24.Wave Optics

T

Chapter Overview ,. • ,. • •

Introduction Newton's Corpuscular Theory Huygen's Wave Theory Coherent Sources Interference of Light

~>

" Diffraction of Light "' Difference between Interference and Diffraction g Polarisation of Light " Doppler's Effect in Light

Introduction

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.

~>

as shown in figure by circles 1, 2, 3 and 4. Rays are radial as shown by arrows.

Direction of propagation

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 p at any instant in the medium affected by the waves originated Rays from a source, on which S -E----+---r----'-- Q each· particle vibrates in same phase is called the wavefront. R The wavefront may also be defined as the S =source of light, AB =wavefront and SP, SO hypothetical surface on and SR are rays of light. which the light waves are in same phase. As the wave travels forward, the wavefront also moves forward . 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

In spherical wavefront Amplitude . I ntens1ty

oc

oc

_!. r

1

2r (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.

Chapter 24 • Wave Optics Resultant wave

A'

937

Second wave

(a) Constructive interference First wave

(b) Destructive interference

B'

(b)

(a)

> Coherent Sources '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.

Conditions of Maxima and Minima Lety1 =A1 sincotandy2 =A2 sin (cot+~) be two sin1ple harn10nic waves of same frequency travelling in tlle same direction, A 1 and A 2 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 = y 1 + Y2 = A 1 sin cot+ A 2 sin (cot+ ~) From this expression, the resultant amplitude is given by 2 A =A~+ A~+ 2A1A 2 cos The resultant intensity I is given by I =A2

I= ! 1 + ! 2 + 2JT;I; cos where The

~is

Resultant wave

the phase difference.

2M cos



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 =I1 +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 unifo rm 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 minimum is called the destructive interference.

So,

I= A~+ A~+ 2A1A 2 cos

2M

I == I 1 + I 2 + cos 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~+ 2A1 A 2

= (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 Imin= A~+ A~- 2A1A 2 = (A1 - A 2 ) 2 So,

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



= ± 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

( a) 18

(b) 16

23

25

25

(d) 23

16

18

(c)

Interpret

.

GIVen

I

_l

I2

A12

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

2

(SlP) = (SlM)

~=~ or

A1

1

=

= (S2 Qi + (PQi = D 2

In b.S1PM,

81

==1 \ A~

A2

2

... (i)

9A2

SzP

+ S 1P = D + D =

2D dx 52 P-51 P=D

..

From Eq. (i), we have

2+ (PM) 2= D 2+ (X- %Y

So, (S 2P) 2 - (S 1P) 2 = 2dx or (SzP + S 1P) (SzP- S 1P) = 2dx Now since, P is very close to 0, we can have to a first approximation assuming that

I max - (Al + Az)2 I min - (Al- Az) 2

+( x +%Y

So, path difference between the interfering waves is dx . D

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

I max

(i) Position of bright fringes or maxima on the screen Constructive interference occurs when the path difference is an even multiple of A/2.

c1oi 25 C8i = 16

ie,

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 (d) 6.441 I

(c) 4 I

or

AD

= 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 5 1 and 5 2 • Slits 5 1 and 5 2 are situated such that the distance SS 1 is equal to the distance 552 • Hence, the waves emanating from the slit S reach 5 1 and 5 2 simultaneously, ie, the waves at 5 1 and 5 2 are in same phase. Infront of the slits 5 1 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.

· (1st bright fringe)

d 2AD

n = 2, x 2 =-d-

~a 2 + b2 + 2 abcos- [S1P + (J..l- 1)t]

dx =(S2P-S1P) -(J..t-1)t=-n -(J..t-1)t

D

.

For maxima dx

_ n

D

-(J..t-1)t=nA xn

Fringe width

~ = xn+l -

D

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

= DA d It means fringe width remains unchanged. Position of maxima in absence of plate. nDA (xn)t=O = -dxn

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, path of ray 1 is more deviated than path of ray 2 by a distance

940

Chapter 24 • Wave Optics Interpret We know that . . AD 6000 X 10-10 X 2 Fnnge Wldth ~ = - = = 0.2mm d 6x 10-3 Position of 3rd maxima 3AD Y3 = d = 3~ = 3 x 0.2 = 0.6 mm

L1x1 = d sin r:i and L1x2 = (J..l 1 -1) t 1 and path of ray 2 is greater than path of ray 1 by a distance L\x3 = d sin~ and L1x4 = J!-!2 -1) t 2 Therefore, net path difference is / LlX = (LlX 1 + LlX2)-(LlX3 + LlX4)

Instance 4 White light is used to illuminate the two slits in a Young's double slit experim ent. 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. (a)

b2

b2 b2

b2 (c)

b2

(b)

D' 3D' SD ...

b2

b2

b2

2D' 4D' SD .. .

b2 (d) None of these

3D' SD' 6D ...

Interpret

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.

According to theory of interference, position y of a point on the screen is given by D y =- LlX

d

c A

For missing wavelengrhs intensity will be minimum ( L1 x = (2n- 1) A/2 D(2n -1)A So, y = - -- - -

=

0) , if

2d

Given, d =band y= b/2

A= Consider a film of uniform thickness t and index of refraction 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. J..l

Instance 3 In a YDSE, if D = 2 m, d = 6 mm, /..

= 6000

the fringe width and the position of the 3'd maxima are (b) 0.8 mm, 0 .1 mm (a) 0._2 mm, 0.6 mm (c) 1.2 mm, 0.2 mm (d) None of these

A,

then

b2 where n = 1, 2, 3, ..

(2n -1)D

So wavelength when n = 1

b2 A---1-

n

=

n

= 3,

b2

A2

2,

(2-1)D

(4-1)D

b2

b2

- - - = - etc will be

(6-1)D absent or missing at point P.

SD

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. 2

(a)

(.J3.J3 +

(c)

(-}2 -1 )2

(b)

1) -1

Interpret

!min

-J2 -1

(d) None of these

-J2 + 1

!max _

(-12 + 1 )2

(.jf; Fz) .jf; -.[!; +

2

Chapter 24 • Wave Optics Interpret

Since,

Here, d = 0.1 mm = 10- 4 m D

··

941

= 0 .5 m, f..=

5000 A= 5.0 x 10- 7 m

Llx =(Xu )dark- (X7 )bright

(2xll-1)AD 2d

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

7/..D

?f..D = 7x5x1o-

d 7

_ x _ m 8 75 10 3

x0.5 2x10-4 2d = 8 .75mm

ilx=

~

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

2. Angular position of nth secondary minima is given by

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

. e =9= n-A. sm 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 = ( 2 n + 1)/... 2a and

x = (2n + 1)DA = (2n + 1) f 2 A. n

1. The source is at a finite

distance. 2. No optical.s are required.

3. Fringes are not sharp and well defined.

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.

Fraunhofer Diffraction 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

,

Slit

Screen

2a

2a

2 3. The angular width of central maxima is 29 = " 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 (e +d) sinS= nf.. where n = order of maximum 8 = angle of diffraction e + d = grating element d = opaque part

942

Chapter 24 • Wave Optics 6. Diffraction of circular aperture When monochromatic

Instance 7 In a single slit diffraction experiment first minima f or

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

A1 = 660 m coincides with fi rs t maxima for wavelength A2 , then A2 will be equal to , (b) 345 nm (a) 240 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

dsin8=1.22A.or sin 8 = l.Z21c d Angular radius of central maximum .

1.221c d

Sill 8 = - -

asin8 1 = 1/....1

2 If 8 is small, sin 8 = 8 = 1. 2"d Diffraction pattern is as shown below

. e1 = -A.! sm

or

a

For the first maxima approximately of wavelength A2

.

Principal maxima

3,

asm8 2 = -~~. 2 2 . e 3/....2 sm z = 2a

Secondary

The two will be considered if, 81 = 8 2 or sin8 1 = sin8 2

A. 3/.... 2 2 2 . = -or /.... 2 = - /.... 1 = - x 660 nm a 2a 3 3 /.... 2 = 440 nm

_l

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.

::~

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. A

---------,o E :

-----r-----7F

Interference

I I

/ /

-----t---G I

a ----- ----cI

Brewster's Law 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) , provided

,

Chapter 24 • Wave Optics 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 From Brewster s

law,~=

siniP

taniP=--. COSIP

and from Snell s

siniP

law,~=-.-

smr cosip=sinr sin (goo iP) =sin r

Hence, or

r=goo iP i p + r =goo

tani =

1

~=--

sinic cot iP = sinic P

or

(i, =critical angle)

law of Malus 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 = cos 2 8 If intensity of plane polarised light incidenting on analyser is ! 0 , then intensity of emerging light from analyser is I 0 cos 2 8

Uses- - ~' t ~ 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. -t

1

.....

'

•

'L

Instance 8 Light reflected from the surface of glass plate of refractive index 1.5 7 is lin early polarised. What is the angle of reji·action of glass ? (a) 32.SO (b) 42S (c) 30.2° (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

~ \·n No::_te:;___,__ _ __ We can prove that when unpolari sed light 'of intensity /0 gets polarised on . 1 passing through a polaroid, its intensity becomes hal f 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

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

Polariser

v

v

Y+-=Y+ - C/Y

A

v' = v

or

,

943

v' = v

(1 + ~) (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'

=

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

~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.

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± ~)

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.

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

Interpret

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,t her.

Here, M

A

0 032 · (postion, as the .shift is towards 100

longer wavelength) Now, Doppler shift is given by

~A= -~A c

V

= - !:iA C =- 0.0 32 X 3 X 108

A =- 9.6

100 x 10 4 ms-1

The negative sign indicates that the star is receding. ,.,

uestions 24.2

,

=

945

Chapter 24 "' Wave Optics

Chapter Compendium 1. Ray of light is

drawn perpendicular to the wavefront.

(v) Intensity variation on screen

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.= D(J.t-l)t d

,

a

p0 = 2,Df::

a phenomenon due to which vibrations light restricted in a parti>b) from,the slits. At a point on the screen dir~ctly in fTont of one of the slits, certain wavel~ngths are missing, figure. Sme of these missing wavelengths are

distance D. The linear width of the principal maximum is equal to the width of the slit, if D equals d2 d (a) 2A

(c) 2A2

2

2b (c) A = 3d

2d 2d .'

'- 3b 2 (d) 'A=4d '.'

j : 5c~-1)~ -~:

(a) (Jl-1)

(b)

(c) 2.5(Jl-1)A

(d) J.5A (Jl-1)

24. The ratio of intensities of successive max1ma m the differaction pattern due to single slit is (a) 1 : 4: 9 (b) 1 : 2: 3 4 4 ' 4 9 (c) 1:-2:--2

(d) 1:2:2

9n 25n n n 25. The equations of displacement of two w~ves an~ given as y 1 = 10sin(3nt + n I 3)

Yz = 5(sin 3nt + J:3 cos 3nt), . then what is the ratio of their amplitude? . (a) 1 : 2

(b) 2 : 1

(c) 1 : 1 (d) None ofthese 26 . .Light of wavelength A is incident on a slit of width d. The resJ lting diffraction pattern is observed on a screen at a

(d) 2A

.

2

23. In a biprism experiment, 5'h dark fringe is ·obtained.' at a point. If a thin transparent film 'is placed in' the path of one of waves, then 7th bright' fringes is obtained at the same point. The thickness of the film in terms of wavelength l ·and refractive index ~ will be 1.5A

<

29. In Young's double slit experimel,lt

0. •-,~-· b . 3b: ' (b) 'A.ot

, 22. A beam of unpolarized light having t1U:X io-3 W falls normally on a polarizer of cross sectional area 3 X 10- 4 m 2 • The polarizer rotateS\,Yith an angular freqilt3rtcy of"31.4 rads-1. The energy of light passing through the polarizer per revolution will be (a) 10- 4 J (b) 10- 3 ,J (c) 10~ 2 J (d) 10- 1 J

.

d ' d 2 7. A glass slab of thickness 8 em contains the same number of waves as 10 em of water when both are traversed by the same monochromatic light. If the refractive index of water is 4/3, the refractive index of glass is (a) 5/4 (b) 3/2 (c) 5/3 (d) 16/15 28. Fringes are obtained with the help of a biprism in the focal plane of an _eyepiece distant 1 m from the slit. A convex lens produces images of the slit in two positions betw.ee:r;J. biprism ~nd eyepiece. Jheqista:r;J.ces,behveen two images of the slit·in two positions are 4.05 x 10- 3 m and 2.90 x 10-3 m~espectively. The distance between the slits will be (a) 3.43 x 10-3 m (b) 0.343 rri (c) 0.0343 m (d) 43.3 m

.' b2 2b 2 (a) A = - d '3d

(b) A

. 0

0

•

~ = 10-4 (d =distance D

between slits, D = distance of screen from the slits). At a point P on the screen resultant intensity is equal to the intensity due to · individmihlit I 0 •• Then the distance of point P from the central maximum is (A,= 600QA) (b) 2 mm (a) 0.5 mm (c) 1mm (d) 4mm 10.. In young's double slit experiment, the intensity of the maxima is I. If the width of each slit is dqubled, the intensity of the r,naxima will be (a) 1!2 . (b) 2 I 0

•

\c). ,~I , _

(d) I

May have Mot:e·Qian One CorrectOpti.o n ',·

31. A light of wavelength 6000 A in air enters a medium of refra.ctive inde,x 1.5. Inside the medium, its frequency is v .al}doit~ ~avelength ~sA..: (a) v = 5 x 1014 I:Iz (b) v=7.5x10 14 Hz (c) A= 4oooA (d) A= 90ooA 32~ In Yotiri:g's•doubTe slit experiment, the distance betWeen the two' slits is 0.1 mm and wavelength of light used is 4 x 1 o- 7 m: lfthe width of fringe on screen is 4 mm, the - distance between screen and slit is (a) 0.1 mm (b) 1 em (c) 0.1 em (d) lm

"

3~. yYit~ a.

monochromatic light, the fringe width using a a ~g = 3 I 2 and '.. rDik _a P.t.,~. 5 / ~-·. {he whole ~~t up is immersed in liquid, then 1 ~.-,.,, _ ~h\\17-f!~: fringe width is , ·:, ' .. kl_i~, r,2 mm (b) 6 mm (c) 6/5 mm {d) 8/3 mm o ·· -34. ·In Young's double slit experiment, on interference, ratio of inten~ities of a brighqJand and a dark band is 16 : 1. The 'ratio of amplitudes of interfering waves is (a) 16 · ' (b) 5/3 (c) 4 ,.,L (d) 1/4 0

,

0

1 ··;; 1 ·~: _' , : ~~ss'bipJ:~~1U setup in air is 2 mm. Given

0

Chapter 24 • Wave Optics 35. Light waves travel in vacuum along X-axis. Which of the following may represent the wavefronts? (a) x = a (c) x = a'

(b) y = a (d) x + y

+z =a

Comprehension Based Questions Passage In YDSE, intensity of light from two slits is proportional to their width, and also to squar~ of amplitudes of light from the slits, ie, II rol a2 I2 = (1)2 = ll At points, where crest of one wave falls on crest of the other and through falls on trough, intensity of light is maximum. Interference is said to be constructive . At points, where crest of one wave falls on trough of the other, resultant intensity of light is minimum. The ·interference is said to be destructive. , ImaJ x (green)

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

'

(d)

x(blue) x (green)

(a)

!_g_

(b)

2

J.g_

.J2

(c) Io

(d) Io 4 62. If the polarising angle of a piece of glass for. green light is 54.74°, then the angle of minimum deviation for an equilateral prism made of same glass iS' (Kerala CET 2007) (Given: tan 54.74° = 1.414) (a) 45° (c) 60°

(b) 54.7° (d) 30°

63. In Young's double slit experiment, the intensity at a point

(c) dD!f..

(a) (b) (c) (d)

60. The necessary condition for an interference by two sources I of light is that (BVP Engg. 2008) (a) two light sources must have the same wave length (b) two point sources should have the same amplitude and same wavelength (c) two sources should have the same wavelength, nearly the same amplitude and have a constant phase difference (d) two point sources should have a randomly varying phase difference 61. In the Young's double slit experiment, the central maxima is observed to be 10 • If one the slits is covered, then the intensity at the central maxima will become (BVP Engg. 2007)

5460 4360

where the path difference is

~(A.

being the wavelength 6 oflight used) is I. IfJ0 denotes the maximum intensity, _!_ Io is equal to (AIEEE 2007) 3 1 (a)

(c)

4 J3 2

(b)

(d)

.J2 1

2

64. The phenomenon of polarization of electromagnetic waves proves that the electromagnetic waves are (Gujarat CET 2007)

(a) transverse (b) mechanical (c) longitudinal (d) neither longitudinal nor transverse 65. In the phenomenon of diffraction of light, when blue light is used in the experiment instead of red light, then (DCE 2007)

(a) fringes will becomes narrower (b) fringes will becomes broader (c) no change in fringe width (d) None of the above 66. Monochromatic light of frequency 6.0 x 10 14 Hz is produced by a laser. The power emitted is 2 x 10-3 W. The number of photons emitted, on the average, by the source per second is (UP SEE 2007) (a) 5 x 1015 (b) 5 x 10 16 (c) 5 X 10 17 (d) 5 X 10 14 67. When an unpolarized light of intensity 10 is incident on a polarizing sheet, the intensity of the light which does not get transmitted is (UP SEE 2007) 1 (a) 2Io

(b)

(c) zero

(d) 10

]:_10 4

Chapter 24 • Wave Optics 68. The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then, (UP SEE 2007) (a) D1 < D2 (b) D 1 = Dz (c) D1 can be less than or greater than D2 depending upon the angle of prism (d) D 1 > D 2

69. In double slit experiment, the distance between two slits is 0.6 mm and these are illirninated with light of wavelength 4800 A. The angular width of first dark fringe on the screen distant 120 em from slits will be (BCECE 2006) (a) 8 x 10- 4 rad (b) 6 x 10- 4 rad (c) 4 x 10- 4 rad (d) 16 x 10- 4 rad 70. A lens is made of flint glass (refractive index = 1.5). Which the lens is immersed in a liquid of refractive index (UP SEE 2006) 1.25, the focal length (a) increases by a factor of 1.25 (b) increases by a factor of 2.5 (c) increases by a factor of 1.2 (d) decreases by a factor of 1.2 71. In Young's double slit experiment, the wavelength of light A. = 4 x 10-7 m and separation between the slits is d = 0.1 mm. If the fringe width is 4 mm, then the separation between the slits and screen will be

(a) 100 mm (c) 106 em

I

957

(b) 1 m (d) 10 A

72. A beam of light of wave~ngth 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is (Karnataka CET 2006) (a) 1.2 em (b) 1.2 mm (c) 2.4 em (d) 2.4 mm 73. In Young's double slit experiment, intensity at a point is (1/ 4) of the maximum intensity angular position of this point is (liT JEE 2005) (a) sin- 1 (A. /d)

(b) sin-1 (A. /2d)

(d) sin- 1 (A./4d) 74. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is (AIEEE 2004) (a) infinite (b) 5 (c) 3 (d) zero 75. If l 0 is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit widt~ is doubled? (AIEEE 2004) (c) sin -l (A. !3d)

(Orissa JEE 2006)

(a) :!_g_ 2

(b) 10

(c) 4 I 0

(d) 2 I 0

~~l&M'1ff11%Pffftf~~:tl'l:%:;ci'::m'dll1Sfal:i1~if~;::!.~;d'ii

w·

Answers Exercise I 1. (a)

11. (d)

·'

21. 31. 41. 51. 61. 71.

(a) (b) (d) (b) (b) (a)

(c) (b) (a) (c) (b) (a) (a)

72. (d)

3. 13. 23. 33. 43. 53. 63. 73.

(c) (a) (b) (a) (c) (a) (a) (c)

(a) (b) (a) (d) (a) (d) (d) 72. (d)

3. 13. 23. 33. 43. 53. 63. 73.

(c) (a)

2. 12. 22. 32. 42. 52. 62.

4. (d)

14. (a) 24. (a) 34. (d) 44. (c)· 54. (b) 64. (b) 74. (d)

5. 15. 25. 35. 45 . 55. 65 . 75.

(a) (a) (c) (d)

(c) (b) (a) (b)

6. 16. 26. 36. 46. 56. 66. 76.

(d) (b) (c) (a) (a) (b) (a) (b)

7. 17. 27. 37. 47. 57. 67. 77.

(a) (a) (c) (b) (c) (b) (c) (a)

8. 18. 28. 38. 48. 58. 68. 78.

(c) (c) (c) (c) (d) (a) (a) (d)

9. 19. 29. 39. 49. 59. 69. 79.

(b) (a) (d) (a) (a) (d) (a) (c)

10. 20. 30. 40. 50. 60. 70. 80.

(d) (c) (b) (c) (b) (b) (a) (b)

6. 16. 26. 36. 46. 56. 66.

(d) (b) (a) (d) (a) (b) (a)

7. 17. 27. 37. 47 . . 57. 67.

(d)

8. 18. 28. 38. 48. 58. 68.

(b) (c) (a) (c)

9. 19. 29. 39. 49. 59. 69.

(a) (b) (b) (c) (a) (c) (a)

10. 20. 30. 40. 50. 60. 70.

(d) (b) (b) (b) (a) (c) (b)

Exercise II 1. (c)

11. (c) 21. 31. 41. 51. 61. 71.

(a) (a,c)

(d) (b) (c) (b)

2. 12. 22. 32. 42. 52. 62.

,

(d) (c) (a) (c)

(a) (c)

4. (d) 14. (c) 24. (c) 34. (b) 44. 54. 64. 74.

(c)

(b) (a) (b)

5. 15. 25. 35. 45. 55. 65. 75.

(c) (b) (c) (a,c) (c)

(d) (a) (b)

(a) (c) (a) (c)

(b) (a)

(c)

(d) (a)

958

Chapter 24 • Wave Optics

Hints &Solutions Exercise I 1. For a given time, optical path remain constant ·· J.l1X1 = J.lzXz or 1.5 x 2 = J.lz x 2.25 l.Sx2 2 20 4 1-lz = 2.25 =1.5=15=3

A2 =(~~)x4358 A2 = 5904A 10. For dark fringe,

xd A -=(2m-1)-

D

2

~p =LillA d

~D = d~p A 10-3 X 3 X 10- 5

rj_l----~--:-1 2

600xl0- 9 0

= 5 em away or towards the slits

7. As we lmow ... (i)

d m=Sx=, 2

Here,

d d A -·-=(2x5-1)2 D 2

... (ii)

and

dz =9A

or

From Eqs. (i) and (ii), •

,·,.

D

pocAoc-1 . J.l

poe~ J.l The refractive index of water is greater than air, therefore fringe width will decrease. B. Here, sin= 8

={i)

~8= ~y

So,

D

Angular fringe width 8 0 =

~8

(Width

80 = l = DA x ~ = !:_ D d D d 80 = 1°= n;/180 rad and A= 6x10- 7 m 180 d = _!:_ = X 6 X 10-7 8a 7r = 3.44 x 10-5 m ~0.03 mm 9. As field of view is same in both cases

n1P1

or

n1 (

or

=nz~z

D~1 )= n2( D~z)

~y =

p)

11. When two cohere!'lt light b~ams of intensities 11 and 12 superimpose, then maximum intensity is cji; +.F z) 2 and minimum intensity is Cji; -ji;) 2 • But when two incoherent light beams of intensities 11 and 12 superimpose, then maximum intensity is'(I 1 + 12 ) and minimum intensity is (11 -12 ) . I~ == 51, Imin ~ 31 12. For incoherent waves, ·.'' Jmax = nl I 32 n=~=-=16 I 2 13. The distance between zeroth order maxima and second order minima is Y1 =%+P=%P

3 =-x0.2mm = 0.3 mm 2 :. The distance of second maxima from point P is . y = (4.8+0.3) mm = 5.1 mm 14. At a point of maxima .. Imax = 410 = 4 wm-2 .. 10 = 1 wm-2

:.' '.

,

= s1s3 -

16. As S3 , &

S2 S3

= 27r & =·0

A

=o

959

Chapter 24 • Wave Optics ·.

13 = ! 0 + ! 0 + 2-JI 0 x I 0 (cos0°)

..

[3 =

21. The position of 30th bright fringe 30AD

4Io

Y3o=-d-

The path difference at 5 4 is

Now position shift of central fringe is

&:' = S1S4 -S2S4 = xd D d AD A

30 ).J) Yo= - d-

=-X-=-

2

D 2d 2n A

,

D

But we know, y 0 =-(f..L-l)t d 30AD D -d-=d(f..L-l)t

!J> = - - = 1 t

A 2 I 4 = I 0 +I0 + 2I0 cos 1t = 0

_2= 4!0 ==

10

(f..L-1) = 30 A = 30 X (6000 X 10- ) = 0.5 t (3 .6 X 10-5 ) f..l = 1.5 22. As velocity of light is perpendicular to the wavefront, and light is travelling in vacuum along they-axis, therefore, the wavefront is represented by y = constant.

0

[4

18. I max= !1 + [2 + 2ji;f; = 4! +I +2.,)2!1 xi

= 9! !min= [1 +[2 -2.,)2!1[2 Imm =I

23. As

19. The intensity of light reflected from upper surface is I 1 = ! 0 x 25% 25 =I0 x 100

~ = ).J) d

:. ~ "" 1 I d

2 4. Fringe visibility (V) is given by ··

v = 1max -

·

I min !max +!min

25. y 1 = a 1 cosmt = a 1 sin(mt + 90°)

= Io

4 The intensity of trans~tted light from upper surface is

y 2 = a 2 sin mt

:. Phase difference = !j> = 90°

I- I - Io- 3Io - 0 4- 4

R=

:. The intensity of reflec;ted light from lower surface is

:. Resultant intensity II= R2 = (a~ + a~ ) .

I

2

= 3I0 X 4

[max --'

.Imin -:

50 100

26. Light appears to travel along straight lines because its wavelength is very small. 27. For destructive interference, path difference = 2:.

= 3I0 8

cji; +JI;t ._ ' Cjf; -Ji;) 2

I.~ JJ¥ Imin- (

fj

+

ftFj'

-P¥J

2

,· ' Therefore, phase difference = t; ''. : ·.•. ~!

. '

-

..

<

xgreen

= ~ = AD = 8 x 10-5 x 2 = 0.32 em d 0.05 30. Optical path for 1st ray = n 1 L 1 Optical path for 2nd ray = n2 L 2

[AE:J"~ [~] = [23.~;,= 2~,'

~ 0.021 . -uoc; . r :. Total number of fringes = 2n 1 + fringe a:t centre = 2n 1 + 1 = 2 x 23 + 1 = 46 + 1= 47 In Young's experiment, the number of fringes sho.uld be

,

xblue

29. Distance between ~uccessive fringes = fringe Width

:. Phase difference, ~$ = 2 7t &: A

20. The number of fringes on either side of centre C of screen is

~d.

2 7t . 2: = n radian = 180°. A 2

28. Distance of n'h maxima, x =nAp_"" A d

··" . ; ,._ ... ..... .

.(~+HJ -(~-HJ n1 =

~ af + a~ + 2a1a 2 cos !J> = ~af + a~

:

21t

= T(nl Ll - n2L2)

31. Wave nature of liglwalone can explain the phenomenon lik.e diffraction. 32. For maximum contrast, ! 1 = ! 2 33,

~

1

= ~13 = SOQO A 6000

X

0.48 = 0.04 em

34. 51P- S2 P = ll(A I 2) = odd integral multiple of A/2

960

Chapter 24 • Wave Optics

35. Let ! 1 = a 2 ,!2 == b2

48.

2 2 2 Imax =(a+b) =a +b +2ab

··

= 11 + I 2 + 2ji; ji; = Cji; + ji;)

~= AD d

P' =AD'

2

d ~ __ P' = A(D -D')

d _5 = AX5x10-2 3 10 10-3

and

x

A =: -3 X w-s = 6 X 10-7 m = 6000A 50

or 37. At the centre, all colour~ meet in phase, hence central fringe is white. 38. Path difference, x = (SS 1 + S 10)-(SS2 + S2 0) If x = n ),, the central fringe at 0 will be bright. If x = (2n- l)A/2, the central fringe at 0 will be dark. 39. Distance between the fringes= fringe width

x=~=

AD

50. When path difference is A., Imax = 4 I = K

When path difference is

!1

!: , phase difference,

4/3

IR

'AD

= I 1 +I 2 +2Nz cos K

X=

(!l-1)t

dD

52. Oil floating on water looks coloured opJy when thickness of oil layer;:;; wavelength of light;:;; 10000 A. 2 X 10-3 X 0.9 X 10-3 (2x 2-1) x 1

2xd (2n -1)D

= 6 x 10-7 m = 6 x 10-5 em b2 b 1 42. max =~=9 or ~=3 lmin (a-bi a-b or 3a- 3b == a+ b or 2a = 4b; I

I a 2 4b 2 __l_=- =-=-4 : 1

b2

53. Amplitude A 1 and A 2 are added as vectors. Angle between these vectors is the phase difference (~ 1 - ~ 2 ) between them

R= ~A?+Ai+2A1 A2 cos(~ 1 ~ ~ 2 )

..

54. Here, n1 = 12,A1 = 600 nm

n 2 = ?, A2 = 400nm As

43. Interference fringes are bands on screen XY running

parallel to the lengths of slits. Therefore, the locus of fringes is represented correctly by W3 W 4 • 44. Distance between two successive maxima

nr Ar = n2A2

n

= 2

n 1 A1 = 12 x 602_ = 18

A2

400

55. d= 0.1 mm = 10- 4 , D = 20 em= !m 5

A= 0.14 m =10-2m 14 '

v=~= 3 x 108 =3x1010 H 2 A

I

a2

10-

9

=-=[2 b2 1

_j_

a b

(a+b)~=(3:1J 3

1

2

= 4 :l

46. Path difference of 3rd bright fringe from central fringe= 3A, so that phase difference= 3(2n) = 61t rad. 47. When one slit is closed, amplitude becomes half and intensity becomes 1/4th ie, I 0 = }:_I or I= 4! 0 4

A= 5460A = 5.46 x w- m 7

Angular position of first dark fringe is 6 =..::_=...?:_= 5.46xloD 2d 2x10-4

7

= 2.73,x 10-3 rad 180° = 2.73 X10-3 X--= 0.156° 1t 56. Here, I 1 =I,I 2 =4I,e 1 =n/2,e 2 =n

3 1

!max= Imin (a-b)

= 1t / 2

=I+I=2I=2 51. For a path difference (!l-1)t , the shift is

4

41. x=(2n-1)2"d-

45.



4

6x10- x2

~· = ~ == -~ = ~x = 0.75x

b2

=-A-=2: d! D e

d

d-= -O.OS x 10-2

x 10-4m = 0.24 em 40. When the arrangement is dipped in water,

!2

~=AD

7

= 24

A=

49.

IA = I 1 +12 +2Nz cos8 1

= [ +4I +2~1x4J COS1t/2'= 51 I 8 = I 1 +I2 +~ cos8 2 = I + 4 I + 2jl x 4 I cos 1t = SI- 41 =I IA-IB=5l-l = 4I

I

Chapter 24 • Wave Optics 57. nAr=(n+l)Ab n + 1 = ~ = 600 = i n Ab 480 5

!=i-l=}_n=4 n 5 4 AD ~d 0 .06 x 10- 2 xl0-3 58. From ~ = d => A= D = 1 = 6 x 10-7 m= 6000 A 5'9. When white light is used instead of monochromatic light, the central bright fringe becomes white, while others are coloured. Hence, distinction is rriade. AD 6000x10- 10 x 2 60 ' ~=d= 4x1o- 3 = 0 .3 x 10-3 m = 0.3 mm 61. From diffraction at a single slit, size of aperture,

A a=-sinO

a= 3141.59 x lo-Io = 18 x 10-6 m--18 11m sin1° 62. For diffraction to be observed, size of aperture must be of the same order as wavelength of light. 63. llv = 1 and ~ta = 1.003

!:x_ = lla = 1.0003 Aa X =

Avn = Aa (n ;- 1)

69. asinO =n A , ax 0.3 x l0- 3 xSxl0- 3 ax =3A or~~.=-=------! 3f 3xl = 5 x 10 7 m = 5000 A 70. To be invisible in vacuum, 11 of medium must be equal to 11 of vacuum, which is 1.

71. Angular spread on either side is e = ~ = _!_ rad a 5

72. Sound waves cannot be polarised as they are longitudinal. Light waves can be polarised as they are transverse . 73. Intensity of light from C2 = 10 On rotating through 60°, I = [0 cos2 60° (law of Malus)

~

104

64. Distance between first and sixth minima S AD X=-d nAD 1 x SOOx10- 10 x 2 a= -x- = O.S x10-3

a= 2 x 10-2 cm = 2 x 10-4m D = SO em = 0.5 m

1

1

2

2

0 =-= -x 2a =a 2 2 76. Here, e = 9.9° , I= 20cm = 0.2m,s = 66°

x =An= 6000xl0- 7 mm x - = 2mm a 3

A = a sinO:. = Ssin30° = 2 .Scm n 1 66. Here, A= 6250 A= 6250 X 10- 10 m

J3 = 1 . 732

75. The intensity of plane pola1i sed light is = 2a2 • :. Intensity of polarised light from first nicol prism

1

65. As, a sinO= nA

=1 0 / 4

"= tani p = tan60°=

1 +- = L0003,- = 0.0003 n n 1 104 n=---=0.0003 3

d = 2.5 x 10-3 m = 2.5 mm

GJ

74. As reflected and refracted rays are perpendicular to each other, therefore, iP = i = 60°

n + 1 = !:x_ = 1.0003 n Aa 1

. ZAD W1'dt h o f centra1mrunmum =a 2 X 6250 X 10- 10 X 0.5 2x10-4 =312.5 x 10-3 em 67. In interference , we use two sources while in diffraction, we use light from two points of the same wavefront. 68. From a sin 8 = nA x nAD 1 x6000 x10- 10 x2 a-=n A a=--=-=--___:__:------;;-D ' x 4 x 1o- 3 = 3 x 10-4 = 0.3 mm

=l 0

llv

961

c = ?c = ~ = ~ = O.O?Sgcc- 1

Is 2 x 66 = 75 gL- 1 • 77. Along the optic axis, 1-lo = 1-le

a 78. We lmow, Am = - and 11 = tan 0 ll

A

A = -a- = A cote m tane a 79. Deviation = iP - r = 20° Also,

iP = r = 90°

Solve to get r = 34° 80. taniP = 11 = J3

:. iP = 60°

962

Chapter 24 • Wave Optics

Exercise II 1. Angular momentum

11. L= nh

21t U = nhv 21t

21t U = Lw L = !!_ (J)

x 5 = nA D = 5 X 6.5 X 10-7 X _!_3

2. IR = 11 + 12 + 2.JI1 12 cos~ I0 =I +I +2Icos0°= 4I When one of the slits is closed, intensity on the same spot =I= I0 /4 3. As x = n1 ~ 1 = n2 ~ 2 = n1A1 = n2A2 :. nz = n1 A1 = 60 x 4000 = 40 A2 6000 4. As sound waves are longitudinal, therefore, polarization of sound waves is not possible. 5. If I 0 is intensity of unpolarized light, then from 1st nicol, I1 = Io 2 From 2nd nicol, I 2 = I1 cos 2(90°-60°)

d = 32.5

10- 4 m AD x 3 = (2n -1)"2d

x5 -

2,

13. From x = n A.P._ d D

d1 =7A1d

(J3J =~I°

I

_1_

2

D

dz =7A2d

8

~=~

= 37.5%

Io 6. From (Jl-l)t = n A.

7. A= 600 nm = 6 x 10-7 m a = 1 mm = 10-3 m, D = 2 m Distance between the first dark fringes on either side of central bright fringe = width of central maximum 2AD 2x6x10-7 x2 10- 3

a

= 24 x 10-4 m = 2.4 mm 8. Angular dispersion of central maximum dispersion of 1st minimum (= 29) . 9 n 2x10-3 . 1 From s1n =-=---,a 4x10- 3 2 9= 30°

A1

4'

n

10.

From~;= .

w, '· tl

'' '

..'

n=-( · 1 '

.- ·, ·, r'

A=~= 0.3xl0-z x2x 1o;}, .,;, ·6 x10-7 m=60ooA. •

•

'~' ~"" ... ,'" 1 '- ' ~' D -..1 ;T-''.

•1~ ··f'4.0-5_x_1_0-:_3:--x-2-.-9 0- x -10-_-=3

1 2

= 3.427 x w-3 m

22. Here, w = 31.4 rads- 1 :. Time period of revolution,

29. The resultant intensity at any point P is 1=410 cos

T = 27t = 2 x 3.14 = 0. 2 s

31.4 Energy transmitted/ revolution

e;

2

(%)

2

(J)

= (IA) T =

Vg

5

b2

For

d2

2W

5 10 7 x - =10-4m= 0.1 mm 5 X 10- 3 20. Fringe width oc wavelength of light. Therefore fringe will become narrower. d=

2 112 +b )

=

~ = 10 = 1 : 1

=m = 6

2

(5~J C~J

25. Here, a= 10,b = ~.-52_+_(_5_.[3=3~)2 = 10

'AD

3

=

. 9n 2 . 251t 2

= ~4 2 +3 2 + 2 x 4 x 3 cosn / 3

5 x 10- =

2 1tJ

= 1 . __±_ . _4_

R = ~a 2 +b 2 + 2abcos

. 19. From

C

10 = 4J0cos / 2

A

or

)r

3

= 0 T = 10- X 0.2 = 10-4 J 2 2 23. For 5th dark fringe ,

'AD 9A.D x =(2n - 1)--= - 1 2 d 2d

If Lll: is the corresponding value of, path difference at P, = 1t(~x) 'A

2

then

d

d

3

[7-~]2

AD d .

As

,

'

2.5'A (!l - 1)

t = ---

-

x=-A.-

or -

,'

I

'

~=_! xd, 3 'A D

'

d .

~ ....

Lll: _'"' r)··~· -·

=(!l-1)t- ' ' d = (I!- 1) t B

'A -

---.x.a... -' .: ,_'

1)

X 2 - X1

21i: ' '

21t

-=-fix.

D ?AD For 7th bright fringe, x 2 = nA. - = - -

This

X =

3d I D 2 inm

= 6 x 1o-7 =Zx1o-3m

_r,

3 x 10-4

·: ·' .-,; -..' ,', ",'

is. the'di~~a~~~ ~f poi~tP- fr~m ;e;{t~al ~aximum.

964 30.

Chapter 24 * Wave Optics

/max=

I= 11

+ lz + 2 ~G

When Wldth of each slit is doubled, intensity from each slit becomes twice ie, !' 1 = 21 1 and 1' 2 = 2! 2 I' max

I' = I' 1 + I' 2 + 2~

=

=2l 1 +21 2 +2J21 1 x21 2

= 2(1 1 +1 2 +2Ji:xl;) = 21 C

o

. 31. \, = 6000A, V a = - =

"-a

3 X 108

· 10 6000x 10-

=5x1014Hz

A.a

6ooo'A

~

1.5

45. The beautiful colours are seen on account of interference

In water, "-m = - = - - - = 4000 A.

32. From

o

'AD P=d

D=

~_:1_ 3

3

4x10- x0.1x10- = m 1 7 4

34.

10-

X

p = -2- = -mm 6 Pt =_a /min-

5/3 5 (a+b)2 - 16 (a-b)2 - l a+b 4

placed in the path of light, interfere constructively at the centre of the shadow resulting in the formation of a bright spot.

1

a-b

4a-4b=a+b or3a = 5b a 5 b 3 35. The wavefront in Y-Z plane travels alongX-axis. Therefore , x = a and x = a' represent such a wavefront.

36. R =~a~+ a~+ 2a 1 a 2 cos 1t I 2 37. In the interference pattern,

light should travel faster in denser mediQ. than in rarer media. It is contrary to present theory of light which explains the light travels faster in air (rarer) and in water (Denser). 47. The waves diffracted from the edges of circular obstacle ,

a~l

]max-

of light reflected from the upper and the lower surfaces of the thin films. Since, condition for constructive and destructive interference depends upon the wavelength of light therefore, coloured interference fringes are observed. 46. According to Newton's to corpuscular the01y of light, the

'A

33.

43. When we seen the painting which is painted by a myriad of small colour dots near our eyes, scientillating colour of dots are visible due to diffraction of light and when we go away from the painting our eyes blend the dots and we see the different colours. This is due to the change is angular separation of adjacent dots as the distance of the object changes. 44. In Young's experiment fringe width for dark and white fringes are same while in the same experiment, when a white light as a source is used, the central fringe is white around which few coloured frings are observed on either side.

=~a~+ a~

_(a+b)2 _(1+2)2 -9·1 /min- (a-b)2- (1-2)2.

48. The waves which consist longer wavelength have more diffraction. Since radio waves have greater wavelength than microwaves, hence, radio waves undergo more diffraction than microwaves. 49. The diffraction of sound is only possible when the size of opening should be of the same order as its wavelength and the wavelength of sound is of the order of 1.0 m, hence, for a very small opening no diffraction is produced in sound waves.

50. For two coherent sources, 1=11 +1 2 +Mcos

Hence, focal length increases by a factor of 2.5.

is the phase difference,

4

Ml

4

_Q_ X _Q_ COS

4

4

1 1 1 1 - = - +- + 2 X- COS 4 4 4 4

or

1 21t cos=- or =2 3 If L\x the path difference betwee11 two waves, then or



L\x

'). _

'). _

21t

'). _

21t

21t

21t 3

'). _ 3

-=-=> L\x=-=-X-.--=For path difference L\x, if angular separation is 8 then d sin 9L\x = 'A I 3

. e ="-- or sm 3d

. -1 ( - "e = sm

3d

J

74. For maxima on the screen

').._

4x10-3 x0.1x10- 3 = m 1 4 X 10-7 72. Distance between first dark fringe on either side of central bright fringe = width of central maximum 2 X (600 X 10- 9 ) X 2 a 10- 3 = 24 x 10-4 m = 2.4 mm 2f...D

=--=

,

~

· _Q_ I = _Q_ I + _Q_ I +2

4

or

D=j3xd

= 11 = 12 and if

..

'AD

71. From 13 = d

4

then I= I 1 +I 2 +2~I1 I 2 cos

/2

f 1 = 0.5 X 1.25 = 2 _5 f2 0.25

10

dsin8=nA. d = slit-width= 2A. 2A.sin8=nA. 2sin8=n The maximum value of sin 8 = + 1 .. n=2 :. Eq. (i) is satisfied by- 2,- 1, 0, 1, 2. 75. The intensity of principal maximum in the single slit diffraction pattern does not depend upon the slit width. Given