'2 112 =p

webave Similarly, Then, were h P=

"2 -"I

n,

.,If, =p .,If, ='2112 =p

"2 -nJ

R

.

Using the above fonnula, give answer the following questions.

F,

, CRB Undentanding Physics Optics and Modem Physics

116

1. The bi·convcx lens ofR.1. n is kept in a medium of R.I. no. The focal length ofthe lens in air isf. Then. the optical power of the lens in the medium oCRI no is: 1 n no d II-no ~ (a) / (b) / (e) (n -1)/

7

""0 ""

()

2. In Q-I, if f = focal length of the lens in the medium ofR.1. =

no. the optical

power of the lens in that medium is:

! /

(a)

(b) no

(d) n - no

(e)n - no

/

(n - 1)/

/

3. If in Q-I , the Rl of the media in both sides of the lens are nl and " 2 respectively, the value of I I I /2=

(a) !!!.

(e)

(b) I

n,

n,n,

(d) n-n, " 2 -11

4. The optical power of the lens is: n (a) II - nl + "2- .

. R,

(e)

n-1I2

RJ

(b)n-nl +n-n2

R2

Rl

n-1I2

Rz

(d)n+nJ -112

+ "-::-'''' Rz

,JR,R,

5. The power of the lens is assume n = 1.5, •

I

•

n, =~.3 R = radius of curvature of all

surfaces = 20 em:

n

",

n

,

(a) 5/3 D (b) 213 D (e) 4/3 D 6, In 0 -5, Ibe focal length oflbe lens is: (a) 0.8 m (b) 0.9 m (e) 0.7 m

(d) 7/5 D ,

(d) 0.6m

•

117

Geometrical {Ray} Optics

Paosage -7 (Slivering ollen..o) Silvering of a lens means mirroring of the lens.1t is done by pasting a metallic coating at one side of the lens. As a result. the lens behaves as a mirror. We can think it as the combination oflens Land M. As the ray is refracted 2 (passing) through the lens twice. we can consider lens twice. Since the 1 M ray gets reflected by the mirror once, we will consider the mirror once, Then. optical power and the focal length of the combination is p = p/+ Pm +PI I I I I And

L

-=-+-+Ii 1m Ii 7 7(;,)+ )m

I

In general,

M

L

=

+

where Ii = focal length of the loth lens (in the medium I) Based on the above theories, give answer the following questions. 1. The surface 2 of the lens is silvered. If R, = 20 cm, R2 = 30cm and n = J.5, the optical power of the lens·mirror combination is:

• 1

(a) 15 D

(b) 25 D 6

N

2

(d) 65 D 6

2. If the concave surface of a plane...concave lens of R = 20 cm and n = 1.6 is silvered, the focal length of the system (mirrored lens) is :

(a) - 25 em

(b)25 em

2

(d) _25 cm 4

• GRB Undentanding Physics Optics and Modem Phyaics

118

3. Two lenses of "I'''' 413 and n2= 1.5 are placed in contact. If the flat surface is

silvered and the radius ofcurvature is 12 em. the focal length of the combination is:

","'"

(a) 25 em

(b) 50 em

(e) 12.5 em

(d) 24 em

Passage - 8 (Magnification and Magnifier) When two rays (at least) get reflected and refracted by mirrors and refracting surfaces, the ratio of size of the final image and object is called magnification given as m = ml )( m2x..... )( m". where mi. m2 etc., are the magnification produced by the optical systems (minors, lenses etc.) 1.2.3 ..... r respectively. For a mirror and a thin lens the lateral magnification is given as

I

m = -;

I

Where p and q are the object and image. However, for refracting surface, the lateral magnification is

The linear magnification of a thin image is basically given as

I

m'

=~!

I

Based on the above ideas give answer the following questions. t. A thin convex lens can produce magnified erect images as shown in this figure when p< f. This property is called magnificalion and then the lens is called a magnifier. The magnification is m = 2. Ifthe eye is at a distance of 5 em from the lens. for distinct region, the focal length of the lens is:

,, I '

'---"'--::::::::: p -

(0) 10 em

q

(b) 15 em

(e) 20 em

(d) 25 em

•

119

2. For a particular object distance two lenses LI and L2 separately produce

magnificationsm, = -2andmz = +3 respectively. If the lenses are placed side by side (in contact), keeping the same object distance, the magnification produced by the combination is: . (a) -5 (e) +1

(b)-I (d)-6

3. The lateral magnification ora thin object due to a concave mirror is 4. The linear magnification is: (a) -2 (b) -16 (e) +8 (d) +16 4. The lateral size of the object is 2 em. The size of the image is:

",s4i3

"2",1.5

, (a) 8115 em (b) +16/ 15 em (d) 15 em (d) +32115 em 16 5. In Q4. the linear magnification is: (b) _16 (a) _32 25 25 (d) +25 (e) _256 16 225 6. The linear magnification of the object due to the thin lens is:

(a) -2 (e) -4

(b) 4 (d) +2

•

120

GRB Undentanding Physics Optics and Modem Physics

(I Subjective Problems l ..el-1 Concept of reflection and refraction, refracttvelndex, Totallntemal reflection I. Find the minimum lime taken by a ray of light to pass through a glass slab of

I•

thickness ~ m.

2. The wave length orlight decreases by a fraction x in a medium (water) say. Find the value of x. 3. A ray oflighl falls at an angle of30° with nonnal to the water surface. What is the angle of deviation of light? 4. Find the permittivity of water for the visible light. S. The wavelength oflighl in a medium (glass) is A. = sooo A. Find its frequency. 6. Find the effective refractive index of the composite slab of two medium placed side by side having R.I. nl and n2. assuming their lengths II and h respectively.

1",1 .. 1 -11--~Images formed by reRection In mirrors 7. A ray oflight is incident on mirror M\ at an angle of300 with vertical. Find the deviation of the ray aftc:r bc:ing rc:Rected by the mirror M 2.

50" M, "

8. If a point object moves with a velocity Vo = 7- 2j + k mls and the mirror moves with vim = -27 mls find the velocity of the image.

I Lx

9. A person can see his total body .in a plane mirror pasted on the wall of a room. Find the position and size of the mirror.

I

1- X-II

10. A man of height h stand at a distance of x' = ~ from the front wall, where x = length of the room. Find the position and height of a mirror pasted on the front wall so that the man can observe the full image of the back-wall behind him.

-

Gcom~tricaJ

121

{Ray) Oprics

I J. A small object is placed in front of a concave mirror of radius of curvature R = 20cm. on its principal axis at a distance of (a) 15 m (b) 5 m from the pole. Find the nature and position of the image. 12. An erect object is placed at a distance of20 cm from the convex mirror offocal length 20 cm. Find the (a) position and nature of the image (b) magnification.

-I

.oem

-

,

J3. A converging beam of light is obstructed by a concave mirror of radius of curvature R = 30 m as shown . Find the position of the image.

--- --- -20cm

1_

Retraction due to prism, deviation and dispersion of light

14. Find the value of i so that the emergent ray emerges the surface S. Assume A = 60° and n =.Ji

15. In 0.14, find the deviation offighl. 16. Find the maximum deviation of the incident ray when it emerges at the face"S" of the prism of angle.

•

•,

122 17. Find the (a rel="nofollow"> minimum value of j so that the ray will emerge at the face 2 of the prism. (b) Following (a> find the corresponding deviation.

18. For what minimum value of R.I. = n, in a prism of apex angle A, any incident ray will not emerge at the opposite face?

• 19. Find the conditions if the ray will: (a) find grace the surface I, (b) be separated at the surface t. (e) be reflected totally at the surface 2.

Images foiilled by r.'.actlon In n.t rehactfng aUIt.ce. 20. When you look into water surface, a fish seems to be at a distance of2 m. Find the actual position of the fish.

t

«;"'"

11. A point object is placed at o below two immiscible liquids. Find the apparent (a) depth (b) shift orlhe object O.

21:. Find the deviation of the ray coming from water to ~ <s" I gass. _

Waler

•

123

Geometrical (Ray) Optics 23. Find the minimum deviation for the prism of apex angle 600 placed air.

R_lon 01 opherlcel ourleceo 24. Find the image of the point object O. TakeR = 20cm forthespherical interface.

25. A point mass is placed on the principal axis at Oatadistance 10cm from the pole. Find the image. Radius of the spherical surface is R = 30cm. "2"'1.2

-26. A point object 0 is placed on the principal axis. Find the image. Assume R =30crn. Ai,

27. A thin beam is coverings at O. Find the image produced by the beam if the radius of curvature of the interface is R = 20 CR1.

• t

124

GRB Understanding Physics Optia and Modem Physics

=

28. A beam of light converges to the point O. If R 30 em at the interface, find the position nature and position of the image fonned by the beam.

-- -- -- --

o

!--5Oan--

29. Two parallel rays are repeated at the curved (spherical) surface of R = 20 em, Find the image.

30. Find the point of inter section (focus) of the rays I and 2 getting repeated at the curved interface dividing the space into two media of R.l nl = 413 and n2 = 1.5. The radius ofcurvature of the swface is 20 em. 1

2

nz-1.5 ' 31. A point object 0 is kept on the principal axis at a distance of p

=15 em from the

spherical surface of R = 20cm. Find the position and nature ot-the image.

n,. '.2 c

0

~I

"2- 1 .5

•

125

Geom,trical (Ray) Opt.. '

32. Find the position and nature of the image if radius of curvature of the spherical interface is R = 20cm.

"1'" 1

"2=1.5

o

33. Find the nature and position of the image if the radius of curvature of the spherical interface separating the two media is R = 30cm.

-11Ocm 34. A point object 0 is placed in a medium (glass) oCRJ nl = I.S on its principal axis. Ifwe see the image from air, find the position and nature of the image. Radius of the spherical surface is 20 em.

o

/-3Oan

..

35. Find the position and nature oftbe image formed by the point object o. Ramus 'or curvature is equal to 20 em.

o

-1-_ .'

,

FWOcm

<

126

GRB Undentanding Physics Optics and Modem Physics

36. A point object is placed at a distance of 20 em from the pole of the interface of

radius of curvature 10 em. Locate the image

-----------

\

n,.'.5

"2* 1,3

,

C

0

~

,/1

7

37. A point object is located at a distance p from the pole such that a point virtual image is fonned at a distance of 90 em from the pole. Find the value of p. R =30cm. nl" 1.5

nr .2 ' C

1-038. A parallel beam of liquid strikes the interface where will it be focused? The radius of curvature of the interface is 30 em. "2"' 1.5

nllaleasa 39. What is the focal length of a thin bi-convex lens of radius of curvature Rl = 20 andR2 = 301 R.I. of the lens material is n = 1.5 •

.to. What is the power of a lens having focal length f = 20 em in air when placed in water?

41. If the point object 0 Coons an image I due 10 the lens in air. find the refractive indexoftheglassifRJ =R2 =12cm. 42. Find the magnification of the object due to the lens of local length or20 em.

.

127 43. Find the image distance of an object which has a magnification: (a)

m=~

(b)m=-~

44. A beam oflighl converges at 0 in the absence of the lens. If the lens is placed as shown, where will the light be focused?

o lOem

-I

45. Two lenses offocalleng1hs 11 = 20 em and f2 = 30 em are placed size by side. If an object is placed at a distance of p = 20 em, find the final image I .

I-

p

.---1

46. Find the (a) power (b) focal length of the composite lens of refractive index nl =nl =I.S,RI =12cmandR2 =lOcrn.

4'. A candle is placed on the principal axis of a thin lens of focal length f = 25 em. Find the (a) lateral (b) linear magnification oftbe image.

i

1 I

•

j •

/-3Oan,-.J'

128

eRB Undentanding Physics Opti" and Modem Physics

48. A bi-convex lens of radii Rl = R2 = 20 em is kept as shown. Find the (a) focal length (b) image, ~~~~

--

C

49. In the previous problem, ifthe objects is placed at a distance of20 cm in LHS of the pole of the lens find the image.

[]I Subjective Problem. Level-2 1. A glass slab of refractive index = 1,5, thickness 3 cm is placed at a distance of 14 cm from a concave mirror of radius of curvature 200 cm. An object 0 is kept at a 9 distance of 4 cm from the slab. Find the distance between final image and the mirror.

,

"

.

,

2. A point object 0 is k'ept at a depth h. Ifwe look at the object with an angleS with vertical, where the image wiU be formed? Refractive index ofliquid is equat"n".

,,

:. -------------- .....- ------------:~---=:i-"--~<- ~=:~---=: 1:::.-_-=-= ------ -..'"':-= -.r=-=-_-:=: ~---

-~-------

h:::::~::-:=::::::

-----------------------j------------------------o

•

Gt"Omelrical (Ray) Optics

129

3. A point object 0 baving object distance p = 20 em moves with velocity ....

~

A

A

0 ::: 3i + 2j - k where as the concave mirror of focal length f ::: IS em moves with 0 V

....

A

A

'L,

~I~==-:p;-:==~

A

a velocity V m =-2i + j-3k. Find the velociry of the image, 4. A long rod of refractive index "I ::: 1.5 is kept in liquid of refractive index ="2 = J .1 . (a) Where will the parallel rays I and 2 to be focused? (b) Jfwc remove the rod fro~ liquid and kept in air find the % age _t-J1t:==:::=~, change In the focal point of the rays in case (a) the radius of curvature is R = 20 em. 2

s.

Two parallel rays coming from infinity a sphere of a transparent material of radius R and refractive index "II". In consequence, as point image J is formed. Find the image ' , d Istance q .

.' 1-

6. A thin rod is kepi at a distance of J5 em from a concave mirror of R = 20 em. A convex refracting surface at R' = 50 em and refractive index n = 1.5 is kept at a distance of 20 em from the

","' ,

mirror. Find the (a) nature and position of the final image

(b)

laleral

magnification of the final image.

1Scm

1-

20an _ _ 1

7. A small object 0 is placed at a distance of5 cm from a spherical refracting surface ofRI. n :: 1.5. The radius of curvature of the refracting surface is R = 20 em. Find the: r - - - - -____ (a) image distance, (b) lateral magnification, (c) linear magnification, diagram.

drawing

the

ray "1= 1.5

-I

Scm

-

, 130

GRB Undentanding Physics Optics and. Modem Physics

8. An objcct 0 is kepi infront of a spherical

"2-1.5

refracting transparcnl surface of a R = 20 em, "2 = 1.5, at a distance of 20 em. The refractive index: orlhc medium of RHS of the cylinder is n'2 = 1.2. Locate the final image.

~!!l

•

C 40em

\

9. A point object is placed at a distance R from the convex surface of a glass of radius of curvature RI = R and refractive index = n = 1.5. The second curve surface is silvered. If the 1 2 radius of curvature of second (silvered) surface is R z =!i, 0 R-ot;; 2 find the position of the final

image from the 1st surface.

n=1 .5 -

,

R

10. A thick glass cylinder of spherical surface I of a radius of curvature R = 10 em. R.I .= " = 1_5 has n nat surface 2. The thickness (length) of tile cylinder is 20 em. If the point object 0 is kept at a distance , I n.,1 .5 2 of 40 cm from the surface I, on the principal axis, find the position of the 0/ image relative to the flat surface 2. --l;;;;;;j~=::-;o;;;;::==r---

~Ocm

.~"

.\'-------' Il. A linear object is placed at a distance of 10 cm from the spherical interface separating two refracting media of refractive index nl = 4/3 and n2 = 1.5. If the radius of curvature of the surface is R = 30 cm, find the:

(a) position of the image, (b) lateral magnification, (c) linear magnification.

•

131

Geometrical (Ray) Optics

12. A linear object is placed at a distance p = 80 cm from a transparent rod of RI 11 = 1.5. The radius of the "t-1 / spherical surface is R =. 30 i'-..... n2=1.5 cm, find the : "0,.5,-=0+1_"'-"i - - - - - - - - + - - (a) image and its position, (b) magnification, \ ' - - -- - - ' (c) ray deviation. 13. A beam converges to point 0 in a medium of refractive index n. A thin ", bi-convex lens of R.I . "2 =. 1.5 cm and ", ~~~~ radii of curvature R\ =. 20 cm and R 2 = 15 cm is placed in the medium. Find the 'Oan (a) focal length of the lens ifthe image 2 is fonned at RHS at a distance q = 230 cm from the lens.

j-p-

-----

---

------h-t----.::..:..;.°

,

I-

"I.

(b) refraction index (c) drawing the ray diagram. 14. An objecl of height h is placed at a distance D from a screen so that its image can be observed clearly on the screen fonned by the lens placed in between the object and screen. The focal length ofthc lens isf· Screen (a) Find the va lue/s of D . (b) If the lens is shifted througb a distance d .l d from position 1 to position 2, we can have lens distinct images of the object on the h •• •• screen, find the value o f f " I+--+-I--+-I--~ • •t '• (c) Find the product of magnification of the : 1 '.' 2 distinct images produced in case (b). : (d) If the heights of the images in casc(a)are~1 . - - - 0--,--_1 ~ and h2 , find". (c) Find the ratio of magnification of the images corresponding to the lens position I and 2 of the object, on the screen. (f) Find the difference in magnification. that is, m) - m2 15. A small object is kept at a distance of 10 cm from a bi-concave tens of R) = IOcm and R2 = IScm. The R.I. of the surrounding medium is nl =. I and that of the lens is n2 =4/3. (a) Find the focal length of the lens. (b) Find the image and its position. (c) What is the lateral magnification? (d) Draw ray diagram.

..

",

",

"-..

•

132

GRB Understanding Physics Optics and Modem Physics

,

16. A beam converges to a point O. A bi-concave lens of focal length f = 20 em is placed al a distance of 30 em from O. (i)Find the:

-L c'

•

300m-I

(a) image, (b) lateral magnification, (e) linear magnification, (d) trace the ray diagram. 17. An object is kept at a distance of 15 em from

the concave mirror of radius of curvature R = 20cm. A lens is kept al a distanceof20 em from the milTor. Find the:

,Scm

(a) location of the image after the reflected rays pass through the lens. (b) magnificatio,"! o r the final image.

_2Ocm-!

18. Three lenses LI . L2 and L) are kept as shown in the figure. The focal length of the

lenses arc

It

= 10 em,

h

= 10 em and

fJ = -10 em respectively. The

surrounding medium is air. Find the position and focallengtb of the equivalent

Jens. L,

L:!

/\

/\ Scm

5I3cm

19. A thin beam having the paraxial rays 1 and 2 are incident on the lens L 1• Assuming 11 :::: 12 :::: 13 = 10 em, find the position of the final image.

~omelrical

133

(Ray) Optics

2

20. Two lenses LI and L2 having the fOCDI lengths L, ~ II =112 1= IOcm Drc separated by a distance of 10 cm. lfa point object 0 is placed at a distance of I 5 em from LI. find the location of the final image. O~~--+-I_ _-J..j_ __ 21. Abi-convexlensL I of radii ofeurvature Rt = 20 cm and R2 = 30cm and R.I. = n = 1.5 is kepi at a ,5cm_!-,0Cm_ separation of 12 cm from a bi-concave lens ~ whose radii of curvature Rj = 12 em, R2 = 24 em and R.I. = ,,' = 4/3. Find the (a) focal length of the effective lens (b) location of the effective lens.

l

n=I .5

n=413

12em

22. Power ofa bi-concave lens LI in air is Pair = -5 D. Its refractive index = n = 1.5. (a) Find its power in a medium of refractive index no = 1.6. (b) Find the powerofanolher lens L2 (bi-concave) in the medium no ,;, 1.6, radii of curvature R I = 30 cm and R2 = 60 cm. (e) If these lenses LI and L2 are kept separated by a distance d = 16 cm in the medium of refractive index. = no = 1.6, find the optical power of the combination. 23. A bi-convex lens L1 of R.I. = 1.5 is placed in contact l, with a bi-concave lens ~ of R.I. = 4/3. The system Of two lenses is placed in air. Find the: (a) optical power of the combination, (b) focal length of the combination.

1.5

1

ABC

134

CRB Understanding Physics Optics and Modem Physics

•• "

24. A convex lens ofrndii of curvature RI, R2 and R.I. ff is kept at a distance I frama plane mirror. In between the mirror and lens. a liquid of R.I. flO is prescnt. Find the optical powerofthecombination. Put n = 1.5. " 0 =4/3 and I = R 30cm.

=

: ••••• : =.',-.::.-.

,

=.-,'.:=.-.:.::._.------:'-.-;: :".';::"..---------------------_ . ... _--- -------

-,-

25. A bi-convex lens of R.t. " 2 = 1.5 is silvered and is placed in a medium of R.I.

"I

=

I (air). The radii of curvature of the lens is Rl == R2

~

20 em. Find the:

(a) focal1cn gth of the silvered lens, (b) image distance for the converging incident rays for the silvered

lens in (a),

"1

"2

(c) trace the ray diagram for (b).

" '"

•

'Oom

•

26. Two thin lenses L\ and L2 are placed in contact in air. The radii of curvature o f the surfaces SI and S2 arc R, = 30 cm and R2 = 40 cm respectively. The R.1. of lenses are nl = 1.5 and n2 = 4/3 respectively. (a) What is the focal length of the combination? (b) Find the focal length of the combination when immersed in a liquid of R.I . =n =1.6.

AI,

AI,

Flat

surfacs

8,

27. A thin bi-convex lens of R.I. n2 separates two media of R.I. n,and n3. If the radii of curvature are R] = 20 em and R2 30 cm, find the : (a) power of the lens in the given system of surrounding media, (b) focal lengths ofthe lens, "3 (c) power of the lens in air,

=

",

(d) focal length of the lens in air. (Given nl = 1.2, n2 = 1.5, n3 = 4/3) System of SUfTOUndiog media

Geometrical (Ray) Optics

135

28. A thin bi-convex lens of R.I. n is kept in a medium ofR'!. nt. its optical power is. PI an~ focallen~th is fl. When the lens is placed in a medium of R.1. "2. lis opucal power IS P2 and focal length is h. Find:

(.)121/, (h)P,IP,.

29. Two lenses LI and ~ are in contact and the lens L,.is silvered. The radii of curvature ofL, are 20 cm and that of the silvered surface is 30 cm. The R.1. of the lenses are 112 = 1.5 and 113 = 1.6. The surround ing medium has R.1. "I =4/3. Find the: (a) effective focal length. (b) optical power of the system.

30. A small thin object is placed in the principal axis of the lens of R.I. n2=1.5. The surrounding media have two R.I. nl = 1.2 and n3"" 4.3. R I == 6cm andR 2 =4 cm. (a) If p = 12cm. findq. "1"2 "3 (b) What is the oplical power oflhe system? (e) Find the lotallateral magnifiealion. (d) What is the over all linear magnification?

1-

p

. 31. Two identical prism of refractive index .Jjare kept as shown in the figure. A light ray strikes the first prism at face AB.

e

0

A~L---------¥C-L--------~E

(a) Find the angle of incidence so that the emergent ray from the first prism has minimum deviation, (b) What the angle of the pri sm DeE should be rotated about C so that the final emergent ray also has minimum deviation? 32. An object is moving with velocity 0.0 I mls towards a convex lens offocallength 0.3 m. Find the magnitude of rate of separation of image from the lens when the object is at a distance of 0.4 m from the lens. Also calculate the magnitude oflhe rate ofchange ofthe lateral magnification.

, 136

GRB Undentanding Ph)'Sic.s Optics and Modem Ph,..,ics

33. Figure shows, a prism of an angle 30° and refractive index J.l p = .Jj . Face AC of the prism is covered with n thin film of refractive index ~ J = 2.2. A monochromatic light of wavelength). = 550nm fllIls on the face AD at an angle of incidence of 60°, Calculate the: A

B'-----'{. (0) angle of emergence, (b) minimum value of thickness I so that intensity of emergent ray is maximum.

34. A thin hi-convex lens of refractive index 312 is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index 413. it is found that when a point object is placed 15cIll above the Jens on its principal axis. the object coincides with its own image. On repeating with another liqu id, the object and the image again coincide at distance 25 em from the lens. Calculate the refractive index of the liquid .

l'-"";<:;;;;;;:>;.......JI 35. The refractive indices of the crown glass for blue and red lights are 1.5 1 and 1.49 respectively and those offliot glass are 1.77 and 1.73 respectively. An isosceles prism of ang le 60° 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. Detennine the angle of the flint glass prism. Calculate the net dispersion oftbe combined system. 36. A convex lens of focal length 15 em and a concave mirror of focal length 30 em arc A kept with their optic axes PQ and RS parallel but separated in vertical ~~~--~-,~-a O.6cm B direction by 0.6 em as shown in the --------·8 figure. The distance between the lens and mirror is 30cm. An upright object AB of 20 em 30"" height 1.2 cm is placed on the optic axis PQofthe lens at a distance of20cm from the lens. If A' B' is the image aOer refraction from the lens and reflection from the mirror, find the distance of A' B' from the pole of the mirror and obtain its

--1"-------

I

Geometrical (Ray) Optics

137

magnification. Also locale the position of A' and B' with respect to the optic ax.is RS.

37. A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a dist..1nce ofmR from it. Find the value ofm for which a ray from P emerge parallel to the table as shown in the figure.

rnA I

A

38. The x-y plane is t,he ~unda!y between two transparent media. Medium·! with z ~ 0 has3 refractive mdex.J2 and medium-2 with z :S 0 has refractive index Jj.

A

A ray of light in medium-! given by the vector = 6J3; + gJj j -10k is incident on the plane of separation. Find the unit veclor in the direction of the refracted ray in mediurn-2. 39. A pin is placed IOcm infront ofa con vex lens of focal length 20cmand madeofa material of refractive index 1.5. The surface of the lens farther away from the pin is silvered a nd has a radius of curvature of 22 cm. Determine the posilion of the final image . Is the image real orvinual? 40. A ray of light is incident at an angle of 60° on one face ofa prism which has an angle of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculale the refractive index of the material ofthe lens_ 41. A prism of refractive index I-' I and another prism ofrcfractive index 1-'2 are stuck together without a gap as shown in figure. The angles of the prisms are as shown. J.i land 1-'2 depend on A, the wavelength of light,

o

60°

40·

ALl.----'-_'\, 4

4

· 120 + 10.8.10 an d 112 = lAS + 1.80 x IO where).. is in nm. aceord Ingtolll =.

A'

A'

(a) Calculate the wavelength AO for which rays incident at any angle on the interface BC pass through without bending at that interface. (b) For light ofwavelengthAo. find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum.

eRB Und~ntandin8 Physics Optics and Mod~m Physics

138

42. A thin equio1:onvex lens of glass of refractive index ~ = 3 / 2 and of focal length 0.3 m in air is sealed into an opening at one end on a tank filled with water ().1 = 4 I 3). On the opposite side of lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in the figure. The separation ~tween the I~ns and the mirror isO.8 m. A small object is placed outside the tank mfront of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to lens) of the image of the object fonned by the system.

•

O.9m

I

Oam

•

"

...

•

........ .... . .

----- - ----~

...................... .... . .. . .. . . . . . . ...... .......... . .. . . .... .. . ............ .........................

43. A thin plano·convex lens o f focal length I is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image planes is 1.8 m. The magnification of the image formed by one of the half·lenses is 2. Find the focal length of the lens and separation between the two halves. Draw the ray diagram for image formation. •

44. 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 ( see figure). The point ofincidence is the origin A (0,0). The medium has a variable index of refraction J.I( y) given by

Y

~(y) = [ky'l2 +1)112, where k = 1.0 (mctre)-3/2

t-1.0m

J

B(x,y) Medium

The refractive index of air is 1.0. ./ (a) Obtain a relation between th~ slope of --l'--: ___::>'A~(~O~,O;;-I- - - - - , the trajectory ofthe ray at a pomtB(x, y) in the medium and the incident angle at that point. (b) Obtain an equation for the trajectory y(x) of the ray in the medium. (e) Detennine the coordinates (Xt. Yl) of the point p. where the ray intersects the upper surface of the slab--air boundary. (d) Indicate the path of the ray subsequently.

Geometrical (Ray) OptiQ

139

45. An image Yis fonned ~fpoinl objeetXby a lens whose optic axis isAB as shown in ~gure..Oraw a ray diagram 10 loc~te Ihe lens and its focus.lflhe image Yofthe object X IS fonned by a .concave mIrror (having the same axis as AB) instead of lens, draw another ray diagram to locate Ihe mirror and its focus. Write down the steps of construction of the ray diagrams.

ox

A---------------------B

0'

46. Light is incident at an angle a on one planar end of a transparent cylindrical rod of refractive index}l. Detennine the least value of~ so that the light entering Ihe rod does not emerge from the curved surface ofrod. irrespective ofthe value ofa.

,, 9Cr-1I ,: _"n' _______ , ~

a

J

,,, ,, ,, ,, ,,

47. A parallel beam of light travelling in water (refractive index = 4/3) is refracted by a spherical air bubble of radius 2 mm siruated in water. Assuming Ihe light rays to be paraxial. (a) Find the position of the image due to refraction at the tirst surface and Ihe position of the tinal image. (b) Draw a ray diagram showing the position of both the images. 48. A right prism is to be made by selecting a proper material and the angles A and B (B S. A). as shown in the figure. It is desired Ihat a ray of light incident on the face AB A,--jf----i-------:,B emerges parallel to the incident direction after two internal reflections. (a) What should be the minimum refractive index ~ for this to be possible? (b) For ~ = 513. is it possible to achieve Ihis with the angle B equal to 30 degree ?

c

, 140

GRB Undentanding Physics Optics and Modem Physics

)

49. Monochromatic light is incident on a plane interface AB between two media of refractive indices III and 1l2(1l2 > JlI} at an angle of incidence 9 as shown in figure. The angle e as infinitesimally greater than the critical angle for the two media so that total intemal reflection takes place. Now if a transparent slab DEFG of uniform thickness and of refractive index ).13 is introduced on the interface (as shown in the figure), show that for any value OfJ.l3 all light will ultimately be reflected back again into medium II. Consider scparately the cases:

°,

Medium I (1-11)

E

1- - - - - - - - - - - - - - ,

, , G f

(113)

'' '

:F

A--~~----~----~---B

•

(a»).13 < 1-11 50. A plano·convex lens has a thickness of 4 em. When placed on a horizontal table, with the curved surface in contact with it, thc apparent depth of the bottom most point of the lens is found to be 3 cm.lfthe 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 be (25/8) cm. Find the focal length of the lens.

141

Geometri<:aI (Ray) Optics

•

•

, ANSlIJEAS ,

Multiple Choice Questions

(A) Only One Choice Is Correct Level-1

I. (c) 6. (d) II. (d) 16. (a) 21. (c) 26. (d) 31. (c) 36. (c) 41. (b) 46. (b)

2. (b) 7. (a) 12. (c) 17. (d) 22. (d) 27. (a) 32. (b) 37. (d) 42. (c) 47. (c)

3. (c) 8. (c) 13. (c) 18. (d) 23. (d) 28. (b) 33. (d) 38. (c) 43. (b) 48. (b)

4. (b) 9. (c) 14. (c) 19. (b) 24. (b) 29. (a) 34. (d) 39. (c) 44. (a)

5. (a) 10. (b) 15. (d) 20. (b) 25. (b) 30. (a) 35. (d) 40. (b) 45. (a)

2. (c) 7. (b) 12. (c) 17. (b) 22. (d) 27. (a) 32. (b) 37. (b)

3. (c) 8. (c) 13. (b) 18. (a) 23. (a) 28. (a) 33. (b) 38. (a)

4. (b) 9. (b) 14. (c) 19. (a) 24. (c) 29. (a) 34. (c) 39. (c)

5. (b) 10. (c) IS. (b) 20. (d) 25. (c) 30. (b) 35. (c)

4. (c,d) 9. (a,b,c,d) 14. (a,b) 19. (a,b,c) 24. (a,b,d) 29. (.,b,c,d)

5,'(b,c,d) 10. (a,b,c,d) 15. (a,c,d) 20. (a,b) 25. (b) 30. (b,c)

Level-2

I. (d) 6. (a) II. (a) 16. (c) 21 . (a) 26. (c) 31. (a) 36. (b)

(B) More Then One Cholcels 2. (a,b,c,d) I. (a.b,c,d) 7. (a,b,c) 6. (a,b,c,d) 12. (a.b) II . (b.d) 17. (a,b,d) 16. (b,d) 22. (a,b,c,d) 21 . (c,d) 27. (b,d) 26. (b,c,d) 32. (b,d) 31. (c,d)

lsi.,.. Correc:1 3. (a,b,c,d) 8. (a,b,c) 13. (a,b,c) 18. (a,b,c,d) 23. (b,c,d) 28. (a,b)

Match the Columns I. a- r ; b-q ; c-s ; d- p

3. a-p,s ; b-q,r ; c-q,r ; d-p,s S. a-p,r ; b-q,s,t ; c-p,r,t ; d-q,s

2. a-q ; b - r; c - p; d- s 4. a-q ; b-r ; c-p ; d-s

,

,

142

GRB Undentanding Physics Optics and Modem Ph)'lics

Comprehensions Palsagf'..l : I. (d) 2. (b) a P ssage-2 : I . (b) 2. (a) Passage-3 : I . (a) 2. (a) Pass.ge-4 : I . (a) 2. (c) Passage.5 : I. (a) Passage-6 : I. (d)

3. (b) 3. (c) 3. (a) 3. (d)

2. (c)

3. (b)

2. (b)

3. (a)

2. (d)

3. (b.c)

2. (d)

3. (b)

4. (b)

5. (a)

4. (d)

5. (a)

6. (b) Passage-7 :

I. (d) Passage.8 : I. (c) 6. (c)

Subject/ve Problem. L.vel-l 1. 2.5 DS

2. -I 4 3.

g - sin-I(~) radian

4. 15.73x l O- 12

S. 4 x 10 14 Hz

6. n = n,n2 nl +n2

7. 1500 .....

.I

A

..

8. u j = -3i - 2j + k 9.

Y= ~andh = ~

10. hi = H ' h' = ~h 4' 4 II . (a) 30 em (real)

(b) - 10 em (virtual)

12. (a) - 10 em (virtual)

(b) !

2

,

143

Geomelrical (Ray) Optia.

13.

~ m. infront of the mirror (real image)

14' . , 1 = 510

15. D =

-1(.J3 -I) -2-

-J)

~+sin-l( .J32

16.
(bJD~ =~+~n-l<.J2~nWJ

17. (a) imin = 30°

(bJ 8 = 8""" = 60·

18. n = cosec A

2

19. (a) Then, if n = cosec A, or the ray will graze the surface. (b) Then, ifn rel="nofollow"> eosec A, the ray will be reflected. (c) Then, if n< cosec A, the ray will be reflected.

20. h = ~ m below water surface 21.
(bJ 0.2 m

=~_sin-l( ~)

23. Dm =2sin-l~-~ . 4 3 24. n = 1.5 25. q = - 8.57 em, the image is virtual. 26. q = -48 cm, the image is virtual. 27. q = 15.4 em, the image is real.

28. q =+30 em, the image is real. 29. q = -40 em, the image is point size fonned at a di~t8.nce of 40 em'in virtual space.

'.

'.

30. q = + J 80 em, the image is real. 31. q = -15.78 cm, the image will be virtual. 32. q = -180 em, the image is virtual. 33, q = -11.74 em, the image is virtual. \ 34. q

=-40 cm. Hence, the virtual image will be fanned.

35. q = - 40 cm, the image will be virtual. 3 36. q = - 23.63 cm, virtual image. 37. p = +450 em

144

CRB Undentandins Physics Optics and Mod~m Ph)"ics

38. q

-90 em, the image is vinual.

=

39. ! = 24em

40. Pwater = ~ 0 41. n = 1.5

42.»1=1 2 43. (a) 10 em

(b)30 em

44. q

=

+5 (real image)

45. q

=

30 em (real image)

46. (a) 9.16 D (b)! = 10.909 em 47. (a) mla! = -5

48. (a) f

49. q

=

(b) mlinear = 25

-20 em (b) q

=00. The image will be ghostly and fonned at infmity.

-IDem (virtual image)

=

Subjective Problem. Leve/·2

1. q=26cm

2. h'

nhcos19

=

(n 2 -sin2 8)3f2

A .

~

.

3. u 1 = -47i -2j - Sk mls 4. (a) q = +55 em, the image is real and point

(b) 27.2 %

~. q = R 2

6. (a) q'

11 40

=

m,.. = 12IT

(c) mlong = - -96 121 -80 em from the 2nd surface in left.

7. (a) q = -Ii (b)

8. q'

(b)m = -20 11

=150 em

9. q' =-2R 10. q'

=

83° em

11. (a) q = - 11. 74 em, the image is virtual. (b) mlal = - 1.043, the image is erect and magnification. (e) mlin ::: - 1.225 12. (a) q = +360 em (real)

(b) m =-3

13. (a)!=20em

(b) nl = 1.05

Geomelrica1 (Ray) Optics

145

(b)/=D'-d'

14. (a) D rel="nofollow"> 41

4D (e) ~ = [D+ JD' -4DI]'

(d) 110 = ..jll,II,

1112

(0 III)

'"

-4-DI"""'

~'::D"

- 1112 =

± -'--,-'--

I

IS. (a) 1= -18 em

16. (a) q

D±~D2 -4Df

=

(b) q = 6.42 em (virtual)

(e)

IIIla! =

9 14

-60 em, the imtlgc is virtual.

(b) mlal = -2 (e) mlineaT = 4 (b) m = 1.5 17. (a) q' = 7.5 cm from the lens. It is a real image. 18. The effective lens must be situated at a distance of6 em from L) towards left. 19. The final image (or focus of the parallel ray incident on LI ) will be situated at a

,.

distance of 130 em towards th~'nght of L). 20. q' = - 20 em

21. (a) I = 48 em

(b) x =24em

22. (a)P, =+10

(b)P, =+0.50

23. (a)p=I~O

(b) I = 90 em

(e) P = 1.9 0

24. 20m -'(~0)

9

(b) q' =: 3.33 em, the imtlge is real.

25. (a)1 =5em 26. (a) I =-72 em 27. (a) P = 3.5 0 (e)/j;" =4.16D 28.

(0/' = (n - n,)n, j,

(n-n,)n,

(b) I = 1440 em (b) I, =28.57emandj, = 34.3 em (d) I' =24em

(b/'=~l PI

n-nl

29. (a) 1= IO.l7 em

(b) 9.8 D

30. (a)q=-160em

(b)p=55 D 6

(d) 1mli=,

160

1= 8I

31. (a) 60° (b) 60° 32. 0.3 per second

(e)m=-12

146

GRB Undtntanding Phytics Optics and Modem Phyaia

33. (a) angle of emergence is zero (b) 12S nm 0 34. 1.6 35. _ 4°, _ 0.04 36. 15cm.~, A' lies 1.5 ern bclowRS.B'lies 0.3 em aboveRS

31. m = 1 3

38. 1,,(37 +4j -5';) 5,2

39. 11 em infront of silvered lens, real

40. 1.132 41. (a)Ao =600nm 42. 0.4 m from the lens 43. Focal length = 0.4 m, separation between two halves = 0.06 m.

44. (.):=COli

(b)Y=(~)'

(c)(4m,lm)

46 • .fi 47. The position of the image due to refraction at the first surface is 6 nun towards left of first surface. Final image formed at 5 mm 10 left of second surface. (b) total internal reflection will occur at AC but not on CB. 48. (a) J2 SO. 75 em

• •

147

Geometrical (Ray) Optics

1I[::::::::::::::~<:JHfii~nlts~A~n~d~s~o~'u~t~i2o~n~,*::::::::::::::11 Multiple Choice Questions

(A) Only One Choice ,. Correct Level-1

sini=~n2-lsinA-cosA or sin30o = ~(n2 -1)sin60D-cos60°

6.

n=~

or

11. Snell's law at 2, for grazing of the ray:

sinS=! n -+

16.

-+

or

n = cosec9

-+

-+

-+

-+

-+

-+

VI =Vim+Vm =-VOm+V", =-(Vo-Vm)+Vm -+ -+ ~ ~ ,.. = -Vo+2Vm = -2; +2{-i) = -4imls

./3 v /2

3. "

--

V

2

....~. t ", ""'3V12 I I I -+-=P q f

4.

Of

-dp/ dt _ dq I dt = 0

p'

q' ~

-tl

or

,m ~-(q) p

or

;:

,

om/p

~

2=V., imlq-

v"m Ifp~15 • L.L!~_L1. q f P 10 15

orq=30m.Then,q/p= 30/15=2·Hencevim =_(2)2 Vom

or

-+ -+ -+ 2-+-+ vi =V1m+Vm =-(2) Vom+Vm

-4[(i) 5.

(-i)l + (I) ~ -9(1) mI,

Sinceq~pordq~dP/n~5/(413)~ 15/4m1, n

dl

de

GRB Undentanding Physics Oplics and Modem Physia

148

..

6. The i'm~ge 1 will be IOca~ed at a height " = x6 = x(n -I)A = 10(\.5-1) 4n

180

I· ' .1 ComprehensIons

Pasaage-2:

1.

P =PI +P,

=(1.5-1\ -;0)XI00+(4I3-1\ +2~)XI00 ~-

10/3 D

1+1 =1.. p q II 1..+1 =1.

2.

30

q

20

q =60cm

1-.

Then, image II behaves as a virtual object for L2

,

Hence,

p' = -(q -d) = - (60- 10) = -50cm

or or Then,

I

1.+1. =_1 p' q' h -1..+1.=1. SO

q'

25

q' = 50/3

•

I

149

Geometrical (Ray) Optics 3. The power of the lenses in water can be given as PI = (1.5-4/3)-'RI

P, = -(1.5 -

and Since and

IS

(PZ)air =(1.5_1)-1-=_100

R,

~

We have

=

Rt

and

~~:::-:. ~:::~:~:::~

3 R, RI

R,

10

100 x2 15

-1- =

_100 x2 10

PI = (1.5-4/3)x I OOx2

Then,

IS

=!x2oo = 20 0 6 IS 9 P, = -(1.5-4/3)x loo x2=_10 0 and 10 3 Hence, the total power is P=Pt +Pz _dP,P2

n

=2~ _ I~-I~/N _13°) =60 90+10=_200 27 27 Passage-3

I.

n~

=ny

51875 = 1.523 +2 1.5145 -I - .

The dispersive power is

4» 1) y

0>=-=

(nv - nr )A (n" -I)A

=1523 -1.5143 = 0.0167 1.51875 -I Passage-4

I.

_1_ =(_1 +_1 )+_1_, wherefm

I",

II Ie

1m

=---m-""---=

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

~)l

= (1.5 _I)~ = 100

(PI )air

= -----"u------ =~---- = ==---~ =::-: ::-:-:. - =:-:~ ::: ::n:: : : :

= .,0

150

GRB Undentanding Phylia Optica and Modem PhYla

or or

f .... =112' Hence ~ ..!. = I

I,

2.

x·

3.

_1-

I",

or or or

=

f ~ =II

x'

or

h

2

2

~2(_1 +1.)+_1 f' h

;.=2(;.+

1m

J}: 1m =00)

_I =1._1.=1._1.=_1. x' II 20 10 20

f'

/'= -20cm _I = _("

f'

_1)(1) R

or

-210=_("-1)(2~13)

or

"-I

or

=!

=!

3

n = 4/3

Pasaage.a

"-"2 p= "1-" + "2-"1 + --

5.

R

=

-R

R

"1 -"-"2 +"1 +"-"2

R

= 2n\ - 2"2

=2"(,--",..,,-;:---,"'0.0)

R

R

= 2(1.5 - 413) x 100 0 20 =5130 !!=p

6.

I or

" 312 9 I=-=-=-m P 513 10

Paaaage-8

4.

16 m,,, = - "",, qp = - 413(12) 312(10) = -15

•

• ~metrical

151

(Ray) Optics

1",I=I-16115x "01 = +32115

II,

s.

(.:'" =2cm)

mlin =--X m1 2 "1 al

= _ 3/2 x l_161' 4/3 · 15 = _3 )(3)( 16 x I6 4 )( 2 15 )( 15

= _32

25

Subjective Problems Leve'·1 ' kent=-=I I I. TImeta u cln

="' = c

-1-

(1.5)(1) 2 3 x to 8

=0.25 x 10-8 =2.S Am um l f um I -= = - =Aa e lf C n I.. -m= -1= -3 4/3 4

2.

'u

or or

ns

::'!!,u,..-::'m!!!. = 1

'u

or

4

x=!

4 3. The angle of deviation is 6 = I-r

N

= ~ _ sin n Stn . =6-

-I( Si~;30o)

-.(3)8 d·

ra Ian

4. The pennittivity t

= COEr =
=8.85xlO-

12

x(~)'

= 15.73 X 10- 12

,

GRB Undentanding Physics Optics and Modem Physics

152 5. The frequency of light is

f ='C = (ct.,) I-

I-

8 = .£.. = 3)(10 " I- (J / 2)(5000x 10- 1°) =4)(10 14 Hz 6. The light taken a time to pass through the composite slab is 11 +12

t =

=11+12

C / II] + C1n2 I _I = _I +_ n nl 112

or

C / I!

or

7. The deviation is

D=30+120 = 1500 as described in the figure.

"' I

60.

", ,

:

,, "

, ,

'",j~D ,,

8. Since,

and we can get

Vi Vi

=Vim + vm

= -vOm ' V, =- Vo m.. + VOx + VO:

"

=-37 -2j +k 9.

y=H andh = H 2 2

153

GcomC'lrical (Ray) OpOet

'0.

Comparing the similar!i.s PQR and PST, hi x' or - = -H x'+x x' xl 3 H or hl = - - H = - · H = x'+x !+x 4 )

h -h'x =-x+x'

Furthennore,

or

h' "" --lE!...-

x+x'

=..!1:!..- = "J. h x+!

4

)

ll. (a) The focal length J=R /2=20 /2=IOcm 1+1=1or p

or

(n)

1+!=1. 15 q

10

q = lO x IS =30em(real} 15 -10

!+1 = 1.

(b)

'2.

f

q

5 q

10

orq = -lOcm(virtual)

1+1=1 p q

J

1.+1 =_1_

20 q

-20

q = -IO(virtual) (b)

The magnification

= m "" _;

=_( ~~O) =!

13. For virtual object p "" -20

Then

1=1_!=1. __'_ q

or

J p 15 (- 20)

q ""..§Q.. = 60 em infront of the mirror(real image) 4+3 7

154

GRB Undentanding Ph)'lics Optics and Modem Physics

i=sin- I (Jn 2

14.

-I

sinA-cosA)

sin -1 [(..J2"=I sin 60°) - cos 60 0 J

=

=gn-'(~ -~)

. -,(,/3-1)

= SIn

-

2

D=(i-~)+(90'-'2)

15.

=90

0

+i - A (0: "I + r:!:

=

A)

=~+Sin-'( .J32-1)_~ D=~+Sin-'(.Jj2-1) 16. The emergent angle

sin

e = -'[nsin(A -9, )) = sin

-'[~sin(72'-42')l

=sin- i ~ 4 (a)

Then, the maximum deviation is , 12 +sm ' - ,3---=-+sm 2. n . - , 3urrax=1t

4

si{60';30')

or

n=

5

10

-'--,"'='''' Sin(~')

= 1/,fi =,fi 1/2

(b)

Drrru

=9Oo+e -

A

=90 o+e_60°

30o+e where e =sin -I nsin(A -eel =

'{6O'-sm. -, J2I)

• -,
=sin - , (Jisin IS') Then,

4

Drru =~+sin-l(.JisinISO)

,

<4ometrica.l (Ray) Optics

155

17. (a) Snell's law atsurface I:

'"

"

1 '

sin; = nsinr r +9 c = A Snell's Jaw at surface 2:

. ac =-1 sm or

imin

",.,-

... (i) .. .(ii) ... (iii)

=sin - J (J,,2 -I sinA-cosA) = sin - 1[./713 -I sin600-co,600] = ,in -'(

3J ·1- 0

=30° (b) 6 = 6 rmx = imi n +90'"-A = 30°+90°-60° =60 0 18. For any angle of incidence the ray will not emerge. It means tha~ at i = 90°. ifthe ray will graze for minimum "n", then for any other values of; < 90°. the ray will be reflected back at the 2nd surface. Hence.r + r ' =9 c +9 c =A

9c = A 2

or Since

n =--1-. sin9 c

we have.

n=-Isin A

2

n = cosec A 2

or

19. For gracing emergence e = 90 0

:"",,, 'n"A~ = 1 sin 900 n n = cosec A Then, if n = cosec A or the ray will graze the

Then,

or (a)

surfaces.

156

CRB Understanding Physics Optics and Modem Physia

(b)

Then, if n > cosec A. the ray will be reflected

(e)

Then, if n < cosec A, the ray will be refracted.

"

Real depth == n Actual depth

20.

Ih i :

2m', , .-

~'

!!. =4/3

or

,

2

or

:

h = 813 m below water surface- _L_ _-'l'

21. (a) The apparent depth is

• h'

=!.!l+ h2 III

).12

= 0.4 +0.3 4 / 3 I.S = 0.3 +0.2 =O.5 m (b) The apparent shift 0.4 + 0.3 - 0.5 = O,2 m 22. The deviation is 5 == i - r

=

..

= nJ4-r

where r is give as r

or

Then, 23.

or

• or

•

~- -

1

I.5XSin1t/6=Sin(ff/ 3+Dm)

2

,

157

Geometrical (Ray) Optics

or

D m=

24. Put llj

=~.1l2

3 "] [2' -14-3" Sin

=1I,p=20,q = - 25.7.R=+20cm N,

............

C;,?,.-;-,-

--- --• III

or

+ '_'2 = 1I::"'-o-:.n:11

p 4/ 3

n

q

R

n-M3

or +-2-0 + =25.7 = :':(-'+2='0::=)

n::::: I.S or I PutnJ =1.5,112 =1.2,p=lOcm.R =-30cm in the expansion 25.

/

J,.., -c ..... o '

n1=1.5

LR nl+~=n2 -nl

or or

or

p q R !2 + 1.2 = 1.2 -1.5 to q -30 1.2=_1__ 1. q 100 20 q

= -8.57 em

The image is virtual.

q

----

r

"

..

n2=1.2

.'

to have

..", '., ;;

\:If' , ' •.1

GRB Undentanding Physics Optics and Modem Physics

158 nl " 2 n2 -nl - +-=

26.

or or

, ,,

R P q 1.5 +1 = 1- 1.5 40 q - 30 -3 -I -I =-+ q 80 60

,

;~

,

, ..

°2- 1

'

UqCr-

A

or

1 = (- 180+ 80) q 60x80

or

q = --48cm

The image is virtual. 27. For virtual object p = -IS

nl

---- ---.

n2 _ n2 -nl

- +--

or or

",-

q

R 4 / 3 + 1.5 = 1.5 - 4/3 p

-IS

+20

q

P2J

1.5 = 0.5 +..i

q

60

or

~1 .5

45

q = IS .4 cm

•

The image is real. ", + _n2 = ", " ,,,,-,.:",,,, 28. p q R

C , • • ~ .....

I-A /

.'

~--- __

0

,~

Put p = -SOcm for virtual object. "I = 1.5. "2 = 1.2 and R = -30cm in the above fonnula to obtain, 1.5 + 1.2 = 1.2 -1.5 -SO

or,

q

q

= +30cm

The image is real.

- 30

159

Geometrical (Ray) Optics 29. Putting n] = 1.5, n2 = I, P = 00, R = 20cm in the fonnuln

--- -- ---

.. _........ n~.1 .5

I--."Ocm--I

n] +n2 =n2 -nl wehave p q R'

1.5+1=1-1.5 00 q 20

or

q = -40cm

The image is point size fonned at a distance of 40 em in virtual space. 30. Put nl = 4/3, P =00, n2 = 1.5 and R =-20 cm in !he fonnula,

..........

~ -=~ "

nl + n2 = n2 - nl to obtain p q R

C

q-+l8Ocm

---I

-

4/3 1.5 1.5-4/3 -+-= 00 q +20

or

q = +180

The image is real.

., "

.'

, 160

GRB Understanding Physics Optics and Modem Physia

31. Putting "I = 1.2,"2 = LS, P = + 15 em and R = 20 em in the fonnula

.

__ N

"" ' ..,.1.5

c

", n2= "2 -Ill ,we have - +p q R

1.2 + 1.5 = 1.5 -1.2 '5 q -20 1.5 _ 1.2 0.3 or -----q '5 20 or q = -15 .78cm The image will be virtual. 32. Putting ", = I, "2 = 1.5, P = 30cm and R = +20cm in eqn. N

I __ .. --- ~ -

,

--- --- --o

3Oan

-p03Oan

.

10-- q.l8Ocm -'--_I "I "2 "2- 11 , -+ -= ,wehave

p

q

R

_,_+1.5=1.5-, +30 q +20

or

1.5 q

'

.,

..

II. _ "',,

=_, _1.

40 30

or q=-180cm The image is vinual. 33. Putting", =4/3.", =1.5. p =+10 em and R = +30 em in the eqn.

"

"2"' ,5

"I +"2 ="2 -nl,wehave p q R 4_ /3 + '_ ,5 = :::1.5",-",4:.:/3 10 q +30

1Ocm11 .7-4cm _

30cm

•,

161

Geomdrical (Ray) Optics

!.1=_23 q 180

or

or q=-11.74cm The image is virtual.

34.

"I

= 1.5, "2 = I, R = -20 em, p = +30 em. Putting the above values in the ,~ , ,

---q=4

fonnula.

II, " " -" , we have -+-= p q R U + !=1-1.5 +30 q -20 or q=-40cm Hence, virtual image will be fonned. 35. Putting nl = 1.5, "2 = I, P = +30 em, R = -20 em in the eqn.

o =401

"I

p

or

+ "2 = "2 -Ill we have q R'

1.5+!=1-1.5 30 q +20 _= __ 1 _1.5 q 40 30

q=-40/3cm or Image will be virtual.

CRB Undentanding Physics Optics and Modem Physics

162 111

36.

p

+n _, = ,":;.'..,-"',,,'1

R

q

.!1. + \.J = \.J -1.5 +20

- 10

q

\.J = +0.2_1. q 10 40

or

q = - 23 .63 em

or

Yirtuallmage ~ + _"2 = ~n',,-=-..:":!.I p q R

37. III = 1.5, "2

= 1.2, R =

+30, q

=- 90

for virtual image

15 + 1.2 =1.2-1.5 p -90 +30 p:;:; +450cm.

38. Putting",

= 1., n2

=

1.5, R

= - 30 and p =

o

IX)

"I ", n., -+-=~

p

q

-R=-nf / -", we have

R'

l+!d =15-1 '"

-30 q =-?Oem

q

Image is virtual. 39.

1 =(~ - 1)(_1 __ I ) f"m

or

R,

",/'

in the formula

-___ q.ooan

n"~,"'7'

R2

~=ei5_1)(;0- _~O)

n• ........... - C

"-

•,

163

Geometrical (Ray) O plics

l=! x 50 2 20 x30 J =24cm

or

J

40. For a convex lens, the optical power is

"m

",

"m

,

.. . (i) . .. (ii)

41.

n = 1.5

or I -20I +-q1 =-20 -

42. or

q= - IO

Then,

m= -; = {~~OH

43. (a) or or or

I I I -+-=P q f

q' +l p

=!!.. J

q = (-m+I)J

=H+')20crn q = lOem

(b) Putting m =

-~,we can set q = 30cm.

• GRB Undentanding Physics Optics and Modem Physics

164

44. p = - 10 em (virtual object). Then applying,

1 +1 =1 p

f'

q

wchavc

_1_ +1 =~ q

-10

or

!O q = +5 (real image)

_1- = _I +_1

45.

II h

loq

=~+.l 20

! +1 =

... (i) ... (ii)

_1_

I"J

q

P

JO

Putting PI = +20 em and using eqns.{i) and (ii), we have

1.+1=1.+1.

20q20JO

or

q =30cm

real image 46. (a) The power of the lens is nl-I

I - n2

+RI

- R2

p=-- + - -

=(1.5 -1 + 1- 1.5)cm-1 +12 -to

( 1 I) -I

= 24+20 em

,

=

44 xlOOD 24x20 ~9.16 D

~ ~(2~ + 210)cm- 1

(b)

f

or

47. (a) or or

= 10.909 em

1+1~1 p

q

I

_1_ + 1=~ +30 q 25

q =30x25=150cm 5

"

,

165

Geomdncal (Ray) Oplics

Then,

mlat = -q = - 150 =-5

P

30

2

(b)

"'Iin

= !L =151 2 = 25 p

~ = -(n -I) ~ = -(1.5 -1)(;0)

48. (a)

f

0'

= -20cm

(b) p = -20 (vinual object) Then putting p = -20./ = - 20, we have _I_ + !=_I_ -20 q - 20 q = co. The image will be ghostly and fonned at infinity. 49. Putting p = +20,/ = -20, we have

-1-+1 =-1-

+20

-20 q = -IO(virtual image)

0'

q

Subjective Problems Leve/-2 I. The apparent shift of the object Oofthe new position 0' due to glass is

given by

-- -

---14cm

OO·=a.-=t(I-~) =3(1- 3: 2) = I cm

The point O' serves as a real object for the mirror to produce the image 0". whose distance q is given as

•

eRB Understanding Phy!ics Optics and Modem Physics

166

LLl f

q'

p

I

=

I

(100/3) 20

N.q =25cm Due to the presence of the glass slab, the final image I will be formed at a distance 0"/

=

t:u:

=

t( 1-~)

from A'

=3(1-2~3) = lcm. Hence, the final image distance is q =q , + O·1=25+1 =26cm 2. In the figure,

,

sinda

=

.,

'.

MN

OM

.::::MPcosa hseca 2

or

do.:::: MPcos . 'i: h a (". Slnoo.

Similarly.

de

=

MPcos h'

2

'i:

... (i)

=00.)

e

...(ii)

Using eqns.(l) and (ii),

or

M

h' = hcos 2 9 do. cos 1 a de

o .. ,(iii)

,

167

Geometrical (Ray) Optics

Snell's law ofrcfraction: .. .(iv)

n = 51na = 51nO

or

cos a

~,,2 -sin 0 = -"''--''-''!!.....!! 2

.. .(v)

n

Differentiating cqn. (iv) ncos a da = cos Oda da = cosO da Ilcos a Substituting :

... (vi)

from eqn. (vi) in eqn.(iv} and putting

~n2 - sin 2 0 = -'-"'---""-"-

cos a

"

From eqn. (v), we have

h'

3

nhcos 9

=

(n2 _ sin 29)3/2

3. Mirror fonnula, )

or

)

)

_

- +- = P q f )dp )dq - _ ___ _ =0 p2 dt q2 dl

2mfs

_3m!, o·

p--

~

or or

vo I ~ -~=- v 'm' p q

, ,

~ =-

V·

Imx

()' ~p

~"'x

Vo

or or ...(i)

or

Y""":::--7

, '.

,

168

GRB Undentanding Ph),!ics Opties and Modem Physics .. ,(ii)

1 = 60=3

Then

P 20

Putting 9..

p

=3 from eqn.(ii) in eqn.(i), -+

v

f;L ..

-47;° mls

=

= _1

Yi

Since

{2 + 5(3) 2Ji

Yo p dYi q dYo -= --dt pdt It means that, ~

and

.

q

~

V I",. = - - V Om . • p •

Hence

~, ~ lily" V m

y -

q(~lIO}. - ~limy

p

)

Ij - 3(2j - Ij) = -2j mls

=

· 1(-; pl '•. - -: . · . . = - 2k - 3[ -k - (-3k)]

= -:", _ _

111 0 )

• = -Sk mI,

Then . the velocity of the image is -)

-+

-!o

-+

Vj=Vi.. = V1y+v i :

=- 471 -2j-Skml' 4. (a) Put nl = l.S,n2 =l.l,p=ooandR =-20cmintheformula n".1 .5

get

1.5 +!1 = 1.\-1.5 <Xl

or

q q

=

-20 +55cm

The image is real and point (b)Putn2 = Ito obtain

I-R

_. t--q.55cm - \

"

Geomelrical (Ray) Optics

169 1.5 + ~ 00 q'

= 1.0-1.5 -20

or q' = 40 em The % change in the focal point is

Iq';qx I00= 1405~5~ x I00 =11. 100 55

= 300 =27.2 % II 5. Refraction at surface I' /'

,/

"'-2

K----12

-I

\

'-....

q' q

./

-

---- ___

p'

n, + _n2 = :n.,2_-.:.n", p

q

R

!+!! = n-I IX) q +R q = nR for image I, fJ -I .

or

The image II serves as a virtual object for surface 2. Then, put p' = -(q -2R) = {nR n-I

-2RJ = {n-2n+2 RJ n-I

=_( 2-n )., n-I/'

For refraction at surface 2: nl

or

or

n2

n2 -n,

-+-= p 9 R n 1 I-n _(2-n). = g' =~ n-I!'

L g'

_n-_1 + :,:n(l::n~-.,:.If)

R

(2-n)R

"

I

, 170

GRB Understanding Physia O plia and Modem Ph)'lia

= (n-I)(I + ~ ) R 2-n I 2(,,-1) q'=(2-n)R

or

q' = (2 - n) R = 2 -1.5 R = R

or

,,(n - I)

2( 1.5

I)

2

6. (a) For the reflection in the mirror R 20 = lOem P = 15, f = -2 =2

I I -I =-+q f p

Then,

1

or

s

= .L.L 10 15 or q =30cm The real image II will be fonned which behaves

s

as a virtual object for the refracting surface I . For the refracting surface I, the

virtual

p'-I

object

distance is nl

= 1.

n2 = loS,

1---

R' = 50 em p' = -(30-20) = -10 em "2 ' -nl, +, P q

_1_ + 1.5 = 1.5-1 - 10 q' 50

or (b)

The lateral magnification between last two images is

, '"Iat = -~!!.!

p n2

=_ (150111) 1 (-10) (1.5) =10

'-

11

q' - - I

, ,

,

171

Geometrical (Ray) Optics

(el

The lateral magnification between 1st two images is ml.al

=_!1. p

= _30

15 = -2 (d)

Then the lolal magnification is

m == mbl ·nilat =

(-2)cn

=_20 II 7.

(a)~+n2

p

=n2 -nl q R

or or (b)

(e)

Ray diagram

..

t--'-~~'~'''«J--

__ N

--",-

./

C ......

121

8. Forsurface I,nl =l,n2 = 1.5,p=-20,R =-20 nl+n2=n2- nl

p

or

q

_I +1.5=1.5-1 20 q +20

R

-

-I

'I

n..=1 ..,

, 172

GRB Undentanding Physics Optics and Modem Physics

or q =-60cm For the surfncc 2, J1 acts as a real " 'L' - r ~-~----object. Thcn put ~ p' = + 100 em, _______ '2-.-.-.-.-.-.-.-."2 ........ C "1 '

#--

n;

i 1:2Dc q=6Ocm

L ni

="2

=

,

T-------.

2Ocm_-2Ocm

q

1.5"2 = I.2,R ' = co in the formula, Iii "2 -_ IIi - ni - +p'

R'

q'

1.5 + 1.2 = 1.2 L5 100 q' co

or

q' = -1.2 x 100 = -80 ern

L5

.

Form the 2nd surface in left. 9. Refraction at surface 1:

__ r---: 1.5

o A

J..--C'--==\o,.:;:£... q' -1--1 _A p'

tIl +"2 ="2 -nl

p

q

R .!.+1.5 = 1.5-1

2R or

R

q

1. = 0

3q

q = +<xl For the reflection at surface 2:

or

l+Ll. p

q

f

•

"2 '=1.2

, •

173

Geometrical (Ray) 'optics

or q = - R12 for 12

or

12 behaves as a real objcct for surface I. For refraction at surface I,

"I = 1.5, 112 = L p ' = +3R. R = -R 2 _III + _" 2 = _n=-,-;-,.."",

p'

or

R

q'

1.5 +~ = 1-1.5 = _1 3R q' -R 2R

2

or

L_I _1

or

q' = -2R

q'

2R

R

10. For the refraction at 1st surface:

", + _/12 = ~",-'-;;-:,""" p

R

q

_I +1.5 = 1.5-1 40 q 10 q = 60 cm

For the refraction at the second surface:

J.-p-

r-1Ocm-

-'o:n

\'-----' or 11. (a) Put III =4/3,112 =1.5,p=lOcm,R=30cminthefonnula, N.........

111:413

n~I .5

,-::::-___ ............ _ i /

' ,,

--:t--:-----~'"~ -- .

, ~ I

~ •

......

J.-P~

...

---

-

, "

GRB Understanding Physics Optics and Modem Physics

174

"\ + _"2 p q or

"

= "":.'..,-,..:",,-, R

4/3 + ~~1.5-4 / 3 10 q +30

or

q = -11 .74 em

The image is virtual (b) The lateral magnification q "1 mlal = - - -

P",

~ _( -11.74 )( 4 /3) I.S

+10

~24/23 ~

+1.043 The image is irrect and magnification. (e) Linear magnification

' '"

m lin= - [ mlal [ -

'"

~ J24)'(~)~ -1.225 \23

12, (a)

4/3

nl +n2 = 112 -II, "

o~

or or

P

R

q

~ + 3_1_2 ~ ~3/-,2,,-:,,1 80 q +30

3

1 ·1 60 80

-~---

2q q

~

+360em(real)

(b) The magnification is

I

m=

_1!!l p"

~ _36O x _l_

80

~' -3

(e) Ray Diagram:

3/2

, "2c1 .5

q

•

, 175

Geometrical (Ray) Optics

13. (a) For virtual object p

= -10 em, q = +20 em 3

!=!+! =_I_+

Then,

f

P q

(-10)

1 (+23°)

f =20em

or

(b) or or "

(e) Ray diagram:

-- --- --

.......... 0

14. (a) Let the initially the lens be at a distance x from the lens. Then, using Anal position . I

Inlllal(

JIO'' ' 'Y' I

Image2

1..

•

LlJ.-2--

·2

d

I

-.JJ_--,1o

Image1

1 1_ _ -1 ,we have -+p q f

!+~=l x D -x f or

x 2 -Dx+DJ =0

•

,',

GRB Undentandin8 Physics Oplics and Modem Physics

176 For real

rOOls

JD' -4Df >0 D> 4/

or

(b)

Solving the above quadratic cqn., D ± J;"DC"-_C"":4D::-: >j x = --'--,;---'2 For two positions Xl and X2 of the object we have two distinct images. Hence, xI

= (D + JD' - 4Df) /2

X, = (D-JD' -4Df)/2

and

... (i) . . .(ii)

Xl ::::: xandx2 =x+ d, we have

Then puuingx, - X2 = (d + x) - x = d Substituting the values of XI and X2 from eqns. (i) and (ii) in eqn.(iii)

We have,

d = JD2 - 4Df

or

D2 _d 2 f= 4D

.. ,(iii)

(e) For two distinct images on the same screen, the lens fannula 1+1 =1 is valid. Hence, when P1 = X, q, = D - x and when p q f P2 = D - x, q2 = x. Then the product of magnification of the images is MI m2

=.!!.ll PJP2

D-x ._x =-_ x

(d)

Pu· hI an d m2 = Ito h, In . mlm2 ttmg mJ = Ito

, (e)

D-x

I/o = Jhlh,

The ratio of magnification is ml =ql lpl m2

q2 / P2

=!l.l P2 q2 PI

=(D;X}(D;X)

have =.,' we

Grom~tricaI (Ray) Optics

177

=(D;X)' =[D'fJD' -4DJ]' D±JD' -4DJ nil _ m2 = D_ -_x __x_ = =2D=(D=,.,-.;.x"),,-.;.D=-'

(I)

x

D-x

x(D-x)

=±JD' -4DJ J 15. (a) For the lens,

c, Putting "2

=

"I

4 /3.

= I.

RJ

= -10

em,

R2 =+15em, we have

I-

", , R, - - /

<-->

~ =( 4:3 -{:o -+:5) J (b)

=

-18cm

The lens fonnula:

1+1=1 p

q

J

Puttingp= +IOem,! =-18em, we have I I = I- =6.42 em _+_ +10

q

(d)

-18

45

or (e)mlat

q

=-p=

q=-7 -(-4517) _ 9 +10 -j4

Ray diagram

I -

P-"",

178

GRB Undentandins Physics Optics and Modem Physics

16. (i)(a) For virtual objctt p = - 30 em; for concave (n2 > "I), f = -20cm. Using

lens formula, we have

!+L! p q I _1_ +1. = _1_ (-30) q -20

or or

q = -60cm

The image is virtual. (b) The lateral magnification is mlat

=_1

p = _ (-60)

"

-30 (e)

=-2 The linear magnification is mlin = (ml at )2

, =(_2)' =4 (ii) Ray diagram

,'

.,

.-.:. ..

.

.. ................ 0

~.

"

-p-I

s.

17. (a) The focal length oflhe lens is given as

~ = (312-1)(~+~)

or

It It = 30cm

30 30

For the mirror. the position of the -~ image JI can be given as !+! =_1 p q 1m

~+!=l 15 q 10 or

!=L~ 10 15

q

~~k-'-:",

•

~a3Ocm

-1,"':--~

_2Ocm~'~_

179

G~metrical (Ray) Optics

or q=30cm The image I, behaves as a virtual object for the lens as its fonned in RHS of the lens. Then, the objcct distance p' = -JOcm. The corresponding imagc distance of the image 13 is given as .L=_L.L=L_I_ q' Ii p' 30 (-10)

or (b)

q' = 7.Scm from the lens. It is a realimage.

The total magnification is m=m, ><m2

=H)+~}(~~O)(- !i~)=1.5 18.

The equivalent foc:allength of the combination LI and £2 is

f'

=

flh fl + h-d =

(10)(10) 20 =-cm 10+10-5 3

scm '

The position of the equivalent lens of the combination LI +£2 is ,x= ,dh = (5)(~ = 10 em

fl + h -d

j

10+10-5

V

1

f_+.20.. 3

'3-- 10

3

The effective focal length of the combination L, + L2 + L3 is

,

f=, f'h

: '

f' +'13 -d = ~(2~0~13~)(,;:-~1O~)-,

.,

(20/3)+(-10)-5

,

.

.

.

=8cm The required position of the final effective lens is

x'=

dh

=

5(-10)

f' + h -d 20/3+(-10)-5

",

=6em

The effective lens must be situated at a distance of6 em from L3 towards left. 19. Refraction at L,: \.

1. + L 1.

i +L q {

co

10

q = 10em

LI

La

...

II

f..

\7

-==t~;~~~-:-~-~-~-~~-~-~~3~-:-::F

-

~5cm

p'

p"

VI ___ ~/c!:==~q~,:::::9::-j---J 110em

12

180

GRB Undentanding Physics Optics and Modem Phyaia The image II behaves as a vinual object for L3' Refraction at L2

p'=-Scmandh =-IOcm

-.L +-L= _' p'

or

_1 +1. -S q'

q'

J'

=_'_

- 10

q'= IOcm

or

The image 12 behaves as a v:~ual objcct for L) . Hence, p" = -5 em, '" = +10

1. + 1. p.

q"

=_,f"

_, +1. =1. -5 q"

or

'0

q" = 10 3

The final image (or focus as (he parallel ray incident on LI) will be situated at a

distance of 1OJ3 em towards the right of L). 20, Refraction due to Lt :

!+1=1 p

q

J

1.+1=1.

or

'5 q '0

°l

~---

'2

:;:=-"

0'-\

q=lO x I5=30cm 15cm_l-l _ _ I - - q - - I 5 Refraction due to L2: The image 1\ behaves as a virtual object for L2. Putting p' = -(30 -10) = -20 (-'Ie sign for virtual object) and f' = h = to em, in the formula

or

=_, p' q' f' _'_+1. =_, -20 q' J' 1.+1

or or

q' == 20cm

..

'

i.

•

Geom~trical

181

(Ray) Optics

21. (a) The focal length of the convex lens is given as ..!. = (1.5 -1)(..!.+..!.) II 20 30 or II =24cm The focal length of the concave lens is given as ),

---',--I

h = -24cm The effective focal length is given as 1:_1 +_1 _ _ d_ I II h nltf, = ..!.+_I__ 12 24 -24 (1)(24)(-24)

f = 48em

(b)

The distance x of the effective lens from L2 is x= df = 12x4H =24cm nil

(1)(+24)

22. (a) Power of a bi-convex lens I in a medium of RI = no is n-no PI

=

Pair

n -I

= I.S -1.6(_5D) I.S -I

= +1 D (b)

Power ofthe lens 2 in the medium

P, =-(n-no'[_1 +_1 ) \RI R, = -(1.5 -1.6' ..!. + ..!.)cm - I '\.30 60

= +OJ x 1- x 100 20

=0.5 D (c)

The total power is P = P, +P, _if.p,P, n =1+1_°.16(+1)(+1) 1.6 = 1.9 D

,,,

-i:

:__.!:I:d=:-I:2cm~,' _____

={~-I)L~+2~)

or

,,

GRB Understanding Physics Optics and Modem Physics

182

23. (a) The power of the system is 15-1 4 /3- 1.5 1-413 _I P=-'- -+ +---cm +45 -30 +60 1 = 90

+I~O +( -1~O)cm-1

P = IO D

9

"1::: n4- 1

"2=' .5 (b)

"3=413

The focal length of the system is /=n\(=11 4 ) :;;; 2. m P 10

=90cm Aliter: The focal length of the lens 1 is

_I = (1.5-1)(_1 +1.)=1.

II The focal length of the lens 2 is

)2

= -(4 /3

=

45 30

36

-1\310+ ~O)

_1.

60

The effective focal length oflhe combination is given as

L_I +_1 =1._1.=_1. or

11,12366090 j=90cm

24. The virtual image L2 of the lens LI due to the plane mirror serves at another lens. Hence. we have a system of two lenses LI and L2 having the intervening medium of R.1. no_ The distance between these lenses is d = 2/. The power of each lens is

n-t· lIo-n P1=--+ RI

Putting RI = R2

R2

= R, we have ::2n:..-...;n"'0'-.---'1 PI =-

R

Then, the power of the system of two lenses is

P

l,

,_

.. _- --

----

~ : :-:. : : :-::::~.

Yfi~h}r~;. .::-:.:.:::: :.:::-::.:

=P, +PI _ dPIPI no

=2n -

,

no -I + 2n - no -I _ ~2/,,(2::n'---...:'''''0'c-....:!.I)_ R R noR2

-<

183

Geometrical (Ray) Optics

Putting I

=

=2(2"-'10 R R, we have p

-ll[I_/(2"-"0 -Il] noR

= 2(2,,- no -1)[1 _ (2,,- no -1)] R

"0 P = 2(211-1l0 -1)(2110 - 2" + 1) noR

=

'2XL5 - (~)-1)(2X~-2XL5+I)

~x30 3 = _I em -I =20/9m- 1(=

45

Ol' . ",

25.- (a) The focal length of the lens without the mirror in airis given as

), =C,;-IVI+i') , =(\5_ I)UO+210) or, /J = 20 cm. The focal length of the mirror is

1m

= R =20 = IOcm

2 2 The ray gets reflectcd one by the mirror and passes twice through the lens. Hence, the focal length of the combination is given as

1=_1 +_1 +_1_ I

j,

j,

1m

j

One reflection

Two reflection

= ~+~+~

20 20 10

I

or

(b)

=5cm

The combination acts as a concave mirror offocallength '5 cm. Applying mirror fonnula,

1.+1.=! p' q' I _1_+1.=! -10 or

q'

5

1.=!+1. q'

5 10

184

GRB Undmtanding Physics Optics and Modem Physics

q' = 3.33 em The image is real. or

(c)

Ray diagram

~q. 26. (a) The focal length of the lens 1 (in air) is given as

1- =(1.5 - 1) ( L1-) II '" 30 or II = -60cm The focallcngth of the lens (2 in air) is given as

), =(j-I)UO- ;0) or

fz ::: +360cm

The focal length of the combination is given as 1= _1 +_ 1 I II h = -1--+ _1_ -60 360 ' or 1= -12cm (b)

The focal length ofthe system when immersed in a liquid ofR.1. = n = 1.6 is given as

···ro

1=; p=

1.5 -1.6 4/3 -1.5 (1.6 - 4/3) 1 _I + + =-cm '" +30 40 900

By using eqn,. (i) and (ii)

1=

1.6 = 1440cm -1/900

.. .

(ii)

• I

r, .1.

185

Geometrical (Ray) Optics

27. (a)The my of light passes through the refracting surfaces I and 2 once each. Hence, the power of the lens is equal to sum of the powers of the interfaces. l'jcos = PI + P2 = "2 -Ill +"3 -"2

Rl -R2 = 1.5 -, + 1.2 -1.5

_l!!.

+ 20 '00

'00 P = 3.5 D

or

~

= ") = PJcns II h

(b)

or

II and

I

(c)

= ~ = _I )(1 00 =28.57 em 1',=

3.5

12 = ..!!.L = 1.2 x 100 =34.3cm 1\=

3.5 Power of the lens in air is ?]~n5

= PI' + Pi ="2 -1+1 - 112 =("2

RJ

- R2

-1)(-'R, +_,R2 )

P,;'" = (1.5 -'(2~ + 3'0)cm-

1

=1. em - I 24

= IOO D 24

= 4.l6D

I

(d)

The focal length of the lens in air is given as nair _ n'

-

f'

-~

or

1. =_, f' 24

or

I' = 24cm

lens

186

GRB Undenlanding Ph)'lics Optics and Modem Physics

28. (a)Thc power of the lens in the medium is P = (1I - nll

(_1 +_1)

R, R, The focal length of the lens in the medium I is

_III = '"P =(.!!.", -I)(_IR +_1) ... (i) R2 J

",

The focal length of the lens in the second medium is

.L = (..'!. _1)(_1 + _I ) fz

... (ii)

R, R2 From cqns. (i) and (ii), "2

h = (n-1II)"2 II (11-112. )111 (b) The ratio of power is Pz= "2 1h = II "2 PI

111 / /1

h

"I

= (n - "2 )111 x"2 _ (n - "1 )n2 "I = n-1I2 n n, 29. (a) The focal length of L] and l..i]. are

), =(:: -IV,

+

i,)

=(!i~ -I )(2~ + 2~) or

_I = 1. 80

I,

",

), =(::-1)(;0 -310)=( !i~ -I )~~ or

_I =_

h

300

The focal length of the mirror is _I

1m

=l=1.

R; 30 Then the effective focal length is

"

",

187

Geometrical (Ray) Optics

1- = 2( -1 +_1 ) + _1 = 2(-.L +-1 )+1I I, h 1m 80 300 30 or

!=_I +_I_ + -.L I 40 150 15 I = 1O.17cm

(b)

p=1-=

I = 100 10.171100 10.17 = 9.80

I

30. (a) For surface I: "I "2 "2 -+-=

- Ill

.. . (i)

P ql RI J behaves as a virtual object for surface 2: n2 +") =n) - 112

-q , q -R2 Eqn. (i) + (ii) yields, ~ + ,_' ) = "2 - "I + e"",2_-..:'",') p q RI R2 1.2 4/ 3 1.5 - 1.2 1.5 - 413 or -+ = + 12 q 6 4

or

... (ii)

n

I-p

q = -160cm

(b) The optical power of the system is p=n2 -III +") -112

RI

R2

.

= ( 1.5 ;1.2 + 1.5 ~ 4/3 ) x 1000

= 55 0 6 (c) The magnification due to surface 1 is q, "I rnl = - - P "2 The magnification due to the surface 2 is q "2 M2 = - - ql n)

Then, the net (total) magnification is

..

m

= mlm2

-ql nl

q 112 (-qil")

=-- x ----

P"2

q

"

188

CRB UndtBlanding Phytia Optics and Modem Physics

po, =

(-I60)xU 12

4/3

= -12 (d)

"I

+ "2 = P

p

or

q

~(-lip)+~(-8q)=0 p2

q2

or

mlin

,

= ~ =( -;) =~

Imlinl (q / p)2(nl / n) n3 Im, .. I' =[(q l p)(n, 1n, )]' =-n, or

Imlinl =lmlatl2 n 3

or

Imli. 1=14/31'

n,

(4/3) 1.2

= 160 81 31. (a) For minimum deviation, the ray will enter the prism parallel to base. B

SO'

D

SO'

A~~------~c~L-------~E

Hence,

Now,

Lr = 30° J.l =

~n i

Sin

r

.Jj = sin; => ;=600 sin 30° (b) Rotation of DeE about 'C so that finally ray emerge out with minimum deviation. Very obvious from figure, the angle by which it must be rotated about C is 600.

189

Geometrical (Ray) Optics BA',-"O_ _ _ _ _-,E

A'-------¥C

32. Vo =0.01 mls.J =0.3m. Vi = ? when u = O.4m. when object is at distance O.4m , then

__ 1 =! v U f 1 =!+1 v f u fu f +u

v = --

Put u = -0.4,J = +0.3

,

Then, v ::::: 1.2 m 1 -1 - -1= v u f differentiating with respect to time, __ 1 . dv+_l . du=O v2 dt

u2 dl

ddlv=u2v2.[dU] dl

:. Velocity of image,

J'imw~

= rl.2]2

LO,4

x O.OJ ===> du = 0.09 mls dl

Also lateral magnification, dm=d(v l u)

dl

dl

dm dl

dv du u' --V dl dl u2

=

.

=-0.4 x 0.09-1.2 x 0.01

. dm = _ 0,3 per second dl

(0.4)2

.

, GRB Und~n landing Physics Optics and Modem Physics

190

•• 33. Angle of prism '= A = 30°

Angle of incidence = 60° ~p ~f

=-13; =2.2;, =550nm

(a)sin60o= .nsin'1 sin '1

=.J3 x _I ::::) "

-13

2

Also

--

•

= 300

1'\ + 'l=A

=0 :. Ray will enter the face AC nonnally. 'l

BL---~

:. Angle of emergence :co O. (b) The intensity of emergent ray will be maximum iftmnsmitted wave undergo constmctive interference. For minimum thickness, 6x = n).

put

.' I.

=l

t

= 550nm = 125nm 2x2.2

~f

34. When water is between the plane mirror and lens. The equivalent power

Peq = 2P4

+2PL~

R

2

R

=[~-II-~- ~]=-3~

PL,

I Pm =-=0.

'"

P
=~ -3~ =;[I-~] =;+3~ =F~

[ where Feq is focal length of equivalent concave mirror] Now,

-I +-I =-I v u F

_1. + _1_ = .±. 15 -15

or

3R

_L.±. 15

3R

R = _60 = -lOem 6

.

"

+ Pm

=[LIIL-I ]=1.R R -R

P4

'0r

I"~

, . ,....

191

Geomelrical (Ray) Optics

: . Radius of curvature oflcns = 10 cm Now when another liquid is poured having refractive index.(ll). Then, Peq =2PL +2PL2 + Pm

= ,rJ.]+2(~ _ pr_1 _1]+0

Peq

110 'L-R 00 2 2(~-1) I = - = for second case. 10 R F",

Now again objeet and image again coincide at 25 cm. !+! = _1- => _1_+_1_ =_1_ u u Feq -25 -25 -Fcq

1. = 1. _2(~ - I) 25

10

10

- ' = 1. _ j..t - I solving j..t = 1.6 25 10 R 35. (n) It is the case of dispersion without deviation, S = (~-I)(A) Thus, B, + B2 =0 or A1(~1 - I)+A,(~, -I) =0

A, = _A1(~I - I) (~, -I)

Now for crown glass, ~1 =1.51+1.49=1.50 '2

For flint glass,

~, =1.77+1.73 = 1)5

Forcrown glass, 6°[1.50-IJ=-40 TI.75 -IJ

2

'r .

'.,

• (b) Dispersion

(9)=91 +9, =(~,, -~,)A1 +(~~-~~)A, 9 = (I51-1.49)(6°) + (1.77 - 1.73)(-4°) 9 =0.12-0.16 e = _0.04 0 36. Refraction from lens, ., I I I --- =v u J

192

GRB Understandinc Physia Optics and Modem Physics

A B'

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

• _Ll~'2om O.6cm B

A'

"'om

A, 1-_1_ = 1. =>

-20 15

11

11

200m

= 60 em

This image out as virtual object for <;onc8ve mirror. Thus, for mirror I I I -+=-

v

U

f

,,

1 1 1 -+=-=> v=-IScm v +30 -30

Magnification produced by lens = 60 =-3 -20 ~ 'fiIcatlOn ' puce rod d y bnurror ' -15 1 an d magnl = +30 =-2 Net magnification =I Ml=s!xl M mirror i= ~

A'B' = 1.2x~=1.8cm Also A' B' will be inverted with respect to AD. Distance of A' from optic axis of mirror = 0.3 em above. Distance of D' from optic axis of mirror = 1.5 em below.

37. Refraction at plane surface, 3 / 2_-L =3 / 2-1 =0 1.1 -mR 00

V=H}R

rnA

Now this image will act as virtual objet! for curve ~urface. Thus, 1 3/ 2 =1-3 / 2

'"

Solving it,

{~mR +R] ....

R

m =4 / 3

38. Incident ray Al = 6Jj i + 8,/3 j + (10) _ k

This x- yplane is boundary separating the two media. The angle made by incident ray with z-axis is Li. Thus,

A

,

' 193

Geome.trical (Ray) Optia y

z

I z+ , ,

,, , ,, ----- -------- -------- ----.x

, cos ;=

y

' -_ _ _ x

=~=! JIO' + (r,J3)' + (8J3)' ./400 2 10

:. L; = 60" Applying Snell's law. J.llsini=J.l2 sinr

.J2 sin 60"= .J3sin r

~nr~!Jx1~ Jz Now, the incident ray, refracted ray and normal, all lie in the same plane, and therefore unit vector in the direction of refracted ray (r).

; ~ I~n

r,J31+8J3j

0

w bere x =

"x) - (Icosr)(k) =

J(r,J3)' + (sJj)'

31+4j

'---c""'"5

I I [31 +5 4j - koJ Hence,ro~ .Ji[x-kJ~.Ji 0

;~

I

0

0

0

0

r.;[3i+4j-5kJ 5,,2 39. Object distance u = -IOcm. Refractive index oflens Ut) = 1.S or

• • ., 1/

Radius of curvature oftbe silvered surface R2 =22 em We can get the equivalent mirror focal length. by using the power formula, PM = 2I\.ms +P", , I I PL=~-~Now,

J

and

Thus,

20 p ~ __ 2_~_L.!. m (-22) 22 II 2 I I I 21 ' PM = - +-=-+-=20 II 10 II 110

o

GRB Undentandins Physics Optics and Modem Physics

194

, ,

PM = 21 110

.

_ _ 1_

=11

F",

110

Using the lens formula,

!+_I_ v

-10

=_11

110

! = _11 +.1

v 110 10 -I = -21+11 = -10 v 110 110 tI = -llcrn Thus, image will be at II em and real. 40, 8 = U+e)-(A)

, •

30° = (60 o +e) - (30 0 ) e = O. Thus emergent ray is perpendicular to face A C. At faceAB. (l)sin60o= {J.l)sin r

A

~ = (Si~60') smr

Also,

So.

~

= (I -8) It = 30° ~ = ~n60' =Jj sin 30 D

50'

BL--------~C

41. (a) Ifno deviatioa occurs, then definitely J.1 I =1J2 Hence. 4 1.20+10.8 x I04 = 1.45-+ 1.80 x 10

A'o Solving, ~ence. .

A'0 , = 9xlO 40 AO I .25

A. 0

= 600 Dm

(b) Angle of incidence for minimum deviation. Ifwe extend the line, AC and BD we get an equilateral 6 (see figure).

195

Geometrical (Ray) Optics

And we know that for minimum deviation, aftcr the refraction at surface AC, the ray mUSl be parallel to base AB. Hence, Lr 30°.

=

Thus,

(I) sin i

=(lll )sin 30°

sini = [1 .20+ 1O.8 XI04] rsin 300] (600)'

or

sin i = [1.20 + 0.3])(! =

2

or

'1'"

..

1.2 2

sin i = 3/4 i =sin - 1[3 / 4J

or

..

42. It is an equiconvex lens,

..

= (~-IJ.L _ I] 'LRI R,

!

f

_I =(LI)x£R

0.3 2 R = 0.3 m

:::::)

O.9m

'.

••

O.8m

,"

.... .... .... . . . - .. , , , , , , , , ...... o

.'::-:: '!l';: .:: '::.::' ,

··2·::·::·::·:o :~m:·:

.1

Refraction from first surface of lens,

'"

~ 2 _ ~l = ~2 -~l

u u R 3/2_...L=3/2 - 1 u -0.9 0.3 u = 2.7 m Thus, fi"om first surface, image will be fonn at 2.7 metre, i,e., behind the plane

mirror. Refractio n from second surface of the lens, ~_~=J.l2 - ll1

u Hc re

+ 1.8 R ->-0.3 Ii ~

u

R •

i r

0196

GRB Understanding Physia Oplics and Modem Physics . III

,

~3 /2

1-' 2 =4/3

Thus,4/3_3/2 = 4 / 3-3 / 2 v +1.8 -OJ ::;. v = 1.2 m i.e., 0.4 metre behind plane mirror. Now, for plane mirror Ihis image will be virtual object Ilt distance (0.4 metre). Hence it will just reflect the rays at 0.4 metre left: from the plane mirror. Now, this 13 will behave like an object for 2nd s urface. Thus, for second surface [surface (2)] 3_1_2 _ _ 4 1_3 =,,-3,-, I 2";;-",4",1,,,3 v -0.4 0.3 1.+ 10=_1 =>1.=10_10=-25 2v 3 1.8 2v 18 3 9 ::;. v = - 0.54 Again refraction at surface I, ! _ 3 / 2 = 1-3 / 2=>!=_1 _~ v -0.54 0.3 v 0.6 1.08 tI = -0.9 m :. 10 em behind the plane mirror, image finally will be fonned . 43, For the first half lens LI , I (v,+d) and and

I

-(u, - d)

= -I "

, .. (i)

f

m=V2+ d =2 Uz -d U2+"'2

... {ii)

=1.8

... (iii)

Also for second lens L2, A f.;::---,__

L,

u,-
,,, ,, ,,

I •

"2

,, • I , ,'. ", a , "

•

1.8m

V2+d

,, ,, ., V2 ' .',,

••

•

Geom~trical

197

(Ray) Optics

_I __I_=!=>_I +_1 v2

-112

f

u2

Solving the four equations, d =0.6 m,f

=1.

u2

... (iv)

f

=O.4m

~ = (ky312 + 1J"2

44. and

k=1

Hence, ~ =

v

[i 12 + 1]1/2

At the point S(x, y), ;+9 =900

T. 1

or 8 = 90°-; Hence, Ian 9 = cot i

A(O, O)

The slope at B(x, y).

dx == tan a = cot( i) dy

Now, (I) sin 90 0 = ~ siner) . .

1

sm I == -

>. ~-COS i = ~1-1 / ~2 =1~1J2

coti=cos; = sin;

- 11I1l

Vi II! .=Vi I I Il

Thus. we get

dY=~=Jll2 +I-1

dx

dy 3/4 dx =y Now we can write it as

f y- Jl4 dy= f dx 4y1l4 =x+C

atx=O-+y=O Hencec=O Thus 4y1l4 = x or y = [x 14]4 is the equation of trajectory.

Now at y= I m-+ x == 4m : . Coordinate of exit point will be P(4, I). Now ~A sin(9O' ) = ~ p sin(ip) sin ip == 11

J2

, GRB Und~nlandin8 Physics Optics and Modem Physiu

198'

~1'=,h312+1=J2 Hence, Jisin ip = (I}sin(r):::::> r = 900

So ray will emerge out parallcl. 45. Bya Lens: (i) Object and image arc above and below respectively on the principal axis. Hence, image is inverted. Thus lens is a convex lens. (ii) Join XY. the ray goes without any deviation. Hence it passes through the optical centre. (iii) Take a ray parallcl to AB from X along X - Z Join ZY,ZY any.

X~:::--fi>l

f

Z

y

x

c

"

B

F

Image formation by a concave mirror: Concave mirror will be placed near X so as to get enlarged image and (i) Join X and Y. Extend it to reach to mirror. (ii) Take midpoint of AC where A is pole and obviously F is midpoint. Join XF and extend it till it rcaches mirror at S. SY is parallel to AB. F lies on AB. (iii) Take a ray xr parallel to AB. II meets mirror at T then join TY drop perpendicular from' ron AB. 46. For no emergence from curved surface, (90 0 -1l) > 9 L

sin(90°-P)

~

sin eL

(cos~) ~ ! ~

~S(co~p)

or For limiting case, Now,

1 cos J3

~=­

(1)sina = j.1sinp

.. . (i)

,

ft

Geometrical (Ray) Optics

199 ,

'R ..t.~ : 7"

'" ,,'S:

a

p_______ J,

sinp

= [Si~a]

a p

or Now if

a = 1t / 2, thcn sinf} = I I ....

~ = [s;~p]

Hence,

... (U)

From eqns. (i) and (ii) limiting cases, we can conclude that p = 45°

~

=41

47. Refracting from first curved surface.

-------------- -- ----------------f-- ---- -- -------

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

1-_ 4 / 3 =1-4 / 3 vJ

--«>

+2

1- = _!

v,

6

=-6 Refracting from lInd surface, 4/3 1 = 4 / 3-1 v -(6+ 4) -2 VI

.i=_1 _ 3v

-6

1 (6+4)

3~ =-[h~J=-!~ = -3~ v

=

-5mm

'.

GRB Undentanding PhYiic! Oplics and Modem Phylia

200

48. AC and CB should be perpendicular ForTIR at D;

A~-+----~~----~7B

; > 9(" Now LA + LB =900 90°> 29 (" : .

i

I

ac < 45°

o

I I

~=_ . l_ =,fi smg e

Minimum R.I. of prism M i t =

.

(b)lf~=5 / 3

Ji

c

LB=30' J.1=l / sin(a)

and

~= I 3 sin(n) .

a

= 37°

Since LB = 30 , TlR will not occur on face CB. 0

Again CA = 90°-30°= 60°. TIR will occur atAe face . : . Thus, TIR will occur at AC but not on CB. 49. Givenll2 < III Cas. (1): Jl3 < Jll At interface AB, 11 2 sin e = 11 3 sin '1 Also sine> sinOe For I- III medium

. and

sin 9 =

,,, ,,

G

~l

!

0 _______ ___ _

A

sin'l=J.12xsin9

1

-

',.

, 1'3

,, ,, 'F B

0,

!:!. ~2

sin '1

Then,

=.e.l > I

which is not possible.

~l

So TIR will occur at interface AB. Case (2):

).11 < J.13

If angle of reference is (9) and (9) is sJightly greater than critical angle. At interface AB, "2sine=~lsin~ III sin '1 = III sin".

and

Now,

l

-

sin'l =

lli:]

= sin".

- -- --

... (i) ... (ii)

201

Geometrical (Ray) Optics

sin '1. = J-ll / J-lJ sin Ii: = I But angle (9) is slightly greater hence no grazing emergence. The ray will get totally internally reflection . But

50. In case of reflection from curved surface, ~2 _ ~=~2 -~I

v u R when curved surface is in contact with table, refraction take place at plane surface only. Thus, I-~ 1 ~ --=--3 -4 co

R" 00 I

13

4

~ =

4/3 Now, when plane surface is in contact, refractive take place at curved surface.

Thus

1 _4 / 3=1-4 / 3 , -25 1& -4 -R R = 25cm y

Now. ifplane surface faces the incident parallel beam. 1 4/3 1-4/3 ----Hence, VI

'"

25 v=-75cm

~

=co, v=!

CRB Understanding Ph),!ics Optia and Modem Phystc.

202

But from plane end = 75 ... 4 = 79cm If curved surface faces Ihe incident parallel beam then 413

1

VI

- <Xl

----::::

4 ' 3-1 ~vl =IOOcm +25

Refracting al plane surface.

L

4_ '_3 : ~I-..,4'-"-,3

tI

+96

00

v = 72 em I = 72 + 4 == 76 em from curved surface.

2

Wave Optics St~dy ~'Points

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10

HuygeR 's Principle Condition oflntcrfcrcnce or Light Optical and Geometrical Path' Phaser Addition of Waves Young's Double Slit Experiment (yDSE)

Intensity Distribution F~nge

Shift due 10 Thin Plates Change in Fringe Width due to Change in Medium

Cl!angc in Phase due to Reflection

.

Interference in Thin Films due to Reflection , ,. . .

t 2.1

,

,

Huygen 's Principle

According to this principle. light is a wave capablc of explaining rectilinear propagation of light. reflection. refraction. interference. diffusion and poI3ris:1Iion .

Wavefront: II is the laws of nit particles of the medium vibrating with same phase

I I

I

i

and the disturbance reach at the same time to the particles in a wavefront. A point some gives spherical wavefront. a linc source yield cylindrical wave front and large flat same radiatcs plane (flat) wavcfront. However, ifthc radii ofa sphcrical or cylindrical wavefront is very largc. a small portion of them behavcs as a flat wavefront. Hu)'gcn's postulales: (i) All Points on a primary wavefront can be considered as the source oflighl which emit diSlurbances (waves) called secondary disturbances. (ii) Tangent envelope to all secondary wavelets gives the position of new wavefront.

GRB Understanding Physics Optics and Modem Physics

204

"uygen's Construction: Originally Huygen derived the idea by observing the wavc pattern in water surface. Ifwe consider the disturbed points AI. B I , C it ... F\ who receive the disturbance at t = II we call it wavefront FI _After a time intervall:!..l = 12 - tl. at t =12 if we draw the tangents to the circular secondary wavelets, we will get another wavefront F2' In this

way, considering each point orany wavefront as the secondary source of waves, we can draw the next waverrents.

A,

..

.,

B, .. B, C, .. c, 0,

0,

E,

E,

F,

F,

Wavefronl- F,

Wavefront- F:

The wavefront F: and F, are separated by a distance \/(l;>:-t,)

2.2

Condition of Interference of LIght

We have learnt that the interference of transverse and longitudinal ehistic wave is possible when the waves maintain a constant phase relationship. If the phase difference remains constant then sources are said to be coherent.

l.1., i,

For the phase difference

=

or, and

'10

to remain constant (or coherent), constant

'20 - constant 001

=002

The above relation tells us that, For coherence, the sources of waves must maintain constant Initial phase relationship and the source must have equal frequencies.

,

r

• Wave Optics

205

In optics, unlike mechanical waves it is not easy to obtain very long (infinite) wavelrains. The emission of light is an atomic phenomena. The excited atoms takes R nearly IO- 5 10 reach the ground sinle to gCI dc-excited 10 emit a W3\'ctrain obeying quantum mechanics.

Hence, the length of the wavetrain is given as 1= Ct. T "" excitation lime. 1ft = 10-8 S, we have / =: 3m. Let us take two sources (atoms) S] and S2. Since the sources (atoms) emit light independently and randomly. they do not bear a definite phase relationship between

them. Over a short lime of 10- 8 any two short wavetrains emitted by the sources S I and

S2 let superimpose at P constructively. in next 10-8 5. other two short wavclrains may

superimpose at P destructively as the emission ofthc wavctrains is random. Hcnce, thc intcrference during IO- ll s can not be detectcd by the eyc. However, due to rapid and random variation of phase of the waves at P. we can detect the average intcns ity as the sum of individual intensities. This is cvident practically in the intcrfcrence of two beams of sunlight. Then we can call the sources incoherent and the superposition is incoherent. Coherence: Ifwe consider two nearest atoms ofa source, the waves coming from them can be treated as whcrent which can be accomplishcd by the pin-hole infront of an incoherent sources such as sunJight, incandescent bulb. Alternately, laser light is coherent over a long distancc and we can produce permanent (noticeable) interference. A noticeable, stable interference panem can be obtained by coherent. monochromatic waves .

..... .....- .1 c=4c:'I!JI.

r

~

.r

Find the resulrant intensity ~ light OJ rwo sources of intensities 10 and 410 if they are (0 incoherent (iI) coherent. Solution: For incoherent waves the resultant intensity 1 10 + 4/0 = 510

=

Ans. For coherent waves the resultant intensity is

1

=(.[10 + J410)'

= 910

Ans.

If the intensities of the sources are 11 and 12, the resultant intensity lor coherent and incoherent sources is I + ..[i2)2 and I ::: I, + 12 respectively.

=(Jii

"

GRB Undentanding Physics Optics and Modem Physics

206

2.3

"

Optical and Geometrical Path

When a light wave enters a medium from vacuum its speed de<::rcases. Since its frequency remains constant. its wavelength is decreased. It means thnt. the length ofthc path covered by light in a medium in a given time known as geometrical path is obviously lesser than that in vacuum (or air) during the given time known as optical path. In other words, (Optical path in a medium) = (Refractive index) (Geometrical path in a medium)

or

O,p, = (G,P,)"

The length of a given wave train in a medium is called g eometrical path and its length

in air is called optical path .

f' =O.P and I'

If

I'

then

=G.P.

=nl ). =

2.4

~

n

Phasor Addition 01 Waves

Any sinusoidally varying electric field can be represented by its amplitude rotating with an angular speed which is equal to the angular frequency of the oscillating electric field covering the phase angle 0 =(wt + cp). Lct the electric ficlds of the interfering waves P be E I = EOI sinwt and £2 = Eo2, sin (rot +4H Then we draw the amplitude phasors Eo] and E02 covering phase angles I = WI and 92 = rot with respect to + x - direction . Now adding the amplitudes by parallelogram law of vectors, the resultant amplitude of the electric field

Eo. _______E,~

,, I$I\Jrq)

Ibt

e

++

is £0 = J£;I + E~2 +2£01 E02 cos¢! and its phase Jl is given as

The resultant ampIiIude Is

/'

Eo,

,

{wH-N

•

Wave Optics

207

n = tan ,.. If Eo] = E02 = Eo. Ermx

_'(_£:::.0,-" S~in2.,~:- ) ". ~

E02 cosq, + Eo]

=2Ecos! 2

and P =

2 2

Then,

E = Ermx sin (ro/ + Jl) or

E = 2£ cos~sin( WI + ~ )as obtain earlier

II more than two waves are lnterfering , we can get the resulting amplitude by writing the individual amplitudes with their respective phasor and adding them by component method or polygon law of vectors.

2,5

Young's Double Slit Experiment (YDSE)

Long before the discovery of Maxwell theory of electromagnctic p wave, in 180 I. Thomas Young proved 5, the wave nature of light through this experiment. He proved the sunlight through a narrow slit (hole) which 5 ,: ;--- - - ---.-jj~ worked as a source of coherent light. The 111~e\:\J C holes S behaves as a of monochromatic light. The spreading spherical wavelets from the slit (hole) S pass through the o equidistant pinholes (slits) S) and S2 to produce two coherent wavefonns atS] and S2. Then the light emitted fromS! and S2 are coherent (because SIS = S2S). Lei at the point P light waves from S, and S2 overlap crest to crest which is known as constructive interference producing bright spots (maxima). At some other points the waves overlap crest to trough producing

t

I

,.

5,

,' ' ,i-o", ,, ,, ,, •

s, \ /'

destructive interference causing dark spots (minima). Between the consecutive bright (or dark) spots. the intensity is distributed from maximum to minimum (zero) which

GRB Undentanding Physia Optics and Modem Ph)'1ics

208

will be discussed in next section. There will be a pattern of continuous variation of intensity bcrn-een nearest maxima or minima which is called fringe. Mulma : LeI two wave superimpose constructively at P at a distance y from the centre C of the screen. For this, the path difference is Ax = SZP ~ SIP = nA

or

dsinO = n'A.

or

sin 0 = nt. d

or

tan 0 = n'A d

i=~ IYn =!JPf; I

or or When n == 0, Yo

~

=

(for small 0, sin 0 :: tan 9)

(tane=~) n = O,I,2 ..

0 ; It is called central maxima.

~).. ; It is called first maxima.

When n = I. YI = When n

(Since D » d, SzP ~ StP :: dsinO)

2, Y2 = ~).. ; which is called 2nd maxima and so on lie up and down

relative to the central maxima.

Minima: At P Let the wave interfere destructively. Then, the path difference is

6.t"=S2P-S1P or

=(n-!}.. .,

dsino= ( n-~))..

or

, .n=I,2, 3 ...

or

When n = 1, YI = W which is called 1Sl minima.

2d

When n = 2, Y2 =

3~ which is called 2nd minima and so on lying up and down

relative to the central maxima.

209

Wave Optics

Fringe width: The distance between two successive maxima (or minima) is called fringe width given as P=Yn+I-Yn for

maxima = (II

+ 1/..0 _ II').J) d

Ip = )~

or

I

d

2nd minima

lsI minIma

'-

The angular fringe width is

lsI minlma'-. r~~

2nd maxima

~ = ~ = (I. Did)

D

D

2nd

r;;'maxima

minima

.-/

I$ = ~I

or

Cenlral maxima

"

.-/

lsI maxima

-'"2ndmuima .-/

o£01::----- 0 ------I} p ---------~~~~

The angular fringe width is defined as the ratio of wavelength of light and the distance between the slits S, and ~ (sources) of coherenlllght.

2.6

Intensity Distribution The light wave has sinusoidally varying electric field given as E = Eo sinwt Let the interfering waves be EI = Eo sin rot

E2=Eosin(ro/+q,)

and

Then the field of the resulting wa ....e at P is

£=£,+£, = Eo (sin rot +sin(rot +q,)} =2Eocos

, Ermx

~ Sin( ro/+~)

Since the intensity of light is

I =

2j..1oc'

p

$,

•

1 .d .f~--

s"O

• Destroctive Interterenee at P 10 Is minimum

GRB Undentanding Physics Optics and Modem Physics

210

the intensity of light at P is given as

10 _E~ -10 - --. £2

p

o

where Errnx = 2Eo cos! 2 and 10 wave at P,

z.

intensity of each interfering

wherc$ =21t Ax = 21tdsin9 ). l-

or

Ie= 41 o cos ,(7tdSin+) I-

lis maximumwhcncosq:, =±1 or q:, = ±2mt;

n=O. 1, 2. 3, ..... .

Then , /!l'WI. = 4/ 0 • corresponding to bright spots (maxima) lis minimumwhencos¢'=O or

q,=±(n-D1t;n =l,

2 ...

Then, J min = O. corresponding to dark spots (minima).

'0

Superposilion 01 two Incoherent source

I

, 2. The graph of '0..... : '0 = 410 cos2~

The energy distribution takes place in accordance with the conservation of energy; the energy disappearing at minima appears at the maxima giving usc to same total energy. In general, if the intensities 01 the interfering waves are I, and '2 at the point of superposition, the resuhing intensity is

I =I, + 12 + 2.[i;i2 COS

+

+= 211,,-1 =1m"" =(Ji + ,[i2)2 when += (211 + 1)•• 1 = 1m," =(,[i; - ,[i2)2 when

211

WaveOptia

In rDSE. are slit is wider. Hence, the amplitlldes a/light reacting the central point o/ fhe screen/rom one a/the slit, acting alone is twice that f rom the other slit when acting alone. Find the ratio resulting intensity at a direction 9 on tlte screen and that at the centre of the screen. Solution: The intensity at the point C is

I , = K{a' + (2a)' +2(a)(2a)co'4»,

~,~'--""'II

where41 =0

.. . (i)

or

At any angular position 9, at P,

d

y

,--- '.

, ,

J c

,

~dSlnO 10 = K{a' + (2a)' +2(a)(20)co'4» = Ka'{5+4co' 4»

= Ka 2{1 + 8cOs2 ~}, • where

'" = 27t.1.x= 21tdsin9

•

).

. A

2 2 10 = Ka ( 1+8cos

or

ndi9)

.. . (ii)

From cqns. (i) and (ii) 10 = !{l +8Cos 2 ( ndSinO)} Ie 9 . A. .

2.7

Ans.

Fringe Shift due to Thin Plates

When we place a plate of thickness t and R1 = n infronl of the slit St, the optical path difference bctween the two waves at P is . M = S,P - {SIA + n(AD) +BP + (AD - AD))

=S,p or For maxima. or

SIP - (n - I)AD

.1.x=dsin9 - (n-l)t .1.x =rnA. i m =0, .. 2 ..... .

d sin8-(n-l)t=m)..

or

dsin9 = mA. + {n -1)1

or

d Y =m). ·+(n - I)1 D

(-:sine~ ;)

212

GRB Undentanding Physics Optics and Modem Physics

p

5, ' ,

,,

,,

y = ~{(n-I)t+m).}

0'

Whenm = 0,

Yo ::; Dt (II - I)

d This means that, the central maxima and hence the entire fringe is shifted up through a distance ~t (II - l)without changing the fringe width when we place the plate of thickness t and RJ = II. Similarly by placing the platc infrant of lower slit S2 the entire fringe will be

shifted down by distance ~t (II - I). Fringe will shift up without changing its width when the plate is placed infrant of the slit 5,. If placed infrant of 52. the fringe will shift down, by a distance

r:;

(n -1).

When we place another slab of thickness I' and RI = n' import of the lower slit, the distance of nth maxima can be given as

k=~{(n-I)t-{n'-I)t'+ml.} If r:--_"""""-c,

~'!!!!:!"''' . .

1(11 - 1)/ - (11'-1)/',1

I

m= 0,1.2 ......

no shift of the fringe takes place.

Monochromalic light of wave length of600 nm used in YDSE. If one oflhe slit is covered with a Iransparent sheet of thickness 1.8 x 10-5 m made ofa malerial of R.Il .6, how many fringes will be shifted?

213

Wave Optics

Solution: The total fringe shift= DA {II -1)t wherc • d A

DI. = FW = n d "

or the numbcr of fringes that will shift = To...,1 shift

FW =D)Jd(n - I)liA

DA d

= (n - I)I A

=(1.6-I)x 1.8 x 10-' .. 18 600 x 10- 9 Ans.

~

Change In Fringe Width due to Change In Medium

When we immense the arrangement of YDSE in a medium of RI = length of the wave in thai medium is A' =~ • II

II,

the wave

where). "" wavelength in air

Then. the fringe width is

or Hence, the change (decrease) in fringe width is

6P=A:(I _ ~) What percentage change of the fringe width when the medium changes from air to waler is YDSE? Solution:

611 = 1_ !

p..

where n=~ 3

= 1- _1 413

=! 4

Then, percentage change is 25%.

Ans.

214

2.9

CRB UndeRtanding Phy1ics Optics and Modem Phy1ia

Change In Phase due to Reflection

Let a light wave of amplitude of the electric field Eo be incident as the interface of two media.

-"

',-

E,"J'VV

The amplitude of the reflected wave can be given as V2 - VI Eo Eo. ==;;'-c;;'VI +V2

EO, ~

~

If n2 > nl, Eo. is - ve. Hence Eo. is antiparallel with Eo . The wave changes its phase by

It

radian when reflected at the boundary whlle

moving from rarer 10 denser medium. ~

~

If 112 < nl,Eo. is + ve, Hence Eo. is parallel with Eo. The wave getting reftected as the boundary experiences zero phase chcinge while moving from denser to rarer medium.

2.10 Interference In Thin Films due to Reflection Let the ray of light I of intensity 10 be reflected and refracted at A and B due to the thin film. In consequence we obtain the rays 2 and 3 after reflcx:tion at A and reflection at C. Since the ray loses a fraction of energy at each reflection. the intensities It and 12 of the interfering waves will not be equal. If the phase difference between the interfering waves is.p. the resulting at P is I = II +12 + 2.Ji1i;cos~;

• •

Wave Optics

215 p

, ,,, ,, ,
I d

,,

\,

: lID

, ,,

A

c'

Q)

"

n,

I

•

n,

The reflected and transmitted wave carrying the Intensities 11 and'2 willinierfere al P

~ = ~1t

(optical path difference) + 0,

where l) = extra phase difference due to reflections of the waves. If

n,

> I1z and "2 > n3. no phase change takes place, then 0= O.

If n, < n2 and I1z < n3 each wave changes its phase by

1t

radian then 5 = O.

If n, ~ I1z and I1z ~ fJJ. 0 :: -n,

If n] = I. 112 =

~

and 11) = I the wave I suffer an addition phase difference

1t at

A

due to reflection whereas the wave 2 does not experience any extra phase while reflected at B. 0 = - 1t Then , The phase difference between the waves 2 and I is

.

I~=¥ax - nl ~

For maxima,

or or

For minima,

=2nn

21t6x _ n =2n n ).

lax = (2n+l)~1 ~=(2n-l)n

. "-.

. .., .

. -. ~

218

GRB UndtntllndinR PhY5ics Optics and Modem Physics

or

2J"t ax _

l.

or where

7t

= (2n - 1)1t

I~ =nl.. I [6.x - 2}ltcos r

I

PrO~'e that the optical path diffe~nce between the wave I a"d 2 is given a.faX = 2}l t cos r, 'where r ::: angle of refraction and}.1 = RI of the glass film in air. ~:t2

:J1c=~~====:::=J Solution: The optical path difference is

o

c

1

I

B

6x=41(AB)-AD = 41(AB) - ACsin i

=[2t~ - ZttanTsini] COST

Since.

tJ sin r = sin; • we have, I!J.x=2}.1ttosr

Proved

217

Wave Optics

_ - ~ , ~ "--"'

i ' "- "- ~

MISCELLANEOUS EXAMPLES

~-

Example I. A thin sheet of glassJ.l = 1.52 is introduced nonnally in the path of one of the two interfering waves. The central bright fringe will be observed to shift to the position of 5th bright fringe. If)" =5890 A. find the thickness of the glass sheet.

Solution: The displacement of the fringes = Sf}

or

D (~ - I)/ = 51.D

or

t = ~ = 5x5890x 10-

or

t = 5.66x 10-4 cm

d

d

~

- I

8

(1.52 - 1)

Ans.

Example 2. Two monochromatic coherent sources of wavelength 5000 placed along the line nonnallo the scrccn. Detenninc the: (a) condition of maxima at P, (b) ordcr of central bright fringe , if d = 0.5 mm andD=lm.

o 5,

5, d

-I

A are P

-D

Solution: (a) The optical path difference is /:!.x = SIP - S2P, Since D » d, /:!.x= dcos6

p

=d(l - sin2~) =

d( 1_ a:) (". a«)

~cP~~~, '(

={-2~") (-:a=;)

I

/"-

0

Y

'~~

s1-------_ d-~

e 0 -

For maxima, /1x = n'A

or

or

d(l- 2D' L)

= nl.

Y=DJ\I-'; )

Ans.

218

GRB Understanding Physics Optics and Modem Physics (b) For central maxima y = 0. or n =!!.

A

=

= 1000

0.5

Ans.

0.5x 10 J Example 3. White light is nonnally incident on a glass plate of thickness 0.5 x IO-6 m and index of refraction 1.5. Which wave length in the visible region (400nm- 700nm) urc strongly reflected by the plale?

Solution: Let the light that is strongly reflected has wave length A. Hence they interfere constructively.

2~1 =( n+~}

For this,

I. = 21lt

or

n+1' 2

where ~t

=1.5,

t = 0.5 x 10-6

Putting the values of Al = 400 x 10- 9 m and A 2 = 700 x lO- 9m . We obtain n

=1.66 which lies between n

= 2 and 3.

Then. putting n = 2 and n = 3 in the c:tprcssion. ). = 21lt

n+1' 2

we have A'I = 600 om and A. ·2

=

429 om are s trongly reflected.

Example 4. A source of light of wavelength 5000 A is placed at one end of a table of 200 cm long and 5 mm above its reflecting surface. Find the fringe width of the interference pattern fonned on the screen. Solution: The image S'behaves as another source then, distance between two sources is d=2d' =2 x 5mm ::o lcm ').::0

and

5000 )( 10-8 cm

D=200cm

Hence, the fringe width is )J) ~= ­

d

Ans.

219

Wave Optics

= ,,-5~x...,10,--;-'~ x :!200", 1 = 10- 2 em

Ans.

Example 5. In the previous example, find the value of y for nth maxima and 11th minima. Solullon: The phase difference between the reflected and direct waves is

or Since ¢I

q,

= 21t(S'p _ SP)

,

=~1tdSine+rr

).

+ 1t

(for reflection) . .. (i)

=2nrr for maxima, using eqn. (i), i1t(d sin 8) + 1t = 2nTt'

or

sin8=(2n - l)l 2d

Since

. 9~ Sin y=

y we have tan 0 = D'

'f:t (211 - I) (or maxima

Ans.

Similarly for minima ¢I = (211 + I)n. using eqn. (i), we have

Y=

n;:

Ans.

for minima

Example 6. The distance between the slit and bi-prism and between the prism 5, and screen are 50 cm each. The angle of bi-prism is 179 0 and its RI is 1.5. If the d· distance between successive fringe IS 0.0135 em, find the wavelength oflight. d'

-a-

1

.-------~--~--------_i

2 -------~--~-O------_i

1_ SOom -I -

,s, •a "

5Oom-

Solution: The deviated rays I and 2 can be used for interference. Hence the images SI and S2 can act as the coherent sources S of monochromatic light. The distance between the sources is d = 2d· =2a8. whereS = Deviation of light rays = (n-I)8

or

d=2a(n-I)8

Since the fringe width is p = ).~. using eqn. (i),

.. :(i)

• , GRB Undentandinl Physics Optics and Modem Physics

220

d =, D =2a(n - I)9

p

, = 2a(n - 1)9P

or

D 0

0

2x50(1 5_1(180 179 )-"-(0 .0 135) . 2 180 100 == 5890 A

Ans. Example 7. In Young's experiment. the upper slit is covered by a thin glass plate of refractive index 1.4 while the lower slits is covered by another glass plate. having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 A.1t is found that the point P on the screen where the central maximum (1/ = 0) fell before the glass plates were inserted now has 3/4 th orlhe original intensity. It is further observed that what used to be the fifth maximum earlier. lies below the point P while the sixth minimum lies above P. Calculate the thickness of the glass plate.(Absorption of light by glass plate may be neglected) Solution: At the point P, the path difference at point P for the wa.ves rcaching fromS] andS2 .. (J..L2

-~I) ·t

=(1.7-1.4)1 =(0.3)(1) It is given that 5th maxima lies below the point P while 6th minima lies above the point P. Hence, at the point P, the path difference should be more than 51.. but less than 11).)2. Thus, (0.3)1 = SA + (""') Now for (6.r),

~

the intensity is th of maximum. Hence,

S1 6th min

-- ------ ----~

IR =/oco,2(~/2) 3/ 0 = /oco"[$ / 2J 4

cos(~ 12) = Jj 2

, 'Now phase difference "" 21C(path difference)

l.

~ = 2"("",) 3 ,

.... 1

221

Wave Optics

6:< =)./ 6

(0.3)(1) = 5A +)./ 6

Hence,

t =

3 1A = 3I x 5400x IO- 10 6 x O.3 1.8

t =9.3 ).lm

Ans.

Example 8. The Young's double slit experiment is done in a y medium of refracti ve index 4/3. A light of600 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 ~m and refractive index 1.5. The interference pattern is observed on a 0 screen placed 1.5 m from the slits as shown in the figure. (a) Finrl the location of the central maximum (bright fringe with 0 zero path difference) on the y-axis. (b) Find the light inlensity at point 0 relative to the maximum fringe intensity. (c) Now, if600 nm light is replaced by while light of range 400 to 700 om, find the wavelength oflhe light that fonn maxima exactly at poinl 0 " [All wavelengths in this problem arc for the given medium of refractive index 413 , Ignore dispersion.] Soluclon: Given d = 0.45 mm 4

:L------::Jj

flm

=:3

}.. = 600nm t = 10.4 jJm ~g =

s,

1.5

(a) Location of central maxima. optical path difference = 0 [S2 P -IJ +[~IJ- SIP = a (S2P -SIP) + (~ - I)I = a Now here central maxima will be shift downward. Hence S2P - SIP will be negative. Thus, [~ g I ~m -I][IJ - (d sinO) = 0

=>

[12. -IJ[ 4/3

3

10.4 x 10-6 J = 0.45 x 10- x sin

~ - p

--------o--

o

e

e

put sin 9 = tan then andtanO = yi D.

= alfJ I m _1)[tJ= ( 3/ 2_ 1) [10.4 X IO-6 x I. 5] y g ~ 4/3 0.45 x 10-3

y = 4.33 x 10-3 m

ADS.

"

• 222

GRB Understanding Physics Oplics and Modem Physics (b) Intensity at 0

6 =

AI 0,

6

[~: -1] (1)

=(!;;

6

-I ) (104 x 10- )

Phase differenceq, : 27[(6.)=

i.

21t x( 3 / 2_ 1)><(l0.4XI0- 6 ) 600 x 10- 9 4 / 3

~ = 13.

3

Now

J = 10 C05 2 (9/2) ::: 10 cos2P~7t]

L=~=0"75

,Ans.

10 4 (e) Those wavelengths for which path difference is Thus 6x =[~g 1 ~m - 1][1] =nA and

where

- I 10.4 >< 10 (43//2) 3

6x= -

-6

n).

will form maxima at O.

6x )=-i.=n

n ::: I, 2. 3, 4, .....

Hence ,AI = ~ = l300nm; A2 = 1300 ::: 650 run;

2

).3 = 1300 =433.33nm 3 OnlYA I. '. 2 andA ) will fonn maxima at O.

Ans. Example 9. In a modified Young's double slit experiment, a monochromatic uniform and parallcl beam of light of wavelength 6000 A and intensity (I DIl't) Wm - 2 is incident normally on two circular apertures A and B of radii 0.001 m and 0.002 m respectively. A perfectly transparent film of thickness 2000 Aand refractive index 1.5 forthe wavelength of6000 A is placed in front of aperture A (see figure). Calculate the

:•• :• ~ ------

.•

",

223

Wave Optics

power (in watt) received at the focal spot Fofthe 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. Solulion : Power received by aperture A = I x 1t

r;

_ _+.

A

,,• ,

• •

• •

----~ - - -

, ,-,

F

,

•

•

--------,,

B

=10 x n(O.OOI)' n PA = 10-5 watt And power transmitted by aperture A = l.Q. x 10-5 100

(PI ) = IO~ wan

In the same way, power received by aperture B, P8 = I x = 10 x 1t(0.OO2)2 = 4 x 10-5 watt n and the power at focal point due to B,

1tr;

(P')=i~ X 4 X IO-S

=4xlO-6 W

•

II.

Now path difference due to film = (jl-I )(/) = [1.5 -1][2000 A] = 1000 A

. Phase difference

6~ =2: (<1<) = ~

. Resultant power at P =PI +P, + 2(J PIP, cos~)

P = (10-6)+ (4 X 10-6) +2JrI0--6-'--x-I-0-6'cos n 13 P =SxIO-6 +2 x I0-6 = 7 x lO-6 wan

Ans. Enmple 10. In a Young's double slit experiment a parallel beam containing wavelengths AI = 4000 A and 1..2 = S600 A incident at an angle ¢I = 300 on a diaphragm having narrow slits at a separation d = 2 mm. The screen is placed at a distance D = 40cm from slits. A mica slab of thickness t = Smm is placed infront of one of the slits and whole the apparatus is submerged in water. If the central bright fringe is observed at C. calculate:

.. 224

GRB Undentanding Physics Optics IUld Modem Physics

Zi--5,\1L!-----IC >,. >,

/1=--0---t

Screen

(a) the refractive index of the slab. (b) the distance of the first black line from C. Bo th wavelengths are in air. Take I-l w =4 / 3. Solution: (a) The mica slab should be placed infront of S2 to obserVe bright fringe at C. In that case net path difference at Cis, tu =

.

(I-lr = I-lSlab )

1)1

~

~=O

For central bright at C,

d~n~ =

or

dsin~ _ (1-1,. U>, -1)/

3 (Il,. - 1) = dsinq, = (2 x 10- )sin30° = 0.2 t S x lO - 3

!-Ir =1.2

or

I-l slab

=1.2

1'.

4 Jl sl2b = 1.2 x Jl w :c 1.2 x - = 1.6

3

Ans.

(b) A black line is formed at the position where both the wavelengths interfere destructively. Distance of nth dark fringe from C, _ (2n-l)AD y2d .. , (2nJ -1)l.'JD = (2n, - IV- ,D For black line, . .. (i)

2d

2d

Here A'I and), Z are wavelengths in water. ).'1 AI I ).1w Al 4000 )., =).,,~. = ).2 = 5600 ·

Substituting these values in Eq.(i). we get 2nl - 1 7 2n2-I = S

For minimum value nl =4 and nz =3. Hence, distance of first black line, y = (2 x 4 - 1)(4000 x 10- 10 )40x 10- 2 x 3 2 x 2 x l0 3 x4 =2.1 x IO""""m c O.2l mm

Ans.

225

Wave Optics y

Example 11. In a Young's double slit experiment sel-Up source S of wavelength ___ L ________ _ _ P 5000A illuminates two slits Sl and S2. which act as two coherent sources. The source S oscillates s about its shown position according to the .--:-::1::-t-------1 equation y = 0.5sin 1Ct, where y is in millimetres ~ and t in seconds. Find ; 1m . . 2m'--..... (a) the position of the central maxima as a function of time, (b) the minimum value of I for which the intensity at point P on the screen exactly infront of the upper slit becomes maximum. ,. Solution :(a) Net path difference at Q, )'d y'd &0=-+ "-' '. D D'

Is

L,

~__

11 IS,

_--y::::/1 J ---f---- ------------ 1 sl Q

sJ,

1. s

-+--

:) .

t--D

0"- - - "

For central maximum,

Ax = 0

or

y = - -y

,

,

D' D

=

-(fYo.5sin 1t/ )

= - (sin1tt)mm (b) y' = ~, at point P exactly infront of S I,

2

&0 =

(~)+ (d:~2)

For maximum intensity,

&0 = nl. O.Ssin1tl+0.25 =O.5n Putting the values, we get 0.5" - 0.25 sin 1t1 = 0.5 or For minimum value of t, n = I. .. sin rtl = 0.5

Ans.

•, 226

GRB Undentanding PhysiC! Optics and Modem Physics

=~ 6 1 ADs. 1=-=0.167, 6 Example 12. Two coherent narrow slits emitting light of wavelength A in the same phase are placed parallel to each other at a small x separation of2>...The light is collected on a screen ~1_____~__ _____ _ ___ ___ S which is placed at a distance D(»).) from the t-2/.---i slit SI as shown in figure. Find the finite distance I_ D---x such that the intensity at Pis equal to intensity at O. S or

ttl

PT

01

Solution: Path difference at 0, SID - S20 = 2A, i.e., maximum intensity is obtained at O. Next maxima will be obtained at point p. where

S,P - S,P or

or or Now in 6SIPO,

= ).

deose = A. (2).)co,9 =).

cosO =! 2 0=60 0 PO = tan 9

S,O x

s.

S,

1+-0-+1

~1'------D'-------1

or

3. = tan 60· = .J3 D

=.J3D

Note: At point O. path difference is 2)., i.e.• we obtain second order maxima. At point P, where path difference is ).(i.e., X = .J3D)we get first order maxima. The next. i.e .• zero order maxima will be obtained where path difference, i.e.. dcose == 0 or e = 90 0 • Ate = 9D o.x =00. So, our answer. i.e., finite distance ofx should be.'C =.J3D corresponding to first order maxima. Example 13. An interference IS observed due 10 two coherent sources S I placed at origin and S2 placed at (0, 3/.., 0). Here i.. is the wavelength of the SOurces. A detector D is moved along the positive x-axis. Find x-coordinates on the x-axis (excluding x = 0 and x = 00) where maximum intensity is observed. Solution: At x == 0, path difference is 3).. . Hence, third order maxima will be obtained. At x = 00, path difference is zero. Hence. zero order maxima is obtained. In between first and second order maximas will be obtained.

I

I

Wave Optics

227 y

T

S

3.

,

L,

-----

-------,

I-

\

,

p

-I

First order maxima:

s,P - s,P=/or

~x'+9/-' - x=/-

or

~x2 +9).2 =x+). Squaringbol~idcs,weget x 2 +9).2 =x 2 +12 +2x)' Solving this,we get . x = 4;1. Second order maxima: S2 P - SIP =

v...

or

Jx 2 +9).2·- x =21

or

~x' +9/-' = (x+2A) Squaring both sides, we get x 2 +9)..2 =x 2 +4).2 +4xA.

Solving, we get

x = ~/- = 1.251. Hence, the desired x-coordinates are, x 1.2S/-and x 4/-

=

=

AnL

Note: (i) As we move on positive x~axis (from origin) order of maxima decreases

fromn=3ton=O.

(i i) Here we cannot take the path difference dcos9 or dsin e. Think why?

Example 14. The interference pattern of a Young 's double slit experiment is observed in two ways by placing the screen as shoWn in figure (a) and (b). The distance between two consecutive right most minima on the screen of figure (a) using light of wavelength A I = 4000 A is observed to be 600 times the fringe width in the screen of figure (b) using the wavelength A2 = 6OOOA. If D (as shown in figure.) is I m then find the separation between the coherent sources SI and S2. Given that d > 3).. I / 2

I S'I

S,r-_______

I

Sc_ (6)

(b)

GRB Understanding Physics Optics and Modem Physics

228

, Solution: P and Q' would have been the positions of first and second minima (last two), had the screen be perpendicular to S2 P . Since. the angular positions of minima do not depend on the position of the screen. Therefore, second minima is formed at Q on the

C n c!.--nmnniO I~

a~........

5'1 I•

screen.

"

1p .... ,'0·

,, ,

.'

"

For right most minima at P, dsin8 1 = /"1 / 2 For small angles,

Seteen

. .. (i)

. e 1 ""tan e1 =d -/ -2

Sin

XI

Substituting in Eq.(i), we get For next minima at Q,

dsin82 . 02

For small angles.

XI

510

Substituting in Eq. Oii), we have

... tan

d' Al

.. .(ii)

=~A.l

.. •(iii)

=-

0 2 =d -/2

x,

d'

. . .(iv)

X2 = -

3AI

2d'

PQ=xl - X2=3).1

In the second case, fringe width

A,D w= - -

Given that.

PQ =6OOw

.

.

d

... (v) ... (vi)

2d' = 600 I.. ,D 3AI d d l =900AIA,D

=900 X 4000 x 6000 xlO- 20 x I =216xlO- 12

..

d=6 x l0-4 m =O.6mm

Ans.

Example 15. In Young's double slit experiment set-up with light of wavelength A = 6000 A. distance between the two slits is 2 mm and distance between the plane of slits and the screen is 2m.The slits arc of equal intensity. When a sheet of glass of refractive index 1.5 (which pennits only a fraction '1 of the incident light to pass through) and thickness 8000 A is placed infront of the lower slit, it is observed that the intensity at a point p. 0.15, mm above the central maxima does not change. Find the value of1\.

229 Solution: Without inserting the s lab. path difference

al P,

.1x=yd =OJ5XIO-3 x2xlO- J

D

2

=1.5 X10- 7 m $=

en

Ax )

=(6000

I

,

)' I.5 XlO-7 ) = ,!!

21t X

'"

s,

Corresponding phase difference at P

J

10-10

2

Lrr 2 4 IntcnsityatP,

/ =4/ocos2t=210

Phase difference after placing the glass sheet,

,

-

2_

¢ = ¢+/:"(jJ - I)1 =

rr + 2

2_

6OOOxlO 10

(1.5 - 1)(8000 x 10- 10 )

II= -

6

TheintcnsityatPisnow.

!1t =2/0

J' = 10 +11/0 +2JnIJ cos 1

Solving this equation, we get, TJ == 0.21.

(given)

Ans.

Example J6. In a Young cx.pcrimenl the light source is at distance /J = 2
12 == 4c.,.tm

wavelength A;:: 500nrn is incidenl on slits separated at a distance JOllm . A screen is placed at a distance D == 2 m away from the

'2

s,

,/'/

#9" d:\.-- -----------

C

slits as shown in figure. 5'1+1·_ _ 0 _ _--1 (a) Find the values of 9 relative to the central line where maxima appear on the screen. (b) How many max:ima will appear on the screen? (c) What should be minimum thickness ofa slab of refractive index 1.5 be placed on the path of one of the ray so that minim~ occurs at C? Solution: (a) The optical path difference between the beams arriving at p. Ax = (/, - /d+dsin9 The condition for maximum intensity is, .1x = nA n=O,±I.±2. " ,

230

GRB Undmtanding Phyeici Optics and Modtm Physia Thus.

sinO = 1[6.< - (I, - h)J= 1[,,1. - (I, - /,)J d d =

1

[nxSOOxlO- 9 - 20xlO -6 ]

lOx \0-"

Hence, (b)

=~:O -I] O=sin-'[{:o- I)] IsinOIS 1

Ans.

-IS~4~ - I]SI

or

Hence.

or

- 20 S (n - 40) S 20

20SflS60 numberofma)(ima = 60 - 20 = 40.

(e) At C, phase difference.

+== C itt Jh-II)

=(

2.

5OO x )0 9

120 x \0-" )

== 80n Hence, maximum intensity will appear at C. For minimum intensity at C. I. ijI - I)/ =2 9 1= ,,,= 500 x 10- = 500nm or 2(~ - I)

2xO.5

Ani.

Ans.

231

Wave Optics

------------_. ....... -.. 111--_ -------_. _____

00

__

-

__ .

ASSIGNMENTS

- - . --..

--.

.- - -. ..- - ....

['J Conceptual Questions I. :Vhen YDSE is conducted inside water, what arc the possible changes in the

2. 3. 4. S.

Intcrference pattern? What is the basic difference between the interference of sound and light waves? Can two different sodium lamp be coherent and monochromatic? What is the best technique to obtain coherent waves for interference? What is the shape of interference pattern in YDSE such as straigbt line, circular etc?

6. Can we conduct YDSE with y-rays? E:ltplain. 7. What will happen if the distance between the slits is (i) larger (ii) smallcr wave length than the light used in YDSE? 8. If one of the slit in YDSE is made bigger, how will it effcct the fringes and interference pattern? 9. A oil drop is put on water. You can see the oil drop fonning a thin film which appears colored and sometime dark . Why docs it happen? 10. Interference effect is well observed in thin films but not thick. What is the limiting thickness of the film to get an interference pattern? 11. In YDSE if we usc white light instead of monochromatic light what will be the change in interference patlem? 12. If we put a thin film (like water etc) on a sheet of glass what will happen to the reflectivity oflight by the glass? 13. When two light waves get superimposed in YDSE, at destructive interference we get dark spots (or fringes). Does it means that the total energy of the interfering light gets destroyed? Explain. 14. What is the basic difference between diffraction and interference of light?

r) Multiple Choice Questions (A) Only One Choice is COmJet Leve/-1 J. If the light have sources same powers, for coherent sources the intensity at the point P is It. If the sources are incoherent, the corresponding intensity is 12 . The valueofl J :}2 is(OP » S'S2): 51

j

o-----~·P

.,1

232

GRB Understanding Physics Optics and Modem PhysiCi

(a) 1 : 1

(b) 4 : 1

(c) 2 : 1

(d) 1 : 2

2. In YDSE if we use sunlight instead of sodium light: (3) the interference pattern will have more fringe width (b) no interference pattern will be fonned (e) the intensity of interference pattern will brighter (d) interference pattern will be produced having variable fringe width having a white central fringe with other colored fringes 3. The color exhibited in the soap film is due to:

\

(a) interference (b) double refraction (e) scattering (d) renection 4. The fom1ation of rainbow in a water drop is due to: (b) reflection (a) interference

(e) double refraction

(d) polarisation

S. Interference of light occurs if: (a) only two light wave arc there (b) the two or more light waves must be coherent monochromatic and move in same direction (c) the light waves are coherent (d) the light waves must be coherent and mono·chromatic 6. Which of the following phenomena can not be explained by wave theory oflight? (a) Reflection (b) Refraction (d) Interference (c) Photo·electric effect 7. The refractive index of a medium: (a) is same for all lights (b) depends on frequency oflight (c) may vary time to time (d) depends on the temperature of the medium 8. The relative pennitivity of a medium: (a) depends on the frequency oflight (b) depends on the temperature of the medium (c) is a constant for all lights

1

(d) none of the above

I

\

9. In VDSE, the interference fringes with two small holes are: (a) elliptical

(b) circular

(c) hyperbolic

(d) parabOlic

10. In interference two waves get superimposcd some where destructively and some where constructively. If the ratio o f intensities of maxima and minima is / mu:/ min = 2, the ratio of the interferring waves is:

(a)3 : 1

(b)4:1

(c)9 : 1

(d)3(J2 + I):1

233

Wa\'e Oplics

II. Which is not the condition of intcrference to produce double interference ... pattern? (a) Waves must move in same direeti'~n (b) Waves should have same intensities(amplitudcs) (e) Wave must have same initial phaseCcohcrent) (d) Wave must have same frequencies (mono-chromatic) 12. 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 13. In Young's double slit experiment,the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is: (a) unchanged (b) halved (c) doubled (d) quadrupled 14. Two coherent monochromatic light beams ofintcnsities I and 41 are superposed. The maximum and minimum possible intensities in the rcsuhing beam arc: (a) 51 and I (b) 51 and 31 (e) 91 and I (d) 91 and 31 15. A beam oflight of wavelength 600nm from n distance source fall s on a single slit I mm wide and a resulting diffraction panern is observed on a screen 2 m away. The distance between the first dark fringes on other side of central bright fringe is: (al 1.2 em (bl 1.2 mm (c) 2.4 em (d) 2.4 mm 16. Light is ineident at an angle~with the nonnal to a plane containing two slits of separation d . Select the expression that correctly describes the positions of the interference maxima in terms of the incoming angle 4land outgoing angle 8 .

-------

(a)sin, +sin8

=( m+~)~

--

GRB Understanding Physics Optics and Modem Physics

234 (b) dsin9 = nil

(c)sin4l-sin9 "" (m+l)?: d

(d)sin~ +sin9 = m~

d 17. In a Young's doublc·slit experiment, the slits nrc illuminated by monochromatic light. The entire SCi-Up is immersed in pure water. Which of the following act cannot restore the original fringe width? (3) Bringing the slits close together (b) Moving the screen away from the slit plane (c) Replacing the incident light by that of longer wavelength (d) Introducing a thin transparent slab in front of one of the slits 18. The intensity ratio of the two interfering beams of light is Jl What is the value of I max -llT'An? Im:u, +/min' (b)

2./fl

I+P (d) I + P

(c) _2_

2./fl

I+P

19. In Young's double-slit experiment, the intensity oflighl at a point on the screen, where the path difference is)'. is I. The intensity of light at a point whcre the path diffe-rcllce becomcs ')J3 is:

£

(b) £

(c) £ 2

(d)/

(a)

4

20. In Young's

3

double~slit

experiment, the angular width of a fringe fanned on a

distant screen is 1°. The wavelength of light used is 6oooA. What is the spacing between the sl its? (a) 344mm (b) O.1344mm (c) O.0344mm (d) O.034mm 21. Two light waves are given by. £, =2sin(lOO1tt -kx+300) and £2 = 3 cos(2001tt - k' x + 60°). The ratio of intensity of first wave to that of second wave is:


(b)~

(c) !

(d) !

3

9

9 3

Wave Optics

235

22. In YDSE. find the missing wavelength at t = d, where symbols have their usual meaning (take D » d),

, 7D ,

(a) d' D

(b)~

(c) 3d' (d)
(a)~.L1

(b)~ =~

(c)~=~

(d)~=1

<, <,

3 4

<, <,

4

5

24. In YDSE, when a glass plale ofrefraetive index 1,5 and thickness t is placed in the path of one of the interfering beams (wavelength A), intensity at the position where central maximum occured previously remains unchanged. The minimum thickness of the glass plate is: (a) V. (b)(2/3)A (c) )J3 (d) < 25. In Young's double-slit experiment, the intensity o f light at a point on the screen where path difference is A, is I, If intensity at a point is 114, then possible path difference at this point arc: (a) Al 2. 1. / 3 (b )U 3.V. / 3 (c) A 13,A 14 (d) 2U 3, Al 4 26. 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 A). the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass-plate is: (a)2A (b) 2A / 3 (c)). / 3 (d) A 27. In Young's double slit experiment, the separation between the slits is halved and . the distance between the slits and the screen is doubled, The fringe width is: (a) unchanged (b) halved (c) doubled (d) quadrupled

Level·2 1. A monochromatic beam oflight falls on YDSE apparatus at some angle (say9) as shown in figure. A thin sheet of glass is inserted in front of the lower slit 52, The central bright fringe (path difference = O)will be obtained:

GRB Understanding Physics Optics and Modem Physics

236

s, 0

0

0

/

S,

5 1 5 2 '" d

(a) at 0 (b) abovc 0 (c) below 0 (d) anywhere depending on angle 9, thickness of plate I and refractive index of glass J.1 2. A plate of thickness t made ofa material of refractive index J.1 is placed in front of one of the slits in a double-slit experiment. What should be the minimum thickness t which will make the intensity at the center oC the pattern zero? (a)(II-I)1,. 2

(b) (p - I)l.

(e) 2(JJl._1)

(d) (p ~ I)

3. In Young's double-slit experiment, how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe (l. = 2000 A)? (a) 12 (b) 7 (c) 18 (d) 4 4. In Young's double-slit experiment. the separation between two coherent sources S]3nd S2is d and the distance between the source and screen is D. In the interference pattern , it is found that exactly on front of onc slit, there occurs a minimum. Thcn the possible wavelength used in the experiment are: 2 2 2 d d d d2 d2 d2 (a)l. = D '3 D'5D (b) l. = D'SD'9D

d' d' d'

(e)l. = D '5D'3D

d' d' d'

(d)l. = 3D'7D'11D

S. In a Young's double-slit experiment, first maxima is observed at a fixed point P on the screen. Now, the screen is continuously moved away from the plane of slits. The ratio of intensity at point P to the intensity at point O(center of the screen):

•

Wav~ Optics

237

I ~---------------I

:

(a) remains constant (b) keeps on decreasing (C) fi rst decreases and then increases (d) fi rst decreases and then becomes constant 6. In a double-slit experiment, the sli ts arc separated by a distancedand the screen is at a distance D from the slits. If a maximum is formed just opposite to each slit, then what is the order of the fringe so formed? d' 2 (n) _(b)2d 2)J) W

d'

d'

(d) 4W

(c)w

7. In a mod ified YDSE, a monochromatic uniform and parallel beam of light of wavelength 6000A and intensity (lOIn )Wim 2 is incident normally on two circular apertures A and B of radii 0.001 In and O.002m. respectively. A perfectly transparent film of thickness 2000A and refractive index 1.5 for the wavelength of 6000A is placed in front of aperture A (see fi gure). Calculate the power, in watts, received at the focal spot F of the lens. The lens is symmetrically placed w.r.t. the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought 10 the focal spot.

• • • • •

•

A

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

F

B

•

(b)l xlO-6 W (d) 7x lO-6 W (c)5xlO-6W 8. Two waves of light in air have the same wavelength and are initially in phase. They can travel through plastic layers with thicknesses of L] = 35 mm and liz = 5.0 nun and indices of refraction n] = 1.7 and n2 = 1.25 as shown in the figure . The rays later arrive at a common point. (n) 3xlO-' W

n, n,

I·

L,

·1

eRB Undentanding Physics Optics and Modem Physics

238

The longest wavelength oflight for which constructive interference occurs at the point is: (.) 0.8 ~m (c) 1.2 ~m

(b) 1.2 ~m (d) 2.9 ~m

9. In a YDSE shown in figure, a parallel beam oflight is incident on the slits from a medium of refractive index nl' The wavelength of light in this medium is AI. A transparent slab ofthickncss 't' and refractive index is put in front of one slit. The medium between the screen and the plane of the silts is n2 . The phase difference between the light waves reaching point ' 0' (symmetrical. relative to the slit) is: (.)

~(n,

I

- n,)1

n' " 3

nlAI

(b)~~(n, (c) 2nnl "2 AI

n,

-n,)1

(nJ _t}

o

"2

2MI (d ) - ( n , - n,)1

Al

10. In the figure. ira parallcl beam of white light is incident on the plane of the slits, then the distance of the nearest white spot on the screen from 0 is: (assume d «

D.A« d]

2d13

o

d

0

(0) 0 (c) dl3

(b) dl2 (d) dl6

11. A ray of light is incident on a thin film. As shown in the figure, M and N are two reflected rays while P and Q arc two transmitted rays. Rays N and Q undergo a phase change of 1[. Correct ordering of the refracting indices is: M •

N

", n, II

", \

';. .; ~{"

~. ,

•r

1,!1

p

0

239

Wave Opba (a) n2 rel="nofollow"> n3 > n, (b) "3 > nz >n, (c) nJ > n, > liZ

(d) none of the above, the specified change cannot occur 12. A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in the figure . Another identical glass plate is kept close to the first one and parallel to it. Each glass plate renects 25% of the light inc ident on it and transmits the remaining light. Then , the ratio of the maximum to minimum intensities in the interference pattern fonned by 2 1 the two beams obtained after one reflection at each plate would be: (a)7: 1 (b) 49: 1 (e) 4:1 (d) 16:9 13. Two plane mirrors are inclined by small anglee. The distance ofsHI from mirror M I is a and distance of screen from mirror M 2 is b. If is the wavelength of light used, find the fringe width.

,,- --

--.

, ,,, , ,,, , • ,,----.:..-" ,, ,, ,, ,

"

.... _-

I-o-I-b-I (b)p=,(a-b)

2a9 (d) p = ,(a - b)

2e th

14. In an interference pattern of a point, we observe 16 bright fringe for AI = 6000 A . What order will be visible if the source is replaced by another bright fiinge, = 4800A? (a) 12 (b) 20 (e) 18

(d) 24

I!. Two identical coherent sources placed on a diameter of a circle of radius R at a separation d«
CRB Undtntanding Physics Optics and Modem Physics

240 (a) 2d + 1

(b) 4d

(c) 4d -2

(d) 4d +2

1

l

1

l

16. In Young' s double-slit experiment, we get 60 fringes in the field of vicw of monochromatic light of wavelength 400oA. If we use monochromatic light of wavelength 6000A, then the number of fringes obtained in the same field of view will be: (a) 60 (b) 90 (c) 40 (d) 15 17. In a Young's double-slit experiment, f... =500 nm, d = I nm, and D = 1m. The minimum distance from the central maximum for which the intensity is halfofthe maximum intensity is: (a) 2 x 10-4 m

(b) 1.25)( lO-4m

(c) 4x 10-4 m (d)2.5x lO-4m 18. Two thin parallel slits that arc 0.012 nun a part arc illuminated by a laser beam of wavelength 650 nm. On a very large distant screen, the total number of bright fringes including the central fringe and those on both sides of it is: (a) 38 (b) 37 (c) 40 (d)39 19. In Young's double-slit experiment intensity at a point is (114) of the maximum intensity. Angular position of this point is:

(a)sin- 1().ld)

(b) sin - 1(). /2d)

(c) sin - I ()./3d)

(d) ~n - l('. / 4d)

20. Monochromatic light of wavelength 400 nm and 560 nm are incident simultaneously and nonnally on double slit apparatus whose slit separation is 0.1 nun and screen distance is I m. Distance between areas of total darkness will be: (a) 4 mm (b) 5.6 mm (c) 14 mm (d) 28 mm 21. In the given diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition in 0 foreonstructive interference at P between the ray BP and reflected ray OP:

a~~O~~~~~~R

c

o , d

A

p

B

(a) cosO = 3l12d (b) cosO =1 / 4d (c)secO-cosO =l l d (d) sec 0 -cosO = 4l l d

r

, 241

Wave Optics

22. In Young's doublc·slit experiment, 12 fringes arc observed to be formed in a certain segment of the screen when light of wavelength 600 om is used. If the wavelength oflight is changed 10400 om, number offringcs observed in the same segment of the screen is given by: (0) 12 (b) 18 (c) 24 (d) 30 23. A thin slice is cut out ofa glass cylinder along a plane parallcl to its axis. The slice is placed on a flat glass plate as shown in figure. The observed interference fringes from this combination shall be:

I I I I

\:2 ,

(a) straight (b) circular (e) equally spaced (d) having fringe spacing which increases as we go outwards

(S) _More Than One Cholcels Is/are Correct 1. The fringe width will change when we change the:

,

,

(a) distance between the sources (slits) (b) medium between slit and screen (c) intensity oflighl (d) wavelength of light 2. If a ray oflight of wavelength A.\ is greater than tha'"ofthe other, that is, A. 2 in two media, then: (a) their frequencies may be different (b) their speeds can be different in their corresponding media (c) their frequencies can also be same (d) their refractive indices can also bc different 3. Which of the following can not be explained by curputuJar theory of light? (a) Polarisation (b) Refraction (c) Reflection and interference (d) Diffractive 4. In a Young's double-slit experiment, the separaton between the two slits is d and the wavelength of the light is t... The intensity oflighl falling on silt 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). (a) If d = A, the screen will contain only one maximum (b) If I. < d < 2A ,at least one more maximum (besides the central maximum) will be observed on the screen

GRB Understanding Physics Optics and Modem Physics

242

(e) If the intensity of light falling on slit 1 is reduced so that it becomes equallD that of slit 2, the intensities of the observed darkand bright fringes will increase (d) lrthe intcnsity of light falling on slit 2 is increased so that il becomes equnllo

that of sl it I, the intensities of the observed darkand bright fringes will increase 5. Two monochromatic coherent point sources S I and S 2arc separated by a distance L. Each source emits light of wavelength I.; where L»).. The line SlS2whcn extended mects a screen perpendicular to it at a point A. Then: (a) the interference fringes on the screen are circular in shape (b) the inlcrferenee fringe s on the screen nrc straight lines perpendicular to the line S tS;?:A (c) the point A is an intensity maxima if L := IIA (d) the point A is always an intensity maxima for any separation L 6. In Young's double-slit experiment. let A and B be the two slits. A thin film of thickness I and refmclivc index ~I is placed in front of A. Let P= fringe width . Then, the central maxima will shill: (a) towards A (b) towards B (c) by lb.

- l)~ ).

(d) by I" ~

).

7. Ifone of the slits of a standard Young's double-slit experiment is covered by a thin parallel sides glass slab so that it transmits only one-half the light intensity of the other, then: (a) the fringe pattern will get shifted toward the covered slit (b) the fri nge pau.:m will get shined away from the covered slit (c) the bright fringe will bet:omes less bright and the dark ones will becomes more bright (d) the fringe width will remain unchanged 8. A parallel beam of light ().. = 500A) is incident at an angle a =30° with the normal to the slit plane in a -;. 151 Young's double-slit experiment. Assume thai the -"';-"~"""l<----l intensity due to each slit at any point on the screen is 10. . S 52 Poin! 0 is equidistant from Stand S2 . The distance between slits is I mm , then : (a) the intensity at 0 is 41 0 (b) the intensity at 0 is zero . (c) the intensity at a point on the screen I m below 0 is 410 . (d) the intensity at a point on the screen 1 m below 0 is zero 9. A light waye of wavelength I.opropagales from point A 10 point B. We introduce in its path a glass plate of refractive index II and thickness I. The introduction of the plate alters the phase of the at B by an angle~. If), is the wavelength oflighl on emerging from the plate, then;

'>--..1

243

Wa\"t Optics

(a) 6$ = 0

(b) 60) = 2ttl

(e) 6$ = 2tt/(! - _I ) A 1.0

(d) 60) =2tt'(n_l)

1.0 ). 0

10. In a Young's double-slit experiment. the separation between the two slits is d and the wavelength of the li ght is A. The intensities of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correCI choice (5). (a) Ii d = )" the screen will contain only one maximum (b) If A < d < V ., aliens! one more maximum (besides the central maximum) will be observed on the screen (e) If the intensity oflighl falling on slit I is reduced so that it becomes equal to ~hat of slit 2, the intensities of the o~servcd dark and bright fringes will Increase (d) Irthe intensity oflighl falling on slit 2 is increased so that it becomes equal to that of slit I. the intensities of the observed dark and bright fringes will increase t I. In an interference arrangement similar to Young's doublc·slit experiment. the slits S I and S 2 are illuminated with coherent microwave sources, each of frequency 10 6 Hz. The sources are synchronized to have zero phase ",,-,e.o- - difference. The slits are separated by a distance d = 150.0 m. The intensity I (0) is measured as a function of9. where ois defined as shown . If J0 is the maximum intensity, then I (9) for 0 :5 e :5 90° is given by: f (a) 1(0) = 10 12, ror e = 30' (b)/(e) = 10 14. rore =90' (e) 1(0) = 10 • rore =(J' (d) J (9) isconslant for all values of9

I

~(f<'

~I

I I i

I

12. White light is used to illuminate the two slits in Young's double slit experiment. The separation between the slits is b and the screen is at a distance de> b) from the slits. At a point on the screen directly ion front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are: ')

\

)

I1 •

2

b2

2

(a)A=b(b))' = ~ (c)A=(d) ). =2b d d 3d 3d 13. In Young's double slit experiment. the imcrference panem is found to have an intensity ratio between the. 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 are 4 units and I unit respectively (e) the amplirude ratio is 3 (d) the amplitude ratio is 2

244

GRB Undentanding Physics Optics and Modem PhYlics •

[)IMatch The Columns I. A beam of light falls nomlal1y ( i := 0) on the transparent film whose RJ is tl2 and its thickness is much smllller than the wavelength oflighl. The light gets reflected and refracted as shown in the diagram . The R.I. of the media above and below the IiIm are III and 113 rcspcctivcJy. The eyes E land £2 of two observer view the film same times bright and same times dark depending

E,

, ' rv-

on the types of interference occuring

4

0,

between the reflected beam 1-2 and refracted (transmitted) beam 3-4. Match the column I with the column II . In column I. we give the conditions and in column 2, the visual effects are given.

Column 1

Column II (P)

E I observes bright film

(q)

E I observes dark film

II)

(r)

£2 observes bright film

< 1'2 and"2 > II)

(s)

£2 observes dark film

(0)

1/\ >"2 and 112

(b)

II]

(c)

II, > "2 and "2 <

(d)

II,

> //3

< "2 and 112 < 113

2. In modified YDSE, we can modify the sct in the fo llowing way as given in RHS column and given the result in LHS column.

If the hole 2 is greater than hole I

-- ,

S 4 C

- - 0 --c3~ ~

•• o

Column II

Column I (0)

\

(p)

The central maxima will be

shifted to position 5 (b)

If the parallel horizontal light (q) becomes slant

There may be 0 change in intensity at the points 3, 4 and 5

Wa~Optjcs

245

(c)

If we place a thin glass plate in front of the hole 2

(f)

There may be a shift of fringes

(d)

If

(5)

There will be the change on fringes width and change in intensity at same point on the screen

decrease the distance between the holes I and 2 we

3. Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S l and S2. In each of these cases SIPO = S2PO.SIP1-S2PI =,. / 4 and S1P2 -S2 P2 =)' / 3, where I. is the wavelength of the light used. In the cases B,C, and D, a transparent sheet of

refractive index Il and thickness t is pasted on slit S2. The thickness of the sheets arc difTerent in difTcrent cases. The phase difTercnce between the light waves reaching a point P on the screen from the two slits is denoted by o(?) and the intensity by I(P). Match each situation given in Column I and Column II va lid for that situati on . Column I

S,

Column II

P, P, P,

I

l-

(a)

s,

(p)

I s,~

(b)

(f.!-1)t ..,)./ 4

. -

(q)

Ii(Pt!

~O

s, I

s,~ (c)

P, P, P,

Ii(Po ) = 0

(.-11' ='·12

s, I

-

P, P, P,

(r)

I(P'I=O

GRB Undcl'5landing Physics Optidl and Modem Physics

246

s,

(d)

()I-l)t"'3A./4~

-

P, P, P,

s,1

(5)

I(Po»

(I)

lIP,)

I(PJl

~

I(PJl

['JComprehensions Passage-l [(Anti-reflection (glare) glasses) Anti - glare or non reflecting glass forspcctac!es arc made by coating the glass with a thin films of suitable transparent materials like MgF2 etc. Generally the t~ickncss of

such coaling (called anti-reflection coating) arc in the order of several hundred nanometers, and hence vcry small compared to the thickness of glass «(2 »> ,,) which is in the order ofmillimclcr. The main principle behind the anti-reflection is destructive interference of light reflected 01 the lop and bottom of the coating. As a result, the intensity of the resulting light after the destructive superposition of the reflected light waves I and 2 will be minimised. 1

2

.L "

"I

,

", ",

Film Glass

1. The phase difference between the rays I and 2 assuming "1 < "2 and nonnal incidence is: 1\ (a) nI2-nl

(b)I"lll -n121

(c) 21t"2tl - 1t/ 2

(d)(~-~'nll)

).

2. The path difference between the rays 2 and 3 (referring Q-1) is: (a) "1/1 +"212 (b)2"1/1 - 112 12 (c) flltl -1I2 t 2

(d)g+1I 1/1 +112'2

_

3. Ifni = 1.38 for MgF3. the thickness II required for making non-reflecting coating is approximately:

• 247 (a) ISO nm (c) 200 nm

(b) 80 nm (d)lOO nm Passage.2

A light of wave length). = 5000 A fall s on two narrow slits placed at a distance of separation tI = 50x 10- 3 em at an angle q, = 30 e relative to the slits as shown in the figure. In front of the lower slit a transparent slab of thickness 0 , t mIn and refractive

. dex In

3 2 IS· PIaced. As a result, the interference pattern is observed on the screen placed

at a distance D = 2 m from Ihe slits .

• --:-_ I

•

------ - ----- C

p

o

t. The angular position 9 of the central maxima is: (a) O· (b)30' (c) 60· (d) 45· 2. The order orthe minima closest to the center c is: (a) 49 (b) zero (c) not defined (d) one 3. The number of fringes shifted up when we remove the slab is: (a) 49

(c) 100

(b) 98 (d) 99

Passage-3

A biconvex lens iscut along its principal axis such that a thin layer of glass of I mm thickness is removed. In consequence, the two halfs Lt and L2 behave as two different lenses as their principal axes are different. A monochromatic point source S is kept on the principal axis which radiates a light of).. = 6000 A The distances d = 15 em and D = 130 em as shown in the figure.

248

GRB

Und~l'$tanding Physics O plit! and Mod ~ m Ph)'S it!

(screen

ilL, s

•

v

o

•

L,

I. The frin ge width is:

(a) 2x 10'" m (c)3 x 10'" m

(b) 10-4 m (d)lO - 3 m

2. If the arrangement is kept in watcr II ::: ~. the foeallength of the lcns will be 20 3 em. Then. the new fringe width will be: (n) 3.8 x 10-4 III (b) 6.3 x 10- 5 m (c}2.4 x IO- 4 m (d)1.8 x I0-4 rn 3. The shin of the third maxima due to the introduction of water is: (a) 3.7xI0-4 m

(b) 4.5 x 10- 4 m

(e) 5.2 x 10-4 m

(d) 4.1 )( 10-4 m

Passage-4

In this arrangement ofintcrference experiment. D » d. ",... Screen

I ==::----l- l---------------_0_

20--_1

'.

0

249

Wa\"e Optics

'. I. The minimum valuc of d to obtain a minima (dark spot) at (a)

J~

a is:

(b) .j2D). /3

(c) ,;W

(d) .jD).t3

2. The position of first maxima for minimum value of din Q-I is: (a) ~ below the point

a

(b) d above the point

a

3;' below the point 0 (d) 3;' above the point a (c)

3. The fringe width is: (a) 3D). (b) 201.

4d

(d) 3W

(c) DI.

d

2<1

d Passage-S

In Young' s double slit experiment as shown in this figure, let us h:we the scope of changing "all" parameters such as d, D,1.. etc. Assume that D » dand the holes (slits) S I and S2 are identical.

L,

,.,....-Screen

s,

• __ .0

•

r

d

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

c

s, o

I. The distance between (n - I }th minima from the zero order maxima is: (a) "W

d

(b) (n-l»).D

U

(c) "W

2d

(d) (n_l)W

d

2. When a glass slab of thickness I and refractive index J.I is placed infront of the slit SI ,the:

GRB Understanding Physics Optics and Modem Physics

250 (a) fringe width will change

(b) fringe pattern will be shifted up by tD (p - 1)

2,J

(c) fringe pattern will be shifted down by tD (j.t2~ 1) (d) fringe pattern will be shifted up by D(Il-1)tJ.. 1

3. If thc parallel beam strikes the slit at an angle & with respect to the y-axi~ so that the principal (1/ =: 0) maximum comes back to the center C of the screen In Q-2. the value of 0 is: _1 ,(" - 1) (a)sin-l/(~I-l) (b) cos d d . - I.JIij (c)sin-I'(~-I) ;;, (d) sm - d vclt

(JI Subjective Problems Level·l Interference of light 1. A beam oflighl of wave length A =: 6000A enters into glass from air. If the RI of glass is 1/ = 1.5, find the (a) frequency (b) wavelength (c) speed oflighl in glass.

2. Find th e resulting intensity of two light of intensities J 0,410 if the sources are (a) coherent (b) incoherent. 3. To coherent monochromatic light beam of intensities 10 and 910 arc super· imposed. Find the mllximum and minimum intens itieso fthe resulting wave. 4. Two light waves of intensities 1 and 41 and same frequency superimpose. Find the ratio of intensities at the two points P and Q where the wave arise with a phase difference of 90° and 380°.

Thin film interference 5. White light is incident nonnally on a glass plate (in aiT) of thickness 500 nm and RI of 1.5. Find the wavelength (in nm) in the visible region (400 om· 700 nm) thaI is strongly reflected. 6. What is the minimum thickness of an oil film floating on water to show color due to interference? 7. A lens is coated with a thin film of RI = 1.25 in order to reduce the reflection from its surface at A = 5000 A. Find the minimum thickness of the film which will minimise the intensity of the reflected light.

i

Wave Optics

251

YDSE 8. Two identical coherent light sources of intensity 10 and wavelength 550 nm can produce an interference pattern on a screen. If the sources are 2.2 mm apart and 2.2 m from the screen, find the (a) fringe width (b) phase difference at a point which is at a distance of y =3p from the center of the screen, where P=fringe width (c) the intensity at the point refered by (b). 9. A thin sheet of glass J.I. = 1.5 2 is introduced nonnally in the path of one of the two interfering waves. If the fringe pattern is shined by a distance of fly = 5P, where p = fringe width, fi nd the thickness of the sheet. (A = 5890A) 10. In YDSE, find the ratio of intensity at a point y from the ecnter of the screen and the intensity of each wave. Assumep =angular fringe width . 11 . In YDSE the slits arc 0.5 111m apart. The interference is observed on a screen of separation I m from the slits. If9th bright fringe is situated at distance of7.5 nun from the 2nd dark fringe from the center of the fringe pattern, find the wavelength oflisht used. 12. In YDSE the ratio of intensities interfering waves is 4: I. Find the ratio of (a) amplitudes (b) intensities of the resuiting wave at maxima and minima. 13. Two coherent monochromatic light sources SI and 52 are placed as shown . If d « D, find the (a) path difference at P (b) maximum wavelength of light to experienee mi nima at P.

:J

0

P

(Jj Subjedive Problems Level-2 1. In YDSE, let us usc two coherent, monochromatic (laser) sources oflisht SI and 52 separated by a d istance d. The distance D(» d) is the distance between source and screen. The intensity of the sources S]and S2 arc 10 and 4/0 respectively. Find the (a) int~ns ity as the function of angular position o.f a po~nt (b) intensity at a point ofa dIstance y fro m t~e center of the ~crcen (c) intensIty at the centcr C (d) position)' of 11th maxIma (e) fn nge wIdth : Assume A = wavelength of light.

c:

51

I 1. S,

d

1

0

~

cSc,een

e

"

252

GRB_Onderstanding Physics Optics and Modem Physics

2. A Source or light ofwnvclcngthA = 5000A is kept infront ora screen ns shown in the figure. Taking the interference of the direct waves and reflected waves at C and D ofthe mirror. find (a) the region where the fringe will be visiblc(b) number of fringes.

s

•

s

I , •• I -- - -----D.~~.;C "''',»''''' ---------- -" " c

lmm

1- San -1- San -1-

1",,",

-I

r

3. Two rays of light of intensity 10 and 4/0 produce interference pattern . Find the (a) resulting intensity at D (y =- P/2). where p = fringe width = D)', (b) ratio of d maximum to minimum intensities (at the desired points P. Q, etc.) on the screen.

0_

-

II

I I I

0_

!

(D»d)

4. Find the maximum radius of the beam oflight that can be focu sed at a distance/. at the poi nt F inside the glass. of RI = 11.

'1--/

" ! -'---1_ 1 _ ' \'----_----1 s. Two rays from the sources SI and S2 are focused at P by a thin lens L. It produces a s"'_-t maxima when the tmnsparent assuming the path difference 6x = nA. If we increase the temperature o f the slab by we again maxima ~-:.._ _ _ _ __ will be produced at P having a path difference

I

l

p

I

253

WaveOplics

!lx' = (11 + I)L If the coefficient of linear expansion of the slab iSCl and its R.1. is ~. find (a) 1/ (b) the wavelength.

6. A ray is i~c ident at an angle i at A of a glass 2 slab of thIckness I and RI = J.l. Find the (a) phase difference between the rays I and 2 (b) value of t for nonnal incidence i = 0 so as to """"T__,*~_--r.~__"~'=_'--, have a constructive superposition of the rays 1 A and 2 (e) if 50 % of light gets reflected at each interface. find the intensity o f light after I superposition of the rays I and 2. assume 10 = intensity of incident ray. (d) minimum B intensity after superposition of the rays I and 2 (put; = 0) (e) condition for minima for nonnal incidence of the light i = O. 7. Two parallcl rays of intensity 10 and 4/0 arc incident on the prism nonnally. The emergent rays arc focused at P. Find the intensity of light at P.

I

1

L

41'o------t---"""--i~ p

8. A point monochromatic source of light of wavelength ). is put above a plane mirror at a distance d. The intensity oflight is 10 . Ifhalfofthe light is reflected,

fi nd the: p

,S

, •• "

e

y

d

L

C

D

(a) intensity ofl ighl at P. (y« d and D» d); 1= 1(9) (b) intensity I at P as the function ofy (c) position of 11th maxima (d) fringe width (e) intensity at the center C.

254

eRB Undentanding Physics Optics and Modem Physics

9. Two coherent and monochromatic sources of light ofwavelengthl are placed at o and Q, OQ =d. Find the points on x·axis where the minimum occurs.

a

, I f p ----'+.=-.c~.=:.c~_, 2

,

10. Three coherent and monochromatic sources 1.2 and 3 of light of wavelength f. arc sepamted at a distance 1 /6 and A/ 4 respectively_If the amplitude of electric field of emw radiated by of each source is A. find the:

(a) resulting amplitude of emw superimposed at a very far point assume 0 = 00 (b) resulting intensity in (a), if! 0 = intensity oreach wavc.

(c) resulting vibration of the electric field assuming ro =angular frequency of oscillating field = 2rrf :: 27t~ .

II. A point soun:c of light 0 (behaving as a point objecl) is placed infrant ora prism at a distance ' 0 ' from il . A screen is placed at a distance b fro m the source is prism. If the angle is very small compared to the angle 0 "":--t-7ilr---i A of prism. find the value of "" .. and 0 assuming n = R.I . o f pri sm and P= fringe width of the interference pattern fonned on a - -._- b • the screen.

e

L

12. Two electromagnetic waves bearing equations of electric field given as YI = A sin rol, Y2 = 2A sin(rot+ Tt/ 3) and Y3 =; cosrol. Find the equation of the resulting wave after all these waves get superimposed.

255

Wave Optics

13. Two thin nngle prisms of RI, ,,] and apex angle Aland A2 arc joined to form a A, bi-prism. Another medium of RI, "2 is attached with the pnsm. A point f-"~'"""">__l_---illuminating object 0 is placcd at a distance n1 a from the bi-prism. Assuming the object emitting monochromatic light, we can have a stable interference pattern on the screen which is at a distance of D » a from the bi-prism. Find the (a) distance between two images due to the bi-prism (b) fringe width. Assume A wavelength of light emitted.

,O""'---__

-.-

=

14. In Young's double slit experiment, a light containing two wavelengths 500 nm and 700 run are used. Find the minimum distance where maxima of two wavelength coincide. Givcn Did :::: 10 3 , where D is the distance between the slits and the screen and d is the distance between the slits.

15. A point source S emitting light of wavelength 600 nm is placed at a vel)' small height Ii above a flat reflecting source AB (sec figure). The intensity of the reflected light is 36% of the incident intensity. Interference fringes arc observed on a screen placed parallel to the reflecting surfaee at a vel)' large distance D from it. p

,,, ,, ,, ,,,

,

,,,

Screen

0

• s A

hi

B

•• c< ••• c C c c
(a) What is 'the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure). (c) If the intensity at point P corresponds to a maximum, calculate the minimum distance through which the reflecting source AB should be shifted so that the intensi ty at P again becomes maximum. 16. A vessel ABeD of 10 em width has two small slitsS t and S2 scaled with identical glass plates of equal thickness. The distance between the slits is 0.8 tnm. POQ is the line perpendicular to the plane AB and passing through 0, the middle point of Sland 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. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. ~ow, a liquid is poured into the vessel and filled upto OQ. The central bright frinp,e IS found to be at Q. Calculate the refractive index of the liquid. .

..

GRB Undmtanding Physics Optics and Modem PhY'ics

256

o

T s,

p

sF~:---: 6---- : •

I

•

2em

•

10em

17. 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 sl.;rfaccs of the la~er and two reflected rays imerfere. Write the condition for their constructIve interference. If!. = 648 om, obtain the least valueofl for which the rays interfere constructively. 18. A coherent parallel beam of microwaves of wavelength I. = 0.5 nun falls on Young's double slit apparatus. The separation between the slits is 1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at n distance of 1.0 m from it, as shown in figure. y

,

d=1.0mm

,-' "

,___~O~=~l~.O~m"-__~ Screen

r

.

(a) If the incident beam falls nonnally 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. 19. In Young's experiment, thc source is red light of wavelength 7)( 10- 7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts by 10- 3 m to the position previously occupied by the 5th bright fringe. Find the thickness of the pl ate. When the source is now changcd to green light of wavelength 5 x 10- 7 m, thc central fringe shifts to a position initially occupied by the 61h 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. 20. In 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 LJ and L z 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

257

Wave Opties distance aI?ng the axi s from S to Lt and ~ is 0 .15 m while that fromLt and L2 100is 1.30 m. screen 31 0 is normat lo SO. (a) If the third inlensity maximum occurs at the poim A on the screen, find the distance GA . (b) If the gap between L1and is reduced from its original value 0.5 nun, will the distance OA increase, decrease. or remain the samc?

The

~

lQI I

:

_~..? mm A T-~:::;;:=----j 0 s

:

4: \ O.ISm

•

1)

-1--

I; Screen

1.3m

Q (separation d) containing radiations of wavelengths 4000A and 5000 A (which are mutually coherent in each wavelength separately) arc incident normally on a prism as shown in figure . The .refractive index of the prism as a function of wavelength is given by the retatlOn,

21. Two parallel beams of light P and

~O,) ~ 1.20 +..£. ).2

where}.. is in A and b is positive constant. The valuc of b is such that the condition for total reflection of the face A C is just satisfied for onc wavelength and is not satisfied for the other. A

p

• sin

e= 0.8

d

90·

a B,LL-----~c' (a) Find the value of b. (b) Find the deviation of the beams transmitted through the face AC. (e) A convcrgentlens is used to bring these transmitted beams into focus. Ifthe intensities of transmission from the faceAC, are 41 and/respectively, find the resultant intensity at the focus. 22. A narrow monochromatic beam oflight of intensity I is incident on a glass plate as shown in figure. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25 per cent of the light incident on it and transmits the remaining. Find the mtio of the minimum and the maximum intensities in the interference pattern fonned by the two beams obtained after one reflection at each plate.

258

GRB Understanding Physics Optics and Modem PhYliCl

1

2

23. A beam of light consisting of two wavelengths, 6500 Aand 5200 A. is used to obtain interference fringes in Young's double slit experiment. (a) Find the distance of third bright fringe on the screen from the central maximum for wavelength 6500 A. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 2mm and Ihe distance between the plane ortne slits and screen is 120 em. 24. In Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen , when a mica sheet of refractive index 1.6 and thickness 1.964 micron is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is Juublcd. 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. c 25. Screen S is illuminated by two point sourees A and B. Another source C sends a par'dllel beam of light towards point P on the screen ( see figure). LineAP is nonnal to the screen and lines AP, BP and CP are in one plane. The • ___ .60:.:...·+~ distance AP, BP and CP are 3 m, 1.5 m and 1.5 m A. P respectively. The radiant powers of sources A and Bare 90 watt and 180 watt respectively. The beam from C is of intensity 2~ watt/m 2 . Calculate the intensity of P on the B S screen. 26. A double slit apparntus is immersed in a liquid of refractive index 1.33. 11 has slit separation of I nun and distance between the plane of slits and screen 1.33 m. The slits are illuminated by a parallel beamoflight whose wavelength in air is 6300A. (a) Calculate the fringe width. (b) One oflhe slits oflhe 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 on the axis.

259

Wave Optics

•

•


Multiple Choice Questions

(A) Only One Choice Is Correct Level-1

I. (c) 6. (c) \I . (b) 16. (d) 2 1. (b) 26. (a) Level-2 1. (d) 6. (a) 11. (b) 16. (c) 21. (b) (B) More Than

2. (d) 7. (b) 12. (a) 17. (d) 22. (b) 27. (d)

3. (a) 8. (a) 13. (d) 18. (b) 23. (a)

4. (c) 9. (c) 14. (c) 19. (a) 24. (a)

4. (a) 3. (b) 2. (c) 9. (a) 8. (b) 7. (d) 14. (b) . 13. (a) \2. (b) 19. (c) 18. (b) 17. (b) 23. (a) 22. (b) One Choice/s is/are Co r ~.ect 4. (a,b) 3. (a,b,c;d) 2. (a,b,c,d) 1. (a.b,d) 9. (c,d) 8. (n,c) 7. (a,c.d) 6. (a.c) 13. (b.d) 12. (a,c) 11. (a,c) M atch the Columns I. a- p,s ; b--p,s ; c--q,r ; d--q,r 2. a--q ; b - r ; c - p,s; d--q,s

3. a- p,s ; b--q ; c-t ; d- r,s,t Comprehension.fI Passage-I : 1. (d)

Passage-Z: 1. (b) Passage-3 : 1. (a) Passage-4 : I. (b) Passagc-5 : 1. (b)

2, (b)

3. (d)

2. (a)

3. (c)

2, (b)

3, (d)

2. (c)

3. (b)

2. (b)

3. (a)

Subjective Problems Leve~ 1 l. (a)5 x I0 14 Hz

2. (a) 9/ 0

(b)4000A (b) 5/ 0

5. (b) 10. (d) IS. (d) 20. (c) 25. (b)

5. (c) 10. (d) I S. (c) 20. (d)

5. (a,c) 10. (a,b)

, -. GRB Understanding Physics Optics and Modem Physics

260

3.16/0.4/0

4.51,71 5.600 nrn, 429 m 6.75 nrn 7.10- 7em

8. (a) p = 5.5 x lO--4m

(bH = 2"

(e)/=/o

3

9.5.66 x 10-4 em

10.1 =410COS2(;) 1I.5000A 12. (a) ~ 1

(b) ~ 1

13. (a)

(b) A = D

, x~L

2D Subjective Problems Level-2 1. (a)q,

d'

=[5+4COs(~7tdSinO)}o (b)J =[5+4COs(2:t)}o

(e) I =910

2. (a) 2 em 3. (a)ID =/0

(d)y = "~D;~ = 0,'2

(c)p = AD

d

(b) 40 (b)4:1

,

4.r(= r=) = U~~:)f 5. (a)" = _1

(b)A = (~ - I)loae

6. (a) ~ = 2t~r~cc2---si~n"; - n

(b) t = (2"2: I)";" = 0,1,2

as

(e) I

= 1;[~-eos(2tJ~' _sin'

i)]

(d)/""

=1:

7.1= 25/ 0 8. (a) = 10[~ -eo{ 4~ sinS )]

1

(e)

y=e"tY':

(b) 1 (d)P =W

2d

=/o(~-eos(~~)1

(e)lcetltre =10

,

2

,

261

Wal'e Optics

.

_ (4d' _ A') (4d' -9A2 ) (4d 2 -25A') 9.x elc SA • 241. • 40, ....

10. (a)[ A: I= (4-.,I3)A and P = lan - t+3.,13) (b)/=/O(4-.,I3)

(c) £ = (4 - .,13)£0 cos[OJ/ + Ian -'( 2 +3.,13)] I1.A="_A(a+b) a(n 12. y(=

1)P

£) = ~29; 4.,13 A cos[OJ/ + tan

-t

+ ;.,13)]

13. (a) d = a(nl - n, )(AI + A,) (b)P =

AD a(lI\ - tl2)(A\ +A2)

14.3.5 nun (b) 1:16 15. (a) Circular 16. (i) 2 cm about point Q on side CD (ii) ~ = 1.0016 17.90 nm 18. (a) ± 0.26 m. ± 1.13 m

(b) 0.26 m. 1.13 m

19. 7x IO-6 m . 1.6, 5.7xIO- sm 20. (a) OA = 10 - 3 m

(b) OA will increase.

(c) See explanation. 15 2 21. (a) b = 8>< 1O- m

(b) 0 =27.20

22.1:49 23. (a) 1.17xlO- 3m 24. 5892 A

25. 13.97 W/m' 26. (a) 6.3 xlO-4m

(c) 300 nm

(c) 91

262

GRB Undtntanding Ph)'1ics Optics and Modrm Physics

1I::::::::::::::~g:!H~in~ts~R~n~d~S~O~I~u~tEio~n~i}~::::::::::~II Comprehensions

Passage·2 1. The optical path difference = 6x

=

dsin¢l +dsin9 -{J.l-I)t.

For central maxima puning /)..r: = O. we have

e

· (~-I)1 SIn =

.... = (3/2 - 1)(OJ) - sm . 30' sm,!,

d or

50xlO 3

sinG=! or9=30°

2

2. The order of minima is

rue = dsin~ -

n=-

A

(~-I)I

A

(50 x 10- 3

G) -

(3/2 - 1)(0.1)

50ooxi0 10

= -50

Hence, at c there will be a maxima. It means, closest to c the order of minima is

49. 3. No. offringcs shifted up is N = (~-I)1 = (3/2 -1)(0.1) = 100 A 5000 x 10- 10 Subjective Problems Leve/·1

f

I. (a)

=.£. =

1.0

3 x 10' 6000 xlO- 1O

=0.5 )( 10 15 =5x 10 14 Hz (b)

(e)

,

_ A,;, _ 6000 _ 4000A - -- n 1.5

""glass - -

, v=f=)x IO :2xI0 8 m/s 11

2. (a)

(b)

I"," =

1.5

(Fa +~410)2 =910

I = 10 + 410 = 510

Wave Oplics

3.

263

1",, = (Fo+.j9/ 0 )' = 16/0 1=, =(.j910-Fo) ' = 410

Ip = 1 +41 + 2Ji · .J4icos90o

4.

= 51

I Q = I + 41 + 2..[1 . .J4i c09,360' -300') = 71

5. For the light of wavelength ).. to be strongly reflected

2~1 = (2" + I)~

/.. =

4!lt . Now putting" = 1,2,3 211+1 We get, ).. = 600 run and 429 nm 3rc strongly reflected in region (400 nm - 700

Thereforc, nm)

6. For constructive interferencc, putting n = 0 in the equation

2~tt

=

(211+1) ~ ,wchave 2

2~Lt =?: 2

t . = 1- = 4(0)( 10nun 4J.t 4 )( 4/3

8. (a)

9

= 7Snm

P= AD d

=

(550 x 10-9 m)(2.2 m)

(2.2xI0 3 m ) = S.sxlO

(b)

~

-4

m

=2·). x = 2lt dsinS ).

~2'd Y

). D

=~'~m

o

, CRB Undmtanding Physia Op,~ and Mod~m Physics

=2'ADd(W) Jd 2,

o-

J (c)

9. or

or

1=10 +10 +2J/olocos2;r; = 10 3 D(~_I)I=5W d d ()l -1)1 = SA S)

l = ~ =

~

10.

5 )(5R90A

-I

1.52 -1

=5 .66)(10

-t

'.

I:: 10 + 10 +2J/o/o cos~

= 4/0 COS--. 2 Where

.... 21t . J sinO""2'lt d Y

Then

iUIl' 4/ cos 1 =410 c05 2 W= o

\2. (a) 01

A

A D

I] 4 AI 2 - = - or - = 12 I A2 I Al + A., 2 + 1 ~A~1-_-CA"'; :: 2---1

.~ 1 (b)

IITfIJI. :: lmin

(AIAI +- Az )' A2

=-9 13. (a) The path difference is

x = S2 P -SI P =JD'+d'-D

2( P~,)

em

Wave Optics

265

(b)

or or

Subjective Problems Level·2 1. (a) The resulting intensity is

I

I, + 12 +~ cos~

=

,,

dI2{

" ,

=10 + 410 +2xJlo x 4/0cos ~

/ = (5 + 4cos ~)/ o The phase difference between the waves is

$ or

~

= 2".6.>-

A

= 21tdsinO

.. .(ii)

).

Substituting $ from eqn. (ii) in cqn.(i)

$ (b)

(d)

= [5+4Co{~"dsine )}o

Putting sin a == Ian a = Y, we have D 1

(c)

= [5+4cof~)}0

Aty=O, 1 = (5+4)/0 = 9/0

cos2~ = 1 or

. .. (i)

y=

n;::n

=0,1;1.

266

GRB UndentAnding Ph)"io Optics and Modem Physics

(e) The fringe width

P= Yn

)J)

-

)'n- i

=d

2. (a) The reflected lights at the ends C and D oflhe mirror will hit the screen at the points PI and P2 respectively. The superposition of the reflected rays and direct rays from S will take place between PI and P2 . Then, the number of fringes lying in this region is

P,

s

P,

,,

,,, , ,, ,

--------T-------d B, ;' '~

, I

,, ...... ;' .... II

--

; "

"

s·~ :

'"

0

.-",,"

.. .. .. "

_________ J!_____________

A

N = P]P2

P,

where p = fringe width = ').J)

.

d

Hence,

N = (P,P2 )d )J)

... (i)

P,P2 = AP2 - API (AS ' ) . 8S ,(AS')8S' = _ _. ~~:::.. 8D BC = (190+5+5)(0.1) _ (I90+5+5){0.1) 5 10 "" 4cm - 2cm = 2cm 2 (b) Putting P,P2 = 2 x 1O- m, d == 2mm = 2 )( 1O-3m, D = (190 + 5 + 5)cm = 200 cm=2m, ). = 5000 x 10- 10 m in eqn. (i), we have ' . N_(2 x I0 -'m)(2 x IO- 3 m) =40 (5000 x lO- JO m) x 2

267

Wan: Optics

3. (a) The path difference between the rays at C = Then, the phase difference

ru: = 0

= $ =21t ru: = 0 ).

Hence, the resulting intensity at Cis

D

fe = 1\ +/2 +2.JI\f2 eosq. A

Ie = 10 +4/0 + 2.j/o (4/0)cosOO Ie = 9/0 Similarly, at D, the path difference between two waves is

ru:' =

d

_-.L

:Co ------

... B

dsin9::::: dtan9 = dy D

=~(~; )=~ Hence, the phase difference is

f

=

2•. 6.<' = 2•. 1, = •

). ). 2 Then, the resulting intensity at D is ID =/\ +/2 +2J!;I2cos$ = /0 +4/0 +2.Jl0 4/ 0cos1t = 10

(b) Now, the slits Cand Dact as two different monochromatic sourcesoflight of intensities Ie = 910 and f D = 10 respectively. The lights emitted from the slits will get superimposed on the screen. The maximum intensity oflight on the screen is

1=,

= (f/C

+.J1D)'

= (.}910 +Fo)2 = 161 0 The minimum intensity of light on the screen is 1",,0 =

(..[1; _.J1D)2

= (.}910 -Fo)2 = 4/ 0

Hence, the ratio of I rrnx and I min is Irrnx 16/ 0 --=-4/ 0 lmin =4: 1

GRB Undentanding Physics Optics and Modem Physics

268

4. To gel focused at a point F. the rays must travel same optical path. Hence, optical J I path ofray2 = optical path or ray I __ I X I x+nPF = nOF 1 ----I'-"-;,;.':e

x+tI~r2+(f-x 2 ) = nf

or

1f~ r2 + (f -

x )2 = II/ - x

ri,

2

or (n+l)(n-l)x2 -2n(n-l)!x+n 2r 2 =0

or

----,>k--+'\_--;-::-~r-f-x-

2 [1+ 1 _ (n+1)r ] 11+1 (11 - 1)/2

x= n!

... (i)

If r is maximum, for any value of X, eJr/ dx = 0

or (n + 1)(11-1)2x dx _ 2n(n - 1)/ dx + 2n2 r dr = 0 dx

or

dx

dn

x ='if -II + 1

... (ii)

Putting x from eqn. (ii) in eqn. (i),

'if = nf [1±1 _ {n+I)Y'] 11+1

11+1

r{= r=) = (

or

(n-I)J2

rn=I)r

V;+'

5. (a) The optical path difference between the rays is Ar = (jllo -10) Ar = {jl-I)lo or For 11 order maxima,

ax = I1A or

(jl -1)10 =

n).

... (i) If the temperature increases by SoC, the length of the glass slab will be 1= 10 (I + as). Then, the optical path difference between the rays is Ar' = {jl - I)I =
For next maxima,

6x' = (n + I»). or (jl-I)lo(l + a9) = (,,+I»). Eqn.{ii) + eqn. (i),

1+a9=,, + 1 n or

,,9 = -I n

... (ii)

t

269

Wave Optics

0' (b)

0'

" "" _I

aO

PUlling 1/ = -' in eqn ,(i), we have

aO (~ - I)/o =_1 .1aO

). = (~-I)/oaO

6. (a) Since there is no phase change at Band C (because 112 > 111 and 112 > 113 ) and ray' is ahead of the ray 2 by a phase n due to the reflection at A. the lotal phase : 1 difference between the rays (as derived earlier in 2 theory section) is :, tfJ = 21lt cos r - 1t = 2v,rJrl- -si-n"-, - n

(.:cosr = "r -·sm.,'--,)

,

l _-

," B

(':v, sinr=sini)

0' (b)

tP

=2r~r~"'-_-~-'-·-n";-i - n

For nonnal incidence. puning ; = 0. we have ~ =~I-n

+

Then, for maxima to occur, = 2mt' or 2v,t - 1t' = 2mt - 012 or t __ (21l+1)n ,, n " 2~

(c) Since 50% energy gets reflected at A, B, and C. The intensities of the rays I and 2 will be 10 10 11='2 and 12=8 Then, the resulting intensity is 1 =1, +1, +z..[T,T;cos~ 10 10 = "'2 +g +2

VoTo "2'gcos*

f

= 10

2

+ 10 + 10 costP 8

2

=1; (~+cos+ )

..

270

GRB Under.;tanding Physics Optics and Modem Physici Putting, = 2/JJ.12 -sin 2 i -1t, we have

1 I; [~-COs(2t~r~,,",---Si~n2'i)J =

(d) The minimum intensity is obtained by putting cosq, = -I lmin = 10

8

(e) Condition for minima is ~ = (2n+I). For nonnal incidence

q, = 21J.l -1t' Then. 2tJ.l -

or

11:

t

= (2n

+ I)n

=(1I+1)n ~

7. The optical path difference between the rays I and 2 is Ax = ,,(DC) - AD f = n(ACsinO) _ ACcosr' Putting r' = 9O o_r, we have

•

At- = .1.C(nsinO - sin.6.) Since IIsinO = sin r (Snell's law)

-- ", A

-

, B

,0

We have tu = 0 Hence, the phase difference between the

0

,

C

P

rays 15 ~ =0

Then,

I = /1 + J 2 + 2.,fiIi2 cos 4>

=1 1 +/2

+2.,fi1i2

= (,[I. +,[1;)'

=(-J4/ 0 +-J9/ o)' = 25/ 0 8. (a) One direct wave goes with intensity 10 . The reflected wave has intensity

I;.

Both

the waves superimpose at P. The resulting

intensity is

10 10 =10 ++2·}0 ·-cosq,

2

2

SI ~

_/-/-../----:::;;;:71

L'

S,

,

--

.

P

Wave Optics

271

I

=/oG+/oeos~)

... (i)

The phase difference ~ is given as

... = 2lt.:\x +7t , 't'}.. \~here I1x = ~ath difference == extra phase dIfference n anses due to reflection of light and at the mirror.

Then q, =

4~ sin 8 + n

. ..(ii)

2.

By using eqns. (0 and (ii),

I =

s, _

~;--~ ;8,..(

lo[~ -10 co{ 4~ sinO )] a = Y,

(b) Putting sinS::: tan

D

1= /O[~-/O'O{:i)] (c) For maxima, cos 4trdy =-1 )J)

or

4rrdy = (2n + I)" )J)

y=en4+1)~

or (d) The fringe width is

f} = Yn - Yn- 1

= ~[(2n + I) - {2(n -1)+ I}J =)J)

2d (e) At the center, put S = 0 to obtain q, = 1t Then, the minima takes place at the center. Hence,

I,

I center =-

2

9. The waves I and 2 will superimpose at P to give the minimum intensity at P. For this to occur, the path difference between the waves is

ax =QP -op =(2n+I)~ 2

GRB Undenl.anding Ph)'1ics Optics and Modem Physia

272

i

a

IIp ---,--;~..o...=.-:?--=-"

i

..._ _ ,

2

,

(~d2+x2 r = [(2n+ l)~ +-tr

or

or

2 d +x2

or

x

=[(2n + l)~r +x2 +(2n+l»)..x

=(

2 d _(2" + 1)2"-: } (211+1)A

Putting x = O,u. ...... we have (4<1' _ I,') (4.1' -91,' ) (4d' -25A' ) x=

8A

'

241..

•

40)"

r

I I

... etc.

10. (a) The phase difference between waves I and 2 is

I

$1 = 2•. ~ I,

I

= 2. , ~

I

). 6 = n/3

I

The phase difference between waves 2 and 3 is

.2

1

= 2n . ,6x ).

:;:; 2n .?=: ). 4

.

A~A A

= !!"

;

2

A_,

By using component method, ~

~

~

Ar = Arz + Ary

.

~

Ar = (A + .if cos60o-Acos300)i + A(cos30o+sin300)j

or

-; = A(3-J3}i+(J3+I}Ai ,

2

2

or

•

273

Wa\'e Optics

and

p = tan -

I(2+J3) 3

(b) The resulting intensity is 1 ==

KA;

= KA ' (4-.Jl) ·

P uttmg KA

,

= 10 , we have

I = /o(4-.j3) (c) The resulting vibration is E

=EOCOs(OlI + P) = (4 - .Jl)Eo cos[001+ Ian

-t

2+3.Jl)]

I!. The object 0 has images II and 12 due to the refraction at two slant surface of the prism. The deviation 8 of the ray is 0 = (0-1)6 Then, d = ao + ao or d = 2a(1I -1)8 I, Ii, Hence, the fringe width is d

;;

t-.:;- 1\

p=W d P=). (a+b) 2a(n - 1)8

or

e=

1.O~ (:1----I,

I- a -

•

•

b

A(a + b) 2a(0 -I)P

Hence, the angle of prism is A = IT-29 A(a+b)

= n - "0(70'-_"\)"'P 12. The amplitude phases can be added vertically. The resulting amplitude is

A:. = (A +2ACOS60)i, + (~ +2Asin60 or

A, dAi + ,i(l + 2.Jl)j 2

y

2A

A

l

90° 60°

A

or L-!.:"":""'-_ .'

I

A,

274

GRB Undmtanding Physics Optics and Mode.m Ph)'lics A,=J29 +4.Jl A

or

2

$ = Ian - , A12(1 +2.Jl) 2A =

tan-Ie+~~ )

Then. the equation of the resulting wave is ~= £)

=

A,cos(",,+~)

=

J29 2

+4Ji A cosoor+an [ I -'(1+2.Jl)1 4 iJ

13. (a) The net deviation produced by the ray I is

/)' = 5\ +02

I,

d,

= (II] -1)A[ - (112 -J)Aj

f--------

A

= (111 - t'2 )A]

. Similarly the deviation of thc ray 2 is o~

., {

=(n ] - n:dAz

The distance between the images 1\ andJz is d = d,' + d z

.0

' \0--------8_

,

=a5' +ao~ =a[(n] - liZ )A] + (III -112 )Az] =a(lI\ - 1/2 )(AI + Az)

(b) The fringe width is

P= AD d

14. IfPlh maxima of A/ coincide with qlhmaxima ofA 2.

Then

•

pp, = qP, Plq=p, /P, =<,/)., =~~=~

Least-distance from centre = 7p\ ,::; SP2 = 7)( D~ 1 9 3 Distance = 7 x 500 x 10- x 10 1

=3.5 rom

"'-------'

Wa\~

O ptics

275

15. (a) Fringes will be circular, (b)

I:: =[.Jll.M +-.J0.} F, ' I ·

Now. intensity of light direct from source = 10 intensity of light after refraction = 0.3610

I min 1=

[Fo Fo

=

JO.36/ 0 }' + JO. 36/ 0

=1. 16

(c) For the same intensity. additional path difference must be there so that minima occurs. If minimum shift of reflecti ng surface is required, then minimum additional path difference to be created = (/.). Thus, path difference will increase by 2(y) if reflect or is shifted by (x). Thus.

2l = lX = 1. /2 = 600 nm 12 = 300 om \6. (0)

c

S]S2 = 0.8 mm,

tan a

= 40cm =! 200 em

.

I I =J26 5.1 5 taoO =(y I O. I)

..

.- ---

I

For central maxima. optical path difference = O. (d) sinO = d sina sinO=sina sinO= tana )' I • -=-

0.1 5

y=.l m = 2cm

50

5

,R

5,

5

sm a = - = -

Also

..

P

y

/0

----- .;y 61---.40 em , .. " 5

':'=• ' ====Jr~5~2~==~• C 2cm

10cm

GRB Undmtanding Phytics Optics and Modem Physics

276

(b) Now the liquid of refractive index (~l) is poured uplC OQ level. Thus, if it is given that at

Thus,

Q. the optical path difference is zero.

(SSI-SS2)=[S2QJ[~I - S,Q

d sina = (j1-I)(IOcm) O.l6mm=(~-I)(IOOmm)

... (i)

From cqn. (i). ~

=1.0016

17. At point A, path difference oD. / 2 will be created but at B, no path difference will ex.ist. Thus. writing the condition of constructive interference,

I {

2~t+~=n')., 2

2J1t

or

=[,,-~}

for minimum t put" = I

2~t =~ 2

t = -I. = I. _)( 1.8 = 648 - mn

or

4~

4

7.2

1=90n01 18. (a) When beam falls normally the path difference at p,

(A) = dsin9

____ :;~.~I ---------::-:----?------~lPy I .-r-'"-'-'~·~"..."'-------_j

____J. , d

M

s,,1 For minima,

11

= (2n -1»)../2

dsin9 = (2n -1)1. / 2

d = Imm;1. =0.5 mm sinS = (2n-I)(0.5) 2x (I)

0

,

2n

Wave Optics

sinO

=

(211-t(OiS)=2114-1

sin e S I => Hence Ilean be I or 2. so if 11 = l,sin9 1 = 1/ 4 n=2;5ina=3 14 and if tana, =1 1,,!i5andtana, =3 1./7 so,

Now from centre '0' location of minim as,

Upside: YI

=(Imm) x ~;y, =(I mm)x ~m viS .

Y1=0.26m Downside: Similarly, Yi = - 0.26 m

v7

y,=-1.I1m

and Y2 = 1.13 m Thus four minimas at locations . p

y

5,

y

GRB Unde~tanding Physics Optics and Modem Physics

(b) When the incident beam makes an angle 0[30° with x·axis. For central maxi ma. optical path di fference must be zero. A, =0::::::> a = 0=30° :. Location of centm! maxima::: Ix Ian 30°= 0.58 m And for fi rst minima dsinO-d sinu =). / 2 dsinO = d sinu +). / 2 sin a = sin a + '). / 2d III =dsin O- dsina 6.2 = d sino. + dsinO

or

= 5;n 300+ 0.5 nun

2xlmm

= 0.5 +0.25 = 0.75 = 3 / 4 Y = (D)tan9 = I dl.fi = 1.13 m For first minima, on either side

t!sinO = ).. / 2 sin e = , .. 12d = O. S nun

2x lrrun

~

=!

4

tan 8 = ~ = 0.26 m

" 15 y= l x tan O=O.26 m Therefore, minima on either side will be at )' = 0.26 m and y = 1.13 m. 19. When red light is used: Shit\(t.) = (II-I)/D = (5P) _ (0) = 5x Dk d d Thus,

... (i)

(1.5 -Ill/liD] = 5 x (D] x 7 x 10- 7 (d) (d]

! =35x I0- 7 2

t = 7 x 1O-6metre

Now when green light is used:

Shit\(t.)=(~ g -I)/D = 6xO)o, d

d

(I-Lg _1) (7xlO-6 =6)( 7)(10-7 I-Lg -I = 0.6

J.1g = 1.6

.. . (ii)

•

279

Wa\'e Optics

Change in fringe widlh due 10 change of lighl : ~,.. = m.• / d, P",,,,, = DA g / d

6~ = ~[I. •

-I'G]

6P=~[7X\o-' - 5x\O-'] It is given that when red light is used with plate insertion, shift occurred was 10metre of ccntral maxima. Thus, 5x DAR =\0- 3 d (D / d)

=

\0-3 5XAR

=

\0-3 5x700xlO 9

20. (a) Each Icns will form the image. Lens L\ at II and Icns L2 at 12 , Thus

,, "

:

Il

........

'

----- :1

I 0.5

1 .... "" I ~""'---------- lI

-- A :0.5 0

s....................

.

0.'

o

,

•

0.15 m

1-! =1 v

II

f

~ - _0 15 = 0\ =>

}

1

I or

m

1 1 =(0 1)-(0. 15)

v = 0.3 metre

· I = -v =0.3 =- - =-2 Th emagm·fiIcatlonm o u -0.15

•

3 Thus. size of image = 05 x 10- x 2 =0.5 mm 2 1,/2 = 05 + 0.5+.05 = 1.5 mm Thus. = 1.5 x 10- 3 metre

Now, /1 and 12 will out as sources for YDSE similar arrangement. Thus D = 1.30 - 0.3= 1 metre

3

GRB Undmtanding Physics Optics and Mod~m Physics

280

Wavelength of the light used = 500 x10- 9 metre Thus, from '0' the distance of third maxima = 3P

=3x

[DI-] = 3 ) d

1x 500x 10-'\ 1.5xW-3

J

L

= I x lQ-3 m = Imm

~ =D)..

(b)

d

As the gap L\L2 is reduced, d will decrease thus it will increase p. Hence, the distance OA will increase if the gap of L,L2 is decreased , 21. (a) Given sin 0 = 0.8 = 415

..

0=53"

~ = 1.2 +.£.. and critical angle Be

sin - 1 (1 / ",)

=

).,

Critical angle will be smaller for 4000 A. b b Thus,~= I.2 + -=I.2+

).,

Hence, 1.2+

.(1,) .~

,

~=~=--'-= _ I =1.25

also

:.

(4000,1,)

sinG e

sine

b (4000 A)'

1.25

0.8

005= b . (4000 x I0- 10 )2 2 b=8xI0- lS m

5000 A wavelenglh beam will strike the force AC. Thus, apply Snell's law ).1

=s~n ~ Sin ,

281 1.2+

b =_s,"_r 1O 0.8 (5OO0xI0 - )'

sin r

=> By lables

= 0.9856

r = 80.3°

Ii ~ [L r- LiJ ~ [80.3 "-53.I " J ~ 27.2"

(e)

I""" ~(jIj +,{0)' ~ [.,f4i + JJJ' ~ 91

1=,

22.

--~

I",n

Thus, and

(jIj +,{0)' (jIj - ,{0)'

91f64

1164 31/16

1,=1 1 4, /2=91 164

91/ 16

114

~: ~(9)

31/4

23. (a) Distance of third bright-fringe

6)~ ~3 p~30)' d 3)(6500)(10- 10 x 120 >< 10-2

2 x 10 3 6Y3 = 1.I7x 1O- 3metrc

(b) Distance where bright fringes due to both the wavelength coincide. Let

maxima of;' , and q Ih maxima oft. 2 are coinciding. Then

p.p,

qP, E ~ (P, 1 P') ~)., ~

q

lA, ~ 4 / 5

Thus, least distance will be 4th bright fringe of 6500 A. Hence,

2

.1)' = 4 x J20 x J0- x 6500 x J0-

2 x 10 3 6y ~ 1.56 x 10- 3 metre

24. Shift in fringe pattern L\x ~ (~ -1)10

d and new fringe width P = ..

).(~)

L\x~p

(~ -1)10 = '})J)

d

d

1

plh

282

GRB Undentanding Physics Oplia and Modem Physia

Hence.

6 ). = (1.6 -1)(1.964 x 10- ) = 5892 A

2 25. Intensity at point P

=fA

Ip

+ 18 + I c

= [ Radiant power x cos 0]

4,.,.' = 90 x cos Oo= (O.79)W / m 2 4rr(3) '

fA

18 = 180 -cos60° = 3.18 W/m 2 4rr(1.5) '

,

I c= IOW / m2

.. /p = O.79 + 3. 18+1O=13.97W / m 2 26.

=1.33

~L

d = l x \O- 3m

D = 1.33 metre A =6300 x 1O- IOm

p = Dp. 1~d = 1.33 x6300x 10- 10

(a) Now

d

1.33 x l x lO- 3

P = 6.3x I0--4 metre

(b) ~ g =1.53 To bring adjacent minima on the axis a path difference (ll, - 1)1 should be created such that shift becomes equnllo 1\ / 2

Th US,

Shift (6) = {J1, - I),D = ~ = 6.3 x 10--4 d 2 2

~

Hence,

= J.1 g r

J.1 L

=[1.53] 1.33

1.53 _ I]x,XI.33 4 [1.33 = 6.3xI0I x 10 3

2

--4 10-3 =1.575xI0-6 1=6.3)(10 xi x m 2xO.20

Particle and Wave Nature of Radiations (Light)

3.1

Maxwell's Laws of Electromagnetic Induction

3.2

Electromagnetic Waves

3.3 3.4

Black Body Radiation : Planck's Law

3.5

X-rays

3.6 3.7

X-ray Diffraction: Bragg's Lnw Mosley's Law and Characteristic X-rays

3.B

Einstein' s Photon Concept and Dual Nature ,o f Light

.

Photoelectric Effect

o.

3.1

-

~

Maxwell's Laws of Electromagnetic Induction

According to Faraday'law of electromagnetic induction. a time varying magnetic field induces an cmfby generating an electric field in space. Clerk Maxwell expected a reverse phenomena on the basis of symmetry of nature. He proposed that. A time va in electric field can roduce a rna netic field in free 5 ace.

Thi s is known as Maxwell's law of electromagnetism just a reverse statement of

Faraday's law of electromagnetic induction . Maxwell got this idea while analysing the current flow in R-C circu its. In R-C circuits. really a current flows even ifthere is no conducting path in between the plates of the capac itor. This surprised Maxwell. He tried to find its cause by a revolutionary concept of "displacement current" which he imagined to flow inside the ~

~

capacitor so as to prove the continuity of both current and fields (£ and B). According

. ..: .;"; '-'-3+=-1Bq~~ -q3\. Both current and" fields {E andB) must be continuous

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284

to Maxwell, the conduction current ic in the wire must be equal to the displacement current id inside the capacitor .that is, ic = id, the name displacement was borrowed ~

from the displacement vector D (related to the displacement of charges between from the pates due to the process of electromagnetic induction). As the current is transient,

q dt

the charge on the capacitor plates changes at a rale of dQ . Then, ic = d , where q = CV.

dt

This gives j

or

c

=d.c.V=d ( "OA. Ed ) drd

dt

liC =;d=COA~ 1

.

Hence, a displacement current, in essence, is a time varying electric field. It 15 not associated with the motion of any charge particles. It is not a conduction current, but equal to conduction current to make the flow of current continuous at the plates. Then, by using the concept of displacement current Maxwell proved that the net flux of magnetic field passing through a surface enclosing each plate of the capacitor is zero to obcy the solenoidal propeny of magnetic field , given as ~

~

rB·dA=O Closed surface

ic=id to make the current continuous

In other words, the magnetic field produced by the conduction current must be equal to that produccd by the displacement current, to satisfy the continuity of magnetic field flux. Hence, we can conclude that, a displacement current has the ability of producing a magnetic field like conduction current. Since a displacement current is ckack with a time varying electric field, we can say that A lime va in electric field can roduce a rna netic field It is reverse phenomena of Faraday's law of induction.

_ "!J'. __ , ~u · 1~1

If the displacement current density inside a ca'Pacitor is 10- 3Alm 2. Find the (i) conduction current density inside the wires, (ii) rate of variation of electric field with time.

Solution: (i) According to continuity of current, J c = Jd = 1O-3 A/m 2

Particle and Wa~ Nature of Radiations (Light)

285

(ii) = 1.13 X IO 'N/C-s

3.2

Electromagnetic Waves Maxwell put all the following basic equations together. -> -> q (i) E· d A = ....E!.(Gauss' law of electrostatics) EO

f

~

~

(ii)

tB.d A =0 (Gauss'llawofmagnetism)

(i i i)

f B· d / = ~ 0 {i + id )(Ampere' s law modified

->

~

by Maxwell's law of induction)

=~O(i+EOd~I£).Whcre $£ =rE.i:4 (iv)

~

~

~B

~

J

~

E·d I =-dt,where$8 = B·d A(Faraday'slawofinduction)

Solving these equations (as shown following differential equations

In

the example) Maxwell obtained the

r-:;----;-, a'E

a'E

dt 2 = £0110 dX 2

a'B

and

at

2 =

I

a'B

£of.lo dX 2

These equations looked similar to the mechanical transverse wave equations.

a' y =[

ax'

~

a' y ="' a' y

ax'

ax'

(.:

"=

f!.)

V~

Then, Maxwell speculated the existence of same sort of disturbance of electric ->

~

field E and magnetic field B. He called it electromagnetic wave (emw). Solving the differential equations of E and B. each case he found the wave moving with a speed

v=

I = I =3xlO'mls .jEW 0 .j8.85 X10-12 x40xlO-1

It is the speed of light, known at that time! lofact this was the most remarkable achievement to fuse radiation (light), electricity and magnetism. in 19th century, according to Albert Einstein. Hence, Maxwell concluded that, All radiations (including visible light) are electromagnetic waves, moving in free space with speed of light. They are originated by the oscillating or accelerating charQes.

GRB Understanding Physics Optics and Modern Physics

286

'l.u~u 2

How dOCJ af! oScillati!lg or accelerating charge prodllce electromaglletic l\'Oves? Solution: l et us accelerate (or oscillate) a charge +q at O. It is a equivalent to a ti me ~

varying current. It produces a time varying magnetic field B _ C at P very ncar 10 0, obeying Ampere' s circuital law. Thi s J ~ . E;. S; lime varying magnetic field HI produces a time varying G electric fie ld E 1(obeying Faraday's law of induct) not at +1 ~ P A same point P, but a little ahead of P at Q. say. This time varying electric fie ld EI produces a time varying magnetic The pulse o/ Iime v~ryi.ng electric and magnolJC fields field 8 2 j ust ahead of Q, at R say. In this way the travel as acombined unil in .

.....

.....

the form 01 emw with a speed

disturbance (time varyi ng fields E and B ) travels with a U::: 3)<108 rrv's in free space. specd of light v = 3 x IO !! mls. In cmw, one fi eld produces . . whcn the other changes not at the same placc o f the fi rst fi eld but j ust ahead of It. ThiS constitutes the propagation of emw, The E- and B- fi elds are no more directly linked with the osci llating (or accelerating) charge, but are caused by it ultim3tcly. Thus, An accelerated char e reduces an electroma netic field,

3.3

Black Body Radiation: Planck's Law

Impericallormula The spcctroscopists in last half nineteenth century gave the black body radiation curve experimentally 3S the monochromatic radiation

energy

density

If ( =

~; )

verses

frequency. It means, If - lor I A- A can represcnt the Black body radiation curve, The biggest challenge was to fi nd the equation of the curve, proposed the cquation using Wein thennodynamics, Maxwell's equations and Boltzmann's statistics, given as

.PL

3 - AT

I,

1 ' ....

, '....

The radiation energy at f, is 11~ decrease when f increases to Infinity or dec.eases to zero,

'/ = AI e This did not match (fit) with the experimental curve at large wavelengths (or low frequencies) Raleigh-Jean gave a fonnula which did not match the radiation curve at lower wavelengths or higher frequencics. Max Planck, in 1900 patched up these two laws at high and low frequencies to derive an empirical formula given as,

287

Particle and Wa\'e Nature of Radiations (Light)

I

_ CI,..-s

}. - c--1_1 ' e }·T

where CI I C2are the empirical fitting constants; C 1 ::: 81tc2 h and C2 = Ire as . k derived later on by Planck . This fonnula sati sfied all points of the experimental black body radiation curve. The real challenge was to derive an exact equation with proper physical interpretation of the fonnula.

Physical Interpretation 01 black body radiation For this purpose, Planck imagined tiny oscillators (called Hertzian oscillators) at the surface of the cavity. The frequency of oscillation of the oscillators mnges from zero to infinite. The cavity surface was imagined as a reflecting surface because the emw radiated by each osc illator get absorbed by the others and radiated by them at the wall of the cavity. This generates a standing electromagnetic waves. The inside temperature of the cavity remains constant in themlal equilibrium. Hence, the energy(emw) radiated by the Electro-magnetic oscillators must be equal to the energy Hertzian oscillations stationary waves the OSCillators are in absorbed (and radiated) by them. In other equilibrium with the in the cavity words, the energy density of the radiators is radiation (emw) emitted equal to the energy densiry of the radiation by them in the cavity field inside the cavity. At any frequency. the monochromatic energy density of the radiation in the cavity is 81[2 / 2 _

If

=

c3

1/

I,

obtained by Planck by using (classical physics) Maxwell's theory of electromagnetism. The real problem was to find ii, that is average energy of an oscillators averaged over all oscillators of same frequency! According to Boltzmann's law of equipart it ion --1L--:!--------:!~-_r of c:nergy, any oscillator can have any amount of /dr energy, that is, a continuous distribution of energy High rrequcncy Low frequency among all oscillators of same frequency. This leads to an average energy per oscillator AI both low and high f~eney dl - ]r df _ O given as

( 6f

ii =

I

;

I:

Ee

EfKT

dE I

I:

e

EfA7

dE = A,7,

which does not depend on the frequency of oscillation. If wc accept ii = KT. we will get the energy radiated at frequency/between frequency range df, given as

GRB Und~nlanding Physics Optics and Modem Physics

288

dl = If

dl = 8./' --u ::: O,whcn f

c'

-t O.

It satisfies the experimental curve because the area orlbe thin strip dA(= I fd/)

tends to zero at low frequency. Let us now apply Ii := A.T for high frequency oscillators. Since the number of modes at a frequency/oftbe standing wave If 8rrj2 that is, 8rr/2 ;c 2 increases, til;=:. - ,-A7 ,, , c ,, will tend to be infinite (which is described

as a ultraviolet catcstrophc as a misnomer of incredible energy). Practically we do not r get incredible energy from a heated object and the high frequency radiations like X·rays and yrays (generally). Furthermore, area of the strip ::;; dA = If dl Allow frequency, J f obeys classical equipartUion tends to zero at high frequency if -t 00). theorem. However at higher frequency If -+ <0 " phYSlca " 11 y sIgn! ""fi h . 81[/2 according 10 equipartition of energy Th IS les t aI, since - -

c'

(the number of modes) increases with frequency, Ii must decrease dominantly with frequency so that Ii -) 0 at f -) 00 to ensure dI -) 0 iff -) 00. Ii tends to zero at high frequency is possible when cOnlinuous (infinitesimal) distribution of energy (according to equipartition theorem) among the oscillators ofa given energy frequency is modified into discrete (finite) energy distribution of energy among the oscillators. This original idea of discrete (or quantum) distribution of energy of Max Planck led to the foundation of quantum physics (even though the presence of discrete energy levels was first proposed by using statistics by Boltzmann in 1857). Hence Max Planck credited his success to Boltzmann.

Planck law of quantisatlon of energy of oscillators

",

To derive the expression ofil Planck established a mathematical node (whose for reaching physical consequence were not known to him at the - - - - - - - - - - - -beginning) that any oscillator of frequency f can posses \, discrete (but not continuous) energy as an integral multiple 3l1.E of a finite energy 6.E, say. It means an oscillator can have 2.1.E energy 0, l'lE.2l'lE.3U ...... and so on, but not in between them. The probability of getting an oscillator having high or energy is lesser (according to Boltzmann's function) given DoE as p = ce - D.£IKT. Then very few oscillators can have higher value of liE. This decreases the average energy pcr oscillator. In consequence the value of Ii drops to zero at Oscillator of frequency f high frequencies. This is equivalent to the case that, if the cut quantised, 0, 610. 26E, etc off (pass) mark of students will be high few student will where 6E = hf score above pass mark and hence the average mark per

°

•

Partid~ and Wave Nature or Radiations (Light)

289

student will decrease. Hence, the oscillators of high frequencies can emil high frequency radiations (emw) and must possess the internal multiple of high quantum of energy 11E. Then we can write the simplest relation between AE and/as 6.£ oc

f

Introducing a constant of proportionality h(= 6.3 x 10- 34 ) known as Planck's universal constants, we have 16.£-lif 1 Hence, we can express the Planck's idea of quantisation of energy of an oscillator as, Any oscillating system of frequency f must posses energy of oscillation which in equal to the integral multiple of hf, Eosc = nhl;n = 0.1,2,3....... laler on Planck proved that the zero-point (minimum) energy of an oscillator is E:: 1f2hf . Hence. E ase = (n + 1/2)hf in general.

By using the revolutionary idea of energy quantisation for the oscillators, Planck could derive the correct expressionrfo"r-",-ta",s'-w~_-,

lif u = ~*­ eli/ItT -1 Putting the value ofil in the expression

S.f'

1/ = --'ii, we have

c'

"/ df

=

dl=

S~[' ( ~

lc

)df

KT - 1

Puttingldfl= ~ d').. (0: A/ = c), we have 12 dl =1 ).d1- =SncS I-

h'l (

-

}

heI

,

CUT -1

Comparing it with the empirical fonnula

h =~JleArd-1)I,

we have

,

he

C! =8n:c lJand C'2 = k'

as narrated without proof at the beginning of this section.

GRB Understanding Physics Optics and Mod~m Physics

290

I! c 1tA!'!1!.U 3

14 " tI t lllree energIes . An oscil/ator /tas frequ encyf = 10Hz. Fmd ,e fIrst t/tat the oscillator can have. Solution: The energy of an oscillator is given as E = IIlif 14 = ,,(6.63 x 10- 34 J _5)(10 ' 5) = (6.63 x 10- 20 )" J

20 Putting n = 0, 1,2 ctc. the energies of the oscillator are, 0, 6.63 x 20- J, 13.26 x I0- 20 J. Ans.

3.4

Photoelectric Effect

Definition and History The emission of electrons from a metallic surface when a ray oflight (photons) fall s on it is called photoelectric effect. It was discovered by Heinrich Hertz while he was producing Metal electromagnetic waves in his laboratory by L - C oscillations. When he directed the ultraviolet light to the The electrons come out melallic sphere the spark current between two sphere 01 the metal when a tight talts on it increased. Hence, he suspected the emission of charge U.V. tight particles (electrons) from the metallic sphere due to the incident light. In the same exp"rimcntal sct. in ""V"V"'v+ one hand Hertz produced electromagnetic waves to ""V"V"'v+ support the wave theory of radiation (light), and in /"" the other hand he discovered photoeiC(:tric effect Spark which was explained by Einstein discarding wave theory of radiation and establishing quantum (particle) nature of radiation. The space In the gap increases when the U.V.light falls on the metallic sphere

«till

Lenard's experiment

In 1897, Thomson discovered electrons and measured its specific charge

!. = 0 .177 x lOll elkg. !n 1902, Lenards conducted the experiments on photoelectric m effect. He calculated the specific charge of the emitted particle same as that obtained by lJ.Thomson. Then he named the emitted electrons as "photoelectrons" because they came out from the metal due to the interaction of radiation particles (photons) with metal. The apparatus shown is a photo cell. It consists of a evacuated glass tube with quartz window. Inside the glass tube there a two plates of same metal. The plate £ is called emitter because it emits electrons when a ray of light falls on it coming from a source (lamp) after passing through the quartz window. Since the photoelectrons are the other plale C, it is called collector. The plates E and C can also be called cathode and anode respectively.

291'

Particle and Wave Nature of Radiation! (Light)

Source Evaluated glass envelope Quartz wlrldoW

c

E

The radiation falls on the emitt&d E. The phot~I'OI'IS are co!leeled at the collector C. The speed of the photoelectron In eonlmtled by retarding potentia l

As the emitter losses electrons it becomes negatively charged. The collcctor gets a positive charge. This induces a voltage which retards the photoelectrons and eventually the photoelectric current slops. Hence. we need to apply a voltage which can be varied its magnitude and polarity to control the flow of electron in the tube.

Stopping potential and saturation current When we kcep the emitter positive relative to the collector, the photoelectrons slowdown (retard). This is called retarding potential. The minimum retarding potential to stop the fastest photo electrons at the collector is called stopping potential, Since the electron's kinetic energy drops from Kroax at the emitter to zero at the collector, the change in kinetic energy of th.e electron is

,

Saturation

(

current

~, ---~-~-~-~-~~,---r-C-•

T

•

,,

,,, ,

--;""';-::;;--::1r--------''----,'-;---::--------,

•

v.!",,'" - Vo 0

vacc. + v (say)

Photocurrent increases from zero

to a constant (saturation value) with increase in accelerating potential

M=-Knnx As the stopping potential pushes the electron back, the work done by the electric field is Wet = -eVstop

GRB Undentanding Physia Optics and Modem Physics

292

Using \,",ork energy theorem, Wei = ll.K -eVsto = - Kiln."

or or

Knux(= ~mu;m )=~

If we reduce Ihe decelerating voltage to zero, the photoelectric current increases from zero to a finite value io because the electrons move from emitter to collector by the virtue orthe inertia of moti on. As we increase the vahage so as to accelerate speed up the electrons, the choose of the cieclrons decrease and more and more electrons strike the collector plate. As a result the photo current in the circuit increases. If we go on increasing the accelerating voltage, at certain voltage +1'. say. the photoelectric current docs not increase further. It is called saturation current i sal . Till the saluration current is attained, the photo current depends on the applied voltage and frequency 01 light for a given metal. After the saturation current is established , it does not depend on the light frequency, applied voltage and the material used. It will onlv deoend linearlv on the intensi'tv of lioht used.

lenard's observations (i) The saturation currenl remains the same for different frequencies of light used keeping the intensity of light constant. The photocurrent depends on the frequency till it gel saturated.

r, r, ~~~~

________________-+ v

More frequency and more photocurrent IOf a given accelerating voltage

(ii) The saturation current fordifTerent intensities will be different. The saturation current increases linearly with intensity of tight of any frequency.

;",

--~~--~--------------------___ v More Intensity, more photocurrent

293

Particle and Wave Nature of Radiations (light)

(iii) The photoelectron emission occurs almost immediately after the light fall s on the metal surface. The tim!! lag is extremely small, that is 3 x 10- 9s. I (Intensity of liquid)

[I

-------

/ ---+'-------'----'--;.., ~"l

Intensity oc (saturation photocurrent) (iv) The photoelectric current do not generally depend on temperature pressure a~d the surrounding conditions. (v) The photoelectrons have wide range of speeds and kinetic energy obeying Maxwell-Boltzmann's speed distribution. No. 01 photoelectrons

_ \ -_ _ _ _ _ _ _ _ _+ Kinetic energy Maxwell· Boltzmann's speed distribution c urve

(vi) The maximum kinetic energy of photoelectron that is, K nn.>; which is equal to d 'Siop does not depend on the intensity oflight.1t increases linearly with frequency of light. It is observed that no photoelectrons will be emitted below a certain frequency of radiation called threshold frequency and it is irrespective -+----~--------I intensity I?f light and accelerating voltage. 10 No. of photoelectrons will be emined below a cutoff frequellCY fo

Failure of wave theory of light In explaining photo-electric·effect Intensity Problem According to Maxwell's electromagnetic wave theory of light the average intensity oflighl is directly proportional to the square of the electric field of the light wave. In more intense light, the electric field intensity is more. It pushes the surface

294

GRB Underst;,nding Physics Optics and Modem Physics

electrons with more forcc . If more force acts in an electron it docs more work increasing the kinetic energy of the cleetron 10 greater exlent. As a rcsu,lt. ,the ,emitted electrons will posses more kinetic energy jfthe intensity of incident rad l~lI o~ IS more. However, this opposes the experimental observat ion that the maximum kmcllC energy ofphotoclectrons docs not depend on intensity ofl ight.

Frequency problem According to Maxwell's wave theory light falls on the metallic surface as electromagnetic wave continuously and speed (distributed) unifonnly. FurthemlOre, the electrons orthe melanic surface will absorb the incident energy continuously sooner or later each electron will accumulate enough energy to overcome the potential barrier of the metal and escape from it. Hence photoelectric effect should occur at any frequency of incident light. Again this rrediction of wave theory of light contradicts the experimental observation that the photoc\C(:tric effect can take place above a cutoff (threshold) frequency of incident light (not at all frequencies for a given a metal).

Time delay problem If the light fa lls continuously and unifonnly and the atoms will absorbs the incident light continuously, each atom wi ll receive light energy proportional to their crosssectional area. Since the cross-section of an atom is very small . each atom will receive very little energy in each second. Therefore. it takes a long time for an atom to accumulate sufficient energy to cjeet an electron from it. However, this prediction also proved to be incorrect because observation says that the photoelectrons start coming out of the metal almost instantaneously after the light falls on it.

Einstein's photon concept explains photoelectric effect Photoelectric is installlaneous. The ~\ instantaneous (or sudden) of electrons ~ emission implies that "some body" is colliding the electrons prompty. That "some body" can be imagined as flash (pulse or packet) of light (or radiation). As light is an light strikes the metal as a electromagnetic wave, we stream 01 photons can imagine the light hitting the metal as electromagnetic wave pulse. Each pulse can be termed as photon (bundles or packet or quartz of radiation). The photeleelric emission stops below certain frequency of incident radiation . This signifies that electron cannot absorb a fraction of an incident photon and cannot increase its kinetic energy by continuous absorption of photons one by one. In other words. each photon must be absorbed completely by an electron inside the metal.

~\

~z\

Particle and Wave Nature of Radiations (Light)

295

The maximum kinetic energy of (or stopping potential for) photoelectrons increases linearly (proportionality) with the frequency of the incident light. It means that the energy of each photon which is converted to (directly proportional to) the maximum kinetic energy oflhe photoelectrons must be varying linearly with the frequency of incident radiation. It can be shown that, the energy of a photon is

I E POO,,"

I

A photon kicks an electron

increasing its KE by hI.

= /if This is expression is called Einstein- Planck law, whose proof is given in an example.

Less photon/s, less photoelectron/s

More photon/so more eleclron/s

If the intensity of light increases. the photoelectric current increases proportionally. This means that each pholon of a monochromatic light must carry equal energy. that is, Iif. If the intensity of light increases. the number of photons striking the metal per second will increase. Eventually, proportionality more electrons will come out of the metal per second. In short. the rate of emission for photoelectrons is directly proportional to the rate of incident photons.

Einstein's photoelectric energy equation

I ,

Let a pholon of energy E ph = 'if collide with an electron in a metal. The electron (since bound in the metal) absorbs the photon completely. As a result, the kinetic energy of the electron will increase by the photon energy Eph = lif. The photon energy may be used in the followin g in three different ways (i) bnnging the volume electrons to make it a conduction electron by doing work WJ against the attraction force of the corresponding atom. (ii) bringing the conduction electrons by doing work \1.'2 when it losses a fraction of its energy is collision with others. (iii) bringing the surface conduction electrons out of the metal by performing a work 1V3 against the combined attraction of all metal atoms. We call it barrier potential energy or work function or threshold energy $: 1\'3 = $. Since an electron does all the above works, the total work done is IV = \\1) + 1\'2 +$. Then the kinetic energy of a photoelectron will be K = Epholon - w

GRB Understanding Physics Optics and Modem Physics

296

Since the electrons absorb photons at a different position, energy lends, \\) and 1l'2 may be different, whereas Q is a constant for a given

metal. Hence, the photoelectrons carry different (variable) kinetic energy given by Maxwell's speed distribution. if an electron is conduction electron and st~ys just at the surface it can get out orthe metal with maximum kinetic energy as it does not waste ?h~l~n's energy given to it in fightin g against the mdlVldual atomic liclds and colli sions with others. It wil\ have to only lose the energy equal to work fun ction of the metal. Hence, the

Kmu

-+____

~--------I

I, Kmu ox freq uency

V,top

maximum kinetic cner of a hOioo is given as, K""" = lif - 01 Putting Krmx = C Slop. we ave

IV,,. =~I -~ ,\ where . ' h = f 0 = threshold frequen cy of I, light for photoelectric eITect. Even though I >10. not all electrons can come out of the Vslep oc frequency 20 metal. If 10 photons fall on the surface. only one electron may came out. Others wi\llose their KE in repeated collisions and never come out of the metal. Thus. photoelectric eITect is a statistical process. ~

...~ ""

..

~ u..~u !:

Why do we write Ephoton = lif ?

Solution: To explain the sudden emission of electrons. Einstein hypothesised the existence of packets or bundle of light energy. Light quanta arc electromagnetic wave pulses. They can be emilled by oscillating charges according to Maxwell's theory. Planck theory says that an oscillator can have discrete energy 0, 'if , 21if, 31if and so on. Hence. the oscillator needs to losses energy at a time n (suddenly) in one quantum jump from higher energy level n to next longer leveln -I, which corresponds hI to the energy ofa photon . ~ Epbolon

= M oscillator = nhi - (n -1 )lif

or

rlE-=-"",,oo-=--:-hf-::1\

0-

1..,.----'--Proved .

Partide and Wave Nature or Radiations (Light)

297

'l1tA1II1!.U ~

An ultra violet light of wave length A = 2000 A/ails on a metal. Th e threshold wave lel/gth for photoe!eclric effcct/or this metal is 4400 A. Find 'he (a) enel1,..'v ofeach photo incidellilight, (b) work fimction. (c) maximum kille/ic: energy ofph010c/cclrollS, (d) stopping potelltial, . (e) power a/the SOllrce 0/11. v. light if it emits 10 10 photon/so Soluhon: (a) TIle photon energy of incident light is E = hc = (6.3 x 10- 3-1 J ·s)(3 x 10 8 m / s)

).

(2000xI0 10 m)

= 9.45 x lO- t9 J ::: 6.0 eV (.: I eV :::: 1.6x 10- 19 J)

Ans. (b)

(e) (d)

(e)

w=hc = 12400 eV = 12400=2.82eV

1. 0 ). 4400 K""" : E - W : 6.0-2.82: 3.18eV II = If' = 2.82 eV = 2.82 volt

c e p : nlif : (10'· Is)(9.45 x 10- " J) = 9.45 x 10 = 94.5 walt

3.5

Ans. Ans. Ans.

Ans.

X-rays

Introduction In photoelectric effect photons collide with metals to emil elecrrons. Since nature loves symmetry we should expect a reverse efTect or inverse photoelectric efTeet, th:n is, emission of photons when eleelrons strike a metal. Infact, it was discoverd by Rontzon in 1895 while he studyi ng the nature of cathod rays. He could see the sample of rock salt glowing very ncar to the cathode ray tube. Then. he speculated that some rays might have caused the glow. He examined the rays and confinned it to be an electromagnetic wave. Since the exact nature of the ray was not clearly known, he called it X·rays. We may call it Rontzon's ray. He received first Nobel prize in physics for the discovery of X-rays in 190 I. When a stream of high energetic electron (cathode rays) strike a metal such as W, Mo etc, X·rays came out of the metal s. X·rays arc electromagnetic waves (radiations) having wave length in the other of I A. X·ray spectrum lies in between ultraviolet and y-rays. There are two types of X-rays is continuous and characteristics.

Coolidge tube; (X-ray Apparatus) The apparatus in which X·rays are produced is called coolidge tube. As shown in the figure. there is a filament F connected to a battery of emf &. When the filament carries a current it becomes hot and emits electron. Thi s is called themlOionic effect and the filament is called hot cathode. There is a metal target (anode) in right which is maintained at a high potentia1 Vrelal ive to the filament(c3thode). These two electrodes

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Glass envelope

x - rays are placed in an evacuated hard Fi!lament (Cooledge tube) (Hot cathode) glass tube. The electrons emitted from the filament with + negligible kinetic energy are E e accelerated in a high voltage V (in the order of several kV). Hence Ihe applied voltage is called accelerating voltage . As a result the electrons collide with the metal target to produce L----t IIIII-----:(.'nod..:'o) X-rays. The glass envelope is \ V Metal target evacuated so that the electrons Accelerating potential will not collide with any gas The electrons emitted from the hot cathode molecules to ionise them.

e

gains KE by accelerating voltage and strikes the metal target to emit X- rays

Continuous X-rays Let an energetic electron gct into an atom of the larget metal. When it moves in the strong electrostatic field of the nucleus, it delivers a fraction of its energy and momentum. Since the nucleus of the melal atom is very heavier than an electron, it practically remains stationary after tbe collision. As the electron slow down while coming out afthe atom, to satisfy Maxwell 's law of radiation it must emit an emw but not continuously, instead discretely while inside the atom according to Planck ~ Einstein's law. In order to conserve the total energy of the

(K)

The bombarding electron accelerates changing Its speed and direction 01 motion In the electric field of the target atom.

system (electron + atom). a photon must be emitted. The conservation of energy gives. 2 2 mo c + K + (Eatom ) == moc + K'+{Eatom)f + Ephoton Since the atom docs not more practically, assuming that it does not excited, (Ea,om ) ; == (Ea1om ) f. Thcn. Ephoton ::::: K - K' According to Planck - Einstein law, the energy of a photon is Epholon

... (i)

= lif = J~c

Using eqns.{i) and (ii), the wave length oflhe X-ray photon emitted is A ~ he K-K'

... (iii)

Particle and Wave Nature of Radiations (Light)

299

The total momentum of the system (atom + electron) can be conserved if we take the momentum ofthe nucleus just after the collision. Conservation of momentum gives -+

-+

-+

-+

.. .(iv) (Pe! ) ; =(Pel )/ + Ppoo!on + P a!om As the electron will be deflected at different angles according ot eqn. (iv), we can have different value of K '. Hence in a single atom different electrons can emit ,,, X~ray photons of different wave lengths ,, using eqn. (iii), Furthermore since same ,,, electron can lose different amount of its kinetic energy discontinuously during --t---~'~ -c-~'~'~-~.---------' The wavelength of continuous collisions with different atoms, it will emil X- ray ranges from '-. to 00. photon of diffcrent wavelength. However, the bombarding electron can lose its total kinetic energy during a single head on collision with an atom (being repelled by the electron cloud of the atom). ln this case K' = 0 . Hence, according to eqn.(iii), the wavelength of the emitted photon will be minimum given as,

',.

A. min == K

or

~

, where

=eV;V = acceleratin

volta e "min = he = 12400 eV

V

in A, where V is in volt. The above expression tells us that.

"m, "m/1Am:!'" AJ If V, > V1 > vJ. we have Al < A2 < A3 & Ami

< "m2 < "m )

The bombarding electrons can lose their kinetic energies as X- radiations ( and other) with a wide range of wavelength having an internal Amin to infinite. This continuous speed of wavelength of X- radiation is called continuous X-ray spectrum. It should nol be misinterpreted as the electrons lose their energies continuously which violates Einstein - Planck quantum law of energy quanta Eph = hI.

The monochromatic intensity of continuous X- ray of wavelength between A and A. + d"A. is g)ven as the area of the curve I A - A..

I,.d"A. = 12.4

kev(t -1 )~

m This is called Kramers' fonnula. The wave length A.' at which I A is maximum is

IA'+"," I

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300

When we increase the accelerating potential maximum KE of the bombarding electron increase. Then, the minimum wavelength of continuous X- ray spectrum will decrease as, )• min I, = -IIC = constanI e However, the wavelength ,o' of most intense X-ray will change (increase) according to the relation J.' JV = constant Thus, the total h -), graph will be propor1ionatcly bigger, tmin will be silln.c~ towards the origin and the peak of I,. '1. graph will also be shifted towards the ongm ifwc increase the accelerating voltage.

I

~r4:!1'~~

I

.

DetC'rmine 'he> clltoff(millinlllm) wal'e/ellg'" of x- rays prodllced by

50 kel' electrons ill a I'ollage IIIbe, Solution: The energy of X- ray photon is E =

...l!2..I. min

)'min

= 12400 =0.248 A 50 x l0 3

3.6

Ans.

X-ray Diffraction: Bragg's Law

In 1912 William Bragg devised a method of calculatin g the wavelength of X-radiation . This method was based on the technique of difTract ion of X-rays by the atoms ofa crystal lattice. The word "diffraction" is confused with the phenomena of " interference". The basic difference between interference and di ffrac tion is simple. In interference two monochromatic waves arc allowed to superimpose constructively or destructively. However, in diffra ction we consider the constructi\'e superposition of many parallcl monochromatic waves being emitted from different sources. A metal or any crystal consists of three-dimensional regular array of atoms. Each atom scatters an c mw(X-ray) falli ng on it Scattering is a process of emission of emw by an atom in all direClions when it receives an emw falling upon it. The process is very simple. When an emw falls on an atom. the electric fi eld of the emw pull s the positive and negative charges of the atom in opposite direction with a force which is much lesser than the force o f attraction between the +vc and -ve charges. The charge separation makes the atom OJ. dipole and induced a dipole moment. This process of inducing a dipole is called polarisation . Thus an atom becomes a dipole which is generated by an emw. When the electric fie ld of the emw oscillates (changes its magnitude altcmating its direction), the- induced dipole will also osci llates with the frequency equal to the frequ cncy of emw. According to Maxwell's law of electromagnelism. an oscillating dipole (charges) must radiate emw. Thus, a dipole radiates emw o f same frequency as that of the inc idem emw causing it . This is called dipole radialion. II takes place in all directions except along the Iinc passing through dipole momcnt(line of oscillation). In

I

r

Partide and Wa,·e Nature or Radiations (Light)

301

this way each atom behaves ns a secondary source of emw of same frequency_ This process is called scattering. The parallel scattered X-rays superimpose constructively if e "" e'

Bragg's plane

2

3

4

Y'

"'-

Scattered X-rays

The X- rays are scattered by all atoms of the Bragg·s plane,

Let us assume that the parallel X-rays. hit the atoms A. B. C and D of the planes 1,2_ 3 and 4 respectively making an angle e with the nonnal YY ". The atom scatter the X-rays almost in all directions. Let us consider four scattered parallel X-rays at the atoms at an 2 angle e' coming out from the plane surface. They will superimpose. Constructively if e = e' and the path difference between any two consecutive parallel scattered rays is given as,

1

lax =2d s;nB = n)..1

where n - O.I~2" ... ... d =atomicspaeing. 11 d B ' r 1 " The scattered X· rays Interlere constructively " Th IS IS ca e ragg s lonnu a lor if they are paranel and ~ "" 2d sinO = n). maxima. The other scattered X·rays will superimpose destructively not giving a trace considerable intensity at the detector. The corresponding angles of constructive superposition or maxima points can be tracted by measuring the angle j3at which the detector records maximum intensity. After knowing

GRB Undentanding Physics Optics and Modem Physics

302

p, we can find e = 90o-p and calculate)." by using the Bragg's fonnula provided the atomic spacing "d" is known.

Detector ~

...... ,

~",

-" ~\

..

p_1___ ,• -:••

Slit

-

,, ,, '

The elec1ric field E Induces a dipole In the atom.

,,

,

,, , ,,

The detector will record maximum Intensity of scattered X-rays at source fixed angles p.

e

--, C

Since = 9' • this satisfies the condition for optical reflection(but this is not reflection), Now, roughly we can call Bragg's diffraction as "Bragg's reflection", The planes at which the reflection takes place arc called Bragg's plane.

--- .....

,'-

,

Representation 01 a photon

as an ideal pulse of emw

X-radiation of wavelength 1.5 A strikes a crystal of interatomic spacing 2.8 A. What is the highest order of Bragg 's reflection possible in this crystal? Solution: From Bragg's fonnula

. e = -"l-

sm Since

sin

2d

e !5: I, n !5: ';!

2x2.8A 1.5 A

or n :5: 3.7 Hence, the maximum possible order is 3.

U

Ans.

MOlley's Law and Characteristic X- raya

Some times a more energetic electron will be able to excite the target atom by knocking its orbiting electrons. Let us assume that the bombarding electron kicks the K-electron of the atom. The atom is now in excited state. Then, the L- shell electron will come to K - shell to fill the vacancy. In this process of de-excitation of atom (or electron) from higher energy of state E2 to lower energy state E I , a X- ray photon is emitted whose frequency is given as

f =£, -£, h

... (i)

Particle and Wave Nature of Radiations (Light)

303

This is accordance with BohrEinstein's law which will be discussed in next chapter. According to Bohr's model, the energy of K-shell is £\ = ll e4 (Z _2)2

+

8&2,,2 11 2

o

where

11

= 1 and ).1= mM ::::m'

M+m ' m = mass of an electron and M = mass of nucleus.

CharaCleris!ic Kn-X-ray The bombarding electron knocks an electron in K- shell. ArI eledron gets de-excited 'rom higher

shell IQ K·shell105ing itS X- rays

or

E, =

me4 (Z_2)2 o£2h2

(0: m«

M)

... (ii)

o

We have subtracted 2e from the nuclear charge +Ze because the + vely charged nuclear is enveloped by a charged shell of a electron cloud of two orbiting electrons in K-shell. Similarly, to find the energy of L-shell, we need to subtract the charges of 2 + 8 = 10 electrons from the nucleons charge as the nucleus ofcharge + Ze is enveloped by two spherical shells of negative charge - 10 e. Then, £2 =

me 4 (z _10)2 222

8E Oh

,wheren=2

11

... (iii)

Hence,

4

or or

f<= me (z-cr)2,whereCJmaybeoneortwo 8E oh2

[Jl = C(z-a),lwhereC = ~ = constant

V&0h2

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304

and 0' is a general constnnt to fit with the experimental result.

If

The above expression is called Mosley's law of which state that,

--~~---------' u

The frequency of the characteristic x-ray directly proportional to the atomic number

of the taraet material.

/fthe Ku X-radiatIOn of Ala _ (2 42)"05 a wavelength of 0.7 1 Ii, find the ware length of Ku X-radiation of Cu(Z = 29). assllming cr = l Solution: According to Mosley' law, =a(2- a) = a(Z-I) '

.Jl

or Then,

I I

I, I

c ' / = -=a-(Z -I) 2 1.

I

,

'C. =(ZMO -I) Zeu- 1 A =AMo(ZMO -1)= 0.7,(\2942 -_I1)' Zeu _)

AMo Cu

=

3.8

1.52 A

ADS.

Einstein's Photon Concept and Dual Nature of Light

The basis of quanlisalion of radiation can be told in two lines: Energy of radiation is quantised (Einstein). Hence, the energy of any radiating system, i.e. , alom, molecule. electron etc. must be quantised(Max Planck and Neil' s Bohr).

We learnt that the particle concept of light was given by Einstein to explain all observations of photoelectric effect. Einstein used the word quanta (or bundle) oflight. In 1923 Lewi's coined the word "photon" which became familiar word meaning a quanta of light.

Properties of a photon A "photon" can be imagined as a quanta Iflash!bundlelparticle of light (radiation) emitted by all fundamental sources like atom, molecule etc. via electro-magnetic intentions. It is an e!ementary(energy clement) particle of energy. It must not be misinterpreted simply as a short puise oflight (or packet of emw) carrying any amount

•

I

I

Partidt and Wavt Natuft or Radiations (Light)

305

of e~ergy . An ideal (monochromatic) photon obtained from _ L=+'n atomic (or other) fundamental process of dc-excitation . A ~ photon must carry an energy equal to lif, where I = dominant frequency of the pulse. It moves with a speed of light. It has no rest mass. Its speed does not depend on any reference. II has . I E It · eqUlva ent mass 111 =- 2' energy E =hi = 1'1 \1' and momentum p =- =T, w. Due to Its c A gravitational mass

m( = ~} it is attracted by gravity. It

carries electromagnetic fields . When it leaves a malter, the energy of matter decreases by lif. The mass of matter

E

I

-. /

n

decreases by c 2 = /...c' The angular momentum of matter changes by II. Hence, the angular momentum ofa photon is

± 'Ii directed along the line of its motion. nie magnitude of

.

/

V -

The Intensity In the elementary volume Is dlr&Ctly proportional to the concentration of phOton in It

spin of a photon is J2 Tt., the spin ofa photon is 1.1t is a stable particle of radiation (mean life infinite). It is denoted as y or lif or 'Ii w. It is diagrammatic~1 given as a short emw pulse (.............-) . It does not decay spontaneously in empty space. It has no electric charge. Photons are emitted by many natural processes. (i) acceleration of a charge particle produce emits synchrotron radiation. (ii) A molecule, atom or nucleus emits photons of various energy when de-excited from higher to lower energy states ranging from infrared to ,,(-rays. (iii) an inhalation of matter, that is, when a particle and its anti particle touch or collide, photon(s) can be emitted. When a photon moves in vacuum its properties, i.e., energy etc. (shape and size roughly speaking) donot change. It can interact with matter or electromagnetic field of nucleus to produce particles and their anti-panicle borrowing the extra energy from the electromagnetic field oflhe nucleus. This process is called materialisation of energy. Thus, photon is the anti particle of itself. A single photon can exhibit interference diffraction, reflection, refraction etc . ~

~

Photon cont~ins a real E and B field.

Wave Interpnltatlon of Intensity The electric and magnetic field at a point (inside an elementary volume) is the average field of the number of photons present there. If E = electric field wave function of any classical emw of Maxwell, the wave equation is given as,

=

E = Eo sin(",' -

kx~

=

where ro angular frequency of the wave and k angular wave number of the radiation (emw). Then, the average intensity of the wave is given as,

GRB Undentanding Physics Optics and Modem Physics

306

.~ ~

. . II.,.= C£oEg I

.

...(il

This is the intensity fannula according to Maxwell's wave theory ofllgh!. Practice Interpretation 01 Intensity If dll photons each of energy lif pass through a given area A during lime dt the average intensity oflighl can be g iven as,

I

nv -

'

Energy incident per second

='7-----;~c_'_,....,...,----

Area (of thc plate) (No.ofphoton incidcnl / s)(Energy of each pholOn) = Arca

I av == eNlif

or

... (ii)

where N = no. of photons per unit volume and C = speed of each photon (light). This is the intensity expression (given by Einstein) using quantum (particle) theory of radiation (light).

Wave-particle duality

•

"

Comparing cqns: (i) and (ii), we have-:

I

IN I~ E'; I I

numbe,r oe PhOlon )

( field

( per wut volwne

)

inten~ ly amplitude

!

!

,(thiS represents the )

this iiv:es (represents») ( the particle nature

! lh.iS is ~iven bY)

I Average electric

wave nature

"

<= ~. _.

( Emstem

! (thiS.iSgiven bY) Maxwell

This tells us Ihat the photon density (concentration of photons), that is, number of photons per unit volume is directly proportional to the square of the amplitudc..ofthe electric field (wave fu nction E). In this way Einstei n could explain (establish) the relation between wave narure and particle nature of radiation. Thus. the apparent contraction and confusion between two highly established laws Maxwell' s law of radiation (cmw) and Einstein's law of quantisarion of radiation was dispelled by the great Albert Einstein. This is called "wave-particle duality ofradiation"- suggested by Einstein . .

,.

, ,.

307

Particle :md Wa\'e Natul-e of Radialions (Light)

Physical significance of wave--partlcle duality; probability density From the above di scussion wc . E, E, understand that the electric or (magnetic) field ofa small region is the average field of all photons present there. The wave-particle duality suggested by Einstein signifies that where thc average electric fi eld is more intense. the density o f photons will be morc. When large number of photons strike a spot on a fluorescent screen it appears as if the spot(small area or volume) is brighter and the distribution A B of light (illumination) seems to be E, > E2 because N, :> N2 ; ' continuous. [t is just like sprinkling of The photon densities in the small rogion color on a paper sheet. If too many A Is more than that in the small region color droplets hit the paper, the B. Henco, E, is greater than E2, where E, and E2 are the average electric lield In the "coloring" appears to be continuous. region. However. sprinkling very few color drops on the paper, we can see the distinct color spots as discrete dots(small marks). Likewise if we decrease the electric field too much, the count rate of photo ns by-a detector (Geigcr Counter or fluorescent screen fitted with a sound recorder via a computer) will be measurable by naked eye! It mealls that we can see few scintillating (illuminating) spots on the screen or photographic plate generated by the striking photons. Since the process of photon emission is statistical. the number of photons crossing through any elementary area (or present in any elementary volume) may change slightly from time to time. Thcrefore, the calculation of exact numbcr of photons in any region is not practically possible. We need to usc the phase "probability of lindine a photon" in any region. Since average number ofpholon present in any region per unit volume is di reclly proportional 10 the probability of finding a photon per unit volume in that rcgion, wc can write

J'

probability offinding ) [Average value of a photon in unit volwne oc electric field , [ = probability density P

amplitude

Ip '" £'1

or Setting the constant of proportionality as "onc", thc probability of finding a photon is a region of vo~lu~m~e,-"d~v~i~s_ _ _ _ _ _ _-,-,:-..., 2 Photon Probability = PdV = E dV

I

I

I

I

The average value of £2 over a finite time is a meaningful measure ofph01on flux (no. of photons hitting per unit lime). Thc photon flux density (or intensity oflight) or conccntration of photon is dircctly proportional 10 the probability dCf!sity.

GRB Undent.anding Ph)'5ics Optia and Modem Ph)'5ics

308

Qucs. Can we detect a radiowave photon? EinsteIn's formula of wave-particle duality of light As we know . reflection, refraction can be explained by both photon and wave concept of light. Interference, di ffraction and polarisation can only be explained by wave concept of light. However, photoelectric effect, compton effect can only be explained by photon concept. A photon has a momentum p

=!i

which was confirmed in Compton efretl of

C scattering of an electron by an X-I"Jy photon. in 1923. Since the energy ofa photon is E = /if = h£. (Planck Einstein's formula) , we have A

h

p=~

where p is particle nature and), is wave nature. This expression relates the particle nature oflight( carrying a momcnlump) 10 the

wave nature of light (carrying a dominant wave length I. ). This equation eltplains wave-particle duality of radiation. It has deduced by Albert Einstein. In later chapter, you will learn that by using thi s relation dc-Broglie found the wave length of maner wave and established wave-particle duality of maner.

~ ~_!'!1L. ~

... .»;ic 9

For a point source of radiation. explain why should the intensity oJ light and corrcsponding electric field of radiating wave decrease with increasing radial distal/ce. Solution: Let the power of the point source be P. As it radiates energy. the spherical wave fronts will move radially away. At any radial distance, the intensity of thc wave is

I = Power Area

=L 4rrr2

As r increases. I decreases obeying inverse square law. Since I cr;: E 2, E will decrease hyperbolically with radial distance. The smaller E corresponds to lesser density (concentration) of photons. It is obvious from the photon concept that. the spherical wavefronls is continuous according to classical At point 1 and 2, Nl > N2 wave theory but discontinuous Hence, El > ~ due to thc presence of discrete ' photons. The laws of all photons (roughly saying at equal radial distance from the source) makes the wavefront. Hence, the concentration of photons will decrease

Particle and Wave Nature of Radiations (Light)

309

according to the relation I = Nhf. As r increases I decreases. Hence, N(= no. or photons per unit volume) decreases.

f!c"*l!U

feJ; (Gravililtional Red-shift) A electromagnetic wal'e (radiation) of frequency fo is emitted bya mass M and radius R. Find itsfrequency

in free space very' f ar from lite star. Solution: Equating the energy of the photon at I and 2, we have 1

/ \

'\

hi,

+( _G~m) = hl ,

mc- = hf o

2

,---,

~

... (i) ... (ii)

By using eqns. (i) and (ii),

,

I

=/o(l_GM ) Rc'

Ans.

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310

-

_ - - ' '- '--'

MISCELLANEOUS EXAMPLES

~

,.;-..-'" - -

Exa mple l. In the given figure J • t , there arc two peaks I and 2. Which corresponds to Ku,. La and which is of Kp? [

2 3

,

,,,

,,, ,

"

,

>,

,, ,

,, , A,

>.

Solution: The wavelength of characteristic radiation is given as

t. = "c

M

"E '

where 6.£ = energy di fference between tv.'o levels of electronic transition. Since A, < '- 2. we have dE] > l1E z. Then. 6£1 must corrcsponds to L\E3---.t (Kp ) and 6 E2 must

-r-+- L_

-1-

corresponds to A E2-+, (K u )' Hence, the peak 1 is of -L---t_ _ K K (I and the peak 2 corresponds to K (l ' Since I. 3 is greatest,

AE3 is least corresponding 10 that of La' Ans. Exa mple 2. The Ka X-ray of tungsten occurs at , .. = 0.02 1 nm. What is the energy difference between K and L level oflh e tun gsten atom? Solution :

6£ = he

1-

= 12400 eV

A = 12400 0.21 = 59047.6 cY

A ns,

E xample 3. A black body radiates heat of 3 x 10 4 \VIm 2. At this temperature.

find the wavelength corresponding to maximum radiancy. Solution: The intensity of radiation of a black body is

I = crT 4 or

Then the wavelength corresponding 10 max imum intensity is

... (i)

Particle and Wave Nature of Radiations (Light)

31.1 ,. -, ' . ,.(ii)

Using cqns.{i) and (ii),

A =_ b_

ml

2.9xIO - 3 K-m

= -=.:.::......~'------;­ I

3xlO' \V I m' )' 8 ( 5.67 x 10- watl / m 2K4 = 3.4xlO - 6 m

ADS.

Example 4. Let us assume that the maximum energy radiation ·takes place by the sun which corresponds to wavelength A = 0.48x lO-6 m. Find the (a) energy lost by sun per second (b) time after which J% of the tolal mass of the sun will be lost. Solution: (a) The temperature of the sun is 2

Ts = ° .29)(\0- = 6042K 0.48 x 10-6

Then, the intensity of radiation of the sun ot its surface is 4

I = aTs

= (5.67x 10-')(6042)' = 0.7555)( 10 8 W 1m 2

•

,

The power (energy lost Is) Joss by the sun is = J 41tR,

I

•• j

, I

,

.;,

"-' .,

= (0.7555 x 10')4 x 2; x (6.95 x 10') ' = 4.5876 x 10 26 watt

(b) The corresponding mass 1051 by the sun per second is ilm dE l dt J = M

Ans. • [: • ,'"

c'

=

4.5876x 10 26 (3xlO')'

=5.lxlO' kg/s The sun will lose 1% of its mass during a time M , x _1_ (1.97 x 10 30 kg) x _I_

100 = 100 tlm l ilt 5.1x 10 9 kg Is =3.863xI0 18 5 = 1.22 x 1011 year

1=

Ans.

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312

Example S. The work function of zinc is "'0 = 3.74 eV (a) Find the threshold wavelength for photo-electric emission for zinc. (b) If a radiation of wavelength A = 250 om is incident on the metal. find the speed of the fastest photoelectron. Solution: (a) The work function arzine is Wo = 3.74 cV The threshold wavelength for photoelectric effect with zinc is Ans. ).0 = 124oo=JJI5A 3.74 (b) The energy oD, = 250 nm = 2500 A is w = 12400=5cV 2480 Then. the surplus energy is l!.E = 5 - 3.74 = 1.26 eV = 1.26 x 1.6)(10- 19 J Then. the max KE of photoelectron is !mv2 = 1.26 x 1.6 x 10- 19 J 2 V=

or

2x1.26)(1.6xlO \9 9.lxlO - 31

= J0.44 x 10" = 0 .67 x 10 6 mfs

Ans. Example 6. When the applied (accelerating) voltage in cooledge tube decreases by a factor n = 2. the minimum wavelength of continuous X-radiation changes by 2 A. Find the accelerating voltage. Solution: The minimum wavelength of continuous X-radiation is ). = he = 12,400 A ... (i) eV V

). + lo). = 12,400 A

... (ii)

VI. Eqn.(ii) - eqn . (i) gives. lo). =

or

(12,4oo)~

V = 12,400 n lo).

-

12,4oo x2

2A

= 12,4oovo\l

~

,

Ans.

313

Particle and Wave Nature of Radiations (Light)

Example 7. A ray orJight of wavelength 0.4 tlm falls on a photocell of sensitivity 3.4 x 10- 3 amp/ watt. Find the no. of photoelectrons emitted per incident photon. Solution: The sensitivity oftne photo-cell is J (given). The power of the incident radiation is P = '~c N , where N = no. of photons striking/so Then, the photo current is i = JP

or

.. .(i)

i=JhcN

A Let II = no . of photoelectrons/so Then, the photo current is j = ne Equating (i) and (ii)

... (ii)

.!!.... = Jhc N Ae

-3 I ) (6.4 x 10-34 x 3 x IO ' ) = (34 . x 10 amp watt (0.4 x 10- 6 x 1.6 x 10- 19 ) = 102 x 1O-3- l4+ 8+ 6+ 19 =0.0102

An••

4 Example 8. A cylindrical beam of light of intensity I = 9 x 10 w/m 2 ralls on a flat reflecting surface at an angle 8 = 45° with normal to the surface. Assuming the coefficient of reflection p = 0.6, find the radiation pressure on the surface. Solution: The change in momentum in nonnal-direttion is !J.Pn =aEcos8(I+ e) c The nonnal pressure acting on the surface is !J.Pn I !J.t Nonnal pressure = P = A


= el!.I

(A' /eo.9)

(.: A' = Aco.9)

=(
or

·-A-

C

P = /(1 + e)eo.' 9 c 4 = 9 x 10 (l +0.6)(00.' 45·)

3 x lO' =2.4 x lO-4 N / m'

Ans.

•

314

GRB Undtntanding Physics Optics and Modem Physics

Example 9. A point source of light of power P is located at an axial distance x from a disc of radius R. P Assuming 100% reflection oflight, find the force exerted by light on the disc. Solution: The nonnal pressure of the radiation at P is p= 2lcos 1 e c The normal force on the circular strip is 0/ = PdA

x--j-,

= 2/ cos 2 9 .2rtrdr

c = 4lrtrdrcos 2 8 c Putting P P x 1 = -- = and ease = r"i'~'i' 4w,2 4n(x 2 +r2) .Jr2 +x2 2 oj = Px rdr we have, c (x2 +r2)2

Integrating the net force acting on the disc is " F = dF ?x 2 R rdr

J

= --;:-

~(X2 + "

,.

)2

ADS.

Example 10. A light of wavelength 1240 A falls on a metallic sphere of radius 1 m and work function Wo = 3 eV. Find the maximum number of electrons left from the " , sphere till the photoelectric effect stops. Solution: As the copper sphere loses photoelectrons, it becomes positively charged. Let it lose" electrons. Then its charge will be nt; .-Hence, + + + the potential of the copper ball is V = _I_ne ... (i) stop

41tZoR

This is the stopping potential which is equal to the maximum KE of the photoelectrons. According to Einstein's photoelectric eqn. (i),

+

+

+"'. . . R

~+

315

Partide and Wa\'e Nalure of Radiations (Light)

hc _WO = !m(l~

or

A 2 he - /" i" "0 = eVstop

... (ii)

By using eqns (i) and (ii),

he _ Wo = J.. or

e(.....L)!!f 4nco R

n = ( he _ Wo )47tC OR J.

c2

12400 3) 1.6 x 10- 19 = ( 1240 )( 1.6 )( 10 19 )2 )(9 )( 10 9 =(1O - 3)x!x 1 9 1.6)( 10- 10

= 4.86)( 10 9 ADS. Example 11. A thin continuous X-ray beam falls on a Nael crystal of inter atomic (planner) spacing d =2.8 A. The minimum angle of Bragg's incidence isS = 4.1°. Find the applied (accelerating) voltagc in the coolcdge tube. Solution: The minimum wavelength of the continuous X-ray is ~ = he A •• •(.) I eV

For smallest angle of incidence for Bragg's reflection, A. = 2dsinO

... (ii)

or

2dsin8 12,400 = ,.......,~':""--:--c:2x(2.8 A)sin4.1' Ans. =31kY Example 12. A beam oflight of energy E falls horizontally on a hanging reflecting mirror or mass m. As a result is receives a momentum and swings to a maximum angle O. If I = length of the flexible mass less string by which the mirror hangs, find the value

oro.

Solullon: The momentum delivered by the light is I>p = 2£ e Then, the momentum gained by the mirror is p=2£ c

61

H

,

2

316

GRB Undentanding Physia Optics and Modern Physics mu =2E c u =2£ or me Consuming energy of the mirror at I and 2,

or

... (i)

IAKI=i AVI I 2 2mu = mg/(I - cos9)

. . ,(jj)

Puning the value of u from eqn.(i) in eqn. (ii), we have.

I

2: m

(2E)2 . 2e ~ =2mg/sm 2:

e= Sin-l(m~)

or

Ans.

Example 13. Find the numberofpholons pcrm) al a point at a distance ofr = 10m from a point source of power 100 wan radiating light of 10000 A. Solution: The density ofpholan at a point (radial distance 1) is given as nCr) = N, N = no. ofpholon lslm 2

e

Putting N =

~

, we have

41'1'r he

nCr) =

PI.

4Ttr2 hc 2

(100)(10000 x 10- 10 )

=-4-X-'-X-(~IO~)~2~~~.3~X=I~0~l~4-)(-3~X-I~08')'2 =

10 12

10-4+ J4- 16-2

41tx6.3x 9

=

41tx6.3 x 9

= 1.4 x lO 9 photons/m 3

Ans.

Example 14. By using Planck's law find the power radiated per unit area of a black body in the narrow range of wavelength. that is 6./.. = I nm. at a temperature T = 3000 K and A. corresponding to maximum spectral radiancy. (Wein's constant' b'= 2.9 X 10-3 m-K) Solution: We know that, u).

I·]

2 2 16. C 1i[

=).s

2PCd1

e AKT -I Then, the intensity is

tV = f u)..6.). 4 Putting the value of!!)..

Partide and Wave Nalu~ of Radiations (Light)

317

,~.

eH.T _ I

SinccA. """"" A.m. we can putA.r =A.trlT I1P = 47t

,, C

!J.).

1;

ASm

,~.

e Putting

=h. Then,

kb

_I

2tr.ch = 2tr. x 3 x 10 8 x 6.3 x 10- 34 12tr. kb 1.38 x to- 23 x2.9xI0- 3

= 4.72 in the above eqn., 11P = 0.312W/ cm' A os. Example 15. When a beam of 10.6 eV photons of intensity 2.0 W 1 m2 falls on a platinum surface of area 1.0 x 10-4 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 cV). (Take leV =t.6 x 10- 19 J) . Solution: Number ofphotoc!cctrons emitted per second = (Intensity)(Area) x 0.53 (Energy of each photon) 100

__ (2 .0)(1.0 x 10-4) x 0.53 (10.6xI.6xI0- ") 100

=6.25 x 1011

Ans.

Minimum kinetic energy ofphotoe!ectrons. Kmin = 0

and maximum kinetic energy is. Krmx

=E -

W = (10.6 - 5.6)eV = SDeV

Ans. Example 16. A monochromatic light source of frequency illuminates a metallic surface and eje<:ts photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the whole experiment is repeated with an incident radiation of frequency ~ j, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215A. (a) What is the frequency of radiation? (b) Find the work function of the metal. Solution: (a) Using Einstein's equation ofphotoeleetric effect, Kmu =hl-W Here, Kmax = l3.6eV ... (i) hi - W = 13.6eV

Further,

h(~

6

1)- W= 12375 = 10.2eV 1215

... (ii)

318

GRB Und~nlanding Physics Optics and Modem Physics

Solving Eqs.(i) and (ii), we have lif

"6=3.4eV

I = (6)(3.4)(1.6 x 10-") = 4.92 x lO" Hz

or

Ans.

(6h3x 10 34) IV = hi - 13.6= 6 (3.4) - 13.6

(b)

[form Eq.(i)] = 6.8eV Ans. Example 17. In Moseley's equation.J7 = a(z- b), a and b are constant. Find their values with the help of the following data . Element

:

Wanlength of Ka X-rays

Mo

42

0.71 A

Co

27

I.7SS A

Solution: .. . (i)

0'

.-

and ... (ii)

From Eqs.(i) and (ii), we have

1

1 ] ., = '(:1 -,,)

./C[ .[[j - ~"

. .. (iii)

Solving above three equations withe = 3.0 X 10 8 mIs, Al = '0.71 X10'- 10 m 1..2 =1.785xIO- IOm. z, =42 and =2 = 27.we get

a =5 XI07(Hz) "2 and b=1.37 .•. ',\; . Aus. " Enmple 18. Ifan 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 ).8 X lOll Clkg.

Solution: de Broglie wavelength when a charge q is accelerated by a potential difference of V volts is . _.(i)

qV = he

For cutofTwavelength ofX-I'llYs, we have

or

).

_ he m - ql'

'm . .. (ii)

•

Particle and Wave

Nalu~

of Radiations (Light)

319

From Eqs. (i) and (ii), For electron q = 1,8 x IOllClkg (given). Substituti~g the values the desired ratio is m 1.8 x lO ll x l0xlO J -,-_~2=-,,-__ = 0.1 Ans. 3 x 10 8 Example 19. In a photocell the plates P and Q have a separation of 10 em, which are connected through a galvanometer without any cell. Bichromatic light of wavelengths 4000 A and 6000 A arc incident on plate Q whose work function is 2.39 eV. If a ullifonn magnetic field B exists parallel to the plates, find the minimum value of B for which the galvanometer shows leTa deflection. .. Solution :Encrgyofphotons corresponding to light of wavelength ).1 = 4000A is E = 12375= 3.1 V I 4000 e and that corresponding to ).:2 = 6000 A is. E =12375=206 eV · 2 6000 As £2 < Wand El >W Photoe!ectric emission is possible with).1 only.

(W'" Work function)

J'T

®B 2,

u

d

Lr---'----L- a 1

•

Photoelectrons experience magnetic force and move along a circu~ path. The galvanometer will 'indlcate 2er~ deflection if the photoelectrons just complete semi· circular path before reaching the plate P. Thus d= '2 r=lOcm.:. r=Scm=O.OSm , mv J2Kj;, Further, r = -8q = 8q I

.J2km .

Bmin

=rq

Here K~EI_W=(3 . 1 ~ 2.39)=0.71.V Substituting the values, wre~h~av::e,,-_ _ _::;;;~:-:-:::--:-:::;;; . J2xO.7 IxI.6xlO- 19 x9.109xlO]I 8 . on" (0.05)(1.6 X 10- 19 )

320

GRB Undentanding Physics Optics and Modem Physics

=5.68 X 10- 5 tesla

ADS. Example 20. In II. photoelectric effect setup, a point source of light of power 3.2 x 10- 3 W emits monoenergetic photons of energy 5 cV. The source is located at II. distance 0[0.8 m from the cCnlre ofa stationary metallic sphere of work function 3 eV and of radius 8 x 10- 3 m. The efficienc), ofpholoelectron emission is one for every 10

6

incident photons. Sphere is initially neutral. and thaI the photoelectrons are instantly swept away after emission. (a) Calculate the number of photoelectrons emitted pcr second. (b) It is observed that the photoelectrons emission SlOpS at a certain time (after the light source is switched on. Evaluate time "('. Solution: (a) Intensity of light at a distance 0.8 m from the source 1= (3.2xI0- ) JIs) .. 4.0xt0-4 walt 4n(O.8)2 m 2 m2

.. Energy incident on the metallic sphere in unit lime, £\ = n(8 x 10- )2 (4.0 x 10-4) = 8.04 x to-8 wan Energy of one single photon. £2 = 5.0 x 1.6 x to- 19 J = 8.0 x 10-19 J Therefore, total number of photons incident on the sphere per second nl

=~=8.04XIO-8 ".10 11 £2

8.0 x 10- 19

Since, the efficiency of photoelectric emission is one for every 10 6 . Hence, total number of photoelectrons per second, nl 1OJ1 S nl

=106 = 106

= 10

(b) Maximum kinetic energy of photoelectrons , Krru = £ - W =2eV Stopping potential, Vo = 2 volt.

Ans.

Ans.

Example 21. Estimate the wavelength of characteristic X·radiation emitted from tUngsten (Z =74) atom in the electronic transition from n = 2 to n = 1. Solution: The formula for the total energy ofhydragen alom lion is

£=_I3.6Z

,

n' The transition n = 2 to n = I leads to an energy difference, M=E2 - EI

1 =(74)' XI3.6(1 - 2 , )=5.59XlO'eV Then,

,

l. = he = 12400 A = 12400 A = 0.022 om AE V 5.59 X 10'

Ans.

Partidt: and Wave

Natu~

. -.. ---- .... ...... ...... -.... - .

321

of Radiations (Light)

ASSIGNMENTS

rJ Conceptual Questions

--

-- ...... ··· ....... -.... .

· - .. -. ._-_

I. Are the magnetic field and electric field in an emw produced by a charge directly? Explain. 2. There are how many types of electromagnetic radiations? 3. What is the relation between infrared and heat? 4. What are the basic assumption of Planck to support his black-body radiation? 5. What is the limitation of We in's exponential law? 6. What is the basis of Kirchotrs law of radiation? 7. What is the basis of Einstein's photon concept? 8. In photoelectric effect, how does the classical wave theory fail to explain it? 9. How does Maxwell-Boltzmann theory of equipartition is violated by Planck to explain the black-body radiation? 10. Any oscillating system does not lose or gain energy continuously, this seems to be superfluous for a macroscopic oscillator such as a swinging pendulum bob or an oscillating spring mass system. Explain . 11. The energy of an oscillating system is consumed. How is it different from the statement that energy of a system is quantised? 12. If a system suddenly lose or gain energy, how cnn we conserve the energy of the system, ete just before and after the event? 13. Is a photon just a fla sh of light? Explain. 14. How can we say that a photon has a panicle and also a wave character? ] S. How can the photon theory explain photoelectric effect? 16. Can we detect a single photon for high frequency emw such as X-ray and 'Y·ray and Jaw frequencyemw (radiation) stich as radio power and microwave? Explain. 17. The

encr~y of an oscillator is E = ( n.+ ~tf but not E = "hf; n = 0,1,2 .. . What

is the significance of the quantity ~ . ]8. How can we imagine the presence ofa spin (,?r spin angular momentum) from a photon? 19. How does the photoelectric current depend on the retarding potential? 20. If N photons hit the surface, N electrons will be liberated. Is it true? 21. Can an object radiate energy at (i) O°C (ij) OK? Explain . . 22. What is the "mechanism of production of continuous X-rays? 23. How is Einstein-Bohr' s law of quantisali,on hold good for production of continuous X-ray? 24. Why do we have a limit to the wavelength .o fa continuous X-r:ays?

322

GRB Undentandins Physics Optics and Modem Phy10ia

25. How does minimum wavelength of continuous X-ray depend on the accelerating voltage? 26. How does the wavelength at which the intensity of X-ray will be maximum depend on the accelerating voltage? 27. Write the equation describing the wave-particle duality ofmdiation.

J

28. What is the physical significance of the tenn probability = E 2 dV ? 29. Can a radiowave emit electrons from a metal?

...

,

<

323

Particlt and Wan! Nature of Radiations (Light)

rJ Multiple Choice Questions (A) Only One Choice ;s Correct

Level·1 1. The voltage applied to an X·ray tube is 18 kY. The maximum mass of phalan

emitted by the X-ray tube will be: (b) 3.2 xlO - J6 kg (d) 9.1 x 1O- 3l kg

(a)2xlO-lJkg

(c) 3.2 x 10-32 kg 2. When an electron accelerated by potential difference U is bombarded on a specific metal, the emitted X-ray spectrum obtained is ShOYlIl in figure. If the potential difference is reduced to U/3, the correct spectrum is:

Am 2.5 ..... x-"'Y w",,~leng1h).

(b)

(a)

(d)

(c)

L--"---L--'_ _ _ 1.

3. Ka wavelength emitted by an atom of atomic number Z = 11 is A . The atomic number for an atom that emits Kr:r.radiation with wavelength 4A. is:

006

~

r•

(~4

(c) II (d) 44 4. In order to delcnnine the value of Eo. a scientist shines photons (light particles) of various energies at a cloud of atomic hydrogen. Most of the hydrogen atoms occupy the ground state. A detector records the intensity of light transmitted

\ 324

GRB Understanding Physics Optics and Modem Physics

1 ~~r

through that cloud (see figure). Next figure is a Laser graph of pan of the - - - - - .

scientist's data. showing the intcnsity of the

_______________ _

~---

transmitted light as a Glass container of function of the photon atomic hydrogen energy. A hydrogen atom's electron is likely to absorb a photon only jfthe photon gives the electron

enough energy to knock it into a higber shell. According to this experiment, what is the approximate value of Eo? (a) 1.6 x 10- 18 ]

1

(b)2. 1x lO- 18 ) (e) 3.2 x 10- 18 J

0.4

0.8

1.2

1.e

Pholon energy (j x 10-18 )

(d) 6.4 x 10- 18 )

5. Electrons with energy 80 keV arc incident on the tungsten target oran X-ray tube. K shell electrons oftungstcn have -72.5 keY energy. X-rays emitted by the tube

contain only: (a) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of - 0.155A (b) a continuous X-ray spectrum (Bremsstrahlung) with all wavelength (c) a continuous X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of - o.lssA and the charactcrstic X-ray spectrum of tungsten 6. Consider a hypothetical annihilation of a stationary electron with a stationary positron. What is the wavelength of the resulting radiation? (a) A

=...!!...

moc

(c) A =

h 2 2moc

(b) A =-1!!....

moc 2

(d) Noneorthese

7_ In a discharge tube, X-rays are produced by applying 40 kV potential difference and by allowing 10 rnA ofcurrent to pass through it. If 1% energy by the electrons is converted into X-rays, the power of X-rays beam is: (a) 0.5 W

(c) 10 W

(b) 1.0 W (d) 4 W

\

Particle and Wave Nature of Radiation! (Light)

325

J7

8. Figure shows Moseley's plot between and 2. where! is the frequency and 2 is the atomic number. Three linesA, Band Cshown in the graph may represent

IT A B

c

"ol----'--<-""'---z (3) K Q , Kp and Kr lines respectively

(b) K r• Kp and Ka lines respectively (c) K"/ , La and M 0. lines respectively (d) nothing 9. Ka wavelength emitted by an atom of atomic number Z = I I is A. The atomic number for an atom that emits Ko.radiation with wavelength 4A, is: (b) 4 (a) 6 (c) \I (d) 44 10. The shortest wavelength produced in an X-ray tube operating at 0.5 million volt is: (a) dependent on the target element (b) about 2.5 x 10- 12 m (c) double of the shortest wavelength produced at I million volt (d) dependent only on the target malerial 11. If the Ko. radiation of Mo(Z = 2) has a wavelength of 0.71 A, find the wavelength of the corresponding radiation ofCu (2=29). (a) IA (b) 2A (c) 1.52A (d) 1.25A 12. In X-ray tube. when the accelerating voltage Vis halved. the difference between the wavelength of Ka1ine 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 13. A Ka X-ray emitted from a sample has an energy of7.46 keY. Of which element is the sample made? (a) Calcium (Ca, Z ~ 20) (b) Cobalt (Co, Z ~ 27) (c) Cadmium (Cd, Z ~ 48) (d) Nickel (Ni, Z ~ 28) 14. 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 the tube (d) the atomic number of the target material

ORB Understanding Physics Optics and Modem Physicl

326

IS. The X-ray beam coming from an X-my tube will be: (a) monochromatic (b) having all wavelengths smaller than a certain maximum wavelength (el having all wavelengths larger than a ccnuin minimum wavelength (d) having all wavelengths lying between a minimum and a maximum wavelength

16. A time varying electric field produces a magnetic field, this is: (a) Planck's law of radiation (b) Faraday's law of electromagnetic induction (e) Maxwell's law of electromagnetic induction (d) none of the above 17. The electric field associated with an cmw is: (a) static (b) conservative (d) associated with charges (e) non-conservative

18. Which of the following is not electromagnetic wave? (a) ( ,ray (b) Infrared (cl Cosmic-ray (d) X-ray 19. Planck's law of radiation states that: (a) an oscillating charge has quantised energy (b) an orbiting charge muSI radiate emw (c) an oscillating charge must radiate energy (d) an accelcr:lIing charge will emil electromagnetic wave 20. A photon is: (a) a quantum of electromagnetic energy (radiation) (b) an arbitrary nash of light (c) not defined (d) the energy radiated by any heated body 21. Photoelectric effect was discovered by: (a) Einstein (b) Hem (c) Maxwell (d) Lenard 22. Photoelectric current : (a) depends on applied voltage forever (b) is independent on applied voltage (e) depends on applied voltage till it is saturated (d) can never be saturated 23. No. of photoelectronls is/arc: (a) proportional to the no. of incident photonfs (b) equal to the no. of incident photon/s (c) proportional to the rrequency of incident radiation (d) proportional to the cutoff voltage

"

327

Particle and Wave Nature of Radialion, (Light) 24. A photon cannot have: (a) energy (e) angular momentum

(b) linear momentum (d) reSI mass

25. The minimum wavelength Amin of continuous X-ray is directly proportional to: (a) applied voltage

(b)

I applied voltage

(e) speed of the electrons (d) none orthese 26. At A', let the intensity of continuous X-radiation is maximum. If V= accelerating voltage. I.' is proportional to: I,

- ¥---!,---- , ! (b) y'

(a) Y

"

(d)

1. Y

27. An X-ray can kick an electron. This show its: (a) wave property (b) particle property (e) both wave and particle property (d) penetrating nature 28. If N = photon density at a point P and Eo == electric field at that point. then Eois proportional to : (a) N

(b)

-IN

(d)

~

N 29. Which of the following graphs correctly represents the variation of particle momentum with associated de Broglie wavelength?

I

(a)

(b)

p

I p

•

I

(e)

(d)

p

•

-

- - -

-

I p

•

GRB Unde:!1tanding Physics Optics and Modem Ph)'SiCi

328

30. In Davison-Genner experiment. the correct relation between angle of diffraction ¢I and glancing angle 0: (0)0 =900

_22

(c)0 = 2

+!

(b) 0 =900

2

(d)O=$

2

31. A proton and an electron are accelerated by the same potential difference. Let f_ ~ and A. p denote the de Broglie wavelengths of the electron and the proton. respectively. Then: (a)Ae = f.p

(b)A«A p (c)A.e > Ap (d) the relation between AI' and A. p depends on the accelerating potential difference 31. An electron and a photon possess the same de Broglie wavelength . If £ (' and E ph are, respectively, the energies of electron and photon and 11 and c are their respective velocities. then E./ Eph = (a) l/ c (b) l n C (c) ul3e (d)
E

(b)

(a) 0

0

0

0

E

(d)

(c)

0

0

• <

- -'

E

0

0

329

Particle and Wave Nature of Radiations (Light)

34. Figure shows the plot ofthc stopping potential versus the frequency of the light used in an experiment on photoelectric effect. The ratio hIe is: 2

,, ,, ,

1.656

1

,,'35 ,,

V, 1 (in volt) 2

35.

36.

37.

38.

39.

3

4

5

(a) 1O- 15 ys (b)2xI0- ll ys (c) 3 x 1O-IS ys (d) 4.14 x 1O- 15 y s The resolving power of an electron microscope opernled at 16 kY is R. The resolving power of the electron microscope when operated at 4 kY. is: (a) RI4 (b) RI2 (c) 4R (d) 2R Find the ratio of the de Broglie wavelength ofa proton and an a-particle which have been accelerated through same potential difference. (0) 2.Jj (b)3J2 (c) 2J2 (d)3.Jj A photoelectric cell is illuminated by a small bright source of light placed at I 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) arc half the previous number (d) are one-quarter of the previous number 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 volts, is: (a) 2 (b) 4 (c) 6 (d) 10 The work function ora substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately: (a) 540 om (b) 400 om (c) 310 om (d) 220 om

330

eRB UnderstAnding Physics Optics and Modem Physia

40. A particle of mass M at rest decays into two particles of masses 1111 and 1112· having non-zero velocities. The ratio of the de Broglie wavelengths of the panicles ). II A2. is:

(a) 1111 / 11/2

(b)m2 Im\

(e) 1.0

(d) ~m2 / ,r.;;

41. A proton has kinetic energy E = 100 keY which is equal to that ofa photon. Th.e wavelength of photon is I. 2and that of proton is A,_ The ratio of A21 1.1 IS proponional to: (a) £2

(b) £ -112

(~£-I

OOE1a

42. In a photoelectric experiment. anode potential is plotted against plate current. Select the correct statement from the following: 1

V

(a) A and B will have different intensities while Band C will have different frequencies (b) Band C will have different intensities while A and C will have different frequencics (c) A and B will have different intensities while A and C will have equal frequencies (d) A and B will have equal intensities while Band Chave different frequencies

J. When the voltage applied to an X-ray tube increases from VI = IOkY to V2 = 20 kY. the wavelength interval between Ka line and cut-off wavelength of continuous spectrum increase by a factor of 3. Atomic number of the metallic target is: (b) 29 (a) 28 (d) 66 (e) 65 2. The power of an X-ray lube is 16 W. If the potential difference applied across the tube is 5 kY. then the number of electrons striking the target per second is: (b) 5 x 10 17 (a) 8.4 x 10 16 (e)2 x 1016

(d)2xI0 19

, Particle and Wave Nature of Radialions (Light)

331

3. Mark out the correct statement regarding X·mys: (a) When fa st moving electrons strike the metal target, they cnter the metal target and in a very short lime span come to rest. and thus an accelerated charged clcclron produces

(b) Characteristic X-ray arc produced due to transition of an electron from higher energy levels to vacant!ower energy level s (c) X-ray spcctmm is a di screte spectra just like hydrogen spectra (d) Both (3) and (b) are correct

4. Electrons in a hydrogen-like atom (Z = 3) make transitions from 4th excited state. The resulting radiations arc incident on a metal plme to eject photoclectrons. The stopping potential for photoelectrons ejected by the shonerwavclcngth is 3.95 V. The stopping pOIen!ial for the photoelectrons ejected by the longer wavelength, is: (a) 2.0 V (b) 0. 75 V (e) 0.6 V (d) none ofthesc 5. If an X·ray tubc 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 )( IO"Clkg: (a) I (b) 0.1 (e) 1.8 (d) 1.2 6. The potential difference across the Coolidge tube is 20 kV and 10 rnA 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 the X-ray beam is P. Then :

(a)P = O.lW (b)P=IW (e)P = 2W (d)P = IOW 7. An X-ray tube is operating at 150 kV and 10 rnA . Ifonly 1% of the electric power is converted into X·rays, the rate at which the target is heated, in cal s - I. is: (a) 3.5 7 (b) 35.7 (e) 4.57 (d) IS 8. When photons of wavelength Alare incident on an isolated spharc suspended by an insulated thread. the corresponding stopping potential is found to be V. When photons of wavelength ), 2arc used, the corresponding stopping potential was thrice the above value. I f light of wavelength A 2is used, calculate the stopping potential for this case.

(al "e[_1 + _1___ I]

e A3

2A.2

"I

(e) he[_1 + _I _ _I ] e A3 A2 A\

(b)

he[_1 +_1 _~] e A3

he[1

2A.2

2).\

I I]

(d)- - - - - e A3 12 AI

GRB Understanding Physics Optics and Modem Physics

332

9. A beam of electron accelerated by a large potential difference Vis made to ~trikc a metal target to produce X-rays. For which ortlle following value of V, will the

resulting X-rays have the lowest minimum wavelength? (a)10kY (c) 30 kY

(b) 20 kY (d) 40 kY 10. In the Bohr model ora 'It-mesic atom a it-meson ofmnsslI and of the same charge as the electron is in a circular orbital of radius r about the nucleus with an orbital angular momentum bl21t . If the radius of a nucleus of atomic on Z for which (coll 2 I TtmC 2 = 0.53 Aand mn I m~ = 264)1t-mesic moms might eltist is: (0) < 105

(b) > 105

(.)<37

(d»37

II. X-rays arc produced in an X-ray tube operating at a given accclernting voltage.

The wavelength of the continuous X-rays has values from: (a) 0 and co

(b) I, min to 00; where).. nUlL: < co (c) 0 to). m:u; where Amu < co (d)ArnintO)_mu ; where 0 <)' min < 00 12. Electrons with energy 80 keY are incident on the tungsten target of an X-raytubc. K-shell electrons of tungsten have -72.5 keV energy. X-rays emitted by the tube contain only: (a) a continuous X-ray spectrum (Bremstrahlung) with a minimum wavelength of - o.lssA (b) a continuous X-ray spectrum (Bremstrahlung) with all wavelength (c) a continuous X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremstrahlung) with a minimum wavelength of - o. lssA and the characteristic X-ray spectrum of tungsten 13. The intensity of X-rays from a coolidge tube is plotted against wavelength A as shown in figure . The minimum wavelength found iSA c and the wavelength ofthc Ka linc is"). k. As the accelerating voltage is increased: I

(a) Ale -").cincreases (c) Ale increases

(b) A. Ie ·A. cdecreases (d) A. k decreases

333

Partide and Wave Nature or Radiations (Light)

14. The potential difference applied to an X-ray tube is 5 kY and the current through it is 3.2 mAo Then, the number of electrons striking the target per second is:

1 (e) I x 10 17

(a)2 x 10 '

(b) 5 x 10' (d) 4 x lOll

IS. Which one of the following statement is wrong in the context of X-rays generated from an X-my tube? (a) Wavelength of characteristic X-ray decreases when the atomic of the target increases (b) Cut-out wavelength of the continuous X-rays depends on the atomic number of the target (c) Intensity ofthe characteristic X-rays depends on the electric power given to the X-ray tube (d) Cut-off wavelength oflhe continuous X-rays depends on the energy oflhe electrons in the X-ray tube 16. The density of photons emined by a point source S at a " / distance r is x. Al a distance 2r, the density of photon is: ;."", (a)~

(b)~

(e)L

(d).L.

2

4

,/2

2,/2

17. 10-3 W of5000A light is directed on a photoelectric cell. If the current in the cell is 0.16 rnA. the percentage of incident photons which produce photoelectrons, is: (a) 40% (b) 0.04% (c) 20% (d) 10% 18. A particle of mass to-llkg is moving with a velocity equal to 10 5 m s-1 . The wavelength orthe particle is equal to:

(b) 6.6 x IO- 'm (d) 1.5 x 10 7m

(a) 0

(e) 0.66 m 19. In the experiment on photoelectric effect. the graph between EK(max) and v is found to be a straight line as shown in figure.

~

•

-

b

4

3.6

2

x

2

2 34

6 8 10 v(10P8 ,,'

The threshold frequency and the Planck's constant are:

GRB Understanding Physi ~ Optics and Modem Ph),!ics

334

(b) 6 x 10 14 5-1,3 x 10-34 J-s 34 (e) 6 x 10 14 5- 1,6 x 10- 34 J-s (d) 2Hz, 6 x 10- J-s 20. Silver has a work function o f 4.7 eV. When ultraviolet light of wavelength 100

(a) 3 x 10 14 5- 1, 6 x 10-34 J-s

mm is incident upon it, a potentia! of7.7 V is required to SlOp the photoelectrons

21.

22.

23.

24.

from rcaching the collector plate. How much potential will be required to SlOP the photoelectrons when light of wavelength 200 mm is incident upon silver? (a) 1.5 V (b) 3.85 V (c) 2.15 V (d) 15.4 V The kinetic energy of an electron is E when the incident wavelength is i.. To increase the KE or the electron to 2£, the incident wavelength must be: (a) 2A (b) 1.. /2 (c) (h eA) / (D. + he) (d)(hc).) / (D. + he) When a metallic surface is illuminated by a light of frequency 8>: 10 14 Hz, photoelectron of maximum energy 2 eV is emitted. When the same surface is illuminated by light offrequencyl2 x 10 14 Hz. photoelectron of maximum energy 2 eV is emitted. The work function is: (a) 0.5 eV (b) 2.S5 eV (e)2.5 eV (d) 1.5 eV The de Broglie wavelength of neutrons in thennal equilibrium is: (a)30.S / .fi A (b)3.0S/.fi A (c) O.lOS/.fi A (d) O.OlOS / .fi A A nozzle throws a stream of gas against a wall with a velocity v much larger than the thermal agitation of the molecules. The wall deflects the molecules without changing the magnitude of their velocity. Also. assume that the force exerted on the wall by the molecules is perpendicular to the wall. (This is not strictly true for a rough wall). Find the force exerted on the wall.

•

I

N

, _I

A

i

1-

(a) Anmv2 cos 2 (I

(b) 2Allmv2 cos 2

(c) 2Allnlv2 sin 2 (I

(d) Anmv2 cose

25. A plane wave ofintensity I = 0.70W em - 2 illuminates a sphere with ideal mirror surface. The radius of sphere is R = 5.0 cm. From the standpoint of photon theory, find the force that light exerts on the sphere. (a) 0.8 ~N (b) 0.2 ~N (c) 0.5 ~N (d) 1.2 ~N

1,

e

,

~

• • •

•

Particle and Wa~ Natu~ of Radiations (Light)

335

26. If the short wavelength limit of the continuous spectrum coming out ofa coolidge tube is loA, thcn the de Broglie wavelength of the electrons reaching the target metal in the coolidge tube is approximately : (a) O.3A (b) 3A (0) 30A (d) lOA 27. In a photoelcctric experiment, with light of wavelength A, the fastest electron has speed v. If the exciting wavelength is changed to 31.. 14. the speed of the fastest emitted electron will become:

(a) v.J3/4

(b) vJ413

(c) less than u..ffj4

(d) greater than v~3 / 4

28. Two identical non-relativistic particles A and B move at right angles to each other, processing de Broglie wavelengths Ai and 1.. 2• respectively. The de Broglie wavelengths of each particle in their C frame of reference is:

(a) l.,+A.,

(b)2l.,l., / (~l.l +l.~) (d)(l., +l.,)/2

29. A sodium melal piece is illuminated with light of wavelength 0.3~m. The work function of sodium is 2.46 eV. For this situation. mark out the correct statement(s): (a) The maximum kinetic energy of the ejected photoelectrons is 1.68 eV (b) The cut-ofTwavclength for sodium is 505 nm (e) The minimum photon energy of incident light for pholoeletric effect to take place is 2.46 eV (d) All of'he above 30. The de Broglie wavelength ofa thennal neutron at 927°C is A. Its wavelength at 327°C will be: (b) l. /.J2 (a) l.!2 (d) 2l. (c) l..J2

31. The radius of second orbit of an electron in hydrogen atom is 2.l16A. The de Broglie wavelength associated with this electron in this orbit would be: (a) 6.64A

(c) 2.116A

(b) 1.058A (d) IJ.2sA

32. The light sensitive compound on most photographic films is silver bromide AgSr. A film is exposed when the light energy absorbed dissociates this molecule into its atoms. The energy of dissociation of AgSr is 10 5J mol-i. For a photon that is just able to dissociate a molecule of AgBr, the pholon energy is: (a) 1,04 eV (b) 2.0S eV (c)3.12 eV (d) 4.16 eV

\

336

CRB Undentanding Ph~iCl Optics and Modem Physics

33. If a star can convert nil the He nuclei completely into oxygen nuclei. the energy released per oxygen nuclei is [ Mass of He nuleus is 4.0026 amu and mass of oxygen nucleus is 15.9994 amu]: (a) 7.6 MeV (b) 56.12 MeV (c) 10.24 MeV (d)23.9 MeV (8) More Than One Cho/cels Is/are Correct I. Which are the laws of electromagnetism? (a) Faraday's law ofinduclion (b) Gauss's law

2.

3.

4.

5.

6.

7.

8.

9.

10.

(c) Ampere's law (d) Maxwe11's law of induction A displacement current: (a) is a time varying magnetic field (b) is a time varying electric field (d) can produce magnetic field (c) can produce electric field An emw has: (a) energy (b) linear momenlum (d) inertial mass (c) angular momentum A photon is effcctcd by: (b) magnetic field (a) electric field (c) gravitational field (d) all of these . When an cmw interacts with a free material particle, which of the following is/are conserved? (a) Linear momentum (b) Angular momentum (c) Kinetic energy (d) Total energy The classical wave thcory of light fails to cxplain which aspects of photoelectric effect (a) intensity (b) frequency (c) time lag (d) none of these An incident photon lights with which of the following factors before causing the ejection ofan electron? (a) Work function (b) Inlemal collision (d) None of these (c) Binding energy ofan electron A photon has: (a) gravitational mass (b) spin (d) energy (c) linear momentum In an X-ray production, energy conversion takes place from: (a) electrical-+ mechanical (b) mechanical-+ electromagnetic (d) all of these (c) electromagnetic -+ mechanical In coolidge tube, the target material must have: (a) high m.p. (b) high m.p. and low conductivity (c) high conductivity and high specific heat (d) high conductivity

Particle and Wave Nature of Radiations (Light)

, 11. When an electron passes through an atom to produce continuous X-rays, whiCh of the following is/are conserved? (a) Linear momentum (b) Angular momentum (c) Total energy (d) Nonc of tile above 12. In a coolidge tube, when the fast moving electrons strike the target: (a) continuous X-rays are produced (b) heat (infrared) is produced (c) visible light is produced (d) characteristic X-rays are produced 13. A minimum (limiting) wavelength of X-rays for a given accelerating voltage is possible due to: (a) wave nature of radiation (b) particle nature of radiation (c) wave and particle duality of radiation (d) penetrating nature of radiation 14. Let). u I. ~ and A~ denotc the wavelength of the X-rays ofthe'K0., K~ and Lulines in the characteristic X-rays for a metal. Then: (ap,~ >/'0. >/.p (b)A~ >A~ >Au (c)_I =_1 +_1_ Ap '-u A~

(d)_I = _1 +_1 Au A~ I.~

IS. An X-ray tube is operating at 50 kV and 20 rnA. The target material of the tube has mass of I kg and specific heat 495 J kg-toe· l . One per cent of applied electric power is converted into X-rays and the remaining energy goes into heating thc target. Then: (a) a suitable larget material must have high melting temperature (b) a suitable target material must have low thennal conductivity (c) the average rate of rise of tcmperature of the target would be 2°C s- I (d) the minimum wavelength of X-rays emitted is about 0.25 x lO- lo m 16. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation: (a) the intensity increases (b) the minimum wavelength increases (c) the intensity remains unchanged (d) the minimum wavelength decreases -

CRB Undentanding Physics Optics and Modem Physics

338

17. The graph between the stopping potential (Vo) and wave number (1O.) is as shown in figure. If/is the work function, then : Vo

Volt

Metal , Metal2 Metal3

(a)h'I>2 : ~) = 1:2:4 (b)h~, : ~ ) = 4:2:1

(e) tan 0 IX he/e, where 0 is the slope (d) ultraviolet light can be used to light photoelectrons from metal 2 and mCln13 only

18. A point source oflighl is taken away from the experimental setup of photoelectric effect. For this situation, mark out the correct statement{s): (a) Saturation photocurrent decreases (b) Saturation photocurrcnt increases

(c) Stopping potential remains the same (d) Stopping potential increases 19. When a monochromatic point source of light is at a distance of 0.2 m from photoelectric field , the cut-off voltage and the saturalion current are, respectively. 0.6 V and 18.0 rnA. If the same source is placed 0.6 m away from the

photoelectric cell. then: (a) thc slopping potential will be 0.2 V (b) the slopping potcntial will be 0.6 V (c) the saturation cunent will be 0.6 rnA (d) the salUration current will be 0.2 rnA 20. The maximum kinetic energy of the emitted photoelectrons against frequency v of incident radialion is plolted as shown in figure. This graph help us in detennining the following physical quantities:

I

~<

.

•

Particle and Wave Nature of Radiations (Light)

339

(a) work function of the cathode·metal (b) threshold frequency (c) Planck's constant (d) charge on an electron 21. Photoelectric efTect supports the quantum nature of light becausc: (a) there is a minimum frequency of light below which no photoelectrons are emitted (b) the maximum KE 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 photoelectrons is quantized 22. When a point light source of power IV emitting monochromatic light of wavelength I. is kept at a distance a from a photo·sensitive surface of work of work function!f! and area S. we will have: (a) number of photons striking the surface per unit time as WAS 14rthca 2 . (b) the maximum energy of the emitted photoelectrons as (1/ A)(hc-A!f!) (e) the stopping potential needed to stop the most energetic emitted photoelectrons as (c / I. )(hc - I.!f!) (d) photo·emission only ift. lies in the range 0 ~ A. ~ (he I,) 23. A collimated only on a 100 mm 2 completely absorbing screen. If Pis the pressure exened on the screen and l1p is the momentum transferred to the screen during a 1000 s interval, then: (a)P=IO-'Nm-2 (b)P=I0--4 Nm -2

"(e) t.p = 10--4 kg m ,-1

(d) t.p = 10- ' kg m ,-1

24. Threshold wavelength of cenain is ,. o· A radiation of wavelength A > A 0 is incident on the plate. Then, choose the correct statement from the following: (a) Initially, electrons will come out from the plate (b) The ejected electrons experience retarding force due to development of positive charge on the plate (c) After sometime, ejection of electrons stops (d) None of the above 25. A laser used to weld detached retina emits light with a wavelength of652 nm in pulses that are 20.0 ms in duration. The average power during each pulse is 6 W. Then: 19 (a) the energy of each photon is 3.048 x 10- J (b) the energy content in each pulse is 12 mJ (c) the number of photons in each pulse is nearly4x lOIS (d) the energy of each photon in nearly 1.9 cV

340

GRB Undtntanding Physics Optics and Modtm Physics

26. The threshold wavelength for photoelectric emission from a materi~1 is 5200~. Photoelectrons will be emitted when Ihis material is iIIummated with monochromatic radiation from a: (a) 50W infrared lamp (b) I W infrared lamp (c) SOW ultraviolet lamp (d) I W ultraviolellamp 27. Photoelectric cffect supports of light below which no photoelectrons are emitted: (a) there is a minimum frequency of light below which no photoelectrons arc emiucd (b) the maximum kinetic energy of photoelectrons depends oolyon the frequency of light and not on its intensity (c) even when the metal surface is faintly illuminated, the photoelectrons leave the photoelectrons leave the surface immediately (d) electric change of the photoelectrons is quanlized 28. When a monochromatic point source of light is at a distance of 0.2 m from a photoelectric ceil, the cut-off voltage and the saturation current are. respectively, 0.6 V and 18.0 mAo If the same source is placed 0.6 m away from the photoelcctric cell, then: (a) the stopping potcnlial will be 0.2 V (b) the stopping potential will be 0.6 V (c) the saturation current will be 6.0 rnA (d) the saturation current will be 2.0 rnA 29. When photons of energy 4.25 eV strike the surface of metal A. the ejected photoelectrons have maximum kinelic energy T" and de Broglie wavelength).A. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA -1.50) eV. If the de Broglie wavelength of these photoeleclrons is ). B = V. A. Broglie wavelength of these photoelectrons is}. 8 :: V. A • then: (a) the work function of A is 2.25 eV (b) the work function of B is 4.20 eV (e) TA = 2.00 eV (d)Ts =2.75eV

341

Particle and Wave Nature of Radiations (Light)

[JIMatch Tho Columns I. Match the Columns I and II . Column II

Column I (a)

Faraday' s law of induction

(P)

Energy of an oscill ator is quantised

(b)

Maxwell's law of electromagnetic induction

(q)

Time varying magnetic fi eld produces an electric fi eld

(e)

Wein's displacement law

(,)

Frequency of most intense radiation is directly proportional to the absolute temperature of the body

(d)

Plank's law of quantisation

(,)

Time varying electric fie ld produces a magnelic field

2. Match the Columns I and II. Column I

Column II

(a)

Photoelectric effect

(P)

Momentum p ;;; ~

(b)

X·rays

(q)

Emission of radiation from healed objects

(e)

Black body radiation

(, )

Emission of energetic radiation when fast moving electron hit the heavy ml!ta\ targets

(d)

Wave- particle duality

(, )

Emission of electrons when light (radiation) falls in a metal.

3. Match the Columns I and II. Column I

Column II

(a)

A charge moving with unifonn velocity

(P)

Can radiate energy

(b)

An accelerating charge

(q)

Ene r~y

remains conserved

342

GRB Undentanding Physics Optics and Modem Physics

(e)

A radio-active material

(,)

Can lose electrons

(d)

An orbiting electron in a hypothetical hydrogen alom in a given (fixed) orbit

(s)

Must have electric field

(')

Can produce magnetic field

rJComprehensions Passage--1 A 0.5 A X-ray strikes a metal, As a result it liberates photoelectrons from K-shell which moves in a circle of radius 23 mm under the action of a magnetic field of 2x 1O-2 T ,

1. The KE of the phOioelectron is: (a) 25.1 keY (h) 16.9 keY

(e) 18.6 keY 2. The energy of the incident X-ray pholons is: (a) 18,6 keY (b) 24.8 keY (e) 6,2 keY 3. The bending energy of the K-shell electron is: (a) 6.2 keV (b) 12.4 keY . (c) 24.8 keY

(d) 20.6 MeY (d) 12.4 KeY

(d) 18.6 keY

Passage--2 light form a hydrogen-gas discharge lUbe is used to emit electrons from sodium metal of work function Wo = 1.82 eV. . 1. If the fastest photoelectron has KE of O.73eV, the incident radiation on the sodium metal is: (a) 5627.63A (h) 4862.74A (d) 6813.18A (e) 4523.83A 2. The transition in hydrogen atom takes place from: (a)nr =4ton2:::1 (b)nr = 4ton2 =3 (c)nr :::3ton2 = 1 (d)nr =4ton2:::2 3. The angular momentum of the emitted radiation is: (a) ~

(h) 2h

(e).!!.

(d) 3h

•

2~

•

2~

Partide and Wa\'e Natu~ of Radialions (Light)

343

Passage.3 Different energy states of an atom are shown in thc figurc. N - - - - - - - - - - - - - E. M

L

_______________________ E,

-------------------------~ (in eY)

K _____________________ E,

I

K.L.M. elc. Icvels have energies E], E 2, E3.etc. respectively. l. The potcnlial difference Ihrough which an electron must lravello knockout the K-electron is proportional to:

~) .Jfi

(a) E,

(e)

.JE2

(d)

,J~{E;"'2--~E~,)

2. Jflhe accelerating voltage for the electron is reduced to : recoil electron is (m = mass of the elcctron);,

= __=

E,' n, the speed of the

(e) P{E, E,) (d) IE," nm '1- m m 3. The X-radiation{let) emitted when an electron from K-shell is knockout in Q.1. TIl en. the wavelength of X-radiation is: (alO

(b) /;E'

he ~) he (e) he £2 - E1 E2 E2 +E\ 4. If the X·ray in Q-3 knocks an electron inM-shell, the KEofthe photoelectron is:

(a)

{a)E,-E,+EJ (e) E, +EJ -E,

{b)E,+E,-E, {d)E,-E, -EJ

P....ge4 When high energetic electron beam, (i.e., cathode rays) strike heavier metal. then X-rays are produced. Spectrum of X-rays classified into two categories: (i) continuous spectrum. (ii) characteristic spectrum . The wavelength of continuous spectrum depends only on the potential difference across the electrode. But wavelength of characteristic spectrum depends on the atomic number (Z).

eRB Understanding Physics Oplic.s and Modem Physics

344

I. The production of characteristic X-ray is due to the: (a) continuous acceleration ofincklcnt electrons toward the nucleus

(b) continuous retardation of incident electrons toward the nucleus (e) electron lransitions betwe..:n inner shells of the target atom (d) electron transitions between outer she\1s of the target alom

2. The production of continuous X-ray is due to the: (a) acceleration of incident electrons by the nucleus of the target atom (b) electron transitions between inner shells of the target atom (e) electron transitions between outer shells of the target alom (d) annihilation of the mass of incident electrons

Passage-S Light from a discharge tube containing hydrogen atoms falls on the surface of a plate 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. t. The energy of the photons causing the photoelectric emission is: (a) 2.55 cV (c) 1.82 eV

(b) 0.73 cV (d) infonnation insufficient

2. The quantum numbers of two levc\ involved in the emission of these phot? ns is:

(0)4-+ 2 003-+2

(b)3 -+ 1 004 -+ 3

3. The change in the angular momentum of thc electron in the hydrogen atom in

above transition is: (a) 211

(b) .!!.

(c) ~

(d) .!!.

,

"

2, 4,

Passage-6

The incident intensity on a horizontal surface at sca level from sun is about tkW m-2 1. Assuming that 5~ per cen~ o.f this intensity is reflected and 50 per cent is

absorbed. detennme the radIatIon pressure on this horizontal surface: (a) 8.2)(10- 2 (b)5 x l0 -6 5 (c)3xlO(d)6 x lO - s 2. FI ind the ratio of this pressure to atmospheric preSSure Po (about I x 10SPa) at sea evel; (a) 5 x 10- 11 (c) 6)( 10- 12

(b)4xI0- 8 (d)8 x I0 - 1I

Partide and Waw! Nature of Radiations (Light)

345

Passag~7

The energy received from the sun by earth and surrounding atmosphere is 2 calcm - 2 min -Ion a surface nonnal to the rays of sun.

1. What is tolal energy received. injoulc. by the earth and its atmosphere? (a) 10.645 x 10 18 J min -I (b) 10.64S x 10 15 J min - I (c) 8.645 x 10 17 J min - t (d) 9.645 x 10 14 J min - I 2. What is Ihe total energy radiated in J min -I. by sun to the universe? (a) 2.3444 x 10 28 J min - I (e) 2.34 x 10 20 J min- I

(b) 2.33 x 10 24 J min- I (d) none of these

3. At what rate, in mega gram per minute, must hydrogen be consumed in fusion reaction to provide the sun with the energy it radiates? (a) 3.66 x 10 14 mega gram min-I (b)3 .66 x lO I6 mega gram min-I (c) 3.66 x lOI S mega gram min - I (d) 3.66 x 1010 mega gram min - t

Passage41 When a high frcquency electromagnetic radiation is incident on a metallic surface. electrons are emitted from the s urface. Energy of emitted photoelectrons depends only on the frequen cy o f incident electromagnetic radiation and number of emitted electrons depends only on the intcnsity of incident light. Einstein photoelectric equation [K IRU ::: hv - ¢ ]correctly explains the PE, where u::: frequen cy of incidcnt light and ¢ ::: work function . I. Light of wavelength 33 00 A is incident on two metals A nnd B. whose work functions arc 4 eV and 2 eV, respectively, then: (a) A will emit photoelectrons but B will not (b) B will emit photoclectrons. but A will not (c) both A and B will not emit photoelectrons (d) neither A nor B will emit photoelectrons 2. Forpholoeleclric effect in a metal. the graph of Vo the stopping potential Vo(in volt) versus 20 frequency v (in hcrtz}of the incident radiation is shown in figure. The work function of the 12 metal (in eV)is : 8 (~12 . 5

(b) 14.5 (e) 16.5 (d) 18.5

4

'-<-_____-..-.4-.:;...i...i.-

GRB Undentanding Physics Opties and Modem Physics

346

3. The slope ofthc graph shown in the figure [here. h is the Planck's constant and e is the change oran electron] is: (a)

~ e

(h) eh

(e) h

(d) ! h

Passage-9

A Cs plate is irradiated with a light of wavelength I. = hcl¢. q, being the work function of the plate, Ii the planck's conSlant, and c the velocity of light in vacuum. Assume all the photoelectrons arc moving perpendicular to the plate toward a YDSE setup when accelerated through a potential difference V. Take charge on a prolon :: e and mass oran elCi:tron = m.

Cs

I 'j ~-------- a I-D-

Read the paragraph carefully and answer the foll owing questions: 1. The fringe width due to the electron beam is: (a) W id (e) hD/(dJ2emV)

(b) W l2d (d) none of these

2. If the wavelength of light used in photo emission is Jess than A, then the fringe width will: (a) increase (b) decrease (c) remain same (d) cannot be decided 3. Instead of moving perpendicular to the plate. if the electrons were moving randomly, then the central maximum would shift: (b) downward (a) upward (c) no shift (d) no fringes will be fonned

P.... g.-10 The figure shows a surface XY separating two transparent media. medium-l and medium-2 . The lines ab and cd represent wavefronts of a light wave travelling in medium-I and incident on XY. The Jines eland gh represent wavefronts of the light wave in medium-2 after refraction.

347

Particle and Wave Nature of Radialions (Light) b

d

,

a

Medium .,

x

y

h

Medium -2

o

g

I. Speed oflight is: (a) the same in medium-I and medium-2 (b) larger in medium-I than in medium-2 (c) larger in medium-2 than in medium·1 (d) different at band d 2. The phases of the light wave at c,d, e and ~ c,~ d,~ t and ~ f respectively. It is giventhat~c *~r (b)~d can equal to~f' (a) ~ c cannol be equal to~ d (e) ($d -~ f) is not equal to (~c -~e) (d)(9 d -$ c) is not equal (~ f . ~, t") 3. Light travels as a: (a) parallel beam in each medium (b) convergent beam in each medium (c) divergent beam in each medium (d) divergent beam in one medium and convergent beam in the other medium

rJ Subjective Problems L..el·l Planck's Law and Thermal Radiation 1. What is the peak wavelength of radiation emitted by human body when the skin temperature is 35°C? 2. Find the peak wavelength of radiation of the sun. 3. A block ofm = I kg is attached to a massless springofforce constant K = 25Nfm. The spring is stretched to 0,2 m from the equilibrium position and released. Find the (a) total energy of the spring (b) value of quantum number n.

348

GRB Undentanding Physics Optics and Modem Ph)'1oia

PhotOQ

4. A radio station broadcasts at 91 .5 megahertz with a power of 4S kW. Find the (a) magnitude of momentum of each photon (b) wavelength orthe emw (e) no. of photons emitted Is.

5. Find the change in momentum ofa photon oD. (a) absorbed (b) reflected by a surface.

0::

5000 A, when it is completely

Photoelectric effect

6. A photon of energy 3.8 cV falls on a metal ofwark function 3.2 cV. Find the (a) kinetic energy (b) speed orthe faslest photoelectron. 7. A light of 5000A fall s on a metal of work function I.SeY.Find the maximum KE orthe photoelectrons. 8. In a photoelectric effcct a reverse potential difference of 1.3 volt can stop the phOiOCUITCnt. Find the (a) maximum KE (b) maximum speed of the

photoelectrons. 9. A sodium metallic foil is irradiated by a light if wavelength 3000 A. The work-function of sodium is 2.46 eV . Find the (a) maximum KE of the photoelectrons, (b) cut off wavelength, (c) maximum speed of the photoelectrons. X-rays 10. The accelerating potential in cool edge tube is 12 kY. Find the minimum wavelength of X-rays produccd. 11. Older telcvision sci B havc cathode ray (picturc) tube which operatcs at25 kV. (a) What kind of X-rays arc produced? (b) Why can't we be affccted by these X-rays? 12. An X-ray lube operated a140 kV emits continuous X-rays spectrum with a short wavelcngth limit 00. min ::: 0.3 loA. Find Planck 's constant. Mosley's law 13. K,. radiation ofMo (Z = 42) has a wavelength ofO.7IA. Find the wavclength of the corresponding X-radiation forCu(Z = 29) target. Bragg's law 14. X-radiation from Molybdenum target (O.626A) falls on a crystal with atomic plane spacing of 4A. Find the three smallest angles at which maximum intensity occurs for the diffracted beam.

(J Subjective Problems Level-2 t. A flat source of powcr P cmits radiation of wavelength )•. Find the
Par1ide and Wa\'e Nature of Radiations (Light)

349

3. Two radiations of wavelengths A I = 1000 A and A2 =2000 A fall on a photo (metallic) cathode of work function Woo If the speed of the fastest photoelectron in one case is double that of the other, find the value ofWo. 4. A ray of light (emw) whose electric field varies as E = Eo (I + Coswlt) 'COSW21 falls on a photo cathode of work function Woo Find the speed of the fastest photoelectron emiul!d from the surface. Put WI ::= 10 ISrad/s, (J,)2 = 0.5 x IOl5 radis and "'0 = I.leV. 5. A phOlo-celi has two electrodes cesium (emitted) and copper (eolleetor).The contact potentials of cesium and copper are 3.24 and 4.24 volts respectively. Ira ray of light of wavelength 2500 A strikes the cesium plalc assuming the work function of cesium as 2.96 eV, find the speed of the fastest photoelectron striking the copper electrode. 6. A beam of light of intensity I strikes a reflecting surface of coefficient of reflection p al an angle ewith nonnal to surface. Find the momentum delivered by light on the surface. [Put p = 0.75, 0 = 300,1 = 0.25 Wk m 2J 7_ A uniform beamoflighl ofinlensity I = 9 x 10 4 W1m 2 falls on reflecting sphere ofrndius R = IOcm. Find the thrust of the light on the sphere assuming 100% reflection of light.

• • •

•

8. A source of laser light emils a light pulse of duration T= 0.15 ms and energy E = 12 J. Find the average pressure eXCited by the pulse when focussed to a spot ofdiameter d( = 2R) = I OJ..lm . Assume the reflection coefficientp =0.5. 9. A cavity of volume V = 2 litre is filled with a thermal radiation at temperature of T= 1000 K. Find the heat capacity of the cavity. 10. The wavelength of the maximum intensity of radiation of two black body I and 2 corresponds to the temperntures Tt and T2« T[). If the difference in these wavelength is ill., Find T2 . Put 6A = 0.5 x 10-6 m, Tl = 2500 K and b =\Vein's constant =' 2.9 x 10- Jm-K. 11. Using Planck 's formula, derive the expansion for the monochromatic energy density for the cases (a) 1100« kT (Ray Leigh-Jeans fonnula) (b) llro » kT (Wein's fonnula). 12. (a) A point source of power Pemits energy ofwavelengthA. At a distance r from the source, find the number of photons crossing per unit area per second (unit time). (b) Find the answer numerically in (a) by putting P = 20 W, I. = 589 nm andr=2m. .

350

GRB Undtntaoding Physics Optics and Modtm Physics

13. Desire the expression for photon density (number ofpholon per unit volume) at temperature Twithin the spectral interval (a) 00, 0) + dO} (b);" ,A + dJ.., by using Planck's law of radiation. 14. Express fill)

the

= h(J)3

n 2 c3

monochromatic

1

energy

dens ity

Planck's

radiation

law

as the function of (a) frequency!(b) wavelength

/100

ciT _ I

IS. A cobalt target is bombarded with electrons and the wavc~c~gth of its characteristic spectrum arc measured. A second, fainter, charactensllC spectrum is also found because oCan impurity in the larget. The wavelength oCthe K a lines are 178.9 pm (cobalt) and 143.5 pm (impurity). What is the impurity? 16. The K-absorption edge oCan unknown element is O.171A. (a) Identify the element. (b) Find the uvcrage wavelengths of the K et • KlJand Kylines. (c) Ifa 100 cVelcclron strikes the targct of this elcmcnt, what is the minimum wavelength of the X-ray emitted? 17. Ultraviolct light of wavelcngths sooA and 700A when allowed to fall on hydrogen atoms in their ground states is found to liberate electrons with kinetic energies 1.8 cVand4.0eV. respectivc\y. Find the value of Planck's constant.

18. The wavelength of the characteristic X-ray, K u line emitted by a hydrogcn-Iike element is 0.32A. Calculate the wavelength of KIJ line emitted by the same element. 19. (a) A stopping potential of 0.82 V is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength 4oooA. For light of wavelength 3000A, the potential is 1.85 volt. Find the value of Planck's constant. [1 electron-volt (eV)= 1.6x 1O- 19joule) (b) At stopping potential, if the wavelength of the incident light is kept fixed at 4oooA, but the intensity of light increased two times. will photoelectric current be obtained? Give reasons for your answer. 20. Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing 0. particles. A maximum energy electron combines with an 0. particles to form a He + ion, emitting a single photon in this process. He + ions thus formed are in their fourth excited state. Find the energies in eV of the photons, lying in the 2 t04 eV range, that arc likely to be emitted during and after the combination. {Takeh = 4. 14 x IO- IS eVs] 21. When a beam of 1.6 eV photons of intensity 2.0 Wm -2 falls on a platinium surface of area 1.0 x lO-4 m 2 and work function 5.6 eV, 0.53% of the incident photons eject p~otoe~e~trons. Find the. number of phOloelectrons emitted per second and their mlOlmum and maximum energies (in eV). [Take I eV =1.6 x lO- 19 Jl

351

Particle and Wave Nature of Radiations (Light)

Multiple ChoIce QuestIons (A) Only One Choice Is Correct Level·1

6. (a) II. (b) 16. (c) 21. (b) 26. (c) 31. (c) 36. (c) 41. (b)

2. (b) 7. (d) 12. (d) 17. (c) 22. (c) 27. (b) 32. (b) 37. (d) 42. (d)

3. (c) S.(d) 13. (d) IS. (c) 23. (a) 2S. (b) 33. (d) 3S. (b)

4. (b) 9. (a) 14. (b) 19. (a) 24. (d) 29. (d) 34. (d) 39. (c)

5. (d) 10. (c) 15. (c) 20. (a) 25. (b) 30. (a) 35. (b) 40. (c)

Level-2 I. (b) 6. (b) II. (b) 16. (b) 21. (c) 26. (a) 31. (a)

2. (c) 7. (a) 12. (d) 17. (b) 22. (c) 27. (d) 32. (a)

3. (b) S. (b) 13. (a) IS. (b) 23. (a) 2S. (b) 33. (c)

4. (b) 9. (d) 14. (a) 19. (a) 24. (a) 29. (d)

5. (b) 10. (c) 15. (b) 20. (a) 25. (a) • 30. (c)

I. (c)

(8) More Than One Cholcels Is/are Correct 3. (a,b.c.d) 2. (b,d) I. (a.b.c,d) 8. (a,b.c,d) 7. (a.b.c) 6. (a.b.c) 13. (b,c) 12. (a.b.c) II . (a,b,c) IS. (a,c) 17. (a,c) 16. (c,d) 23. (b,d) 22. (a,b,d) 21. (a,b,c) 2S. (b,d) 27. (a,b,c) 26. (c,d)

Match the Columns I. a-q ; b-s ; c-r ; d-p 2. a--s;b - r;c--q;d-p 3. a-q, s, t; b-p, s, t ; c-p, r ; d- p, q. s, t

4. (c) 9. (b) 14. (a.c) 19. (b,d) 24. (a,b) 29, (.,b,c)

5. (a.b,d) 10. (a.d) 15. (a.c) 20. (a,b,c) 25. (a.b,c,d)

352

GRB UndC'ntandins:'Physics Optics and ModC'm Physics

Comprehensions

Passage-I: 1. (e) Pauage-2 : I. (b) Passage-3 : 1. (a) Passage-4 : 1. (e)

2. (b)

3. (a)

2. (d)

3. (a)

2. (e)

3. (a)

4. (d)

2. (a)

Passage-S : 1. (a) Passage-6 : 1. (b) Passage-7 : 1. (a) Passage-8 : I. (b) Passage-9 : 1. (e) Passage-IO: 1. (b)

2. (a)

3. (e)

2. (a) 2. (a)

3. (b)

2. (e)

3.
2. (b)

3. (e)

2. (e)

3. (a)

Subjective Problems Level·1 1. 9.4 x IO - 6 m ( Infrared) 2.0.5 x IO-6 m, this is the centre of the visible spectrum (yellQw-grcen)

3. (a) 0.5 J

(b) 9.478 x 10"

4. (a) 2.02)( 10-34 kg mls S. (a) 1.3252 x 1O - 21 kg mls 6. (a) 0.96 x 10- 19 J

(b) 3.278 m (e) 7.422 x 103Sphoton Is 21 (b) 2.65 x 10- kg mls (b) 0.459 x 10'mls

7.I.088x l O- 19 }

8. (a) 2.08 x 10- 19J

(b) 0.676 x lO'mls

(b) 5040.65 A (e) 0.766 x 10'mls 9. (a) 1.67 eV 10. 1.033A fl . (a rel="nofollow"> 0.496A (b) These X-rays are absorbed by the shielding (coating) material. 12.6.6 1 x 10-34 J-s 13. 1.52A

r "

'

,

i

353

Particle and Wave Nature of Radiations (Light)

Subjective Problems Level-2

•

I. (a) f'). he

2.fo(I+ ~)

3.12.4 eV

5.1.03 x 106m1s 7.9.42 x 1O- 6 N

6.0.1267 x IO-4N-mls 8. 0.509 x IOlON/m2 10.2.9 x 1O- 3 m_K , 0.5 x 1O-6 m

9.6.048xlO- 9 JfK 2

J) hro (b)/lu = ~e-K T

II.(a) ~ 2 KT n e

n2c 3

PI.

12. (a)

,

(b) 1.18 x 1014photonslcm 2 /s

41t'r2hc 2

13. (a) n(",)d", = ~ (b) n(A).
= 8"


14 a n

_ 16tcll/) ( I ) 3 2Jtlrf/ KT c e -I 2 (b) II } = 16n ell 1 2McI . AS (e ').KT - I)

. () f -

15. Zx = 30; the impurity is zinc.

16. (a) Z = 74, the element is tungsten.

(b)AKe =0.182 A

(cP. e = 124.31 A 17. 6.57 x 10- 34 J-s 18.0.27 A 19. (a) 6.592 x 10- 34 J-s (b) No. because stopping potential does not depend on intensity of incident light 20. Energy of photons emitted arc 2.64 eV and 3.86 eV. 21.6.25 xiOIl. 0, 5 cV

.'

354

GRB Understanding PhysiCli OptiOi and Modem PhYlia

1I::::::::::;c:JH~i~n~t!S1R~n~d!}S~O~lu~t~l~oEniP~:::::::::l. Comprehensions

Passage -1 1. TheKE of the photoelectron is k = ~m(l2 2

mu = euB R

u = E. BR m 1 ,2 S 2 R 2

or

K =-~

By using cqns. (i) and (ii),

2 2. The energy of the incident photon is :

m

.. ,(ii) J = 18.6keV after estimation.

£ = 12400 = 24.8 keY

0.5 3. The B.E. is If = E -K =24.8 - 18.6 =6.2keV Passage -2

.r

1. The energy of the incident function is : E(= hf) :..:: Krmx + Wo = O.73eV +1.82cV =2.SSeV " the wavelength oflhc incident radiation is

l. = I;~~o = 4862.74 A 2. The quantum number associated in the electronic tranSltton is from EI = - O.8SeV to £2 = 3.4 eV. Hence the transition takes place from n = 4 to n=2

3. The angular momentum of the emitted radiation is:

Lnd = 6.Lalom = 6n" = -h 2. • Passage -3 ,J• . To reserve theelcctron fromK-sbell, we need the energy E. Then. the speed of the bombering electron just before collision is given as: !m(l2 = eV = E\ or V <>< E\ 2 2. If the accelerating voltage releases by T\ , the energy of the electron will be less than the critical. Hcnce the electron will recoil back with samc speed /J'

=

.J2El / llm .

3. Referring Q-l, the X-ray has the wavelength).,. which corresponds to the encrgy difference D£ = E2 - EI. Then,

Ihe =(E, -Ell

, Partidc and Wave Nature oC Radiations (Light)

355

A=

he (E , - Ell 4. The energy of X-roy is l1/:." = £2 -£\

The binding energy of the electron is Eb = £ 3' Then the KE of the ejected electron is K = l!.E-Eb = £2 -£1 -E]

Subjective Problems Leve/-1 1. According to Wein 's displacement law;

AntT = b _E_2.898xlO- 3 m-K

A

or

-r-

m

)08K

= 9.4 X 10-6 m (Infrared) 2. Wein's displacement law gives A m

3 =£:=2.898 x lO- m-K · T . S800K =0.5 X 10-7 m

.'

"

This is the center of the visible spectrum (yellow-green):~'

3. (oj

E=!~' 2

4. (aJ

.

'

E ' I (Ii orn=hf' wheref=2nV;

(bJnhf=E or

= 2! X25 ' (0.2/= 0.5J

n=

E

.!!.. (Ii 2ft V;

=

0.5

/2S

6.626 x 10 "

•. I

•

= 9.418 x 10 32

'IT

2x3.14

E -/if p= -= c

C

=

6.626 x 10-34 x91.Sxl0 6

3 X 10 8 = 2.02 x 1O-34 kg_mls A =£ = 3xlO

(bJ

f

8

n=

poh~er = '1

, r •• '..r

I'

...

f,

=3.278m

91.5 x 10'

nhf = Power

(cJ or

"

45 x 10

3

6,626xlO 34 x91.5 x 10 6

., = 7.422 x 10 29 photonls

GRB Undentanding PhysiC5 Optics and Modem Physia

356

s. (,)

IIp "" §.. = !Je/,.

,

,

= 6.626 x 10-

=!!). 34

5000 x 10- 10

= 1.3252 x IO- 27 kg _mls

tlP r::. 2E =2.6SxIO - 27 kgmfs

,

(b)

Kmu. = Eph::l'Gn - "'0

6. (a)

= (3.8 - 3.2):V = 0.6 cV

= 0.6)( 1.6)( 10- 19J = 0.96 )( 10- 19 J

~mu2 = 0.96 )( 10- 19

(b)

or

11 _

2xO.96xIO9.1)( 10- ) ]

19

= .J0.2109)(10 6

= 0.459)( I06m/s .

7. The energy ofthe photon is

E = 12400 ::: 2.48 5000 The threshold energy = 1.8 eV The max. kinelic energy of the photoelectron is Kim)!' = 2.48 - 1.8 = 0.68 eV '" 0.68)( 1.6 x 10- 19 J = LOSh 10- 19 J 8. (a) Krnu = eV = e(1.3) = 1.6)( 10- 19 )( 1.3 =208x 10 - 19 J (b)

u = 2)( 2.08)( 10- 19 9.lxlO 31

or

= 0.676)( I06mfs

9. (a) Tbe pholon energy is E =

Ii:: =4.13 eV

The work function = IVo = 2.46 eV Then. Krrw. = E - IVo = (4.13 - 2.46) eV = 1.67 eV (b»). = 12400 = 12400 = 5040.65 A IVo 2.46 (c)!mv2 = 1.67)( 1.6 x 10- 19

2

or

v = 1.67)(2)(1.6)(106 =0.766)(10 6 mls

9.1

Particle and Wave Nature of Radiations (Light)

=12400 =

A

10.

12400 12 x to3

=1.033 A

V II. (a) The minimum wavelength of continuous X~rays is 12400 A A min = 25000 = 0.496 (b)

Thcse X.rays are absorbed by the shielding (coating) material. Amin = he

12.

cV

or

h(Amin)= eV

e

(0.310 x 10-10 ro)( 1.6 x to- 19 C)(40 x 10 3 volt) = (3x10 8 m ts) =6.6 I xiO- 34 J_s 13. Mosley's law gives f(=

k)~ (Z _I) '

Ac. (Z". _I)' (42 _I)' - - = = AMo

(ZCu _1 )2

(29 _ 1)2

AC. = (07lA>(;~)' = 1.S2A Subjective Problems Level·2

1 (a) The energy of each photon is

E _ he

- A

Let N = no. of photonfs emitted by the source Then, the power of the source is p= Nhc

A or

N = PI..

he

(b) The momentum of each photdn is h

P='i.

The momentum delivered by the photon is

~=~(I+P)

Hence, the force exened by the beam is

357

• <

GRB Und~rslanding Ph)'5ics Optics and Modem Phy1ics

358

, I

F = o.p · N

=~(I + p) . p). ).

he

F - = E(I + p)

e (e) The radiation pressure is Pressure

=F - = P(I + P' )' A cA = £(I + p)

e

2. Consuming the t~tal energy of the photon at points 1 and 2, hfo + mgh 1

y, .

.

Particle and Wave Nature of Radiations (Light)

or or

(2)'

359

=U>Wo H{~ -IVo)

4 he _ he = 3W

"2

or

AI

Wo

="c( ~

__I)

1· 2 1..1

= 12400(~ - _I_ ) ev 2000 1000 = 12400 x _1- = 12.4 eV 1000 4. The electric field of cmw incident on the metal varies as, .. E = EoO +COSOOlt)cOSOO2 t =

Eo COS 002 t

=

Eo Eo COSOO2t + T[cos(wi + 002 }t + COs(WI - 002 )t]

+ EO COS 00 1, -cosOO2t

The maximum frequency of emw is (001 +002 )/2l't. Then, The maximum frequency component of emw has energy

E = /if = 11«" , +0),) = h(oo, +00,) 2. The maximum Kinetic energy of photoelectron is K= = /if-IVo

or or 34 2[6.4 x 10- [10" + 10" ) -1.2 x 1.6 x 10- 19 2x3.14 2 , " , . 9.1xlO 31

2x(3.05 xlO 19 - 1.92 x 10 19) 9.1 x 10-31 =

2xl.l3x1O 9.1

12

=0.5xI06m/s

.

GRB Undtntanding Physics Optics and Modem Physia

360

S. Let the potentials of cesium and copper be VI and V2fCspectively. If the fastest

electron leaves the cesium plate with a speed

VOl

the maximum speed of the

electron striking the copper plate is fires by consuming the energy as ce!o\utf\

Copper

00r'" ", v,

00

(I/,
!mv~ +(-eVd = !mv2 + (-eV2 )

2

,

2 or ! mo2 :;: ! mVo2 + e(V2 - VI) 2 2 Einstein's photoelectric eqn. gives

! mv 2

2

:;:

0

.. ,(i)

... (ii)

he _ Wccsium A

By using eqns.(i) and (ii),

,= M(~-W,,)+e(V2 -Vil] =

2 (12400 _ 2 .96 )+e(4.24 _ 3.24) m 2500

= ~2[2e+e] = V;;; We m = 6)(1.6><10 9 x 1O- 31

19

= ~1.066 )( 1012

= 1.03xI0 6 tri/s 6. The change in momentum of tbe radiation along x-axis is

APx = PXf -PXl

=( ;>JC~S6)_( +Jc~e) = (I +p)Lcose c

The cbange is momenrum of the radiation along y-axis is APy = PYf - Py;

=( _Pl~ne)_( Jsi;e) ine

= (I-p)L s

c

Particle and Wave Nature of Radiations (Light)

361

Then. the total change in momentum of the radiation is IAPI = IAPx + APyl =

= ~J(I +p)' cos' 0 + (I -p)' sin '

£JI +p' +2pcos2e c

=O.25xl04 3xlO' = 0,25 x 10-4

3

1 +( ~)2 + 2x~ xcos600 4

4

JI + (0,75)' +0,75

= 0, 1267 x 10-4 N- mis

7. The nonnal pressure is

,, : ,,, ,, ,, ,, ,,

.sF

p=21cos 2 0. c The net force acting on the ring is 5F = PtiAcosO The net force acting on the sphere is F =!OF

= !PdAcosO= -D

= 4/j \ f l

or

rJ 2 21

f

2

-cos O,(2!trRdO)cosO

o c

2 n/ 2

f sinOcos 3 dO

c 0 F=1tR 2 £ c

=22 X ( IO)2 x9xlO4W/m2 7

100

3x10'

= 22 x3xlO -2+4-8

7 = 9.42 x IO-6 N

.

(.: r = RsinO)

0

GRB Undentanding Physics Oplics and Modem Physics

362

8. The momentum of the incii:lcnt radiation is P, =E c

--

P,

The momentum of the reflected radiation is

P,

P, = _ Ep c

I

A

Then, the change in momentum is E t,p =P, -P, =--(I+p) c The average force exerted on the disc is Fai'

!:J.p III F =E(l+ p) av cT

or

=

=

_E(I+p) ---"c-;;;-_ T

Then, the average radiation pressure is Pal' = Fa\' = £(1 +p) I cT A

1tR'

or

=-

12(1 +0.5)

-77-"-'--"=;------::;.6 8

'J'tX (5 x 10- )2 x 3 x 10 x 0.15 x 10-6

=0.509xIOION / m 2

9. The intensity of radiation is 1 = £'11, 4 u = volume energy density

where Then,

U

41 =-

c According to Stefan's law 4 I =oT Using eqns. (i) and (ii) •

4cT' c The total internal energy is U = u(volume oflhe cavity) 4cT' ·v =-c u= - -

... (i)

... (ii)

363

Particle and Wave Nature of Radiations (Light)

Then, the heat capacity is C

_(flU) aT

I' -

I''''C

= 1OOT ' V c = (16) x (5.67 x 10- 8 )(IO )' (2 x IO-l ) ' 8 3 x 10 9 = 6.048 x 1O- 11K

10. The wavelength corresponding to maximum intensity for the black bodies are,

=£.

Al

TI

A, = J!.. T, A, -AI = b(_1 --'-) T2 Tt AA=_I _-'b T, TI _I =-,-+AA TI b

or

T,

.,

r _

Tlb , - b+TIAA

or

= 1747K(by putting TI = 2500 K, b =2.9xlO-l rn - K andAl. = 0.5x lO"'rn)

11. The monochromic energy density is given as

:~:[~

u, =

e lT

]

..

_I

'00

(a) When hro« kT, IIm « I. Then ekT z 1+ 11m

kT

kT

hro

hro Hence , elT -I z -kT . So, we have. I ' u, = n'c (hrol kT)

hro

' =-kT ro'

1[

2C3

I· ,"

...

.

'",'

•

GRB Undentandirtg Physics Optics and Modem Physics

364

.-

no» > kT nw» 1· Then e lT » 1

(b) When

• kT

1100 1100 iT e - I :::: e kT . So, we have 3 _ /niJ u,. = /too e kT n2 c 3

Hence,

12. (a) At a radial distance r, the area ortbe spherical surface is

A = 4nr2

----

Then, the intensity at P is J = Power

Area

=--L 4nrl Each photon has energy E = hF = he

A

If N = number ofpholons flowings /per unit area,

Then.NE = / N

or

=L E

= pi 4 . . ' =

PI.

he

4rt1!2hc

).

(b) Evaluating, we have, N = 4.18 x 1014ph~tonlcm 2/s. 13. (a) Each photon has energy E = hw

The energy of the spectrum in between ro. ro + d w. is u(ro. T) .J E tOlal = uCi)

hro

Then, the number of photon per unit volume is

n(ro)dro = "(co, T) dro

hro

, n(ro)dro = ro

Putting u(w, T) from Planck's law, we have

I

2 3 hrolb

n c e

(b)

Using the expression

we have,

n(ro)dro = -n(A)d).. n().).tf).. = 8n d'A. ).4

27t1r" e UT -I

-1

•

Particle and Wave Nalut'!: or Radiationl (Light)

365

14. (a) Planck's radiation law is given as u(j)

=___ "W, ,.-!-_ Jt2c3 hw elcT _ I

Since Since

fI(Jl d oo = /lIdI .

uw d(2nf)

0)

=

2rr./. we have

=U f d f

or or

1 -I /I

(b) Since

we have,

or or

.

or

. .. (i)

is eqn.(i) to have

put

15. Using Moseley's law and putting ciA. for II (and assuming b = I ). we obtain,

~

C

A. Co

= aZco - a

and

J

c = aZx - a A..1"

,,

GRB Undentanding Physics Optics and Modem Ph)'5ic.s

366

II:

Z. -I V~ =Zeo- 1

Dividing yields, Substituting gives us,

'"17"8" ,9'::pm= Z. - I 143.5pm = 27 - I

Solving for the unknown, we find Zx = 30.0; the impurity is zinc. 16. From Moseley's law, the wavelength of K series of X-rays is given by taking 0" = 1 in modified Rydberg' s fonnula given as:

1 = R(Z _1)2 ( 1__I ) A.

,,2

for K lines, where n ::: 2.3,4...

(a) For K~ absorption edge, we put n =00 in above expression to gCI (Z -I) = ~ or

Z=

.

CO.71 x 10- 10 )(1.097 x to 1 )

+1

=74

The element is tungsten. (b) For KG line: _1- = ,), Ku

~

AKa

R(74 -1)'[1 -1,] 2-

=O.228A

For Kp line: _1_ = R(74 _1)2 [ 1 _ _I ]

A...

=>

3'

AK~ =O.l92A

I ForK'f line: - -=R(74-1)2 [ 1-_I] "Ky ,

I

"

"

42

=> AK =O.182A 1 • " (e) The shanes! wavelength corresponding to an electron with kinetic energy 100

eV is given by A, = he = 12431 A = 124.3IA . . E · 100 17. The energy of incident photon = (he I A) IfW; is the ionization energy and E,t the kinetic energy of the emitted electrons, then we have

he = ~Y; +EKI

A.I

!-

,>1

:. For incident of wavelength A.\ =:800 A = 8)( IO:,8m, _

-he =W; + E KI A,

.. .(i)

367

Particle and Wave Nature of Radiations (Light) And for incident photon of wavelength A2 = 700 A = 7 x 10- 8 m,

he = W. + EK

A2

I

Subtracting eqn.

Ii =

or

... (ii)

2,

0) from (ii), we get (_ I -_I ) = EX2 A,

),'

EXI

(Ex, -EXI )AIA2

-

ciA, -A,) 19

Here,

EX1 =I.SeV = I.S x I.6xlO- J

and

EX2 =4.0eV= 4.0x 1.6xlO-

19

J

Substituting given values, we get 8

h = (4.0-I.S)x 1.6 x 10- 19 x S x 10- x 7 x 10-

8

3xI0'(8x I0 '-7x lO ') = 6.57 x 10-34 J-s 18. For hydrogen-like element

k=Z'{,I( II~ J

(* -~) R(* -~) "

ForKalinc,_I- = Z2 R Aa

For Kp line, _I = Z2

Ap

1

1

... (i)

2

3

... (ii)

,

, .

L Ap _I'

_I I 1 2' _ -4_3 / 4 _ 3x9_27 .. on dividing, Aa - 1. __1 -1_! -S/9 -4 xS-'32

I', 3'

9

An =271.. =27 x0.32A = 0.27 A "3232 ,

19. (a)

--.

",

he = W +eVI

;f:

A,

'.

..

(

, . )

\

... (ii)

he =W+eV2

A,

I

"

he - he = elV, - Vil A, At _, -t Here' c=3xJ ms 0 . .. ' 7 . , .. A, =4oooA=4xlO- m, Vt =0.82V Subtracting,

,

GRB Unde~tanding Physics Optics and Modem Physics

368

,,=3000A=3xlO-7m,V, =1.85V ..

" X3X10'(

1

_

3xl0 7 Ir x 3 x l0 8

or

1

)=1.6XIO-19(1.85-0.82)

4 x l0 7 1

12x10-7

: 1.6 x 10- 19 x 1.03

. . Planck's constant, II ", 1.6 x 10- 19 x 1.03 x 12x 10- 7

3 x 10 8 =6.592 x I0- 34 J-s (b) No, because stopping potential does not depend on intensity of incident light. 20. From Einstein's photoelectric equation, maximum kinetic energy of emined electrons EK =hc _W I.

h = 4.14xlO- ls eV-s

I

= 4.14 x IO- IS x 1.6 x 10- 19 J-s =6.624xI0- 34 J-s

..,

E

= Irc =6.624 x 10- 34 x 3 x 108 A

\

,

4oo x 10-9

= 4.968 x 10- 19 J_s =4.968xlO- 19 1.6 x 10-19

=3 .leV

Ek = 3.1 eV - 1.9 eV = 1.2 eV e+He-+He+ + Photon

.,

Energy of He atom in their fourth (n = 5) excited state (for He+ ion Z = 2). Eo=

Z'Rhc=_(2)'XI3.6=_2.176eV (5)'

n'

. From conservation of energy, 1.2 eV + 0 = - 2.176 eV+

Ey

Energy of photon during combination,

Ey = 1.2+ 2.176 = 3.376 eV Energy of helium ion, En

=_ Z2~hc =_4 )( ~3.6 n

= - 54.4 -;;;: eV, n = 1.2.3..... .

n .

=- 54.4 eV, - \3.6 eV, - 6.04 eV, - 3.4 eY, - 2.176eY, - 1.51 eY

[

r

Particle and Wave Nature of Radiations (Light)',

369

Difference of energies lying between 2 to 4 cV is - 3.4 + 6.04 = 2.64 cV and - 2.176 + 6.04 = 3.86 cY. Energies of photons emitted are 2.64 cV and 3.86 eV.

21. Energy incident on surface per second P = fA

=2.0xI.OxlO-4 =2xlO-4 J Energy of each photon = 10.6 eV = 10.6 x 1.6 x 10- 19 J :. Number afphatons incident on the surface =

-

2x 10-4

"-"-"'--= 19

IO.6x 1.6x 10 -

Number of photoelectrons emitted = 0.53 x

100

2 x 10-4 = 6.25 x 1011 10.6 x 1.6 x 10 19

According to Einstein's photoelectric equation, maximum KE of photoelectrons E K = E - 11' = 10.6 cV - 5.6 cV = 5 eV Minimum kinetic energy of photoelectrons = zero.

.,

,.

Particle and Wave Nature of Matter

4

•

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17

4.1

Introduction

Compositi on of an Atom Thomson's Model ( 1897) Rutherfo rd 's Model ( 19 11) (l. Particle Experiment Atomic Spectrum and Energy Levels Bohr' s Model (Idea of Quantisation by Using Balmer's Fonnula) Mechanism of Emission and Absorption: Atomic Hydrogen Spectrum Bohr's Correspondence Principle Frank·Hertz Experiment: Confirmation of Atomic Energy levels dc-Broglie 's Matter Waves I Experimenta l Confinnation of Wave Nature of Matter by Davison-Gcnner and George D.Thomson ( 1927) Double Slit Interference ("If Electrons Wave fu nction tV : Max Born Inlcrpretalion of'J' Schrodingcr's Wave Equation Particle in a Box Heisenberg's Uncertainty Principle

Wave-Panicle Duality of Matter and Radiation; Complemental) Principle (1927·28)

Introduction

.. Malter is divisible" is the basis of creation and dcstruction of different o bjects. Since time immemorial, the vedic texts and then the greek phi losophers proclained the grainyness of matter. Lavisier and his wife believed that the conservation of maller in chemical reaction could bc possible on the bas is of atomic (panic le) nature ofmatlers. Dalton speculated the ato micity of matter in the multiple proportion of compounds. Avogadro believed that due to atomic ity (particle nature ofmalter) all pure gases al same temperature and pressure have same number of molecules in a unit volume. Maxwell a nd Soltzman explained the macroscopic quantities like pressure and

f

I

371

Partide and Wavt! Nature of Malter

temperature by postulating the random molccularspccds of ideal gas. We learnt this as kinetic thcoryofgascs. Jean Pcrsin and Albert Einstein performed experiment and gave theory of molecular motion in fluids called Brownian motion. that is. zig-zag motion of small suspended particle under the collision forces of molecules of liquid.

4.2

Composition of an Atom

An atom is comprised of electrons, protons and neutrons. The electrons arc negatively charged e = - 1.6 x 1O - 19 C. very light m = 9.1 x 10-31 kg. The protons is positively charged e = + 1.6)( JO- 19C and neutron is neutral. The mass ofa neutron is

slightly

greater

than

mass

of

a

prolon,

"'p

= 1.672 x 10- 27

kg

and

mil = 1.674 x IO - 27 kg . The electron was discovered by J. J. Thomson in 1897 from his cathode-ray experiment. The proton was discovered by Ernest Rutherford in 1918 (though the concept had been earlier suggested by the work of Eugene Goldstein) and long after. in 1931 neutron was discovered by chadwick. The protons and neutrons remain in a compact form in a spherical region. ca lled " nucleus" . The existence of the protons and neutrons inside the nucleus is due to a strong short·range force acting inside the nucleus. It is called nuclear force . We will discuss this in nuclear physics. The "nucleus" was discovered by Rutherford in 191 I. The electrons revolve around the nucleus in the fonn of electron clouds or (waves) and distributed in different orbits and orbitals obeying quantum mechanics. We will discuss few simple atomic models evolved historically from "Thomson" to "Bohr".

4.3 Thomson·s Model (1897) The conduction of electric current in solid, liquid and gas is attributed to the ions and electrons. The law of electrolysis of Michel Faraday in 1833 gave us a bint of elementary charges in matter. The cathode ray was identified as a stream offas! moving negatively charged particle who was influenced by electric and magnetic field . Thomson in 1897 measured elm of the cathode rays emitted from different metal filaments and found a unique value. He called these panicles electrons Millikan measured the charge of electrons in his famou s oil drop experiment By using!!. value

m

given by Thomson, Millikan measured the mass oran electron is 1000 times lesserthan that of a hydrogen atom. Thomson by his time believed in three following points regarding an atom. (a) Atom should be electrically neutral because it is the • • constituent of neutral matter. • • • (b) Atom must be stable. • • • (c) Atom radiates electromagnetic waves in the fonn of • • • • various lights. • • • Keeping all the above factors in front he proposed a model The positive and negative of an atom, in which the positive charges (not known that time) charges are distributed In and negative charges (electrons) are distributed uniformly an atom likes the plums in possessing spherical symmetry, This satisfies the condition (a) • p
1, 372

GRB Undentanding Physics Optics and Modem Phyt;ia

that atom is neutraL He argues that. the nct force acting on each charge is zero because it is acted upon by attractive force of opposite charges and equal repulsive [orce of like charges. This satisfies the stability condition (b) Next. in order to radiatc cmw. he

proposed that the electrons vibrate about their mean position under the action of restoring coulombic (electrostatic) force. As the charges arc distributed like plums in a pudding, you can call it plum-pudding modd. Let us take Ihe case of a hydrogen atom in which an electron vibrates about the

mean position(ccnlcr of the spherical cloud of positive charge).

'{.;;;"1!.£;J.

Accordillg to n,omsoll 's lIIodel.find lIre frequenc)' alld wQ\'elclIg'" of radiation cmil/cd by (III atom.

Solution: The uniform spherical charge distribution of positive charge inside an atom gives the electric field E

=L

r, the restoring force "cting

3'0

on the electron is

.\

:~~~-¢

F = - eE. = _P£'r 3<:0 This accelerates the electrons towards the mean lK>sition with an acceleration a=F = __P_, III 3mco

Substituting

, ,#p.--, ,

/. ,

'-_..... E' •

a = - ro~sc '" we have wose

.\

'··",,·1 \: . 1-~ :'1 ,.~

, I

i

Since E oc r inside the atom, the electron oscillates

=J3m£0 P

Then. the frequency of oscillation of the electron is

f = l.~

p -wherep = (.2 21t 3m£0' 11t R)

3

or

1_ _ e,-2~ 21t 4l'tEomR 3

f =1

=-2x22 7

(1.6xlO 19)2 x9x 109 9.lxlO]) x( IO - IO»)

( .: .....L=9.10 4ncO

9)

(m=9.l x lO-] l and R=lO-I O m ) or

r JOse

=2.SxIO l' Hz

Ans. Hence, the frequency of· radiation lose equal to f rad = f ose to satisfy Maxwell's theory afemw. Then. the wavelength of radiation is

Particle and Wave Nature of Maller

373

A=---"--= 3xlO' m= 1200A An5. lose 2.5 X 1015 Thi s wavelength corresponds to ultraviolet region. Since the hydrogen atom mostly radiates in vis ible range in addition to ultraviolet region, we can not accept this model. FurthemlOrc Ihe frequency of radiation depends on vol ume charge density, Hence. all hyd rogen like atom should emi t radiation of same frequency which really does nol happens in practice.

4.4 Rutherford's Model (1911) a -Particle Experiment Ernest Rutherford. a student of J.J.Thomson proposed a model to avoid the li mitations orhis teacher's proposed model. It was based on the concept ofa "centra l massive positively charged panicle nuc leus that is. ofan atom. Nucleus was discovered by Rutherford in his famous. gold-foil scatteri ng experimcnt. He bombared a -particlcs obtained from radioacti ve materi als (a lready discovered by 8equcrrel) onto a Ihin gold foi l. Hc found that: (a) Max imum Cl- part icle went undcviatcd. Hcnce hc confi nncd that majority of spacc occupied by matter (atom) is "accum. (b) FcwCl- partic les deviate wi lh large and small angles. Thi s made him to belicve that, thc central particlc of the atom is highly positively charged, The central massive positively cha rged which could deOect thc +vely (X- pan icle by the gold nucleus deflects few d-particles Iremendous electrostatic force of repulsion. (c) Very few (1:10 (' ) (X- panicles retraccd their paths. This con fi nned that the central panicle is not on ly slrongly +vely charged but also having huge mass compared to (.( particle. From this go ld foil experiment. Rutherford di scovered a massive positively charged part icle at the center of each atom . He eallcd it "Nucleus". Thus Rutherford discovered the "nucleus" and using this entity he proposed an atom ic model as fo llowi ng.

Planetry Model I. As the atom needs to be neutral, th e charge of the nucleus is equa l (positivc) to the tota l charge of the electrons o f each aloms and opposite in nature. 2. As each atom is mechanica ll y stable. the electrons should movc in circular path around the nucleus. The electrostatic pul l Fe!. is the centripetal force providing necessary centripetal acccleration. 3. For the atom to emit radiation. the orbiting IS an accelerating electron. electron a = r(J)2 = 182510 / 5 2 hence it must rad iate

v

err,w",

An electron radiates an continuously wavelength and spirals onlO the nucleus

Partid~ and Wave N ature

or Mauu I.=...£....= 3xlO'

373

m=1200A

Ans. fo~ 2.5 x lOI S Thi s wavelength corresponds to ultravi olet regi on. Since the hydrogen atom mostly ~diates in visible range in addition 10 ultraviolet region. we can not accept tillS model. Furth cmlorc the frequen cy ofmdiation depends on volume charge density. Hence. all hydrogen like alom should emit radiation of same frequency which really does not happens in practice.

4.4 Rutherford's Model (1911) a-PartiCle Experiment Emesl Rutherford. a student of J.J.Thomson proposed a model to avoid the limitations of his teacher's proposed model. It was based on the concept of a "centrol massive positively charged panicle nucleus that is. of an alom. Nucleus was discovered by Rutherford in hi s famous. gold-foil scattering experiment. He bombared a.·paniclcs obtained from radioactive materials (already discovered by BequeTTel) onto a thin gold foi l. He found that : (a) Maximum a- particle went undeviated. Hence he confinncd that majority of space occupied by mrtlter (atom) is vacculll. (b) FcwCt- particles deviate with large and small angles. This made him to believe that, the central particle of the atom is highly positively charged, t maS5ivepos~ivelycharged which could deflect the +vcly a- particle by the gold nuc!eus dallecls few dpartidas tremendous electrostatic fo rce ofrepul sion. (c) Vcry few (1:10 6) a- particles retraced their paths. This continued that the central pat1i cle is not only strongly +vcly charged but also having huge mass compared 10 a part icle. rrOll1 this gold foil experiment. Rutherford discovcred a mass ive positively chargo:d particle at the center of each atom. He cal1cd it "Nucleus". Thus Rutherford di scove red the ··nucleus" and using this cmily he proposed an atomic model as followin g.

Planetry Model

, I

, ,,

l

I. As the atom needs to be neutral, the charge of the nucleus is equal (positive) 10 the total ch3rgc of the electrons of each atoms and opposite in nature. 2. As each atom is mechanical1y stable, the electrons should move in circular path around the nucleus. The electrostatic pull F e1. is the centripetal force providing necessary centripctal acceleration. 3. For the atom 10 emit radiation. the orbiting eleclron IS an accclerating electron. 2 a = rtt)2 = 1825m / 5 hence it mllst radiale

v a ,

An electron radiates an continuously wavelength and spirals onto the nucleus

, ;

374

GRB Undentanding Physics Optics and Modem Physics

electromagnetic wavc, but we have a serious problem here. This is thaI. according classical theory of emw proposed by Maxwell. an orbiting electron (charge) must "continuously" radiale emw offrequcncy which is equal to frequency of its revolution.

f rad :: JrC'o' Since the electron loses electromagnetic inner energy continuously, its total mechanical energy must decrease. As a result. its radius of orbit decreases. Therefore. the frequency of revolution increases as it revolve more quickly. Eventually, the frequency of radiation increases gradually to an incredible frequency. Furthennore, the intensity of radiation also increases as the effective area of the sphere (radius) decreases. This leads a "catastrophic collapsc" on the nucleus, releasing a continuous electromagnetic wave of evcr decreasing wavelength and increasing amplitude but this does not happens in practice. Furthennorc, Ruthcrford could not explain why the protons should lie in a nucleus inspite of tremendous elcctrostatic rcpulsion between them. Atleast, another scrious question arises about the mass of the nucleus which contains both neutrons and protons. Rutherford could not explain about the measured mass of the nucleus which was different from the "theoretical mass". '!;;;~;k,.,

rc~l«Zi

Find the clo.test distance ofapproach ofon alpha particle scattering. if jt rna\"(' towards the gold foi/wjth a hnc/jc energy K 0. = 7.7 MeV. Solution : The initial kinetic energy of the a-particle decreases due to the strong repulsion of the gold nucleus. Hence, the a- particle comes to rest momentarily. Let the minimum distance between the gold nucleus and a particle be 't}. We call it closest distance of approach work-energy theory states that,

!!K a = Wei O-Ko. =U~l - Ko. = - !!UeI

Then, or or

-=====::::":;)j.Io..I

Ci>

-K{l=KQq( ~-!)

or

KQq

0'

where Then,

'l>

=K'

•

K= _ I_=9xI0 9.Q=Ze=79e, q=2e

4""0 '11 =

(9 x 10 9 )(79)(1.6 x 10 -19) x 2(1.6 x 10- 19 )

(7.7xI0 6 xI.6xlO- t9 ) =29.5xlO- 15 m

This tell us that, the size ofa nucleus is in the order of to- 15 m.

4.5

ADS.

Atomic Spectrum and Energy Levels

As discussed in previous section(or chapter), a heated solid emit emw of all frequencies. On the other hand, when a low pressure gas (preferably monoatomic) is subjected to a high electric field. some of the outer electrons will be detached from thin respective atoms. Then, these free electrons accelerate in the electric field and gain

Particle and Wa\"e Nature of Matter

375

kinetic energy. Thcse Slit energetic electrons Photographic plate col!ide with the gas atoms. Absorbing energy (discontinuous obeying quantum physics) the gas atoms get excited from ground state. In returning to their ground state again, the cxcitcd atom give up Prism their excess energy. This results an emining emw or light Una .spectrum photons. This radiated Une spectl\Jm of a light emitting gas atoms Is energy (wave) is seen as distil"lCtlines on the photographic plate collimated by a slit and then it passes through a prism (or a diffraction grating for better resolulion), where it is broken up (split) into scveral wavelengths. The monochromatic light strike the screcn or pholographic plate as different lines. This discrete emission (and absorption) spectrum of a gas in the fonn of a line in a pholographic plale is called line spectrum (contrary to continuous spettrum). The study of line spectrum of different atoms leads 10 their chemical analysis and called atomic or molecular spectroscopy. In between 1860·1885, many spectroscopists ranging from Kirchoff. Bunsen, Frawnhofcr etc experimented on the radiation from heated objects. The hydrogen atom being the simplest one, many gilled spcctroscopistlike Balmer. Paschen elc., paid their attention in the hydrogen spectrum and 410.2 pro
I

).(incm)=c2 (

n'

n 2 _22

); n=3,4,5......•

where). = wavelength emitted in em and C2 == 3645.6 cm, a constant called convergent limit because it gave the wavelength for the largest value of n(n ~ -).

GRB Understanding Physics Optics and Modem Physics

376

By putting" = 3. 4,5.6 in the above formula we have)., for HQ.H~. Hyand 1/(, respectively. The above fannula can also be given as 2

- 10

).=3646 2n 2A, whcreA = IO m. " -2 "A" is the unit of length tlamed afler the famous physicists "Angstrom" as mentioned earlier. Then. other ten lines in the violet and uhraviolet region had been measured. All the wavelength of these ten lines were satisfied by the above formula by a slight

modification done by

Balm).er=a~:O(llo..:~ng. );11 = 4.5.6..... . ,, -_3 3

and

A=C4( II ,"'_ 4 2 );,, =5.6.7......

In 1890, Rydberg felt that hydrogen - spectrum could be beuer expressed in tenns of wave number (reciprocal of the wavelength). He modified the Balmer's formula which incorporated all the series of hydrogen spcctrum givcn by Balmer.

~ =* = RU(212 - n12 }

,, = 3. 4.5

where R" = Rydberg constant for hydrogen =1.097xi0 7 m- 1 More generally,

- (I fIrI) ).=RII - - , "1 2

where nf and ni8re integers. For a given series. ni = IIf + I, nl +2. ..... . ~

Bohr's Model (Idea 01 Quanllsatlon by Using Balmer's Formula)

In 1913, Danish physicist Neils Bohr who worked with both J.J Thomson and Rutherford presented his atomic model by carefully combining classical and quantum physics which gave correct derivation of ahe Balmer's formu la of hydrogen spectrum. How did Neils Bohr get a clear picture of his model after seeing the Balmer's formula (on request of his friend), lct us scc. Balmcr's formula which gives the wavclength of four visible radiation emitted by

hyd,ogen ",om can be

wrik': : C~ = .1, ).. = ], H

4. 5.....

Bohr argued that according to Balmer's fonnula, the wavclengths of light emitted by hydrogen atom is discontinuous or discrete. As pcr Planck-Einstein's

Particle and \Vave Nature of Matter

equation(formula), the energy ofa photon oflight (radialion) ofwavelenglh)", emitted by a hydrogen atom is Ephoton(=

hfrod}:=' hi

Then, he went on arguing that, the atom must lose its internal energy from E to Eosay, after emitting the photon. Since the atom takes a negligible recoiling kinetic cneq,,), due to the impact of the escaping photon. using conservation of the energy, we can write

E

or

:=

v-

E

Eo + Ephoton = Eo + ~c

,.

E-Eo = lie

,.

Substituting the value of 1 from Balmer's formula,

E-Eo =hcRH _hcRH 4 ,,2 This tclls us that, the energy of an atom (electron) is quantised

! E ~ -~

!

Then, Bohr suggested that, not only as Planck's oscillator but also an atom should change(deerease or increase) its internal energy stepwise. The internal energy of an atom is mostly due to the motion of the electron and the potential energy of interaction between the electron and proton. For the sake of simplicity, we disregard the motion of the nucleus (prolon) because its mass is 1836 times that of an electron. Then, the energy ofa hydrogen atom is given as

!E=_ke2r2 I

Comparing these, last two equations, Bohr understood E - - , - - that the electron can have some specific orbits. In these orbits. the electron need not energy ifit accelerates. In other ways the total energy of an electron remains stationary (constant) with time. Hence we call these orbits stationary orbits. The corresponding energies arc called energy levels EO _-"L_ (states). The electron needs to jump from one The energy level diagram of the stationary orbit to another stationary orbit (not continuously but discontinuously (suddenly) within de-excited atom (or electron) which loses its e~cess energy lJ.E s E - Eo (1O- 8 s) to emit a photon. For the first time Bohr had drawn the horizontal lines to represent the energy levels. The spacing

in the form of a packet 01 rad iation (photon) of frequency f

•

378

CRB Undenlanding Physics Optics and Modem Physics

between two horizonlallines gives nn amount of the r, energy difference which will be equal to the energy of the emitted pholun . \37. Not to be confused with the two horizontal bars \ photon representing the energy levels and two concentric circles representing the circular orbits. we emphasise on the fa ct that . the distance between two circles The electron Jumps from higher 10 lower increases as we go rad;ally away where as the orbit releasing its excess ene~ as a photon distance between two horizontal bars (energy levels) corresponding to the difference in energy in these orbits decreases. The energy of the

~-';====1~

r -

orbits increases as -£0,- E: • _ E90 ..... and the corresponding radii of the orbits increase as lb. 4,.,. 9't) ... .. . Following these analysis, Bohr suggested, that, not only the energy of an electron, but also the corresponding radii of the orbits and the orbital speed etc will be quantised. Putting both position radius (r) and velocity (I) together, we can say that the angular momentum of an electron in stationary orbits must be quantised. By using the correspondence principle which states that the frequency of radiation o f an orbiting electron will be equal to frequency of it revolution, Bohr proved that, the angular momentum of the electron of mass //I moving in a stationary orbit of radii rwith a speed

u is equal to integral multiple of l!. .

r=2""_ _- T < \ L(= /III/v) =

n~

puuiog all these ideas generated by Bohr using two classical laws (i,e., Newton's 2nd law of motion and Coulomb's law) and two quantum laws (i.e., Planck' s law of quantisation and Einstein - Planck law of photon emission) alongwith the two new quantum laws suggested by him (such as law of quantisation of energy and angular momentum of an electron), he tried to find the value of RH(Rydbcrg's constant) theoretically. Before proceeding for derivation, let us summarize the postulates of Bohr.

Bohr's postulates I. The orbits and the energies of an electron are quantised so long as an electron moves in these quantiscd orbits, it does not radiate energy. Its total energy remains constant. Then the electron is said to be in stationary state.1f it jumps from one stationary state to the other, it emits a photon (emw). However. during transition (At = 10- 8s ), no emission takes place. This postulate avoids the continuous emission of emw and collapsing ofthe electron, giving the mechanical stability to the atom. 2. It obeys the classical mechanics (Newton' s 2nd law and coulomb's law) to revolve in a specified orbit with a specified speed (which is discarded by Heisenberg's uncertainity principle. However, it disobey the classical theory of emw proposed by Maxwell not to radiate emw(or photon). It radiates energy obeying Planck- Einstein's quantum law of photon emission.

379

Particle and Wave Nature of Matter

3. The electron jumps from higher energy state(level) E; to the lower energy E/ decreasing the energy of the atom by M(= E j - Ef). This difference of energy will be carried by the emitted photon. Hence, the emitted photon has frequency f(or wavelength;l..) given as

fH) =~ or E,- Ef =hf ...(i) 4. The stationary orbits are characterised by the quantised angular momentum L(-- mvr )_nh --

... (ii)

2n

where n = angular momentum quantum number. Let us calculate thc energy of the stationary states following thc above postulates. Since the electron moves in a circle of stationary radius r, the centripetal acceleration is given by

,

,

Coulombic force Fe\ = k ~. Applying Newton's 2nd r

law,

The electron moves in 8 circuiar path with constant speed under the centripetal force (Coulomb's force) F_ke 2,,2

ke 2 = ma, where a =u2

F =-

r

r

v = Jke'

or

mr

... (iii)

Eliminating u from eqns. (ij) and (iii) n 2h 2 r=--2' n = 1,2,3...... mke Putting n = 1, the smallest aHowed radius is I1J

= -h'- =0.529 A

mke' This is called Bohr' s radius denoted as 't). Then the allowed radii (for no-radiation of the orbiting electrons) are ~_---,,.., 1r = 'lln' ,I n = 1,2,3....... r= Ib, 41b .9't)andsoon. or The radii of radiation orbits are discrete or uantised

,

The total energy of the atom E :::: total energy of the electron =K+U 1 , ke' =-mu - 2 r

GRB Understanding Physics Optics and Modem Physics

380

~ !m(ke' )_ke' mr

2

r

ke 2 2,.

E~--

Substituting r = 'bn2. we have

r=~--o--"

E~Je' (_1 ). 2'b ,,2 ,

where

You may put k = _ 1- and 'tl = ~ in the above expression to gct 4ncO mke 2

r-"""'------,---, E=_

me'+

8£~h2112

Comparing the obtained energy equation with the energy obtained by Rydberg equation

(9.1 x IO - li )( \.6 x 10- 19 )'

~--~~~~~~~~

8(8.85xI0 - ")'(6.3xI0 34)3(3xlO') = 1.097 x 10 7m

This was an excellent agreement with the experimental value with a precision of

1%.

Inserting the numerical value,,';:.'----c:::-;--,

IE =

-7 ev·1 ,,=

1.2,3 .....

called principal quantum number. Putting 1/ = I, we have, E = - 13.6 = - Eo say. This minimum energy is called

. E ground state energy. Puttingn =2,3etc,wehavcEl = -3.4eV=- 4°, £2 = 1.59 eV etc. E] and £2 are called the energy of first and 2nd excited states etc. The horizontal bars up representing the energies are called energy level (state) diagram. We can see that the spacing between the energy levels decreases to zero at n = 00, which corresponds to zero energy state. At this state the electron becomes free from the nucleus. Then the electron can have any kinetic energy. Therefore Levels will be continuous after the energy (above) zero energy state and all the . stationary states are discrete below ene.rgy.state.

1.

381

Particle and Wa,"e Nature of Mauer Continuous energy states with positive

y~

""9y(KE,O)

E=O

Eground = -Eo Discrete energy stales, the spacing between " adjacent states decreases"

f or heavier atoms, removing all electrons except one, we can make it hydrogen like. for hydrogen like 3tom (ionised heavier atoms). Bohr modified his formu la by putting the atomic number Z in the force equation F = _ I _Q ,Q2

41tEo

r

where q] = Zeand q2 = e 41tEor

=-r

... (i)

the quanlisation of angular momentum:

nh 2.

mvr=-

... (ii)

Solving these equations, we have the foll owing equations r

= 'tl1l2. 'tl = rll in ground state V =VoZ

n ' Vo = speed ofthe electron in ground state hydrogen atom 2 4 ""E = _ mz e = EoZ2 "- ..... . 2 2 2 8£.2 h n n o

Eo = -13.6 eV

..

I)

1 = RZ2[_1 __ A

n2 ,

112 2

'.

GRB Understanding Physics Optics and Modem Ph .

382

","

Referring Bohr's theory, filld (a) mininmm energy of a hydrogen atom (b) minimum \l'm'deflgrll a/light emilled by a hydrogen atom (c) radius oflSI excited orbit a/hydrogen alOin (J) angillar momentu", of ground state electron in hydrogell atom (e) ratio o/maximllm speed

and speed in 2nd excited orbit 0/ an electron

Solution: (a)

(b)

or

(d)

An!.

0-(- 13 .6eV)=~ " . = mm

he = 12400 A 13.6 eV 13.6 =91I.76A :=

'bn2. For 1st excited orbit n

Ans. :=

2,

Then. , = (2)' '" = 4", =4(0.56 A) =2.24 A L := nh. where n = I for ground slale

Ans.

2.

or (e)

n'

r

(e)

hydrogen atom.

E = _13.6 eV

Putting n := 1, Emin = - 13.6 eV 6E=hf he or E '"' - E groWld = i"

or

jn

L = l!. = 6.3 x 10-

21t

34

:::.10-34 ).s

2 x 3.1428

Ails.

v = tlo. For maximum speed n = J

n

v max

= Vo

.. .(i)

v v;

For 2nd excited orbitn = 3. Then, =

.. •(ii)

Using eqns. (i) and (ii), unm. !u =3

Y

Mechanism of Eml •• lon and AblOrptlon: Atomic Hydrogen Spectrum

Fnlquency of radiation Bohr's theory suggests the existence of stationary states (energy levels) in an atom. When an atom loses its internal energy corresponding to the higher energy stale E2 to the lower energy state E I. it emits one quantum o(radiation(pholon). The energy orlhe emitted photon will be equal 10 the difference in energy between the two levels. if the

Particle and Wave Nature of Matter

383

atom is keP.t fixed or moved with constant velocity or jf the atom wdl take a negligible fraction of the excitation ~nergy 11£(= E2 = E 1). Then the frequency of radiation

E'Z----,---

" A photon Is emined when the atom jumps from higher to lower energy states

Excitation and Excitation time

To emit radiation, an atom needs to stay above the ground (minimum) energy level Eo· Then the atom can have highcr cnergy states E I, E2..... etc. A particular atom does not always stay in same higher energy state. It comes down to lower energy (or ground) !.tatc to emit radiations. The emission of photons is a statistical phenomena(process). The probability of finding an atom in an energy state E above the ground state Eo is given by Boltzmann's function

IF = ce -~, ~here c =constant

The excited atom can stay in a higher energy level for an average time AT = 10-8s. Since Ihis time is million times greater than the period ofrcvolulion of the electrons, we can say that the excited energy states arc stationary slates which should not be interpreted as the pennanent state the ground slates is the _~ f~.. _ p only pennanent state. The ground state is the only -"::",.~L~ pennanent statc. Howcver. according to Bohr's theory, electrons are allowed 10 stay in stationary(radiations less) The angular momentum olltle photon may be directed parallel orbits for a long time. Even though the excited electron is or antiparattel with its veleocity in the radiation less orbit. after a time compared to the period of revolution of 1O- 8s it will come down to the ground or (lower level). The transitions are allowed belween the stationary levels. No photon is emitted during transition. Just after transition (quantum jump), .. a photon" is emitted. As the electron comes down to the next lower energy orbits its angular momentum decreases by n. Hence, to conserve angular momentum the emitted photons must carry the angular momentum of same magnitude in opposite direction. You should remember that the angular momentum of a photon may be parallel or anti parallel with its linear momentum (direction of motion) having magnitude n. In other words, like electrons etc, a photon has a spin

±~ .

It is often argued that the conservation of energy does not hold good (violated) during the transition. but it is not true.lnfact to obey Heisenberg. uncertainty principal we can not measure the momentum and energy precisely during the transition. This makes it difficult to account for the unknown history of the quantum phenomena. Hence. we can not just say that the conservation law are violated.

GRB Understanding Ph}'1LCS O ptics and Modtm Physics

Method's of excitation Now a quc~tion arises, how does an atom get higher energy levels? This is done by two wa~~. One IS by colliding a photon with an atom. Secondly an atom can be excited by colhdlng a material particle (i.e., electron, prolon, neulron, alom etc) with it.

@ ~I,oc;~@ A

Photon

~r.-.. I, V

=l

(<0

=l

@~I

I Methods of excitation 01 an alom by coItlslon with a photon or atomiC particle

.After absorbing the photon or a fraction ofthc kinetic energy of the electron, say the mterna) energy of the atom A increases and get excited to the higher state. In this process, the eM of the syslem (colliding photon or electron plus alom) moves and the mternal energy (KE + PEl of the system relative to the eM incrcases obeying laws of quantum mechanics. Thus. the atom A gets exciled to its higher energy le\'cls to become A (excited atom). The excited alom gets dc-excited (corning 10 the ground Slate) and releases a pholon or rrequency f. The rrequency of the emined pholon is decided by the rrequeney of in ciden I photon or KE of the bombarding electron and the direction or collision. It should be noted that the kinetic energy orlhe colliding particle (photon or electron etc) must be greater than the threshold value. Othel'\vise the collision will be clastic and no excitation of the atom takes place . During the process or excitation and dc-excitation. conscrvation of lincar rnomenlum. angular momentum and total energy takes place. However, the rrequency of the absorbed and emitted photon will be equal for the fixed atom is lattice structure. Recapitulating, To excile an atom, there must be some minimum energy Kmin 01 the colliding photons or matter particles. 11K < Kmln, the collision will be elastic and the atom does not absorb energy. The colliding particles will scatter just after the collision. 1\ K > Kmin, the energy absorbed by the atom obeying Bohr' atomic model mechanical law and quantum given a s 6E :::

Rh~ - ~t. During excitation and de-excitation 01

the alom, the momentum (linear and angular) and energy of the syslem (atom and the colliding pholon etc) remains consumed. Hence, the trequency ot absorption and emission need not be same.

Hydrogen spectrum Bohr predicted the existence of wave series (spectrum) in hydrogen atom which come in the form of five emission spectrum's as given below. Transition from higher orbits to orbit tI = L. tI = 2, tI = 3,11 = 4,11 = S give rise to Lymen, Balmer. Paschen Bracket and p·fund saw rcspectively. Each series has a minimum wavelength and maximum wavelength. Maximum wavekngth corrcsponding to the transition between lowest (for that series) and next higher level. The minimum wavelength of a series

385

Particle and Wave Nature of Malter

corresponds to the transition between the lowest (for that series) and zero energy level (n = 00). Lymen series lies in the ultraviolet region. Balmer series lies in the visible spectrum. Pnschen, Bracket and P-fund series lie in infra-red zone.

n_·

Lymen series

,

II = 2.3.4 ..... )1• = R( l -~); 11 -

,

, ,, , ,,

Balmer series

, ,, ,•

,

)1. =

R(-\-2- -~);" = 3. 4.5..... n-

J

!tlr(p '!Und)

n.s

(Slackel)

n:4

(Paschen)

til

• •• •

-L (Balmer) H.. ,H",H"H, .

pr }

Paschen Series

n.' 0_'

,,

-K IL",,"'I

•

LR(-I __I )'.=4.5.6... _ 32 ,,2

I..

•

Bracket Series

f

n'2}"

= 5.6,7.....

i Rel2- nl2 } "

= 6, 7,iL ..

= RC l2 -

P-Fund Series =

The spacing between any two lines is propor1ional to the energy difference between the corresponding levels. (a) Filld the ionisation energy of hydrogen atom. (b) What is 'he minimum wa\'elellgtlJ oflight emitted by hydrogen atom in Balmer's series. Solution: (a) Ionisation energy corresponds to the energy difference between n = I and n = 00. E,,=I

=_13.6 = _ 13.6 =-13.6 eV .'

(1) 2

E,,=f2J = _1 3.6 = _ 13.6 = 0

.'

Thcn. the ion isation energy is

'"

Eion = ErI;) - Eo = 13.6 cV

An!.

GRB Understanding Physics Optic" and Modem Phy,ics

386 (b)

For Balmer series "I 2 and 112 = 3. 4, ..... co, The minimum wavelength of radiation corresponds to maximum energy of radiation in Balmer's series which - ' - - - - 0.2 corresponds to the tmnsitions from "2 -)0 C(l to II / = 2. 1 2Then, aE = ) ::0:

I

13.6L ~2

-----0_'

= 3.4 eV

The minimum wavelength of radiation is 0 =;: 1240 A. V in volt ) -min V = 12400

(V = 3.4 voh)

3.4 = 3647 A

4.8

Bohr's Correspondence Principle Summarizing all the experimental evidences. Bohr tried 10 interlink quantum

physics and classical physics. He suggested Ihal, The prediction of quantum theory for the behaviour of any physical system can be equal 10 that of the classical theory irthe quantum number is too large.

Symbolically (CorrnpondsIO)

Limit quantum theory

;::::>

Classicallheory

11-)000

The above statement is called Bohr's correspondence principle. A simple pendulum oscillates with a frequency f = S hz . lIs total energy is 10 J. Find the quantum number. Solution: The energy orthe oscillator is E = "hf n =£.

hJ

= _----'IO~­ (6.3 x W" )(5)

=3. 17xlO)3 The osci llator loses a very small amount of energy of3 1.S x 10- 34J at a time 3.17x lOB limes so as to come 10 rcsl.This quantum number is so large that, our eye cannot catch the small change in energy of oscillation and corresponding height. It appears as if the pendulwn bob slows down continuously to our classical cye.

r •

387

Particle and Wave Natu~ of Matter

4.9 Frank~ Hertz Experiment: Confirmation of Atomic Energy Levels In 1913, Bohr presented his atomic model which suggested about the discrete energy levels in atoms. In 1914. Frank and Hcrtz(Ncphcw of Henerich Hertz) experimentally proved the existence or discrete energy levels in mercury atoms. They bombarded electrons by passing there in an electric field with the mercury atoms (in mercury vapour) with a kinetic energy 4.9 eV (or greater). The low energetic electrons CQuid excite mercury atoms to their higher energy states through the process of inelastic collision which occured at same panicular kinetic energies of the bombarding electrons above 4.9 cV. This proved the existence of same particular (specific) energy levels in mercury atoms. The dc-excited mercury atom emitted ultraviolet light. This is infact the process of getting light in our mercury vapour lamps or tube lights in our houses.

; )

\

i, I

!

I I

I

I

I I I

)

l

Find the wavelength oJtlzeemitted photon ill Frank-Hertz experiment Jor the electrons accelerating throllgh a potential difference of 4.9 \'Olt . Solution; We know that, tlE = h! Bohr Einstein's fonnula

0'

4.ge '" hc ),

). =

-.!!£ = 6.3 x 10-34 4.9 e

x 3 x 10

8

4.9x 1.6 x 10 19

=241 nm Ans. This was exactly equal to the wavelength confinned by the spectroscopic measurement oftbe emitted radiation. It falls in ultraviolet region.

4,10 de-Broglle's Matter Waves By 1920, physicists were aware of various ways of emission and absorption of light. Einstein and Planck proposed that a black body would emit radiation discontinuously due to the oscillating charges in solids. According to Bohr and Einstein. a revolving electron can also emit (and absorb) radiation in a atom. Furthennore, an electron can emit continuous X-ray when accelerating in the electric field of an atom. In all the above methods we can get radiations. According to Maxwell all radiations arc emw. We measure the radiations intenns of wave numbers in spectrometers. This proves the wave nature of radiations. According to Einstein, radiation is quantised. In other words, radiation is emitted and absorbed by matter (atoms, molecules. nuclei etc) discontinuously in the from of energy packets (photons). Thus. radiation has both wave and particle nature as suggested by Einstein. Now a question arises, why should an electron accelerating in circular path or any arbitrary curve emit radiation possessing a wave naturl! (having a wavelength and frequency) and particle naturc? In more straight fomlcd way. we can ask. who makes the radiation "wavy" and grainy? Bohr could not answer this question. In 1923.

388

GRB Understanding Physics Optics and Modem Physics

dc-Broglie presented his idea of "matter wave" to answer the above question. This is based on the following logic. In wave mechanics we learnt that a vibrating tunning force can produce transverse waves in a string and longitudinal waves (sound) in air. Hence, only a vibrating system can produce waves. According to Maxwell's law of electromagnetism an oscillating charge can radiate emw continuously. However, Max Plank and Einstein quantised radiation. saying that, any vibrating system must lose or gain energy not cofinuously but discontinuously or packet wise. This make the thing more clear that "3 vibrating system of charges can produce (give) wave nature to radiation . This is a rough answer to the first question "who gives wave nature to radiation". Now come to the second question to enquire about the cause of "grainyness of radiation. To answer this question. de-Broglie took the example of a vibrating streched string. A strcched string can not vibrate with any frequcncy (or energy). It can absorb the waves of energy or frequency equal with one of the resonance frequency. Likcwise it emits the waves (sound) of frequency equal to one of the resonant frcquency , This made him (dc-Broglie) to think that. just like the string contains stationary waves emitting and absorbing waves of discrete frequencies. thc elcctrons might be associated with same sort ofwavcs fonning a stationary wave an atom. This might be the cause of emission and absorption of radiation (energy)in discontinuous (discrete) fashion, so also the cause of providing a frequcney to the emitted (or absorbed) radiation, whose frcquency is equal to thc frcquency of the associatcd waves of the electron. In other words," the wave nature of radiation com.:s from the wave nature of matter", Thi s concept modernised the quantum physics. Matter and radiation are two counter parts orthc nature, as described by Einstein in 'this theory of relativity. Matter and energy arc equivalcnt. de-BJOglic finnly believed the mass energy equivalence between matter and radiation. Based on the symmetry of nature and mass-energy equivalence. de-Broglie argucd that. since radiation has both wavc and particle propeny according to Einstein. mattcr should also posscss a wave charactcr in addition to its most revcalcd paniclc nature. dc-Broglie imagincd some sort of wave associated with mattcr called "mailer wavc". We should not misunderstand the mattcr wave as waves in mattcr such as transverse wave or longitudinal waw. Thcn, what exactly is waving in matter wave'! fnfact, this qucstion baffled thc physicists de· Broglie, Schrodinger etc, AI last Max Boon gave the correct interpretation of the matter wave asa "probability wave", which is just a mathemalical one. We will discuss it in latcr scctions. Again we suggest the reders not to think a moving particle as a group of waves guiding or piloting it, the way people used to think at the beginning of the invcntion of this concept. According to Einstein wave-particle duality of radiation possesses a particle nature (having momentum P) and a wave nature (have a wave length A.) given as

Ie radiation

::::: n Ii l"radiation

Sinee matter and radiation are equivalent, the wavc-particle duality of matter can also be given by the above fonnula. Hence we can write

389

Particle and Wa"'e Nature of Maller

, fo.

where

PfT\1\t\!r

matter wave.

matter =

h

pII'I3l1cr =momentum of the moving particle and ). mailer

=wavelength of

This equations is called Einstein-de-Broglie's matter equation. Any particle of rest mass 1110 moving with a speed v has momentum P = m".where m =

R' mo

1--

c'

I) -~~

Then,

= -,-~ c_-

"maner

1II0!)

The frequency oflhe corresponding matter wave is f =E

" " ~l-

= ",c

2

(1110 I

=

u2 ; c 2 )('2

or ,.....--~

~

~%4I111'U'!.,

Why

realise any wave nature of a hall of mass 200 gm movillg wilh a speed oJ30 ml s? COII't we

' olution : The dc-Broglie wavelength is

'A. =!!.=!1... p

mu

6.3>:: 10-34 200 x30 1000 = 1.05 x 10- 34 m =

This wavelength is so small compared to the size ofa nucleus ::::10- 15 m, that the nucleus can not different this incredibly small wavelength. Hence. the

wave nature of macroscopic objects are unnoticable.

I

GRB Understanding Physics Optics and Modem Ph),'Sics

390

4.11 Experimental Confirmation of Wave Nature of Matter by Davison· Germer and George D. Thomson (1927) In doctoral thesis of de-Broglie he suggested the wave nature of electrons. This b3ffied his superiors 3 S they did nOl accept this proposal because it was nol experimentally verified. Then. they called Ein stein to decide whether the doctoral degree should be confomlcd to dc-Broglie based on his proposal or not. Einstein could realise the impor1ancc of the idea of dual nature of matter and advised the guides 10

award the degree. However. he speculated that the molecules, atoms elc should exhibit diffraction to possess wave character. In Rellected olectron 1925, dC-Brogl ie already proposed that the waves electron beam could be diffracted. This was infaci proved concept by Davisson, Genner in Bell laboratory atom. e USA and independently by George P. Thomson son o f 1.1. Thomson in U.K. They proved the diffraction ofa beam of electron ""Surface atoms at metallic surface and inside the metals. d The electrons in the beam behave like a The scattered electrons maner wave from the surface atoms Interfere wave when interacting with the surface to Pfoduce interference cons\nJctiYely atoms of a crystal. If the spacing between (or diffraction) panem similar to two nearest atoms is d o f the crystal. the X' ray ditfraction panern scattered electrons (diffracted) waves superimpose constructively to produce maxima at some points distructively to produce minima at some other points of the photographic plates. Here. Nichel is the target at which the diffraction occurs. For constructive interferences. the path difference between two adjacent rays is

,

I

Ax

=n~~

or d sin9 = nl Putting d = 2.15 A, n == I and 9 500 (as recorded by the detector). Darrisson and Genner, got ,. = 1.65 A for the electron wave. Then they found the wavelength of the electron by using de· Broglie· Einstein's fonnula h A=p'

=

where P = ~2mk

).=-".J2mk' where K = KE of the electron = eV. V = potential difference through which the

or

electron wave accclemted. Then, where V

=54 volt ). =

A =-Fh=;; .J2meV

6.3 x 10-34

~2 x 9. l x lO-31

x 1.6 x 10- 19 xS4

= 1.67

A

391

Particle and Wave Nature of Mauer

This was an excellent agreement with the theoritical result of de·Broglie and experimental result of Davisson, Germer. It is interesting to note that the experiment of Davission and Germer was accidental, Furthermore, son G.P.Thomson discovered wave nature of matter and his father discovered the particle nature of matter. Later on, all particles: neutron etc., were proved to be having wave properties.

We should note that for fast moving electrons, we should use Bragg's formula; for maxima 2d sin e (instead of d sin 9) "" ni~ This is because, energetic electrons get scattered not by surface atom but at the bulk atoms \ 10/ 1

'\ iY /

'\i/ •

4.12 Double Slit Interference of Electrons We know that light

--

}d Screen

produces interference and diffracLlOn. Like wise, a D (detector) beam of electrons after -' ,/ passing through the slits ,/ (holes)! and 2 will ,,' produce similar effect "," (interference pattern). No\v a question arises, o 0 1 .. d how can the electrons being the particles o 0 0 superimpose to produce interference patterns like a 0 2 .. ,, waves? Let us assume that an o 0 electron is a classical \ particle like hard tiny Electron elastic ball. Then it can be.m pass through any onc hole (lor 2) at a time and scatters in Curve a and b Intensity distribution due to the electron passing different directions and finally through holes 1 and 2 respectively. Curve-c gives the intensity hit the screen at different distribution due to the superposition of curve a and b. CurvlHi points. Statistically, all electrons have equal chance gives the observed intensity distribution when both holes are (probability) to pass through ope'. each hole. If half of the total

......

-...... ---

-

-

...... -- ...... -- - --------:7""-.---------1."--;-+.,... __ ------------------------------17:':----_ --

392

GRB Undentanding Physics Optics and Mooem Physics

numberofelcctrons pass through the hole I. they will produce diffraction pattern given by curve-a. The other half of the electrons pass through the hole 2 will produce the diffraction pattern given by curve-b. Since both occur simultnncousJy(cquaJ number of electrons are assumed to pass through both holes). we should expect an interference pattern (curve)c. as the superposition of the Clln'cs a and b. However. in practice we observe a different interference pattern similar to that obtained by a beam of ligh!. given by the curve -d. To produce the interference pattern given by curve -d. a light wave or single photon passes through both the holes simultaneously. Hence. each electron must pass thro ugh both holes simultam:ously to produce similar imerfcrcnce pattern. It is unbelievable if we think an electron as a classical panicle. but true as they are quantum.

Any atomic particle has dominating wave property in addition to Its particle property. However, a macroscopiC particle has dominating partiCle property than wave

orooortv.

4.13 Wave function '1'; Max Born's Interpretation of 'V Probability Density The intensity of an interference pattern is the mC3sure of brightness of the fluore scent screen (or photogmphic plate). Each electron will hit thc screen ofa point producing a flash of light (scintillation) or the electron can hit the photographic plate leaving a mark on it. Alternately you can move a dctcclOr Minimum probability which records the striking sound of an electron as a "click". density If at any position of the detector the number of"clieks" per second (arranged over a long time) is more. we can say that the probability (or chance) of reaching an electron on the screen (where detector is present) is more. The total number of electrons hitting the detector (or any small area of screen) per second, divided by the total numbers of electrons hitting the screen per second is called "probability of finding an I I electron" at that area (or volume) . This probability divided by the area of the detector (on which the electrons hit) is defined as probability density. denoted as P. This is a measure of intensity of the pattern (diffraction or • interference of electrons). If the probability density P is more on the elementary area dA . thai area looks brighter on the scintillaling screen (or darker with more black spots on the The Intensity distribution white photographic plate). Then, we can write. given as P =1 1fII2 - II graph

393

Particle and Wa\'e Nature of Mauer

In a nut-shell for diffraction and interference, Brightnes.<; (Intensity) oc Probability density Let us understand the above expression more physically. When we look at the TV screen (or a black photographic plate), at the brighter region we have greater chance (probability) of finding an electron hitting the screen. At dark spots there is a little possibility of finding on electron landing on the screen. As the brightness of the screen (or plate) changes (varies or waves) from point to point (or time to time), we can say that it is the "probability" (chance) of finding an electron at a point" which is The electron interact WAVING. In other words, de-Broglie's "maner wave" is a with the screen to "probability wave". A matter wave, here after, must not be misunderstood as a some sort of displacement waves associated produce a scintillation with the particle, as de-Broglie himself imagined.

I

I

Max Born Probability Wave Function (t925-1926) Schrodinger invented a wave function IV ' to describe the matter wave. He felt that the amplitude of the electron vibration could be given by the wave function IV like dc-Broglie. Funhennore he (wrongly) assumed IIV1 2as the electric charge density which was rejected by Bohr (please refer next section). In the same year. 1925, Max Born correctly interpreted IIVI2 as probability density of electron following the probabilistic approach of Einstein to interpret wave-part icle duality of radiation. Max Born argued that Y can not be measured as displacement of electron (or matter) wave because the displacement or position of a moving microscopic particle can not be measured precisely. When we pin down an electron in a small region (volume) to find its position more precisely, its momentum will be tremendously increased (disturbed). This is the idea behind Heisenberg'S uncertainty principle. More tin Ii hter articles occu

reaters ace Qwin to their 'ark orwa

motion.

Einstein suggested that, the intensity of the screen described by the probability of finding a photon on the photographic plate is directly proportional to the square of the amplitude of wave function E or B ~

~ where Eo = amplitude of wave function E. Following the footsteps of Einstein, Max Born argued that, the intensity which is given as the probability of finding an electron striking spot on the TV screen or photographic plate should be directly proportional to tbe square of the amplitude of probability wave function because both the cases, we obtained similar interference pattern. So,

GRB Und~ntanding Physics Optics and Modem Physics

394

I p "' ~~·1

where 'V 0

=amplitude of wave function 'V .

Diagrametric comparison between wave functions for water wave, emw(Ught) and matter (electron) wave:

The wave function y is real measurable and physical for water wave. y= Yl + Y2.atP. "

5,

The wave function E is real, measurable and physical for cmw. E=E 1 +E2,tltP.

5,

.mw

p

I"::))IL. .4---H;;;f1 s,

The abstract (probability) matter wave function measured for mattcr(elet:tron) wave. \jI = 'VI

\jI

is imaginary. c.1nnot be

~

+'V2. atP .

. ... L

, .-----'~-;J4 , " o , _<:I , ,, ," ... 0

p

0

Electron wave 1

Normalisation Then, the probability of finding a particle between x = a and x = his Probabilily = Pdx

J

= J:I~12 dx = Area under graph l'l'!2-x.

•

,, i

I

Particle and \Va\'e Nature of Matter

395

is definitely present between x = hSinceltohoeo~article (= 1) probability

can ave

-00

andx

=+00

on x.axis we

1+1 '

/0

,

IProb ~ C ;1 ljfj'
So,

Th~ ~bovc expression is the condition for no~ahsatiOn.ofthe wave function'll. It states that the particle descnbed the wave function must be present somewhere on the x·axis. In . ~ state ~f well defined energy level, the pro~ablhty denSity of finding an electron, i.e., 10/ 1 does not change with time. This idea gives birth of various quantum state or shell-sub shells of different shapes, sizes i.e., oval, dumb·bell etc. In three dime?sio~s, th.e probability of finding a particle anywhere 10 thiS umverse is "one". Then. Ja1\space l'V12 dv = I

'I' - x graph and 1",1 2 - x graph 01 an electron bound In an atom

The orbitals represent a probability distribution 01 electrons

Expected values of'll

The electromagnetic wave function E satisfies Maxwell's wave equation. The matter (probability) wave function'll satisfy Schrodinger's wave equation. E can be measured whereas 0/ cannot measured. Even then knowingth wave function'll we can calculate all physical quantities like energy and momentum of a particle. lfwe know the wave function'll = I(x, t)ofa particle moving alongx·axis, we can calculate the average position of the particle after experiment trials. The average position is called the "expected value" ofx. It is given as

or

< x>~

f.: xjIjf j'

I

~ I' :xjljf j'

<x >


I

(',' I' :jljf j'
This expression states that, if a particle is in a definite energy state, the probability density of finding the particle does not vary with the time. Properties of Ijf To prove a physical interpretation of the solution ' of Schrodinger equation, 'V should possess the following properties: L 0/ should contains all information of the properties of the particle that can be measured in the experiment.

,

1 GRB Undentanding Physic! Optia and Modem Ph),!iu

396

2. \V has no physical significance as it is a mnlhcmnticnl quantity to describe a non· physical " probability wave" (like "center of mass", where no real mass need to be present). 3. 'V must be single valued and continuous. OW () 2\V 4. -a ' --2 etc .• must be single valued and continuous. x 5. 'V must be nonnalisablc ; \111 2 dx = COJlstant, etc.

ox

I':I

6. 'If must have +vc and - vc sign in order to explain the fonnation of maxima and minima in interference pattern and fomKltion of molecules by over lapping of orbital s

probability distribution). 7· 1'Va 12 has a physical significance which explains the real positive mathematical quantity "probability", 8. 'If must be given by a complex number. Hence. 'V. 'V. =l'l'l~.

4.14 Schrodlnger's Wave Equation Let the probability wave function'll represent the motion of a free particle along x-axis. Sinee Fexl =O, the momentum P of the panicle of mass m remains constant. The total energy of the particle is also constant.

'"

A

}---'.,-/----'''-,''--'''7'----

,.1

-!tEI - P);}

Probability function 1¥ '" As II

particle carrying an energy E and

E=K+U or

E = ~+ul

of a moving

momentum P

... (i)

The wave function in complex fonn can be given as \jI

where A

= Ae - i ( 1I1-h),

=probability wave amplitude.

Putting w =

~ and k = ~ • rl------:---, '" = Ae

_i.. (£t-P,l ) \ 1i _

... (ii) The above expression describes the probability wave of the particle carrying energy E and constant momentum P along x-axis.

Particle and Wa\"e Nature

or Matter

397

Differentiating'll twice wn x,

r=,;--~;---,

8'IjI

P'

ox2

n2

-=--

Differcntinting~ one wrt

t,

1jI

" .(iii)

r . ; - ----, Oljl = _ iE IjI

... (iv)

01

n

By using the eqns. (i), (iii) and (iv) ' +u~ in 8-.! = __1;,_' 8-..1 at

... (v)

2m ax2

In three dimensions,

[0'

' ~ +----'f a' + 8' in 8-.! = -~ ----'f +u~ at

2m ox2

ay2

0=2

J

... (vi)

,

The egns. (v) and (vi) are Sehrodmger s tIme-dependent equations to one and three dimensions respectively, where U = potential energy of the particle is the function of x,y.= and I . .If E is constant, put; n

mvDt = E,

2 2 a 2 + a + a ) = H (Hamiltonian), in the eqn.(vi) to obtain [ ax 2 0,2 0='2

I'H-'-IjI-=--=E'-IjI," where IV = region function, H = Hamiltonian operator and E = tOlal energy of the particle. This fonn is a general foml (or can be used as a steady state fonn) of Schrodi nger's equations. It helps us to find the wave function if we know the variation of total energy of the particle with distance. Schrodinger equations solves the wave function for probability distribution of electrons in hydrogen. other atoms and molecules. It is not necessary to impose quantisation as Bohr did in his atomic theory. The quantisation comes in a natural way while we solve the Schrodinger's equation. This equation predi cts the probable position of an electron and other microscopic particles ifinitial condition and boundary conditions are givcn. Thi s equation cannot be derivcd from any other basic principles like Newton 's laws because it is a new concept in itself. The new mechanics of atomic particles arises from Schrodinger's equation is called quantum wave mechanics. Despite of its quick recognition, its wave function IV was not properly interpreted by Schrodinger. Later on in 1926. Max Born interpreted the wave function ~ as probability wave function ( as described in last section). According to Schrodinger l~o l2 was the charge density. Max Born argued that, when we get a fractional electric charge at a point, it is physically impossible because electron is present with its total mass and charge completely (but not fractionally) at any point of time and space. Hence, probability of finding a fraction of electron at a point became meaningless. Then, hc interpreted IV as the probability wave function whose amplitude square, ;.c.,

GRB Undtntanding Physics Optics and Modtrn Physiq

398

10/ 0 (orA )12 represents the probability density 0 r finding "an electron but not fraction of electron" at a point. This concept was received by majority of phy~ici sts excl~~i~g Schrodinger, Einstein elc, as they believed in determinism (opposmg probabilistic approach).

4.15 Particle In a Box The term particle in a box technically means "an object moving in between 1\"'0 rigid walls", Classically speaking. a particle i.e., n ball or an electron. say moves back and forth between two rigid elastic walls with any amoun t of momentum and kineti c energy. Howewr, any atomic particle (i.e .• an A classical particle electron, say) owing to its dominating wave nnlme move in moves back and forth such a way that its probability wave [anns a stationary wave after repe ated collision between the rigid walls. It can be compared with the stationary between two rigid walls waves generated by the superposition of transverse progressive waves in a string rigidly fixed at both Y >12 its ends. Furthennore, we can compare the ~~".. ,. ...... stationary matter (probabi lity) wavc with thc ~ ___ , ____ ~ __ _ stationary emw fanned between two rigid supports ' .. . ,/' ' .. . ",/ (mctallic cavity) and stationary sound wave s a Stationary transverse wave cavity (drum etc). Taking the simplest case of stationary wave in ,. .. a string of length I. the necessary condition is Ihat " ' «the length of the string must be equal to the ,.. . ", -' integral multiple of the half wavelength (loop Stationary matter (probability) length)", wave 01 a quantum particle Symbolically,

_I

1_

,

:-1)J2 1-__

"'..

n~ = I

,

•

where n = no. of loops, ~ = loop length.

I)· =~ I

or

.. ·0)

Some expression holds good for matter wavc. Hence, the momentum of the particle is given by de-Broglic's equation

.

en ~

Puning A. from eqn. (i) in eqn. ('i.t I D .

nh

p = 21

Then. the kinetic energy of the particle is

p' 2m

k =-

... (ii)

Parlid~ and Wave Nature of Malt~r

399

__

n2 112

K(=E)=s;;;;>' n=I,2,3. ... ..

or

From the above expression, we learnt that, the momentum and kinetic ener

of a article movin in a box are uantised.

By using Schrodinger's equation and solving it, the wave function of the probability wave can be given as

, ---,0--,

10/ = ~Sin Txl - - - - - - - E3

It is important to note that, a particle cannot be at rest in a box. Its speed increases if we decrease the length of the box . The particle cannot have any value of momentum and KE. It will have only some allowed (discrete) values of energy and E, momentum. The energy levels are not equally spaced like that in oscillators. The electrons in a The spacings of the energy levels of a metal is the ideal example of "particle in a box" . particle In a ben: are not uniform.

ea-
~&!A-1!li8j

An electron is in a box 0.1 nm ofwidth which is the order ofmagnitude ofatomic dimensions. Find the pennitlcd energies.

Solution: From the previous fonnula, n2 h-2 E= 8mP n 2 (6.3 x 10-]4 J -5)

= --'-;;----'-;;--;; 8(9.lxI0 "kg)(O.lxIO-"m)' = 5.45 x 10- 18 n 2 J

=34n'eV

(.; leV

=1.6x 10-19 J)

The pennitted energies of the trapped electron can be obtained by putting

n=I,2. .... E1 = 34eV,E2 =136eV,E J = 306eV and so on.

4,16 Heisenberg's Uncertainty Principle In 1927, Heisenberg analysed the wave-panicle duality of both matter and radiation presented by the leadership of Neils Bohr. He found that, to express the "wave and nature and radiation" are must need a "wave fonn", which must be "confined" and "very". This could be given as a "wave pulse".A ware pulse has a limited length (size). Inside the wave pulse either we can see lot of smaller waves of amplitudes varying from zero (at the ends P and Q) to maximum (at the middle R).

eRB Und~rstandinB Physics Optics and Modem Physics

400

-I)' 1-

Let us assume a matter (probability) wave pulse associated with an electron moving along x-direction . Let us try to locate

J--

the position of the electron relati ve to the 0 origin 0. Since the

II: - - -••

p

•

Q

R

electron probability wave pulse has width • I:!.x. the position of the electron can be located between x and x + I:!.x. In other terms, the exact position of the e(cciron is The position of the particle P electron can be meaningless. It is most likely to find as the located some where between II: and II: + Ax mid point R. lcast likely to slay at the ends P and Q of the wave puise. 11lis means that, the electron is likely to stay in side the pulse (of probability wave) between x + Ax. Hence, the inaccuracy (or uncertainty) in location (or position) • of the electron is the width.1x oflhe pulse. It is mathematically proved (as Fourier theorem) that "a wave pulse" can be created by superimposing many waves of wave numbers ranging from kl to k2. The range of spread of the wave number that is 6.k(= k2 - k 1 ) decides the width .1x of the pul se. Fourier theorem, states that product of the width of the pulse and thc spread 6.k otlhe wave numbers (or numbers of different waves The pulse is the combinations of waves 01 angular wave numbers superimposed) is a constant which can be ranging from k, to ~ approximated as onc. " .(i) Furthermore, it can be proved by Fourier theorem, that the minimum value of M 6. x

is~: then. ".(ii)

.Since, each superimposing infinite harmonic sine waves have momentum P

=:

= lIk( 1;

=;~ ). by considering many hannonic waves to create a pulse, a

variation of momentum of the particle occurs from P t = lIk\ to corresponding 10 fi rst wave k\ to P2 = nk2 corresponding to last wave k 2 . Then. the inaccuracy (or uncertainty) in calculating momentum of the particle (electron) is llP =P2 - PI = 1i (k2 - kd M' = t: l>k or l>k=M' or .. ,(iii)

n

Substituting 6.k from eqn.{iii) in eqns. (i) and (ii), we have

II I, I

Partid~ and Wa\'e Nature of Malter

401

61'-6-, =

~

!1P . ax > -.!l

and

- 2

The above expressions lell us that the product of uncertainty of measuring momentum P and position xof a particle (of radialion or matter) at a given time in the same experiment can never be greater than

~

• or nearly equal to 71 .

In the course of derivation, we can see thaI these uncertainties arises not from the defect of the Illcasuing devices (like microscope etc) or set up in the laboratory. They (AP and 6.:r) arise due to the wave-particle duality of matter or radiation. To be more clear. Ax arises due 10 the finite (but not zero). To be more clear. Ax arises due to the invol\'ement ofntany hannanic waves to create the pulse. ![we want to make the pulse shorter to decrease the uncertainty in position l1x the number of waves resulting the pulse must have to increased. This incrcases the increase in llP uncertainty of measuring momentum. In short, if we try to increase one, the other will decrease due to the vary nature of the wave-pulse generated to represent wave particle duality of nature. Hencc. this uncertainty relation is a fundamental law of nature, according Heisenberg. However, Einstein opposed it by saying that the uncertainty might be caused by the ignorance. The future process of refining the science and philosophy could tell more about it rccapilutaling Heisenberg wrote:

-

One can never know with perfect accuracy both of those two important factors which , . determine the movement of one 01 the smallest particles-its position and its velocity. It is Impossible to determine accurately both the poSition and the direction and · speed of a particle at the same instant. Mathematically, ~

!J.x·tJ,P =h

--,

~.:!t,!'_li!.,

Derive the uncertaint), relation beh\'een time and ('lIergy. Solution: Let us assume thaI an oscillator is switched on. It takes a small time to . build up its nonnal amplitude from zero, oscillates for a finite time llt then switched off " to de<:rease ils amplitude to zero rapidly as shown in the figure, Hence, the vibration is .not completely hannonic. 1t is a periodic wave pulse oftime llt and dominant frequency , • ,..' nearly equal to the frequency of the oscillator. Let us assume that during time lll, the oscillator .vibrate n tim~ during timet-Ail then the average (or dominant) frequency is ;:.

.

-

r

II

,

• ,

CRB Understanding Physia Optics and Modem Physics

402

•

•

I~

"". f

::c

•

. ''''' "".

•

•

•

nih

oyc<e

.!!..

M Then, taking differential of both sides,

~J =~n ~t

or

Sj-SI=Sn

Since. the minimum value of difference in no. of oscillation is I , we have

Sf ·5tlmin = 1

I !'.f · 61> I I

0'

for a photon

f :: E.

given as the

I1f = l1E . Substituling value of AI in the above eqn., we have

h

Hence, the uncertainty of frequency and energy can be

h

[,v;.6t>h

I

4.17 Wave-Particle Duality 01 Maner and Radiation Complementary Principle (1927-28) The radiations(visible light, heat etc.} were known as waves to the physicists like

Huygen. Young, Kirthhoff etc. In 1867, Mal(well suggested that all radiations are electromagnetic waves. In 1888. Hertz proved the physical existence of electromagnetic waves. From 1885 to 1913 the particle (quantum) nature of radiation was established by Max Planck. Einstein and Bohr. By this time the particle nature of matter was confirmed by the discovery of protons, electrons, nucleus by Gold strain, J.J.Thomson and Rutherford respectively. In 1923 Compton proved that an X-ray photon could kick as free electron by delivering a momentum like a particle. In t 923, Einstein established the dualism of radiation suggesting that, radiation has both wave and particle properties. From 1923 to 1927 de Broglie. Schrodinger. Heisenberg. Bohr and Born proved the wave like character ofmatler particles such as electrons, protons, neutrons etc. At least. in 1927. Neils Bohr unified the dualism of radiation and matter in a single theory, called "principle of compie men tory". This slates that. nature (matter and radiation) can be better understood by taking its both (wave and particle) character into a~count. The wave and particle : haracter cannot be revealed at a time in a single expenment. However. wave and particle nature do not oppose (contradict) each other. Rather they complement each other to glorify the nature.

403

Partide and Wave Nature of Malter

A particle of radiation (photon) or a particle of matter (elcetron etc.) having momentum P and energy E has wave property whose wavelength). frequency I are given a single set of fonnulae,

Thc wave-particle duality of matter and radiation can be graphically explain by a wave pulse. In cose of E,B radiation, a photon can be given as an electromagnetic wave pulse having energy E = hI and momentum p

= ~.

In case of matter particle, the wave pulse is a matter (probability) wave pulse having momentum p = mv =

f

and energy E = me 2 =hi. The localisation of the wave pulse in each case gives the particle nature possessing energy E and momentum P. The phase waves in the pulse gives the wave nature possessing A and I. Any wave pulse is subjected to Heisenberg's uncertainty principle, as described in later section. We cannot measure energy, momentum and position with 100% accuracy due to the inherent wave-panicle duality ofmattcr and radiation. According to Einstein, we cannot find the fraction of a photon. The probability (chance) of finding a photon is more where intensity is more or electric (or magnetic) field is more. P(
-

1\ f\ q,1-I--Hrf''-'-_X,1

V V4X

_

•

• 1\ f\ q,1-I--Hrf',-,-_x,1 _

V V4X

-

•

Repralentatlon of ~ nature of taliatlon and matter by a wave puIN having wave func:tIoi .. E and 'f' respect/Wt)'

Likewise, according to Born. we cannot find the fraction of an electron (or any particle of matter). The probability of finding an electron is more when the intensity is more or \jI is more.

GRB Undentanding Physics Optics 4lld Modem Ph"ics

404

Born's interpretation

Einstein's interpretation

Probability density of an electron is

Probability density ofa photon (radiation particle) is

P ocj'l'ol'

Poc£2(orB2)

\jI 0

E and B arc cmw function they are real

and measurable

/'--

.--..,...."

~~~oj

is a matter probability wave function. It is not physical wave. Hence,o/ is a complex quality and cannot be measured experimentally

, I

Find the momentum ofa photon of wal'elength 6000 A.

Solution :.p = ~

6.3)( 10-34 J - 5 = ~6~OO~O~X~I-o--;,",o~m = 1.05 x IO-27 kg mls . Ans. ,. Find the wal'elength ofan electron \l'Ol·c.if the electron has speed of 10 6 m/s.

.

Solution:

(6.3)(10- 34 J_5) = -(9-.I~x~I-'-O~ - 'C;''kg C-)-(I-0<'~m/ s)

,

,. .,: '..!~ .

=6.92x.1O- l om

Ans. ~,

, ,

','

Partid~ and Wave Nature

~b-"

or Mall~r

405

MISCELLANEOUS EXAMPLES

.-" " ~ ~ ~

Example 1. The angular momentum of an electron in a hydrogen like ion of atomic number Z. is L. Find the energy of the electron. Solution: The angular momentum is

L:: or

nl!..

2n

n :: 2~( = ; )

Then, the energy of the electron is

E

= mZ 2 e4 =-",m~Z~2~c~';&:on 2 112

8E: on2 Jt2 4 mZ 2e

==

_,)2

&;2 (2~

= Jt2

Ans. 32J'[2c2L2 0

o " Example 2. Find the (i) magnetic dipole moment J.I (ii) ratio of magnetic dipole moment and angular momentum of the orbiting electron in nih energy level. Assume, m = mass of the electron and e = charge of the electron. Solution: (i) The current produced by the orbiting electron is

• •

• •

i=! = _e_ =E!e= ~

T

2rt/00

21t

2w

The magnetic dipole moment is Ii = iA = i(nr 2 ) (0: A= nr 2 = Area of the current loop) 2 eu =nr -

2nr

or

Ii =~

. .'

2

... (i)

Bohr's quantisation of angular momenrum teUs that

L == mvr = nh

o ••

2n

Eliminating v, from eqn. (i) by putting vr =

2':::n from eqn. (ii)

(ii)

GRB Undcntanding Physics Optics and Modem Physics we have

(ii)

Il :::! nh ::: neh 221tm 41tm The ratio ofll and L is neh I 4run

e. : : L

nh 2.

~ =.£.. L 2m This is called gyro-magnetic ratio,

or

AnL

..£!!.. = 1 Bohr magneton ::: Il B ::: 9.27 x 10-24 A -rn 2 4"", Enmple 3. Find the magnetic field induction due to the orbiting electron in Bohr's atom. Solution: The current produced by the orbiting electron is j=

ef

~

=

2.

The magnetic induction due to the circular current loop is B=lJoi=J.to~ 2r 2r 21t

(-: ro.;) , me in eqn.(i)

lJoeu B.--

or

4",'

e'

Put v ::: -,-->---,;- and r = 4JU:o 1i 1I2(4ru:o) to obtain,

J.tom 2e' B.~'7i~ 2S6l't 4 £lo 1r5 = 12.56 T putting the values of fundamental quantities.

,

Eumple 4. A particle of mass m moves in a circular path unifonnly under a potential field U =

~

,where r :::: radial distance from the center of circular path.

Using Bohr's quantities condition, find the pennissible radii of the orbits and energy levels oftha! particle.

Kr'

Solution: The potential energy ::: U = T

Then. the force of the field acting on the particle is F •

or

F

.

, _.

.

_r!Y.. • dr

.-Kr

_.- ..

'

. .

-

_.!L(Kr') dU 2

Particle and Wave Nature of Malter

407

Newton's 2nd law applied on the particle yields

F =ma

"r

Kr=mmv = (~mK)r

0'

... (i)

Quantisation of angular momentum yields mur = n1t Using eqns. (i) and (ii)

r=~

"If .JmK

u=Jnn

and

... (ii)

.JmK l m

Example 5. A charge particle of charge qt. mass ml moves in a straight line with a constant kinetic energy K by the side of another charge of charge qz and mass m2.lf the charge qz is free to move find the kinetic energy q, k gained by the charge q2. Assume that qz does not --Oo-----tr--move parallel to ql as qz passes by it. Put qt = +eand ml b q,! = -e, ml :: mp andmz = ml'.

1"'2

·2

Solution: The net displacement of qz alon~ x-axis is zero. Hence. Px = Fxdt = O. Then, the momentum is delivered to q2 along y-axis which is given as

J

Py =JFydt =

=

Fsin9dt

J qlq,

.!!dt 41t£orz r

J

= qlq,b dt 4m::o r3

Putting x = bcot e. fix = - bcosec 2SdS, r = bcoscc 9 and dx = vdt P

,

d

•

y

= qlq2 rsin9da

41t£oub

e

Then, the kinetic energy gained by m2 is

•

P

•

K' = ;

2m,

eRB Understanding Physics Oplia and Modem Physics

408

,

=

IIIp(!

:::

3.82 cV

Ans.

2

(4nco) 2 Kb mc

I .... to ground stale Example 6. A helium atom is dc-excited frolll an energy Ieve " . to emit two consecutive photons ofwavelcngths 1085 A and 3040 ~. ~md II. Solution: let the energy of the initial (exc ited level) be E. It IS Siven as

i

Z,

E = -13.6 -

,,'

::: - 13.6x ~::: _5\4 cV

1

1/ 2

11In coming to STound state, the lotal energy di(Tcrcncc is M = E-Ej

=(

_5:;4 )_(-5i.4)

. dE :::54.{1- nl2

)cv

... (i)

The total energy of the emitted photons is Eh = hc+hc p

'.\

"2

: : 12400(_I _ +_I_)ev 1085 304 or Eph ::: 52.08 cV Since 6E ::: Epb.

54.,1 -

... (ii)

nIl) = 52.08

1_ _1 = 0.96 ,,2 n =5

Ans.

Example 7. The binding energy of a ground electron in neutral Helium atom is 24.2 eV. Find the total energy required to remove both electrons from ground state He-atom. Solullon: The binding energy of the electron in Bohr's hypothetical atom is E = 13.6Z 2 eV To remove one electron from He-atom, energy the required Ebj = 24.6 eV To remove another electron from He + ion, put Z = +2

EIlz. =13.6x4=54.4eV Then, the total energy required to remove both electrons is

Eb = Ebj + EIlz. = 24 .6 + 54.4 = 79 eV

ADS.

Particle and Wave Nature of Maller

409

, Example 8. A light ofwavclent,>1h 124 A falls on a singly ionised fix.ed Helium atom. Find the velocity of photoelectron if the photon knocks the electron of the He + ion onto fit. Solution: The binding energy of the electron with the atom is 2 Eb = 13.6Z = l3.6x (2)2 = 54.4 eV The energy of the incident photon is Eph ~ 12400 ~ 12400 ~ 100eV ). 124 The K.E. of tile photoelectron is

K:::Eph-E b ~ (I00 - 54.4)cV

~mv2

= 45.6 eV = 45.6x 1.6x 10- 19 J

45.6x1.6xlO 19 x2 9.lxlO 31 ::::4 x 10 6 mls

V=

or

Ans. Example 9. The difference in first line of Lyman and first line as Balmer series of a hydrogen like ion has energy equivalent to the photon of wavelength 373 A. Find the atomic number of the atom. Solution: The first line of Lyman series,~] = 1,';2 = 2

Thcnt>E J

~ +13.6Z'(I- 21, }~ ~ 1O.2Z'CV

For first line of Balmer series, nl = 2, n2 = 3 I'lE2 = 13.6Z

(2'2 - 3~ )

2

I1E2 ~ 1.89Z 2eV

The difference in energies is tlE = !1E1 - tlE2

•

r, 'I

M ~ 12400 ~ (10.2 -1.89)Z'cV ~ 8.31 Z'eV 3730 Ans. or Example 10. Let an electron or neutron hit a stationary hydrogen atom with KE(= K] ). Find the (a) condition and types of collision (b) minimum K] to excite a hydrogen atom (e) minimum speed to excite the hydrogen atom. Assume that the atom iskeptfixedandK] =12eV. Solution: (a) Let E 2bc the 1st excited. state and EI be the ground state. Let the neutron have KE

If the atom is kept fixcd.

c

410

CRB Undentanding Physics Oplics and Modem Physics

®

n

0-

~

M

If K J > (E2 - Ed, the KE oflhe bombarding particJe decrease by £2 - EI and the collision is said to elastic. If K\ < (E2 - £\). the KEofthe bombarding particle remains consumed as it is reflected and collision is elastic.

K{=~mnU2)=(E2 - Ed+K2

(b)

or

K2=K,-IlE.

where

AE

0'

(e)

(bccauscKatom = 0)

= 13.6(1-~) = !o.2eY

K2 =12eV - IO.2=1.8eV (Kdmin = 10.2 eV

~mnv2

Ans.

= 10.2 eV

v = 12(10.2 eV) m.

x-,,IO;,-"_ = .1"20".4o.x=I.,,6.:: 1.67 x 10 31 =4,42xI0 6 m!s

Ans. Eumple 11. Find the de Broglie's wavelength of a proton accelerated from rest in a potential difference of 100 V. Solution: The KE of the proton after moving through potential difference V is K = eV = e(1 00 volt) = 100 e volt = IOOxI.6xlO- 19 J The momentum of the proton is p

=.J2mii

= J2x1.67x 10-27 xlOOx1.6 x 1O 19 = 2.31x IO- 22 kg mls

The de Broglie's wavelength is

1.=E. p

= 6.3xIO-34 m 2.Jlx10 22 = 2.72 x·IO- 12 m

•

Ans.. .

Particle and Wave Nature or Malter

411 ~

~

Eumple 12. (a) Inhe moments of two particles are PI and P2 wrt ground, find the de BrogUe's wavelength of one relative to the other (or wrt the CM). (b) If the de Broglie's wavelength of two particles moving perpendicular to each other)., 1 and ).,2. find the de Broglie's wavelength of each wrt their CM. ~

~

Solution : (a) When PI moments in eM frame will be ~

and P2 arc the moments of two particles their

~

~

1P'I =t p.~I-~P2:::... 1 , -_ __ =

JP~ + pi -2P1P2 cosa

Then the de Broglie's wavelength of one observed t the other will be (same), given as

h =""7"7==7~==

Jp~

(b)

If 9 = 90°. PI

+

pi -2PIPlcos9

An••

=.l!.... P2 = .!... we have ).,1

,=

).,2

2).1'2 1).,2+).,2 V I 2

An•.

[umple 13. ·Find the de Broglie's wavelength ora hydrogen molecule 27°C at its most probable speed. Solution: )., de BrosJie

h =m,mp .~

...(0 .. .( ii)

Then,

)., de Broglie

=

m[f{

6.3xI0-34 =rr====¥~--x l.38 x lO 23 X300) X2 X1.67 X10-27 ( 2xl.67x]0-27 = 1.2 x to- 7m

AnL

[umple 14, A beam of electrons hit the slits of separation d after passing through a potential difference V. They will form interference panern on a screen situated at a distance D from the slits. (a) Find the distance Pbetween two adjacent maxima. (b) Putting d =50 x 10-6 m. D = I m and V =25 volts. find p.

412

GRB Und~r1tanding Physics Optics and Mod~m Physics

l

'

...., ,"

."

-~~:::~~~::::§:;

"

"

M

.... , "

,

....,

A

, D.

-----------

'. ".

I Solution: (a) The distance between two adjacent maxima is

~= AD

... (i)

A= ~

... (ii)

d The de-Broglie's wavelength of the electro'ns is

•

. .,

p

The momentum of the electron just before hitting the slits is p=J2mK

... (iii)

The KE of the electron just before hitting the slits is

K =cV

... (iv)

d'

Using all eqns.,

"= I-' (b)

A ns.

Dh

d.J2meV (lm)(6.3 x 10-34 J. s)

~= --~~~~~~~~~==== (50 x IO-6m)~2 x9.1 x 10-31 kg x 1.6 x 10 19 C x 25 volt =4.7xl0-6 m Ans. Example 15. A thin beam of mono-encrgelic electrons strike nonnally on a

metallic plate. The fourth order of maximum Bragg's refraction takes place at an angle 9 = 55° with the nonnal to the surface, Iflhe kinetic energy of each electron is K = 180

eV, find the inter planner distance. Solution: The clectron wave gets reflected at the atomic sites A and B. The reflected waves 2 and 3 get superimposed constructively to fonn a maxima on the detector. For the maxima. the path difference between the waves is given as 3 6X=d case

2

, A

" '.

-d-

B

"

Particle and Wave Nature of Malter

413

IJx = nA

ax = dcos9 +d = dcos 2 ~ = nt. 2

or

d=

1/). cos 2 Q

... (iJ

2

A=

h .J2mK Then. by using eqns. (i) and (ii), 2 d = nhsec (9 / 2) .J2mK Putting n =4,K =180x 1.6x 10 19 J, where

... (iiJ

, •

.,

9=55°, m=9.lxlO "' 3I kg.wehave

d=O.232x10 9 m Ans. Example 16. Applying uncenainty principle. calculate the minimum, kinetic energy of an electron confined in a region of length I = 0.2 nm. Solution: The uncertainty in position of the electron is !u = I ., , Then, the uncertainty in momentum of the electron is. p>_h_

2,!u

'; :"

p';?

or

. '"

,

2nI.

Hence, the minimum momentum is Pm = JL 2,1 Then, the minimum KE of the electron is

K

or

"',

•

-:. I

J!.... .

".

\

,, .. ,

. ,:

,.

.

; ." .... .J--

."

2

= Pm 2m

,

',.

•• 1

,

.1

'11' .-,

• ',.:"

,~'

".J .

.: Ans. Example 17. An electron being accelerated from rest through a potential difference of V = 4 volts trapped in a tube a length 1= IO-6 m. Using uncertainty principle, find the relative uncertainty in its velocity. Solution: The uncertainty in position of the electron is !u = I The minimum uncertainty in momentum of the electron is <

CRB Undentandins Physics Optics and Modem Ph)'lics

414

or

The uncenainty in velocity is Au = IIp

m I1v = ....!.!.-

or

... (i)

2"",[

The speed of the electron is u = ,l2mK ...(ii) from eqns.(i) and (jil, the relative uncertainty in the velocity (speed) is, h

Putting

V=4

volt,

e=t6)(1O- 19 C.

m=9.lxl0-3!kg,

h = 6.3 x 10-14 J 5, we have I1v :::t. 10-4

1=IO-6 m, ADS.

u

Enmple 11. By uncertainty principle. find the (a) Bohr's radius for hydrogen atom (b) minimum (ground state) energy of the electron in hydrogen atom. SoIullo.: (a) The total energy of an orbiting electron is E=K+U

p' , E=-- e 2m 4m:or The uncertainty in position for the particle is bx = r The uncertainty in momenWnl is llp~...§.

or

""

11 h !Jprrin = -; = 21tT i

... (i)

... (ii)

By using, cqns. (i) and (ii),

E~L- i lmr2

... (iii)

41t&or

Since dE = 0 for minimum E; dr 0= l..(2r- J )-...!.-(-r-') 2m 40£0

,

,

Particle and Wave Natuff; of MaHer

415

~,

or (b)

r = -=0.53A me'

,

Ans.

Then, by r :::: 11 2 in eqn.(iii},

me E min ::::- me

4

:::: -13.6 V

2);'

Ans.

Example 19. Find the speed of the electron in Bohr's nth orbit taking the motion of nucleus into account. Solution: The angular momentum of the system is

L = Mm r.2 (i) M+m H 'II

Then,

=(I+Z~

L:::: Mm ( 1+ .!! )2 r.2C!) M+m M 0 m(M + m) 2 "" M roC!) L = mrn( l+

or

VI =

L

m'b(l+

; )VJ

Z)

In general,

Example 20. A gas of hydrogen like ions is prepared in sucb a way thal1he ions are only in the ground state and the ftrst excited state. A monochromatic tight of wavelength 1216 A is absorbed by the ions. The ions are lifted to higher excited states and emit radiations of six wavelengths, some higher, some lower or some greater than the incident wavelength. (a) Find the principal quantum number of the final excited state.

416

GRB Understandins Physici Optics-and Modem Physics

,. (b) Identify the nuclear charge on the ions . • (e) Calculate the value of the maximum and minimum wavelengths. ", , Solution: (a) Energy corresponding to A = 1216A is, E = 12375 eV = 1O.177eV

1216 Now. total six wavelengths arc obtained in the emission spectrum, hence from n(n - l) = 6 2. "

We have n = 4, i.e., after excitation the single electron jumps 10 3 rd excited state or

n = 4.

. (b) Now it may jump either from n = 1 to" = 2. · ~TT.----n

I

.

. Ans.

""

---,1'tI-TT-n =3

--'-I---'-I-lr- n =,

.. _-'-_l.L_ =, n (.j

(bj

(oj

If it jumps from n = I, 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 or greater than the wavelength of absorbed photon. While in the question it is given Ihallhe emitted wavelengths are either less than or greater than or smaller than the wavelength ofabsorbcd photon. Which is possible only in the second case, i.e .• when electron jumps from n = 2 to" = 4. Hence, £4-£2 =1O.177eV

-13.26z' _(-13.6Z') = 10.177

or

4

22

Solving this, we get z .. 2. Ans. (c) Maximum wavelength corresponds to minimum energy, i.e., ~ transition from n=4t011=3. . Thus, ll£min = £4 - £3

,-,./

= (- 13.6)(2)' _ [(-13.6)(2)']=2.64ev (4)'

(3)'

A~ = 12375 = 4687A ADS. 2.64 , •• _ :J •• • • • • , ., .Minimum wavelength corresponds to maximum energy, i.e., transition from n = 4 ton=1. Hence, ,

.,

417

Particle and Wave Nature of Maller

~ (- 13.6)(2)'

_ [(-13.6)(2)'] (I)'

(4)'

==SI.OcV ). . _ 12375 nun

-""5i'J) ~

242A

Ans.

=

•

Example 21. A hydrogen like atom of atomic number is in an excited state of quantum number 2n. It can emit a maximum energy photon of204 eV. If it makes a nansition to quantum state II , a photon of energy 40.8 eV is emined. Find n. z and the

ground state energy (in cV) 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: Given,Eln .

- E, =204eV.

(13 .6):' ( 1_

4~' )~ 204

... (i)

13.6:'(~'~ ~"): :~88CV 2 n

... (i1)

4,,2

Solving Eqs.(i) and (ii), we have" == 2 and z == 4.

E,

~

Ans.

,

(-13.6):-cV

~ (- 13.6)(4)'eV = - 217.6eV During dc-excitation, minimum energy emitted is. Emin = E2n - E2n-l = £4 - E J

~

Ans.

217 .6 _ ( 217.6) 42

3

2

~

IO.s8eV Ans. Eumple 22. The energy levels ofa hypothetical one electron atom are given by En =_18.0 eV

"'

where n =1,2.3..... (a) Compute the four lowest energy levels and construct the energy level diagram. (b) What is the cxcitation potential of the stage n =2? (c) What wavelengths (in A) 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) If these atoms are in the ground state. can they absorb radiation having a wavelength of 2000A? (e) What is the photoelectric threshold wavelength of this atom?

418

GRB Und~ntanding Physics Optics and Modern Physics

SoluHon : (a)

- IS 0 E\ =-2· = - IS.OeV (I) E2 = - IS.O = - 4.5eV

(2)2 E 3 = - IS .O = - 2.0eV (3)2

and

£4

-IS 0 =--i =-1.l25eV

(4)

The energy level diagram is shown below. - - - - - - E " = -1 .125 ell - - - - - - E3 .. -2.0 ell - - - - - - E2 :::-4.5 ell

------E,"'-18.0ev (b) The excitation potential ofstagen = 2 is, IS.O - 4.5 = 13.5volt. ADs. (e) Energy oflhe electron accelerated by a potential difference of 16.2 V is 16.2 eV. With this energy the electron CAn excite the atom from n "" I to n = 3 as

and

E4 - E\ = -1.125 - (- IS.O) = 16.S75 eV > 16.2eV E3 - E\ = -2.0 - (-IS.O) = 16.0eV < 16.2eV

Now,

1.32 = 12375 = E3 - E2

12375 2.0 - ( 4.5)

= 4950A 1.3\ = 12375 = 12375 = 773A E3 -E\

and

1.2\ = 12375 = E2 - E\

16

Ans.

Ans.

12375 -4.5 - (- IS.0)

=917A

(d) No, the energy corresponding to A = 2000A is, E = 12375 = 6.1S75eV 2000 The minimum excitation energy is IJ.SeV (n = 1to n = 2).

Ans. AD•.

(e) Threshold wavelength for photoemission to take place from such an.atom is, 12375 A A";n = 18 = 6S7.5

ADs.

Example 23. A doubly ionized lithium atom is hydrogen like with atomic nwnber 3. Find the wavelength oflhe radiation 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.

419

Particle and Wave Nature or Matter

Solution:

.' n'

E, = -=-(l3.6eV)

By putting z::: 3, we have

E" = _122.4 eV

n'

E,

=_122.4 = -122.4eV (I)'

E, = _122.4 = -13.6eV

and

(3)'

I1E=E, - E, =\OS.SeV The corresponding wavelength is

..

,,=

12375 A = 12375 A l1E(in eV) \OS.S =1\3.74A

ADs.

Enmple 24. 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 like ion x. Calculate energies of the first four levels of x. Solution: Wavelength ofthe first line of Lyman series for hydrogen atom will be given by the equation

1.. R(L _I ) =3R

... (i)

:E

)..,

12

22

4

The wavelength of second Balmer line for hydrogen like ion X is _I =

Rz'(_1 __I ) = 3Rz'

'"

2' 4' or

Given that,

i,e.,

... (ii)

16

1. =_1 )..,

).2

3R = 3Rz' 4 16

:=2 i.e.,.r ion is He ++ . The energies of first four levels of x are, E, = -(13b)z' = -54.4eV

E, =.E!.. = -13.6.V

(2)' E, =.E!..=..{;.04.V (3)' £4 =

~ = -3.4eV (4)'

GRB Undentanding Physics Optics and Modem Physics

420

--_ ...... _-------_._-----------.--_ Ili --------(JI

ASSIGNMENTS

.._---_. - ........

-.--

-------.. .... ........

Conceptual Questions

I. Can an electron be subdivided into fundamental particles ? 2. What nrc the drawbacks of Thomson's model? What arc the main observati ons of Rutherford's gold foil experiment? What is the limitation of Rutherford's pianalcry model? How docs Balmer fomlUla give a clue of discrete nature of radiation ? An accelerating cl ~tron should radiate cmw, where as Bohr tried to centradici this is his radiation less circular orbits. Comment. 7. What is meant by stationary orbit? Does it meant that the electron energy remains constant throughout?

3. 4. S. 6.

,8. When an electron gets excited from one lower stationary state to another (higher) stationary, can it stay in the higher stationary state forever? lfno. why should we call these stationary state, as poised in Q-101 9. What are the quantities in Bohr's atom quantised?

10. Is the fonnula f = £2 ; EI always valid for excitation or

de-cxcit~tio'ri of any

radiating system? 11. Is there any difference between the formula (i) M = lif (ii) E pMton = lif (iii)Eosc = ttll/,! 12. What is the difference between line and band spectrum? 13. What is the differencc betwecn absorption and emission spectrum? Is it always true that no of emission and absorption lines arc samc?

14. Whc"n an atom in energy Slale III = 2 get excited to n2 = 5 how many lines of emission spectrums are possible? 15. In spcetroscopy, wavelength is more convenient than frequcncy in spectrometers, why? . " 16. When wc say the wavelength ofa light is A = 5500A what does it mean? Can we gct a pure or true monochromatic radiation? 17. Is it expected that an atom always oscillate to emil radiation continuously? lfwe accept this suggestion what will be the effects?

18. If any radiating system absorbs and radiates energy discontinuously, why is not poss ible to see the grainy-ness of radiation in naked eye? 19. In our TV screen electrons hit and make scintillation. explain the process of production of scintillation by using Bohr's theory ofrndiation. 20. The chance ofge"ing an atom (or any particle) in higher energy state is more. Is it true? Comment. '

Particle and Wave Nature of Matter

421

21. When an electron go away from a nucleus the energy of an electron increases why? ' , 22. In Bohr's theory the kinetic and potential energy of an orbiting electrons are equal. why? . .

23. What is Bohr' s correspondence principle? 24. Is it possible to conserve energy and momentum during excitation and de-excitation of atoms? 25. What is the outcome of Frark-Hertz experiment? 26. What was the basis of de Broglie's hypothesis? 27. de Broglie' s pilot wave signifies that a displacement wave is associated with matter. Is it true ? Ifnot. what exactly mcan physically? 28. The velocity of matter wave can be greater than that of light. How do you reconcile this fact? 29. What is cause of great resolution of electron microscope? 30. Can we accelerate the electrons to any speed to get any value of wavelength we want for electrons microscope? To put it in a straight way. is there any limit to resolution. 31. Double slit interference experiment of electrons suggest's that a single electron can pass through both the slits at a time. How can we digest this seemingly constructively statement? 32. What exactly is waving in matter wave? Is it the matter that is waving (oscillating) in material body? Explain. 33. What was Schrodinger's original interpretation of wave function 'V? 34. "Giving a periodicity to matter" (electrons ray) is the basis of maUer-wave proposed by de Broglie. Then how did de Broglie interpret matter wave at that time? 35. The probability density P oc EJ for radiation and P ocl1.V12 for matter. £ois a physical quantity where as \II is non-physical . Then, is there any physical interpretation of \II or I\Vl 2 . 36. Schrodinger's equation is most basic and can not be derived from another basic ideas. How is it? 37. The probability density of an electron in an orbital at any fixed point is constant, explain this physically. 38. How did de Broglie explain the quantisation of angular momentum of an electron in an alom? 39. How does a micro particle in a box behave? 40. What is a quantum dot? 41. How can an LED emit light? Can it be explained by the concept of particle in a box?

422

GRB Undentanding Physics Optics and Modem Ph)"K:s

42. When the box size goes in decreasing. how does the behaviour of the particle trapped inside change? 43. What is the root cause of Heisenberg's uncertainty principle? 44. An electron can stay "pennanently" only in the ground stale (n =1) but not in other stationery states. Explain this by using Heisenberg's uncertainty principle.

423

. Particle and Wave Nature of Matter

(J Multiple Choice Questions (A) Only One Choice Is COt'J"8Ct

Level -1 1. The ratio of minimum to maximum wavelength in Balmer series is:

(.)5 :9

(b)5 :36

(e)I:4

(d) 3:4

2. The frequency of revolution of an electron in n1horbils is

In . If the

electron

makes a transition from nth orbits to (n - 1) th orbit, then the relation between the

frequency (v) of emitted photon and In will be:

(a)v=f,'

(b)v=H,

(e)v=;,

(d)ll=f,

3. Figure shows five energy levels of an atom, onc being much lower than the other four. Five transitions between the levels are indicated. each of which produces a photon of definite and frequency .

~=i=PRR::=:

.,+++-+-+E'~-4--+--L--+---

E, -1__L--L____-1___ Which onc of the spectra below best corresponds to the sct of transitions indicated? (Low)

F,.quency (High)

I I I I I I I I I I I I

I II

(.) (b)

II I I I II

(e) (d) (e)

II I II

II

GRB Und~rstanding Physics Optics and Modem Physics .

424

4. Figure represents in simplified form some of the lower energy levels of the hydrogen atom.' If the transition of an electron from £4 to £2 were associated with the emission afhlue light, which one of the foll owing transitions could be associated with the emission of red light? E4 - - - - - - - - - - - - - - - E3 - - - - - - - - - - - - - - - Energy

~---------------

E, ________________

(a) £4 to £1 (b) E3 to £\ (e) E3 to £2 (d) £, to £) S. The orbiting speed Un of e- (the same as for e+ ) in the 11th orbit in case of positronium is r·fold compared to that in nth orbit in a hydrogen atom, where x has the value: (a) 1 (b) ./2 (c) 1/ ./2 (d) 2 6. When a electron jumps from a level n = 4 to n = 1. the momentum of the recoiled hydrogen atom will be: (a) 6.5 x lO- 27 kg m s- I

(b) 12.75 x 10-'9 kg ms- ' (e) 13 .6 xIO- 27 kg ms- 1 (d) zero 7. When a hydrogen atom is raised from the ground state to fifth state:

(a) both KE and PE increase (b) both KE and PE decrease (c) PE increases and KE decreases (d) PE decreases and KE increases 8. Consider a spectral line resulting from the transition n = 5 to n = 1. in the atoms and ions given below. The shortest wavelength is produced by: (a) helium atom (b) deuterium atom (c) singly ionized helium (d) ten times ionized sodium atom 9. Hydrogen atoms are excited ground state to the state of principal quantum number 4. Then, the number of spectral lines observed will be:

(a) 3

(b) 6

(c) 5

(d) 2

Particle and Wave Nature of MaHer

"

425

10. The e~ergy of an electron in an excited hydrogen atom is -3.4 eV. TIlen, according to Bohr's theory, the angular momentum of this electron , in los, is l4 (d) 0.5 xlO-l4 (a) 2.11 x lO(e) 2 xlO-l4 (b) 3 xlO- l4 t t. The potential energy of an electron in the fifth orbit of hydrogen atom is: (a) 0.54 eV

(b) - 0.54 cV

(e) 1.08 cV

(d) - 1.08 eV

12. In fig. EI to £2 represent some of the energy levels of an electron in the hydrogen atom.

E. E.

-0.38 aV -0.54 aV

E.

-O.SSeV

E3

-1.5 aV

-3.4eV

E, __________________________ -13.6 eV

Which one of the following transitions produces a photon of wavelength in the ultraviolet region of the ci«:tromagnetic spectrum? [I cV~ 1.6 XIO- 19 j (a)E,-E, (b)E, - E, (e)E, - E, (d)E,-E, 13. The frequency of emission line for any transition in positronium alom(consisting of a positro n and an electron) is x times the frequency for the corresponding line in the case of H atom, where x is : (a).Ji (b) 1I.Ji (e) 112.Ji (d) 112 14. The ratio of maximum to minimum possible radiation energy in Bohr's hypothetical hydrogen atom is equal to: (a)2 (b) 4 (e) 4/3 (d) 3/2 15. Ifelements with principal quantum number n > 4 were not allowed in nature, the number possible elements would have been: (a) 32 (b) 60 (e) 64 (d) 4 16. In order to break a chemical bond in the molecules of human skin, causing sunburn, a photon energy of about 3.5 eV is required. This corresponds to wavelength in the: (a) infrared region (b) x- ray region (c) visible region (d) ultraviolet region

GRB Undentanding Physics Optics and Modem Physics

426

17. The angular momentum of an electron in a hydrogen atom is proportional to: (a) 11$

(b) II,

(e) $

(d) ,2 18. In hydrogen spectrum, the shortest wavelength in Balmer series is A. The shortest wavelength in Brackett series will be: (a) 2).

(b)4k

(e)9k

(d) 16k

19. The angular momentum of an electron in first orbit ofLi++ ion is: (b) 9lt (e) l!. (d) l!. 21t 21t 27t 6n 20. A sample of hydrogen atom is in excited state (all the atoms). The photons

(a) 3"

emitted from this sample are made 10 pass through a filler through which light having wavelength greater than 800 om can on ly pass. Only one type ofphatons are found to pass the filter. The sample's excited state initially is ; [Take he = 1240 cV -om, ground stale energy of hydrogen atom = - 13.6 eV.] (a) 5 th excited state (b) 4th excited state rd (c)3 excited Slate (d) 2nd excited state 21. A neutron having kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision: (a) must be clastic (b) must be completely inelastic (c) may be completely inelastic (d) information is insufficient

22. The ionization energy of the ionized sodium atom Na to+ is: (a) l1.6eV

(b) 13.6xlleV

(c)(ll.6/II)eV (d) 13.6X(lI')eV

23. Magnetic field at the center (at nucleus) of the hydrogen like atoms (atomic number = z ) due to the motion of electron in nth orbit is proportional to: ,

(a) !!... z3

,,4

(b) -

(e) Z

.. 2 =nl

(d)

.. 3 =n5

24. Magnetic moment due to the motion of the electron in nth energy of hydrogen atom is proportional to: (a) n

(d) n 3

(b) nO (e) n' 25. As per Bohr model, the minimum energy (in eV) required to remove an electron, from the ground state of doubly ionized Li atom (Z = 3) is: (8) 1.51 (b) 13.6 (e) 40.8 (d) 122.4 26. Imagine an atom made up of prolon and a hypothetical particle ofdouble the mass of the electron but having the same charge as 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 A. (giving in tcons of the Rydberg constant R for the hydrogen atom) equal to: (a) 9/(SR) (b)36/(SR) (e) 18/(SR) (d) 4/R

i I

l

!

I I

427

Particle and Wavc Nature of Matter

27. The electron in a hydrogen atom makes a transitions from an excited state to the ground state. Which of the following statements is true? (a) Its kinetic energy increases and its potential and total energies decreases (b) Its kinetic energies decreases and its potential energy increases, and its total energy remains the same (c) Its kinetic and total energies decrease and its potential energies increases (d) Its kinetic, potential and total energies decreases 28. The transitions from the state n = 4 to n = 3 in a hydrogen-like atom results in ultraviolet radiation . Infrared radiation will be obtained in the transition: (a) 2 .... I (b)3--> 2 (e) 4--> 2 (d) ' S-->4 29. If the atom 100 Fm 257 follows the Bohr model and the radius of 100 Fm 257 is n times the Bohr radius, then find n: (a) 100 (b) 200 (e) 4 (d) 114 30. The de Broglie wavelength of a charge particle moving through a potential difference V is proportional to: I

!

(a) V (b) V - ; (e) V' (d) V-I 31. The de Broglie wavelength of a body falling under unifonn gravitational field g through a height h is proportional to:

(b)..{h

(d) h

32. Heisenberg's uncertainty principle is due to: (a) particle nature of radiation (b) wave nature of matter (c) wave-particle duality (d) defect of the instruments 33. The width of the ground state energy is: (a) undefined (b) zero (e) infinite (d) non-zero but finite

Level-2 1. A stationary hydrogen atom of mass M emits a photon corresponding to the first line of lY,J113n series. If R is the Rydberg's constant, the velocity that the atom acquires IS: (a) ~Rh

4M

(b)~ 4M

(d) Rh

M

2. Transitions between three energy levels in a particular atom give rise to three spectral lines of wavelengths. in increasing magnitudes, A I. A 2 and A 3. Which one of the following equations correctly relates A I. A2 and A3? (a) AI = A, -A) (b) AI =1.) -A,

GRB Under1landing Physics OptiC! and Modem Physics

428 (e) _I = _I + _1 AI

),,2

A3

1

1

1

(d)r;=[J+"

3. An electron in H atom makes a transition from n = 3 to momentum of H 310m will be:

,,= I.

The recoil

(a) 6.45 x IO- "N s (b) 6.S xlO-" N s 24 (e) 6.45 xlO- N s (d) 6.S xlO-24 N s 4. The force acting on the electron in a hydrogen alom depends on the principal

quantum number as: (a)F=n'

(b)F ~ _1 ,,2

(e)F~,,4

(d)F~J, n

5. An alpha particle of energy 5 MeV is scattered through 180 by a fixed uranium nucleus. The distance of the closed approach is of the order of: (a) 1O- ljcm (b) IO- Il cm (e) to- 12 cm (d) to- 19 cm 0

6. Monochromatic radiations of wavelength A. arc incident on a hydrogen sample in ground state. Hydrogen atom absorbs the light and subsequently emits radiations of 10 different wavelengths. The value of A is nearly:

(a) 203 nm (b) 95 nm (e) SO nm (d) 73 nm 7. Hydrogen atoms in sample are excited to n =S state and it is found that photons of all possible wavelengths are present in the emission spectra. The minimum number of hydrogen aloms 11 the sample would be: (a) 5 (b) 6 (e) 10 (d) infinite 8. tflhe average life lime of an excited stale of hydrogen is of the order on 0- 8s. the number of revolutions an electron will make when it is in n =2 state before coming to ground state will be : [Take ao = 0.53 A and all standard data if requiredJ (a) 10'

(b)SXIO· (e)2xlO' (d) noneorthese 9. The retoil speed of hydrogen atom after it emits a photon is going from n = 2 state to" = I stale is nearly [Take R... = 1.0 X 10- 7 m s-1 and h =6.63 x10- 34 J s] : (a) 1.5 m, - t (b)3.l ms - t (e) 4.5 ms- t (d) 6.6 ms- t

10. A proton strikes another prolon at rest. Assume impact parameter to be zero. i.e., head-on collision. How close will the incident proton go to other proton?

(aJ

e3 1tEom2vo

e3 (b)="1tEomVo

(d) none of these

•

429

Particle and Wave Nalurc of Malter

J I. If elements of quantum number greater than clements in nature were have been:

II

allowed. the number of possible

(a)

~ n{1I + 1)

1 (b) ["("2+ )]'

(e)

~"("+I)(2n+l)

(d) ~ n(n + 1)(2" + I)

12. Let VI be the frequency of the series limit of the Lyman series. v2 be the frequency of the first line of the Lyman series and v 3be the frequency of the series limit oflhe Balmer series. Then: (a)vl -v 2 =v3 (b)v2 -vI =vJ 1

(e)v) ~2(Vl +v,)

(d)VI +v2 =v3 13. In an excited state ofa hydrogen-like atom an electron has a total energy of - 3.4 eV . If the kinetic energy ofthc electron is E and its de Broglie wavelength is A, then :

(a) E ~6.8 eV: l. ~6.6 xlO- II m (b) E ~ 3.4 eV: l. ~ 6.6 x lO- II m (e) E =3.4 eY; A. = 6.6 x lO- 8 m

(d) E ~6.8 eV: l. ~ 6.6 xlO-'m 14. The ratio between total acceleration orlbe electron in singly ionized helium alom (both in ground statc) is: (. (a) 1 (b) 8 (e) 4 (d) 16 15. A hydrogen atom is an excited state of principal quantum number n . It emits a photon of wavelength A when it returns to the ground state. The value of n is : (a) JAR(AR

I)

(b)t(~I)

(e)

~AR~I

(d) Jl.(R

I)

16. A neutron moving with a speed u makes a head·on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision will take place in (assume that mass of proton is nearly equal to mass of neutron): (a) 10.2 eV (b) 20.4 eV (e) 12.1 eV (d) 16.8 eV t 7. In hydrogen and hydrogen-like atoms, the ratio of difference of energies E4n-2n and E2n-n varies with atomic number Z and principal quantum Dumber n as:

z2 (a) n2

Z4 (b) - 4 n

Z (e) n

(d) none oflbes.

, CRB Undc~Wldin8 Physics Optics and M<xkrn Physics

430

18. The mtio of the maximum wavelength of the Lyman series in hydrogen spe<:trum to the maximum wavelength in the Paschen series : 3 6 52 7 (a) 105 (b) 15 (e) 1" (d) 108

19. When an electron in the hydrogen atom in ground state absorbs a photon of energy 12. t cV. its angular momentum: (a) dct:reases by 2.11 x lO- 34 J-s (b) decreases by 1.055 X 10- 34 J-s

(c) increases by 2. 11 x lO- 34 J-s (d) increases by 1.055 X 10- 14 J-s 20. An electron and photon have same wavelength. Ifp is the momentum of electron and Ethe energy ofphelon, the magnitude of piE in S.I. unit is: (a) 3.04 xlO' (b)3.33 x10- 9 (e) 9.1 x IO- JI (d) 6.64 xIO- 34 21. An electron in a Bohr orbit of hydrogen atom with quantum number nz has a

angular momentum 4.2 x lO-34 kg m 2s-1 . If the electron drops from this level to next lower level, the wavelength of this line is: (a) 180m (b) 18.7 om (e) 1876A (d) 18.76x104 A 22. A hydrogen atom and Li ion are both in second excited state. If III and fu are their respective electronic angular momenta and E Hand Eu their respective energies. then: (a) IH > lu andiEHI> IEul

=

(b) IH lLi and I EHI < I Eli I (e) 'H = lLi and lEHI > IELi I (d)IH lLi andIEH I< IELi I

=

23. The electric potential between a proton and an electron is given by a Y =1'0 In' I lb. where Ib is a constant. Assuming Bohr's model to be applicable, write variation of 'n with n. n being the principal quantum number? (a) 'n oc n

I (b) 'n oc -

n

(c) 'n

OC"

, n·

(d) 'n oc -

I

.'

24. A photon collides with a stationary hydrogen atom is ground state inelastically. Energy of the colliding photon is 10.2 eV. After a time interval of the order ofa microsecond. another photon collides with a energy of 15 eV. What will be observed by the detector? (a) One photon of energy 102 eV and an electron energy 1.4 eV (b) Two photons of energy of 1.4 eV (e) Two photons of energy 10.2 eV (d) One photon of energy 10.2 eV and another photon of 1.4 eV

,

431

Particle and Wave Nature of Maller

25. 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 (e) 1882 nm (e) 1648 nm 26. Electrons with the de Broglie wavelength A fall on the target in an X-ray tube. The cut-ofTwavelength of the emitted X-rays is:

(a)"o

=2me..

2

h

(b)"o

= 2h

me

(e)

"0

2 2

=2m e

h2

..,

(d)"o

=A

27. A proton and an alpha particle move with same KE. The ratio of this de Broglie's wavelength is: (a) I : 2 (b) 4 : I (e) 2 : I (d) noncoflhesc

28. If the excitation time is T = 10- 85 fora decay, the width of the line, that is, AI is: 8 (a) 8 x 10 6 Hz (b)3 x 10 Hz 8 (d) 6 x 10 (e) 10 8 Hz

(8) More Than One Cholce/$ lsi.,. Correct

1. According to Bohr's theory of hydrogen atom, for the electron in the

nih

pennissible orbit (a) linear momentum oc

!n

(b) radius of orbit oc n (c) kinetic energy oc

~ n

(d) angular momentum oc n 2. Continuous spectrum is produced. by : (a) incandescent electric bulb

(b) sun (c) hydrogen molecules (d) sodium vapour lamp 3 A of mono atomic hydrogen is bombarded with a stream of electrons that have • ~ accelerated from rest through a potential difference of 12.75 V. In the emission spectrum. one can observe lines of:

(a) Lyman series (b) Balmer series (e) Pascheo series . (d) Pfund series

432

GRB Undentanding Physics Optics and Modem Physics

4. Which of the following statements about hydrogen spectrum are correct? (a) All the lines of Lyman series lic in ultraviolet region (b) All the lines of Balmer series lie in visible region (e) All the lines of Paschen series lie in infrared region (d) None of the above

5. The third line of the Blamer series spectrum of a hydrogen like ion of atomic number Z equals to 108.5 om. The binding energy of the electron in the ground state of these ions is Ea _Then: (a) Z = 2 (b) 54.4 eV (e) Z = 3 (d) 12.24 eV 6. In Bohr's model of the hydrogen atom : (a) the radius of the 11th orbit is inversely proportional to n 2

(b) the total energy of the electron in nIh orbit is inversely proportional to n (c) the angular momentum of the electron in an orbit is an integral multiple of

" 121.. (d) the magnitude of potential energy afthe electron in any orbit is greater than its KE 7. The mass number ofa nucleus is: (a) always less than its atomic number (b) always more than its atomic number ", (c) sometimes equal to its atomic number (d) sometimes more than and sometimes equal to its atomic number 8. The electron in a hydrogen atom makes a transition nJ -+ n2. where nl and "2 are

the principal quantum numbers of the two states. Assume the Bohr mOdel to be valid. The time period aftne electron in initial state is eight times that in the final state. The possible values of and"2 are: ' ,

n,

(a) n, (b) "I (e) 'II (d)1I1

= 4, n2 =8.112 =8. =6,

II,

,I,

=2 =2 =1 =3

9. ·Which of the following is/are of the Heisenberg's uncertainty principle?

(a)AE·"'I~h

(b)Ap·/lx=h

(e)IlL·"'~~h

(d)

/lx.Ilk=~

10. Which will from the basic of Heisenberg's uncertainty principle for a wave pack arising from superposition of two waves of different frequencies? (a)"'Ol·Ilt~1

(b)/lx · Ilk=1

(e)AE·"'I~h

(d) None of these

Partide and Wave Nature

or Malter

433

11. Electron microscope is based on the property: (a) wave property of electron (b) particle property of electron (e) wave~particle duality of electron (d) all of the above 12. In low energy electron diffraction in a crystal, sin ¢l is proportional to, where ¢l =

angle between incident and diffracted beam n = order of maxima t K "'" KE of electrons, d = atomic plane spacing: 1

(a) n

(b) ,fK

13. An atom emits a photon offrcquency 10 frequency of emission can be:

(d) K 2 when it is kept fixed. It if moves, the

(a) fo (h) > fo (e)
(d) none of lhese

~M.tch The Columns 1. Match the Columns I and

n.

Columa J

Columa

I

n

(a)

Photoelectric effect

(P)

Momentum P =

(b)

X~rays

(q)

Emission of electromagnetic wave

(e)

Black body radiation

(r)

Emission of elettrons

(d)

Wave~particle

(s)

Planck's law

duality

.

GRB Undentanding Physiu Optics and Modem Physics

434

2. Match the Columns I and II. Column II

Column I (u)

Wien's dispiocemcnt law

(p)

E = IIlif for osc illat ion

(b)

Wien' s exponential law

(q)

Radiation from hcr,:cd solids

(e)

Bohr's law of quantisation

(r)

Radialion from heated gases

(e)

Planck's quantisation law

(s)

Af'/T = b(constant)

3. Match the Columns I and II . Column 11

Column 1 (0)

Heisenberg's uncertainty principle

(P)

~oscillalion =

(b)

de Broglie's maucr wave

(q)

dETC"
(e)

Planck-Einstein's fannula

(r)

). =.!.. mu

(d)

Bohr's quantisation fannula

(s)

t.P~IL

hi

= hf

lu

Column I (Energy of a system)

(b)

Column 1I (Proportional to tbe integer "n")

Energy ora moving particle in a box (conduction electron in metals)

(p)

Energy of revolving electrons is

(q)

(d)

• oscillating charge Energy of an Frequency of revolution of an electron

-I

.' -I

.'

an atom (e)

!, •

I

•

I

!, ,I, I

4. Match the Columns I and II.

(0)

j

(r)

.'

(s)

•

Particle and Wave Nature of Malter

435

r J Comprehensions Passage-l The energy level s of a hypothetical one electron atom are shown in figure. n =_ n=5 n~

OeV - 0.80 eV - - - - - - - -____________ -1 .45eV

n=3 --------------_______ - 3.08 eV

n=2 - - - - - -_____________

-5.30eV

n=l

-15.6 eV

1. Find the ionization potential of the atom. (a) 11.2 eV (b) 13.5 eV (e) 15.6 eV (d) 2. Find the short wavelength limit of the series tcnns n = 2 (a) 3256 A (b) 2339 A (e) 2509 A (d) 3. Find the excitation potential for the state" 3. (a) 14.64 eV (b) 9.93 eV (e) 12.52 eV (d) 4. Find the wave number of the photon emitted for transition n

=

(a)2.23 x lO" m- J

12.6 eV 3494 A 10.04 eV = 3 to n = l.

(b) 1.009 x IO " m- J

(e) 3.005xlO'm - J (d) 0.432 xlO'm - J S. Ifan electron with initial kinetic energy 6 eV is to interact with this hypothetical atom, what minimum energy will this electron carry after interaction? (a) 2 eV (b)3 eV (e) 6 eV (d) 0 eV 6. The initial kinetic energy of an electron is II eV and it interacts with above said hypothetical one electron atom, the minimum energy carried by the electron after interaction is: (b) 0.3 eV (e) 0.9 eV (a) 0.7 eV (d) 1 eV

Passage-2 Consider a hypothetical hydrogen-like atom. The wavelength, in spherical lines for transitions from n = p to II I are given by:

=

A=

1500p' 2

wherep=I.2,3.4•....

p -I

1. Find the wavelength of the most energetic photons in this series:

(a)ISOOA ' . (e)1300A

(b)1500A (d)1650A

A.

for the

GRB Undentanding Physics Optics and Modem Physics

436

•

2. What is the ionization potential of this element? (0) 3.96 V ···· (b) 9.23 V (c) 6.34 V

(d) 8.28 V Passage-3

In a mixture of H _ He + gas ( He + is singly ionized He atom), H atoms and He + ions arc excited to their respectively 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. I

He+ atom

Hatom

n _. n.3

•

n _2

O.85eV

-3.4 eV

-1 .51 eV

-6.04 eV

•

-3.4 eV

- 54.4 eV

-1 3.6eV

n ",I

-13.6eV

1. The quantum number n of the finally populated in the He + ions is:

002

003

(c) 4

(d) 5

2. The wavelength of light emined in the region by He + ions after collisions with H atoms is: (a) 6.5 x 10- 7 m (b) 5.6 X 10- 7 m (c) 4.8 x 10- 7 m

(d) 4.0 x 10- 7 m 3. The ratio of the kinetic energy of the n = ~.electron forthe H atom to that afHe + ion is: . . . (a)

!

4 (c) I

(b)! 2 (d) 2

Partide and Wave Nature of Mauer

437

Passage-4 (Bohr's model) Some energy levels arc shown for a hypothetical one electron atom.

J. [fthe lowest energy state has energy Eo, the value of EOin terms ofAois: (a) 4he 3~o

(b) he

(e) lhe

3~o

4~o

(d)

he

2~o

2. The value of A is: 27

( d) 27 ~ 32 0 3. Jfa radiation which carries an energy equal to ionisation energy ofthe atom fall on a photocathode of threshold wave length AO, the kinetic energy of fastest photoelectron is: (a) i6~o

3he (a ) 4AO

(d) 3he 2~O

Passage-5 The wavelength of radiation emitted by a hydrogen like atom when de-excited from nl = n to n2 = I is given as kn' ~ =--:;--

n' -I

1. Then the ratio of maximum wavelength to minimum wavelengths of radiation is: (a) 2:3 (b) 4:3 (e) 6:5 (d) 3:2 2. The ionisation potential of the atom is 54.4 eV. Then the value of k is: (a) 12400A (b) 3100 A (e) 228A (d) 1377.8A

I

GRB Und~nlandin8 Physics Optics and Modem Physics

438

3. 111e ratio of frequency of resolution of an electron is n = 100 and that of the

radiation from n = 100 to 1/ = 101 is approximately: (a) 1:2 (b) 1:1 (e) 2:1

(d) 100:101

Passage-6 In a series afline spectrum formed by an ionist:d atom obeying Bohr's model. the shortest and largest wavelengths arc 227 .9A and 410.2 A respectively

1. Then, the atomic nQ. Z and the lowest level n of de-cxcitation are: (a)Z=4.,,=2 (b)Z=3.n=2 (e)Z=2,n=3 (d) Z=2.//=2 2. The minimum wavelength of a radiation to ionise the atom by knocking its an innennost electron is: (a) 35.29A (b)40.16A (e) 56.98A (d) 65.34A 3. The ratio of the radius corresponding to the lowest level of de-excitation nand that orlhe hydrogen atom is: (a)3:5 (b) 2:3 (e) 1:2 (d) I: I

4. The series is: (a) Lyman

(b) Balm«

(c) Paschen

(d) P·fund

Psssage--7 The energy level diagram of a hypothetical hydrogen like ion is given . The stationary ion receives a blow from a bombarding particle to gain an energy

AE=0.2IE.

EJ4

_ _ _ _ _ _ _ _ _ n.2

E

1. The change in angular momentum of the electron is: h h 3h (a) 2n (b);; (e) 2n

(d) 3h n 2. If the energy ofn = 3 is 13.6 eV, the energy absorbed IlE by the ion is:

(a) 25.7 eV

(b) 122.4 eV

(e) 30.4 eV

(d) 91.8 eV

3. The no. of emission lines corresponding to Q-2 is: (a) 8

(b) 10

(e) 15

(d) 6

439

Partide and Wave Nature of Matter

Passage-8 1. A light of 10.02 eV fall on atomic hydrogen gas. The hydrogen atom will radiate an energy wavelength A. The value of A is: (a) 121 S.6SA (b) 121.56A (c) 1256.6JA (d) IJI5 .6SA 2. The energy emitted by H·atom in Q·I falls on fixed He +ions. The lotal no. of e:mission of He +ions will be: (b) 5 (c) 6 (d) S (a) 4 3. If the energy radiated by H. atom in Q·I fnlls on a fixed Li +2 ion, the excitation takes place from:

=I ,

=

=

=

=

n 4 (b) II 2 , n 4 (c) n 3 -+ n 4. In Q·3 , the total no. of emission lines will be: (a) "

(a) S

(b) 6

(c) 10

=4 (d) II =I ,

" =5

(d) IS

Passage-9 (Atomic spectrum, emission and absorption) A fixed atom in ground state must absorb some fixed amount of energy from a bombarding atom, ion or any particle to get excited to the corresponding levels, obeying Bohr's postulatcs. The de-excited atom will emil radiations of fixed frequencie s given by I:!.E = ~c . The energy Icvels of a hydrogen atom must be quantised according to the relation E = _ 13 6 eV. Based on the above infonnation

ni

answer the following questions. 1. A gas of mono atomic hydrogen is bombarded with an electron beam which have been accelerated through a potential difference of 12.75 Volts. The numbers of spcctrnlline should be emitted is: (a) 2 (b)J (c) 4 (d) 6 2. A continuous band of radiation having wavelengths ranging from 1000 A to lOoooA is passed through a atomic hydrogen gas. Wh ieh lines will be significantly depicted in the spectrograph?

(a) Lyman

(b) Balmer

(c) Paschen

(d) Brackel

3. A sample of atomic hydrogen gas absorbs a radiation of wavelength 103 nm which of the possible transition will occur?

Ic,

1°, n=2

6,

(a) AI

(b)B,

(c) C,

(d) D,

440

GRB Undmtanding Physics Optics and Modem Physics Passage-10 (Wave nature of matter)

. Matter possess wave character like radiation was proposed by de Broglie in his piiol wave theory similar to wave theory of radiation. In a wave piloting a matter

particle. I. The product of group and phase speed is:

(a)
(b)=C' (c»C' 2. In a pilot wave, the wave funclion is:

(d)
(a) displacement (b) velocity

1

(el energy

I

(d) probability of finding the particle 3. Probability wave was suggested by:

(a) Heisenberg (b) Max-planer (e) Max Born 4. Heisenberg uncertainty principle arises due to the: (a) defect of measuring instruments (b) defect is our senses (e) wave particle duality (d) limiting value ofspccd of light

(d) de Broglie

I I I

I I

I

rJI Subjective Problems L..el-1 Atomic models and spectra 1. Sodium alom emits a yellow light of A. = 5896 A.What is the difference between two levels of energies of transition orthe atom. 2. Find the wavelength offslline of Balmer series. 3. Find the wavelength of 2nd line orLymen series of hydrogen atom. . 4. Find the frequen cy of radiation emitted when a hydrogen atom is de--excited from ooton=3. S. Find the angular momentum of an electron in hydrogen atom having energy -1.5 I eV. 6. When a hydrogen atom is bombarded, it gets excited to its higher energy states. As it comes back to its lower energy states, it radiates lights. Find the three longest wavelength due to the transitions to n = I. 7. What is the maximum wavelength of a radiation when collided with a fixed hydrogen atom will ioni se it?

,,=

8. Find the radius of 1st excited state ofLi++. 9. Find the 1st three energy states of doubly ionised lithium. 10. In the previous problem, find the corresponding longest wavelengths.

..

r

t

441

Particle and Wave Nature of Matter

I I. Find the mdius of 3rd electron orbit in He + ion. tZ. Find the energy of I st line of Balmer series afHe + .

~

I I I

•

13. Find tI~c value of first four energy levels of: H including the reduced mass correction. 14. Find the ratio of velocity of an electron 2nd excited orbit of He + to that of 1st excilcdorbitofU++ . 15. Estim:nc the energy nceded to remove an electron with" = 1 from a lead atom. Find the corresponding wavelength of x-rays required 10 do this (Z = 82). Matter waves 16. Find the wavelength ora neutron waveofeners)' I keY. 17. A prolon is accelerated through a potential difference of2 kV. Find the de Broglie wavelength of the proton beam. 18. Find the de Broglie's wavelength orana-particle moving with a speed ofl0 6m/s. 19. An electron accelerates from rest through a potential differencc of 100 V. Find its dc Broglie length. 20. de Broglie's wavelength of an electron is IA. Find the speed and KE of the electron. Electron diffraction 21. An electron beam having the electron speed of400 mfs strikes a crystal. Find the minimum interatomic (planner) distance d so that a strong beam of electron will come out as a diffracted maxima at an angle of 25 0. 22.

An electron beam strikes Nael crystal as shown such that

~= ~.

a

~S

Find the

minimum potential difference through which the electrons be accelerated so that. they can come out with a strong reflected beam. Put a = 5.63A.

-L.:J 0

•

v

•

0

0

0

0

0

0

0

". '<

~

Particle In a Box 23. An electron having energy of6 eV moves between two rigid walls of I nm apart. Find the value of the quantum number n (energy states) that the electron can have. 24. A ruby laser emits 694.3 nm light. If this light is assumed to have been caused by the transition of an electron in a box from n = 2 to,J = 3, find the width ofthe box. 25. An electron is confi ned to a tube oflength I. The potential energy of the electron in one half of the tube is zero and in the other half of the tube is 10 eV. The

GRB Undmlanding Physics Optics and Modem Physics

442

electron has total energy of 15 cV. Find the ratio of de Broglie's wavelengths in the two half of the tube. Sketch the wave function of the electron in the tube. Heisenberg uncertainty principle

26. If the uncertainty in position of a moving particle is equal to its

ue Droglie

wavelength, find the uncertainty in its velocity.

27. The radius of a typical atomic nucleus is about 5 x JO-1Sm. The uncertainty in position of a prolon in the nucleus is 5 x IO- IS m. What will be the lc~st uncertainty in proton's momentum. 28. In the previous problem. find the smallest uncertainty in proton's KE is cV.

29. A ball of mass 100 gm is confined in a room of 15 m side. Iflhe ball moves with a 30.

31. 32. 33.

velocity v = 2 mi s, find the spread of its speed. Although an excited atom can radiate at any time between t = 0 and t = -. the average time after excitation a! which a group of atoms radiate is called life time r. 8 1ft = 10- 5, find the (a) line width Ilf oflight (b) Fractional broadening of light if A = 500nm. A light source is used to locate an electron in an atom upto a precision ofO.5A . Find the uncertainty in the speed of the electron. Find the uncertainty in position of a stone of mass m when it fall s under gravity through a small distance H. The uncertainty is time during which an electron remains in an excited state is 7 10- s. What is the least uncertainty in the energy (in eV) of the excited state?

34. From the relation l:!.p.l:!.x ilL ·Il& ~

1~,

<:!:

A,

4.

show that for a particle moving in a circle

whcre ilL = uncertainty

In

angular momentum and 119

=

uncertainty in angle.

(JI Subjedive Problem. Level-2 1. Calculate the electron'S speed and angular momentum in ground state and 4th excited atom of hydrogen atom taking the motion of nucleus into account. 2. Prove that the energy of one electron atom or ion is E =

-p.z2 e4 2 8E oh 2 n 2

.

3. Using the reduced mass ofhydragen atom as described in last example, find the (a) minimum Bohr's radius (b) 5th Bohr's nucleus.

4. A particle of mass 1111, charge ql is projected with a kinetic energy Ko towards a stationary free particle of mass m2and charge Q2 . Find the : (a) minimum distance 'b between the particles (closet approach),

(b) value of'b, taking mJ asa· particle and m:zas lithium nucleus, (c) value of 'b . taking m] as a-particle and m2as a fixed lead nucleus. [Assume KO = OAM eV for (b) and (e)]

443

Particle and Wave Nature of Matter

S. In the previous problem. find the minimum distance between a·particle and Hg-nucleus. 6. An a - particle gets deflected byS = 90° while moving under the electric field ofa stationary Hg nucleus. If the initial speed of the a-particle is K = 0.5 MeV, find the minimum radius of curvature of its trajectory. 7. For an oscillator (spring mass system, say), prove that, the ground state energy or minimum energy of the particle oscillating with a frequency f = -'

fK, where 2It'l{;;;

K = stiffness oflhe spring and m = mass of the particle. is Emin =hf. Compare it

with the semi classical value zero and detailed quantum mechanical value ~ lif, S. A particle of mass m is trapped in a cube ofiength I. Using uncertainty principle, find the force exerted by the particle on the wall ofthe cube. 9. A free electron is confined in a region of linear dimension 1=0.1 nm. Using uncertainty principle, find the time over which the width of the corresponding probability (matter) wave trains will be II = to times larger. 10. A thin beam of mono energetic electrons fall on a metal sunace at a glancing angleS. The distance between the Bragg's plane is d. The mirrorreflection(called Bragg's reflection) will be maximum at Vo and NVo. Find the value ofVo. Put 9 = 60° ,d =0.2nmandN =2.25. 11. (a) Find the de Broglie's wavelength of an electron whose KE is equal to its rest mass energy. (b) Find the frequencyofmaUer(cleetron) wave in (a) 12. (a) Find the change in KE of an electron of mass m when its de Broglie's wavelength decreases from A1 to A2. (b) What is the potential difference through which the electron will pass inCa)? (c) putting A1 = 100 x 1O-12 m and A2 = 50 x 1O-12 m , find the results in (a) and (b).

13. A proton is projected with a KE 20 eV towards a free stationarya·partic1e. Find the de Broglie's wavelength of both particles they become closest. 14. Find the minimum (a) KE (b) speed ofa hydrogen atom to excite a stationary free hydrogen atom. 15. A fixed He + ion emits a photon corresponding to first line of Lyman series. This photon knocks the electron from a fixed hydrogen atom in ground state. Find the maximum speed of the photo electron. 16. The third line of Balmer series ofa hydrogen like ion has wavelength 108SA, find (a) the ion, (b) the radius of the atom in its setond excited state. (c) What is the binding energy orthe electron?

eRB Undmtanding Physics Optics and Modem Physics

444

=

=

17. A photon emits from a fix.ed hydrogen alom when it de·excitcs n 2 to n I. The frequency o f photon is f [fthe hr~gen atom is made free, the frequency ofthe emitted photon becomesf'. Find

fft

18. A stationary free hydrogen atom dc-excited from n = 2 to n = I in 1st line of Lyman series. Find the recoil speed of hydrogen atom. 19_ A proton o f mass m and charge (' moving with a kinetic energy Ko is deflected by the electric fi eld of a Gold nucleus of charge Ze. If the Gold nucleus does not move, assuming the ai ming parameter, find the magnitude of change in momentum of the proton. 20. According to Maxwell theory of electromagnetic radiation (classical theory of radiation) an electron moving with an acceleration a radiates its energy at a rate. dE e 2a 2 dt ::; - 6rt£o c1

21. Show that an electron in classical hydrogen atom spirals into the nucleus at a rate at: dr -=

dt

e4

12rt 2£2r2m2e3 o ,

22. Find the time it takes the electron to reach the nucleus from It) = 2 x 1O-1 0m. 23_ The general expression for the energy levels of one electron atom and ion is 222

E /I

__ lJ.Ke Q,Q2 2 1j2 n2 '

Mm

where J.1 ::; reduced mass of electron-nucleus system = 'M~:":: +m 24. Find the wavelength of the light emitted by (a) positronium atom when it is de· excited from n = 3 to n ::; 2 Note that. a positronium atom consists ofone electron and a positron but no nucleus, (b) singly ionised helium atom. 2S. Prove that for large quantum number n » I, the frequency of emitted photon is equal to the frequency of revolution of the electron. The quantum theory approaches 10 classical theory when II» I, in accordance to Bohr's correspondence principle. 26. Calculate the mass ratio of deuterium and hydrogen if their H"" lines have wavelengths 6561.01A and 6562. 8A respeclive\y. Consider the mOlionofnuclei. 27. Show that the quantisation o f angular momentum in Bohr's postulates gives rise to the quantisation of radii. velocity and energy of hydrogen alom.

r

I

I

I

I

Partid~ and

Wa\"e

Natu~

or Matter

445

28. A beam of a mono energetic electron passes through a square hole oflength d, gets diffracted and fonn a diffraction pattern on the screen . The width of the Electron beam b central diffraction pattern is c and the 1 distance between slit and screen is D . Find the speed of the electron in the ••_ _ beam. Putd = 1O-6m, c =O.36mm, D =

1------1-1

~~

I f I

I I

I

I I

1

I

! I

I

0 __ •

c

1-

O.5m 29. A single electron orbits a stationary nucleus of charge +Ze • where Zis a constant and e is the magnitude of electronic charge. It requires 47.2 cV to excite the electron from the second orbit to third Bohr orbit. Find: (a) the value of Z. (b) the energy required to excite the electron from the third to the fourth Bohr orbit, (e) the wavelength of electromagnetic radiation required to remove the electron from first Bohr orbit to infinity, (d) find the KE. PE and angular momenrum of electron in the 1st Bohr orbit. (e) the radius of the fi rst Bohr orbit [The ionization energy of hydrogen atom = 13.6 ey. Bohr radius = 5.3 x 10- 11 m , velocity of light "" 3 x lO IS ms- I ,

Planck's constant ;;; 6.6 x 10-34 J·s] 30. A particle of charge equal to that of an electron, - c and mass 208 times the mass of electron (called a Il·meson) moves in a circular orbit around a nucleus of charge +3c (take the mass of the nucleus to be infinite). Assuming that Bohr model of the atom is applicable to this system. (a) Derive and expression for the radius oflhe nth Bohr orbit. (b) Find the value of n for which the radius oflhe orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom. (c) Find the wavclength oflhe radiation emitted when the Il· meson jumps from the third orbit to the first orbit. (Rydberg's constant;;; 10967800 m) 31. A moving hydrogen atom makes a head--on collision with a stationary hydrogen atom. Bdore collision. both atoms are in ground state and after collision they move together. What is the minimum value of tile kinetic energy of the moving hydrogen alom. such that one oflhe atoms reaches one orthe excitation state. 32. A neutron moving with speed. umakes a head--on collision with a hydrogen atom in ground state kept at rest . Find the minimum kinetic energy oflhe neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen "" 1.67 XlO-27kg. 33. How many head·on, clastic collisions must a neutron have with deuterium nucleus to reduce its energy from I MeV to 0.025 eV?

446

GRB Understanding Physies Opties and Modem Physies

34. A gas of identical hydrogcn.like atoms has some atoms. in the lowest (ground) energy level A and some atoms in a particular upper (excited) energy level ~nd there arc no atoms in any other energy leveL The atoms of the gas make transition to higher energy level by absorbing monochromatic light of photon energy

S.

2.7eV. Subsequently, the atoms emit. radiations of only six different photon energies. Some of the emitted photons have energy 2.7 eV, some have energy more and some have less than 2.7eV. (3) Find the principal quantum number of the initially excited level B. (b) Find the ionization energy for the gas atoms. ·(c) Find the maximum and 1he minimum energies of the emitted photons. 35. A hydrogen·like atom (atomic number Z) is in a higher excited state of quantum number II. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV, respectively. . Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.954 eV, respectively. Determine the values ofn and Z(ionization energy of hydrogen atom = 13.6 eV). 36. Electrons in a hyc;!rogen·like atom (2 =3) make transitions from the fourth excited state to the third excited state and from the third excited state to the second excited state. The resulting radiations are incident on a mctal plate and eject photoclectrons. The stopping potential for photoelectrons ejected by shorter wavelength is 3.95 eV. Calculate the work functi on of the metal and stopping potential for the photoelectrons ejected by the longer wavelength. 37. The radiation emitted when an electron jumps from n =3 to n =2 orbit in a hydrogen atom falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of(1 /320) T in a radius of 10-3 m. Find (a) the kinetic energy of the electrons, (b) work function of the metal and (c) wavelength ofradiatlon. 38. The ionization energy ofa hydrogen·like Bohr atom is 4 rydbergs. (a) What is the wavelength of radiation emitted when the electron jumps from first excited state to ground state? (b) What is the radius of first orbit for this atom? (Given that Bohr radius of hydrogen atom.: 5 x 10- 11 ...10 and 1 rydberg "" 2.2 x 10- 18 J.) 39. A doubly ionized lithium atom is hydrogen like with atomic number 3. (a) Find the wavelength of the radiation required to excite the electron in Li++ from the first to third Bohr orbit (ionization energy of hydrogen equals 13.6

eV).

•

Partide and Waw: Nature of Matter

447

(b) Ho~ many spectral lines are observed in the emission spectrum of the above excited system? 40. The energy ofan electron in an excited hydrogen atom is _ 3.4 eV. Calculate the angular momentum of the electron according to Bohr's theory. Given:

R =1.09737 X 10-7 m -I , Planck's Rydberg's constant 34 8 h = 6.626176 X 10- )·s, speed oflight c =3 X 10 m s- l .

constant

41. Two hydrogen·like atoms A and Bare of different masses and each atom contains equal numbers of protons and neutrons. The difference in the energies between the ~Irst B.almer lines emitted by A. and D, is 5.667 cV. When atoms A and e, moving wllh the same velocity, slrTke a heavy larget, they rebound back with the same velocity. In the process, aton~ B imparts twice the momentum to the target than that A imparts. Identify the atoms A nndB. 42. An electron in a hydrogen·like atom i5 in an excited state. It has a total energy of - 3.4 eV . Calculate: ~ (a) the kinetic energy and (b) the de Broglie wavelength of the electron.

(11= 6.63 x 10- 34 J·s) 43. Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one·dimensional array wi.th nodes at each of the atomic siles. II is found that one such standing W;1 VC is formed if the distance between the atoms of the array is 2A. A similar standing wave is again formed if d is increased to 2.sA but not for any intermediate value ofd. Find the energy of the electrons in electron volt and the least valu\! of d for which the standing wave of the type described above can form . (h =6.63 x 10-34 J-s) 44. A gas of hydrogen· like ion is prepared in such a way that the ions are only in the ground state and the first excited state. A monochromatic light of wavelenbrth 1316A is absorbed by the ions. The ions arc lifted to higher excited states and emit radiation of six wavelengths, some higher and some lower than the incident wavelength. Find the principal quantum number of the excited states. lndentify the nuclear charge on the ions. Calculate the values of the maximum and minimum wavelengths.

45. An electron is hydrogen·like atom makes a transition from nth orbit ~nd emits radiation corresponding to Lyman series. If de Broglie wavelength of electron in nth orbit is equal to the wavelength of radiation emitted, find the value ofn. The atomic number of atom is II. 46. A single electron orbits around a stationary nucleus of charge + Ze. where Z is a constant and f! is the magnitude of the electronic charge.lt r~quires 47.2 eV to excite the electron from the second Bohr orbit to the third Bohr orbit. Find: (a) the value of Z, (b) the energy required to excite the electron from the third to the fourth Bohr orbit,

448

GRB Undentanding Physics Optics and Modem Physics (c) the wavelength of electromagnetic radiation required to remove the electron from first Bohr orbit to infinity, (d) the kinetic energy, potential energy and the angular momentum of the electron in the first Bohr orbit, (e) the radius of the first Bohr orbit. [The ionization energy of the hydrogen atom = 13.6 eV, Bohr radius = 5.3 x 10- 11 m, speed of light =3 x 10 8m s-I,

Planck's constant =6.63 x 10-34 J·s] 47. Light from a discharge tube containing hydrogen atoms falls on the surface ora piece of sodium. The kinetic energy of fastest photoelectrons emined from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find: (a) the energy of the photons causing the photoelectric emi ssion. (b) the quantum numbers of the two levels involved in the emission of these photons, (c) the charge 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 10 be at rest before the transition. [Ionization potential of hydrogen is 13.6 eV] 48. An electron in the ground state of hydrogen atom is revolving in • anticloekwise direction in a circular orbit of radius R . " (a) Obtain an expression for the orbital magnetic dipole moment of the electron. ~

30"

(b) The atom is placed in a unifonn magnetic induction B such that the plane nonnal to the electron orbit makes an angle of 30° with the magnetic induction . Find the torque experienced by the orbiting electron. 49. The radiation emined when an electron jumps from

Ii

= 3 to

Ii

-,, -R

= 2 orbit of

hydrogen atom falls on a metal to produce photoeleclrons. The electrons emitted from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of(1/320) T in a radius of 1O- 3m. Find: (a) the kinetic energy of the electrons, (b) work function of metal, (c) wavelength of radiation.

[Planck's constan~ h ~ 6.63 X 10-34 J·s]

\

Partid~ and Wa\~

-449

Nature of Matter

•

•

'RNSWERs)

Multiple Choice Questions

(A) Only One Choice Is Correct Level-1

1. (n) 6. (a) 11. (d) 16. (d) 21. (n) 26. (e) 31. (c)

2. (d) 7. (e) 12. (a) 17. (e) 22. (d) 27. (a) 32. (c)

3. (e) 8. (d) 13. (d) 18. (b) 23. (d) 28. (d) 33. (b)

4. (e) 9. (b) 14. (e) 19. (e) 24. (a) 29. (d)

5. (a) 10. (a) 15. (b) 20. (e) 25. (d) 30. (b)

2. (c) 7. (b) 12. (a) 17. (d) 22. (b) 27. (c)

3. (a) 8. (b) 13. (b) 18. (d) 23. (a) 28. (a)

4. (d) 9. (b) 14. (b) 19. (e) 24. (a)

5. (c) 10. (c) 15. (c) • 20. (b) 25. (b)

Level-2

1. (a) 6. (b) 11. (d) 16. (b) 21. (d) 26. (a)

(8) More Than One Cholcel. la/Ire CGmIcI 4. (a,c) 3. (a,b,c) 2. (a,b) 1. (a,c,d) 9. (a.b,c) 8. (a.d) 7. (c,d) 6. (a,c,d) 14. (a.b,d) 13. (a.b,c) 12. (a,c.d) 11. (a)

5. (a,b) 10. (a.b)

M6fch the Columns

1. a- r ; b-q ; C-
l. a-q,s ; b -q; c - r, d- p.q.s 4. a- r ; b-p ; C--5 ; d- q

Comprehen61ons

Pa.sage--l : 1. (c) 6. (a)

2. (b)

3. (e)

Passage-2 : 1. (b)

2. (d)

Passage--3 : 1. (c)

2. (c)

3. (a)

2. (d)

3. (b)

Passage-4 : 1. (a)

4. (b)

5. (c)

GRB Undmtanding Physics Optics and ModemPhysics

450

Passage-S : I . (b) Passage -6: 1. (a)

2. (e>' ~

3. (b)

2. (e)

3. (d)

2. (a)

3. (b)

2. (e)

3. (b)

1. (d)

2. (a)

3. (b)

Passage -10: 1. (b)

2. (d )

3. (e)

Passage -7: 1. (e) Passage -8: 1. (a) Passage -9:

4. (b)

4. (d)

,

Subjective Problems Leve/-1 I. 2.1 eV 2.656.3 run 34 5.3.16)( 1O- J_5

•

4. (e) '

3.1 025A

•

6. 122 nm, 102 nm, 97 nm

7. 910A 8. 0.53A 9. - 122.4 eV. -30.6 cV, - 13.6 cV. - 7.65 eV 10. 13.5 nm, 11.3 nm. 10.8 nm II. 2.385 A 12. 7.56 eV 13. - 13.598 . V, -3.3995 eV. - 1.5 109 eV, -11.8449 eV 14.

1

9 17. 6.41)( to- 11 m 19. O. 123 y' )O-9 m

21. 4 .3 x IO-6 m

15. 9.1 keY, 0.01 36 nm

23.=4 25. 1.73

26. tw 2:: ..£.

27. 6p 2:: 1.05 x 1O- 20 kg mls

28. BK ;:: 414 keY

31. .6t: = 1.1 6 x 10 6m/s

16.906 fin

18.9.9 19 x to- 14 m 20. 7.28x 10 6 m/s, 151 eV

22. 7.44 voll 24.7.95A

29. 1.8 x 10- 35 30. (a)8x 10 6Hz

•

.

"

4.

(b) 1.3 x 10-8 h

32. 6.< ~

=

m,2gH

33. AE=O.33x I0-8 eV Subjective Problems Letfft/-2

I. 21 89 kmls, 437.8 km/s, 2.108 x 10 - 34 J-s, 10.541 x 10- 34l -s 3. (a) 0.5296A (b) 13.24A

r..

Particle and Wave Nature of Matler

ZIZZC2( 1+

ml)

m, 41ttoKo

4. (a) 'I> =

451

(b) 'I> = 0.034xI0- 12 m

(c) 'I> = 0.591 x 10-I'm

5. 'I> =0.557x I0 - I'm 6. Po =0.23IxI0- I' m

7. Emm

=

hi h'

8. F =--"~ 4'/t 2 ml3

9. ':: 10- 15 S \0. 45.94 volt 11. (a)0 .67x IO- I' m

(b)2.6x IO'OHz

12. (a)M="'(_1 __ I) 2m)..2 ,

(c)

h'(1

I)

(b)l>V=- - - 2me)..2 , 2

)..2

I

)..2

I

l>K = 451 eV ,l> V = 451 voll

11 13.0.76 x lO- 11 m,3.04 x lO- m

6

\4. (a) 20.4 cV

(b) =2.68 x 10 m1s

\5. 3.092 x 10 6 m1s \6. (a) Z = 2

(b) 2.385A

(c) 54.4 eV

17. 0.55 x lO-6%

\8. 3.3 mls ,---r-------,;, \9.

4=2 2mK 'll+(2bK;:;EO)),

22. 0.846 ns 24. (a) 1.35 ~m

(b) 1.64 nm

26. ~ =0.499

MD

I

I,

28. 1.92 x 106 m1s 29. (a) Z = 5 (d) 1.05 x 10-34 J·s 30. (b)n=25 31. 20.4 eV 32. 20.4 cV

(b) 16.53 eV (e) O.l06A

(c) A = 0.548 A

(c) 36.5A

452

GRB Undentanding Physics U lltics and Mod~m Physics

33. n=8 34. (a)n8 = 2

(b) 14.4 eV

(e) 0 .7 eV

37. (a) 0.86 eV

(b) 1.04 eV

I . . 6542A

38. (a) JOl.4A 39. (a) 114.26A 40.2 .1102x )O- 34J_s

(b) 2.5x lO- ll m

35. n =6and Z = 3 36. 0.75 V

(b) 3

41. Atom B is singly ionized helium. 42. (a) +3.4 cV 43. O.5A 44. O.24 x 1O- 7 m

(b) 6.66A

45. n= 25 46. (a) Z = 5 (b) 16.53 cV (e)36.056 A (d) 1.0546 x 10-34 l·s (e) 1.06 x 10-

47. (a) 2.55 eV (b) The transition is specified by nl (c) ~

=4 ~ n2

11

m

•

--+ 2

== !! (d) Recoil speed ofH atom = 0.814 m s - I n

= ehB newton-m~trc 4Jtm 8M1 49. (a) 0 .86 cV (b) 1.02 eV (e) 6602 A

48. (a} M

= ~A _ m2 (b)t

•••

•

,. ,.

\

\

,.

453

Particle and Wave Nature of Mauer

1IC:::;:~::::::::~C~Hmi~n~t~S:R~n~d~S~D~I~u~t~iD~nis}~:::::::::::::11 Comprehensions

Passage -4 Eo _ Eo = he 4 Ao

I.

Eo = 4he

or

3Ao Eo _ Eo = he 9 A

2.

4he )(~=he

or

3Ao 9

A

A = 27). 32 0

or •

Krrex

3.

= Eo -IV = Eo _ he = 4he _ he ). 0

3). 0

). 0

= he 3).0

Pas.age -5

I lim ~ = k

1. A ---+ Amu when n ---+ co .Then A =

II-+~

.

1- - 2

n k =3 4k lim --I

A ---+ A max when n ---+ 2. Then AtmX =

J

11-+2

1-4

Henee,Annx :Amn =4:3 . 12400 2. Puttmg n ---+ 00, Amax = k. Then, Eion = -k- = 54.4

3.

,

or k=228A fITS ::; f md . Bohr's theory of correspondence.

Pas.age -6

= lowest energy state. Then n2 = (n + I) state corresponds to

1. Let n]

n2 ---+

IX)

Hence,

Amu and

corresponding A. min '

-1!£.. = 13.6Z 2( _12 _

A. max

n

1 ) (n+I)2

... (i)

GRB Undmtanding Phy,ics Oplics and Mod~m Physics

454

. he =13.6Z'(-1 Amin n2

_1) 00

puningAmin =227.9 A andA nm = 410.2 we have, Z=4andn=2

2. ).' =12400 = 12400 i'.E I3.6Z'

=

3. Radius r = 'bn2 /Z

,

12400 13.6(4)'

' ... (ii)

A

=56.98 A

.!:. =n'/Z = (2) =1:1

or

4

'1l

Passage -7

which corresponds the transition n = 2 to n = S. -d! \4_l) 25 llL =An.!!. =(5 -2).!!. =3h

t. dE = 0.21E = Hcnc~, . " .

, .' "

2ft

2ft

2ft

2

. 2. 13.6 ~ = 13.6. where n = 3. Then Z = 3

n'

Hence, E=-13.6Z 2 = -t3.6x3 1 = -122.4eY Then, AE = 0.21E = 0.21 x 122.4eV= 25.704 cV 3. n = 2 --+ n = 5 transition corresponds to the total emi ssion lines SC2

r

=.2L=1O 2! x 3!

Passage -8 J. Hydrogen will be excited from n = I to n = 2 releasing some energy. Hence, the corresponding wavelength of radiation emitted by hydrogen is

12400 = 1215.68A 10.2 2. 10.02 eV energy will be absorbed by He + and excited from n = 2 to n = 4. Then no. of emission lines will be 4c 2 = 6. ).. =

n.4 _ _ _-'--__ -O.85eV n:3 n=2

-1 .51 eV -3.4eV

n=1 - - - - - -13.6eV H·atom

455

Particle and Wa~ Nature of Matter

3. By hit and trial method, we can see that 10.02 cV energy can be absorbed by U 2+ to get excited from n = 4to n = 6. Then, the no. of emissions will be (3.4 eV) (-7.65 eV)

======

0=5 n=6 n=4

(-3.4 eV) (- 6.04 eV) n=2 (- 13.6 eV)

(-13.6eV) _ _ _ __

0=3 (- 30.6 eV) _ _ _ _ _ _ 0=2 (- 122.4 e V ) - - - - -

0=1

0:1 (- 54.4 oV) He+ ·Ion

Passage ·9 1. If an atom absorbs 12.75 eV, it rises from ground state n = I £1 = -13.6 eV to £2 = - 0.85 eV corresponding to 11 = 4 while coming back to ground slate it may excite three Lyman, two Balmer and one Paschen as shown in the figure. Hence the t(\tal no. of spectral lines possible will be six . Z. In the range of 1OOOA·l ooooA, there are two lines, i.e., 1215 A, 1025.7 A in Lyman series. Apart from this, the total Balmer series and 1st four lines of Paschen series may be included. However, the lyman series excitation will be dominant because the atoms arc in ground state before adsorbing the photons. 3. 103 nm corresponds to from" = 1to

11

II~~~O = 12.03 eV which corresponds to the excitation

=3.

Subjective Problems Leve/·1

,

I ' =,r=hc=12400=2IeV ),5896 ' I) whereR=1.097xI0 7 m - 1 1 R(I 2. 1: = 2' - 3' .

or ), = 656.3 run 3.

In~ =1.097XI070-3~) 7 l. = 9x10- = 1025 A 8 x 1.097

or 4.

f

=

f = RC(3I, - ~) = 1.097 X10 7 x3 x 10' = 3.291 X10" Hz

,, ,

456

S.

GRB Unde~tllnding Physics Optics and Modem Physics

_1.51 = -13.6 or n=3

,,'

Then,

6.

,,=2-.11=1 t.E

= 13.6(1-~)=10.2ev

,,=3~n = 1

I

t.E = 13.6( I- i) = 12.I.V "

n = 4-+1I = 1

AE = 13.6( 1- I~) = 12.75 eV The corresponding wavelengths are)..

= hcl 6.E = 12~OO •

122 nm, 102 nmand 97 nm respectively 7. The maximum energy corresponds to the ionisation energy = l1E i.e,

Then, l. = 12400 = 12400 A = 91OA t.E 13.6

8.

r=ron2 , Z

= (O.53A)(2)' 14 = 0.53A

9.

E = - 13.6Z

,

.'

- 13.6(3)'

=

.' = 122.4 eV "'have E = -J22.4eV Puttingn = t, 2,3, we

.

t

E, =-30.6.V,E, =- 13.6.VandE, = - 7.65eV. 11.

"'

r='tIZ

= (0.53)(3)' A

2 = 2.385A

=

13 .6 eV

Particle and Wave Nature of Matter

12.

t.E = 13.6Z'(_1 __ I 22 3 2

457

)ev

=136(2)'(H)eV = 7.56 cV

13.

E, =(E,)O = -13.6 =-13.598eV I+E!. 1+ _ 1_ M 1836 E, = E, = - 3.3995 eV . 4

E,

E, = 9 = -1.5109 ,cV E, = 14.

E, i6 = -

0.8449 eV

V=VO~

n

VHe (l +_ I) = _ZUe x _nLi = _2x _ VLi ZLi nHe 3 (2 + I)

=1 9

16.

1.=L_h_ .J2mk

p

6.626 x 10-34 =r===~~~=~ ~2x1.67xlO 27 xl0 3 x1.6xlO 19 =906fm / 17.

h

h •

" :' .J-2-m-k =

r..J2~m::e"V

6.626 x 10-34 = --;======¥.~==;;;,= ~2x1.67x10 31 xl.6xl0 '9 x 2000 = 6.41 x IO-lI m 18.

" =.l!. = mfl

34

6.626 x 104x1.67xJO 27 xJ0 6

=0.9919x 10- 13 = 9.919x 10-l4 m

,.

GRB Und~ntanding Physics Optics and Modem Ph)'lics

458

, =~=

19.

..J2mk

6.626 x 10- 34

~2x9. l x IO-]1 x 1.6)(10- 19 xloo

= O. 123 x lO-9m

d sin 9 = nA for maxima

21. Bragg's law gives

By putting n = I and A = .lL . mv

we have,

dsin9 = 2!..

mv

•

d=

h = 4.3xtO-6 m mtJsinG 22. Bragg's law: UsinG = nJ... for longest wavelength, pUlling n = I A = 2dsin9 or

d =!!..... = 5.63 x 10-

.J5

10

.J5

,=E. =

.. .(i)

m

. . .(ii)

h

... (iii)

p .J2meV By using all the eqn., we have V = 7.44 volt

23.

P = ~ ... (i)

.,

n' = I ...ClII T

and

,

K=L

2m

.. .(iii)

These three eqns. gives 2 2 K=p2 =1l h

SmP ;:g:X:"'9-.I-X-I-=0-='>;'' -x-(::I-=-O::< -''')''-x-6:-x-:I:-.6-:x- 10::-:;0"

2m

or

(6.626xlO 34)' =

10-

9

~8)(9.lxlO

6.626x 10 34 = 26.43 x 10-9+ 34- 25 6.626

31 x6x I.6 )(IO-19

=3 .9g9~4

2

flE=(n 2 -n 2 )h-

24.

I

hC=(2'

,

2

8mP

-I,)L gmt'

'=-""""""-:-:::;""""-:--::-:--:-::::0

or

I=lh'= 3x6.626x10 " x 694.3xlO' =7.9SA 8rne 8><9.1xIO 31 x 3xlOR

459

Partide and Wave Nature of Matter

E=-

nZh Z

25.

8mP

By putting I = d = 20x 1O- 14 m and m = 1.67 x 1O-27kg we can get the energy for n = 1.2.3 respectively. 0.513 MeV. 2.05 MeV and 4.62 MeV. Yes

26. Since, the wavelength of the electron wave is given as

."

v =o

10---

V::: 10 eV

f ---4.1_.- - f - - -

nnd

p = .J2mK.

we have

~=t2 = ~=I.73 K, ~15=Iii

1.2

27. Heisenberg uncertainty principle gives
> .fr.-

4.

or or u

or

6u~-

4.

/!.p . tu 2:

28.

(dP)5 x 10-15

or 29.

or or Since

or

II

4.

,.'

"

.,

> 6.626 x 10-3•

4. dP> 1.05 x 10- 20 kg mls m/!.u 2: 1.05)( 10-20 1.67 x 10-27 dU > 1.05 x 10-20 &> > 6.3 x 10'mls

K = !mu 2 2 6K =

! . m2v6v = mvi}v

2

.. . (i)

,

,, GRB Undentanding Physics Optics and Modem Physics

460

;, (1.67 x 10-27 )(6.3 x 10' )(6.3 x IO')J ;, 414 keV

!J.Px 2:

30.

or

• uUx

'.

.!!..

4.

2: -Ir- =

41tml

I.

-,6".6",2",6"X~IO;--_34_ •

4x TtX 100 x 15 1000 =3.5 x to-35 m/s.

• ~

Then the spread of speed is 3.5 x 102

dUx =

"x 31. (a)

or

t::.E . !:J.t

35

= 1.8 x 10-35(1t is a negligible quantity),

.!!..

O!:

4.

1.6/ - 61;' ~

4.

6f ~

or

1

4rr· l1t

I

=

=8xto6Hz

4nxIO 8

III =8x106 f cl')..

(b)

6 8xlO 3x10 8

=

= 1.3 x 10-8

500 x 10-9

6p>_h_ 4n6x

32.

or

D,vO!:

It

4ron6x

=I.I6xlO~mls

33. The speed aquired by the stone is,

"= J2gh The momentum of the stone is p

= m~2gH

Then, the uncertainty in position is, 6x>

h - mJiiii

6£·61

34.

or

6£

>~

4.

>.JL = 6.626 x 10- 34 J 47t61

4ltx 10 7

I

.,

-,

I

I

I

Partid~

461

and Wave Nature or Mauer = 0.528 x 10- 27 J

=

0.33 X IQ-8cY

35. Since the particle moves in the circle, the Heisenberg's principle of uncertainty can be applied along tangential direction along S·direction, the uncertainty in position is dS = RdB equal the unccnainty in momentum is !lp = mAY.

Hence,

Ap .1!!.S :2:: .!!..

4n

or

mdv , RI1S 2: 41t

or

lI(mvR)IIS ~ 4n

. r Odirection

"

v

"

m~

Putting nluR = 1.. we have h

l\L ·IIS

..

,

~-

4n

(Proved)

..

Subjective Prob/ems "Level·2 1. The speed of an electron in ground state

""I -_...!JL -_ e/137 m 1+ M

1+

I 1836

=2189km1s

(e = 3 x lO'm ls)

The speed of the electron in 4th excited stale is

V, =V, =2189=437.8km1s

n 5 The angular momentum for n = I is

L=nh=!!=2.108 X IO- 14 j.s

-

For

n

n

..

L=5h=IO.54lxlO-14 j.s n

n=5,

2. The coulmbic force is

0-, J.'m

· '7

I

I

II I

I

(.e)(e)

4)'[£0

r2

F---

••'.

... (i)

The angular momentum is 2

L = Mml CJ) M+m

... (ii)

GRB Und~nlanding Physics Optics and Modem Physics

462 Newton's 2nd law,

F I) =

= rna = m,}002

distance of m from = -,--"M,,-,_ M+m

... (iii)

eM of the system ... (iv)

3. (a) The reduced mass is J.l =

Mm M +m

~=(I+;}n =

(I + 18136)<°.5293 A)

=0.5296 A (b) 'I = n'~ = (5)'(0.5296 A) = 1J.24 A

4. (a) Since both the particles are free to move and their field is electrostatic, using classical mechanics (disregarding energy radiation). caD conserve mechanical energy and linear momentum.'o fthc· syst em ql ~ of the panicles. 0 Vo 0 _ _ m _ r _1m2

We

11

Pj =Pj

(... at the time of minimum separation)

mlVo = mlV +m2 V

or

V =

mlVO

m, +m2 or closest approach VI = V2 = V " E j =Ef , 2 !m1V2 + qlq2 =!mlV +!m2V2 + Q,Q2 2 0 41t£or 2 ' 2 41t£o'b

Since r is very large initially q, q2

.- t

41t&or

!m

V2

2 I 0

-l(m +m 2

J

2

...{i)

,

co.

o-v

q'l-

""

0ro -I~

0, then.

)V 2 = Q,Q2

.:" .'

. ,.~

,1'1

_( .. ) '"

4ru:o'b

II

Eliminating V from eq. (ii) by eq. (i), '1)=

qlq2(ml +m2)

41t£°Omlm2Vo2 )

V' - K 0 ) ( ..'21m10 -

or

,

463

Partide and Wave Nature of Matter

(b) By putting =) =2,z2 = 3,ml = 4,m2 = 7,K = 0.4 x 10 6 x 1.6 x 10- 19 j 1U = 0.034 x 1O- 12 m. (c) The lead nucleus is kept fixed; m2 -+

,

ex>

Hence, put ml -+ O. Then,

""

'll ~ "Z-"Iz:;,..:e" 41t&OK

Put

ZI = 2, Z2 = 82 to obtain

'll ~ 0.591 x 10-l2 m .

5. Referring previous problem, solving eqns. (i) and (ii), 'll

ZJ z 2 e

~

2

',

cot~ -

2

.. .(i)

41(£0./2mK Vo

Substituting min eqn. (ii), we have K

~ !mVo' +./2mK Vo lan~ 2

2

..

or

. ..(ii)

Putting Vofrom eqn. (ij) in eqn. (i) and simplifying the factors, we have

m = (Zlz2e2

18tt&OK\ I

+cosec~)

Puttingzi =2,z2 =80,K =O.S x t06 x1.6xlO- 19 j we have, 'b = 0.557 x 10-12 m.

6. At the point A, let the a-particle move with a' speed Vo perpendicular to the position vector,1t>(= minimum distance). Since the electric field is conservative,

n

K =!my2 + qlq2 2

0

41(£0'll

.. . 1

Since net torque acting on the a -particle is zero (": the electrostatic force is repUlsive and acts away along the line joining the a-particle and nucleus), we can conserve the angular momentum ofa -particle

v, A b

mVinitialb

Thus,

= mvo'b. where mVinirial = (.J2~)b ./2mK b ~ mVO'll

..... ,

]-----------"-~ + ... (ii)

GRB Undel'5landins Physics Optics and Modem Phyaia

464

Newton's

2nd

law

at

A

. .. (iii)

gives.

(p = minimum radius of Cl1I"VtllurC)

2KJ} = q]q2 Po 4ru:o

, ,,(iv)

we know that.

,

ZI=2e2 COil

or

Po =

.

.. ,(v)

~ 2

8ncoT Putting z, :: 2, Z2 = 80 and 9 = 900, we have Po :: 0.231 )( 1O- 12 m"

7. The total energy of the particle is E=K+U

,

= L+!Kx2 2m 2 K = moi. we have

,

Putting

E = L +!mro 2x 2 2m

2

Since the .uncertainty in position is XI the uncertainty in momentwn is

l>p~ ~ x

or

Prrin

.. ,(ii)

=J1... =!L 21U x I

I ," .

.

Using eqns. (i) and (ii),

,

E 2: ~ + !mro2x2 2nu1

2

dE =0

For minimum E.

dx

. ' (-2x') + 1";;'" (2x) =0

2m

2

x', =moo~

""

Putting

xmin

= -1i,tn eq. ("') III

moo

.. .(iii)

465

Particle and Wave Nature or Matter E

I

E.. or

E

~

-J_......J_____ ,

tiro

Emin = 11ro = l!. .2rtf

2.

or

£min =

hi

The minimum energy of an oscillator can never be zero. Hence the particle cnn never be brought to rest. I A morc rigorous quantum mechanical calculation gives Errin = - Iii (ground

2

state energy).

8. The uncertainty in position is l>x = I The uncertainty In momentum is

l>p~~ l>x

or

l>p ~

!!... I

In ground state IIp:= p. Then. the total energy. that is K.E of the particle is

p'

E=2m

=

(It I I)'

or

j 9. The uncertainty ~n position of the electron is l>x = I =0.1 run

The uncertainty in speed of the electron is

GRB Undmtanding Physics Oplics and Modem Ph)'lics

v~

II 2ron.6x

or

"=-"21t1ll1

The time during which the probability elcctron wavc train will spreadoul and to a distancc d :: II/ is r = !!.

" :: ...!!L h

21tl1l1 I

or

\

,

= 21tnm/ ~

"

Putting m::9.1xlO- J1 kg, n = 10. I = O.l xtO-9 m and olhcrvaluc, we havc 1= 10- 15 s. 10. For nth order maxima in Bragg's law is givcn as 2dsinO:: n}' 1

... (i)

For next maxima, 2dsinG = (11 + 1)A.2 eq. (ii) + eq. (i) yields,

... (ii)

n)'1 = (n+ I»'2

.?:! =" +1

f. 2

Since

).,::!! =

... (iii)

II

P .J2meV

~:: ),2

Thcn,

II

Jii2:: JNVO

.,

'IV;

Vo

1::l = ,fN

... (iv)

Solving eqns. (iii) and (iv),

,fN=n+1 n

or

n=

I

(,fN - I)

... (v)

Puning n from eqn. (v) in eqn. (i),

2dsine=~')., 1 N -I

Putting

)., I::

h

2meVo

in eqn. (vi),

... (vi)

I 467

Partick and Wave Nature of Malter

h'

~ = --~~~~~~ 2

8med sin 2 9(JN _1)2

(6.3 x 10-34)'

=

8 x 9 x 10-31 x 1.6 x 10 19 x (0.2 x 10- 9 )2 x sin 2 600(~2.2S _1)2 =:

, I

I I

45.95 volt

11. (a) The energy of a particle orrest mass m moving with a speed v is E = me2 The rest mass energy is U = moe 2

Then. the K.E is K=E - U=(m-mo)c 2

. Since K::: U or or

(m - mo)c 2

= moc2

m= 2m o

or

or

F{

Then the momentum of the particle is p=m"=

=

mo

4mo/\

./3 = (

(I-W

2 .

=2.[lm oc =2./3 x 9.1 x 10-31 x3xlO' = 9.456 x 1O-22 kg mI, Then,

'dB =-Ph ,,

= ..:6".3",x",I",0,--'-,4"

9.456 x 10. " =0.67 x IO-12 m .

l

GRB Understanding Physics Optics and Modem PhysiCl

468 (b) The total energy is

E = me 2 = 2mo c2 =2x9.1xlO-31 x(3xI0 8 )' =

163.8 x 10-" J

Then thc frequency of the matter (electron) wave is /=E

=

"

163.8)( 10- 15

i

6.3 x 10- 34 = 26)(10 19

= 2.6x10 20Hz 12. (a) The momenlum of the particle of mass 11/ and D.B. wavelength). is

"

p=~

Then, KE of the particle is

r

p2= _ h 2_ K =2m

The change in KE is

2m').,2

1lK-,,'r-1__IJ 2m\}.~

A~

(b) The potential difference through which the electron will move is IlK ~V=e

=;:t\ -,~J (c)PuttingA2 =SOxlO- 12 m,Al =IOOxlO- 12 m

wehave, M = 451eVand l1V=45Ivolt. 13. When they will be closest both of them will move with same velocity gives as V=

mpVp

mp +ma 1.67 x 10 27 + 4 x 1.67 x 10- 27

= ~2)(20xI.6xlO 19 xI.67)(10-27 5)(1.67xlO-27

Particl~

and Wave Nature ofMalt~r .

469

= 10.338 x 10-23 5xl.67x1O 27 =1.238xl04m1s

Then,

Aa

=

34

" muVa

6.3 x 10=---''''-'~---""7 4 27 4x 1.67 x 10-

x 1.238 x 10 = 0.76xlO - ll m

Ap- - -" - -- }'a -rna =lI.ax , 4 =. 304 x 10- 11 m

mpVp

mp

14. (a) We know that, a stationary hydrogen atom can be excited from fll absorbing a minimum quantum of energy

= 1ton2 = 2

1 /l.E = 13.6(1- 2 , )=1O.2ev.

This energy is delivered by the colliding hydrogen atom to the stationary hydrogen atom in enelastic impact. The KE oflhe system (two atoms) relative to their eM is K' =: ! m V 2, 2 re

or

::; ! E!.:.!!!. V 2 2m+m

0

K'::; !mV 2

4 0 This energy will be used in exciting the stationary atom, as both atoms will move with same speed Then,

11lK'I=imVi

conserving energy w.r.t. the eM oflhe system M.'= o.E

imVo2 ::; 1O.2eV or or (b)

Ko ::;1O.2eV 2 Ko ::: 2004 eV

r------.;; x 20.4 x 1.6 x to t9 9.1 x 10

31

=2.68xI0'mls 15. The energy of 1st line of Lyman series is

I

I f

! (

1 o.E = 13.6Z'( 1- 2 , ) = 1O.2Z'eV = 10.2 x (2)' = 40.8 eV

470

GRB Understanding Physics Optics and Modem PhYSK:a

The photon corresponding to the energy 40.8 cV will eliminate the e,lectron fro~ the hydrogen atom by spending Eb = J3 .6eV. The surplus energy will be used In increasing the KE of the knocked electron.

Then, KE oflhc photoelectron is K=(40.8 -13 .6) eY or

or

lmu =27.2cV = 27.2 x I.6 x IO- 19 2

27.2 )( 1.6 )( 10- 19 x 2 9. l x 10 31

= A5 6 x10 1' = 3.092xlO'mls 16. (a) For 3rd line of Balmer series, nj =Stonj =2 Then , or

1:1£ = 13.6Z 2( _1 __ I J OY 22 52 OE = 2.856Z 2eV AE=Ephoton = 12400= 12400=11.43eV , 1085

By using eqns. (i) and (ii), 2.85 6Z 2 = 11.43 or Z=2 Then, the ion is single ionised helium. (b)

r= 1JJ

=

n' -Z

1JJ(3)'

(n = 3 for 2nd excited state)

2

= (0.53 A)~ = 2.385 A (c)

Eb = 13.6Z'

= 13.6(2)' =54.4 eY

17. The recoil speed ofthc nucleus when the photon comes out is given as vH=3h RH '

4M The KE of the recoiling nucleus is p' I ,

KE=-2m =iMvH 9h'R'

= -=-:-,Ho..

32M

... (i)

... (ii)

471

Particle and Wa\'C Nature of Matter

The energy of the emitted photon in 2nd case is E'=E-KE 9h 2R2 E-E'=KE= H or 32M Then,

E-E'

E

9,,2R~ 3

:::: 32M 14Ru "c 3hRH

I1H

8Mc

2c

=--= or

f - f' :::: 3.3 x 100:::: 0.55 x 10 - 6% f 2x3xlO 8

18. We can conserve the total energy and linear momentum ofthe system:

E; ::::Ef

p' E[ :::: Eo + Pr e +~, where Prand PH arc the linear momenta ofphoton 2M and hydrogen alom respectively.

,

PJ/

... (i)

E[ -Eo :::: Prc+2M E[ -Eo =RllhC( t - nl2 )=RhC(I- 212

orE! -Eo :::: ~Rhc

) =~RHhC ... (ii)

Pi ::::Pf

O=-PH+Pr or

PH :::: Pr By using eqns. (i), (ii) and (iii),

P~ +2MPHC -~hRHCM =0 Since P~ « MpHC, we have PH = ~hRH

4

or

= 3hRH = 3.3 mI, UH = PH M 4M

... (iii)

471

Particle and Wave Nature of Matter

The energy of the emitted photon in 2nd case is E'=E-KE

9h R~ E - E' = KE =-;;-;;-;-¥32M 2

or

9h 2R 2

I

Then,E-E= H / ~RHhc E 32M 4 3hRH vH =8Mc 2c

=

I

or

f-f' = 3.3 x I00=0.55xlO- 6% f 2x3xlO8

18. We can conserve the total energy and linear momentum of the system:

Eo

E, E, =Ef

E]

=

P'

Eo + Pre + 2Z" where Prand PH are the linear momenta of photon

,

and hydrogen atom respectively.

£1 -Eo

=

PH Pr c + -

... (i)

2M

EI - Eo =RHhC(I -_1 )=RhC(I-_1 n2 22

orE,-Eo =~Rhc 4

)=~RHhC 4 ...(ii)

Pi =Pr

O=-PH+Pr or p}f = Pr By using eqns. (i), (ii) and (iii),

P~ +2MPHC-~hRHCM =0 Since P~ « MPH,c. we have PH or

~hRH 4

... (iii)

472

GRB Understanding Physic.s Optic.s and Modem Physics

19. The magnitude of momentum dose not change because the KE is conservedjusl before and after the collision. Then,

Pi =Pr =p,say The change in momentum is 6 1 11 = 11;-",.-P",-"-;--:_-----, , 2 = Pf + Pi -ZpiPr cosO

=~p2 + p2

---- ... x-axis

-Zp2eosO

l~ = Zpsin~. where P = ~2mKo or

li.\~ = 2~2mK 0 sin Q

... (i)

2

Since, we know that tanfl =

2

Ze

2

8ru: obT'

using eqn.(i),

16ij =22mK'H2hK;:;C Ol )' 25. The frequency of revolution is

freY =

2 2 4 41t K me 33

n h The frequency of emitted radiation is fmd

= ~=CRro( (n -I)I 2 -- cR•

-J,) n

2n-1 n 2 (n _ l)2

, whenll»l,n-l:::::n.then!rad =!re\'

(Proved) According to thc classical theory of (Maxwell's electromagnetic induction) radiation. the frequency of radiation is equal to frequency of revolution. This classical theory matches with quantum theory for large quantum number. 26. From ·the modified fonnula,

I (IRroZ'Z)(In~ II;I)

): =

fA

A 1/

D

+

-

=1+mIMII~(I+-2!L)(I_...!!L) l+ml MD

Mlf

MD

: : : (1+ M: - :D)

Particle and Wave Nature of Matter

<

473

6S61DIA ( 1 1 m ) 6562.8 A = +1836 - MD or

M': =0.499

27. Referring the theory,

•••(1)

mv' =-4ncor2 r

... (ii)

Using these two equations we can get the relevant expressions. 28. For the center maximum diffraction the position of the 1st minima is given as dsinS = tzA,

, where sinS::::- tanS =fl D =..E.... andn = I 2 2D de =A

then,

2D h A=mv

.•• •(1)

... (ii)

Using eqns. (i) and (ii), v=2hD mcd

2x6.3xlO- 34 xO.5 (9.1xI0-31 )(0.36xI0 ')(10 ') .

=

=1.92xI0 6 m1s 29. The energy required to excite the electron from n1 to n20rbit revolving round the nucleus with charge +Zeis given by E

... =Z2XI3{_1 _ _I ] electron volt nz -E", 22 n n 1

2

(a) Since 47.2 eV energy is required to excite the electron from nl orbit, therefore

47.2 = Z'

X31.t~ - 31,]

Z2 =47.2 x 36=25 or Z = 5 13.6x 5

=2 to n2 =3

474 •

GRB Undt~landing PhY1ics Optics and Modem Physics

=

(b) The energy required I0 excite . tee h . by celron fromi nl = 3 to" nz 4 orbit IS gIVen £4 - £3 = 25)( 13.6><

rli'J.. __4' 1 I

= 25)( 13.6 )( 7 = 16.53 cV

144

«C))~hc ~nergy required to remove the electron from the first Bchrorbit to infinity C(I IS gIven by Eo:! -£1

=13.6 XZ2[1~ - ~2 l = I3.6X25 ev

In o~cr to calculate 'the wavelength of radiation, we usc Bohr's frequency relation

hi = ~c = 13.6 x 25 x (1.6)( 10- 19 ) J 'I.

or

~

_

-

(6.6x 10- 34

»( (3xI0 8 ) = 36.5 A

13.6 x 25)«1.6)(10 19)

(d) KE = - Tolal energy = -(-13.6 x 5') x 1.6 x 1O- t9 J =5.44 x IO-

17

J

PE:: -2 x K.E = - 2 x 5.44 x 10- 17 = 10.88x to- 11 ) Angulnr momentum = "Will = hl2rr; = 1.05 x 10- 34 J·s

' (e) The radius '1 of the fi rst Bohr's orbit is given by 1O fz2 f} = £o 1 = 0.53 x lO- =O,106 x IO- lO m = O.106A rcme 2 Z 5 .: £oh2 = 0.53 x 10- 10 ) (

nme'

30. (a) We have rndius of nth orbit of a hydrogen atom as rn =

,,2,,2

2

4n: 2Kze m If el~ctron is replaced by a heavy particle of mass 208 times that of the el~ctron. then the radius is given as 112,,2 n2 h 2 .. .(i)

r,

=4.'K(3).'(208m) =2496.'Kze'm

Here, we have not used reduced mass because it is given that mass of nucleus is assumed to be infinite and here we take z = 3. (b) The radius offitst Bohr orbit is given as ~ =

h' --:;"---0;,2

41t2 Ke m

475

Particle and Wave Nature of Mauer '·

From equation (n, we have n2,,2 ,,2 = 2496n 2Kc 2 111 4n 2Ke 2m or 112 ::: 624 or n~25 (c) Rydberg constant for hydrogenic atom is 2 2 4 R :::2tt K e m

ell 3 Now, when electron is replaced by ll-mcson, Rydberg constant will charge as 2 3 4 R' =2n K e (208m) chl t or =: 208R =: 208 x 1967800 m Now, the wavelength A. of the radiation emitted when Il-meson makes a transition fromn2 =3 ton=lis

I]

1 = R'Z'(_ I2 __ A

0' 0'

nI

1 = R'(3{ I..

112 ,

1

I' _3'

']0'1I.. ~ 8R'

I.. =_ 1 8R' A=

I m- 1 8 x 208 x 10967800

0'

I.. = 0.548 A Let K be the kinetic energy of the moving hydrogen atom and K' the kinetic 31. energy of combined mass after collision.

t.E:10.2 eV n=1

From conservation oflinear momentu",inc,'=,-, p = p' or= ,J2Km = j2K'(2m)

0'

K =2K'

... (i)

476

CRB Undentanding Physic. Optics and Modem Physia

From conservation of energy, K = K'+tJ.E

... (ii)

Solving equations (i) and (ii). we get /j£ = K. 2 Now, minimum value of I!E for hydrogen atom is 10,2 eV. or liE ~ 10.2 eV ..

~>1O.2 K <e. 20.4 tV

Therefore. the minimum kinetic energy of moving hydrogen is 20.4 eV. 32. Suppose the neutron and the hydrogen alom move at speeds 1'1 and \/2 after the collision. The collision will be inelastic if a part of the kinelic energy is used to excite the atom. Suppose an energy l!.£ is used in this way. Using conservation of

linear momentum and energy. mil = mill + mill

!mu 2 = !n/v 2, +!mv,'

and

, , ,

2

2

From cqn. (i).

II

From eqn. (ii),

v

2

'"

2

. .. (i)

+M:

.. ,(ii)

v t +11 2 +2vlv2.

=v~ +ui~ m

Thus,

Hence, As (VI -v2) must be real. v2_4M~O

m

!mv 2 >2U

or

2

The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited slale is 10.2 cV. Thus, the minimum kinetic energy of the neulron needed for an inelastic collision is !mu2 . = 2xlO.2eV-20.4eV

2

~"

33. Let mass of neutron"" m and mass of deuterium - 2m. Initial kinetic energy of neutron = Ko · Let after first collision kinetic energies of neutron and deuterium be K I and K 2· Using conservation of linear momentum along direction of motion. mu= mUI +mu2

•

Particle and Wave Nature of Matter

•

477

As

~ Hence

U=~=f~' .J2mKo = .J2mK I + .Jr.4n-'''K~2

Velocity of separation = velocity of approach As we know, 1'2 - vI

e=

III -"2

In clastic collision, e = I 1)2 -vI

Hence,

=

"I -112

J4mK2

J2mK,

J 2mKo

2m

m

m

Solving equations (i) and (ii), we get KI

... (ii)

=~o:

Loss in kinetic energy after first collision,

AX,

=Ko -XI

llK\ = Ko After second collision, llK2 =K, :. Total energy loss, llK = tJ(\ +~K2 +· ·· ·+M" 8 As. M = Ko+·····+-Ko

9"

=KO( 1+....+ 9~-1 )

M

M Ko

1- _1 8 9" =1_1-

=9 I-!

9"

9

Here, Ko

= IO'eY, M

(10' -O.025)eY

Ko-M _ O.025

1 9"

=

=

Ko

- 106

9"".4xl0' or Taking log on both sides and solving. we get n ". 8.

478

GRB Undentanding Physic. Optics and Modem Physics

34. (a) Since only six different transitions take place, the final state is " = 4. The energy levels of hydrogen atom are given by

En = _ 13.6 cV

II'

If 118 is the principal quantum number of the initially excited state B. then

E, -

Eo. = _13;6 _ (_ 13;6) 4

=

"B

IJ", ~l 1 B

16

E4 -Ens =2.7eV

27 =

13{,~ - i6]

which gives"8 = 2 (b) The transition energy is numerically equal to the ground state energy E\ of level A.

2.7eV~ - 2E,

16 £\ = - 14.4 eV Thus. the ionization energy of the given atom is 14.4 eV. (c) Maximum energy of the emitted photon is for the electron transition" = 4 to n=l,i.e.• E, E . 15 E E4 - E, = 16 - ,= -16 '

=~x(14.4)=13.5eV 16

'

Thus, the maximum energy of the emitted photon is 13.5 eV. Minimum energy of the emitted photon corresponds to the transition n = 4!o n = 3, i.e.,

E4 -E3 =E,_E, =-2E, 16 . 9

=

144

_2 x (- 14.4) = 0.7 eV 144

Partid~

and Wa.ve Na.lure ofMall~r

35, The electronic transitions in a hydrogen·like atom from a state 112 to n lower state 1/1 are given by

AE = 13.6Z'[J, - J,] "2

111

For the transition from a higher state 1/ to the lirst excited state "I = 2. the total energy released is 10.2 + 17.0 eV or 27.2 eV. Thus, for 6£ = 27.2 eV, "I = 2 and " 2 = II, we have

27.2=13.6Z'[L_I]

., .

n'

4

For the eventual transition to the second excited stale 1/1 ·= 3, the total energy released is (4.25 + 5.95) eV or 10.2 eV. Thus,

10.2= 13.6Z'[L _I ] 9

n'

,

Dividing the two equations. we get 27.2 = 9/1 - 36 10.2

4n 2 - 36

-,

Solving. we gel n 2 = 36 or II = 6 Substituting n = 6 in any of the above equations, we obtain Z2 =9 or Z=3 Thus, n=6andZ=3 36. Energy of a photon corresponding to transition from II = 5 (fourth excited state) to 11 = 4 (third excited state) is hv= 13.6(3)'[_1 __ I] =2.75eV 4 2 52

... (i)

Similarly, energy of a photon corresponding to transition from 11 = 4to 11 = 3 is

hv = 13.6(3)'[_12 _lc]=5.95CV 2 3

I,

I

... (ii)

4

from Einstein's photoelectric effect equation, hv = f + KEma., KEmax =eVs =hv-$ , The shorter wavelength corresponds to greater energy difference between energy levels involved in the transition. So, shorter wavelength photons are emined for transition 11 = 4 to 11 = 3. Thus, we have

3.95 = 5.95 -~ =>~ =2eV and for longer wavelength photon, eV, = 2.75 - 2 = 0.75 eV

480

GRB Undtntandins Ph)'lics Optics and Modern Physics

So, stopping potential = (O.7~

eV )=0.75 V

37. (a) Radius ofthc circle traced by a charged particle in a magnetic field is given by R = !E!!. = J!... = ..JfmK Bq

Hence,

Bq

Bq

K = (BqR) ' = [10- 3 x 1.6 x 10- '9 x (1 / 320)]'

2m 10- 17

2 x 9.1x )0-31

10- 17 -x eV=0.86eV 72.8 1.6 x 10 19

= -J = -

72.8

(b) Energy of the photon emitted when the electron makes a transition from n = 3 to n = 2 energy level is hv = £ ) -£2 = 13,J..1 _ _1 ] cV => 1.9 eV 32

122

From Einstein's photoelectric: effect equation. hv = + +KEmu: So, • = (1.9 - 0.86) eV = 1.04 eV (e) Wavelength of radiation is

A = he = (6.63 x 1O-"X3x 10')m =6542 A E 1.9 )( 1.6)(10 19

38. The energy of electron in hydrogen-like atoms in nth orbits is En

=

2 Z Rhc

n' We have Rhc = I rydberg. The ionization energy E~ - E, =Z2 Rhc =4 rydberg

...

Z, = 4 rydberg = 4 rydberg = 4 Rhc 1 rydberg

Z=2 (a) The energy required to excite the electron from n =1to n = 2 is given by

E, -E, = Z'RhC_( _Z'RhC)

2' =

I'

Z'RhC(I-~) = iZ'Rhc

=-34 x 4 rydberg =3 rydberg

•

Particle and Wave Nature of Malter

481

IfI.. is the wavelength of radiation emitted, then he· , he ;:- - 3db ry erg. I.e.,h = (3 rydberg) A =6.63 x lO- 34 x3xlO8

•

3x2.2 x lO 18 = 301.4xlO- lO m =301.4A (b) Radius of first Bohr orbit,

r,

"(c,,,o,,-h'--,I,;:ltm=e'...!.) Z

=

"

= Radius of first Bohr orbit of hydrogen = 5 x 10 - 11 Z

2

= 2.5 x to - 11 m 39. The energy of electron in nth orbit of hydrogen-like atoms is

En

Z2 Rhe

=

n'

,

Here, Z =3 and Rile = 13.6 eV (a) The energy required to excite the electron in Li++ from n = 1to n = 3 is

.

E, -E, =_Z'RhC_( Z'RhC) 32

}2

.

= Z2Rh{I-~) = ~Z2Rhc ,...

Eo

e, -,--*--+-E, __

~

____

~

.

. ____ __

= ~x(3}2 xI3.6eV=8xI3.6eV 9

=8x'13.6 x l :6 x lO- 19 J '

The wavelength of incident radiation required for excitation is given by

hC=E, -E, A

or

he

A= E3

6.63 x 10- 34 x3xl0 8 8x13.6xI.6xlO- 19

=~~~~~

EI

= 114.26x 1O-lOm = 114.26 A

.,

GRB Understanding Physics Optics and MOdem Ph)'lia

482

(b) The possible lines in the emission spectrum of this excited system arc

[n(n -1) = 3(\-1) = 3} 2

Three in number, represented in figure. 40. The energy ofan electron in

nIh

orbit ofhydrogcn atom is

=_Rile

Ell

n'

Rhc = 1.09737)( 10 7 x 6.626)( 10-34 )( 3)( 10 8 joule En = 1.09737)(10 7 x 6.626x 10-34 x3x10 8 cV

•

1.6 x 10-19

=13.6 eV From equation (il, - 3.4 eV .,, 13.6 eY

,

,,2 ,

n 2 :::: 13.6=4,i,e.,n = 2 3.4

According to Bohr's quantum condition, angular momentum, mur = n.E.. = 2x 6.626 x 10- 34 =2.1102 )(10-34 J5 2n:

2)(3.14

41. Lei ZA and ZB be the atomic numbers' ana mA and rns be die m'ass numbers of hydrogen-like atoms A and B. respectively.

Energy of 11th stale of hydrogen-like atom is En = _Z2Rhc=_Z2x13.6 eV

n2 n2 Energy emitted for I line of Balmer series for 8tom A. AEI =_;2 A

x

1_1 __I ) ev

13.

122 32

I

Energy emitted for I line of Balmer series for a~~m B!

l1E, = Z,

x

B

Given llEl -llE2

1-_I )CV . 13.1122 32

••

.

:..~.

=5.667 eV, therefore

S.667eV-(Z' -Z')X I3.6(_1 __ I )ev B If 22 32 : . Z, -Z' =S.667x36~3 B

A

13.6x5

Let u be the initial velocity of each from A and B and m the mass of target.

483

Partidt and Wave Nature of Matttr

Iful and U2 arc velocitics oftargct after collision with A and B, respectively, then according to principle of conservation of momentum for A,

mAli = MUI-mAIi .. . (ii) or Mu] = 2mA It ... (iii) Similarly, for B, MU 2 = 2mB II Given. MU2 = 2mu], therefore 2mB 11 = 2(2mA II) . .. (iv) This gives, mB = 2mt! As number of protons and neutrons in each of A and B is same separately, .. mB = 2ZB andmA = 2ZA Substituting this in eqn. (iv), we get ... (v) 2ZB =2(2ZA ),i.e, ZB =2Z,4 Solving eqos. (i) and (v), we get ZA = I and ZB = 2 i.e., atom A contains 2 protons and 2 ne!.ltrons. Hence, atom B is singly ionized helium. 42. (a) Energy of electron in hydrogen-like atom, 2

En = _ Z Rhc =_ 3.4 eV

n2 Kinetic energy of electron in hydrogcn-like .atom is equal to negative of total energy, i.e., Ek =-En =-(-3.4 eV) +3.4 eV (b) The de Broglie wavelength of eJectron. )..=E.= p

h

,j2mE,

,

,

6.63 x 10-34

= -'==~~~=7.C' ~{2x9.1xlO-3] x3.4x1.6xIO 19}

=6.66 A 43. For standing wave, the' separation 'between consecutive nodes (or antinodes) is ').)2.

. .

,I' .:

d = nA. , where n is integer 2 ..

For next node,

D' = (n + l)~

2

d'-d=~

2 ).. =2{d' - d)

.

~,

GRB UndtBtanding Physics Optics and Modem Physics

484

Here,d=2A,d' =2.5A ). = 2(2.5 A-2 A) = 1 A According to de Broglie hypothesis.

).=~= mtl

h .J2mE

•

or :. Energy of electrons,

E = -h'2m)" 2

Here

h = 6.63)( 10- )4 J.5 m = 9.1 x 1O- 31kg

A = IA = IO- I om. E=

(6.63)( 10-34 )2 ) 2><9.1><10 31 x(1O 10)2

,

= (6.63)· )( 10-17 joule =2.415>< 10- 17 J

2)( 9. t

=2.415)(10-

17

1.6)( 10- 19

eV:=o15! eV

For least value of d, n :: I. Therefore. from eqn. (i) d,.;, =~=!A - O.5A 2 2

44. The energy state of hydrogen-like atoms is given by En

=_lL

n'

where B = Z2 Rhc. The emission of six wavelengths implies

that the higber excited state corresponds to quantum number " = 4. As some wavelengths are lower and some wavelengths are higher, it implies that the initial excited stales are n = 2 Thus, the

, '. '"

principal quantum numbers of excited states are n = 2,12 = 3,n = 4.

hC = S( .,L_I

l.

2'

4'

)=> hc=3S ).

~A.

16

'. '.

no.

.

n-3

n.,

Particle and Wave Nature or Matttr

8

485

=16 x 6.62x 10- 34 x3 x 10 8 3<(1216<10- 10 )

=8.7 1x 10- 18 J If Z is atomic number of nucleus, Ihen 8 = Z2 Rhc. ..

Z2

=..!l..... =8.71 x 10- 18 J Rhc

13.6eV 18 = 8.7 1xlO=4 19 13.6x 1.6 x 10-

Z =2 The maximum wavelength of emitted radialion corresponds to transition n =4 to n = 3 and is given by

...!!£...- =B(_I - _I ) 32

Anux

42

AITI1J[ = 144hc =144 x 6.62 x 10- 34 x 3 x 108 78

7x8.7I x IO- 11i

= 4.69 x 1O- 7m The minimum wavelength of emitted radiation corresponds to transition n = 4 to n = 1 and is given by

...!!£...- =

A. min

B(L_I ) 12

42

16hc 16 x6.62 x I0- 34 x3xl08 =Amin 158 = 15 x8.7 Ixl0- 18 = 0.24 x 10-7m

=

45. If A is de Broglie wavelength, then for nth stalionary orbit2rvn = nA .

.

th.

orbIt rll =

where rll IS radIUS of n

2ft

eO h ' n 2

"",Ze

2

h'n') =nA. = -1 =-mm~Ze:.' 2 A 2eoh2 n «ronZe

,

-... (i)

0

.,

For Lyman series of hydrogen like atom

1=z2R(L_I) ).

}2

From eqns. (i) and (ii),

Z2 R(I-_1 n2

)= 2£mZe'h2n o

n2

... (ii)

GRB Undel'5landing Physics Optics and Modem Physics

486

Ryberg constant,

R 4 Z2 mc ( 1__ 1 2 8t~ch 3 tl (

me 8£2 c1l 3 o

)=mZc2l

2I;. oh n

1- _I ) = tl

4

=

2

=

4£oc" ne 2Z

4 x (8.85 x 10- " ) x (3 x 10 8 ) x (6.62 x 10- 34 ) n )« 1.6 x lO- 19 ) 2)( 11

2

n -1=25n

11=

25 n

=-

n 2 -25n - I = 0

or

25± ~(-2S)2

+4xlxl

2

25±J625

=~7!==

2

As negative n is inaccessible. so n =25.

46. The energy of the electron in nth orbit ofhydrogen-likc alom is given by

En

=_Z2Rhc

... (i)

,,'

(a) The energy required to excite the electron from second {n = 2)10 third (n =3) orbit is given b y liE = EJ -£2 ::::: _ Zl Rhc _ (_ Z 2RhC)

32

32

1):: 36~Z2 Rhe

= Z2 RhC( ! 4 9

Given, ionization energy ofhydragen at~m = Rile =13.6 eV andl1E = EJ -£2 =47.2eV Thus we have, 47.2 eV = 2. Z 2 x 13.6 eV

,

36

or Z2 = 36)( 47.2:::: 25 ::::) Z=S 5 13.6 (b) The energy required to excite the atom from third to fourth orbit is given by

I)

E4 -E, = _Z2RhC_ ('_ Z 2RhC) = Z'RhC(_ 1 __ 2 2 4 3 32 4 2

= 52 x (13 .6 eV) x2 = 16.53 eV 144

(e) The ionization energy of atom or the energy required to remove the electron from first Bohr orbit to 00 is '

Partide and Wave Nature

or Matter

"

.

Z2Rhc_(_Z2RhC)= Z2 R hc (.,)' I'

If A is the wavelength of corresponding electromagnetic radiation, then he = Z2 Rile ).

~e = 52 x 13.6 eV= 340 eV

i.e.,

A=

=6.63xI0-34x3xI08 340 eV 340x1.6xlO 19 he

= 36.056 xlO-lOm = 36.056 A (d) Kinetic energy of electron in first Bohr's orbit, EKt = EI = 340eV Potential energy of electron in first Bohr's orbit, VI =2E I =-2x340eV = -680eV Angular momentum of electron in first Bohr's orbit 34 = n.E... = I.!!.. = 6.63 x 1021t 21t 2n 34 J-s = 1.0546 x 10(e) Radius of first Bohr orbit for given atom

= (£.oh2

=[£.Oh2n;) runZe =

~rune2)

1:=1

Radius offirst Bohr orbit of hydrogen

Z 1I

47.

=5.3 x lO- m= 1.06xlO- 1l m 5 EK =0.73 eV, W = \.82 eV Ionization energy of H atom = 13.6 eV (a) hv = W + EK = \.82 eV + 0.73 eV =2.55 cV (b) The ~Iectronic energy levels ofH-atoms arc given by E =_Rhc=_13.6 eV n , ,

n

n

Forn=l,EJ =-13.6eV

Forn=2.E2 =-3.4eV Forn=3,E3 =-1.51 eV For n = 4,E4 = -0.85 eV Clearly, E4 - E, = - 0.85 eV -(- 3.4 eV) = 2.55 eV,

487

GRB Und~rstandin8 Physics Opria and Modem Physics

488

i.e., quantum numbers involved in the photons of energy 2.55 eV arc 2 and 4. The transition is specified by 111 = 4 -+ n2 = 2 (e) The angular momentum of electron in H atom J =

nlL

2.

Forn=4. JI =4.!!..=2h

2.

Forn

•

=2, J, =2!! =2h

2. • : . Change in angular momentum, t.I=J\-J, =2h_~=~

•••

(d) According to conservation of momentum. hv +m!J = O e v = _hv = _ 2.55 eV em (3 x lO')(1.67 x lO 27) 2.55)( 1.6)( 10-

V=

19

= -0.814 ms-1

3xI08)(1.61)(1O 27

:. Recoil speedofH atom = O.814ms- 1 48. (a) If f is the frequency of orbital motion, then the orbital magnetic dipole

moment of circulating electron M = iA =/eA According to Bohr's condition of stationary orbits.

. ,,(i)

MvR = n..!

2.

For ground state, n = I and area A = 1tR 2 Also, II = ro.> = R 2rrf . m(R 2nf)R =

I!!

2.

h / =-,:'''-: 2 4n mR2

" Fromeqn. (i),M=(

,h 41t mR2

L., y... ,

M=eh A _m 2 4nm (b) The direction of magnetic moment is aiong the normal to the plane of electron . , orbit, i.e., along;' '

Particle and Wa~ Nature or Matter ~

489

~

Anglc between M and B is 30°. :. Torque on orbiting electron, 't=MBsino=( eh lDsin30°

.

4~;V

· 't = -ehB I.e.,

newtron x meter 8... 49. (a) When charged particle moves perpendicular to a magnetic field, then magnetic field provides necessary centripetal force for circular path of radius r is givcn by mu'

-=quB r

mu = qBr As momentum, p = mu = ~r.2-m""E~K. where Exis kinetic energy, therefore

=>

,J2mEK = qBr

··.

EK = (qBr)' 2m 1 = {1.6XIO- • x (Jto)XIO-'}' J 2x9.lxlO-31

= 1.374xIO- 19 J = 1.374xlO-

19

1.6 x 10- 19

eV

=O.86eV (b) Energy of photon released due to transition from n = 3 to n = 2 in hydrogen atom, £ =U

=Rhe[_1 __ I )=(13.6,V)(_'--_I) 2 n2 1

22

n2 ,

3

= 1.88 eV Woridimction of metal W = £ = EK = 1.88 ~ O.86 = 1.02 eV (c) Wavelength of emitted radiation (photon) is given by U = he )"

),,_ he _ 6.62xI0-34 x3 x I0

-u-

1.88 x 1.6 x 10 " = 6.602 x 10-7m = 6602 A

8

Nuclear Physics

-.

•

Stildy_~ Points 5.1 5.2 5.3 ·

5.4 5.5 5.6 5.7 5.8

5.1

11

""t'

Nucleus

, Nature ofNuCtear Forces Binding Energy and Stability ofa Nucleus Law of Radioactive Decay , . _ . ... 'Successive Disintegration and Radioactive Equ.i1ibrium ' Secular and Transient Radioactive Equilibrium ~. :'\' a-decaf ~-dccay

5.9

y-decay

5.10 5.11 5.12 5.13 5.14

Nuclear Reaction ',f. -' Conservation Lawsin Nuclear.Reactlons ., Endoergic Reactions in Laboratory: Reaction Threshold . Nuclear F'ission . ~'. '., ~.; I -

",

Nuclear Fusion

.~

,

Nucleus

Rutherford discovered the nucleus as the spherical region of diameter which is equal to one tcn-thousandth the diameter of the atom, where almost all the mass of the atom is concentrated. A nucleus is comprised of protons and neutrons, called nucleons. Both particles have spin ~ and obeys Pauli's exclusion principle. Ifwe assume Z = no. of protons (atomic number), N = number of neutrons, the total number of nucleons is given as A := N + Z, is called mass num~r. Atoms of different mass number and same atomic number arc called isotopes, different atomic number and same mass number arc 31 called isobars. mp ~ mn = 1.67 x 10- kg (but mn > m p ). q p = e = 1.6 x 10- 19 C and neutron is electrically neutral.

Nudear Physics

491

Nuclear force: As the nucleus is imagined as the cluster of • protons and ncutro~s. we ~ust imagine a strong force of attraction ~ n~ p between any two nelghbounng nucleons to overcome the coulombic • n force of repulsion between the protons for stability of the nucleus. .fi . II d I r . . . Cluster of protons h· T IS oree IS ea e nue ear IOrcC which eXists strongly In the range of nd eu1 - 15 . a n rons 1 - 2 fim (I fm= 10 m). It IS not perfectly controlled.

Size: Due to strong nuclear force the nucleus at the interior of the nucleus almost touch each other. Hence. the density of the nucleus at its bulk is approximately a constant. As the surface · ··R··· nuc1:ons arc vibrating by the net inward pull of the nucleus, the 0.+-----'--''4 r density p of the nucleus decreases as r. Then. the average radius of R - roA'f3 the nucleus can be given as

I R = '!JAVJ,I where 'Y 0 is the emperical constant for all nuclei = 1.2 x 10- 15 m. Then, the average density ofa nucleus of mass number A is Mass P Volume (m )A = 4 P • where R = 1t)A II3 _1tR J

=

3

The average density of a nucleus is independent of its mass number.

Angular momentum: Inside a nucleus each nucleon spins (rotates) about its centroidal axis possessing a spin angular momentum. Further more the eM of each

z L

..,

.... 0"

/0.

,

:0 '.

G

,... . y

"

•• .P-.'...........

"

Nucleus

Nuclear spin

x nucleon moves inside the nucleus under the combined attractive force field of the other in a specified orbit possessing orbital angular momentum. The vector sum of spin and orbital angular momenta of all nucleons is defined as the angular momentum of the nucleus given as

492

GRB Undenlanding Physics Optics and Modem Physics L = ~jU + I)i~. wherej = nuclear spin quantum number (or nuclear spin) The component of L along z-axis is h

L:=

mj

2Tt

when In j = magnetic spin quantum number 1. Non-zero nuclear angular momentum of a nucleus is technically called nuclear spin even though it is the combination of orbital and spin angular momentum. 2. Non-zero nuclear spin does not mean that the nucleus is actually spinning like a top. 3. mj =

±(i -1). ± (j -

2) •............ 0 or ± ~ depending on whether j is integer or

half integer. 4. If A is even j is an integer, when A is odd j is a half integer.

5. L = 0 if both Nand ZaTe even.

Magnetic moment: Since the proton has +vc charge. its magnetic dipole moment J.l p is parallel to its angular momentum. Since the surface of the neutron has - ve charge and its interior has equal +ve charge, its magnetic dipole moment ~n is opposite to its angular momentum. Experimentally it is verified that,

~pl, ~2.79282'" mp

I

j.l n z -=:: 1.913 2'"

mp

.'

Hence, a nucleus possesses a magnetic dipole moment given as

~""'''"' = ~/(l + I)" Its component along z-axis is j.l nucleus Iz -=:: It!

5.2

Nature of Nuclear Forces

(I) Nuclear force is a short range force existing within 1-2 fm. (2) Nuclear force is same between any two nucleus, F independent of charge. (3) Nuclear force depends upon the relative orientation of the spins ofthe interacting nucleons. Nuclear force favours the opposite spin between a pair of same nucleons and vice-versa. (4) Nuclear force is not purely central as we can not conserve the orbital angular momentum of two nucleons about theirCM.

"

.

.. Nuclear Physics

493

(5) Nuclear force is repulsive for r < lb . which confirms that the nuclear volume is in compressible.

5.3

Binding Energy and Stability 01 a Nucleus

BE: The energy that binds the nucleus is called its BE. In other words, the minimum energy given to the nucleus to just

2

separate (free) the nucleons is called BE. BE is numerically equal to the internal energy oflhe nucleus. If the internal energy is -ve or BE is + ve, nucleus is stable. if BE is zero or - ve, the nucleus becomes unstable. A nucleus is more stable if the BE per nucleon is more. It Fnet = 0 on the nucleus 1 . "I"Ity 0 r a nue 1ellS IS " deel"dedby thc BE per Fne t"Oonthenucleus2 means that, t he staOI nucleon (not the total BE oflhe nucleus). The BE pcrnucleon.lhat is BElA can be given

as

Eb) BE l A (=-A =

0,

(2)

(I)

(3)

ai-AU] - 0 3

(4)

:(: - 1) A 413

- a4

0

(5)

(A - 2z)' A

(±,O)~ 3f4 A

where

al "" 14. 1 MeV, a2"" 13 MeV, Q3 "" 0.595 MeV, a4 = 19 MeV, and as = 33.5 MeV. The above formula is called scmi·empericlll BE fonnula. The expressions (1), (2), (3), (4) and (5) are volume energy, surface energy, Coulomb's energy, assymetry energy and pairing energy respectively. The graph E b / A waves A is given as following E';A

Volume energy

15+-------~~~~------

10

5

-5 -10

, ' ,, ''

50 1

Total energy

:60 100

150

200

250

ts*~~A Surface energy

Coulomb's energy

Taking all the effects into account, we can find that,

E: is maximum atA =60. This

means that from A = SO - 60, the nuclei are mostly stable (more tightly bounded). This tells us that,

In the fUSion of two lighter nuclei.to a middle - mass nuclei (A = 50 - 80). the BE Is released. In the fission of a heavy nucleus to smaller l'1JCIel, the BE Is also released. In both the cases the daughter nucleus are shIfted towards the higher stability zone.

GRB Understanding Physics Optics and Modem Physics'

494

Mass· Energy Equivalence : When a particle (system) of rest energy E, its mass is given ns

m~ss

rno gains

m=mo+~. c

where

~ is called man equivalent of energy E and hence III is called relativistic c-

mass. Then, when a system gains (or losses energy. its relativistic mass increases (or decreases) by a factor Elc 2 .

I

I

E tn-rno =-

or,

I

c'

or,

IE = (6m)C',1 where Ilm == mass defect (change in rest mass of a system) BE(E b ) and mass defect: Let the rest mass ofnucJeus be m,.,clem . The minimum energy required 10 just split the nucleus inlo the corresponding nucleus is called binding energy. According to mass-cnergy equi\'alence, the mass of the system after gaining the BE Eb is

II I I

,I

@ @

@

®

+Eb -

@

@'

I

® Free nocIeons

NUCleus

Eb

m = mnucleus + 2' C

where m "" sum aCrest mass of all nucleus :: l:.m j Eb

(Un, - m!Ulcus) == 2'" c The term in LHS is the difference in rest mass of products and reactants, called mass defect" !!.nI" or, IEb = (6m)c'i .. Hence, or,

The rest mass of an alom (or nudear) is atways ~ss than that of its condition.

'1_r' vl~

Find the expression for binding energy ofan atom.

Solution: The BE of a nucleus is BE = (6m)c'

,

= [Iml - mrucleus]c

I

I

I

• ,

Nuclear Physics . .

495

= [Nm" + 2mp

- mnuclear Jc2

= [ NIII" + 2mp -(matorn - Zm t )Jc2

:: [Nm" + Z(m p + 1111') - 1II'lIcm ]c2

+ Zmll - ma10m )c 2

or,

I BE = (Nm"

t!;"~ 2

PrOl'e that BE is approximately I % of the rest mass energy 0/ a

Ans.

1

nue/ells.

Solution:

=

BE I nucleofl'i

Rest mass energy I nucleons

8 MeV = 0.009

930MeV Proved.

:::: 1%. ti .. ' ? .

238 ~~ ~ Find the BE of 92 U. ...-....,

Solution: BE = (dm) 93l.5 MeV = ( Nm" + ZmH - matom}931 .5McV

- (146 x 1.00866 +92 x 1.007825 - 238 .0508) 931. 5 MeV

::::: 1801 MeV Stability : For the number of protons less than 10 (2 < 10) in the lighter nuclei, N = Z because the addition of move neutron proportionately

nullifies

Ans. z

"

the

coulombie repulsion by providing nuclear attraction.

When move neutrons arc added proportionately to fonn a heavier 10

nucleus, U coulorrb oc A2

U nuclear

IX

A

(Coulomb

but effect

,

-l~~:::'----------~ N 10

dominates the volume effect). In other .'; ',. ·1 S ~ 1.6 ' words, coulomb (tepulsion) energy \ ', dominates nuclear (negative) energy measuring the total energy of the atom to a positive blue. This makes 'the hucleus unstable. Further inclusion of neutron will again increase the internal energy leading to unstability of a nucleu"s"'.::-_ _ _ _ _ _-;;-_ _ _ _---. . ;, !!. 51.6 for Z e:: lOand !!. 2: I for Z < 10

(!!p)S

.

~.~

p

M

p

Law of Radioactive Decay In the binding energy graph. we observe that most of the nuclcides are unstablc. They become stable by releasing the binding energy and some rest masses in the fonn of a, ~ and 'Y~rays. Let us take I mg of 238 U consisting2.S x 10 t8 atoms. Not all atoms at 8 lime will emit a~rays .

GRB Undentanding Physics Optics and Modem Phyaia

496

Let in one second 12 nuclei of decay of 238 U to emit an a - particle we can not exactly predict which nucleus emit a-particle. Disintegration (ons'.n!: Each nucleus has equal probability (chance) ofemitting an a - particle during onc-second. that is , 12 • onc chance in 2 x 10 17 . 2.5x 10 18

Generally speaking, ira sample has N nuclei. dNnuclei decays in a time dt, in one second the probability of disintegration of each nucleus is dN

~

= A. (0 constant),

). = disintegration constant.

We give negative sign as N decreases with time to get,

I

dN = - AN dl

I

... (1)

dN= - A.dl N During time t, the total nuclei (not decayed) decreases from No to N i then,

or,

IN dN = - A.J'dl No N 0 In N I

or,

Zo = - ')..J ... (2)

From equation (I) we can say dN - - = AN dl

The quantity [ -

d:] gives the number of decays

per unit time and is caUed.the

activity ofttle sample. Thus, the activity of a radioactive sample is A = ). N. From . equation (2) •

.

. .. (3)

Unit 01 ActIvity The activity of a radioactive material is measured in terms of the disintegrations per unit time. Its Sf unit is bacquerel wbich is the same as 1 disintegration/second. It is denoted by the symbol 8q. However, the popular wit of activity is curie and Rutberford defined respectively as. I curie = 3.7 x 1010 disintegrations/s

I rutherford = 10 6 disintegrationsls The unit curie is represented by the symbol Ci. while unit rutherford is represented byRd.

497

Nuclear Ph)'5ics

The activity pcr unit mass is called specific activity. Half-life period : Hence, the number of undecayed ncuc1ei decreases exponentially. The time during which halfofthe undecayed (initial) nuclei get decayed is called half-life period TII2 . N

No

z

o

T,. N '" Notfxl

Putting N =

~0,

= TJ12 in the above expression, we have

I

TJ/2 = In2 = 0,693 Putting

,

/I.

A In2,

=-1'

10 t

A h

r-----~1~/2~f--, :. =

,N

e equatJOn N

=e-
0

GY'12

Mean-life: The mean-life Tmean ,generally called "life time" ofa nucleus is given

as Tm:an

I

=): = ~"'=---,­ (0 ,693/TJ/2 )

Of,

T..,n

I TJ/2 =I=0,693

Hence, mean-life is directly proportional to halflifc.

~'",, ~,-~ ~,,·u "-.-~ The mean lives of a radio-active sample for a and Il-decays are T, = 1620 yrsand T2 = 405 yrs respectively. Find the time after which 75% of the sample will be decayed. , ' fater w h'Ie h N N o = 100100 SolutIOn: The time _ 75 = 4 WI'II be

GRB Understanding Ph),sic!'Oplics and Modem Ph)'5ia

498

1 =2.303 10gNO A(1' 1' N =2.303 - - 1og 2

•

Aa \·

= 4.606 10g2. where

).01' =

Am. Aa + Ap

=_1 + _1 Ta T~

= (_1_ + ...L)ycnr -1 = 3.08 X 10- 3 (ycar) - J

1620 405 t = 450 .13 year

Then,

5.5

ADS.

Successive Disintegration and Radioactive Equilibrium Consider radioactive decay of 2§~ V (referred as A) to 2J)'Th (referred as B) which

funher decay 2J J4 POl (referred as C); such decays are called successive disintegrations. Here A will be called parent nucleus and B daughter nucleus of A. Any two adjacent nuclei may be considered as pan~nt and daughter nuclei. Suppose Nl and N2 are the numbers of two nuclei A and B alli me I, which undergo decay, with decay constants AI and A 2 respectively, according 10 the following reaction.

A~B~C Rate disintegration of A = d; I = -A[ N [. which is also rate of fonn3tion of B. · . . 0f B Rate dIsmtegratlon

dN, =- '11.2 N2 =dt

... (.) I ... (ii)

The rate ofincrease of the (stable) nuclei C is equal to the rate of decay of nuclei B. Thus.

tiN dt ) = ).,N,

Number of nuclei of A at time I, N I = N oe-

... (iii) .. .(iv)

).]t.

where No is the initial number of nuclei of A at t Substituting N 1 in Eq. (ii), we get dN 2 + N21.2 = A1Noe-).lt dl Multiplying both sides by ,/"2( .we get

=O.

dN, + N 211.2 ' ) eJ..~I 'N Oe -{).1-).2)1 ~ = 11.\ ( -dl-

(

499

Nuclear Physics

or d N 2 e h21 + N 2 A 2e 1'2 1= A t Noc-{ht - h2)1 dt ; (N 2C"21) = AI N oe-{J· I - A2)1

or

t

N,C

No

Integrating above equation. we get N., c )·21 =

AI N O c-(i' I- A2)1 +C

-

A2 AJ where C is constant of integration; at t = O. N 2 = 0, hence

Tim e -

.. . (v)

Hence,

This gives number of daughter atoms at any instant I. thus N2 depend not on ly on its ovm decay constant )'2 but also on parenf s decay constant A] . On substituting the value of N2 in Eq. (iii) and using condition thai at t = O,N 3 =Oweget N 3 = No [1 + A A\ e- A21 _ A.2 e- A1 1] 2 1 A2 - AI After time t. the ratio of activities is A2 N 2 = A2 (l _ e-(A21- )'I I» A]N ] A] A2

5.6

,I ,1

)

... (vii)

Secular and Transient Radioactive Equilibrium

(a) In a successive di sintegration when the parent has a very·vcry long half·life. a state is reached when daughter products arc fonned at the same rate as they decay. At this stage the proportions of the different radioactive atoms in the mixture is constant, i.c., do not change with ti me. In this situation the parent is said to be in ' secular radioactive equi librium ' with its daughter produclS. Suppose that the parenl atom I has a much longer half-life than any of the decay product (Tt » T2) . Then. I /...] ",, 0. and Al «)..2 , A] «A3 substituting A =:: 0 in Eq. (iv) we get N1 = No l ase- All "" l when At ... 0] SubstilUling AI «A2 in Eq. (v); we get

A, N (I - A,, ) N 2 = A2 0 - e -

=~Nt(l - e- h21) A,

I

.. .(vi)

GRB Understanding Physics Optics and Modem Physics

500

For t tending to infinity, e- A21 becomes negligibly small. Al Hence, N 2 =-N , A, or A\ N J = A2N2 Thus, after a sufficient time. the activities of parent and daughter become equal. This condition is known as "secular equilibrium".

For a series disintegration AINI = A2Nz = A) N) = .. . in lenn of half·lives

(T = JOg,f}

•

NI N2 N3 - = - =-= T,

T2

T)

Example of this equilibrium can be seen in the uranium series. Uranium has half-

life of 4.5 x 10 9 years which is very large so that it takes 50 X 10 6 year for its quantity in a rock to change by 1%. (b) The parent is longer-lived than the daughter 0.., < A2), but the half-life orthe pareO! is nol vcry long (Tt '"" TZ)' After t becomes sufficiently large. negligible compared with e the equation [from Eq.(v)]

N

)'l'

t

e- )' 2

I

becomes

so that the number of the daughter atoms is given by

AI N - AI' 2- A2 Aloe

Comparing it with Eq. (iv) and noting that AI < A2. we find that the daughter eventually decay with the same half· life as the parent. Since.

Noe- A I I =Nl . wehave

N,=,

Al

A2

oc

Nt N2

=

) N, ~1

AI 1..2-"1

After a sufficient time the ratio of parent atoms to daughter atoms becomes constant and both eventually decay. Thi s condition is called 'transient equilibrium'. (c) When the parent has a shorter half·life time than the daughter (TJ < T2 0r A I > "2), no state of equilibrium is attained . If initially we have only the parent atoms, then as the parent atoms decay, the daughter atoms increase in number, pass through a maximum and evenrually decay with their own half·life.

•

501

Nuclear Physics

5.7

a - Decay

An alpha particle is a nucleus of a helium atom consisting of 2 neutron and 2 protons having zero spin. possessing extra-ordinary binding energy. The large unstable nuclei emit a -particles decreasing its mass by 4 ami change by 2, moving loser to the most stable region to the Scgre's chart. This is called a-decay. Let a stating parent nucleus X emit an a- particle to get decomposed to a daughter nucleus y. A Z

X -)

A- 4 Y Z- 2

+a

Then, the energy of reaction is

m.

o --_. Vr-O mr

m"

,

o-v"

Q={m,.r - (mr+ma)} c· It must be positive for spontaneous a.-particle emission.

a.- particle decay is possible if the rest mass of the original (parent) nucleus is grealer than Ihe sum 01 the rest masses 01 a-particle and the daughter nucleus; Q> 0.0decay occurs if Ihe nucleus is 100 heavy 10 be slable.

To conserve linear momentum, the total momentum oflhe recoiled nucleus "( and the alpha particle is zero. or

Po = P"I

Then, the energy of reaction is Q=Ka +Ky p,2 p,2

=...1L+_Y_

2nla

2my

P~ (1+""' ) my

2ma 2

where

Po - K

2ma -

a

Hence, the (X- particles come out from the nucleus with a KE

where A;;;: mass number of the parent nucleus Since Q is same for a particular nucleus (element) aU a-particles coming out from same sample must possess same kinetic energy.

502

GRB Understanding Physics Optics and Modem Physics

The maximum K£ carried by a·particlc is Ka:1nw:= Q = {IIIX - (1111 + l1Ia)} X 931.5 MeV = {238.0508 - (234.0436 + 4.0026)} 931.5 MeV =4.28 MeV For 238 U. Q =4.1 MeV . Hence maximum KE oCa-particles is 4.1 MeV whereas

,

the potential barrier of the nucleus lies in the mnge of25 - 30 MeV. U(r)

•

30 MeV ____ .,__

5 MeV --- - ----------~~_ _

-t--+---------------~,

I

• -40 MeV 1---'

1

Potential energy (U ru:1ar + UQl'IUI.."m) v6rses distance for an alpha particle inside a nucleus

Then. how can an a-particle escape from the strong pOIcmial barrier of the nucleus? We can not answer it by

classical physics. We have to resen to quantum mechanics. According to the concept of dc-Broglie 's matter wnvc any

material particle can be imagined as a wave whose wave function 'I'

Inside the nucleus

!

,,~_

OulSi the nudeus

~-

(probabilities of finding the panicle) is V not zero anywhere. In other words, wave L._ _ _ _~ function the a-particle inside and The probability of finding a-particle trapped in outside the nucleus is nOI zero. It means the (potential well) nucleus outside the nucleus is not lero. that an a-particle can have a finite chance (probability) of tunneling the nucleus. The probability increases rapidly with a little increase in energy of a-particle inside the nucleus. For this we can imagine that the escaped a-particle would have 38 rattled back and forth as a rale of 10 times for 2J I! U for its half live period of 9 TII2 = 10 yrs. before coming out or the nucleus.

f!r ..!!1!..~~ Ca/clI/ate the (a) ellergy released ill a-decay o/238 U (b) maximum KE 0/ the emitted a -particle. The atom 2 masses a/thorium. uranium and a-particle are 234.04364. 238.05084 and 4.0026).1 respectively. Solution : The reaction can be given as 238 U ~ 234 Th + a (X )

(y)

•

Nuclear Physics

503 •

(a) The energy of reaction is

• •

_t:.- • •

Q = [mx - (Illy + "Ia )]931.5 MeV;~: = [238 .0508 - (234.0438 + 4.0026)] x93 1.5 MeV = 4.28 MeV (b) The KE o f thea-panicle is , Ka =

1111

Q

+nlu 234 .0438 = 234.0438 +4.0026 ( 4 .28)MeV = 4.03 MeV

5.8

my

Ans.

Ans.

P-Decay

When there is an imbalance of number of neutrons and protons for stability of a nucleus, a Il-particle (fast moving electron or its antiparticle. that is, position e +) is emittcd from thc decaying nucleus along with a neutrino v or anti-neutrino V. This is called ~decay. ~plus decay: When the number of protons is more than the required for stability,

e

p+ -decay occurs. In positive fJ-decay, a prOlon becomes a neutron at the expense of the binding energy of the parent nucleus releasing 3 positron and a neulrino. p-----+ n +e+ + y Since. the mass o fa proton is lesser than that ofa neutron, Q is negalive. Hence a free proton can nOI emit a position. Then. the proton must lie inside the nucleus and usc the rest energy (BE) of the nucleus 10 release n positron. When a nucleus undergoes p+-decay the atomic (proton) number decre3ses by one but the 10lal numbers of nucleons remains constant . Let a neutral parent atom X undergo 11 + -decay to become a daughler nucleus Y. A A Y +e+ +v z X ~ Zl The energy of reaction is Qlc 2 = m(.x ) - (m (Y) + m,.}

= m( ~ X) - Z me - [m(1 _1Y)- (Z - 1)me + me] or

Q = [m(~ X) - m(~_1 Y) - 2me]c2

P+ -decay takes place if the mass of the parent neutral atom is alleast two electron mass largers than the final (daughter) atom. This occurs when ~ ratio is lesser than

p

theJ~ui(ed

for stability,

Jl-mlnus decay: When the number of neutrons are more thnn the required for stability. the 11- -decay occurs. In negative beta decay. a neutron becomes a proton releasing an electron along wilh an anti-neutrino v as the following scheme.

GRB Understnnding Physics OptiC! and Modem Physics

504

p+e- +v Since the energy of reaction Q(= mil - mp 111 ... ) > O. n free .nc~lron cnn also emit a beta-particle with TII2 16 mins and Tol' = 10 mins. Hnvcncs. mSlde a nucleus, 11~

-"1" -

=

we can write this reaction as A X A Z ~ Z+ l Y

+ e -+v

In -vc !3-decay, atomic 2 (proton) as number increases by onc as neutrons arc

transfonncd isoprotons where as the total nuclear number remains constant. The energy of reaction is Q = [m(X) - m(Y) - me]c 2 A = [melA X) - Z meJ - [m(l+1 Y) - (Z + l)meJc-'

= [111(1 X) - II/(~+l Y)]c 2

.

-ve p·decay takes place when the mass of the original nucleus atom is greater than that of the final alom . This occurs when!:! is greater than the required for stability. p

Electron - capture: As we learnt, a proton inside the nucleus can emit a positron in +p -decay at the expense of nuclear BE. Sometimes. the nucleus does not give required amount of energy to the proton to emit +{>-particles. In this case. the crazy ploton catches the K-shell electron s flying very close to the nucleus just like n lizard catches the orbital in sects. This is called K - electron capture. Ju st after capturing the orbital electron. the K-shell remains vacant. Then. an electron from higher orbits comes to the K-shcll radiating the difTerence in energy of these shells in the fonn of X-rays which arc characteri sed by the daughter clement (but not the parent one) as the atomic number decrr;-ases by one after the electron capture. The equalion of K-e1ectron capture is (;;ven by the following scheme. p+e- ----+ Il+V For instance,

K- electron capture occurs when the numbers of protons is greater than that is required for stability for heavy atoms. In heavy atoms, the K-electrons orbit very

close to the nucleus makes this phenomena more common than the other ~ (+ve and -vel decav. Inverse jl-decay : We have learnt that a proton becomes a neutron by electron capture and p+ emission, each time associated with a neutrino. If this happens. onc should expect a reverse of this. Since the absorption of anti-neutrino is equivalenllo emission of a neutrino, a proton can also be decayed to a neutron by absorbing an anti-neutron followed by the emission of a positron as per the following scheme.

p+v----+ ,,+e+ . ~imilnrly; a neul~on can be transfonncd to a proton by absorbing a neutrino emlttmg an electron gIven as

Nudrar Physics

505 II+V---+ p+e

For inslance, V + ~jCl---+ ~J Ar+e~ As we know the neutrinos are very rarely interact with matter. Hence the above reactions are extremely rare. You can call it neutrino.capturc! Neutrino and conscn'ation law's: Pauli theorietically proposed the existance of a particle called neutrino or its anti panicle anti·neutrino which rnean "little neutrol me" to sati sfy the conservation laws following the X perirental facts, in Il-dccay. Let a nucleus X disintegrate to Y via Il-decay.

X---+ y +c-

+v

or

if/P "

-->

Since the nucleus X is stationary Px = O. Then to conserve the linear momentum of the system Py -->

-->

...

Py + P c+ Pv = O. This means that the p. particle need not

':--0

e

0

y

move just opposite to the motion orthe recoiled nucleus }' evident from the experiment. Since the recoiled nucleus is much heavier thnn the 1\+ Pe+ Py ::::: Otosalisfy electron (or positron) if cassies negligible production of the conservation of linear energy ofrcaclion. Then, momentum

...

--

No. of electrons

-f--------------------+KE Since Ke + Kv :; Q emitted Il-particles can have variable KE

Ke +Kv =: Q Since the neutrino can earring variable energy, we can satisfy the experimental evidence of variable KE ofP - particles from a P - decay. Since the neutron and proton have spin halfn). each can not be decayed to yield two particles of half spin. Hence we need to accept the third particle (neutrino or anti·neutrino) of spin half to consetve the spin-angular momentum.

L, =L/ p--->~ e++v+n

GRB Understanding Physics Oplics and Modem Physics

506

"

~

p+v +e-

H)+D+D Forexample.the~-decayin i:C~

iiC+p- +v

~4 C has 8 neutron and 6 proton, hence spin is zero.

~4 C has 7 neutron and 7 proton, hence spin is ~ + ~ = 1. To causes the angular momentum we can sum up the spin of p- and

(- ~) +

v to

gel

(-D= - I. This gives us zero spin angular momentum ofRHS which is equal

to that in LHS.

Properties of a neutrino : Keeping the experimental evidence and the conservation laws, Pauli attribute the following properties to neutrino which were confirmed experimentally after its disearsy in laboratory. Charge - zero (to satisty the charge conservation) 2.8 eV mass :::: zero (mv « me = , to conserve energy) cspin =! ( to conserve spin angular momentum) 2 speed :::: c (to acertain no rest mass for mass conservation) Interaction -+ very weak (as it is non-electromagnetic) Nature -+ non-electromagnetic Penetrating power - Highest (As it is non-electromagnetic if can panetrate 100 light year thickness of lead or iron without any noticable intereaction) Possible interaction - Only it interacts with matter via inverse j3-dccay which is extremely rare In + va or - ve p-decay. positrons and electrons are formed inside the nucleus at the expense of nuclear binding energy or rest mass energy. Hence they are not planetry (orbital) electrons. In p-decay aU conservation lawes hold good by the introduction of neutrino and its anti-particle. In p-decay neutron prolon ratio is shifted towards the most stable zone.

507

Nuclear Physics

~u",t!..t~~ Write the decay equations and expressions/or the disintegration energy Q of the follolVing decay: decay. + clecay, electron capture.

p-

p

Solution: In all the cases discussed here, we neglect any ncutrino mass. 2 decay: M nllel (~X) = M nuc l (~+1 D) + lIIe +Q I c

P-

where nlnucl indicates the nuclear mass. In order to convert it to atomic masses we add Zmc to each side above.

mnucl(~X)+Zme

= mnucl

~+ID+(Z+l)me+Qlc2

Since, atomic binding energies arc less than nuclear binding energies, they arc

neglected on the two sides of the equation.

m(~X)= M(~+lD)+Qlc2

Q for 'his equation = [M(; X) - M(j., D)]c2 (P- decay)

p+ deca~' :

MmlCl<1X) = Mnucl(~_ID)+mr +Q l c

2

We again add Zme to each side, in order to convert to atomic masses.

M nud (ZA X) + Zme:::; Mnucl (ZA -I D) + (Z + 1)11I r +Q I c 2 In this case only (Z -I)m is needed for the daughter atomic mass, which gives us a remaining mass Of2l11 c.

M(~X)+ ~of( ~_ ID)+2me +Qlc 2 Q for thi s equation :::; [M(~ X) - (M Z~ID) -2me]c2

(Jl+ decay) 2

Electron capture: Mnucl(~X) =Z11I e = Mnucl(~ _ ID)+QI c We add (Z - l)me to each side above and obtain 2 M nucl X) + Zme = Mnucl (~-1 D) + (2 -l)me +Q f c

<1

In this case electron masses arc balanced. i.e., cancel out in equation, M(1 X) = M(1_,D)+Q I

c'

Q for this equation = (M<~ X) - M(~_l D)]c2

(Electron capture)

~~Mtll ?i Find whether alpha decay or any o/the beta decay are allowed/or

, 22A

c . Solution: Our first step will be to write the reaction, then to find the disintegration energy Q. If Q > 0, the decay is allowed. ~

Alpha decay:

226Ac-7222Fr+a 89 87

Q = [M(i;' Ac) - M(if Fr) - M(' He)]c2 = 5.50 MeV ( Alpha decay is allowed)

5.9 y-Decay When an excited nucleus reaches its ground state it radiates y-mys . This process is called y-ray.

GRB Undenlanding Physics Optics and Modern Physics

508

Like quantised atomic (electronic) energy states, a nucleus possesses quantised energy level s. Whcn the internal energy of a nucleus is greater than the minimum (ground state), it is said to be in excited state. A nueleus attains an excited state when it collides violently with other particles slIch as other nuclei, neutron etc. or \\';lil.il is just undergoes a - and P- decay. The excited daughter nucleu s reaches the ground state with 10 in 10- s radiating the ditTerence in energy between two states which is in the order of I MeV. Hence, the frequency of electromagnetic radiation is given as I =

il/~ which

falls with in thc y-ray spectrum. This process is called y-radiation which follows a - and

p - radiation.

y - ray emission is a two step process. The parent nucleus X Is converted to an

•

excited daughter nucleus Y after a or p - emission .

•

Then, the excited daughter nucleus Y comes to its ground state Y after radiating a "I-photon.

•

Y----4 y+y

Sometimes the emitted y-photon in y-decay knocks off an orbital electron. In cousquence, the electron leaves the atom receiving KE equals to the difference in energy of y-photon (nuclear excitation energy) and the atom binding energy of that electron. This phonomena is called interval photo-electric effect as the interval proton (y-ray) knows the orbital electrons. Then, 10 file Ihe vacany oflhc knocked electron, higher energy level electrons can dc-excite giving up the characteristic energy an X~ray photon.

~;;;;u" S·Di.~c!lss the possible decays between X Solution: Since zAX • A and Z+IY we can write X

)

•

~+/+e-

= ~2 Band Y

= ~2c.

+v

A

Z+IY +Y, )

y+~+v

The nucleus X emits a ~- particle followed by an anti-neutrino. This leaves the

•

daughter nucleus Y in excited state. The daughter nuclear lose its energy hi from excited state E2 to ground stale Eo to because neutral (ground state) nucleus Y . 6E = lif, where ilE = E2 - Eo

t

Nudear Physics

509

E , - - , - - - , - -______ X

-+__-'-______

E, _ _

y.

j@-v

e-(fl-ray)

E,

Y

Sometimes the parent nucleus X decays directly to ground state Eo losing the energy !1E = E I - Eo which is all used in the p. decay. Sometimes the nucleus emits p -ray to reach two possible energy stales having

•

energy £2 and £3. Then the excited daughter nucleus}' emits Y-photons of energy tJ.E I = E. 2 - Eo at a time to come to the ground state or emits two protons succcstively of energIes.

tlE2 = £ 3 - £2 and!lE 3 = E3 - Eo respectively.

-,-__-----------E,

--~-1--------E,

----------"----------------- E,

@_l!.E,

5.10 Nuclear Reactions The simplest nuclear reactions involve rearrangement of nucleons between different nuclei leading to nuclear transmutation. The number of protons and neutrons remains separately constant in such reactions. An example of such a reaction is a-decay: 92U238 ~ 2He

4

+ 9QTh 234 + Energy

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510

A second type of nuclear reaction is one in which a neutron gets converted to a proton and vice-versa. In this type of reaction, the total number of nucleons is maintained constant, e.g., in jl-dccay. 27

c0 oo

~

28 N·I

60 +

_ Ie 0 + Energy

The basic reaction in~dccay involves the decay ora neutron into a proton, electron and an antineutrino:

o1/ 1------7

- 0 E lP I + _Ie 0 + OVe + nergy Energy is released in all these reactions and hence these are spontaneous. Reactions in which energy has to be added arc not spontaneous; but can still be conducted in the laboratory. Reactions can therefore be classified as: (i) Exoergic. where the total kinetic energy oflhe reactants is less than that of the products. (ii) Endoergic, where the total kinetic energy of the reactants is greatcr"than that of the products. This seems to raise questions on energy conservation. It is observed that in exoergic reactions, the total mass of the reacting nuclei is greater than the mass of the products. Consider the nuclear reaction: A + B~ C + D+ Q(Energy) where Q is the excess kinetic energy of the products over the reactants, Q=(Kc +Ko)-(KA +K B ) ... (i) Q is positive for exoergic reactions, while it is negative for endoergic reactions . For the masses of the nuclei, the reverse is true: 11m = (m.4 +mo)- (me +mD) >0 Further, it is observed that, Q=l1mc 2 .. . (ii) where c is the velocity of light. This mass discrepancy and corresponding energy evolved are related by the equation, 2 6E=Q=l1mc,

leading one to suggest that mass is being converted to energy and vice-versa as given by Einstein's theory of relativity. • We can rewrite equation (ii),

Q = (n/A +mo

-me -mD)c2

••

'1~

Combining this with equation (i), we get,

... (iii) .,

~+~_~_~=~+~_~_~)c2

-me

2

or Ke +KD -KA -Ko = (rnA +mo -mD)c Defining as the total energy, with as the 'rest mass energy'. we get, EA +EO =Ec +ED

... (iv)

i,

Nuclear Physics

511

5.11 Conservation Laws In Nuclear Reactions •

The conservation laws, useful in the study of nuclear reactions, falls into two classes: those in volving discrete variables and those involving continuous variables. (A) Di screte Variables I. Conservation of charge. 2. Conservation of nucleon number (mass number). (B) Co ntinuous Variables I . Conservation of momentum. 2. Conservation of mass-energy (in view of the equivalence of mass and energy). We give a few illustrations of the application of these laws.

5.12 Endoerglc Reactions In Laboratory: Reaction thresholds Endoergic reactions are usually conducted in the laboratory by reacting a projectile nucleus with a static target nucleus. The projectile is a very light nucleus which is prepared by stripping the (corresponding) atom of its electrons and accelerating it through a very large potenlial difference so that it moves with high energy. The target is a heavy nucleus. which is prepared as a thin sheet: thin enough to eli minate multiple scattering and other interference. thick enough to pennit an observable rate of reaction. The reacti on products arc ana lysed by using specially dcsigned experimcntal apparatus with associated electronics. The energy momenlum, charge and nature of thc products are usually observed. Endoergic react ions, not being spontaneous, have to be studied in the laboratory. It is observed that in cndoergic reactions, the projectile has to have a certain minimum threshold energy above the Q-valuc (i n magnitude), for thc reacti on to begin . This is explained by considering the kinematics of the reaction. Consider the reaction: A +8 ~ C + D + Q(Encrgy), where Q < O. Suppose that the threshold energy for the reaction is E 1h , i.e., the reaction j ust proceeds in the laboratory if this energy is supplied to the projectile. Nuclear reactions almost always proceed through an intennediate state X, which is a compound nucleus consisting of all the nucleons within A and B. The compound nucleus then breaks up into the product nuclei, C and D. Projectile Target Intermediate Products A + B X C+D

-- --

o Kinetic energy Momentum ~2mA Elh

Ex

o

Conservation of momentum requ ire.;.s~.--00.J2mAEIh =.J2mxEx or nJ,fElh =mxEx

... (i)

',. :'- '

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512

The kinetic energy of X, Ex is necessary for momentum ,conservation. in the laboratory frame. The excess cn;rgy over Ex is used to dnve the reactIOn (at threshold): ... (ii) E,h -Ex =IQI Substituting for Ex from eqn. (i),

Eth(l - ~) mx = I QI mx E'h = -/Ilx -"-IQI I1lA

or Ax

Ax -AA

IQI, where the masses have been substituted by the mass numbers.

+AB IQI_ (I+AA) IQI E,,, -_AA AB As .

... (iii)

Eth can be measured directly from experiments.

This expression allows one to measure the Q~valuc of cndoergic reactions, especially the ones whose products arc short-lived. In case only one of the product nuclei is short-lived, this gives us a method to detennine the mass oftha! short-lived nucleus accurately.

5.13 Nuclear Fission We have seen that a heavy nucleus has larger rest mass energy than that of its two middle·weight fragments. It is thus energetically favourable for the heavy nucleus to break into two middle· weight nuclei. However, before finally breaking into two parts, the heavy nucleus has to undergo a distortion which gradually increases to break the nucleus. The situation is shown in the figure.

,

0. ",

i.

. '

.

Energy = £1

~ ~

Energy = £2 Energy = £3 £2 >£\ >E3 The rest mass energy £\ of the heavy nucleus is larger than the combined rest mass energy E3 of its fragments but the energy £2 in the intermediate state happens to be larger than £\. Thus, it is not simple for the heavy nucleus to break sponta~eously ..In fact, according to classical physic.s, the process is impossible unless energy IS supplied to the heavy nucleus to reach the mtermediate state. Once it reaches the intennediate state, i~ can break into two parts and release energy. But the amount Er EI has t~ be supphed to the heavy.nucleus so that it may reach the intermediate state. Left to Itself, the heavy nucleus Will not break. according to classical physics.

t

513

Nuclear Physics E,

-------Q'O

E3 _______ __ _____________ "__...Ll.O_________ ..

Initial

Intermediate

Energy released

Final

The world of subatomic particles is much different from that of our common day experience, According to quantum mechanics, if the final state has lesser energy than the initial energy, process will take place even if the intermedi ate stale has energy greater than the initial one and no energy is supplied externally (a s shown in figure), Such processes arc called harrier pelletratioll. The energy seems 10 be created out of nothing. a violation of energy conservation! BUllhis is a fact oflhe physics of small particles. The energy conservation in the usual sense may be violated for 'short times' , The amount of energy seems 10 be created and the time for which il is created are related through Heisenberg uncertainty relation, DE . DI » hl2p where" is the Planck 's constant. The breaking of a heavy nucleus into two or more fragment s of comparable masses. with the release of tremendous energy is called as nuclear fi ssion. The most 235 typical fi ssion reaction occurs when slow moving neutrons strike 92 U . The following nuclear reaction takes place. 92 U 235 + 0111--+ 56Bal41 + 36Kr92 +30,,1 + 200MeV

Chain Reaction: Shortly after nuclear fission was discovered, it was realized that, the fission neutrons can cause further fission of 235 U and a chain reaction can be maintained .

f

Neutron

" -, 'U 235j

~--\"--o F;ss;on

F;sslon fragment

,_1

'-;>.

IU"~

\__

fragment

" lU23S;

cI; '\, (f'" \\,

}\, ,/\,

, ' ' 235> ' ' ' 235> :U235j IU ' : u~ IU , " ~ - -" " ' --'" ~- -" " ~--'

[n practice only a proportion of the fission neutrons is ~vailable for ne~ .fissi~ns since, some are lost by escaping from the surface of the uramum before collldmg with another nucleus. The ratio of neutrons escaping to those causing fission decreases as

GRB Under..tandins Physics Optics and Modem Physics

514

the size oflhe piece ofuranium-235 increases and there is a critical size (about the size

of a cricket ball) which must be attained before a chain reaction can start. In the 'atom ic bomb' an increasing uncontrolled chain reaclion occurs in a very short time when IWO pieces ofuranium-2 35 arc rapidly brought logether to fonn a mass

greater than the critical size. Nuclear Reactors: In a nuclear reactor the chain reaction is steady and controlled so that on average only one neutron from each fission produces another fi ssion. The reaction rate is abjusted by inserting ncutron-absorbing rods of boron steel into the uranium 235 . Graphite core is used as a moderator to slow down the neutrons. Natural uranium contains o\'cr99% or 238 U and less than 1% of 2J5 U. The rormer captures the medium speed fi ssion neutrons without fi ssioning. It fissions with very fast neutrons. On the other hand 235 U (and plutonium-239) fi ssions with slow neutrons and thc job of modern tor is to slow down the fissi on neutrons very quickly so that most escape capture by 238 U and then cause the fi ssion of 235 U.

Graphite

co.. UranIum

,ods

----,,/~.....~~rl--

""

-t-

,"",,:J;~~T'--

Steet

Concrete shield

Boron steel control rods

Nuclear reac10r

A bombarding particle gives up most energy when it has an elastic collision with a particle of similar mass. For neutrons, hydrogen atoms would be most effective but they absorb the neutrons. But deuterium (in heavy water) and carbon (as graphite) are both suitable as moderator. To control the power level control rods arc used. These rods are made of materials such as cadmium. that arc very efficient in absorbing neutrons. The first nuclear reactor was built by Enrico Fermi and his team at the University of Chicago in 1942.

I

I I

Nud~ar

Ph}'1ics

515

~"'1!..k 91t is proposed to lise the IfIlciear fusion reaction: I H2 +

IH 2

= 2He4 ill

a /luclear MW ratmg. . If tJIe energy from abot'e reaction is used '.I . J 25 (>reactor, or200 Will a % efficienc}' in IIIe reaclor, IlOW many g ralllS of deuterium will be needed per day ? (The musses of I H2 alld 2He 4 are 2.0141 and 4.0026 amll respectively.) Solution: Let us first calculate the Q value of nuclear fusion. 2 Q =flmc =flm(93 I) MeV => Q = (2 x 2.0141- 4.0026) x931 MeV = 23.834 MeV = 23.834 x lO'eV Now efficiency of reactor is 25%. So, effective energy used = (25 / 100) x 23.834 x 10' x 1.6 x I 0- 19 J = 9.534 X 10- 13 J Now9.534 x IO- IJ J energy is released by fusion of2 deuterium. ~ (9.534 x 10- 13 ) 1 2 J/deuteriull1 is released. Requirement is 200 MW = 200 x l 0 6 J I s x 86400 for Iday. 6

::)no. of deuterium nuclei required = 200 x 10 x 86400 = 3.624 x 10- 25 9.534 x 10- 13

2 Number of deuterium nuclei = ~ x6 x 10 23 2

2' =f 6 x 10 23 => =2x3 .624 X I0 ' =12083 =>3624xI0 . 2x g "'3 . gm Id ay. 6xlO~;-_ r; --~ Q 92 U 235 Illldergoesjissiofl due to absoption of thermal neutrons (where

K.E.can be ignored) g;vblg the nuciei 39R 95 and 5.3 J 139, plus h,·o neutrons:

on' +

92U23 5~ 39R95 + 5/139 +20nl.

The kinetic energy of the incoming lIelilrOIl may be ignored. (a) Write down the intermediate nllclells (compound nucleus) formed in the reactioll. !fllle mass ofthis lIucleus;s 236.045611. lind itsfissiollthreshold is 5.3 MeV, what is the excitatioll energy of this Illicieus ? Is it possible to explain why fission occurs?

Below: m(n) = 1.00901/ m(U 235) = 235.04391/

m (R 9' ) = 94.9058 u m(l139) = 138.9061u 2

and II/ = 931.5 MeVIe .

.

(b) Find the energy released in the reaction. WhatfractIon ofthe mass of U COIll'erted to ellergy ?

235 . IS

GRB Understanding Physics Optics and Modem Physics

516

Solution:

236 (a) The intennediate excited nucleus fonned in the reaction is 92 U . 235 The total mass of the nuclei in the initial state is 111(92 U ) + men) = 236.0529 u and therefore the excitation energy of 92 U236 is 6.£' = /1mc' = (236.0529 - 236.0456) x931.5 MeV = 6.8 MeV

Ans.

This is greater that the fission threshold of 5.3 MeV and hence fission occurs. (b) The total mass of the final products is : . mfhul = 138.9061u+94.9058u+2xl.OO90u = 235.8299u The energy released in the reaction is : Q = (236.0529 - 235.8299) x 931.5 MeV = 0.223 x931.5 MeV = 207.7 MeV.

The fraction of the mass of U235 converted to energy is :

f = °i;~3 x 100% = 0.95%. ~~~u 1~ A neutron with a kinetic ellergy of T

ADS.

10 MeV activates a

=

nuclear

reaction 11+ 6C12~ 4Be9

+(1.

whose threshold ('I/ergy is Tlh=6.17 MeV. Find flw kiffetic effcrgy of the a~partic/es ifa~parljc/e I1WI'e al riglll allgle 10 the direction o/neutron. Solution: The Q~valuc of the reaction can be found from the threshold energy:

m(6C12) 12 12 Tth =-13x6.l7MeV = 5.695MeV ... (i) m(6C ) + m(ff) Let the kinetic energy of the a-particle be Ea,that of the products be Ef. Assuming the mass of the neutron to be mn,that of the a-particle to be maand the mass of Be as mae,we get,by applying conservation of momentum, Po

Q= -

a

Be Initial

Rnal

J2"1rJ. xEa =~2mBe(Ef -Ea)sinS From the equation of conservation of mass-energy, we get,

or

10-E[ =5.695 E[ = 4.305 MeV

Pea

... (ii)

517

N udea.r Ph~iC!\ Squaring and adding the pair of equations. we get, I 4 10. + Ea 9 = 4.305 - Ea

9

Solving, we get, Ans.

Ea =2.21 MeV Filld the (a) energy ofreactioll, (b) minimllm (threshold) KE of the proton reaction given as p+ 7Li~ 7Be +11.

10

initiate the fission

Solution: The energy of reaction is 2 (a) Q = [nip + mLi - (mac + Ill n )]c = [(1.0073 + 7 .0144) - (7.0147 + 1.0087)] x 1.6 x 10-

I

:::: - 2.5xlO- 13 J (b) The threshold energy is nlBe + mp

Kp

~

~

mu

27

x (3 X10')' Ans.

(-Q)

7.0147+1.0087 10- 1l 70 {-(-2.5x

»)

. 144 ;;>;3xI0- 13 J

Then

Kthreshcl ld =3 X 10-

13

J

Ans.

5.14 Nuclear Fusion When the light nuclei (A < 20) X and Y combine, a more tightly bound bigger nucleus Z is fanned. We call it nuclear fusion given as X+Y~Z

BElA

-+--~2~O--~OO~------------~A

The energy of reaction is Q = (mx +my - mz)c 2 (>O)

=BEz - (BEx +BEy),(>O)

GRB Understanding Ph~iC5 Optics and Modem Physics

518

Since, the sum of the rest masses "'x -+ my of the lighter nucl~i X a~d Yis.lFater than that of the heavier nucleus Z. that is mz. the energy of reaction Q I~ poslttve . ~n othcrwords. the difference in binding energy is positive. Hence. energy IS released In nuclear fusion In nuclear fusion of two lighter nuclei, the re-emiting heavier nucleus is sh.if~ed towards the more stable region of BE curve. For nuclear fusion, the colliSion between the nuclei is periectly inelastic.

,.

~~~~ 1~ (Fusion

Dellterons) (a) Find the barrier of coulombs potclltial betweell two de/Ilerolls. (b) Estimate Ihe minimum temperature required /or jusion 0/11":0 dellterons. (c) If we take thl! lunflelling of deuterOIlS through the coulomb potential. does the result ill (b) remain affected? Solution: (a) The coulomb neutral energy of the two deuterons is O/Iwo

U

= qlq2 • 47tto r

where ql = q2 = eand r = 10-

14

m.

9 (1.6XIO- 19 )2 or,

U =9x10

(10-

14

)

= 2.3xlO- 14 J Ans.

= O.14McV (b) The average thennal energy ora deuteron is

l V/!! = '2 KT The couldmb potential energy per deuteron is

, U 0.14 U =2"=""2 MeV = 0.07 McV. -= 1.1 XIO- 14 J To over come the potential barrier. U th > U'. or,

~ KT = 1.1 X 10- 14 J, where K = 1.38 X 10- 23 JIK

or,

Tmin =5.3x I0 K

8

Ans.

(c) Yes. when the tunnelling effect is taken into consideration the minimum 8

temperature becomes 4 x 10K.

.

Ans.

7 At the centre 01 the sum temperature is T ='1.5 x 10 K < Tmin for fusion of deuteron which corresponds to average energy Uav = 1.9 keV. However. the deuterons having peak energy Umax(> l.Icoulomb) can overcome the coulomb·potential bamer for fusion .

Nuclear Physics

519

~JI:!''''''r... 1~ (Thermo ~ ,,,,clcar fusion in SUit) Describe 'he mail! sOl/rce of energy prOdJ/Cliol! ill Slll/. Solution: Mainly the energy is produced in the sun by nuclear fu sion of highly energetic (thennal) protons. This is called thenno-nuclear fusion of two protons or proton-proton cycle. Two fast moving protons (I in 10 26 ) combine to form deuteron, a positron and a neutrino.

IH+ IH ---> 'H+e+ + v (Q=0.42McV) The positron gets nnitilatcd quickly with the true electrons in the care of the sun to produce two flashes of light (y - pholons)

e++e----+2y (Q=1.02McV) Then a deuteron collides with a proton within few seconds to fonna helium isotope J He and a y - photon. 'H+ IH---> J llc + y

(Q=5.49 McV)

Two isotope of He. that is. 3 He (T112 = lOS years) combine to form a more tightly, bound He nucleus (TIf2 = 109 years). 3 He + 3 Hc ~ 4 He + 'H+ lH (Q= 12.86 MeV) Combining the above equations,

14:H~

;He+6y + vl

The energy of reaction

.

Q={4m
MeVl"

= {(4.007825)- (4.002603)} 931.5 MeV = 26.7 MeV. In cach cycle of p - p fusion . y -photon and neutron have zero and negligble rest mass respectively. Hence, we can nOI consider their masses to calculate Q-value oflhe fusion. Since two particles of neutrino arc fonned in each p -p cycle. due to its high panetrating power, it escapes the sun without any interaction with mattcr canying the energy 0.5 MeV. As the neutrino docs not take part in increasing the internal energy of the sun, we have Q' = Q ~ 0.5 MeV = 26.7 - 0.5:: 26.2 MeV available inside the sun, which is called "heat of combustion" of nuclear fusion (burning) of hydrogen into helium.

ORB Under.ltanding Physics OpliC$ and Modem Physics

520

_

~ < ~p MISCELLANEOlJS EXAMPLES

[<

<~~

Example 1. Find the minimum kinetic energy orana-particle 10 cause the reaction 14 N (a. p)17 O. TIle masses of 14 N . 4 He. 1Hand 17 0 arc respectively 14.00307 u, 4.00260 u, 1.007 .83 u and 16.99913u,

Solution:

Since. the masses

3fC

given in atomic mass units. it is easiest to

proceed by finding the mass difference between reactants and products in the same units and then multiplying by 931 .5 MeV/u. Thus. we have

Q = ( 14.00307u + 4.00260u - 1.00783u -16.99913U>(931.5 M;V) "" - 1.20 MeV Q value is negative. It means renction is endothcnnic. So, the minimum kinetic energy ora-particle to initiate this reaction would be,

K.,;, =IQI(mu

m ,\ '

+1)

= (1.20) (4.00260 + 1)

1400)07

1.S4McV Ans. Example 2. Derive an expression for average life and find the ratio of rav and =

Ttnl f ·

Solution: lei dN = Numbers of nuclei dissociated in lime dt after a time I. Then, (dN)t = total life span o f all tiN nuclei because No - N nuclei have bet:n dissociated. Now, the totallifc of all nuclei is N, Total time = Jt dN

o =itdN . dt o dt d (Noe- " )dt = -I t-

o dt

The total time period = -N 0 AJI e - At dl

... (i)

Then, the average time period is T. v = Total time span :a Total no. of nuclei

" =_+.0__ -Nol./ te- dt

No

0'

Ans.

521

Nuclear Physics I

The ratio of Tav and T1/2 is, Tay = Thalf

i"

0.693

= 1.44

Ans.

'. l. Example 3. A radio nuc1eide ofhalflifeT is fonned in a reactor at a constant rate R. Find the time from the beginning of production after which the activity of the radio nuc1eidc will be equal 10 A. Solution: Let after a time I, the activity is A. The supply (production) rate is R. The dissociation rate is R'= -A N. Then the rale of change of the total number of radio-nuclcide is Rnc:1 = R+R' dN=R_l.N or. dt

or,

dN R -l.N

dt

or, or.

Then. the activity,

Ans.

or,

Example 4. A proton strikes a stationary lithium nucleus 7Li. It initiate a nucleus reaction causing a nucleus 7Be and a neutron. Find the minimum kinetic energy orthe proton. _ Solution: Let the KE orthe bombarding proton be 0 ~ @0 K. Conservation of momentum,

@

!

r

~

~2mpK i = PIk

•

I

... (i)

Conservation of energy. ...(ii) ~

Eliminating Pee from eqns. (i) and (ii)t

521

Nuclear Physics I

The ratio of Tall and TI I2 is,

=

Tav

Thalr

l0.693 = 1.44

Ans.

J Example 3. A radio nuclcide ofhalflifcT is fonned in a reactor at a constant rale R. Find the time from the beginning of production after which the activity oflhc radio nucleide will be equal to A. Solution: Let after a time I, the activity is A. The supply (production) rate is R. The dissotialion rate is R'= -).. N. Then the rate of change of the total number of radio·nucleide is Rnd = R+R' dN or. -=R-).N

dr

dN =dt R-).N

or.

At time t = O,N = O,AI time t, N==N 01.

-i

01.

1 dNAN =jdr oR

In(R -

0

AN) I:

N =~(I_e-).J)

01.

I I

).

Then, the activity.

[

r l

1

i

I

I

dN A = -=AN

dr

=).[~(I-e-")l

I II

=I

A=R(I - e-")

r=Lln(I_A)

01.

In 2

Ans,

R

Example 4. A prolOn strikes a stationary lithium nucleus 7Li. It initiate a nucleus reaction causing a nucleus 7 Be and a neutron. Find the minimum kinetic energy of the proton.

Solution : Let the KE of the bombarding proton be

-0 @

~

@0

K.

Conservation of momentum•

...

~2mpK i= PBc

Conservation of energy, ... (ii)

...

Eliminating p&from eqns. (i) and (ii),

GRB Understanding Physics Optics and Modem Physics

522

or,

where Q = [niB' + mil - (nip + niL; )le or

2

=- 1.64 MeV

K=1.9IMcV

Ans.

Example 5. A nucleus X-initially at rest, undergoes alpha-decay, according IfI the

•

equation, ~2X -+22~ Y +a (a) Find the value of A and z in the above process. (b) Thea-particle in the above process is found 10 move in a circular trock of radius 2 1.1 xlO m in a unifonn magnetic field of 3.0 xlOJT. Find the energy (in MeV) released during the process and binding energy of the parent nucleus X.

Given: my =228.03 amu, n'a

=4.003 amu. m(~n) = 1.009 amu,

md H)

=1.008

amu, I amu = 1.66 x 10-27 kg = 931.5 MeV / c2 • Solution! (a) The given cqualion is.

AX -+ 228: y +24 He A=228+4 = 232 92 =Z+2, :. Z =90 92

and (b)

l.lxl0 2 x2xJ.6xlO- 19 x3xlO 3 4.003x1.66 x IO 27 = 4.0x10 6 m I s From conservation of linear momentum, mava. = nl).l' y

=

vy

"'a va

(4.003)(4.0XI0 6 )

m),

(228.03)

=--=

=7.0xI0 4 m/s Therefore, energy released during the process I

1

1

=2:[n'a.va. +m}'vy]

Ans. ADS.

t

Nuclear Ph}'5ia

523

(I 66xI0- ") =. 13 [(4.003)(4.0 X 10')' +(228.03)(7.0xlO')') MeV (2xI.6 x I0- )

=0.34MeV =9~i~ amu = 0.OOO365amu Therefore, mass of

;~2 X = III Y + Ina +0.000365 = 232.033365u

mass defect.

Alii

Ans.

=92(1.008) + (232 - 92)(1.009) -232.033365

= 1.962635 amu Binding energy = 1.962635 x 931.5MeV = 1828.2 MeV

Ans.

Example 6. An excited 108 Ag nucleus emits a y. photon with energy of87 keY. It knocks a K·electron having binding energy of 26 keY. Find the speed of the photoelectron. Solution: The KE of the photoelectron is K = Ephoton - BEofthe electron =87 keV - 26keV = 6lkeV The resl mass energy of an electron is U= mec- =0.5 1IMeV Then. the lolal energy of the photoelectron is E(= mc l ) = 0.061 + 0.511 = 0.572 MeV

,

mo

Since

mo -E--;;;

U_

m=-2' I- ~

R" - c'

c'

or

1_

i. = (0.572) - '

. c2

0.511 v ::: 0.449 c

0'

Ans.

Example 7. An excited stationary nucleus X with excitation energy E emits a y·photon.lfthe mass ofthc nucleus is M. find the fraction of change (loss) in energy of the emitted y·photon. Assume X =191 Irand E = 129keV. Solution: Since the excited energy!1E = E. the energy ofy·photon can be E approximately (neglecting the react of X). Then the momentum of y-photon Pr = E . Conservation oflinear momentum:

c

Px or

=;(=

Px = -

c

EI c) ...(i)

The charge in energy of the y·photon due 10 the react of the nucleus X is equal to recoil KE of X.

524

GRB Understanding Physics Optics and Modem Physics

Then.

.. ,(ii)

Using eqos. (i) and (ii),

My = E' --

... (iii)

2Me' l?e excitation energy is Q = K"( + K X :::: K r

(-:Kx «K"f)

Q=E

or, Using eqns. (iii) and (iv),

.,,(iv)

6.Er

::::......!i.-

Er

2Mc 2

•

129xlO-3 = ------''-=-'-''-'-'',-oc-2x 191x(3x 10 8 )2 X 931.5

=4.03 x 10- 24

Ans. Example 8. In the chemical analysis ora rock the mass ratio of two radioactive isotopes is found to be 100: I. The mean lives of the two isotopes arc 4 x 10 9 years and 2 X10 9 years respectively. If it is assumed that at the time of fonnalion the atoms of

both the isotopes were in equal proprotionai, calculate the age orthc rock. Ratio orlbe atomic weights orlhe two isotopes is 1.02: 1. Solution: At the time of observation (t = f). nil = 100 (given) m, 1 AI

Further it is given that,

1.02

~ =-

A,

N=m

Number of atoms,

A N] m] Az 100 -=-x-= ... (i) N2 m2 AI 1.02 Let No be the number of atoms of both the isotopes at the time affonuation, then NI Noe-).\I = = /}'2- A\)t ... (ii)

Nz

Noe

A21

Equating (i) and (ii), we have

e

0,2-),.\)1_100

-1.02

or (A, - AI)t

=In

Substituting the values, we have t = 1.834 x 101o year

(100) -In(1.02)

Ans.

525

Nuclear Physics

Example 9. A ncutron col\idcs elastically with the stationary deuteron with a kinetic energy Ko. In consequence, the neulron is scattered by 90°, Find the kinetic

energy of neUlron and deuteron just aftcr the collision. Solution : MomcnlUm conservation yields. poi = Pill + Pd p 2 = pz + p 2 or

... (i)

dO'

Conservation of KE yields, pl p l p2

~=

';:1 + ;;"

0--' Po

...(ii)

Eliminating Pd from eqns. (i) and (ii),

Pg =[pi+PJ]+p; 2m

or or

2m

2M

P2Mm ; (~+.!)=p'0 (_12m2M __I ) , P"(M+m)= Po2m (M-m) 2m M M

and

K = M - mK = 2- I K =JK /I M + m 0 2+ 1 0 3 0 K d= Ko - K n

or

Kd = M+mKo = 2+1= j KO

or

2m

,

2( 1)

Ans.

2

Ans.

Enmple 10. Neon-23 decays in the following way 2J

23 loNe~1I

0

-

Na +_1e+v

Find the minimum and maximum kinetic energy that the beta particle L~ e) can

!

I

have. The atomic masses of 23 Ne and 23 Na arc 22.9945 u and 22.9898 u, respectively. Solution: Here, atomic masses arc given (not the nuclear masses), but still we can use them for calculating the mass defect because mass of electrons get cancelled both sides. Thus, ~m = (22 .9945 - 22.9898) = 0.0047u mass defect Q = (0.0047u) (931.5 MeV I u) = 4.4 MeV Hence, the energy of beta particles can range fTOmO to 4.4 MeV. Ans. Example 11. Natural uranium is a mixture of three isotopes ~r U. ~s U and ~~8 U with mass percentage 0.01%, 0.71% and 99.28% respectively. The half-life of three isotopes are 2.5 x lOS years, 7.1 X10 8 yean and 4.5 xlO 9years respectively.

Determine the share of radioactivity of each isotope into the lotal activity of the natural uranium.

GRB Undc:ntanding Physio Optio and Modem Physics

526

Let RJ, R2 and R3 be \he activities . . . 0 f U13' • U135 and U'"

· SoIUt Ion:

respectively. Total activity Share of U 234 ,

R=RI + R l +R 3

A,N,

R,

R = AI N I + A2N Z +A 3N 3

Let m be the total mass of natural uranium.

om

Then, nil = 100 nI

.

m,

11/2

0.71

= 100 11/

N, _

Now, N 1= MI '

99.28

and

m,

Ml

nl3

=100111

"" M,

N3 = -

and

where M I, M 2 and M 3 are atomic weights.

RI

Ii = nil I .-

M\ T\

(;Jil ml

+-

I

m]

.- + -

M2 T2

.-

I

M 3 T]

(0.0 II I00) x _ --'-.,--_ 234

2.5 x I 05 year

=~--~----------~~~=----------100)( I )+(0.711 100)( I ) ( 0.011 234 2.5 x lO s 235 7. 1x 10 8 +

(99.2238 81100)( 4.5 xI 10 9 )

= 0.648", 64.8% U2) 5 = 0.01 6% Similarly. share of U2)8 = 35. 184% andof Ans. Example 12. A proton is bombarded on a stationary lithium nucleus. As a result of the collision two a·particles arc produced. If the direction of motion of the a-particles with the initial direction of motion makes an angle cos -\ (1 / 4), find the kinetic energy of tbe suiking proton . Given binding energies per nucleon of Li 7and He 4 are 5.60 and 7.06 MeV respectively. (Assume mass of proton .. mass of neutron) Solution: Q value of the reaction is. Q =(2 x4 x 7.0 6-7 x 5.6)McV =17.28 MeV p

,,~-.

u'

a 'IiIt' "'____ a _____ _ a

•

Applying conservation of energy for collision, K p+ Q=2Ka ... (i) (Here Kp and Ka are the kinetic energies of proton and a-particle respectively) . From conservation of linear momentum,

527

Nuclear Physics

~2t11pK'J =~2n'o.Ka cos 9 2

Kp = 16Ka cos 9 = (16Ka)

... (ii)

ur

~ = Kp

.. .{iii)

Solving eqns. (i) and (iii) with Q = 17.28 MeV. We get. K p = 17.28 MeV.

Ans.

Example 13. Suppose a nucleus initially at rest undergoes a- decay according to equation.

2~iX~Y+a. At t = 0, the emitted a-particle enters in a region of space where a uniform -+

~

-t

~

magnetic field B = 80; and electric field E = Eo; exist. Thea-parcticle enters in the region with velocity

V = voj from x = O. At time 1 "" J3 x 101

"'0.£ sec, the particle

q.

0

was observed to have speed twice the initial speed vo. then find : (a) the velocity ora-particle at time t, (b) the initial velocity "0 of the a-particle, (c) the binding energy per nucleon of a-particle. (Given that. til (Y) = 221.03 u. m (a) = 4.003 u, m (n)= 1.000 u , m(p) = 1.008 u, Mass of a-particle "'0. = ~ x 10-

26

kg,

2 charge on a-particle qa = 3.2 x 10- 19 C and 1 u = 931MeV I e )

Solution:

(a) Magnetic force on a-particle, (at t = 0) -+

-t

~

-t

~

Fm =q{vxB)=q.[{voj)x{Boi)] = -lJavOBO k

Force due to electric field (at any time t) ~

~

.

y

F e =qE=qaEoi

Hence, the particle will move in a circular path in y -z plane due to magnetic field and at the same time it will move along x-direction. The resultant path is therefore, a helix. with increasing pitch. Hence, velocity o f particle at any time t can be written as, ~ £0 ). • • v = ~ t ; +vocos8j-vosin9 k

(q

I ,

I I

I

Here,

8 = tot = BOqa t "'a

,,,

'.

1------1I---z

Ans.

GRB Undentanding Physics Optics and Modem Physics

528

(b) Speed of particle at any time t would be, v=

Given \' = 2"0 at I

(q:otf +v~ =

(../3 x JO 7 )~ , we gCI qaEo

(2vo)' = (JixI0 7 )' +v~ Vo =10 7 m / s

Ans.

When an a-particle is emitted with velocity Vo from a stationary nucleus X, decay product (nucleus y) rcacoils. Hence, m yv)' = "'aV()

..

Vy

=nlo:vo =(4.003 )(107) my

221.03

=1.8lxlOSm /s

Total energy released during a-decay of nucleus X. £ = KE ofnucJeus Y+ KE of a-particle _I 2 I 2 - 2m ),,,y +2"'0.1'0 27 = l.66 x IO[(221.03) ( 1.81 x lO ')' +(4.003) (l 0 7)'J 2 x 1.6 x 10- 13 = 2.IIMeV

Hence,

mass lost duringa-decay = 92j:.~ u = 0.0023 u

Mass of nucleus X; Mass defect in nucleus X ,

..

mx = (m y + I7Ia + 0.0023)u = 225.0353 u

6.m = 92mp + (225 -92)m/l - mx = 1.898u

Binding energy per nucleon =1.898 ;;31 .5 MeV 2 = 7.86MeV

A D ••

•

Nuclear Physics

...... -. .......... ..... .... ..... ...... .. .

ASSIGNMENTS

rJ Conceptual Oue5tion5

....... -

529

.- ....... . -_ ......_. ...-._ .. .

J. Neutrons are electrically neutral but still possess a magnetic dipole moment. why? 2. What is thennal neutron?

3. Why neutrons arc used as bettcr projectile in nuclcar fission? 4. Aftcr neutron capture why should tbc binding energy of 238 U or U 235 increase and makl! il unstable?

5. What is annihilation of mailer? How much minimum energy a released while a

,

proton and an electron get activated? What is the energy of each gamma ·photon? 6. A neutral pion decays into two gamma photons as,

It° -) Y +"(_ 7. 8. 9. 10.

11. 12.

13. 14.

Why can't wc get a single photon? When a nucleus undergoes a charge so that it gets transfonned to another nucleus. What happens to its orbital electrons? In MRI why do we use prOlon spin rather than protons? When the ncutron number increases gradually what happens to the binding energy and bind ing energy per nucleus'! Stabi lity is characteri sed by Ebl A but not Eb, Why? What are the other factors influencing binding energy? Why can't the mass of a nuclcus is equal to sum of masses of all nucleus? What arc the si" known elements possess ing ("Ira-ordinary stability like ., He" ? What is the cause of this'! Why these value of Z corresponds to " Magic numbers"? Thc nuclear binding cnergy per nuclear does not val)' too much . Can we apply same concept for atomic electrons'! When an alpha and a !3-particle have same energy. The penetr.lting power of

P- -panicle is move, why?

I

I \

I

IS. In natural radioactivity, a heavy nucleus most likely emit an alpha-particle but not smaller particle like protons, neutrons etc, why? 16. Can the chemical composition, external conditions affect a,p, y decays and electron-capture? 17. What is internal photo-electric effect? 18. Can we observe emission of X-my following a nuclear reaction? 19. In each second, the number of nuclei decaying is negligibly small compound to the total number to nuclei of an element. What does it signify physically? 20. In fusion reaction, energies libemted. why is so?

aRB Und~ntandin8 Ph),!ics Optics and Modem Physics

530

21. In fission reaction, why should we expect buge relcase of energy? 22. What arc energy of reaction Q. energy of excitation, thrusold kinetic energy and binding energy? Explain. Establish the relation between them. 23. What is the meani ng of I amu and its encrgy equivolume. 24. In nuclear reaction mass is convened into energy, comment. 25. In a-decay, all a-particle have same KE whereas in ~decay the ~par1icles have different KE. 26. How does the neutrino hypothons can help to conserve energy and momentum in r>decay? 1

rJ Multiple Choice Questions (A) Only One Choice ;s Correct

Level·1 1. An clement A decays into an element C by a two-step process:

A -4 B+Hej andB ~ C+2e~1 Then: (a) A and C arc isotopes (b) A and C are isobars (d) A and B are isobars (c) Band C are isotopes 2. Consider t\vo arbitrary decay equations and mark the correct alternative(s) given below: ( i) 230 U ~ 11 + 229 U 92

92

U ~ P + 229 Pa ( ii) 230 92 91

[Given: M(2~f U) = 230D33927u,

M(2t: U) = 229.o3349 u , mn = 1.008665 u, M(2Jr Pol) = 229D32089 u . mp = 1.007825 u, I a.m.u. = 931.5 MeV.] (a) Only decay (i) is possible

(b) On ly decay (ii) is possible (e) Both the decays arc possible (d) Neither of the two decays is possible

3. In a sample of rock. the ratio of 206 Pbto 238U nuclei is found to be 0.5. The age or thc rock is: (a) 2.25 x 10' year

(b) 4.5 x 10 9 In 3 year

,I

Nuclear Physics

(e) 4.5 x 10 9

531

In

(l)

2 year

In2

(d) 2.25 x 10 9 ln

(D

year

4. Let EI and E2 be the binding energies of two nuclei A and B. It is observed that two nuclei of A combine together to fonn a Bnucleus. This observation is correct only if:

5.

6.

7.

8.

(.)E1>E, (b)E,>EI (c) E2 > 2EI (d) nothing can be said A radioactive sample decays by 63% of its initial value in lOs. It would have decayed by 50% of its initial value in: (a)7 s (b) 14 s (e) 0.7 s (d) 1.4 s Which of the following statements is incorrect for nuclear forces? (a) These arc strongest in magnitude (b) They arc change independent (c) They arc effective only for short ranges (d) They result from interaction of every nucleon with the nearest limited number of nucleons In an a-decay, the kinetic energy of a-particle is 48 MeV and Q value of the reaction is 50 MeV. The mass number of the mother nucleus is: (assume that daughter nucleus is in ground state) (a) 96 (b) 100 (c) 104 (d) none of these A radioactive nucleus 'X' decays to a stable nucleus' Y'. Then, time graph of rate of fonnation of' Y' against time 't' will be:

" (a)

"1----

t

(b)

t

-I

-I

"1----

" (e)

t

(d)

-I

t -

1

GRB Understanding PhysiC5 Optics and Modem Physics

532 9. A I. )8 n )C T =0 No 0 0 T N, N, N3

The ratio of N 1to N 2 when N 2 is maximum is: (a) at no time this is possible (b) 2

(e) 1/2

(d) In2

2

10. The mcan·lifc time ora radionuclidc. ifits activity decreases by 4% for every I h. would be : [product is non-radioactive. i.e .• stable] (0) 25 h (b) 1.042 h (e)2h (d)30h 11. On an average, a neutron loses half of its energy per collision with a quasi-free photon. To reduce a 2 MeV neutron to a thcm131 neutron having energy 0.04 eV, the number of collisions required is nearly: (0) 50 (b) 52 (e) 26 (d) 15 12. If the Q value of an cndothennic reaction is 11.32 MeV, then the minimum energy of the reactant nuclei to call)'out the reaction is: (in laboratory frame of reference) (n) 11.32 MeV (c) grealer than 11.3 MeV

(b) less Ihan 11.3 MeV (d) exactly equal 10 15.4 MeV

13. 1.00 kg of 235 U undergoes fi ssion process. If energy released per event is 200 MeV, then the total energy released is: (o)5.12xI0" MeV (b) 6.02 x 1023 MeV (e)5 . 12 x IO ' 6 MeV (d) 6.02 X 10'6 MeV 14. Mark out the incorrcct statement: (n) A free ncutron can transform itself into proton (b) A free proton can transform ilselfinto neutron (c) In beta minus decay, the electron originates from nucleus (d) All of the above 15. A radioactive substance is being produced at a constant rate of I s-I. After what time will the number of radioactive nuclei become l00? Initially, there were no nuclei present. . (0) Is

1

I (b)ln(2)s

(e) In(2) S (d) 2 S 16. 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 Ihis constant is: XY (a) In (2) (b)XY

•

t

I 1

Nud~ar

533

Physia

(e) (XY) In (2)

(d) X

Y

]7. Half~life of a rad ioactive substance A is two times the half~lifc of anothcr radioactive substance B. Initially. the number of A and Bare N A and N 8. respectivcly. After three ha lf~livcs of A, numberofnuclci of both are equal. Then, the ratio N If I N B is: (a) 1/4 (b) 1/8 (e) 1/3 (d) 1/6 18. If 92 U 238 changes 10 85 At 210 by a series oro: and ~ decays. the number ofa and

{l- decays undergone is: (a)7 and 5 (b)7 and 7 (c)5and7 (d)7and9 19. Thc activity ora radioactive substance is RI al time Iland R2at time 12 (> II)' Its decay constant is A. Thcn: (b) R2 = Rl e}.(t1-12) (a) RI/) (c)

RI -R2

= constant

(d) R2 = Rle}.(12 - 1J}

'2 - II

20. In the previous question. number of atoms decayed between time interval Iland 12are : In(2) ,R,) (a) T(R

(b)R,e- '"' - R,e- '"1

(e) A(R, - R,) 21. The ratio of molecular mass of two radi oactive substances is

~ and the ratio of

their decay constants iS~. Then. the ratio of their initial activity per mole will be:

(a) 2 (e)

1 3

(b) ~

9

(d) 2

8

22. N Iatoms ofa radioactive clement emit N 2 beta particles per second. The decay constant of the element is (in s -I ):

N,

(a) N,

(b) N,

N,

(e) N ,ln (2) (d) N, ln (2) 23. The binding energies of nuclei X and Yare Eland £ 2. respectively. Two atoms of X fuse to g ive onc atom of Yand an energy Q is rel eased. Then :

(a) Q=2£, - £, (c)Q<2£ ,- £,

(b)Q=£,-2£, (d)Q>£, -2£,

534

GRB Undmtanding Physics Optics and Modrm Physics

24. The half-life of a radioactivc decay is x times its mean-life. The value of x is:

(a) 0.3010

(b) 0.6930

(c) 0.6020

(d) 0.6~lO

25. Neutron decay in the free space is given as follow s: 0 III -71Hl +-1 eO + [JThen, the parenthesis represents: (b) graviton (a) photon (c) neutrino (d) antincutrino 26. In the disintcgration series, 238

92 U

p- , Ay

Q)X-- z

the values of Z and A, respectively, will bc: (a) 92, 326 (b) 88. 230 (c) 90, 234 (d) 91, 234 27. The minimum frequency of a y-ray that causcs a deuteron to disintegrate into a proton and a neutron is :( md =2.0141 a.m.u.• It1p =1.0078 a.m.u., mn =1.0087 a.m.u.)

(a) 2.7 x \020 Hz (c) 10.8x10 20 Hz

(b) 5.4 x \020 Hz (d) 2 1.6 x 10 20 Hz

28. A stationary thoron nucleus (A = 200. Z = 90 ) emits an alpha particle with kinetic energy. What is the kinetic energy of the recoiling nucleus?

Ea

(a) \08 (c)

~~

(

b) E.

(d)

ilii

~;

29. If 10% ofa radioactive substance decays in every 5 years, then the percentage of the substance that will have decayed in 20 ycars will be: (a) 40% (b) 50"10 (c) 65.6% (d) 34.4% 30. The radioactivity ofa sample is Alat time (land A2at time 12' If the mean-life of the specimen is T, the number of atoms that have disintegrated in the time interval ofl2 -I], is:

(b) A, - Al

T (c) (A, - A,)T

I

(d) A,I, - A,', 31. Plutonium has atomic mass 21 0 and a decay constant equal t05 .8 x I 0- 8s -1. The number of a-particles emitted per second by I mg is: (Avogadro's constant = 6.0 x 10 23 )

I I

Nuclear Physics

535

(a)I.7x!O' (b)I.7xI0 11 (e)2.9xI0 11 (d)J.4XIO' 32. A radioactive nucleus can decay by two different processes. The mean value period for the first process is'1 and that for the second process is'2' The effective mean value period for the two processes is: (a) /I;'2 (e)

(b) II +12

.ftIi2

33. The nuclear radius ofa nucleus with nucleon number 16 is 3 x IO- lsm. Then, the nuclear radius ofa nucleus with nucleon number 128 is: (a) 3 x 1O - IS m (b) 1.5 xlO - 1Sm IS (c) 6 x 1O- m (d) 4.5 xlO- lsm 34. A star has 10 40 deuterons. It produces energy via the process:

\H2 +t H2 ~IH3

+P

I H2 + IH3~2H3 +n If the average power radiated by the star is 10 6 W, the deuteron supply of the star is exhausted in a time of the order of: [The masses of the nuclei arc as follows: M(H2) = 2.04 a.m.u., M(P) = 1.007 4

a.m.u., M(n) =a.m.u. and M(He ) =4.001 a.m.u.}

(a) JO's

(b) J08 s (d) 10 16 5

(c) 10[2 5 35. In fusion, the percentage of mass converted into energy is about: (b) 1% (a) 10% (c) 0.1% (d) 0.01% 36, In the nuclear reaction:

IH2+ IH2 ~ 2He3 +onl If the mass of the deuterium atom = 2.014741 a.m.ll., mass of 2 He 3atom = 3.016977 a.m.ll., and mass of neutron = 1.008987 a.m.u., then the Q value of the reaction is nearly: (a) 0.00352 MeV (b) 3.27 MeV (e) 0.82 MeV (d) 2.45 MeV 37. Consider the following reaction 4 JH2 + IH2 ~ 2He +Q If m(1 H2) = 2.0141 u: m(2 He 4) = 4.0024 u, the energy Q released (in MeV) in this fusion reaction is:

536

GRB UJldmtaJldiJlg Ph), ics Optics and Modem Physics

(a) 12 (b) 6 (c) 24 (d) 48 38. A graph is plotted between rate of disintegration and time of a. radioactive material. Which ortlle following correctly represents the form or tills plot?

dNidt

(a)

dNIdt

,

(b) t

dNidt

(c)

dN/dt

(d)

!

• 39. Rank the following nuclei in order from largest to smallest value of the binding t"s m. (") tOO Hg, () energy per nue Ieon: (") I 4H 2 c. ("") 11 "c 24 r, (""") III 62 IV 80 V "'cr 92 . (a) E(v) > E(iv) > E(iii) > £(ii) > £(i) (b) E(i) > £(ii) > £(iii) > E(iv) > £(v) (c) E(ii) > E(iii) > E(iv) > E(v) > E(i) (d) E(i) = E(ii) = E(iii) = E(iv) = £(v) 40. The half-life of radioactive radon is 3.&days. The timc at the end of which 1120 th of the radon sample will remain undecaycd is: (given log 10 e =0.4343)

(a) 3"8 day (b) 16"5 day (c) 33 day (d) 76 day 41. Beta rays emitted by a radioactive material arc: (a) electromagnetic radiations (b) the electrons orbiting around the nucleus (e) charged particles emitted by the nucleus (d) neutral particles

42. The equation 4: H ~ 2He 4 + 2i + 26 MeV represents: (a) ~ - decay (b) y - decay (c) fusion (d) fission 43. During a negative beta decay: (a) an atomic electron is ejected (b) an electron which is already present within the nucleus is ejected

,

Nuclear Physics

537

(c) a neutron in the nucleus decays emitting an electron (d) a part of the binding energy of the nucleus is convened into an electron 44. During a nuclear fusion reaction:

45.

(a) a heavy nuclells breaks into two fragments by itself (b) a light nucleus bombarded by thennal neutrons breaks up (c) a heavy nucleus bombarded by thennal neutrons breaks up (d) two light nuclei combine to give a heavier nucleus and possibly other products The decay constant ofa radioactive is A. The half~ life and mean-life of the sample are, respectively. given by: (a) ItA and (ln2)tA (b) (In2)A and ItA , (c) A(ln2 ) and ItA (d) )j(ln2) and ItA , Fast neutrons can easily be slow ~d down by: (a) the u sc of lead shielding (b) passing them through water (c) clastic collision with heavy nuclei (d) applying a strong electric field Consider a- panicles, ~ - particles and y - rays, each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are: (a)a.p.y (b)a,y , p (c) P. y,a (d) 'l.p.a Masses of two isobars 29Cu64and 30ZnMare 63.9298 u and 63 .9292 u, respectively. It can be concluded from these data that: (a) both the isobars are stable

.

46.

41.

48.

(b) Zn Mis radioactive, decaying to Cu64 through ~ - decay (c) Cu 64 is radioactive, decaying to Zn 64 through y - decay (d) Cu 64is radioactive, decaying to Zn 64through ~ - decay 49. Order of magnitude of density of uranium nucleus is: [mp = 1.67 x 10- 27 kg] i7 (a) lO 20 kgm - J (b)IO kgm- J (c) IOi4 kgm -J

(d)IOiikgm - J

50. 22Ne nucleus, after absorbing energy, decays into two a- particles and an unknown nucleus. The unknown nucleus is: (a) nitrogen (b) carbon (c) boron (d) oxygen 51. Binding energy per nucleon vs. mass number curve for nuclei is shown in figure. W. X. Y and Z are four nuclei indicated on the curve. The process that would release energy is :

538

GRB Understanding Physics OpfiC! and Modem Physics

y

8.5 ----------/-T'-~

~.~ ========-t ==~ 5.0

0

-------

/

0

w

Z

30

120

90

Mass number 01 nuclei

(a) Y -. 22 (e) W-.2Y

(b)W-.X~2

(d)X-.l'+Z 52. Which of the following is a COrTeet statement?

(a) Beta rays arc same as cathode rays (b) Gamma rays arC high·cnergy neutrons (c) Alpha particl es arc singly ionized helium atoms (d) Protons and neutrons have exactly the same mass 53. The electron cmined in beta radiation originates from: (a) inner orbits of atoms (b) free electrons exisling in nuclei

(e) decay of a neutron in a nucleus (d) photon escaping,from the nucleus

Level·2 I. A star initially has 10 40 deuterons. II produces energy via the processes ~ H +~H --+ + p and H H --+ ;He + n. If the average power radiated by

lHe

f +l

the star is 10 6 W. the deuteron supply of the star is exhausted in a lime of the order of: [Given: M(' H)=2.014 u.M(n)= 1.008 u. M(P)= 1.008 u and M(' He) =4.001

uJ 0010', 0010 ', (e) 10" , (d) 10", 2. A radioactive nucleus is being produced at a constant ralea per second. Its decay constant is A. If No arc the number of nuclei al lime I = O. then the maximum number of nuclei possible are:

a

(a) !! y

(b) N o -

(e) No

(d) - +No

y

y

a

, I

I

I

539

Nuclear Physics

3. A radioactive substance X decays into another radioactive substance Y. Initially. only X was present. Ax and A y are the disintegration constants of X and Y. N x and N y will be maximum when: (a)

Ny = A)' Nx - N y Ax A y

Nx

(b) Nx

N)

=

Ax Ax - A )'

(c)A y N )' =").x N x (d)AyNx =AxN y 4. There are two radio nuclei A and B. A is an alpha emitter and B a beta eminer. Their disintegration constants are in the ratio of I :2. What should be the ratio of number of atoms of A and B at any rime t so that probabilities of gening alpha and beta particles are same at that instant ? (a)2:1

(b)I:2 (d)e- I

(c)e

S. Stationary nucleus 238 Udecays by a emission generating a total kinetic energy T 23 8U ~ 234Th+4a

92

90

2

What is the kinetic energy of the a- particle? (a) Slightly less than TI2 (b) TI2 (c) Slightly less than T (d) Slightly greater than T 6. AS X 10-4 A photon produces an electron-positron pair in the vincinity ofa heavy nucleus. Rest cnergy of electron is 0.511 MeV. If they have the same kinetic energies. the energy of each particle is nearly; (b) 12 MeV (a) 1.2 MeV (e) 120 MeV (d) 1200 MeV 7. 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 the source is: (a) 6 h (b) 12 h (e) 24 h (d) 128 h 8. Half-lives of two radioactive substances A and B are, respectively, 20 min and 40 min. Initially. the samples of A and B have equal number of nuclei. After 80 min, the ratio of the remaining number of A and B nuclei, is: (a) 1:16 (b) 4:1 (e) 1:4 {d) 1:1 9. Assuming that about 20 MeV of energy is released per fusion reaction 4 IH2 + IH2 ~ onl + 2He

then the mass of 1H 2 consumed per day in a fusion reactor of power 1 megawatt will approximately be: (b) 0.1 g (a) 0.001 g (d) 1000 g (e) 10.0 g

GRB Understanding Physi~ OptiC5 and Modem Ph)'lics

540

236 10. Ifmass ofU 235 = 235.12142 a.m,ll., mass of U = 236.1205 a.m.ll. and mass of ncutron = 1.008665 n.m.ll., then the energy required to remove one neutron from

the nucleus U236 is nearly about: (a) 75 MeV (c) 1 eV

(b) 6.5 MeV

(d) zero

4 11. The binding energies per nucleon of deuteron (I H2) and helium ( 2 He ) atoms arc 1.1 MeV and 7 MeV. If two deuteron atoms react to fonn a single helium atom, then the energy released is:

(a) 13.9 MeV (b) 26.9 MeV (c) 23 .9 MeV (d) 19.2 MeV 12. Why is a He nucleus more stable than a j Li nucleus?

i

(a) The strong nuclear force is larger when the neutron to proton ratio is higher (b) The laws of nuclear physics forbid a nucleus from containing more protons than neutrons (e) Forces other than the strong nuclear force make the lithium nucleus less stable (d) None of the above 13. What is the power output of 92 U 2JS reactor if it takes 30 days to use up 2 kg of fuel and if each fission gives 185 MeY of usable energy? Avogadro's number 6.02 x 10 26 per kilomole. (a) 45 megawatt (b) 58.46 megawatt (e) 72 megawatt (d) 92 megawatt 14. A radioactive nuclide is produced at the constant rate of 11 per second (say. by bombarding a 1¥get with neutrons ). The expected number N of nuclei in existence t s after the number is N ois given by:

=

(a) N == Noe- i.J

(b) N ==!.! + Noe- 'N

(C)N == ~+(No - ~}-)J

(d)N=~+(No+~)e-'N

I.

15. A rndionuclide A[with decay constant )..[transform into a radionuclide A2with decay constant A 2. Assuming that at the initial moment the preparation contained only the radionuclide A! . then the lime interval after which the activity of the radionuclide A2 reaches its maximum value is: (a) In(A,

n.Jl

1.,-1.1

(b) In(AI IA.,)

1. 2 -1. 1

(c)ln(l._2-).,\) (d) none of these 16. A free nucleus of mass 2~ a.m.u. emits a ga~a photon. When initially at rest. the energy of the photon IS 7 MeV. The recOil energy of the nucleus in keY is:

(a) 1.1 keV (c) 3.3 keV

(b) 2.2 keV (d) 4.4 keV

(

f f

Nuclear Physics

541

17. The half-life of 13 11 is 8 days, Given a sample of 1311 allime t = 0, we ean assert that: (a) no nucleus will decay before t = 4 days (b) no nucleus will decay before 1 = 8 days (c) all nuclei will decay before 1 = 16days (d) a given nucleus may decay at any time aftcr t = 0 18. In hydrogen spectrum. the wavelength of Ha linc is 656 nm. whereas in the speclrum of a distant galaxy, Ho: line wavclength is 706 nm. ESlimated speed of the galaxy with respect to earth is: (a)2xI0 8 ms- 1 (b)2xI0 1 nu- 1 (e)2xI0 6 ms- 1

(d)2xlO s ms - 1

19. The half-life pcriod of n radioactive elemcnt X is same as the mean-life time of another radioactive clement r. Initially, both of them have the same number of atoms. Then: (a) X and Y have the same decay ratc initially (b) X and Y decay at the same ratc always (e) Y will decay at a faster ratc than X (d) X will decay at a faster rale than Y 20. Two radioactive matcrials X t8nd X z have decay constants lOA and A, respectively. lfinitially they have the same numberofnuclei. then the ratio ofthc numberofnuc1ei of X Ito that of X zwill be l ie after a time: 1 1 11 1 (b) iII (a) 10l. (c)lOl. (d) 9).

21. A radioactive samplc consists of two distinct species having equal number of atoms initially. The mean-life time of one species is t and that of the othcr is 51. The decay produclS in bolh cases arc stable. A plot is made ofthe total number of radioactive nuclei as a function of time. Which of the following figures best represenlS the form of this plot? N

N

(b)


,

t

N

N

(c)

(d)

542

GRB Undentanding Physics Optics and Modem Physic.

22. The half-life of 215 At is 1001-1s. The lime taken for the radioactive ofa sample of 21 SAt to deeay to 1/ 16th of its initial value is:

(b) 6.3~s 40~s (d) 3OO~s 23. Which of the following processes represents a 'Y-decay?

(a) (e)

400~s

(a) A Xz + 1' -+ A XZ_ 1 +a+b

(b) A Xl + 1"0 -+ A-J XZ -2 +c

(c) AXZ-4AXZ (d) A X z +c- I

+/

-4 A X Z _t

+g

24. For uranium nucleus, how does its mass vary with volume? (a) m oc: V (b) m oc: IIV

(c)m~Jfi

(d)m~V2

25. A nucleus with mass number 220 initially at rest emits an ex- particle. If the Q value of the reaction is 5.s MeV. calculate the kinetic energy of the a- particle. (a) 4.4 MeV (b) 5.4 MeV (e) 5.6 MeV (d) 6.5 MeV 26. A 280 day old radioactive substance shows an activity of6000 dps. 140 days later its activity becomes 3000 dps. What was its in it ial activity? (a) 20000 dps (b) 24000 dps (e) 120000 dps (d) 6000 dps 27. If a star can CODvert all the He nuclei completely into oxygen nuclei. the energy released per oxygen nuclei is: [Mass of He nucleus is 4.0026 a.m.u. and mass of oxygen nucleus iIhs.9994 a.m.u.] (a)7.6 MeV (h) 56.12 MeV (e) 10.24 MeV (d) 23.9 MeV

28. 2ii Ra is a radioactive substance having half-life of 4 days. Find the probability that a nucleus undergoes decay after two half-lives: I 3 001 OO~ ~)i

29. In the oplions given below, lei E denote the rest mass energy ofa nucleus and n a neutron. The correct option is: (a) E('Jt U) rel="nofollow"> E('ji I) + E(!lY) + 2E(n) (h) E('Jt U) <

E('ji I) + E(!l Y) +2E(n)

(d) E('Jt U) < £('nBa) + E(f. Kr) + 2E(n) (d) £('Jt U) =

£('i~ Sa) + £(f.Kr) + 2E(n)

Nuclear Physics

543

30. A radioactive samplc S I having an activity of5~Ci has t\\'icc the number of nuclei 3S another samplc S2which has an activity of1O~Ci. The half-lives of S]and S2can be: (3) 20 years and 5 ycars, respectively (b) 20 years and 10 years, respectively (c) 10 yC3rs each ' (d) 5 years each (8) More Than One Cholce/s Is/are Correct

1. Mark out [he (:orrect statement(s):

(a) Highcr binding energy per nucleon means the nucleus is more stable (b) If the binding energy ofnuclcus were zero, then it would spontaneously break apan

(c) Binding encrgy ofa nucleus C3n be negative (d) Binding energy of a nucleus is always positive 2. Mark out thc correct statements (5): (a) In alpha-decay, the ent:rgy released is shared between alpha particle and daughter nucleus in the fonn of kinetic energy and share of alpha particle is more than that of the daughter nucleus (b) In beta-decay, the energy released is in the form of kinetic energy of beta particles (e) In bcta-decay, the energy released is shared between electron and antineutrino (d) In gamma-decay, the energy released is in the form of cnergy carried by photons termed as gamma rays 3. During ~deca y (beta minus), the emission of antineutrino particle is supported by which of the following statements(s)? (a) Angular momentum conservation holds good in any nuclear reaction (b) Linear momentum conservation holds good in any nuclear reaction (c) The KE of emitted J).particle in varying continuously to a maximum value (d) None of the above 4. A nuclide A undergoesa-decay and anothernuciideB unde rgoes~decay. Then: (a) (b) (e) (d)

all the a.-panicles emitted by A will have almost the same speed the a-particles emitted by B will have widely different speeds all the 13 -particles emitted by B will have almost the same speed the ~ -particles emitted by B may have widely different speeds

S. In a nuclear reaclor: (a) the chain reaction is kept under control by rods of cadmium. which reduces the rate (b) the thick concrete shield is used to slow down the speed of fast neutrons

544

GRB Underllanding Physics Optics and Modem Physics (c) heavy water (or graphite) moderate the activity oflhc reactor

(d) out ofU 238 and U235natural uranium has less than 1% ofU 235 6. A radioactive sample has initial concentration N oofnucJci. Then : (a) the number of un decayed nuclei present in Ille sample decays exponentially with lime (b) the activity (R) of the sample at any instant is direclly proportional to the number of undecayed nuclei present in the sample at that lime (e) the number of decayed nuclei grows exponentially with time (d) the number of decayed nuclei grows linearly with lime 7. From the following equations. pick out the possible nuclear fusion reaction: (a) 6CIJ

+ \Ht ~

(b) 6C l2 l4 (e) 7C

+ JHI ~ 1C IJ +2 MeV

(d) 92 U

I

• I

6C l4 +4 .3 MeV

+ JHI -+ gOl 5 + 7.3 MeV

23S

+ 0,,1 ~ S4Xe I().I + 3SSr 94 +onl +onl + Y + 200 MeV

•I

. 8. Which ofthe follO\ving statement(s) is (are) correct? (Il) The rest mass ofll stable nucleus is grcaler than the sum of the rest masses of its separated nucleons (b) The rest mass ofa stable nucleus is greater than the sum of the rest maises of ils separated nucleons (c) In nuclear fis sion, energy is released by fusing IwO nuclei of medium mass (approx.imatcly 100 a.m.u.) (d) In nuclear fis sion, energy is released by fragmentation of a very heavy nucleus

9. Let IIIpbc the mass of proton, mnthe mass of a neutron, M Jlhe mass of a ~g No nucleus and M21he massofa ~gCa nucleus. Then: (Il) M2 = 2M,

(b) M2 > 2M, (e)M2 <2MI (d)MI <1O(mp+mn) 10. Assume that the nuclear binding energy per nucleon (BIN) versus mass number (A) is as shown in figure. Use this plot to choose the correct choice(s) given below; BIN

,,

I

4

~:t:!.--.---,. 100 200

•

(a) Fusion of two nuclei with mass number lying in the range of 1'< A'<SO will release energy

I ,I ,

Nudear Physics

545

(b) Fusion of two nuclei with mass number lying in the range of 51< A< IOO will release energy (c) Fission of a nucleus lying in the mass range of 100< A< 200 will release energy when broken into equal fragments (d) Fissio n of ;l nucleus lying in the mass range of 200< A< 260 will release energy when broken into equal fragments

t'lMatch The Columns 1. Match the Columns I and I1. Column II (Properties)

Column I (Rays) (a)

y

(P)

Highest penetrating power

(b)

~

(q)

No rest mass

(e)

a

(r)

Most ionising power

(d)

Neutrino

(s)

Electromagnetic wave

(I)

Associated with electromagnetic field

2. Match the Co lumns I and II .

-

Column II

Column I (a)

Gravitationa l

(P)

Long range force

(b)

Electromagnetic

(q)

Short range force

(e)

Nuclear

(r)

Weak force

(d)

~decay

(s)

Strong force

-

3. Match the Columns I and II. Column II

Column I

I

(a)

Cyclotron

(p)

Magnetic confinement of plasma

(b)

Betalroun

(q)

a·decay

(e)

Smoke detector

(r)

(}-particlc accelerator

(d)

Tokomak

(s)

Prolon accelerator

GRB Undenlanding PhysiC! Optics and Modem Physia

546

4. Match the Co lumns I and II.

Column II

Column I

(a)

Nuclear fission

(p)

It is followed by the production of X-rays

(b)

Nuclear fission

(q)

Atom-bomb

(e)

Internal photoelectric effect

(r)

y-cmission

(d)

K -electron capture

(s)

Energy liberated in Sun

5. Match the Columns I and 11. Column U iscovcr)

Column I (Discover) (a)

Fermi

(p)

Nuclear model

(b)

Rutherford

(q)

Discovered natural radioactivity

(e)

Yakara

(r)

Fusion of urnnium nucleus

(d)

Bequcrral

(s)

Anificial nucleus reaction

6. Malch the Columns I and II.

Column II

Column I (a)

Fission

(P)

(b)

Fusion

(q)

Involves weak nuclear forces Can occur in atoms of high

atomic number (e)

Jl-dceay

(r)

Can occur in atoms a ftaw

atomic number (d)

Exothermic nuclear reaclion

(s)

Matter energy

547

Nuclear Physics r J Comprehensions

Passage-1 (Neutrino-hypothesis) As we have discussed in p - decay, three "neutrino" saves us from isolating the conservation laws. The neutrino have spin. energy and momenta. I. The energy of react ion for a ~tlccay is Q.lrthe neutrino carries an energy Ko, the energy carried by the ~particl es is : (a)~Q - Ko (b)
~

--jo

1~ 1 Y + p. + v, for A and 2 being even, if the spin

~ . the spin of" u"is (A = 14, 2 = 6): (b) zero

1 (e) - 2

(d) indetenninate

3. The neutrino : (a) does not have its anti-particle (b) cannot move with the speed 01 light (c) is non-electromagnetic (d) has a rest mass Passage-2

Natural radioactivity involves the emission of A.p. y and followed by neutrinos. The natural radio-activity is a spontaneous process and it is statistical in nature. The number of nuclei remaining after a time' is given as N = Noe- 'At, where A = decay constant, where No = no. of nuclei at , = 0. 1_ The member of nuclei disintegrated after three half-life is:

(a) No

(b) No

(e) 7No

2 8 8

(d) 3No

8

2. The ratio of activity of the sample at'l = 2T and'2 = 3T is: (a) 2:1 (b) 3:1 (e) 1:2 (d) 2:3 3. The over under N-r graph isX. The average life of the sample is: N

(a) NoX

(b) No

X

(d)

L

No

548

GRB Understanding Physics Optics and Modem Physics Passage- 3

voO

In artificial nuclear reaction, the projectile X panicle strikes the nucleus }'. For a particular minimum KE of the projectile X, the m nuclear reaction takes place. we call is the sold KE in Ihi s process of O~ collision. the excitation energy can be increased. Assuming head on collision, assumes the following questi ons. 1. IrQ = reaction energy, the threshold KE is :

(.) M;;'"Q (c)

'"

M+m

(b)

M

(I + ;;)<-Q)

Q

2. The excitation energy of the nucleus M after capturing the projectile m is : (a) Q

(b)Q+~mu~

(e) Q +

A:~'I1/J

(d)

Q(l+ ;;)

3. If the kinetic energy of the projectile is less than threshold kinetic energy: (a) the collision is cnelastic (b) the collision completely cilclastic (c) the collision may be enclastic (d) the nucleus Y may be excited

Passage-4 (Binding energy and reaction energy) The binding energy is that which binds a systcm. Thc rcaction energy of a grang of reacting nuclei is equal to difference in binding energy. Then, answer the following questions. 1. The binding energy increases due to: (a) pairing effeet (b) addition of neutrons (c) addition of protons (d) small size of the nucleus 2. For low mass number A < 25, for the stability, !!: p

(.) = 1

(b»1

(c) 1.6

(d)
3. In fusion of two lighter nuclei and fission of a heavier nucleus: (a) binding energy is released (b) reaction energy is equal to BE of the reactants (c) decreases in binding energy occurs (d) excitation energy is equal to the reaction energy

Passage-5

p-decay, Let a nucleus X goes to an excited state of nucleus Y releasing • an electron. X ~ Y + e+ I1E I ... (i) In

•

Then the excited nucleus Y gets de-excited to release y-ray

549

Nuclear Physics

•

. .. (ii)

Y--+ Y+y+.6.£2 I. The mass defect in thc above p-decay is:

(a)Mx - My ~ lIIe (b)Mx - My - 2me (c) M x - My (d) M X + M)' - 2me M x, My, and /lie arc the atomic makes of X,Yand an electron respectively 2. In eqn. (i), the energy released fi£! is: (a) = .6.mc

,

,

(b) < .6.mc-

(c) > .6.mc 2 (d) none of these (where dill = mass defect) 3. In the total p-decay reaction, the energy lost is: (0)

t..E, + t..E,

(c) > .6.£1 +.6.£2

(J

(b) M, - M, t..E, +t..E, (d) 2

Subjective Problem, Level-l

Properties of nucleus 1. Find the ratio between the radii of 7 Li and 239 Pu nuclei.

2. A nucleus ~ X undergoes K-electron capture. Find the new nucleus and the other

3. 4. 5. 6.

radiation. What is nucleus magneton? Compare bohr magneton with nucleus magneton. A spinning proton is kept in a magnetic field B = 2T. Find the minimum magnetic energy possessed by thc proton. In Q-5. if the proton flips from its + ve to -ve energy state. find the energy and frequency of radiation emitted by it.

Binding Energy, mass defect and stabl11ty of a nucleus 7. Find the electrical repulsion energy per nucleon for Z ,,; 2 and Z = 92. Assume R = 2 x 1O- 15 m and 8 x JO-1Sm for Z = 2 and Z = 92 respectively.

8. Find the binding energy of ana-particle. 9. Find the average BE per nucleus ofi~Caand ~~7 Au nuclei.

10. A neutron brakes into proton and electron. Find the energy of reaction. II. What is the maximum wavelength of a "'(-ray that could break a deuteron into proton and neutron in the process of photo-disintegration. 12. Find the maximum kinetic energy of the J>-rays emitted in neutron decay.

GRB Undentanding Phy5ics Optics and Modem Physics

550

Natural Radlo-actrvity 13. What is the decay rate of I gm of radio·activc strontium? (T1f2 = 29years) 14. A count rate ora sample is 4750/min al time 1. Five minutes later the count rate is 2700/min. Find (i) decay constant (ii) half li fe or the sample. 15. In a uraniUIll arc the ratio of 238 U nuclide to 236 Pbnuclidc is 128: liJ~vatualC the age of the one. assuming that 10la1 uranium will get ponsfonned to Pb. TII2 for uranium is 4.5 x 109ycars. 16. The hatfHfe orthe radioactive nucleus rndium -226

i;6R is 1.6 x t0 3ycar. (a)

What is the decay constant? (b) If the sample contains 3 x IO l6 nuc i ci at t = 0, fi nd the initial activity . 17. In an wooden artifact. 10.5 (disinlcgralion/hr) gm is found . Calculate the age of the artifact.

Nuclear fission and fusion 18. In the fissio n reaction ; H + ; H -+ ; He. (a) calculate the energy of reaction. (b) Find the amount of deuteron consumed I day assuming 25% ofthe reactor.

19. Find the Q.value of the nucleus reaction 4 He + t4N -+ 170+ lH 2

7

8

1

20. In the previous questi on find the minimum KE energy of the bombarding a·particlc to initiale the reaction.

21. What is the power adjust of a 2JS Un reactor, if it takes 30 days to usc 2 kg of fue l. Assume that each fi ssion gives 185 MeV of usable energy. 22. Calculate the energy released in the (n, /I) reaction for ~ Be.

(J

Subjective Problem. Level-2

I. Find the minimum kinetic energy required for a proton to split a deuteron whose binding energy is Eo = 2.2 MeV. 2. A proton strikes a stationary lithium target (nucleus) with a kinetic energy K = I MeV. As a result the lithium nucleus is split into two a· pan.icles. If the a· particles ny symmetrically with the initial direction of the bombarding proton, find the kinetic energy of the a· particles and the angle made by the a· particles with each other. 3. In atomic mass units find the mass of 8 Li atom whose nucleus has binding energy 41.3 MeV. 4. Find the mass of tOe atom in amu of its nucleus has BE pernuc1cus is 6.04 MeV. S. Find the difference of BE of neutron and proton in a nucleus A ZX.

551

Nudu.r Physics 6. Find the energy required to separate gOl6 into four identical pieces. Given 6111a

=

0.0026 omu and 6mo1 6 = 0.00509 amu.

7. An (X.particle strike a stationary 6 Li nucleus elastically with kinetic energy K= 7 MeV. As a result thea·particle and the 6 Li nucleus move in different directions. If the angle between their lines of motion isS = 60°, find the kinetic energy of the scattered nucleus. , 8. A neutron collides a t2C nucleus at rest with a KE = K = 10MeV. It initiates the underrcaction as a unreallywhich Be will be fonned along witha·particlcs. If the threshold energy for the reaction is K lh = 6.17 MeV. find the KE of the (X·particles. Assume that the (X·particle moves perpendicular to the collid ing neutron. 9. Find the kinetic energy of the recoiled nucleus in the positron decay of 13 N nucleus for the case if the energy of positron is maximum. 10. A radioactive sample of decay constant A.I of total number ofnuclci Nl nat time t = 0 decays \0 fonn anolher radioactive material o f decay constant A2'(0.) Find

the number N 20f nuclei of sccond radioactive material at time I (b) After what time the value ofN 2wi ll be maximum. II . An alpha strikes a stationary 9 Be nucleus with kinetic energy K = 5.3 MeV. This initiates a nuclear reaction ofenergy Q = 5.7 MeV produc ing a neutron and a 12C nucleus. If the neutron moves perpendicular to the bombarding (X.particle just after the collision. find the kinetic energy or neutron. 12. An alpha particle strikes a stationary 14 N nucleus with a kinetic energy Ka = 4MeV. As a result, the nuclear reaction occurs and after the a 11 o nucleus and a proton arc formed. It the KE of the proton is = 2 .® MeV. Find the energy of re.action assuming that the proton moves as an angle of S = 60° with the direction of motion of thea-panicle. 13. In the forcgoingproblem. findO.

K"

552

GRB Understanding Physia Optics and Mod~m Physics

Multiple Choice Questions (A) Only One Choice Is Correct

Level·1

1. (a) 6. (b) 11. (c) 16. (a) 21. (c) 26. (d) 31. (b) 36. (b) 41. (c) 46. (b) 51. (c)

4. (c) 9. (b) 14.(.) 19. (b) 24. (b) 29. (d) 34. (c ) 39. (c) 44. (d) 49. (b)

5. (a) 10. (a) IS. (c) 20. (d) 25. (d) 30. (c) 35. (c) 40. (b) 45. (b) 50. (b)

I. (c) 2. (a) 3. (c) 6. (b) 7. (b) 8. (c) 11. (c) 12. (c) 13. (b) 17. (d) 16. (a) 18. (b) 21 . (d) 22. (a) 23. (c) 26. (b) 27. (c) 28. (b) (8) More Than One Cholce/s Is/are Correct

4. (a) 9. (b) 14. (c) 19. (c) 24. (a) 29. (al

5. (c) 10. (b) IS. (d) 20. (d) 25. (b) 30. (a)

2. (a,c,d) 1. (a,b,d) 7. (a, b,c) 6. (a,b,c) Match the Columns

4. (a,d) 9. (c,d)

5. (a,d) 10. (b,d)

2. (d) 7. (b) 12. (c) 17. (b) 22. (b) 27. (b) 32. (d) 37. (c) 42. (c) 47. (c) 52. (a)

3. (c) 8. (c) 13. (c) 18. (b) 23. (b) 28. (d) 33. (c) 38. (b) 43. (c) 48. (d) 53. (e)

level-2

I. a-q,s,t ; b-t ; c-r,t ; d-p,q J. a-s; b-r ; c--q ; d- p S. a- r ; b--s; c- p: d-q

3. (a,b,c) 8. (a.d)

2. a-p; b-p; c-q,s ; d-r 4. a-q,r ; b - r,s; C - p,T; d-p 6. n-q,s ; b- r.s ; c- p; d-q,r,s

Comprehensions

Passage..1 : 1. (c)

2. (aJ

3. (c)

2. (al

3. (d)

Passage-2 : 1. (c)

Nudear Physics

Passage-3 : I. (b) Passagc-4 : I . (a) Passage-5 : I. (a)

553

2. (e)

3. (b)

2. (a)

3. (a)

2. (b) 3. (e) Subjecllve Problems Level· I I. 3.2

2. The new nucleus of z~land the other radiation is neutrino. 3. 5.05 x l 0- 2S Jrr 4. ~ 1 836 5. -2.82 x 10- 24 J 6. +5.64 X 10- 24 J, 0.895 X 10 10

Hz

7. 0.72 MeV. 753.5 MeV 8.28 .5 12 MeV 9. 4.819 MeV, 3.254 MeV 10. 0.73 1 MeV It. 5.575 x 10- 13 12. 0.78 MeV 13. 5. 1x 10 12 Bq (~ I disintegrationsls) 14. 0.1 129 / min. 6. 14 min

15.1.98 x 10 9 year 16. (a) 1.4 x IO· II /s 17. 36000 year 18. (a) 23.834 x IO'eV 19. - 1.1 9 MeV 20. 1.S3 MeV 21. 58.46 MeV 22. 5.7 MeV

(b) 4.2 X IO '/s (b) 120.83 gm/day

Subjectlv. Problems Leve/·2 1.3.3 MeV 2. 170.530 3. 8.02248 amu 4. 0.0 1678 amu 5.[l> Y' -l>Y]c'

6.4 .946 MeV

554

7.6 MeV 8.2.21 MeV 9.0.111 keV AI N I 0 «(,-).2/ _

10. (a) ).

e - All)

I -).,2

11. 8.52 MeV

12. -1.19 MeV --{(mN

+ rna) -(mo + mp )}c2

+Kp(l + ~)+ Ka(~ -I)

2J:= KpKa

555

Nuclear Physics

1IC:::::::::::::::C~H~i~n~t!sJR~n~d~S~O~lu~t~i~o~ni,*~::::::::::~II Comprehensions

Passage-'

,

;

1. The energy ofrcaction is Q = K,c:coil + Kel + Kneutrino'

or Q=::'Kel + Kn or, Kcl :::.Q-Kn 2. For A and Zbeing even, Nand Z arc cven, the total spin of Xis 0, of Yis +1 . Since the spin of clectron is + ; then to conserve angular momentum, the spin of the

.

neutrino must be

,

2

3. It is non-electromagnetic. v = c and have zero rest mass. Passage--2 +3T

I

1. The no. of nuclei disintegrated is

. . .IS Th e no. 0 f nuc ,el. remaining 2. The ratio of activity

N N~nY No(D t =

=

=

~o

7N N'-_No_No 8 _ 8o

(dN) t(dN) = ANIf)..N 2 dt1dtz

= 'Vj IVz

.2T

Z: =((~r =1.~ =2.,.

NI Nz = N2 No

1

)+3T T

8

2

-x= JON tdN I No =T. v No 0

3.

Subjective Problems Level-1 I

I.

r= 'bA3,

Since

I

I

~ =(APu)j =(239)j tti

AU

7

=3.2

2. After capturing a K-c1ectron, the atomic number decreases by one where as atomic mass no remains the same. Hence, A A zX +e-'+Z_I

Y +V

The new nucleus of ~_I Y and the other radiation is neutrino.

GRB Undentanding Physiu Optics and Modem Physics

556

3. It is the unit of magnetic dipole moment of nucleus. . ~ n = ell = 5.05 x 10-" )rr One nuclear magneton IS 41tmp

4. One Bohr magneton is J.l b = 4 eh

,m, eh

J.l b = 4Jtme Il l! ~

= mp :::: 1836 me

4Rmp

S. The minimum magnetic energy possessed by the proton is U = -j1 pB = -(2 .7928 ~" )(2T) = -S.S8J.ln = -5.58 x 5.05 x 10- 25 = -2.82 x 10- 24 J 6. = E,

E,,,,,,," I E, - I

= 1(-2.82 X10-24 ) - (+2.82 X 10-24 )1 +5.64 x 10-24 J The frequency of radiation =

f = Epholon h 5.64 x tO- 24 = 6.3xlO 34

=0.895 x 10 10 Hz 7.

(2-1),'(1) 2 R

U =_1_ 2

41t£o

=9XI092(2-1)(1.6XIO-19)'( 1 ) 2 2xlO- IS = 0.72 MeV U =9 x 10 9 (92)(92-1)(1.6 XIO- 19 )' ( 2

8.

1

)

8xlO- IS

= 753.5 MeV BE = [2mp +2mll -ma]cz = (2!!J, P + 2.1." - Aa) x 931.5 MeV

= [2(0.007277) + 2(0.00866) - 0.00 1265]931 .5 =28.512MeV

NudearPh~ia

10.

557 Q=[m n -(m p +m c »)c2

= (1.6747 x 10- 27 -1.6725 X10-27 -9 X10- 31 )(3 x 10' )' J : 0.731 MeV 11. ~ d-t p+1I . 2 he The energy of reactIOn Q = (nld - mp + mn )c =

r:

= 5.575 x 1O- 13 m

h

A:

(1110. -mp + mn)c n -t p+e- +V

12.

Q=[mn-(m p +me)]c

2

" =0.78 MeV Since mp » me. practically the proton will be at rest Then, Kt.lrmx ==1 Q =Kp +Ke +K vQ == K e +Kv Since Kv =0. Kt'lmu =Q = 0.78 MeV 13. The atomic mass of stromium = 89.9 gm I gm of S" ~ (1/89.9)(6.02 X 10") atoms = 6.7 x 1021atoms I

dN 0.693 N oe-","Th e dccayrate=--=-dl

TII2

0.693

=

x6.7x102 1c-O

(': t =0)

(29x3.16xlO')

5.1 X 10 12 Bq(= Idisintegrationls) Then, I gm of strontium emit 5.1 x 10 12 j3.rays (particJesls). =

(~),

14,

dN) ( 71, or

AN,

No

= AN = N No: -4750 - =475 N 2700 270

.. ,(i)

I

:0 =ur

N 270. (.) b ' put 1=5 min, and No = 475 m eqn. I to 0 tam

.. .(ii)

• Nuclear Physics

557

Q ={nr"

10.

- (nip

+me)]c2

= (1.6747 x 10-27 -1.6725 x 10-27 _9xI0- Jl )(3xlO')'J = 0.731 MeV ~ d~ P+II

11.

The energy of reaction Q = (nld - mp + m" )c2

1. = 12.

(nra -

h nip

+ mll}c

= '~

= 5.575 x 10-ll rn

n-+ p+e- +v

Q=[m,,-(m p + mr )]c 2 =0.78MeV Since nip »me. practically the proton will be at rest Then, K f! IInLX == 1 Q

= Kp +Ke +Ky_

Q == Ke +Ku Since KIJ =0, Kl'lmu =Q =0.78 MeV 13. The atomic mass of strontium =89.9 gm 1 gm of S ,. ~ (1/ 89.9)(6.02 X 10 23 ) atoms

,

dN 0693 -1be decay rate = - - = ...:.....- Noe TII 2 dl

TI /2

=

x 6.7x I02I e~

0.693

(":1=0)

(29x3.16 x lO')

= 5.1 x 10 12 Bq(= 1disinlegrationfs} Then, 1 gm of strontium emit 5.1x 10 12 ~rays (paniclesls).

(~)I

14.

( or

.ti -. , .

ANo

No

dN) = l. N

=Ii

dt,

No

.!

4750

475

Ii = 2-7-00 = 2-7-0

,

:0 =ur put t = 5 min, and

~.

:0 = ~~~ in eqn. (i) 10 obtain

. .. (i)

.. . ( ii)

558

GRB Understanding Physics Optics and Modem Physics

I

). =0.1I 29/min

r =6.14min

and

II

16. (a) . ).=0.613= 0.693 =1.4xlO- I' /s 1.6xl0 3

Til,

Ro =d; =).No

(b)

= (1.4XIO - ")(3xlO 16 )

=4.2xlO s /s 17. The time of the artifact is given as 1=_1 1n N

). NO Put No = 13.6 (disintegration/min) gm of the contemperary sample (wood), N = 10.5 disintegration gmlhr and TII2 = In 2 / 5730 year N =36000 year No Then, the plant died 36000 year ago. 19. Q = (nlHe +mN -rno -mH)c 2

we have, t

= _5730 In 1n2

,

= (4.002603+ 14.0030704 -16.999133 -1.007825) x 931.5 = -1.19 MeV Kth

20,

=mHe +mN {_Q} mN

~

= 4.002603 + 14.003074 H _1.19)] MeV 14.003074 . = 1.53 MeV

22. The reaction is :Be+i He--+~2 C+~ n+Q Q = (""')c' = (0.006121 anru)931.S MeV/amu

=5.7MeV Subjective Problems Level-2

1.

0-

0

=9

po Since

K

ddted""""", .

~ (I + :: )i-QI. puning the values

we have Kmin = 3.3 MeV

559

Nuclear Physics

2. The reaction is given as I p +7 U -)2 4 He

0

--------------~ ~. PLIO ,

0-

a

Conservation of linear moments Pp = 2Pa eosS Conservation of energy p2 2l~2 -.L + Q : ---.!!.. = 2K 2mp

or

2ma

a

Ka =K+Q 2

where

.. .(i)

... (ii)

Q = (d}! + du -2dHe)C 2 = [0.00783 + 0.01601- 0.OO52]c2

=O.OI864amu xc 2 = 17.37 MeV Then.

Ka =1+1;.37 =9.18MeV

From eqn.(i). 9=cos =cos

-I

-1

Pp 2Pa

,j2mp Kp J2n1a Ka

=co, - 1 J::..:..WhereK=IMeVand Ku =9.18MeV

Then the angle between the line of motion of the a- particles is $ =29 =170.53'.

3. The BE of an nucleus X of an atom ~ X is or

Eb = 2mH + Nmn - M alom Eb = Zt.H + (A - Z)t.. - t. 41.3 MeV=3t.H + (8-3)t.. -t.

560

GRB Undenlanding Physics Optics and Modem Physics 41.3 = 3 X 00 931.5 .0783+5xO.00867-11 A =0.02248 amu

or

Hence the atomic mass of the atom Li 8is 8.02248 amu. 4. The binding energy of 10C nucleus is Eb =6.04xlO=60.4MeV

= :30i~ = 0.06488 amu Then, using we have

Eb =ZLlH+(A - Z)6.. n

-a,

II = lllB +(A -Z)II, - Eb =6AH +(10-6).6./1 -Eb = 6An +4.6./1 -Eb =6 x 0.00783 + 4 x 0.00867 - 0.06488 =O.oJ678amu

5. For a neutron,

.

1X 1- Y + n --jo

1

(E b ), = [m(X) - m(y)]C'

Then,

. .. (i)

Ax Foraproon, t z -joZA-I - l Y' +p Then,

(Eb)p =[m(X)-m(Y')]c'

Hence. from eqns. (i) and (ii), (Eb)/I - (Eb) P

= [m(~=: Y') -

... (ii)

=[m(r') -

m(Y)] c 2

m(1=: y)]C'

= [t.Y' - t.y]c' 6. Required Energy+Ol6

--jo

4a(ideally), Let.

Then the energy required is E=m(4<»-m(016)

4Au - .6.016 = 4(0.00260) - 0.00509 = 0.00531 amu =4.946 MeV 7. Let thea-particle initially move with a momentum =

~o= ~2mo.Kai Then. applying conservation of momentum

.,.

p}=P;

-=> t'a +Pu = Pao ~

•

= .J2mo.K i

0-0 (Just before collision)

561

Nuclear Physics

P(l, + p' U +2P(lPU cose =2"'o,Ko.

or

P:/"" "" (P 0.)0

... (i)

...

.......

------------- ,

, ,,

,, , ,,

(Just after collision) Since the collision is elastic, the conservation of kinetic energy yields K(l +KLi = Ka

or

/,;'

p'

~+--1L=K 2"'0, 2mu

.. .(i i)

pi +~p' =2m".K LI

mLi

eqn. (ii) - eqn.(i), PL'(I- m". )+2Po.Pu coso =0 mLj

1

or or

PLi

(I-~) = -2Pa coso mLi -2Pa. cose

PLi = (

m". ) 1-mLi

Then. the kinetic energy of 6 Li nucleus is

pl KLi =---L

,

2mu

=

or

-2P(l cose

12mLi

\_ m". mLi

K b - M and m Li _L;= 4K(cos' O)m".m u ,wenma.Ka (m~; -

m.i)

GRB Undentanding Physics Optics and Modem Physics

562

Ku

or ~~;;- = ---!'----Ka + K L " ( 111- M)' , =-,-:-"'-'- sec 8 + I

4Mm

or

KLi ::; - --

1+

K

.!>.--;:--

(M _Ill)'

Putting the moves M

4MIII

sec

-

'8

=4, 111 =3,8 =60° , K = 7MeV

we have

8. Conservation oflinear momentum:

o

12C

or

tie = 2m" K" + 2nla Ka

" ,(i)

P

Conservation of energy gives p 2

p, 2

13,2

_n_=~+~+1QI

" ,(ii)

2m" 2i1la 2mBc Substituting POc from eqns. (i) and (ii), Pill ~ Pel + 2m"Kn +2ma.Ka + Q 2m" 211'0: 2mac or

or

or

III

m-

K=Ka + K-" +Ka-u-+Q mac

nIBc

K(I-!'!mBe >!...) =Ko (I + mac ~) +Q K(I- !'!>!. . )- Q mBe

,'

.. ,(iii)

.. ,(iv)

563

Nuclear Physics Substituting

IQIfrom eqn. (iv) in eqn. (iii),

9. Thep + decay is given as

Nil ~Cll +e+ +v

Conservation of linear momentum yiclds

i1- =p,.+?; +~ . Fr +?; +~ =0

or

where Px

=0 .. .(i)

The conservation of energy yields.

Q -=. K y + K c + K v. where Kv = 0 because the /l-particle carries maximum energy. Since the p + particle too much lighter than the nucleus Y.

Q=K, or

Q = [m(N") -m(C") - 2m, Jc' Q = 1.71McV (afterevaluationr'

By using eqns . (ii) and (iii),

K, Since,

Pv

->

=1.71 MeV

=0, eqn. (i) Yields ->

Py = -Pe Then, the kinetic energy of the receiling nucleus Y is

K e =_I_ (1.7 1xI0 6)eV x my 1836

=l1Ic

or

K y =O.llIkcV

... (ii)

... (iii)

Nucltar Physics

563

Substituting \QI from eqn. (iv) in eqn . (iii),

K.

=

K(I-

"I, )_ ('h ) 1[1 + lila ]=2.21 MeV "'Be

1 + '" ---.!!.

"'lie

lIIe

9. The ~ + decay is given as NI3 -+ CI3 +e+ +v

Conservation of linear momentum yields -=> "'"e +P"'"v , Px = -=> f'y +P where Px = 0

-=> d 11' +Pe +Py = 0

~

or

... (i)

The conservation of energy yields.

Q -:: K y + K e + Ky . wherc K y = 0 because the f}-particle carries maxImum energy. Since the ~ + particle too much lighter than the nucleus Y. Q '" K , Q or

... (ii)

=[meN ") _ m(e") -2m,]c'

Q = I. 7 t MeV

By using eqns. (ii) and (iii),

K e = 1.71 MeV Since,

Py

->

=0, cqn . (i) Yields ->

Py = -Pe Then, the kinetic energy of the receiJing nucleus Y is

=meKe =_I_ x (1.71 x I0 6 )eV my

or

1836

Ky=O. llIkeV

."

(after evaluation)

... (iii)

GRB Undentanding Physics Optics and Modem Physics

564

10. (a) At time

No. of nuclei ofIst material (I) No. of nuclei of 2nd material (2)

1=0

N10

0

LetAttimet=t

NI

N2

Then, at time I, the rate of decay of I is dN I -

dl

... (i)

= -).,INI

The rate of fonnation of 2 at time t is Then the net rate of decay is dN, ---;j( ).,IN I - )",N,

A.I N Iand

rate of decay of 2 is -A. 2N 2· . ... (iI)

=

Usingeqn.(i). NI =

.. ,(iii)

NlOe-). 11

Putting N1from eqn.(iii) in eqn.(ii), ~ N2 dt = A1NIOe -1.)1 -fl.2

dN 2

or

dN, ), -All ( Tt+A2N2 =1I.\NlOe

Multiplying both sides by e X21 and integrating both sides w.r.t. time, we have

f (d~, +)."

N,} +A21 dl =

Since, ; (N2e X2t ) =

I(d~, +

I)", N'oe-(AI- A2" dl

i. 21 ( N2A2 + d~2 }

)",N, )eA" dl

By using eqns. (iv) and (v).

=N,eA"

... (iv)

we have ... (v)

565

Nuclear Physics

or or

e-).,2 1

xI

e-).,\t

=x2

e().,\-),2)t

=

A.]

A, In(AllA, )

or

I

= (AI -A,)

11. Be+o,-)C+n The energy of reaction is

Q+(Ka) =K, +K,

p' Q+K=-'-+K, 2m,

... (i)

G.!5..0 a Be

The conservation of momentum yields, -7 -> -> Pa = I{ +PH or ,J2=-ma-K"i = +..j-2=m-,K",-i

or

P; 1Fcl' = (,J2ma K )' + (,J2m,K,)\

,

Pc =2ma. K ,+2m nK n Substituting P/from eqn.(ii) in eqn.(i),

or

or

Substituting all values. we have K, = 8.52 MeV

... (ii)

GRB Understanding Physics Optics and Modem PhysiC!

566 12. The reaction is given as,

a.+14N~170+p

The conservation of linear momentum, --> --> --> Pa. == Pp+Po or Po2 =Pa2 +PI'2 -2Pp PoaCOS e

or

•

N

Ka

0'"

.. .(i)

0

Th~ energy of reaction is

... (ii)

Q=Kp+Ko-Ka Substituting

= pJ 12mo from cqn. (i) in cqn. (ii),

KO

,

i I,

\ Q=K

P; ?pPa cose - Ku +-+--

,, ,

pJ

p

2mO

2mo

I

rna

~cc2m-p-;:K'-p~2"'aKa cosO Q=Kp+Ka-+Kp '~m -Ku

"'am

ma

Q=K

P(

ma

m) ('"" ) . 1+1 +Ka - - I -

rna

0

2-,~2m;p"'a~~K.c:a~K~p_CO_se

ma

Putting the values of the quantities K p. Knetc, we have

Q

= -1.19 MeV

rna

i I

Nuclear Phy1ics

567

-Q+K P(l+mp)+Ka(n'a -I) mo mo

or cos 8 = ----'--;=,,;,'==,...:.--"--'-

... (i)

nip

2 - KpKa "'a

As we know ,Q = [( mil + "'ex) - (mO + mp )}c2

... (ii)

Then, -[(m N +111a)-(mo +ml)}c 2

8=cos- 1

+Kp (mp) t+- +Ka (n'a --I) mo mo

,

Understanding Physics

Optics & Modern Physics i!I Ii Ii

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