24.Wave Optics

T

Chapter Overview

,. Introduction

• Newton's Corpuscular Theory ,. Huygen's Wave Theory • Coherent Sources • Interference of Light

~>

Introduction

" Diffraction of Light

"' Difference between Interference and Diffraction

g Polarisation of Light " Doppler's Effect in Light

The nature of light has been the source of one of the longest debates in the history of science for centuries. Different scientist's theories like Newton's corpuscular theory, Huygen's wave theory, Planck's quantum theory, Maxwell's electromagnetic theory, Einstein's photoelectric effect etc presented light as waves and particles. Finally the quantum electromagnetic dynamics completely explained the light and electromagnetic interactions. According to this theory, light shows wave like behaviour in certain situations eg, interference, and diffraction and it behaves like particles (photon) in other circumstances (eg, photoelectric effect).

In this chapter we deals about wave nature of light interference and diffraction.

R

>

Newton's

Corpuscular Theory

In 16 7({>, Newton propounded the corpuscular theory of light to define its nature. He assumed that

936

,

I

Chapter 24

•

Wave Optics

(i) Light is composed of nearly massless tiny particles called corpuscles.

(ii) The corpuscles travel in a straight line with a tremendous speed in a homogeneous medium.

(iii) The speed of corpuscles changes when they travel from one medium to the other.

(iv) The corpuscles oflight of different colours are of different sizes.

(v) When the corpuscles strike on retina, we get the sensation of vision.

~>

Huygen's Wave Theory

In1678, a Dutch scientist, Christian Huygen propounded the wave theory of light. According to him

(i) Light travels in the form of waves.

(ii) These waves travel in all the directions with the velocity of light.

(iii) The waves of light of different colours have different wavelengths.

(iv) Initially the light waves were assumed to be longitudinal. But later on while explaining the phenomenon of polarization the light waves were considered to be transverse.

(v) The whole universe with all matter and space is filled with a luminiferous medium called ether of very low density and very high elasticity.

(vi) Huygen's theory could explain reflection, refraction, diffraction, polarization but could not explain photoelectric effect and Compton's effect.

(vii) Wave theory introduced the concept of wavefront.

Wavefront

A surface drawn at any instant in the medium affected by

p

the waves originated Rays

from a source, on which S - E - - - - + - - - r - - - - ' - -Q each· particle vibrates in

same phase is called the wavefront.

The wavefront may also be defined as the hypothetical surface on which the light waves are in same phase.

As the wave travels forward.

R

S =source of light, AB =wavefront and SP, SO

and SR are rays of light.

forward, the wavefront also moves The perpendicular to the surface of a wavefront gives the direction of light ray. The disturbance from the source propagates in all directions, with the same speed c. Hence, after a time t, all the points on the hypothetical sphere of radius ct, with the point source Sat the centre, are in the same phase of vibration. Hence, this hypothetical sphere acts as a spherical wavefront of light. The energy transfers along the direction of rays.

Wavefronts in Different Cases

(a) Light emerging from a point source Wavefronts are spherical and eccentric with point source at their centre

as shown in figure by circles 1, 2, 3 and 4. Rays are radial as shown by arrows. In spherical wavefront Amplitude oc

_!.

r

I ntens1ty . oc ₂ 1 r Direction of propagation

(b) Light coming out from a line source Wavefronts 1, 2 and 3 are cylindrical and co-axial with the straight source as their common axis.

(c) When light sources is emitting parallel rays, or when the light is coming from a very far off source Wavefronts will be planes as shown in figure.

Plane wavefronts

Huygen's Principle of

Secondary Wavelets

In 1678, in order to explain the propagation of wave in a medium, Huygen propounded a principle known as Huygen's principle of secondary wavelets by which at any instant we can geometrically obtain the position of wavefront. He made following three assumptions

(i) Every point on a given wavefront (called primary wavefront) can be regarded as fresh source of new disturbance and sends out its own spherical wavelets called secondary wavelets.

(ii) The secondary wavelets spread in all directions with the velocity of wave (ie, velocity of light).

(iii) A surface touching these secondary wavelets, tangentially in the forward direction at any instant gives the position and shape of the new wavefront at that instant. This is called secondary wavefront.

A'

B' (a)

>

Coherent Sources

(b)

'TWo sources are said to be coherent if they have the same frequency and the phase relationship remains independent of time. In this case, the total intensity I is not just the sum of individual intensities !₁ and !₂ due to two sources but also includes an interference term whose magnitude depends on the phase difference at a given point.

I= !₁ + !2 + 2JT;I; cos<j> where ~is the phase difference.

The

2M

cos <1> averaged over a cycle is zero if (a) the source have different frequencies.

(b) the source have the same frequencies but not constant phase difference.

(c) for such incoherent sources I =I ₁ +I _{2 ,} where ~ does not change with time, we get an intensity pattern and the sources are said to be coherent.

In practice coherent sources are produced either by dividing the wavefront or by dividing the amplitude (as in the case of thin films, Newton rings etc) of the incoming waves.

·

>

Interference

of Light

When two light waves of exactly equal frequency having phase difference which is constant with respect to time travelled in same direction and overlap each other, then intensity is not uniform in space.

At some points the intensity of light is maximum (more than the sum of individual intensities of those waves), while at some points the intensity of light is minimum (less than the difference of individual intensities of those waves).

Thus, formation of maximum intensity at some points and minimum intensity at some oilier points by the two identical light waves travelling in same direction is called tlle interference of light.

The interference at the points where the two waves meet in same phase, ie, the intensity of light is maximum, is called the constructive interference while at the points where the two waves meet in opposite phase, ie, the intensity of light is

Chapter 24

•

Wave Optics

937

Resultant wave Second wave

(a) Constructive interference First wave Resultant wave

(b) Destructive interference

Conditions

of Maxima and Minima

Lety1 =A1 sincotandy2 =A2sin (cot+~) be two sin1ple harn10nic waves of same frequency travelling in tlle same direction, A1 and A2 are tlleir amplitudes, ~ is tlle initial phase difference between tllem and co/2n is the common frequency of the two waves.

By the principle of superposition the resultant displacement is y

=

y1 + Y2

=

A1 sin cot+ A2 sin (cot+ ~) From this expression, the resultant amplitude is given by

A2 =A~+ A~+ 2A₁A₂ cos<j> The resultant intensity I is given by

I =A2

So, I= A~+ A~+ 2A₁A₂ cos<j>

I == I 1 + I 2 +

2M

cos <I>

Thus, the resultant amplitude (or the resultant intensity) at a point depends on the phase difference at that point between the two superposing waves.

(i) For constructive interference (maximum intensity) The intensity I is maximum, when cos ~ = + 1, ie, when phase difference ~ = 2nn; n = 0,1,2, ... etc,

n = 0 stands for zero order maxima n = 1 stands for Ist order maxima n = 2 stands for lind order maxima Path difference = n'A

So, I max= A~+ A~+ 2A₁A2

=

(A1

+ A

2

i

(ii) For destructive interference (minimum intensity) The intensity I is minimum when cos 4> = - 1

ie, when phase difference

~

=

(2n -1)n; n

=

1, 2, ... etc.

n = 1 for first order minima, n = 2 stands for second order

minima

'A

path difference

=

(2n -1)-2

Figure shows the variation of intensity I with the phase difference !]> due to the superposition of two waves of equal amplitudes. The graph is called the intensity distribution

curve. , .' . '. . ,

2 . when!]> = 0, ±21t,±41t, ... , !max= 4A and when <1> = ±1t, ±3, ±51t , ... ,!min= 0

Instance 1 Light waves from two coherent sources of having

i11tensity ratio 81 : 1 produce interferenr;e. Then the ratio of maxima and minima in the interference pattern will be

( ) 18 a (b) 16 23 25 25 (c) 16 Interpret or (d) 23 18 . I A2 81 GIVen _l = - 1 ₌ -I2 \ A~ 1

~=~

A2 1 A1 = 9A2 I max - (Al + Az)2 I min - (Al-Az)2 From Eq. (i), we have

I max (9Az + Az)2 Imin = (9Az-Az)2

c1oi

25

C8i = 16

... (i)

Instance 2 Light waves from two coherent sources having intensities I and 2I cross each other at a point with a phase difference of 60°. The intensity at the point will be

(a) 4.414 I (b) 5.455 I

(c) 4 I (d) 6.441 I

Interpret Here, I₁ =I, 1₂ = 21 and !]> = 60° we know that the amplitude A of resultant wave is

A=

~a

2 + b2 + 2 abcos<jl A2 =a2+b2+2abcos<jl We also know that

I oc A2

:. Resultant Intensity, I= 11 + 12 + 2ji'J"; cos<jl = I + 21 +

z.J

I x 21 cos 60° = 4.414I

Young's Double Slit Experiment

In the given figure, Sis a slit illuminated by a monochromatic light. Light waves from the slit S are incident on slits 51 and 52 • Slits 5₁ and 5₂ are situated such that the distance SS1 is equal to the distance 55_{2 •} Hence, the waves emanating from the slit S reach 5₁ and 5₂ simultaneously, ie, the waves at 51 and 52 are in same phase. Infront of the slits 5₁ and 5_2, there is a vertical . screen P. We are to find the positions of bright and dark fringes

on the basis ofYoung's experiment.

Suppose the distance of the point P from centre of the fringe pattern be x then in b.S₂PQ. (S2P)2

=

(S2

Qi

+ (PQi

=

D 2

+(

x

+%Y

In b.S1PM, (SlP)2 = (SlM)

2

+ (PM)

2

= D

2

+

(X-

%Y

So, (S₂P)2 - (S₁P)2 = 2dx or (SzP

+

S1P) (SzP- S1P) = 2dx

Now since, P is very close to 0, we can have to a first approximation assuming that

SzP

+

S₁P = D

+

D = 2D dx

.. 52P-51P=D

So, path difference between the interfering waves is dx . D (i) Position of bright fringes or maxima on the screen Constructive interference occurs when the path difference is an even multiple of

A/2.

xd

A

ie, -=2n-=nA

D 2

or X = - -

nAD

d (where n = 0, 1, 2, 3 , .. ) When n = 0 gives central fringe, hence,

x = 0, ie, x0 = 0 AD n = 1, x₁ = -d 2AD n = 2, x2

=-d-nAD

n=n,xn = d If ·(1st bright fringe) (2nd bright fringe) (n'h bright fringe)

(ii) Position of dark fringes or minima on the screen Destructive interference occurs when the path difference is an odd multiple of

A/2.

ie, or xd

A

-=(2n-l)-D 2 (2n-l)AD X= 2d (where n = 1, 2, 3 ... ) Let the positions of the minima be represented by x~ instead of xn (so that there is confusion as xn represents the'position of nth maxima). When n = 1,

xi=

AD 2d n = 2 x'

=~AD

, 2 2 d , (2n-l) AD n = n, xn = - - 2 - d (1st ~nima) (2nd minima) (n'h minima)

Fringe width The distance between two consecutive bright or dark fringes is called fringe width.

Let the distance of nth and (n

+

l)th bright or dark fringes from the central fringe are xn and xn+l' then

and nW X = -n d (n + 1)W xn+l

=

d

:. Distance between (n + 1)th and nth 'fringes

(n + 1)W nW ')..])

x - x = -- - = -[n + 1-n]

n+l n d d d

')..])

or _{xn+l -xn}

_=d

:. Width of bright or dark fringe,

~

= ')..])

d

Import

ant Features

1. Consider two coherent sources 51 and 52 • Suppose two

waves emanating from these two sources superimpose

at point P. The phase difference between them at P

is ljl (which is constant). If the amplitudes due to two individual sources at P are A₁ and A_2, then resultant

amplitude at P, will be

s1

p

A=

~Af

+ Ai + 2A1A2 cos<!J

Similarly, the resultant intensity at P is given by

I= /1 +12 +2M cos<!J

Here, 11 and 12 are the intensities due to independent

sources. If the sources are incoherent then resultant

intensity at P is given by

I = Jl

+ l

z

2. For two coherent sources the resultant intensity is given by Putting, We have, l=l1+12+2Mcos<!J ll=lz=lo I= !0 + !0 + 2~1

0

x /0 cos<!J

Simplifying the above expression, we get

I=

4J

0

cos

2

~

2

3. Resultant intensity due to two coherent sources is given by

I =!1 +!2 +2M cosljl

!max= cJf; +

_,

ji;)

2 where cos <I>·=

+

1

Fz )

Chapter

24

•

Wave Optics

939

If both the slits are of equal width, 1₁ = !₂ = 1₀ (say) I max = 4!0 and Jmin = 0

Thus, I = I cos2

!

max

2

4. Optical path length We can show that a thickness t in

a medium of refractive index J..l is equivalent to a length J..lt in vacuum (or air). This is called optical path length.

Thus, optical path length= J..tt.

5. If the whole YDSE set up is taken in another medium

then A. change, so~ changes 1 eg, in water A _w

= 2::9_

f..lw

I

=?

~w

=fu

f..lw

6. Displacement of fringes If a thin transparent plate of thickness t and refractive index J..l is placed in the path of one of interfering waves (say in path S₁P), then effective path in air is increased by an amount (J..l- l)t due to introduction of plate. Effective path difference in

air For maxima =

S.J>-

[S₁P

+

(J..l- 1)t] dx =(S₂P-S₁P)-(J..t-1)t=-n -(J..t-1)t D . dx _ n -(J..t-1)t=nA D D xn =d[nA+(J..l-1)t] Fringe width

~

= xn+l - xn = DA d

It means fringe width remains unchanged. Position of maxima in absence of plate.

nDA (xn)t=O

=

-d-Displacement of fringes L'.x = xn - Cxn )t=O ' D nDA

=-[nA+(J..t-1)t]--d

d

= E.cJ..t-1)t =

~CJ..t-1)t

d

A

7. In the problems of YDSE our first task is to find path difference. Let us take a typical case. In the figure shown,

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Chapter 24

•

Wave Opt

ics

L1x₁ = d sin

r:i

and L1x2 = (J..l1 -1) t1 and path of ray 2 is greater than path of ray 1 by a distance

L\x3 = d sin~ and L1x4 =

J!-!

2 -1) t2

Therefore, net path difference is/

LlX = (LlX1

+ LlX

2)-(LlX3

+

LlX4)

Interference in Thin Films

Interference effects are commonly observed in thin films

such as thin layers of oil on water or the thin surface of a soap bubble.

c

A

Consider a film of uniform thickness t and index of refraction

J..l as shown in figure. Let us assume that a monochromatic ray of light AB incident at angle i, on the upper surface PQ of the film. The ray of light will suffer reflection and refraction. The rays BC

and FG are said to interfere in reflected system and the rays DE

and HI are said to interfere in transmitted system. It can be shown that

(i) In the reflected system For constructive interference

2J.!t cos

r

= (2n - 1) A/2

For destructive interference or 2!-lt cos

r

= n'A.

(ii) In the transmitted system For constructive interference

2!-lt cos r = nA.

For destructive interference

2J.!t cos

r

= (2n-1)A/2

Obviously, conditions for thin film to appear bright/dark in reflected system are just the reverse of those in the transmitted system.

Instance 3 In a YDSE, if D

=

2 m, d

=

6 mm, /.. = 6000

A,

then

the fringe width and the position of the 3'd maxima are

(a) 0._2 mm, 0.6 mm (b) 0.8 mm, 0.1 mm

(c) 1.2 mm, 0.2 mm (d) None of these

Interpret We know that

. . AD 6000 X 10-10 X 2

Fnnge Wldth ~ = -= ₃ = 0.2mm

d 6x

10-Position of 3rd maxima 3AD

Y3 =

d

= 3~ = 3 x 0.2 = 0.6 mm

Instance 4 White light is used to illuminate the two slits in a Young's double slit experiment. The separation between the slits is b and the screen is at a distanceD > > b from the slit. At a point on the screen

directly infront of one of the slits. Find the missing wavelengths.

b2 b2 b2 b2 b2 b2 (a) (b)

D' 3D'

SD... 2D' 4D' SD ... b2 b2 b2 (c) 3D' SD' 6D ... (d) None of these

Interpret According to theory of interference, position y of a point on the screen is given by

D y = -LlX

d

For missing wavelengrhs intensity will be minimum ( = 0), if

L1 x = (2n-1) A/2 D(2n -1)A So, y = - --- -2d Given, d =bandy= b/2 b2 A

=

where n = 1, 2, 3, .. (2n -1)D So wavelength when n = 1 b2

A

-1 -(2-1)D b2 n = 2, A₂ (4-1)D n = 3, - - - = -b2 b2 etc will be (6-1)D SD

absent or missing at point P.

Instance 5 The intensity of the light coming from one of the slits in

a Young's double slit experiment is double the intensity from the other

slit. Find the ratio of the maximum intensity to the minimum intensity

in the interference fringe pattern observed.

(a)

(.J3

+

1 ) 2

.J3 -1

(c)

(-}2

-1

)2

-J2

+

1 (b)

(-12

+

1 )2

-J2

-1 (d) None of these Interpret !max __

(.jf;

+

Fz)

2 !min

.jf;

-.[!;

Chapter 24

•

Wave Optics

941

Since, Interpret Here, d = 0.1 mm = 10-4_m

Instance 6 In YDSE, the two slits are separated by0.1 mm and they are 0.5 mfrom the screen. The wavelength of light used is 5000

A.

What is the distance between 7th maxima and ll th minima on the screen ?

(a) 5.65 mm (b) 6.75 mm (c) 8.75 mm (d) 7.8 mm

D

=

0.5 m, f..= 5000

A=

5.0 x 10-7_m

· · Llx =(Xu )dark-(X7 )bright

(2xll-1)AD 7/..D 2d d ilx=

?f..D

= 7x5x1o-7x0.5 8_75x10_3 m 2d 2x10-4 ~ = 8.75mm

lntext Questions

24.~.1

(i) What is the geomenical shape of wavefront of light emerging out ofa convex lens; when point source is placed at

(ii) Can twoi,J:tdependent sources of light be coherent ? What happens to the interference pattern, if the phase diffel1enc:j'! blemreeJ!l.

sources connnuously varies?

(iii) Why does an excessively thin film appear black in reflected light?

(iv) What is the effecjpn the i.J:!,terference pattern in Young's double slit experiment due to each of Jhe following ot)eratio:ils ? (a) The width of the slits are increased equally.

(b) If the source is not exactly on the centre line between the slits.

>

Diffraction of light

Diffraction of light is the phenomenon of bending of light around corners of an obstacle or aperture in the path of light. This bending light penetrates into the geometrical shadow of the obstacle. The light thus deviates from its linear path. This deviation is more effective when the dimensions of the

aperture or the obstacle are comparable to the wavelength of

light.

The phenomenon of diffraction is divided mainly in the following two classes

(a) Fresnel diffraction (b) Fraunhofer diffraction

1. The source is at a finite distance.

2. No optical.s are required.

3. Fringes are not sharp and well defined.

Fraunhofer Diffraction

The source is at infinite distance.

Optical.s are in the foi:m of collimating lens and focusing lens are required.

I

Fringes are sharp and well 1 defined.

1. Fraunhofer's arrangement for studying diffraction at a single narrow slit (width = a) is shown in adjoining figure. Lenses, L1 and L2 are used to render incident light

beam parallel and to focus parallel light beam

,

2. Angular position of nth secondary minima is given by

.

e

A.

sm =9=

n-a

and linear distance

X = De = nDA. = nf2A.

n a a

where

f

₂ is focal length of lens L_{2 and} D = f_2. Similarly, for nth maxima, we have

sine= e = (2n + 1)/... 2a

and x = (2n + 1)DA = (2n + 1) f2A. n 2a 2a

3. The angular width of central maxima is 29 = 2" or linear

a

width of central maxima = 2Df.. = 2f2 /..

a a

The angular width of central maxima is double as compared to angular width of secondary diffraction

maxima.

4. Condition of diffraction minima is given by

a sinS= nf.. where n = 1, 2, 3, 4, ...

But the condition of secondary diffraction maxima is

a sinS= (2n + 1)2:. 2 where n = 1, 2, 3, 4, ...

5. Diffraction at plane grating When polychromatic

or monochromatic light of wavelength A. is incident normally on a plane transmission grating, the principal

maxima are where (e +d) sinS= nf.. n = order of maximum 8 = angle of diffraction e

+

d = grating element d

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Chapter 24

•

Wave Optics

6. Diffraction of circular aperture When monochromatic

light of wavelength A is incident on a circular aperture of diameter d, then angular width of dark fringe

dsin8=1.22A.or sin8= l.Z21c d

Angular radius of central maximum

. 8 1.221c Sill = -d If 8 is small, sin 8 = 8 = 1.2 2"-d

Diffraction pattern is as shown below

Principal maxima

Secondary

3/Je

•

Difference between lnterterence and Diffraction

1. Interference is the superposition of waves from two different wavefronts, whereas diffraction is due to superposition of wavelets from different points of the same wavefront.

2. Interference all maxima are equally bright and minima equally dark (perfectly black). In case of diffraction maxima are of decreasing intensity, minima are not perfectly black.

3. In case of interference, there are large number of maxima and minima, but in case of diffraction bands are few.

Interference

,

Instance 7 In a single slit diffraction experiment first minima for A₁

=

660 m coincides with first maxima for wavelength A_{2 ,} then A₂

will be equal to ,

(a) 240 nm (b) 345 nm

(c) 440 nm (d) 330 nm

Interpret Given, A

=

660 nm

Position of minima in diffraction pattern is given by a sin 6

=

n'A

For first minima of A1 , we get

asin81 = 1/....1

or sm

.

e

₁ =-A.!

a

For the first maxima approximately of wavelength A2

. 3,

asm82 = -~~.2

2

. e

3/....2 sm z = 2a The two will be considered if,

8₁ = 8₂ or sin8₁ = sin8₂ A. 3/.... 2 2 . _ l = - -2 _{or /....} 2 = - /....1 = - x 660 nm a 2a 3 3 /.... 2 = 440 nm

::~

Polarisation of light

Light is an electromagnetic wave in which electric and

magnetic field vectors vary sinusoidally, perpendicular to each other as well as perpendicular to the direction of propagation of wave of light.

The phenomenon of restricting the vibrations of light (electric vector ) in a particular direction, perpendicular to the direction of wave motion is called polarisation of light. The

tourmaline crystal acts as a polariser.

Thus, electromagnetic waves are said to be polarised when their electric field ve~tor are all in a single plane, called the plane of oscillation/vibration. Light waves from common sources are oscillation/vibration. Light waves from common sources are unpolarised or randomly polarised.

The plane ABCD in which the vibrations of polarised light are confined is called the plane of vibration. The plane EFGH which is perpendicular to the plane of vibration is defined as the

plane of polarisation.

Brewster's Law

A

_---,o

E

:

-

-

-

--r----

-

7F

I / I /

---t-

--

G

I I

a

----

-

----c

According to this law, when unpolarised light is incident at an angle called polarising angle, iP on an interface separating air from a medium of refractive index !-! , then the reflected

light is fully polarised (perpendicular to the plane of incidence),

This relation represents Brewster's law. Note that the parallel components of incident light do not disappear, but refract into

the medium, with the perpendicular components.

Note

If ~l =tan iP then iP + r =goo

siniP

From Brewster s law,~= taniP = - - .

COSIP

siniP

and from Snell s law,~=-.-smr

cosip=sinr sin (goo iP) =sin r Hence, or or

law of Malus

r=goo iP i _p + r =goo 1 tani = ~= --P sinic

cot iP = sinic (i, =critical angle)

When a beam of completely plane polarised light is incident on an analyser, the resultant intensity of light (I) transmitted from

the analyser varies directly as the square of cosine of angle(8)

between plane of transmission of analyser and polariser.

ie, .I= cos2 8

If intensity of plane polarised light incidenting on analyser is !_{0 ,} then intensity of emerging light from analyser is

I₀ cos2 8

~

\

·n

No

::_

te

:;___,_ _ _ _

_

We can prove that when unpolarised light 'of intensity /0 gets polarised on

f. 1 passing through a polaroid, its intensity becomes hal 1e, I= ~/₀

2

Polaroids

Polaroids are thin and large sheets of crystalline polarising

material (made artificially) capable of producing plane polarised

beams of large cross-section.

A polaroid sheet can also be obtained from a sheet of

polyvinyl alcohol. When such a sheet is subjected to a large strain, the molecules get oriented in the direction of the applied

strain. When the sheet is impregnated with iodine, the material

becomes dichroic. The polaroid sheet so obtained is called

H-polariod.

If the stretched sheet of polyvinyl alcohol is heated in the

presence of a dehydrating agent such as HCl, it becomes strongly

dichroic as well as very stable. Such a polaroid sheet is called K-polaroid.

A polaroid has a characteristic plane called transm1ss1on

plane. When unpolarised light falls on a polaroid, only the

vibrations paralled to the transmission plane get transmitted

Polariser

Chapter 24

•

Wave Optics

943

Uses-- ~' - t t 1 ... ' • ~

'L In sun glasses Polaroids are ·use-d· in sun glasses. They protect the eyes from the glare.

2. In window panes The polaroids are used in window P'lnes of a train and especially of an aeroplane. They help to contr;ol the light entering through the windows. 3. In three dimensional motion pictures The pictures

taken by a stereoscopic camera, when seen with the help

of polaroid spectacles, create three dimensional effect.

4. In wind shield in automobiles The wind shield of an

automobile is made of polaroid. Such a wind shield protects the eyes of the driver of the automobile from the dazzling.light of the approaching vehicles.

In addition to the above uses, polaroids are used as filters

in photography to produce and detect the plane polarised light

in the laboratory and to study the optical properties of the metals.

Instance 8 Light reflected from the surface of glass plate of refractive index 1.5 7 is linearly polarised. What is the angle of reji·action

of glass ? (a) 32.SO

(c) 30.2°

(b) 42S

(d) 44S Interpret Given that ~t = 1.57

According to Brewster's Law

~l =tan iP = 1.57, iP = tan-1 _{(1.57) = 57.5°} As r = 90° ~ iP orr= 90° ~ 57.5° = 32.5°

>

Doppler's

Effect in light

The phenomenon of apparent change in frequency

(or wavelength) of the light due to the relative motion between the source of light and the observer is called Doppler's effect in light.

When the source of light and the observer approach each

other, the frequency of the light appears to be more than the actual

frequency of the light emitted from the source (or the wavelength of the light appears to be lesser than the actuial wavelength). On

the other hand, when the source of light and the observer recede

from each other, the apparent frequency of the ljght decreases or the apparent wavelength increases. The change in frequency (or the wavelength) of the light for the given relative velocity of the source and the observer is independent of the fact, whether the source is in motion.or the observer is in motion.

Consider that a source of light emits light waves of frequency

v and wavelength A. If the light travels with velocity c, then

c = ,. A.

A stationary observer will receive v waves per second. In case, the observer moves towards the source of light with velocity v along the direction of propagation of light, then the number of waves received per second ie, apparent frequency of

light will be

v' = v

+

number of waves contained in distance equal to v

v v Y+-=Y+

-A

C/Y v' = v

(1

+

~

)

or v' = v

(1+~

)

•

944

Chapter 24

•

Wave

Optics

It follows that v' > v ie, when the source and the observer approach each other, apparent frequency of the light increases.

When the source of light and the observer move away

from each other, then the apparent frequency of light can be obtained from above equation by changing v to -v. Therefore,

when the source of light and the observer move away from each other, the apparent frequency of the light is given by

v'

=

v(l-~)

It follows that v'

<

v ie, the apparent frequency of light aecreases, when the source of light and the observer move away from each other.

For apparent frequency of light may be written in the combined form as

v

'=v(l±

~)

Doppler shift. The apparent change in the frequency (or the wavelength) of the light due to relative motion between the source and the observer is called Doppler shift.

The apparent change in frequency is given by v'-v = ±~v

c

The apparent change in frequency v'- vis denoted by ~v and is called Doppler shift. Therefore,

v

~v = -v c

The positive sign is to be considered, when the source and the observer approach each other and the negative sign, when the two move away from each o,ther.

,., by

Doppler shift in terms of the wavelength of the light is given

~A.=

±~A.

c

In equation, the negative sign corresponds to the case,

when the source of light and the observer approach each other.

The negative value of M indicates that the wavelength of the light decreases, when the source of light and the observer move

towards each other. It is called blue shift.

On the other hand, in equation the positive sign corresponds

to the case, when the source of light and the observer move

away from each other. The positive value of D.A. indicates that the wavelength of the light increases, when the source of light

and the observer move away from each other. It is called red shift.

Instance 9 The spectral line for a given element in light received

from a distant star is shifted towards the longer wavelength by 0.32%. Deduce the velocity of star in the line of sight.

Interpret Here, M = 0·032 (postion, as the .shift is towards

A

100

longer wavelength)

Now, Doppler shift is given by

~A= -~A

c

V = - !:iA C =-0.032 X 3 X 108

A 100

= -9.6 x 104 _ms-1

The negative sign indicates that the star is receding.

uestions 24.2

Chapter 24 "' Wave Optics

945

Chapter Compendium

1. Ray of light is drawn perpendicular to the wavefront.

2. Huygens' Piinciple

(i) Each pomt on given Or primary wavefront acts as

a.

source of secon~ary:vavelets, sending out disturbance~

all directions in a similar manner as the origil1al sogrce oflight. ·

(ii). The new positi<m of the wavefront at any instant is the em,.~lope of the secondary wavelets at that instant.

3. If two coherent waves with intensities 11 and lz are

superimposed with a phase difference off, the resulting wave

intensity is

I= 11 + 12 + 2~I

1

12 cos$

For maxima, x"' .n'A (A.= optical path difference)

Forminima,x=(rt-~)A.

or

(n+~)A.

2 2

Phase difference,

q,

=

27t (x)

'A

Ratio of maximum to mininmm intensity

lmax _ Cal + az)2

[min - (al-

azP

f1

P1

a~

Also, Iz =

Pz

=a~

4. Young's double slit experiment

(i) .. For maxima ,

or y

=

(ii) For rninirna,

DA- 3DA

Y=+-_{- 2d ,}+ -_{_ 2d} ·- _?· so on

(iii) Fringe width,

p

= AD

d

(iv) When a fil:nl of thickness t and refractive index

f.t

is

introqw:ed in the path?f one ofth~source's light;then fringe shift occurs as the optical patl;ldifference changes.

Optical path. difference at P

L'lx

=

S₂P"-(S1P·qtt t]

= S₂P-5₁ P-(!+ -l)U y. (d/D)--; (Jl -l)t

The fringe shift is given by, 4)>.= D(J.t-l)t

d

(v) Intensity variation on screen

IfJ

₀

~. the intensity of light beam coming from each slit,

then the resultant intensity at a. pqi.):lt where they have

a phase difference

pf

¢ is given by I== 41₀ cos2

%,

_where

_q,

₌ 2

7t(d~ine)

5 .. Diffraction is the phenomenon of bending ()flight round the sharp corners and spreading into the regions of

.geometrical shadow is called diffraction. . 6. Diffraction from a slit

(i) Angular position of the nth secondary minimmn

nA.

a

(ii) Distance of the nth secondary

nDA.

centre of the screen xn =

-;-maximum from the

(iii) Angular positions of the nth

9, _ (2n + l)A.

secondary m imum

n··-2a

(iv) Distance ofttte n'h secondary maximum from the c~ntrt=

of the screen x' = (Zn + l)DA.

11

2a

( v) Width of a secondary maximWP-or mininlurn

p

= (vi) Width of the centraLmaximum

p

₀

=

2,Df::

a

a

phenomenon due to which vibrations light

restricted in a parti<;ular plane is called polarisation.

8. Brewster's Law states that when light is incident at polarising angle, the refl:ected and refracteq rays are perpendicular to each other.

Mathematically J.l=tan iP

9. Law o~Malus states that when a completely plane po~ris~d

light beam is incident on an analyser tlie intensity of tht= emt;rgent: light varies as the square pf the cosh1e of the angle between the plane of trans:rni5~ion of ·the analyser and the polariser

Mathematicap.y

e

10. Doppler shift.

(a) Llv=±~v

Example 1 Ifamplitude rqtio of two sources producing interference is 3 : 5, th,e ratio pfinten,sities bfmaxima cirid'minima is

(a) 5 : 3 (b) 16 : 1

(c) 25 : 16 (d) 25 : 9

Solution We know in ~terlere~ce, ,

1max

=

c.ji;

+

JI;Y

while 1min

=

.

c.Ji;

~

J[;y

I • J , I . 1 I l • • ~ '' ' ' I;,~ _ C.ji; +

.ji;y

_

(A

₁ + A₂)2

!min-

(.ji;-;~1

2

)

2 -:-'(Al -A2)2

According to problem . ·. ·. ·. · ' 1 ~·, rllr ~~ ·~lr{,ax ~ (3+5)2 ; 16, -1 '. .·.: .... ·'·' flz'

·

s

'dmin (3'- 5)2 1 • r ~ t, ''~.: ·· ' t • ' 1 •• - ... r, _I:.' ·' ' . ' ' ' .

Examp'le 2 . 'Monochromatic green light; of wavelength 5 x 10-7_m

illuminates a pair of slits 1 mm apart. The separation of bright lines on

the interference pattern formed •on a screen 2m away is

· (a) Ldmm1

• • • (b)

o.b1

mm

(c) 0.1 mm (d) 0.25 mm

Solution Separation ofprig~t line.$ o.r.fl;i.nge width

D'A Sxl0-7 _{x2 _}

3

•,· ..

·-.~=~-=.

·

···

·

:3·

·

m=lO m=l.O

·

mm

. .

·,·.

·

!J -~~~-r11~ ~fll ,'lQ,- • J I 1 1 , · ; · 1 \ .~,

Example 3 Young's double slit

eX:p~r

iment

is'·ccirried out wing

microwaves of wavelength A "''-·3 em. Distance between the slits is

d

=

5 em and the distance between the plane of slits and the screen is D

=

100 em. Then ·what is the number 6pin'ax:imas 'and their positions

on the screen? '

-·•

·

(a)

,)'

3

/

0

1 Qnd

±

'

7s

~J1i' ~ r - I (c) 3, 1 and± 55 em Solution

·

(b}

4;1

and-±

6'0 d;z

(d) None of these

PT

s1

Y

s2

~---

---

Jl

I I

The maximwn path difference that can be produced

=

distance

between the sources ie, 5 em.

Hence, in this condition we can have only 3 maximas one . central maxima and two on its either side for a path difference of/.. or3 em.

Now, for maximwn intensity at P

S2P-S1P

='A

,

or

~(y

+ d/2)2 + D2 -

~(y-

d/2)2 + D2 =

i

Putting, d

=

5 em, D

=

100 em

and A = 3 em, we get

y

=

± 75 em

Thus, the three maximas will be at

y

=

0 andy

=

± 75 em

Example 4 In Young's double slit experiment, an interference

pattern is obtained for A

=

6000

A,

.coming from. two coherent sources

51 'and 52 . At certain point P on the screen third dark fringe is formed.

Then the path difference S₁P- S₂P in microns is ' (a) 0.75 (b)o 1.5

'<'(c) 3.0 ·fd) 4.5

· •' ' · • (2n-1)/..

Solution For dark fringe at-P, ,S₁P- S₂P

=

--~

2 5/.. S1P-S2P = -2

=

5 x 6000

=

15000-

=

1.5 micron 2

Exam pi~ 5 · Tw.o slits. are separated.by a distance of 0.5 .mm and illuminated with light of A

=

6000

A.

If, the screen is placed 2.5 m

from the slits . .The ~tance .of the third bright image froTI! the centre

will be

(a) 1.5 mm (b) 3 mm

. (c) 6 mm (d) ,9 mrn.

Solution Distance of n'h bright fringe from the centre

nD'A

Yn=d

Y3

=

3 X 6000 X 10-lO X 2.5 0.5 X 10-3 ---~9 x 10-3_m

=

9 nun. 0 I • ~

Example 6 In a double slit arrangement fringes are produced

using light of wavelength 4800

A.

One slit is covered by a thin plate

of glass of refractive index 1.4 and the other with another glass plate

of same thickness but of refractive index 1. 7. By doing so the central

bright shifts to original fifth bright fringe from centre. Thickness of

glass plate is

(a) 8 ~tm (b) 6j..i.m

(c) 4!lm (d) 10j..i.m

Solution When one slit is covered by a·thin plate of glass, then

when another slit is covered by a thin plate of glass of different

refractive index, then shift

~ ~Yz = i(llz -1)t Net shift ~y = ~y₂- ~y₁ ~ = i(llz -111) t Here given, t..y=5~ 5

~

=

t

(Jlz -Ill) t

Example 7 What is the minimum thickness of a soap film needed for constructive interference in reflected light,

if

the light incident of the film is 750 nm? Assume that the index for the film is f.l

=

1.33

(a) 282 nm (b) 70.5 nm (c) 141 nm (d) 387 nm

Solution Here, 2 f!t

=

A/2

A

.

tmin =-=141 nm

411

Example 8 A beam of light of wavelength 600 nmfrom a distant

soilr'Ge faits on a single slit 1.0 mm wide and the resulting diffraCtion '

pa~tern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe ?

(a) 1.5 mm (b) 1 mm (c) 1.2 mm (d) 2.4 mm

Solution For the di.ffiaction at a single slit, the position of mininla is given by

a sin 8

=

n/..

For small value of 8

,

sin 8 == 8

=

y/D D

a (y!D) =/..or y=-A a

... l

Chapter 24

•

Wave

Optics

947

Substituting the values, we have 2x6x1o-7

y = = 1.2 x 10-3m= 1.2 mm 1 xl0-3

. . Distance between first minima on either side of central

maxima

=

2y

=

2.4 mm

Example 9 A parallel beam ofmenochromatic light ofwaveiength 450 nm passes through ~~siit of ~i~tlf. ~r.2

mrri.

·Find th~ q'ngular

divergence in which most of the light is' diffracted. (a) 4.5 x 10-3 _{rad (b) 5.5 x 10-}3 _rad

(c) 3.2 x 10-2 _{rad (d) 3.5 x 10-}3 _rad

Solution Most of the light is diffracted between the two first order mininla. These minima occur at angle given by,

. · .· A A5ox·lo-9,',',.

asm9=±nA,sm9=±-=± ,

·' a 0.2x10~3 _' '=·±

2.~5x10-

3

rad

:. angular divergance = 4.5 x'10-'-3 'riid

Example 10 · Two poldr~ids are ~rosseil to each 'other. Now, one of them is rotated through 60°. What percentage of incil}.ent unpolarised

light will pass through the system ?

(a) 37.5 % (b) 45.6%

(c) 52.3% (d) 66.7%

Solution Let I0 be intensity of the incident unpolarised light on

the system. As the in<;ident .light unpo~aris~q, the intensity of light on being transmitted from the first polaroid will become

· I - I

cos;~,Ei

'--'- Io (:.' eos2

8=_!_)

'

- a

-2

. .

2

If 8 is the angle between the trans~sion planes

pf

the_ two

polaroids, then intensity of light on passing through the s~ond

polaroid is given by

·f=Jc6s2_9=··10 cos2

e

)

.

·

..

~

....

2

Since, the ··!Wo polaroids were initially cros~ed; -the angle between the transmission planes of the two' polaroids on r~tating thr~mgh 60° will become , , .. , . , , , , ,

,8

=

·

90° -:-60° = 306 _{· ' · •}

.

_{'• ·I}

.

(

~)2

' .

0 -·-·2••• I v3

·---. , _' J'c=_'' .-XCOS _{2'' I ,} 30"=_Q__{• 2} X _' -_{2 •'} =0_{' . '} .37510 , · ·

·.

Hence, per.centage of tncident light that pass~s through :the '

system, · ' · ~ ' I_ ' ~ ~ I' -x100

=

0.375 X 100

=

37.5% Io ' I ' ' 1' ... , • • •• ~ • ~ ' ' ' ' .\

.

.

~ .. . I I ,~ l J l . r · : I ! '1.;,..;, f.( r.t ' ' ,· ,o ~ I

'·

Chapter Practice

Exercise I

Theories

of

Light and Interference

1. Light propagates 2 em distance in glass of refractive index

1.5 in time t_{0 .} In the same time t_{0 ,} light propagates a

distance of 2.25 em in a medim:n. The refractive index of

the medium is

(a) 4/3

(c) 8/3

(b) 3/2

(d) None of these

2. The wave front due to a source situated at infinity is (a) spherical (b) cylindrical

(c) planar (d) None of these

3. Which of the following is not an essential condition for interference?

(a) The two interfering waves must be propagated in

almost the same direction or the two interfering

waves must intersect at '\'ery small angle

(b) The wave must have the same period and

wavelength

(c) The amplitude of the two waves must be equal

(d) The two interfering beams of light rriust originate

from the same source

4. In Young's double slit experiment with monochromatic light of waveiength 600 nm, the distance between slits is 10-3 _{m. For changing fringe width by 3}

_x

_10-5 _{m .}

(a) the screen is moved away from the slits by 5 em

(b) the screen is moved by 5 em towards the slits (c) the screen is moved by 3 em towards the slits

(d) Both (a) and (b) are correct

5. In Young's double slit experiment, the distance between

slits is 0.0344 mm. The wavelength ~f light used is

600 nm. What is the angular width of a fringe formed on a distant screen?

(a,) 10 (c) 3°

6. A Young's double slit experiment uses a monochromatic

source. The shape of the interference fringes formed on a

screen is

(a) hyperbola (c) straight line

(b) circle

(d) parabola

7. If Young's double slit experiment, is performed in water

(a) the fringe width will decrease (b) the fringe width will increase

(c) the fringe width will remain uncha'nged

(d) there will be no fringe

,

8. In Young's double slit experiment, the spacing between the slits is d and wavelength of light used is 6000

A.

If the angular width of a fringe formed on a distance screen is 1°, then value of d is

(a) 1 mm

(c) 0.03 mm

(b) 0.05 mm

(d) 0.01 mm

9. In Young's double slit experiment, when violet light of wavelength 4358

A

is used, the 84 fringes are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen _in the field of view, the wavelength of sodium light is

(a) 6893

A

(b) 5904

A

Cc) 5523

A

Cd) 6429

A

10. In a double slit experiment, 5th dark fringe is formed opposite to one the slits. The wavelength of light is

d2

d2

(a) - (b)

-6D 5D

11. Two non-coherent sources emit light beams of intensities I and 4I. The maximum and minimum intensities in the resulting beam are

(a) 9I and I (b) 9I and 3I

(c) 5Iandi (d) 5Iand3I

12. The maximum intensity in the case of n identical

incoherent waves, each of intensity 2 wm-2 _{is 32 wm-}2

• The value of n is (a)

4

(c) 32 (b) 16 (d) 64

13. In an interference pattern the position of zeroth order maxima is 4.8 mm from a certain point P on the screen. The fringe width is 0.2 mm. The position of second maxima from point P is

(a) 5.1 mm (b) 5 mm

(c) 40 mm (d) 5.2 mm

14. 5₁ and 5₂ are two coherent sources. The intensity of both sources are same·. If the intensity at the point of maxima is 4 wm-2

, the intensity of each source is

(a) 1 wm-2 _{(b) 2 wm-}2

(c) 3 wm-2 _{(d) 4 wm-}2

15. n incoherent sources of intensity I0 are superimposed at a point, the intensity of the point is

16.

(a) n !₀ (b)

!.9_

n

(c) n2_I

0 (d) Noneofthese

In the given figure, C is middle point of line S₁S2• A

monochromatic light of wavelength A is incident on slits. The ratio intensity of S₃ and S₄ is

(a) zero (c) 4; 1

~c~: ~~

.}J._o---(b) 00 (d) 1 : 4

17. In the given arrangement, S1 and S2 are coherent sources

(shown in figure). The point Pis a point of

ft

·IP

Sz

(a) bright fringe (b) dark fringe (c) either dark or bright (d) None of the above

18. When two coherent monochromatic light beams of intensities I and 4 I are superimposed, what are the maximum and minimum possible intensities in the resulting beams?

(a) SI and I

(c) 91 and I

(b) SI and 31

(d) 91 and 31

19. A parallel beam of light of intensity 10 is incident on a

glass plate, 25% of light is reflected by upper surface and 50% of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is

(a)

[~+If]z

2-H

5 (c) -8 (b) (d)

~

5

20. In Young's double slit experiment, the length ofband is 1 mm. The ring width is 1.021 mm. The number of fringes is

(a) 45 (b) 46

(c) 47 (d) 48

21. In a Young's experiment, one of the slits is covered with a transparent sheet of thickness 3.6 x 10-3 _{em due to which}

position of central fringe shifts to a position originally occupied by 30th fringe. The refractive index of the sheet, if A = 6000

A,

is

(a) 1.5 (b) 1.2

(c) 1.3 (d) 1.7

22. Light waves travel in vacm1m along the .Y-axis. Which of

Chapter 24

•

Wave Optics

949

(a) y = constant (c) z = constant

(b) x = constant (d) x

+

y

+

z

=

constant 23. The correct curve between fringe width ~ and distance

between the slits (d) in figure is

(a)~~ ~)~~

~

- d - d

~1

/

1

~

(c)

j

L__

(d)

~

L_

- d - d

24. The correct formula for fringe visibility is

(a) V = [max-[min (b) V = I max+ !min

[max +[min [max - [min

(c) V =I max _(d) V =I min

I min [max

25. Two coherent waves are represented by y₁ = a₁ cos rut and y2 == a2 sin rut, superimposed on each other. The resultant intensity is

(a) (a₁

+

a₂₎ (b) (a_{1 -} a₂₎

26. Light appears to travel in straight lines since (a) it is not absorbed by the atmosphere

(b) it is reflected by the atmosphere (c) its wavelength is very small (d) its velocity is very large

27. The equations of two interfering waves are y₁ = b cos rut and

Yz

=

b cos (rut

+

<!>). For destructive interference the

path difference is

-(a) 0° (b) 360°

(c) 1806 _{(d) 720°}

28. The Young's double slit experiment is performed with blue and with green light of wavelengths 4360

A

and 5460

A

'··respectively. If x is the distance of 4th maxima from the

· central one, then' ' (a) x (blue) = x (green)

(b) x (blue) > x (green)

(c) x (blue) < x (green)

(d) x (blue)/x (green) = 5400/4360

29. In a double slit interference experiment, the distance

between the slits is 0.05 em and screen is 2 m away from

the slits. The wavelength of light is 8.0 x 10-5 _{em. The}

distance between successive fringes is

(a) 0.24 em (b) 3.2 em

(c) 1.28 em (d) 0.32 em

30. Two light rays having the same wavelength A in vacuum

are in phase initially. Then the first ray travels a path L₁ through a medium of refractive index n_{1 ,} while the

950

Chapter 24 • Wave Optics

to observe interference. The phase difference between the

two waves is

(c) 2rr (n2LI - ni L2) (d) 2rr (LI - L2 )

A . A ni n2

31. Through quantum theory of light we can explain

a number of phenomena observed with light, it is necessary to retain the wave nature of light to explain

the phenomenon of.

(a) Photoelectric effect (c) Compton effect

(~) Diffraction

(d) Black body radiation 32.-If the intensities of the two interfering beams in Young's

double slit experiment be II·. and I2 , then the contrast

between the maximum and minimum intensity is good

·,·.when

(a) II is much greater than I₂ (b) I I is much smalle'r than I 2

(c) II = I2

(d) Either II = 0 or I2 = 0

' 33. The fringe width a:t' a distance of 50 em from the slits in Young's experiment· for light of wavelength 6000

A

is 0.048 em. The fringe width at the same distance for

A=

sooo A

will bE~,

35. 36. (a) 0.04 em (c) 0.14 em (b) 0.4 em (d) QAS em

Two waves originating from sources SI and 52 having zero phase difference and common wavelength A will show complete destructive interference at a point P, if

(S₁P-S₂P) = r.

-(a) 5 A (b) 3A

4,

(d) llA

2 '.·

Two coherent sources of intensities II and I2 produce an interference pattern. The maximum' intensity in the ·interference pattern will be '.

(a) I1 .+·I2 (b)' I~ +I~''

(c) CII +I₂

i

(d)

·

cji;

+fi;f

In Young's double slit experiment, the s;v€mth maximum with wavelength A1 is at a distance di and the same

maximum with wavelength A2' is at a distance d2 . Then

dl/d2 =

) AI

(a - · '·A2 r c)

Ai

,

A~

)._2

(d)

A.~

I

37. When monochrbm:atic light is replaced by white light in Fresnel's qiprism arrangement, the central fringe is

(a~ col.0ured1 · • ' • (b) white (c) dark (d) None·ofthese

38. In i:he

~ei:

up shown in figure, tlie

tw~

slits, 5₁ and 52 are

not equidistant from: the slit S. The central fringe at 0 is,

then , ·

I

sLf--- o

~

s2

I

(a) always bright (b) always dark

(c) Either dark or bright depending on the position of S (d) Neither dark nor bright

39. In a double slit interference experiment, the distance between the slits is 0.05 em and screen is 2 m away from the slits. The wavelength of light is 6000

A.

The distance between the fringes is

(a) 0.24 em (c) 1.24 em

(b) 0.12 em

(d) 2.28 em

40. The separation between successive fringes in a double slit arrangement is x. If the whole arrangement is dipped under water, what will be the new fringe separation? [The wavelength of light being used in

soooA]

41.

42.

(a) 1.5 X (b) X

(c) 0.75 x (d) 2x

In a Young's experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one meter away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used will be

(a) 60 x 10-4 _{em (b) 10 x 10-}4 _em (c) 10 x 10-5 _{em (d) 6 x 10-}5 _ern

In the Young's double slit experiment, the interference pattern is found to have an intensity ratio between bright and dark fringes as 9. This implies that

(a) the intensities at the screen due to two slits are 5 units and 4 units respectively

(b) the intensities at the screen due to two slits are 4 units and 1 unit respectively

(c) the amplitude ratio is 3

(d) the amplitude ratio is 2

43. Interference fringes are being produced on screen XY by the slits 51 and 52• In figure, the correct fringe locus is

S1

!

w1

_, _,---W3 /

s

\ ' Q Sz! _Wz '

_----·

_w4

(a) PQ (b) W₁W₂ (c) W₃W₄ (d) XY

44. Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance of 0.14 m. The frequency of transmitter is

(a) 1.5 x 1010 _{H (b) 10}10 _H

(c) 3 x 1010 _{H (d) 6 x 10}10 _H

I

I

45. Two waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to

(a) 10 : 8

(c) 4 : 1

(b) 9 : 1

(d) 2 : 1

46. In Young's double slit experiment, phase difference between light waves reaching 3rd bright fringe from the central fringe when A = 5000

A

is

47. 48. 49. 50. 51. 52. 53. (a) 6 n (b) 2 1t (c) 4n (d) zero

In an interference pattern produced by two identical slits, the intensity at the slit of the. central maximum is I. The intensity at the same spot when either of the slits is closed is I0 • Therefore

(a) I = 10

(b) I= 2 I0 (c) I = 4 I0

(d) I and I0 are not related to each other

In a two slits experiment with monochromatic light,

fringes are obtained on a screen placed at some distance ·from the slits. If the· screen is moved by 5

x

10-2 m towards the slits, the change in fringe width is 3

x

10-5 _m.

If separation between the slits is 10-3_{m, the wavelength of}

light used is (a) 4500

A

(c) 5000

A

Cb) 3ooo

A

(d) 6ooo

A

In Young's double slit experiment, let 5₁ and 5₂ be the two slifs, and C be the centre of the screen. If LS₁CS₂ = 8 and A is the wavelength, the fringe width will be

A • (a)

e

(c) 2 A

e

(b) A8 (d) _!:____ 28

In Young's double slit .experiment, tl).e intensity on screen at a point where path difference .is

A

-

is K. What will be intensity at the point where path difference is A/4? (_a) K/4 (b) K/2

(c) K (d) zero .

In the Young's double slit ex'Perimefif, a· mica slip of thickness t and refractive index 11 is introeluced in the ray from first source 51'. By how much distance fringes

pattern will be displaced.

d (a) -C~-t-1)t D d (c) (f..L-1)D D -(b) -C~-t-1)t d . . -D (d) d(f..L-1)

Oil floating on water looks coloured due to interference of light. What should be the order of magnitude of thickness of oil layer in order that this effect may be observed?

(a) 10,000

A

(b) 1 em

(c) 10

A

Cd) 100

A

" ' !')'l

Two waves y₁ = A₁ sin(wt-~₁) and .h .= A2 sin(wt-~2) superimpose

to

form a resultant wav~·wnose amplitude is (a)

~~

+ 2A₁ A₂ cos(l3₁-

-

13

_{2 )}

(b) ~A~+A~+2A

1

A

2

sin(l3

1

-13

2

)

(c) A1

+

A2

Chapter 24

•

Wave Optics

951

54. In Young's double slit experiment, 12 fringes are. obtained to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by

(a) 12 (b) 18

(c) 24 (d) 30

55. In Young's double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 em. Wavelength of light used is 5460

A.

Then-angular position of first dark fringe is

(a) 0.08° ~b) 0.16° (c) 0.20° (d) 0.32°

56. Two beams of light having intensities I and 4I interfere to produce a fringe ,pattern on a screen. The phase difference between the beams. is n/2.at point A and 1t at point B. Then the difference between the resultant intensities at A andB is

(a) 2I (b~.4I

(c) 5 I (d) 7 I

57. In a Young's double slit experiment usilig red and blue lights of wavelengths 600.nm and 480 urn respectively, . · .. the value of n from which the nth red fringe coincides with

(n

+

1) the blue fringes is

-(a) 5 (b) .• 4. , , "~"-'

(c) 3 (d) 2

58. In Young's double sfit experiment, distance between sources is 1 mm and distance betw~en the screen and source -is 1m.

-If the frin.ge width on the screen is 0.06 em, •then A is (a) 6ooo

A

Cb) 4ooo

A

(c) 1200

A

Cd) 24oo

A

59. In Young's double s.lit experiment, the central bright fringe can be identified

(a) as it has greater: intensity than the other bright fringes (b) as it is wider than the other bright fringes

(c) a,s ,it is 11arrower than the other bright :(ringes

,_: , (d) by using white J,ight inst~iid of ffi9nochromatic light 60. Two slits, 4 mm apart are illuminated by-light of wavelength

600

A.

Wnat Will be the fringe width on ·a screen placed 2 m from the slits ?

(a) 0.12 mm (c) 3.0 mm I i ' , ~ ' ' (b) 0.3 mm' · (d) 4 .. 0.11llT]-.

.

Diffraction of Light

,

. '

61. A parallel beam of light of wavelength 3141.59

A

is incident on a small aperture. After passing through the aperture; the beam is no longer parallel but diverges at 1° to the incident direction. What is the diameter of the aperture?

(a) 180m (b) 18 ~-tm

(c) 1.8 m (d) 0.18

:U

62. To observe diffraction,• the size of. an fiperture

(a) should be ofthe same order as wavelength should be much larger than the wavelength

(b) should be much larger than the wavelength (c) have no relation

t9

wavele~gth ' ' -- .