Wiley Chap 10

Wave Optics 10TARGET CBSE

C H A P T E R R E V I E W

1. According to wave optics, or physical optics, light is a form of energy that travels through a medium in the form of transverse wave.

2. Particles of a light wave which are equidistant from the light source and vibrate in the same phase constitute a wavefront.

3. Depending on the type of light source, we generally have three types of wavefronts – spherical, cylindrical, and plane wavefronts.

4. A spherical wavefront is formed by a point source of light, for a linear source of light we get a cylindrical wavefront, and for a point source located very far away we get a plane wavefront.

5. According to Huygens principle, each and every point on the given wavefront, called “primary wavefront,” acts as a source of new disturbances called “secondary wavelets” that travel in all directions with the velocity of light in the medium. A surface touching these secondary wavelets tangentially in the forward direction at any instant gives a new wavefront at that instant, which is known as the secondary wavefront.

6. The wavelengths of light in two media are proportional to the speeds of light in those media, which is expressed as

1 1

2 2.v

v=l

l

7. Snell’s law of refraction: 1 1 2 2sin sin .n n=q q

8. The fractional change in frequency Δn n/ due to Doppler effect is given by

radial ,vc

−Δ = −nn

where radialv is the component of the source velocity (relative to the observer) along the line joining the observer to the source.

9. Whenever there is relative motion between a source of light and an observer, the apparent frequency of light received by the observer is different from the true frequency of the light emitted from the source. This phenomenon is known as Doppler’s effect in light.

10. Another form of Doppler’s effect: When the direction of�n is directly away from the source, then

01

,1

−= +b

n n b

wheren 0 represents the proper frequency of the source (frequency that is measured by an observer in the rest frame of the source),n represents the frequency detected by an observer moving with velocity v

� relative to that rest frame, and / .v c=b

When the direction of v�

is directly toward the source, we must change the signs in front of both b symbols.

11. Low-speed Doppler’s effect:

( )= − + <<20

11 (where 1)2

n n b b b

CHAPTER

Wiley-Chapter 10.indd 613Wiley-Chapter 10.indd 613 12/26/2011 3:24:53 PM12/26/2011 3:24:53 PM

Chapter 10 | Wave Optics614614

12. Astronomical Doppler’s effect:

0

| |(radial speed of light source, ),c v c

Δ= <<ln l

where the difference Δl is the wavelength Doppler’s shift of the light source.

13. The relation between phase difference f and path difference of x of two waves is

2 .x= pf l

14. Constructive interference will result whenever the distances traveled by the waves are the same or differ by any integer number of wavelengths. If this difference is denoted by x, then = lx n , where n = 0, 1, 2, 3, …

15. Destructive interference will take place whenever the distances traveled by the waves differ by any odd integer number of half-wavelengths, that is,

1 ,2

x n⎛ ⎞= +⎝ ⎠ l

where n = 0, 1, 2, 3, …

16. In Young’s double-slit experiment, the condition for bright fringes is given by

sin , for 0, 1, 2,d m m= =q l …

where d is the slit width. For dark fringes,

( )1sin , for 0, 1, 2,2

d m m= + =q l …

17. If D is the distance between the slits and the screen, the fringe width is given by

m 1 m and angular width is .Dy y yd d+Δ = − = =l lq

18. Given the phase difference f between the waves at the screen, the intensity of the interference pattern is given by the relation

204 cos

2I I= f ,

where I0 is the uniform intensity on the screen when one of the slits is covered up.

19. When source and observer approach each other (i.e., they come toward each other) the change in frequency is positive. Thus the apparent frequency as observed by the observer increases or the apparent wavelength decreases (frequency and wavelength are inversely related). This is known as blue-shift.

20. When source and observer recede from each other (i.e., they move away from each other) then change in frequency is negative. Thus the apparent frequency as observed by the observer decreases or the apparent wavelength increases (frequency and wavelength are inversely related). This is known as red-shift.

21. According to superposition principle, when two or more waves traveling through a medium superimpose on each other, a new wave is formed in which the resultant displacement at any instant is equal to the vector sum of the displacements due to the individual waves at that instant. Mathematically, superposition principle is given as

1 2 3y y y y= + + +�

22. Sources of light that emit continuous light waves having same wavelength, same frequency, and in same phase or having a constant phase difference are known as coherent sources of light.

23. Two independent sources of light cannot be coherent. Two coherent sources of light can be obtained from a single source of light by reflection, refraction, etc.

24. For constructive interference at a point, the phase difference between the two waves reaching that point should be zero or an even integral multiple of 2p. In other words, the path difference between the two waves reaching the point should be zero or an integral multiple of wavelength (l).

25. For destructive interference at a point, the phase difference between the two waves reaching that particular point should be an odd integral multiple of p. In other words, the path difference between the two waves reaching the point should be an odd inte-gral multiple of half-wavelength (l/2).

26. The redistribution of light energy on account of superposition of light waves from two coherent sources of light is known as interference of light.


Chapter Review 615

27. In Young‘s interference pattern, dark fringes are situated in-between bright fringes and vice versa. All the bright and dark fringes are of equal width.

28. The pattern of bright and dark fringes on the screen is called an interference pattern.

29. When we use white light instead of monochromatic source of light then the interference fringes will be colored and the central maximum will be white in color.

30. The phenomenon of bending of light around the corners of an obstacle placed in its path, on account of which it penetrates into the region of geometrical shadow of the obstacle, is known as diffraction of light.

31. The condition to obtain minima in a single-slit diffraction pattern is

sin , for 0, 1, 2,a m m= =q l …

where a is the slit width and q is the angle to the central axis.

32. By Rayleigh’s criterion, two objects that are barely resolvable must have an angular separation qR given by

R 1.22 ,d

= lq

where d is the least distance between the two objects so that their diffraction images are just resolved.

33. For a microscope,

= 2 sin1 ,1.22dm q

l

where q is the semi-vertical angle of the cone in which rays of light from an object enter the objective lens of the microscope m is the refractive index of the medium, and l is the wavelength of the light used to observe the objects.

34. For a telescope,

1 .1.22

D=q l

35. When source of light is monochromatic, the diffraction pattern consists of alternate bright and dark bands of unequal widths.

36. When we use white light, diffraction pattern is colored. Central maximum is white and the other bands are colored. The band with higher wavelength (red) is wider than the band with smaller wavelength (violet).

37. In a single-slit diffraction experiment, dark fringes are produced where the path length differences (a sin q) between the top and bottom rays are equal to l, 2l, 3l, …

38. The power or ability of an optical instrument to produce distinctly separate images of two closely spaced objects is known as resolving power of the optical instrument.

39. According to Rayleigh, two point objects A and B will be just resolved when the central maximum of diffraction pattern of object B lies on the first secondary minimum of diffraction pattern of object A.

40. Resolving power of a microscope is defined as the reciprocal of the least separation between two close objects so that they appear just separated when seen through the microscope. Resolving power of a microscope is given by the relation

2 sin1 ,1.22d

= m ql

where d is the least distance between the two objects so that their diffraction images are just resolved, q is the semi-vertical angle of the cone in which rays of light from an object enter the objective lens of the microscope m is the refractive index of the medium, and l is the wavelength of the light used to observe the objects.

41. Resolving power of a telescope is defined as the reciprocal of the smallest angular separation between two distant objects (such as stars) so that they appear just separated when seen through the telescope. Resolving power of a telescope is given by the relation

1 ,1.22

D=q l

where, q is the angular separation between the two stars, l is the wavelength of the light in which the two stars are observed, and D is the diameter of the objective lens of the telescope.

42. The phenomenon of restricting the vibrations of light to a single plane is known as polarization of light.

43. If both polarizer and analyzer are rotated with same angular velocity in the same direction, no change in intensity of transmitted light is observed.



44. According to law of Malus, when a beam of plane polarized light is incident on the analyzer, the intensity of light I transmitted from the analyzer is directly proportional to the square of the cosine of the angle q between the planes of transmission of the polarizer and analyzer. If the transmission axis of an analyzer is oriented at an angle q relative to the transmission axis of the polarizer, Malus’ Law is given by

20 cos ,I I= q

where 0I is the average intensity of the light entering the analyzer.

45. The reflected light is completely plane polarized in a direction perpendicular to the plane of incidence.

46. The angle of incidence at which the reflected light gets completely plane polarized is called polarizing angle.

47. According to Brewster’s law, when unpolarized light is incident at polarizing angle on the interface separating air from a medium of refractive index n, the reflected light is fully polarized provided the refractive index of the medium is equal to the tangent of the polarizing angle. Brewster’s angle qB is given by

21B

1tan ,n

n−=q

where n1 is the refractive index of the medium through which the incident and reflected rays travel and n2 is the refractive index of the medium from which the light reflects.

W O R K E D P R O B L E M S

TEXTBOOK EXERCISES1. A double-slit arrangement produces interference fringes for sodium light (l = 589 nm) that have an angular separation of 3.50 × 10–3 rad.

For what wavelength would the angular separation be 10% greater?

Solution

The angular positions of the maxima of a two-slit interference pattern are given by d sin q = ml, where d is the slit separation, l is the wavelength, and m is an integer. If q is small, sin q may be approximated by q in radians. Then, q = ml/d to good approximation. The angular separation of two adjacent maxima is Δq = l/d. Let l’ be the wavelength for which the angular separation is greater by 10.0%. Then, 1.10l/d = l’/d. or

l′ = 1.10l = 1.10(589 nm) = 648 nm.

2. Suppose Young’s experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits. How far apart are the bright fringes near the center of the interference pattern?

Solution

The condition for a maximum in the two-slit interference pattern is d sin q = ml, where d is the slit separation, l is the wavelength, m is an integer, and q is the angle made by the interfering rays with the forward direction. If q is small, sin q may be approximated by q in radians. Then, q = ml/d, and the angular separation of adjacent maxima, one associated with the integer m and the other associated with the integer m + 1, is given by Δq = l/d. The separation on a screen a distance D away is given by

.Dy Dd

Δ = Δ = lq

Thus,

93

3(500 10 m)(5.40 m) 2.25 10 m 2.25 mm.

1.20 10 my

−−

−×Δ = = × =

×

3. In a double-slit experiment, the distance between slits is 5.0 mm and the slits are 1.0 m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480 nm, and the other due to light of wavelength 600 nm. What is the separation on the screen between the third-order (m = 3) bright fringes of the two interference patterns?

Solution

The maxima of a two-slit interference pattern are at angles q given by d sin q = ml, where d is the slit separation, l is the wavelength, and m is an integer. If q is small, sin q may be replaced by q in radians. Then, dq = ml. The angular separation of two maxima associated with different wavelengths but the same value of m is

Δq = (m/d)(l2 − l1)


Worked Problems 617

and their separation on a screen a distance D away is

2 1

9 9 53

tan ( )

3(1.0 m) (600 10 m 480 10 m) 7.2 10 m.5.0 10 m

mDy D Dd

− − −−

⎡ ⎤Δ = Δ ≈ Δ = −⎢ ⎥⎣ ⎦⎡ ⎤= × − × = ×⎢ ⎥×⎣ ⎦

q q l l

The small angle approximation tan Δq ≈ Δq (in radians) is made.

4. The distance between the fi rst and fi fth minima of a single-slit diff raction pattern is 0.35 mm with the screen 40 cm away from the slit, when light of wavelength 550 nm is used. (a) Find the slit width. (b) Calculate the angle q of the fi rst diff raction minimum.

Solution

(a) The separation between the first (m1 = 1) and the fifth =2( 5)m minima is calculated as follows:

2 1sin ( ).m D Dy D D m m ma a a

⎛ ⎞Δ = Δ = Δ = Δ = −⎝ ⎠l l lq

Solving for the slit width, we obtain

62 1( ) (400 mm)(550 10 mm)(5 1) 2.5 mm.

0.35 mmD m m

ay

−− × −= = =Δl

For m = 1,

−−×= = = ×

641(550 10 mm)sin 2.2 10 .

2.5 mmm

alq

(b) The angle of the first diffraction minimum is q = sin–1 (2.2 × 10–4) = 2.2 × 10–4 rad.

5. A plane wave of wavelength 590 nm is incident on a slit with a width of a = 0.40 mm. A thin converging lens of focal length +70 cm is placed between the slit and a viewing screen and focuses the light on the screen. (a) How far is the screen from the lens? (b) What is the distance on the screen from the center of the diff raction pattern to the fi rst minimum?

Solution

(a) A plane wave is incident on the lens so it is brought to focus in the focal plane of the lens, a distance of 70 cm from the lens.(b) Waves leaving the lens at an angle q to the forward direction interfere to produce an intensity minimum if a sin q = ml, where

a is the slit width, l is the wavelength, and m is an integer. The distance on the screen from the center of the pattern to the minimum is given by y = D tan q, where D is the distance from the lens to the screen. For the conditions of this problem,

−−

−×= = = ×

×9

33

(1)(590 10 m)sin 1.475 10 .0.40 10 m

malq

This means q = 1.475 × 10–3 rad and y = (0.70 m) tan(1.475 × 10–3 rad) = 1.0 × 10–3 m.

6. In Fig. 10.1, initially unpolarized light is sent into a system of three polarizing sheets whose polarizing directions make angles of q1 = 40°, q2 = 20°, and q3 = 40° with the direction of the y-axis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.)

q3

q1

q2

y

x

Figure 10.1

Solution

Let I0 be the intensity of the unpolarized light that is incident on the first polarizing sheet. The transmitted intensity is =1 0(1/2) ,I Iand the direction of polarization of the transmitted light is q1 = 40° counterclockwise from the y-axis in the diagram. The polarizing



direction of the second sheet is q2 = 20° clockwise from the y-axis, so the angle between the direction of polarization that is incident on that sheet and the polarizing direction of the sheet is 40° + 20° = 60°. The transmitted intensity is

2 22 1 0

1cos 60 cos 60 ,2

I I I= ° = °

and the direction of polarization of the transmitted light is 20° clockwise from the y-axis. The polarizing direction of the third sheet is q3 = 40° counterclockwise from the y-axis. Consequently, the angle between the direction of polarization of the light incident on that sheet and the polarizing direction of the sheet is 20° + 40° = 60°. The transmitted intensity is

2 4 23 2 0 0

1cos 60 cos 60 3.1 10 .2

I I I I−= ° = ° = ×

Thus, 3.1% of the light’s initial intensity is transmitted.

7. In Fig. 10.2, a beam of light, with intensity 43 W/m2 and polarization parallel to a y-axis, is sent into a system of two polarizing sheets with polarizing directions at angles of q1 = 70° and q2 = 90° to the y-axis. What is the intensity of the light transmitted by the two-sheet system?

q2

q1

y

x

Figure 10.2

Solution

The angle between the direction of polarization of the light incident on the first polarizing sheet and the polarizing direction of that sheet is q1 = 70°. If I0 is the intensity of the incident light, then the intensity of the light transmitted through the first sheet is

2 2 2 21 0 1cos (43 W/m ) cos 70 5.03 W/m .I I= = ° =q

The direction of polarization of the transmitted light makes an angle of 70° with the vertical and an angle of q2 = 20° with the horizontal. q2 is the angle it makes with the polarizing direction of the second polarizing sheet. Consequently, the transmitted intensity is

2 2 2 22 1 2cos (5.03 W/m )cos 20 4.4 W/m .I I= = ° =q

8. The largest refracting telescope in the world is at the Yerkes Observatory in Williams Bay, Wisconsin. The objective of the telescope has a diameter of 1.02 m. Two objects are 3.75 × 104 m from the telescope. With light of wavelength 565 nm, how close can the objects be to each other so that they are just resolved by the telescope?

Solution

According to Rayleigh’s criterion, the minimum angular separation of the two objects is

( )±±×≈ = = ×

97

min565 10 m(in rad) 1.22 1.22 6.76 10 rad.

1.02 mDlq

qmin y

L

Earthobserver

Object 1

Object 2

Figure 10.3

Therefore, from Fig. 10.3, the separation y of the two objects is

4 7min (3.75 10 m)(6.76 10 rad) 0.0254 m.y L ±= = × × =q

9. Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of 4.2 × 1017 m from the Earth. One planet is believed to be located at a distance of 1.2 × 1011 m from the star. Using visible light with a vacuum


Worked Problems 619

wavelength of 550 nm, what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

Solution

At the aperture the star and its planet must subtend an angle at least as large as that given by the Rayleigh’s criterion, which is qmin ≈ 1.22 l/D, where l is the wavelength of the light and D is the diameter of the aperture. The angle qmin is given by this criterion in radians. We can obtain the angle subtended at the telescope aperture by using the separation s between the planet and the star and the distance L of the star from the Earth. The angle in radians is qmin = s/L. Using Rayleigh’s criterion, we have

9 17

111.22 1.22(550 10 m)(4.2 10 m)1.22 or 2.3 m.

1.2 10 mLs D

L D s

−× ×≈ ≈ = =×

ll

10. A diff raction pattern forms when light passes through a single slit. The wavelength of the light is 675 nm. Determine the angle that locates the fi rst dark fringe when the width of the slit is (a) 1.8 × 10–4 m and (b) 1.8 × 10–6 m.

Solution

This problem can be solved for the value of the angle q when m = 1 (first dark fringe).

(a) When the slit width is W = 41.8 10 m−× and 9675 nm 675 10 m,−= = ×l we find

91 1

4675 10 msin sin (1) 0.21 .1.8 10 m

mW

±− −

±⎡ ⎤×⎛ ⎞= = = °⎢ ⎥⎝ ⎠ ×⎣ ⎦

lq

(b) Similarly, when the slit width is W = 61.8 10 m−× and 9675 10 m−= ×l , we find

91

6675 10 msin (1) 22 .1.8 10 m

±−

±⎡ ⎤×= = °⎢ ⎥×⎣ ⎦

q

SECTION I

➥ Light as Wave ➥ Huygens Principle ➥ Refraction and Reflection of Plane Waves using Huygens Principle ➥ Doppler’s Effect in Light

(1) The wavelengths of light in two media are proportional to the speeds of light in those media; that is

1 1

2 2,v

v=l

l

(2) Snell’s law of refraction: 1 1 2 2sin sin .n n=q q(3) The fractional change in frequency Δn n/ due to the Doppler’s effect is given by

radial ,vc

−Δ = −nn

where radialv is the component of the source velocity (relative to the observer) along the line joining the observer to the source.(4) Doppler’s effect formula in another form: When the direction of v

� is directly away from the source, then

01 ,1

−= +bn n b

wheren 0 represents the proper frequency of the source (frequency that is measured by an observer in the rest frame of the source),n represents the frequency detected by an observer moving with velocity v

�relative to that rest frame, and / .v c=b

When the direction of�v is directly toward the source, we must change the signs in front of both b symbols.

(5) For low-speed Doppler’s effect: ( )= − + <<20

11 (where 1)2

n n b b b

(6) Astronomical Doppler’s effect: 0

| |(radial speed of light source, ),c v c

Δ= <<ln l

where the difference Δl is the wavelength Doppler’s shift of the light source.

FORMULAS AND UNITS



Conceptual Problems

1. When a wave undergoes reflection at a denser medium, what happens to its phase?

Solution

When a wave undergoes reflection at a denser medium, its phase changes by p or °180 .

2. (a) If a wave undergoes refraction, what happens to its phase? (b) What is the shape of wavefront emitted by a light source in the form of a narrow slit?

Solution

(a) If a wave undergoes refraction, no phase change occurs during refraction.(b) Cylindrical shape.

3. How is a wavefront related to the direction of corresponding rays?

Solution

Wavefront related to the direction of corresponding rays is always normal to the rays.

4. What is the phase diff erence between any two points on a wavefront?

Solution

Phase difference between any two points on a wavefront is zero because according to the definition of wavefront “wavefront is a collection of points vibrating in same phase or having constant phase difference”.

5. Does the speed of light in vacuum depend on relative motion between source and observer?

Solution

Speed of light in vacuum does not depend on relative motion between source and observer. Speed of light in vacuum is constant.

6. Is the speed of light in glass independent of color of light?

Solution

The speed of light changes with color because different colors have different wavelength and the refractive index varies with wavelength. According to Cauchy’s relation, the refractive index and wavelength are inversely proportional to each other:

1 .∝m l

Therefore, the refractive index of glass for red color will be less than the violet color because wavelength of red color is more than wavelength of violet color. Therefore, v r .>m m Now the relation between speed of light and refractive index is

.cV

=m (1)

From Eq. (1), we conclude that the refractive index is inversely proportional to the speed of light (V). Therefore, speed of light for violet color will be more than speed of light for red color. Hence, v r .V V>

7. When light travels from a rarer to denser medium, it loses some speed. Does the reduction in speed imply a reduction in the energy carried by the light wave?

Solution

When light travels from a rarer to denser medium, the reduction in speed does not imply a reduction in the energy carried by the light wave because the total average energy density associated with an electromagnetic wave (i.e., light wave) is given by the relation

20 2

ave 0 00

1 ,2 2B

E= = ∈m m

where mave is the energy density, 0B is the amplitude of magnetic field vector, and 0E is the amplitude of electric field vector. From this expression of energy density, it can be concluded that the energy density is independent of the speed of light. Hence, the reduction in speed does not imply a reduction in the energy carried by the light wave.

Additional Problems

1. In Fig. 10.4, a light wave along ray r1 reflects once from a mirror and a light wave along ray r2 reflects twice from that same mirror and once from a tiny mirror at distance L from the bigger mirror. (Neglect the slight tilt of the rays.) The waves have wavelength 620 nm and are initially in phase. (a) What is the smallest value of L that puts the final light waves exactly out of phase? (b) With the tiny mirror initially at that value of L, how far must it be moved away from the bigger mirror to again put the final waves out of phase?


Worked Problems 621

Lr2

r1

Figure 10.4

Solution

The fact that wave W2 reflects two additional times has no substantive effect on the calculations, since two reflections amount to a 2(l/2) = l phase difference, which is effectively not a phase difference at all. The substantive difference between W2 and W1 is the extra distance 2L traveled by W2.(a) For wave W2 to be a half-wavelength “behind” wave W1, we require 2L = l/2, or L = l/4 = (620 nm)/4 = 155 nm using the

wavelength value given in the problem.(b) Destructive interference will again appear if W2 is 3/2l “behind” the other wave. In this case, =′ l2 3 /2L , and the difference is

3 310 nm.4 4 2

L L− = − = =′ l l l

2. In Fig. 10.4, a light wave along ray r1 reflects once from a mirror and a light wave along ray r2 reflects twice from that same mirror and once from a tiny mirror at distance L from the bigger mirror. (Neglect the slight tilt of the rays.) The waves have wavelength l and are initially exactly out of phase. What are the (a) smallest, (b) second smallest, and (c) third smallest values of L/l that result in the final waves being exactly in phase?

Solution

We consider waves W2 and W1 with an initial effective phase difference (in wavelengths) equal to 1/2, and seek positions of the sliver that cause the wave to constructively interfere (which corresponds to an integer-valued phase difference in wavelengths). Thus, the extra distance 2L traveled by W2 must amount to l l …(1/2) , (3/2) , We may write this requirement succinctly as

2 1 where 0, 1, 2,4

mL m+= =l …

(a) Thus, the smallest value of l/L that results in the final waves being exactly in phase is when m = 0, which gives = =l/ 1/4 0.25L .(b) The second smallest value of l/L that results in the final waves being exactly in phase is when m = 1, which gives = =l/ 3/4 0.75L .(c) The third smallest value of l/L that results in the final waves being exactly in phase is when m = 2, which gives = =l/ 5/4 1.25L .

3. In Fig. 10.5a, a beam of light in material 1 is incident on a boundary at an angle of 30°. The extent to which the light is bent due to refraction depends, in part, on the index of refraction n2 of material 2. Figure 10.7b gives the angle of refraction q2 versus n2 for a range of possible n2 values, from na = 1.30 to nb = 1.90. What is the speed of light in material 1?

30°

q2n2

n1

40°

30°

20°na nb

n2

q2

(b)(a)

Figure 10.5

Solution

Note that Snell’s law (the law of refraction) leads to q1 = q2 when n1 = n2. The graph indicates that q2 = 30° (which is what the problem gives as the value of q1) occurs at n2 = 1.5. Thus, n1 = 1.5, and the speed with which light propagates in that medium is

88

1

2.998 10 m/ 2.0 10 m/s.1.5

c sn

×= = = ×n

4. In Fig. 10.6, assume that the two light waves, of wavelength 620 nm in air, are initially out of phase by p rad. The indexes of refraction of the media are n1 = 1.45 and n2 = 1.65. What is the (a) smallest and (b) second smallest value of L that will put the waves exactly in phase once they pass through the two media?



L

The difference in indexescauses a phase shift

between the rays

n1

n2

Figure 10.6

Solution

(a) Since a half-wavelength phase difference is equivalent to a p radians difference, the smallest value of L that will put the waves exactly in phase once they pass through the two media is set as

min2 1

620 nm 1550 nm 1.55 m.2( ) 2(1.65 1.45)

Ln n

= = = = μ− −l

(b) Since a phase difference of 3/2 (wavelengths) is effectively the same as what we required in part (a), the smallest value of L that will put the waves exactly in phase once they pass through the two media is set as

min2 1

3 3 3(1.55 m) 4.65 m.2( )

L Ln n

= = = μ = μ−l

5. In Fig. 10.7, two light rays go through different paths by reflecting from the various flat surfaces shown. The light waves have a wavelength of 420.0 nm and are initially in phase. What is the (a) smallest and (b) second smallest value of distance L that will put the waves exactly out of phase as they emerge from the region?

L

L

L

LRay 1

Ray 2

Figure 10.7

Solution

(a) We note that ray 1 travels an extra distance 4L more than ray 2. To get the least possible L that will result in destructive interference, we set this extra distance equal to half of a wavelength:

420.0 nm4 52.50 nm.2 8 8

L L= ⇒ = = =l l

(b) The next case occurs when that extra distance is set equal to (3/2)l. The result is

3 3(420.0 nm) 157.5 nm.8 8

L = = =l

6. Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm. (NCERT)

Solution

Aperture width is a = 4 mm = 4 × 10−3 m; wavelength of light is l = 400 nm = 400 × 10−9 m. Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation and is given by the relation,

2

F .aZ = l

Substituting the values, we get3 2

F 9(4 10 ) 40 m.400 10

Z−

−×= =

×

Therefore, the distance for which the ray optics is a good approximation is 40 m.


Worked Problems 623

7. The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth. (NCERT)

Solution

Wavelength of Hα line emitted by hydrogen in a star is l = 6563 Å = 6563 × 10−10 m; Red-shift observed in star is (l′ − l) = 15 Å = 15 × 10−10 m; Speed of light c = 3 × 108 m/s.

Let the velocity of the star with which it is receding away from the Earth be n. Using the relation for red-shift, we get 8 10

510

3 10 15 10( ) 6.87 10 m/s.6563 10

cc

−

−× × ×− = ⇒ = × − = = ×′ ′ ×

nl l l n l ll

Therefore, the speed with which the star is receding away from the Earth is 6.87 × 105 m/s.

8. Explain how corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confi rmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment? (NCERT)

Solution

According to Newton’s corpuscular theory of light, when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.

If i is the angle of incidence, r is the angle of reflection, c is the velocity of light in air, and n is the velocity of light in water, we can write the expression

=sin sin .c i rn (1)

The relation for relative refractive index (m) of water with respect to air is

= .cnm

Hence, Eq. (1) reduces to

= =sin.

sini

c rn m (2)

But for water, the refractive index is greater than 1 or >m 1. Therefore, it can be concluded from Eq. (2) that speed of light in water will be greater than the speed of light in air or n > c. This is not possible because according to experimental result, the speed of light in air is greater than the speed of light in water or c > n. Alternative picture of light which is consistent with experimental results is the wave picture of light.

9. You have learnt in the text how Huygens principle leads to the laws of refl ection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror. (NCERT)

Solution

Let an object at O be placed in front of a plane mirror MO′ at a distance r (as shown in Fig. 10.8).

X

rO′

M

Y

Plane mirror

O

Figure 10.8

A circle is drawn from the center (O) such that it just touches the plane mirror at point O′. According to Huygens principle, XY is the wavefront of incident light. If the mirror is absent, then a similar wavefront X′Y′ (as XY) would form behind O′ at distance r as shown in Fig. 10.9.



OO′r r

X

Y

X′

Y′

Figure 10.9

X′Y′ can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).

10. For sound waves, the Doppler’s formula for frequency shift differs slightly between the two situations: (a) source at rest; observer moving and (b) source moving; observer at rest. The exact Doppler’s formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light traveling in a medium? (NCERT)

Solution

Sound waves can propagate only through a material medium because they are material waves whereas light waves are electromagnetic in nature and can propagate through vacuum also. For sound waves, the Doppler’s formula for frequency shift differs slightly between the two situations: (a) source at rest; observer moving and (b) source moving; observer at rest because the motion of an observer relative to a medium is different in the two situations.

In case of light waves, which can travel in a vacuum the frequency shift is same for source at rest; observer moving, or source moving; observer at rest. This is because in a vacuum, the speed of light is independent of the motion of the observer and the motion of the source. Also the formulas are not strictly identical for the two situations in case of light traveling in a medium and in case of sound traveling in a medium because when light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

11. What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm? (NCERT)

Solution

The change in wavelength of the sodium line is Δ = − = +l 589.6 nm 589.0 nm 0.6 nm. Let the speed with which the galaxy should move with respect to us be radial.n Therefore,

Δ Δ Δ= − = − ⇒ − = −

Δ ⎛ ⎞⇒ = × = × × = + × =⎝ ⎠

radial radial

8 5radial

0.6 3 10 3.06 10 m/s 306 km/s.589.0

c c

c

n nn l ln l l

ln l

Therefore, the galaxy is moving away from us because radialn is positive.

12. (a) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why? (b) When light travels from a rarer to a denser medium, the speed decreases. Does the reduction in speed imply a reduction in the energy carried by the light wave? (c) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What feature determines the intensity of light in the photon picture of light? (NCERT)

Solution

(a) When monochromatic light is incident on a surface separating two media, both reflected and refracted light have the same frequency as the incident frequency because reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light equals the frequency of incident light.

(b) The reduction in speed when light travels from a rarer to denser medium does not imply a reduction in the energy carried by the light wave because the total average energy density associated with an electromagnetic wave (i.e., light wave) is given by relation

20 2

ave 0 00

1 ,2 2B

E= = ∈m m

where mave is the energy density, 0B is the amplitude of magnetic field vector, and 0E is the amplitude of electric field vector. From the above relation of energy density it can be concluded that the energy density is independent of the speed of light. Hence, the reduction in speed does not imply a reduction in the energy carried by the light wave.


Worked Problems 625

(c) For a given frequency, intensity of light in the photon picture is determined by the number of photons crossing a unit area per unit time.

Previous Years’ CBSE Problems

1. (a) How is a wavefront different from a ray? Draw the geometrical, shape of the wavefronts when (i) light diverges from a point source and (ii) light emerges out of a convex lens when a point source is placed as its focus. (b) State Huygens principle. With the help of a suitable diagram, prove Snell’s law of refraction using Huygens principle. (2005)

Solution

Locus of the point in the wave, which oscillates in phase is called wavefront; otherwise, wavefront is also defined as the surface of constant phase. Figure 10.10a shows the diverging wavefront and Fig. 10.10b shows converging wavefront.

BA

PointfocusB

b

(b)(a)

Converging pencil (+)

Figure 10.10

According to Huygens principle, “each point of the wavefront is the source of a secondary disturbance and the wavelets emanat-ing from these points spread out in all directions with the speed of the wave. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres, we obtain the new position of the wavefront at a later time,” which is depicted in Fig. 10.11. With O as center, F1F2 represents the spherical wavefront at t = 0. The envelope of the secondary wavelets emanating from F1F2 produces the forward moving wavefront G1G2. The backwave D1D2 does not exist.

O

vt

vt

D1 D2

A

A′

vtG1

F1F2

G2

Figure 10.11

Snell’s Law of Refraction: Let PP′ represent the surface separating medium 1 and medium 2, as shown in Fig. 10.12. Let v1 and v2 represent the speed of light in medium 1 and medium 2, respectively. We assume a plane wavefront AB propagating in the direction A′A incident on the interface at an angle i as shown in Fig. 10.12. Let t be the time taken by the wavefront to travel the distance BC.

= 1 .BC n t

To determine the shape of the refracted wavefront, let us draw a sphere of radius n2 from the point A in the second medium (the speed of the wave in the second medium is n2). Let CE (Fig. 10.12) represent a tangent plane drawn from the point C on to the sphere. Then, AE = n2t and CE would represent the refracted wavefront. If we consider two triangles ABC and AEC,

1 2BC AEsin and sin .AC AC AC AC

v vi r= = = =

Therefore,

1 1 21 2

2 2 1

sin sin sin .sin

v v ni c n i n rr v v c n

= = × = ⇒ =



Incident wavefront

i

i

Medium 1

Medium 2

P A

A′

P′

B

r

r

E

Cv2t

v1t

v2 < v1

v1

Refractedwavefront

Figure 10.12

SECTION II ➥ Coherent and Incoherent Addition of Waves ➥ Interference of Light Waves and Young’s Experiment

(1) The relation between phase difference f and path difference of x of two waves is

2 .x= pf l

(2) Constructive interference will result whenever the distances traveled by the waves are the same or differ by any integer number of wavelengths. If this difference is denoted by x,

x = nl, where n = 0, 1, 2, 3…

(3) Destructive interference will take place whenever the distances traveled by the waves differ by any odd integer number of half-wavelengths. That is,

1 , where 0,1, 2, 32

x n n⎛ ⎞= + = …⎝ ⎠ l

(4) In Young’s double-slit experiment, the condition for bright fringes is given by

sin , for 0,1, 2,d m m= =q l …

where d is the slit width. For dark fringes,

1sin , for 0,1, 2,2

d m m⎛ ⎞= + =⎝ ⎠q l …

(5) If D is the distance between the slits and the screen, the fringe width is given by

1 .m mDy y y

d+Δ = − = l

(6) Given the phase difference f between the waves at the screen, the intensity of the interference pattern is given by

204 cos ,

2I I= f

where I0 is the uniform intensity on the screen when one of the slits is covered up.

FORMULAS AND UNITS

Conceptual Problems

1. Suppose that a radio station broadcasts simultaneously from two transmitting antennas at two different locations. Is it clear that your radio will have better reception with two transmitting antennas rather than one? Justify your answer.


Worked Problems 627

Solution

A radio station broadcasts simultaneously from two transmitting antennas at two different locations. The radio that receives the broadcast could have better reception depending on the location of the receiving antenna. According to the principle of linear superposition, when the electromagnetic waves from the transmitting antenna arrive at the same point, the resultant wave is the sum of the individual waves. The amplitude of the resultant wave depends on the relative phase between the two waves. The relative phase between the waves depends on the path length difference between the two waves. If the two waves arrive at the receiving antenna so that the path length difference between them is equal to an integer number of wavelengths, the two waves will be in phase, they will reinforce each other, and constructive interference will occur. The resulting amplitude of the radio wave will be larger than it would be from either transmitting antenna alone; therefore, the radio reception will be better. If the two waves arrive at the receiving antenna so that the path length difference between them is equal to an odd multiple of half of a wavelength, the two waves will be exactly out of phase, they will mutually cancel, and destructive interference will occur. The receiving antenna will receive little or no signal in this situation. As a result, the radio reception will be very bad. Thus, having two transmitting antennas does not necessarily lead to better reception.

2. Two sources of waves are in phase and produce identical waves. These sources are mounted at the corners of a square. At the center of the square waves from the sources produce constructive interference, no matter which two corners of the square are occupied by the sources. Explain why.

Solution

Constructive interference occurs between waves from two in-phase sources when the difference in travel distances from the sources to the point where the waves combine is the same or equal to an integer number of wavelengths. For a square the distance from any corner to the center is the same. Hence, constructive interference occurs at the center.

3. (a) How would the pattern of bright and dark fringes produced in a Young’s double-slit experiment change if the light waves coming from both slits had their phases shifted by an amount equivalent to a half-wavelength? (b) How would the pattern change if the light coming from only one of the slits had its phase shifted by an amount equivalent to a half-wavelength?

Solution

(a) The light waves coming from both slits in a Young’s double-slit experiment have their phases shifted by an amount equivalent to half a wavelength. Since the light from both slits is changed by the same amount, the relative phase difference between the light from the two slits is zero when the light leaves the slits. When the light reaches the screen, the relative phase difference between light waves from the two slits will be the same as if the phase of the waves had not been shifted at the slits. Therefore, the pattern will be exactly the same as that described in the text.

(b) Light coming from only one of the slits in a Young’s double-slit experiment has its phase shifted by an amount equivalent to half of a wavelength. Now, there is a relative phase diff erence of one half of a wavelength between the light leaving the slits. When the light reaches the screen, there will be a relative phase diff erence due to the fact that light from each slit, in general, traveled along diff erent paths. In addition, there will be the initial phase diff erence that is equivalent to half of a wavelength. Therefore, the pattern will be similar to that described in the text; however, the points of constructive interference will be points of destructive interference, and the points of destructive interference will be points of constructive interference. In other words, the positions of the light and dark fringes are interchanged.

4. In Young’s double-slit experiment, is it possible to see interference fringes when the wavelength of the light is greater than the distance between the slits? Provide a reason for your answer.

Solution

The angle q for the interference maximum in Young’s double-slit experiment is given by sin /m d=q l where m = 0, 1, 2, 3,… When the wavelength l of the light is greater than the distance d between the slits, the ratio l/d is greater than one; however, sinq cannot be greater than one. Therefore, it is not possible to see interference fringes when the wavelength of the light is greater than the distance between the slits.

5. When sunlight refl ects from a thin fi lm of soapy water, the fi lm appears multicolored, in part because destructive interference removes diff erent wavelengths from the light refl ected at diff erent places, depending on the thickness of the fi lm. As the fi lm becomes thinner and thinner, it looks darker and darker in refl ected light, appearing black just before it breaks. The blackness means that destructive interference removes all wavelengths from the refl ected light when the fi lm is very thin. Explain why.

Solution

When sunlight reflects from a thin film of soapy water, the film appears multicolored, in part because destructive interference removes different wavelengths from the light reflected at different places, depending on the thickness of the film. As the film becomes thinner and thinner, the path length difference between the light reflected from the top of the film and the light reflected from the bottom of the film becomes closer to zero. The light reflected from the top of the film undergoes a phase change equivalent to half of a wavelength. Therefore, as the thickness of the film approaches zero, the light reflected from the top of the film is almost exactly out of phase with the light reflected from the bottom of the film. Therefore, as the film becomes thinner and thinner,



it looks darker and darker in reflected light. It appears black just before it breaks, because at that point destructive interference occurs between the light reflected from the top and the light reflected from the bottom of the film.

6. A camera lens is covered with a non-refl ective coating that eliminates the refl ection of perpendicularly incident green light. Recalling Snell’s law of refraction, would you expect the refl ected green light to be eliminated if it were incident on the non-refl ective coating at an angle of 45° rather than perpendicularly? Justify your answer.

Solution

A camera lens is covered with a non-reflective coating that eliminates the reflection of perpendicularly incident green light. If the light was incident on the non-reflective coating at 45°, rather than perpendicularly, it would not be eliminated by the coating and an observer would see it.

If the refractive index of the film is less than the refractive index of the lens, the minimum non-zero thickness of the non-reflecting coating is given by = lcoating /4.t This result follows from the fact that the extra distance traveled by the wave that travels through the coating is 2t when the wave strikes the film at normal incidence.

If the wave strikes the thin film at 45°, then the extra distance traveled by the wave that travels through the coating is no longer 2t. Instead, it is 2d, where d is the distance shown in Fig. 10.13 at the right. The value of d depends on the value for q2, the angle of refraction. This angle, in turn, depends on the refractive index n1 for air and n1 for the lens coating, according to Snell’s law. To eliminate the light that is incident at 45°, the coating would have to have a minimum non-zero thickness such that = lcoating /4d . Since the coating is designed such that = lcoating /4t , it does not eliminate the light incident at 45°.

d dt q2

q1 = 45°

n2

n1

Air

Lens coating

Figure 10.13

7. Two pieces of the same glass are covered with thin fi lms of diff erent materials. The thickness of each fi lm is the same. In refl ected sunlight, however, the fi lms have diff erent colors. Give a reason that accounts for the diff erence in colors.

Solution

Two pieces of the same glass are covered with thin films of different materials. The thickness t of each film is the same. In reflected sunlight, however, the films have different colors. This could occur because, in general, different materials have different refractive indices. The wavelength lair that is removed from the reflected light corresponds to a value of lfilm in the film that satisfies either of the following two conditions:

= =film2 , 0, 1, 2, 3,…t m ml film glassn n> (1)

= + …film2 [ (1/2)] , = 0, 1, 2, 3,t m ml film glassn n< (2)

However, the value of the wavelength lfilm of the light in the film depends on the index of refraction of the film. We have=l lfilm vacuum /n, where n is the index of refraction of the film. If the index of refraction of the two materials is different, Eqs. (1) and

(2) will be satisfied for two different values of lfilm. These values of lfilm correspond to different values of lair, and, therefore, different wavelengths will be removed from the reflected light.

8. A thin fi lm of a material is fl oating on water (n = 1.33). When the material has a refractive index of n = 1.20, the fi lm looks bright in refl ected light as its thickness approaches zero. But when the material has a refractive index of n = 1.45, the fi lm looks black in refl ected light as its thickness approaches zero. Explain these observations in terms of constructive and destructive interference and the phase changes that occur when light waves undergo refl ection.

Solution

A thin film of material is floating on water (n = 1.33). When the material has a refractive index of n = 1.20, the film looks bright in reflected light as its thickness approaches zero. But when the material has a refractive index of n = 1.45, the film looks black in reflected light as its thickness approaches zero.

When the material has a refractive index of n = 1.20, both the light reflected from the top of the film and the light reflected from the bottom occur under conditions where light is traveling through a material with a smaller refractive index toward a material with a larger refractive index. Therefore, both the light reflected from the top and the bottom undergoes a phase change upon reflection. Both phase changes are equivalent to half of a wavelength. Therefore, the reflections introduce no net phase difference between


Worked Problems 629

the light reflected from the top and bottom. As the thickness of the film approaches zero, the path difference between the light reflected from the top and the bottom of the film approaches zero. Since the path difference is close to zero, and there is no relative phase difference due to reflection, the light reflected from the top will be in phase with the light reflected from the bottom of the film; constructive interference will occur, and the film looks bright in reflected light.

When the material has a refractive index of n = 1.45, only the light reflected from the top of the film is traveling through a material with lower refractive index toward a material with a larger refractive index. Therefore, only the light reflected from the top undergoes a phase change upon reflection. This phase change is equivalent to half of a wavelength. Therefore, there is a net phase difference equivalent to half of a wavelength between the light reflected from the top and bottom of the film. As the thickness of the film approaches zero, the path difference between the light reflected from the top and the light reflected from the bottom of the film approaches zero. Since the path difference is close to zero, and there is a relative phase difference due to reflection that is equal to half of a wavelength, the light reflected from the top will be exactly out of phase with the light reflected from the bottom of the film, and destructive interference will occur. Therefore, the film looks black in reflected light.

Additional Problems

1. In the double-slit experiment shown in Fig. 10.14, the electric fields of the waves arriving at point P are given by

151

152

(2.00 V/m)sin (1.26 10 )

(2.00 V/m)sin (1.26 10 ) 39.6 rad ,

E t

E t

⎡ ⎤= μ ×⎣ ⎦⎡ ⎤= μ × +⎣ ⎦

where time t is in seconds. (a) What is the amplitude of the resultant electric field at point P? (b) What is the ratio of the intensity IP

at point P to the intensity Icen at the center of the interference pattern? (c) Describe where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. In a phasor diagram of the electric fields, (d) at what rate would the phasors rotate around the origin, and (e) what is the angle between the phasors?

D

y

P

S1

S2

b

d

r1

r2

q

Incidentwave

S2

S1

r2

r1

q

q

q

bd

Path length difference ΔL

The ΔL shifts onewave from theother, which determines theinterference

(a) (b)B C

Figure 10.14

Solution

(a) We can use phasor techniques or use trigonometric identities. Here, we show the latter approach. Since sin a + sin(a + b) = 2 cos(b/2)sin(a + b/2), we fi nd

1 2 02 cos( /2)sin( /2),E E E t+ = +f w f

where E0 = 2.00 μV/m, w = 1.26 × 1015 rad/s, and f = 39.6 rad. This shows that the electric field amplitude of the resultant wave is

02 cos( /2) 2(2.00 V/m)cos(19.2 rad) 2.33 V/m.E E= = μ = μf

(b) The equation I = 4I0 cos2(1/2)f leads to = =f20 04 cos ( /2) 1.35I I I at point P, and 2

center 0 04 cos (0) 4I I I= = at the center. Thus,= =center/ 1.35/4 0.338I I .

(c) The phase difference f (in wavelengths) is obtained from f in radians by dividing by 2p. Thus, f = 39.6/2p = 6.3 wavelengths. Thus, point P is between the sixth side maximum (at which f = 6 wavelengths) and the seventh minimum [at which f = 6(1/2) wavelengths].

(d) The rate is given by w = 1.26 × 1015 rad/s.(e) The angle between the phasors is f = 39.6 rad = 2270° (which would look like about 110° when drawn in the usual way).

2. In the double-slit experiment shown in Fig. 10.14, the viewing screen is at distance D = 4.00 m, point P lies at distance y = 20.5 cm from the center of the pattern, the slit separation d is 4.50 μm, and the wavelength l is 580 nm. (a) Determine where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. (b) What is the ratio of the intensity IP at point P to the intensity Icen at the center of the pattern?



Solution

(a) Referring to Fig. 10.14a makes clear that

q = tan−1 (y/D) = tan−1(0.205/4) = 2.93°.

Thus, the phase difference at point P is f = d sin q/l = 0.397 wavelengths, which means it is between the central maximum (zero wavelength difference) and the first minimum (1/2 wavelength difference).

(b) From the equation I = 4I0 cos2(1/2)f (with the value f = 0.397 × 2p = 2.495 rad), we get

20 04 cos ( /2) 0.404I I I= =f

at point P and

2center 0 04 cos (0) 4I I I= =

at the center. Thus, = =center/ 0.404/4 0.101I I .

3. A thin fl ake of mica (n = 1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the viewing screen is now occupied by what had been the seventh bright side fringe (m = 7). If l = 550 nm, what is the thickness of the mica?

Solution

Consider the two waves, one from each slit, that produce the seventh bright fringe in the absence of the mica. They are in phase at the slits and travel different distances to the seventh bright fringe, where they have a phase difference of 2pm = 14p. Now a piece of mica with thickness x is placed in front of one of the slits, and an additional phase difference between the waves develops. Specifically, their phases at the slits differ by

2 2 2 ( 1),m

x x x n− = −p p pl l l

where lm is the wavelength in the mica and n is the index of refraction of the mica. The relationship lm = l/n is used to substitute for lm. Since the waves are now in phase at the screen, we have

2 ( 1) 14x n − =p pl

or9

67 7(550 10 m) 6.64 10 m.1 1.58 1

xn

−−×= = = ×− −

l

4. In Fig. 10.15, two isotropic point sources S1 and S2 emit light in phase at wavelength l and at the same amplitude. The sources are separated by distance 2d = 6.00l. They lie on an axis that is parallel to an x-axis, which runs along a viewing screen at distance D = 20.0l. The origin lies on the perpendicular bisector between the sources. The figure shows two rays reaching point P on the screen, at position xP. (a) At what value of xP do the rays have the minimum possible phase difference? (b) What multiple of l gives that minimum phase difference? (c) At what value of xP do the rays have the maximum possible phase difference? What multiple of l gives (d) that maximum phase difference and (e) the phase difference when xP = 6.00l? (f ) When xP = 6.00l, is the resulting inten-sity at point P maximum, minimum, intermediate but closer to maximum, or intermediate but closer to minimum?

P

d d

S1 S2

D

Screen

0x

Figure 10.15

Solution

(a) We note that, just as in the usual discussion of the double-slit pattern, the x = 0 point on the screen (where that vertical line of length D in the picture intersects the screen) is a bright spot with phase diff erence equal to zero (it would be the middle fringe in the usual double-slit pattern). We are not considering x < 0 values here, so that negative phase diff erences are not relevant (and if we did wish to consider x < 0 values, we could limit our discussion to absolute values of the phase diff erence, so that, again, negative phase diff erences do not enter it). Thus, the x = 0 point is the one with the minimum phase diff erence.


Worked Problems 631

(b) As noted in part (a), the phase difference f = 0 at x = 0.(c) The path length difference is greatest at the rightmost “edge” of the screen (which is assumed to go on forever), so f is maximum

at x = ∞.(d) In considering x = ∞, we can treat the rays from the sources as if they are essentially horizontal. In this way, we see that the

difference between the path lengths is simply the distance (2d) between the sources. The problem specifies 2d = 6.00 l, or 2d/l = 6.00.

(e) Using the Pythagoras’ theorem, we have

2 2 2 2( ) ( )1.71,

D x d D x d+ + + −= − =f l l

where we have plugged in D = 20ld = 3l and x = 6l. Thus, the phase difference at that point is 1.71 wavelengths.

(f) We note that the answer to part (e) is closer to 3/2 (destructive interference) than to 2 (constructive interference), so that the point is “intermediate” but closer to a minimum than to a maximum.

5. A double-slit arrangement produces bright interference fringes for sodium light (a distinct yellow light at a wavelength of l = 589 nm). The fringes are angularly separated by 0.30° near the center of the pattern. What is the angular fringe separation if the entire arrangement is immersed in water, which has an index of refraction of 1.33?

Solution

The angular positions of the maxima of a two-slit interference pattern are given by sinL d mΔ = =q l, where ΔL is the path-length difference, d is the slit separation, l is the wavelength, and m is an integer. If q is small, sin q may be approximated by q in radians. Then, q = ml/d to good approximation. The angular separation of two adjacent maxima is Δq = l/d. When the arrangement is immersed in water, the wavelength changes to l’ = l/n, and the equation above becomes

.d′Δ =′ lq (1)

Dividing Eq. (1) by Δq = l/d, we obtain

1 .n

Δ ′ ′= =Δq lq l

Therefore, with n = 1.33 and Δq = 0.30°, we find Δq’ = 0.23°.

Note: The angular separation decreases with increasing index of refraction; the greater the value of n, the smaller the value of Δq.

6. In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. (NCERT)

Solution

Distance between the slits in Young’s double-slit experiment is d = 0.28 mm = 0.28 × 10−3 m; Distance between the slits and the screen in Young’s double-slit experiment is D = 1.4 m; Distance between the central fringe and the fourth (n = 4) fringe in Young’s double-slit experiment is y = 1.2 cm = 1.2 × 10−2 m; Order of fringes is n = 4.

Let the wavelength of light used in Young’s double-slit experiment be l. In case of a constructive interference, the relation for the distance between the two fringes is

2 371.2 10 0.28 10 6 10 600 nm.

4 1.4ydDy n

d nD

− −−× × ×= ⇒ = = = × =×l l

Hence, the wavelength of the light is 600 nm.

7. In Young’s double-slit experiment using monochromatic light of wavelength l, the intensity of light at a point on the screen where path diff erence is l, is K units. What is the intensity of light at a point where path diff erence is l/3? (NCERT)

Solution

Let I1 and I2 be the intensity of the two light waves used in Young’s double-slit experiment. If f is the phase difference between the two waves and we know that for monochromatic light waves, =1 2I I . Their resultant intensity can be obtained from the relation

1 2 1 2 1 12 cos 2 2 cos .I I I I I I I= + + = +′ f f (1)

The relation between phase difference and path difference is given as follows:

2Phase difference ( ) Path difference.= ×pf l



In this case, the path difference is l, therefore, the phase difference is calculated as

2

2 .

×

⇒ =

p llf p

(2)

Substituting Eq. (2) in Eq. (1), we get

1 1 1 1 12 2 cos 2 2 2 1 4 .I I I I I I= + = + × =′ p (3)

According to question the resultant intensity is

I′ = K (4)

Substituting Eq. (4) in Eq. (3), we get

1 .4KI = (5)

When path difference is l/3, the phase difference is given by

2 2 2Path difference .3 3

= × = × ⇒ =p p l pf fl l

Hence, the resultant intensity is calculated as

R 1 1 1 1 1 1 12 12 cos 2 2 .3 2

I I I I I I I I⎛ ⎞= + + = + − =′ ⎝ ⎠p

Using Eq. (5), we have

R 1 .4KI I= =

Hence, the intensity of light at a point where the path difference is l/3 is K/4 units.

8. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? (NCERT)

Solution

Wavelength of the one light beam used to obtain interference fringes is l1 = 650 nm; Wavelength of another light beam used to obtain interference fringes is l2 = 520 nm.

Let the distance of the slits from the screen be D and the distance between the two slits be d.

(a) Distance of the nth bright fringe on the screen from the central maximum is calculated using relation

1 .Dx nd

⎛ ⎞= ⎝ ⎠l

For the third bright fringe, n = 3. Therefore,

3 650 1950 nm.D Dxd d

⎛ ⎞= × = ⎝ ⎠

(b) Let the nth bright fringe due to wavelength l2 and (n − 1)th bright fringe due to wavelength l1 coincide on the screen. We can equate the conditions for bright fringes as

2 1( 1) 520 650 650 650 130 5.n n n n n n= − ⇒ = − ⇒ = ⇒ =l l

Hence, the least distance from the central maximum is calculated as

2 5 520 2600 nm.D D Dx nd d d

= = × =l

9. In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3. (NCERT)

Solution

Distance of the screen from the slits is D = 1 m; Wavelength of light used is =l1 600 nm; Angular width of the fringe in air is= °q1 0.2 ; Angular width of the fringe in water is q2 ; Refractive index of water is 4/3.


Worked Problems 633

The angular width of the fringe if the entire experimental apparatus is immersed in water is

12 1

2

3 3 0.2 0.15.4 4

= ⇒ = = × =qm q qq

Therefore, the angular width of the fringe in water will reduce to 0.15°.

10. In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits? (NCERT)

Solution

Wavelength of light used is l = 6000 nm = 600 × 10−9 m; Angular width of fringe is given by

3.140.1 0.1 rad.180 1800

= ° = × =pq

Angular width of a fringe is related to slit spacing (d) by the relation q = l/d. Therefore,

94600 10 3.44 10 m.

(3.14/1800)d

−−×= = = ×l

q

Therefore, the spacing between the slits is −× 43.44 10 m .

11. Answer the following questions: (a) When a low-flying aircraft passes overhead, we sometimes notice a slight shaking of the pic-ture on our TV screen. Suggest a possible explanation. (b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justifica-tion of this principle? (NCERT)

Solution

(a) When a low-flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen because weak radar signals sent by a low-flying aircraft can interfere with the TV signals received by the antenna. As a result, the TV signals may get distorted.

(b) The principle of linear superposition of wave displacement is essential to our understanding of intensity distributions and interference patterns. This is because superposition follows from the linear character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the second-order wave equation, then any linear combination of y1 and y2 will also be the solution of the wave equation.

12. What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations when (a) the screen is moved away from the plane of the slits; (b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength; (c) the separation between the two slits is increased; (d) the source slit is moved closer to the double-slit plane; (e) the width of the source slit is increased; (f ) the monochromatic source is replaced by a source of white light? (In each operation, take all parameters, other than the one specified, to remain unchanged.) (NCERT)

Solution

(a) Angular separation of the fringes remains constant (=l/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits.

(b) The separation of the fringes (and also angular separation) decreases. (However, see the condition mentioned in part (d) below.)

(c) In a Young’s double-slit experiment, if the separation between the two slits is increased, the separation of the fringes (and also angular separation) decreases. (However, see the condition mentioned in part (d) below.)

(d) Let us consider s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition < l/ /s S d should be satisfied; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e., the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed.

(e) Same as in part (d). As the source slit width increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition ≤ l/ /s S d is not satisfied, the interference pattern disappears.

(f) The interference patterns due to different component colors of white light overlap, incoherently. The central bright fringes, for different colors, are at the same position. Therefore, the central fringe is white. At a point P, for which S2P – S1P lb /2, where

≈lb( 4000 Å) represents the wavelength for the blue color, the blue component will be absent and the fringe will appear red in color. Slightly farther away, S2Q – S1Q = =l lb r /2 (where r( 8000 Å)≈l is the wavelength for the red color), the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.




1. What are coherent sources? Why are coherent sources required to produce interference of light? Give an example of interference of light in everyday life. In Young’s double-slit experiment, the two slits are 0.03 cm apart and the screen is placed at a distance of 1.5 m away from the slits. The distance between the central bright fringe and fourth bright fringe is 1 cm. Calculate the wavelength of light used. (2007)

Solution

Two wavefronts that has a constant phase relationship is called coherent source of light. Since the interference patterns are visible only when the phase difference of the interfering waves are constant, only coherent light sources can produce a visible interference pattern. The colors appearing in the soup bubble and oil sleek in water are two examples of interference of light in daily life.

The fringe width of the given pattern

distance between central and fourth fringe 1 cm.4 4

= =b

Also

.Dd

= lb

Therefore

2 27

1 10 0.03 104 5 10 m.

1.5d

D

− −−

× × ×= = = ×bl

2. What is interference of light? Write two essential conditions for sustained interference pattern to be produced on the screen. Draw a graph showing the variation of intensity versus the position on the screen in Young’s experiment when (a) both the slits are opened and (b) one of the slits is closed. What is the eff ect on the interference pattern in Young’s double-slit experiment when (i) screen is moved closer to the plane of slits? (ii) separation between two slits is increased. Explain your answer in each case. (2006)

Solution

Interference is the phenomenon of redistribution of energy due to the superposition of waves. When two constant initial phases superimpose each other, they produce a pattern consists of periodic minima and maxima, which is called interference pattern.

Condition for Interference: The two sources should have equal intensities, otherwise the minima will not be zero and there will be general illumination. The two interfering beams must be coherent and of same wavelength. That is the two sources should emit light in same wavelength and their initial phases should remain constant. If the initial phase is not constant the phase relation between the two interfering waves vary continuously as a result the resultant intensity at any point will vary with time and so the pattern become unobservable.

(a) If the two slits are open, interference pattern appear on the screen as shown in Fig. 10.16a.(b) If the only one slit is open, diffraction pattern appears on the screen as shown in Fig. 10.16b.

0 p−p−2p 2p

I

f f0 2ll–l–2l

I

(a) (b)

Figure 10.16

The fringe width of Young’s double-slit diff raction is given by

.Dd

= lb

(i) Therefore, the decrease in separation between the screen and the slit decreases the fringe width. (ii) The fringe width is inversely proportional to the separation between the slits. Therefore, the fringe width decreases on increasing the separation between the slits.


Worked Problems 635

3. (a) In Young’s double-slit experiment, deduce the conditions for (i) constructive and (ii) destructive interference at a point on the screen. Draw a graph showing variation of the resultant intensity in the interference pattern against position x on the screen. (b) Compare and contrast the pattern which is seen with two coherently illuminated narrow slits in Young’s experiment with that seen for a coherently illuminated single slit producing diff raction. (2005)

Solution

Let the two slits be separated by a distance d and a screen be placed at a distance D from the slits (Fig. 10.17). The slits S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2. Thus, the two sources S1 and S2 will be locked in phase.

The constructive interference happens when the wavelets emitting from both sources are in phase, that is, S2P – S1P = nl, where n = 0, 1, 2, … Then,

( ) ( )1/22 2

22

1 ( /2)S P 1 .2 2d x dD x D

D⎡ ⎤ ⎡ ⎤+= + + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Similarly,

( )2

11 ( /2)S P 12

x dDD

⎡ ⎤−= +⎢ ⎥⎣ ⎦

G

G′

D

S1

S2

d

O

x

P

x

zy

z

Figure 10.17

Therefore, the path difference is given by

2 2

2 1[ ( /2)] [ ( /2)]S P S P .

2x d x d xd

D D+ − −− = =

Hence, point P will be in maximum intensity if

xd nD

= l

or

.n Dxd

= l

The condition for destructive interference is given as follows:

1Path difference .2

n⎛ ⎞= +⎝ ⎠ l

Therefore,

( )12

xd nD

= + l

or

(2 1) .2

n Dxd

+= l

The resultant intensity distribution is as shown in Fig. 10.18.



0 1 2 3 4–1–2–3–4–5

–1

1

2

x

Intensity

Figure 10.18

SECTION III ➥ Diffraction ➥ Resolving Power of Optical Instruments

(1) The condition to obtain minima in a single-slit diffraction pattern is

sin , for 0, 1, 2, ,a m m= =q l …

where a is the slit width and q is the angle to the central axis.(2) By Rayleigh’s criterion, two objects that are barely resolvable must have an angular separation qR given by

R 1.22 ,d

= lq

where d is the least distance between the two objects so that their diffraction images are just resolved.(3) For a microscope,

2 sin1 ,1.22d

= m ql

where q is the semi-vertical angle of the cone in which rays of light from an object enter the objective lens of the microscope m is the refractive index of the medium, and l is the wavelength of the light used to observe the objects.

(4) For a telescope,

1 .1.22

D=q l

FORMULAS AND UNITS

Conceptual Problems

1. Whenever single-slit diffraction is discussed, generally, we consider the ratio of the wavelength l to the width W of the slit. We ignore the height of the slit, in effect assuming that the height is much larger than the width. Suppose the height and width were the same size, so that diffraction in both dimensions occur. How would the diffraction pattern in Fig. 10.19 be altered? Give your reasoning.

Figure 10.19


Worked Problems 637

Solution

Suppose that the height and width are of the same size, the diffraction occurs in both dimensions. The resulting pattern changes in the following way: the height of the horizontal fringes is reduced. Furthermore, there are two sets of fringes. One set is along the horizontal axis, and the other set is along the vertical axis. The two sets share a common central maximum. The intensity of the central maximum is the greatest. The intensity decreases on either side of the central maximum, in both the horizontal and the vertical directions. Figure 10.20 shows the central maximum and the less intense higher order on either side of the central maximum in both dimensions.

Figure 10.20

2. Yellow light is used in a single-slit diff raction experiment with slit width of 0.6 mm. If yellow light is replaced by X-rays, how will the diff raction pattern be aff ected?

Solution

Yellow light is used in a single-slit diffraction experiment with slit width of 0.6 mm if yellow light is replaced by X-rays no diffraction pattern will be observed because wavelength of X-rays is very small as compared to the size of the slit.

3. How does resolving power of a telescope change on decreasing the aperture of its object lens? Justify your answer.

Solution

Resolving power of a telescope is given by

,1.22

Dl

where D is the aperture of the objective lens and l is the wavelength of the light. From the above relation we conclude that the resolving power is directly proportional to the aperture of the objective lens. Therefore, on decreasing aperture of the objective lens, the resolving power of the telescope decreases.

4. How does the resolving power of a compound microscope change on (a) decreasing the wave length of light used and (b) decreasing the diameter of its object lens?

Solution

Resolving power of a microscope is given by

2 sin

,m q

l (1)

where l is the wavelength of light used, q is the semi-vertical angle, and m is the refractive index of the medium. From Eq. (1), we conclude that

(a) the resolving power of a microscope is inversely proportional to the wavelength of the light; therefore, on decreasing wavelength of light, resolving power will increase and

(b) the resolving power of a microscope is directly proportional to the semi-vertical angle. Therefore, on decreasing diameter of objective lens, semi-vertical angle q decreases and hence the resolving power decreases.

5. How will the resolving power of a compound microscope be aff ected, when (a) the frequency of light used to illuminate the object is increased and (b) the focal length of the objective is increased. Justify your answer in each case.



Solution

Resolving power of a compound microscope is given by

2 sin

.m q

l (1)

Since wavelength and frequency are related by the expression c = nl, Eq. (1) is given by

2 sin

,c

n m q (2)

where n is the frequency of light and m is the refractive index of the material.

(a) From Eq. (2), we conclude that resolving power of a compound microscope is directly proportional to the frequency of the light used. Therefore, when the frequency of light is increased, resolving power will increase.

(b) When the focal length of objective is increased, distance between object and objective will increase. It will lead to a decrease in the value of angle q and from Eq. 1 we conclude that resolving power of a compound microscope is directly proportional to the angle q hence resolving power of the compound microscope will also decrease.

Additional Problems

1. Manufacturers of wire (and other objects of small dimension) sometimes use a laser to continually monitor the thickness of the product. The wire intercepts the laser beam, producing a diffraction pattern like that of a single slit of the same width as the wire diameter (Fig. 10.21). Suppose a helium–neon laser, of wavelength 632.8 nm, illuminates a wire, and the diffraction pattern appears on a screen at distance L = 2.60 m. If the desired wire diameter is 1.37 mm, then what is the observed distance between the two tenth-order minima (one on each side of the central maximum)?

LWire-making

machine

Wire

He–Nelaser

Figure 10.21

Solution

From y = mlL/a, we get

( ) (632.8 nm)(2.60) [10 ( 10)] 24.0 mm.1.37 mm

m D Dy ma a

Δ = Δ = Δ = − − =l l

2. Light of wavelength 633 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.20°. What is the width of the slit?

Solution

The condition for a minimum of a single-slit diffraction pattern is

sina m= ,q l

where a is the slit width, l is the wavelength, and m is an integer. The angle q is measured from the forward direction, so for the situation described in the problem, it is 0.60° for m = 1. Thus,

95633 10 m 6.04 10 m.

sin sin 0.60ma

−−×= = = ×°

lq

3. The full-width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern (Fig. 10.22). (a) Show that the intensity drops to one-half the maximum value when sin2 a = a2/2. (b) Verify that a = 1.39 rad (about 80°) is a solution to the transcendental equation of (a). (c) Show that the FWHM is Δq = 2 sin−1 (0.443l/a), where a is the slit width. Calculate the FWHM of the central maximum for slit width (d) 1.00l, (e) 5.00l, and (f ) 10.0l.


Worked Problems 639

05101520 5 10 15 20

a = 5l1.0

0.8

0.6

0.4

0.2

Δq

Relative intensity

q (degrees)

Figure 10.22

Solution

(a) The intensity for a single-slit diffraction pattern is given by2

2sin

mI I= aa

where a = (pa/l) sin q, a is the slit width, and l is the wavelength. The angle q is measured from the forward direction. We require I = Im/2, so

2 21sin .2

=a a

(b) We evaluate sin2 a and a2/2 for a = 1.39 rad and compare the results. To be sure that 1.39 rad is closer to the correct value for a than any other value with three significant digits, we could also try 1.385 rad and 1.395 rad.

(c) Since a = (pa/l) sin q, we have

( )1sin .a

−= alq p Since a/p = 1.39/p = 0.442, we get

1 0.442sin .

a− ⎛ ⎞= ⎝ ⎠

lq

The angular separation of the two points of half intensity – one on either side of the center of the diffraction pattern – is

( )1 0.4422 2sin .a

−Δ = = lq q

(d) For a/l = 1.0, we have

1 0.4422sin 0.916 rad 52.5 .1.0

− ⎛ ⎞Δ = = = °⎝ ⎠q

(e) For a/l = 5.0, we have

( )1 0.4422sin 0.177 rad 10.1 .5.0

−Δ = = = °q

(f) For a/l = 10,

1 0.4422sin 0.0884 rad 5.06 .10

− ⎛ ⎞Δ = = = °⎝ ⎠q

4. Light of wavelength 440 nm passes through a double slit, yielding a diffraction pattern whose graph of intensity I versus angular position q is shown in Fig. 10.23. Calculate (a) the slit width and (b) the slit separation. (c) Verify the displayed intensities of the m = 1 and m = 2 interference fringes.

7

6

5

4

3

2

1

0 5

Inte

nsi

ty (m

W/c

m2 )

q (degrees)

Figure 10.23



Solution

(a) The first minimum of the diffraction pattern is at 5.00°, so

0.440 m 5.05 m.sin sin 5.00

aμ= = = μ°

lq

(b) Since the fourth bright fringe is missing, d = 4a = 4(5.05 μm) = 20.2 μm.(c) For the m = 1 bright fringe, we have

sin 5.05 m sin1.250.787 rad.

0.440 ma ( μ ) °= = =p q p

a l m

Consequently, the intensity of the m = 1 fringe is

⎛ ⎞ ⎛ ⎞= = × =⎝ ⎠ ⎝ ⎠

2 22 2

msin sin 0.787 rad

7.0 mW/cm 5.7 mW/cm .0.787

I Ia

a

which agrees with Fig. 10.23. Similarly for m = 2, the intensity is I = 2.9 mW/cm2, also in agreement with Fig. 10.23.

5. If you look at something 40 m from you, what is the smallest length (perpendicular to your line of sight) that you can resolve, according to Rayleigh’s criterion? Assume the pupil of your eye has a diameter of 4.00 mm, and use 500 nm as the wavelength of the light reaching you.

Solution

1.226.1 mm.

LD

d= =l

6. Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects? (NCERT)

Solution

Distance between the towers on top of two hills is d = 40 km; Height of the line joining the two hills is d = 50 m, thus, the radial spread of the radio waves should not exceed 50 km. Since the hill is located halfway between the towers, Fresnel’s distance can be obtained as

ZP = 20 km = 2 × 104 m.

Aperture is a = d = 50 m Fresnel’s distance is given by the relation

2

P ,aZ = l

where l is the wavelength of the radio waves, which is calculated as

2 24

4P

50 1250 10 0.1250 m 12.5 cm.2 10

aZ

−= = = × = =×

l

7. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the center of the screen. Find the width of the slit. (NCERT)

Solution

Wavelength of light beam is l = 500 nm = 500 × 10−9 m; Distance of the screen from the slit is D = 1 m; For first minima n = 1; Distance between the slits is d; Distance of the first minimum from the center of the screen is x = 2.5 mm = 2.5 × 10−3 m. Therefore,

94

31 500 10 1 2 10 m 0.2 mm.

2.5 10d n Dn x dD x

−−

−× × ×= ⇒ = = = × =

×ll

Therefore, the width of the slits is 0.2 mm.

8. In deriving the single-slit diff raction pattern, it was stated that the intensity is zero at angles of nl/a. Justify this by suitably dividing the slit to bring out the cancellation. (NCERT)

Solution

Consider that a single slit of width d is divided into n smaller slits. Therefore, width of each slit is

.ddn

=′

Angle of diffraction is calculated as

( / ) .d dd d

′= = ′l lq


Worked Problems 641

Each of these infinitesimally small slit sends zero intensity in direction q. Hence, the combination of these slits gives zero intensity.

9. Assume that light of wavelength 6000 Å is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch? (NCERT)

Solution

Diameter of objective of telescope is given by

= = × 2.54 cm2 100 inch 100 inch = 254 cm.

1 incha

Wavelength of light used is − −= = × = ×l 10 56000 Å 6000 10 m 6 10 cm.Let the limit of resolution of the telescope be Δq . The limit of resolution of telescope is calculated as

571.22 1.22 6 10 2.9 10 rad.

2 254a

−−× ×Δ = = = ×lq

10. For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm? (NCERT)

Solution

Width of the aperture is −= = × 33 mm 3 10 m;a Wavelength of light is −= = ×l 9500 nm 500 10 m.Let the distance for which ray optics is a good approximation be zF, which is obtained as

2 3 2

F 7(3 10 )= = = 18 m.

5 10az

−

−××l

From the above calculations, it can be concluded that even with a small aperture, diffraction spreading can be neglected for rays many meters in length. Thus, ray optics is valid in many common situations.


1. In a single-slit diffraction experiment, when a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. Explain why? State two points of difference between the inference pattern obtained in Young’s double-slit experiment and the diffraction pattern due to a single slit. (2009)

Solution

A bright spot appears in the middle of the shadow due to the diffraction of light at the edges of the tiny object.

Difference Between Interference and Diffraction: (1) The interference pattern has a number of bright and dark fringes of equal intensity. The diffraction pattern has a central bright fringe of maximum intensity and intensity falls as go away from the center. (2) Interference is due to the superposition of two waves coming from two coherent sources whereas diffraction is the superposi-tion of secondary wavelets coming from different parts of same wavefront.

2. Defi ne the term “resolving power” of an astronomical telescope. How does it get aff ected on (a) increasing the aperture of the objective lens? (b) increasing the wavelength of the light used? Justify your answer in each case. (2007)

Solution

The resolving power of a telescope is the ability to distinguish between two objects separated by a small angular distance a.

0.25 .D

= × la

Therefore, increasing the aperture size of the telescope increases the resolving power of the telescope. Increasing the wavelength used decreases the resolving power of the telescope.

3. State the condition under which the phenomenon of diff raction of light takes place. Derive an expression for the width of the central maximum due to diff raction of light at a single slit. A slit of width a is illuminated by a monochromatic light of wavelength 700 nm at normal incidence. Calculate the value of a for position of (a) fi rst minimum at an angle of diff raction of 30° and (b) fi rst maximum at an angle of diff raction of 30°. (2007)

Solution

Diffraction occurs when a obstacle is placed in the path of the beam and it is prominent if the size of the object is comparable to that of the wavelength of the light incident on it.

Let a source of monochromatic light be incident on a slit of finite width a, as shown in Fig. 10.24. In diffraction of Fraunhofer type, all rays passing through the slit are approximately parallel. In addition, each portion of the slit acts as a source of light waves according to Huygens principle. For simplicity, we divide the slit into two halves. At the first minimum, each ray from the upper half is exactly 180 out of phase with a corresponding ray form the lower half.



a/2

a/2

q

sin qa2

Figure 10.24

The condition for the first minimum is

sin2 2a = lq

or

sin .a

= lq

Applying the same reasoning to the wavefronts from four equally spaced points a distance apart, the path difference is d = a sin q/4 and the condition for destructive interference is

2sin .a

= lq

The argument can be generalized to show that destructive interference will occur when asin q = ml for distructive interference.

a

L

sin–1 q

l/a

2l/a

–2l/a

–l/a

q O

Figure 10.25

The intensity distribution of the diffraction pattern is given in Fig. 10.25. We have

.sinma = l

q

Therefore, at 30°, we have

9700 10 1.4 m,sin 30

a−×= = μ

and at 30.5°, we have

9700 10 1.37 m.sin 30.5

a−×= = μ°

4. What is diffraction of light? Draw a graph showing the variation of intensity with angle in a single-slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern. How would the diffraction pattern of a single slit be affected when (a) the width of the slit is decreased? (b) the monochromatic source of light is replaced by a source of white light? (2006)


Worked Problems 643

Solution

The process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced (Fig. 10.26).

a

L

sin–1 q

l/a

2l/a

–2l/a

–l/a

q O

Figure 10.26

In double-slit diffraction, the separation between two nearby maximas are constant and the fringe width is constant whereas in single-slit diffraction pattern, the central maxima has the maximum width and the separation between the adjacent maxima (or minima) are not constant. Also the intensity of the maxima in interference pattern is constant but in diffraction patter the central maxima has the maximum intensity.

If white light is used in single-slit diffraction pattern the central maxima appear white and other bands appear colored. The width of the first maxima is given by

= 2sin .alq

Hence, it is concluded that the width of the slit is decreased when the deviation increases.

SECTION IV

➥ Polarization ➥ Polarization by Scattering ➥ Polarization by Reflection

(1) If the transmission axis of an analyzer is oriented at an angle q relative to the transmission axis of the polarizer, Malus’ Law is given by

20 cos ,I I= q

where I0 is the average intensity of the light entering the analyzer. (2) Brewster’s angle qB is given by

21B

1tan ,n

n−=q

where n1 is the refractive index of the medium through which the incident and reflected rays travel and n2 is the refractive index of the medium from which the light reflects.

FORMULAS AND UNITS

Conceptual Problems

1. Sound waves cannot be polarized. Why?

Solution

Sound waves cannot be polarized because only transverse waves (the waves in which direction of propagation of wave is perpendicular to the directions in which particles oscillate of particles) can be polarized and sound waves are longitudinal waves in which the particles oscillate parallel to the direction of propagation of waves.



2. Does the value of wavelength of light have any role in polarization?

Solution

No, the value of wavelength of light does not have any role in polarization. Electromagnetic waves of all wavelengths can be polarized.

3. Which plane is defi ned as the plane of polarization in a plane polarized electromagnetic wave?

Solution

In a plane polarized electromagnetic wave the plane perpendicular to the plane of vibrations is called plane of polarization.

4. A beam of unpolarized light is incident on the boundary between two transparent media. If the refl ected light is completely plane polarized, how is its direction related to the direction of the corresponding refracted light?

Solution

If a beam of unpolarized light is incident on the boundary between two transparent media and if the reflected light is completely plane polarized, then the reflected ray of light is perpendicular to the refracted ray as shown in Fig. 10.27.

Incident

Refracted

Medium

AirReflected

Figure 10.27

5. When light is polarized by refl ection, what is the plane of vibration of the electric fi eld vector in polarized light?

Solution

When light is polarized by reflection, the electric field vector is perpendicular to the plane of incidence.

6. Two polaroids are crossed to each other. Now, one of them is rotated through 60°. What percentage of incident unpolarized light will pass through the system?

Solution

The angle between the two Polaroids that are crossed to each other is q = 60°. Using the Malus’ law, we get2

02 20 0 0 0 0

1cos cos 60 0.25 25% of .2 4

II I I I I I⎛ ⎞= = ° = = = =⎝ ⎠q

Additional Problems

1. In Fig. 10.28, unpolarized light is sent into a system of three polarizing sheets. The angles q1, q2, and q3 of the polarizing directions are measured counterclockwise from the positive direction of the y-axis (they are not drawn to scale). Angles q1 and q3 are fixed, but angle q2 can be varied. Figure 10.29 gives the plot between of the intensity of the light emerging from sheet 3 as a function of q2. (The scale of the intensity axis is not indicated.) What percentage of the light’s initial intensity is transmitted by the three-sheet system when q2 = 90°?

q2

q1

q3

y

x

Figure 10.28


Worked Problems 645

60°0° 120° 180°

I

0 q2

Figure 10.29

Solution

We note the points at which the curve is zero (q2 = 60° and 140°) in Fig. 10.29. We infer that sheet 2 is perpendicular to one of the other sheets at q2 = 60°, and that it is perpendicular to the other of the other sheets when q2 = 140°. Without loss of generality, we choose q1 = 150°, q3 = 50°. Now, when q2 = 90°, it will be |Δq | = 60° relative to sheet 1 and |Δq ′| = 40° relative to sheet 3. Therefore,

f 2 2

i

1 cos ( )cos ( ) 7.3%.2

II

= Δ Δ =′q q

2. In Fig. 10.30a, unpolarized light is sent into a system of two polarizing sheets. The angles q1 and q2 of the polarizing directions of the sheets are measured counterclockwise from the positive direction of the y-axis (they are not drawn to scale in the figure). Angle q1 is fixed but angle q2 can be varied. Figure 10.30b gives the intensity of the light emerging from sheet 2 as a function of q2. (The scale of the intensity axis is not indicated.) What percentage of the light s initial intensity is transmitted by the two-sheet system when q2 = 90?

90° 180°q20

I

(b)

q2

q1

y

x

(a)

Figure 10.30

Solution

We examine the point where the graph reaches zero: q2 = 160°. Since the polarizers must be “crossed” for the intensity to vanish, then q1 = 160° – 90° = 70°. Now, we consider the case q2 = 90° (which is hard to judge from the graph). Since q1 is still equal to 70°, then the angle between the polarizers is now Δq = 20°. Accounting for the “automatic” reduction (by a factor of one-half ) whenever unpolarized light passes through any polarizing sheet, then our result is

21 cos ( ) 0.442 44%.2

Δ = ≈q

3. A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through 360° while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of 5.0 during the rotation, what fraction of the intensity of the original beam is associated with the beam’s polarized light?

Solution

Let I0 be the intensity of the incident beam and f be the fraction that is polarized. Thus, the intensity of the polarized portion is fI0. After transmission, this portion contributes fI0 cos2 q to the intensity of the transmitted beam. Here q is the angle between the direction of polarization of the radiation and the polarizing direction of the filter. The intensity of the unpolarized portion of the incident beam is (1 – f )I0 and after transmission, this portion contributes (1 – f )I0/2 to the transmitted intensity. Consequently, the transmitted intensity is

20 0

1cos (1 ) .2

I fI f I= + −q



As the filter is rotated, cos2 q varies from a minimum of 0 to a maximum of 1, so the transmitted intensity varies from a minimum of

min 01 (1 )2

I f I= −

to a maximum of

max 0 0 01 1(1 ) (1 ) .2 2

I fI f I f I= + − = +

The ratio of Imax to Imin is

max

min

1 .1

I fI f

+= −

Setting the ratio equal to 5.0 and solving for f, we get f = 0.67.

4. Unpolarized light of intensity 10 m W/m2 is sent into a polarizing sheet as in Fig. 10.31. What are (a) the amplitude of the electric fi eld component of the transmitted light and (b) the radiation pressure on the sheet due to its absorbing some of the light?

Incident light ray

Vertically polarized light

Polarizing sheet

Unpolarized light

The sheet’s polarizing axisis vertical, so only vertically

polarized light emerges

Figure 10.31

Solution

(a) Since the incident light is unpolarized, half the intensity is transmitted and half is absorbed. Thus the transmitted intensity is I = 5.0 m W/m2. The intensity and the electric field amplitude are related by = m2

m 0( ) /2 , soI E c = m2m 0( ) /2 , soI E c

7 8 3 2m 02 2(4 10 H/m)(3.00 10 m/s)(5.0 10 W/m ) 1.9 V/m.E cI − −= = × × × =m p

(b) The radiation pressure is pr = Ia/c, where Ia is the absorbed intensity. Thus,

3 211

r 85.0 10 W/m

1.7 10 Pa.3.00 10 m/s

p−

−×= = ××

5. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.) (NCERT)

Solution

Refractive index of glass is m = 1.5; Brewster angle is q.Brewster angle for air to glass transition is calculated as

1tan tan 1.5 56.31 .−= ⇒ = = °q m q

Therefore, the Brewster angle 56.31°.

6. Unpolarized light is incident on a plane glass surface. What should be the angle of incidence so that the refl ected and refracted rays are perpendicular to each other? (NCERT)

Solution

Reflected and refracted rays are perpendicular to each other; therefore, +i r = p/2.

Refractive index of glass is m = 1.5

Let the angle of incidence be ip. The angle of incidence is equal to the Brewster’s angle for air to glass interface. Therefore, using Brewster’s law, we get

m = tan ip ⇒ 1.5 = tan ip ⇒ ip = tan−1 1.5 ⇒ ip = 57°.


Worked Problems 647


1. What is an unpolarized light? Explain with the help of suitable ray diagram how an unpolarized light can be polarized by reflection from a transparent medium. Write the expression for Brewster angle in terms of the refractive index of denser medium. (2010)

Solution

An unpolarized light is one for which the displacement is randomly changing with time though it always be perpendicular to the direction of propagation. It can be thought of as the superposition of two polarized waves whose planes of vibration are perpendicular to each other.

An unpolarized light can be polarized by reflection from a transparent medium such as water, as shown in Fig. 10.32. The dots and arrows indicate that both polarizations are present in the incident and refracted waves. Consider the case when the reflected wave travels at right angles to the refracted wave. The arrows are parallel to the direction of the reflected wave and motion in this direction does not contribute to the reflected wave. The reflected light is thus linearly polarized perpendicular to the plane of the figure (represented by dots). When an unpolarized light in incident on the boundary between two transparent media, the reflected light is polarized with its electric vector perpendicular to the plane of incidence. Thus, when reflected wave is perpendicular to the refracted wave, the reflected wave is a totally polarized wave. The angle of incidence in this case is called Brewster’s angle and is denoted by iB.

We have,

B .2

i r+ = p

Therefore, by Snell’s law, we have

−= = = ⇒ =−B B B 1

B BB B

sin sin sin= tan tan .

sin sin[( /2) ] cosi i i

i ir i i

m mp

For angles of incidence other than Brewster’s angle, both components are present but one is stronger than the other. There is no stable phase relationship between the two perpendicular components. This kind of light is called partially polarized.

Incident ray(Unpolarized)

Partiallypolarized

Medium

Air Completelypolarized

Figure 10.32

2. If the angle between the pass axis of the polarizer and the analyzer is 45°, write the ratio of the intensities of original light and the transmitted light after passing through the analyzer. (2009)

Solution

The intensity of the light after passing through the polarizer a

20 icos ,I I= q

where angle qi is the angle between the initial polarization and the axis of polarizer. Let I0 be the intensity of the incident light; therefore, after the passing through the analyzer, the intensity becomes

0 .2pI

I =

After passing through the analyzer,

0 02 1cos 45 .2 2 4a pI I

I I= ° = × =

Therefore, the intensity of the light becomes one-fourth of the initial intensity after passing the analyzer.



1. What is the wavelength range of visible radiation?

2. Name the colors visible to human eye.

3. What is the velocity of light in km/s?

4. What is a light year?

5. Name the scientist who made the first successful attempt to measure the speed of light.

6. Define a wavefront.

7. If light waves are diverging from a point source, what is the shape of the wavefront?

8. For a point source of light at the focus of the convex lens, what is the shape of wavefront of light emerging out of lens?

9. Are light waves longitudinal electromagnetic in nature?

10. What is the velocity of light in vacuum?

11. Can two wavefronts ever cross each other? If not, why?

12. Mention the shape of the wavefront in each of the following case: (a) When a small piece of stone is dropped over a water surface. (b) When a wave originates from a point source and travels around in a homogeneous medium. (c) When the origin of a spherical wavefront is at infinity.

13. Draw the type of wavefront corresponding to a beam of light from (a) light emerging out of a convex lens with point source at its focus, (b) light diverging from a point source, and (c) the portion of the wavefront from a distant star inter-cepted by Earth.

14. Name and draw three different kinds of wavefronts.

15. What is the phase difference between any two particles on a wavefront?

16. If a wave undergoes refraction, what happens to its phase?

P R A C T I C E P R O B L E M S

VERY SHORT ANSWER TYPE

Huygens Principle

Interference

17. What are non-coherent sources? Can they produce sustained interference?

18. Is it possible to obtain two coherent light sources from one coherent source? Give one example.

19. Is the time of persistence of vision less or more than the time of coherence?

20. Can a laser beam produce coherent light?

21. Give the relation between phase difference and path difference.

22. Is headlight of a car (a) plane polarized and (b) highly coherent?

23. What are coherent sources?

24. What are the values of path difference for obtaining a constructive and a destructive interference?

25. Two coherent sources of light have an intensity ratio of 64 : 25. What is their amplitude ratio?

26. Write equations that give the positions of the nth bright and dark fringes and intensity at point on the screen during interference.

27. What do you mean by the superposition principle?

28. State the conditions which must be satisfied for two light sources to be coherent.

29. Why cannot we obtain interference using two independent sources of light?

30. The widths of two slits of Young’s experiment are in the ratio 9 : 4. Calculate the intensity ratio in the interference pattern.

31. The intensity ratio in the interference pattern is 4 : 1. Calculate the ratio of slit widths.

32. What happens to interference pattern when one of the slits in Young’s double-slit experiment is closed?

33. What is the main condition to produce interference of light?

34. If Young’s double-slit experimental setup is immersed in water, what will be the effect on the fringes?

35. Name the source that can produce stationary interference.

36. Name the phenomenon that explains the colors of thin films.

37. Do we still observe interference if the films are quite thick or wide?

38. In Young’s double-slit experiment, will the interference take place using (a) red light and (b) laser light?

39. Can we produce interference with sound waves?

40. Can interference be produced with X-rays?

41. In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

42. In Young’s double-slit experiment, three lights; blue, yellow, and red are successively used. For which color will the fringe-width be a maximum?

43. Give a formula for finding the width of the central maxima.

44. When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle. What is it due to?

45. In Young’s double-slit experiment, what is the effect on fringe width if (a) light of smaller frequency is used and (b) separa-tion between the slits is increased?

46. What is fringe width? Write an expression for it.

47. What are the factors on which fringe width depends?

48. Give examples illustrating the phenomenon of interference of light.


Practice Problems 649

49. The arrangement in Young’s double-slit experiment is immersed completely in a liquid of refractive index m. What is the effect on width of fringes?

50. Why should we have two narrow sources to produce good interference fringes?

51. No interference pattern is detected when two coherent sources are infinitely close to one another. Why?

52. Interference pattern is not sustained if the phase difference between the two sources varies continuously. Why?

Diffraction

53. Why cannot we get diffraction pattern from a wide slit illuminated by monochromatic light?

54. Can we produce diffraction using radio waves?

55. For what size of the obstacle does light undergo diffraction?

56. In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the diffraction band?

57. Why is diffraction more frequently experienced with sound waves than with light waves?

58. Draw an intensity distribution graph for diffraction due to a single slit.

59. If the slit is made narrow, will the width of the central maxi-mum increase or decrease, in diffraction?

60. What should be the approximate size of slit to observe diffraction?

Polarization

61. What is Brewster’s angle?

62. Name two commonly used devices which use polarized light.

63. Does the intensity of light vary with polarization? If so, how?

64. Draw a graph to show the variation of intensity of polarized light transmitted by any analyzer.

65. The polarizing angle for a medium is 60°, find the refractive index of the medium.

66. Is sunlight polarized or unpolarized?

67. Is the headlight of car (a) plane polarized and (b) highly coherent?

68. Why cannot longitudinal waves be polarized?

69. What are polaroids? Give its two uses.

70. Is the head light of a car plane polarized?

71. What is the value of refractive index of a medium of polari-zing angle 60°?

72. Does the value of polarizing angle depend on color of light?

73. What is the angle between the reflected and refracted rays at polarizing angle?

74. Write some important uses of polaroids.

75. Which among X-rays, sound waves and radio waves can be polarized?

SHORT ANSWER TYPE

Huygens Principle

1. What is a wavefront? State the postulates of Huygens theory.

2. Draw the diagrams to show the behavior of plane wavefronts as they pass through (a) a thin prism and (b) a thin convex lens.

3. State the postulates of Huygens wave theory. Sketch the wavefront that corresponds to a beam of light (a) coming

from a very far away source and (b) diverging radially from a point source.

4. State Huygens postulates of wave theory. Sketch the wavefront emerging from a (a) point source of light and (b) linear source of light like a slit.

Interference

5. Briefly describe the theory of interference. Hence, deduce the conditions for constructive and destructive interference.

6. What are coherent sources of light? Why is it not possible to observe interference with non-coherent sources?

7. In Young’s double-slit experiment, the slits are 0.03 cm apart and the screen is placed 1.5 m away. The distance between the central bright fringe and fourth bright fringe is 1 cm. Calculate the wavelength of light used.

8. A screen is placed at a distance of 1.5 m from a narrow slit. The slit is illuminated by light of wavelength 5000 Å. If the first minimum on either side of the central maximum is at a distance of 5 mm from it, find the width of the slit.

9. Distinguish between interference and diffraction of light.

10. What is superposition of waves? Define the principle of superposition of waves. Does this principle still hold if the waves are not simple harmonic in nature?



11. In a Young’s double-slit experiment, the distance between the slit and the screen is 1 m. If the distance between the slits is 5 mm, and the separation of successive maxima is found to be 0.1092 mm, calculate the wave length of light used.

12. Find the ratio of intensities of two points P and Q on the screen in Young’s double-slit experiment when waves from sources S1 and S2 have phase difference of p/3 and p/2, respectively.

13. Draw the (a) intensity distribution and (b) fringe pattern graphs for Young’s double-slit experiment.

14. State and explain the conditions for interference of light waves.

15. In Young’s double-slit experiment, the slits are separated by 0.28 nm and the screen is placed 1.4 m away. The distance between the fourth bright fringe and the central bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

16. The two slits in Young’s double-slit experiment have widths in the ratio 1 : 16. Deduce the ratio of intensities at the maxima and minima in the interference pattern.

17. The two slits in the Young’s double-slit experiment have widths in the ratio 9 : 4. Find the ratio of the light intensity at maxima and minima in interference pattern.

18. If the path difference produced due to interference of light coming out of the two slits for yellow color of light at a point on the screen be 3l/2, what will be the color of the fringe at the point? Give reason.

19. The phase difference between light waves from two slits of Young’s experiment is p radian. Will the central fringe be bright or dark? Explain your answer.

20. If the two slits in Young’s experiment have widths ratio 1 : 4, deduce the ratio of intensity at maxima and minima in the interference pattern.

21. In Young’s experiment, the width of the fringes obtained with light of wavelength 6000 Å is 2.0 mm. What will be the fringe width if the entire apparatus is immerged in a liquid of refractive index 1.33?

22. Prove that the law of conservation of energy is obeyed during interference of light.

Diffraction

23. Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?

24. Differentiate between interference and diffraction of light.

Polarization

25. (a) What is polarization? (b) Is sunlight polarized or unpolari-zed? (c) Distinguish between unpolarized and plane polarized light.

26. Define the terms plane of vibration and plane of polarization. On a diagram represent an unpolarized and polarized light.

27. Explain the polarization of light by scattering.

28. What is the polaroid? Give a few important applications of polaroids.

29. Define Brewster’s law. What will be the value of refractive, index of a medium of polarizing angle of 60°?

30. Let I0 be the intensity of polarized light. The angle between the axes of the polarizer and the analyzer is 60°. What will be the intensity of the outgoing light?

LONG ANSWER TYPE

Huygens Principle

1. (a) The refractive index of a denser medium is 1.732. Calculate (i) the polarizing angle of medium and (ii) the angle of refraction. (b) Discuss the refraction of a plane wavefront across a plane boundary and prove that Snell’s law of refrac-tion is followed in such a case.

2. Explain how a plane wavefront gets reflected from a plane boundary following the laws of reflection.

3. The critical angle between a transparent medium and air is denoted by ic. A ray of light in air medium enters this trans-parent medium at an angle of incidence equal to polarizing angle ip. Deduce a relation for the angle of refraction rp in terms of ic.

4. Prove Snell’s laws of refraction on the basis of Huygens principle.

Interference

5. In Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is l is K units, l being the wavelength of light used, (a) Find the intensity at a point where the path difference is l/4. (b) Two sources of intensity I1 and I2 in Young’s experiment undergo interference.

Show that = + −2 2max min 1 2 1 2/ ( ) /( ) ,I I a a a a where a1 and a2

are the amplitudes of two sources S1 and S2.

6. A monochromatic light of wavelength 600 Å illuminates two narrow slits 0.3 mm apart producing an interference pattern on screen 75 cm away. Calculate the separation between



(a) the second bright fringe and the central bright fringe and (b) the second dark fringe and the central bright fringe.

7. Define interference of light waves. Derive an expression for resultant displacement of a wave. Hence, obtain expression for intensity and amplitude of resultant wave. Also define constructive and destructive interference.

8. Describe Young’s double-slit experiment and obtain expression for fringe width.

9. Explain Young’s double-slit experiment of interference of light waves. Calculate the path difference between interfering waves and give conditions for maxima and minima.

Diffraction

10. Obtain the conditions for maxima and minima in a single-slit diffraction. Draw the intensity graph for the diffraction pattern obtained.

11. Define diffraction of light. When does this process become more pronounced? Represent diffraction on a diagram. How can we say that diffraction is a kind of interference?

12. Explain the generation of polarized light by reflection. State and prove Brewster’s law.

13. (a) How does diffraction limit affect the resolving power of an optical instrument? (b) In a Young’s double-slit experiment,

the distance between the slits and the screen is 160 m. When light of l = 5.89 × 10–7 m falls on the slits, the distance between the center of the interference pattern and the fourth bright fringe on either side is found to be 1.60 mm. What is the slit separation?

14. Explain diffraction due to a single slit. Obtain an expression for width of central maxima and prove that it is twice of any secondary maxima.

Polarization

15. What do you mean by polarization of light? How are polarized and unpolarized light represented? How will you prove that light waves are transverse in nature?

16. State and explain Brewster’s law.

17. State and explain law of Malus.

18. What is the phenomenon of polarization? Derive the rela-tion connecting the polarizing angle of a medium and its refractive index.

CHALLENGE – HOTS PROBLEMS

1. The two slits in Young’s experiment are 3 mm apart illuminated by a light of wavelength 480 nm. The screen is at 2 m from the plane of the slits. Find the separation between eighth bright and third dark fringe with respect to central bright fringe.

2. What is the effect of the following operation on the inter-ference fringes in a Young’s double-slit experiment? (a) The screen is moved away from the plane of the slits. (b) The (monochromatic) source is replaced by another (mono-chromatic) source of lesser wavelength. (c) The separation between the slits is decreased. (d) The width of the source slit is made wider. (e) The distance between the source slit and the plane of two slits is increased. (f ) The widths of the two slits S1 and S2 are of the order of the wavelength of the light source. (g) The monochromatic source is replaced by a source of white light.

3. A screen is placed at a distance of 1.5 m from a narrow slit. The slit is illuminated by light of wavelength 5000 Å. If the first minimum on either side of the central maximum is at a distance of 5 mm from it, find the width of the slit.

4. In a Young’s double-slit experiment, the distance of the screen from two slits is 1.0 m. When light of wavelength 6000 Å is allowed to fall on the slits, the width of the fringes obtained on the screen in 2.6 mm. Determine (a) the distance between the two slits and (b) the width of the fringe if the wavelength of the incident light is 4800 Å.

5. The two slits in Young’s double-slit experiment are 0.03 mm apart, an interference pattern is produced on a screen 1.5 m

away. The fourth bright fringe is at a distance of 1 cm from the central maximum. Find the wavelength of light used.

6. Geometrical optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the geometrical optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

7. A glass plate of refractive index 1.5 coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength l traveling in air is incident normally on the layer and the two reflected rays interfere. Write the condition for their con-structive interference. If l = 648 nm, obtain the least value of t for which the rays interfere constructively.

8. A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. The distance between the slits is 0.8 mm. POQ is the line per-pendicular to the plane AB and passing through O, the middle point of S1 and S2. A monochromatic light source is kept at S1 40 cm below P and 2 m from the vessel, to illumi-nate the slits as shown in Fig. 10.33. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. Now a liquid is poured into the vessel and filled upto OQ. The central bright fringe is found to be at Q. Calculate the refractive index of the liquid.



A

S1

P

S

40 c

m

2 m 10 cmB C

S2

O

0.8 mmQ

D

Figure 10.33

A N S W E R K E Y

Very Short Answer Type

25. 8 : 5

30. 25 : 1

31. 9 : 1

32. Diffraction occurs

66. 1.732

Short Answer Type

7. 5 × 10−7 m

8. 7.5 × 10−3 m

11. 5.46 × 10−7 m

12. 6 × 10−13 m

16. 25 : 9

17. 25 : 1

18. Dark (no color)

19. Dark

20. 9 : 1

21. 1 : 5

30. 1/5

Long Answer Type

3. (a) (i) 60°, (ii) 80°6. (a) 3 × 10−3 m; (b) 2.25 × 10−3 m

13. (a) Resolving power = D/1.22l, where D is the aperture of the objective. More the aperture, more is the resolving power and hence less diffraction of waves through the opti-cal instrument; (b) d = 0.236 mm

Challenge – HOTS Problems

1. 3.52 mm

3. 0.015 cm

4. (a) 3 × 10−4 m; (b) 1.6 × 10−3 m

5. 5 × 10−7 m

7. 90 nm

8. Central maxima forms at a distance of 2 cm from point Q. When liquid is poured, the central bright fringe shifts to point Q; Refractive index is 1.0016


Worked Problems 653

TARGET COMPETITION

C O M P E T I T I V E E D G E

1. All points on a wavefront serve as point sources of spherical secondary wavelets. After a time t, the new position of the wavefront is that of a surface tangent to these secondary wavelets.

2. When a wave gets refracted into a denser medium (v1 > v2), the wavelength and the speed of propagation decrease but the frequency remains the same.

3. When a convex lens focuses light to form a real image, although the ray going through the center traverses a shorter path, but because of the slower speed in glass, the time taken is the same as for rays traveling near the edge of the lens.

4. Doppler’s effect for light waves depends on only one velocity, the relative velocity between source and detector, as measured from the reference frame of either.

5. According to the principle of linear superposition, when two identical waves arrive at point in phase with one another, or crest-to-crest the waves mutually adds and constructive interference occurs. When two identical waves arrive at point out of phase with one another, or crest-to-trough the waves mutually cancel and destructive interference results.

6. Interference cannot be observed unless the two light sources are coherent.

7. The phase difference between two waves can change if the waves travel paths of different lengths.

8. In Young’s interference pattern, dark fringes are situated in-between bright fringes and vice versa. All the bright and dark fringes are of equal width.

9. Diffraction occurs due to mutual interference of secondary wavelets starting from portions of the wavefront which are not blocked by the obstacle, or from portions of the wavefront which are allowed to pass through the aperture.

10. In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.

11. The minimum angular separation of two point sources is such that the central maximum of the diffraction pattern of one source is centered on the first minimum of the diffraction pattern of the other, a condition called Rayleigh’s criterion for resolvability.

12. Light scattered in a direction perpendicular to the incident light is always plane polarized.

13. The reflected light is completely plane polarized in a direction perpendicular to the plane of incidence.

14. Diffraction phenomena define the limits of ray optics. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light.

15. Most interference and diffraction effects exist even for longitudinal waves like sound in air. But polarization phenomena are special to transverse waves like light waves.

W O R K E D P R O B L E M S

1. The initial shape of the wavefront of the beam is

(a) planar (b) convex(c) concave (d) convex near the axis and concave near the periphery (AIEEE 2010)

Solution

For parallel cylindrical beam, the wavefront is planar.

Hence, the correct option is (a).

2. A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the

Wiley-Chapter 10a.indd 653Wiley-Chapter 10a.indd 653 12/26/2011 3:24:27 PM12/26/2011 3:24:27 PM


third bright fringe of known light coincides with the fourth bright fringe of the unknown light. From this data, the wavelength of the unknown light is

(a) 393.4 nm (b) 885.0 nm (c) 442.5 nm (d) 776.8 nm (AIEEE 2009)

Solution

For constructive interference, path difference is given by nl. Since path difference are equal for two wavelengths, we have 3l1 = 4l2. Therefore,

2 13 3 590 nm 442.5 nm.4 4

= = =l l

Hence, the correct option is (c).

3. In a Young’s double-slit experiment, the intensity at a point where the path diff erence is l/6 (l being the wavelength of the light used) is I. If I0 denotes the maximum intensity, I/I0 is equal to

(a) 1/ 2 (b) 3 /2 (c) 1/2 (d) 3/4 (AIEEE 2007)

Solution

Phase difference = = =2 (path difference)3

p plΦ

And intensity at point

020

0

3 3cos .2 4 4

I II II

= = ⇒ =F

Hence, the correct option is (d).

4. A Young’s double-slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is

(a) hyperbola (b) circle (c) straight line (d) parabola (AIEEE 2005)

Solution

Hyperbola.

Hence, the correct option is (a).

5. In a Young’s double-slit experiment, the separation between the two slits is d and the wavelength of the light is l. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).

(a) If d = l, the screen will contain only one maximum(b) If l < d < 2l, at least one more maximum (besides the central maximum) will be observed on the screen(c) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark

and bright fringes will increase(d) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark

and bright fringes will increase (IIT-JEE 2008)

Solution

For a minimum of one maxima, sin q = l/d and the second maxima occurs when path difference is l. If d = l ⇒ sin q = 90° = 1 or second maxima occurs at infinity. If l < d < 2l, then q is finite and second maxima occurs.

Hence, the correct options are (a) and (b).

6. In a Young’s double-slit experiment an electron beam is used to obtain interference pattern. If the speed of electrons is decreased then

(a) no interference pattern is observed(b) distance between two consecutive fringes decreases(c) distance between two consecutive fringes increases(d) distance between two consecutive fringes remains the same (IIT-JEE 2005)

Solution

Fringe width,

Dd

= lb

and the wavelength and speed are related by

= ,hm

l n



where h is Planck’s constant. So wavelength is inversely proportional to the velocity; hence, the fringe width increases with decrease in electron speed.

Hence, the correct option is (c).

7. In a Young’s double-slit experiment (Fig. 10.34), the angular position (q) of a bright fringe having intensity one-fourth of the maximum intensity is given by

(a) sin−1 (l/4d) (b) sin−1 (l/3d) (c) sin−1 (l/2d) (d) sin−1 (l/d) (IIT-JEE 2005)

Imax

4q

Figure 10.34

Solution

The intensity distribution of interference fringes is given by

( )20 cos .dI I y

D= p

l

Given that

= ⇒ = =0 .4 3 3I d DI y or y

D dp p ll

But y/D = sin q. Therefore,

( )1sin = sin .3 3

ord d

−= l lq q

Hence, the correct option is (b).

P R A C T I C E P R O B L E M S

MULTIPLE CHOICE TYPE

One Option Correct Type

1. Two waves are represented by the equations y1 = a sin wt and y2 = a cos wt. The first wave(a) leads the second by p(b) lags the second by p(c) leads the second by p /2(d) lags the second by p /2

2. When a light ray is incident on glass from air at an angle 57°, the reflected ray is completely polarized. Then Brewster’s angle q for glass is

(a) =q 57° (b) >q 57°(c) <q 57° (d) =q 90°

3. In Young’s double-slit experiment, the two slits act as coher-ent sources of equal amplitude A and wavelength l. In another experiment with the same setup, the two slits are of equal amplitude A and wavelength l, but are incoher-ent. The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is(a) 1 : 2 (b) 2 : 1(c) 4 : 1 (d) 1 : 1

4. For a wave propagating in a medium, identify the property that is independent of the others:(a) Velocity(b) Wavelength(c) Frequency(d) All these depend on each other

5. A Young’s double-slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is(a) hyperbola (b) circle(c) straight line (d) parabola

6. In Young’s double-slit experiment, the y-coordinates of central maxima and 10th maxima are 2 cm and 5 cm respec-tively. When the Young’s double-slit experiment apparatus is immersed in a liquid of refractive index 1.5 the corresponding y-coordinates will be(a) 2 cm, 7.5 cm (b) 3 cm, 6 cm(c) 2 cm, 4 cm (d) 4/3 cm, 10/3 cm

7. In the setup shown in Fig. 10.35, the two slits 1S and 2S are not equidistant from the slit S. The central fringe at O is then



(a) always bright(b) always dark(c) either dark or bright depending on the position of S(d) neither dark nor bright

S1

S O

S2

Figure 10.35

8. The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young’s double-slit experiment is:(a) infinite (b) five (c) three (d) zero

9. The intensity ratio of the two interfering beams of light is b. What is the value of max min max min/I I I I− +

(a) b2 (b) +bb

21

(c) + b2

1 (d) + b

b12

10. A parallel beam of light of intensity 0I is incident on a glass plate, 25% of light is reflected by upper surface and 50% of light is reflected from lower surface (Fig. 10.36). The ratio of maximum to minimum intensity in interference region of reflected rays is:

(a) ⎡ ⎤+⎢ ⎥−⎣ ⎦

2(1/2) ( 3 / 8 )(1/2) ( 3 / 8 )

(b) ⎡ ⎤+⎢ ⎥−⎣ ⎦

2(1/4) ( 3 / 8 )(1/2) ( 3 / 8 )

(c) 58

(d) 85

Air

Figure 10.36

11. Two light rays having the same wavelength l in vacuum are in phase initially. Then the first ray travels a path L1 through a medium of refractive index n1 while the second ray travels a path of length L2 through a medium of refractive index n2. The two waves are then combined to produce interference. The phase difference between the two waves is

(a) −pl 2 1

2 ( )L L (b) −pl 1 1 2 2

2 ( )n L n L

(c) −pl 2 1 1 2

2 ( )n L n L (d) 1 2

1 2

2 L Ln n

⎛ ⎞−⎜ ⎟⎝ ⎠pl

12. 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 very 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 must originate from

the same source

13. If the wavelength of light used in an optical instrument are1 4000 Å=l and =l2 5000 Å, the ratio of their respective

resolving powers (corresponding to l1 and l 2) is(a) 16 : 25 (b) 9 : 1(c) 4 : 5 (d) 5 : 4

14. In Young’s double-slit experiment, a third slit is made in between the double slits. Then(a) fringes of unequal width are formed(b) contrast between bright and dark fringes is reduced(c) intensity of fringes totally disappears(d) only bright light is observed on the screen

15. In Fig. 10.37, 1S and 2S are coherent sources. The intensity of both sources is same. If the intensity at the point P is 4 W/m2, the intensity of each source is(a) 1 W/m2 (b) 2 W/m2

(c) 3 W/m2 (d) 4 W/m2

P

a a

aS1 S2

Figure 10.37

16. Ray diverging from a point source form a wavefront that is(a) cylindrical (b) spherical (c) plane (d) cubical

17. A beam with wavelength l falls on parallel reflecting planes with separation d as shown in Fig. 10.38. The angle q that the beam should make with the planes so that the beams reflected from successive planes may interfere construc-tively (where n = 1, 2, . . .) is

(a) ( )1sin nd

− l (b) ( )1tan nd

− l

(c) ( )1sin2n

d− l (d) ( )1cos

2n

d− l

d

qq

Figure 10.38

18. When an unpolarized light intensity I0 is incident on a polari-zing sheet, the intensity of the light which does not get transmitted is



(a) I0 /2 (b) I0 /4(c) zero (d) I0

19. When a thin metal plate is placed in the path of one of the interfering beams of light, then the(a) fringe width increases(b) fringes disappear(c) fringes become brighter(d) fringes becomes blurred

20. Interference was observed in an interference chamber when air was present. Now, the chamber is evacuated and if the same light is used, a careful observation will show(a) no interference(b) interference with bright band(c) interference with dark bands(d) interference in which breadth of the fringe will be

slightly increased

21. Ratio of intensities of two waves are given by 4 : 1. Then, the ratio of the amplitudes of the two waves is(a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4

22. In a Young’s experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed 1 m away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be(a) 60 × 10−4 cm (b) 10 × 10−4 cm(c) 10 × 10−5 cm (d) 6 × 10−5 cm

23. Colors appear on a thin soap film and soap bubbles due to the phenomenon of(a) interference (b) scattering (c) diffraction (d) dispersion

24. What happens to the interference pattern if the two slits in a Young’s experiment are illuminated by two independent sources such as two sodium lamps S and S’ as shown in Fig. 10.39?(a) Two sets of interference fringes overlap(b) No fringes are observed(c) The intensity of the bright fringes is doubled(d) The intensity of the bright fringes becomes four times

S

S1

S2

S′

Sheild

Screen

Figure 10.39

25. The idea of the quantum nature of light has emerged in an attempt to explain(a) interference(b) diffraction

(c) radiation spectrum of a blackbody(d) polarization

26. In Young’s experiment (Fig. 10.40), monochromatic light is used to illuminate the two slits A and B. Interference fringes are observed on a screen placed in front of the slits. Now, if a thin glass plate is placed normally in the path of the beam coming from the slit, then (a) the fringes will disappear(b) the fringe width will increase(c) the fringe width will decrease(d) there will be no change in the fringe width but the

pattern shifts

A

B

C

Figure 10.40

27. Two waves having intensity in the ratio 25 : 4 produce inter-ference. The ratio of the maximum to minimum intensity is(a) 5 : 2 (b) 7 : 3 (c) 49 : 9 (d) 9 : 49

28. Ratio of intensities of two waves is 9 : 1. If these waves are superimposed, what is the ratio of maximum and minimum intensities?(a) 9 : 1 (b) 3 : 1(c) 4 : 1 (d) 5 : 3

29. In Young’s double-slit experiment, the intensity is I at a point where the path difference is l/6, where l is wavelength of light used. If I0 denotes the maximum intensity, then I/I0 is equal to (a) 3/4 (b) 1/ 2(c) 3 /2 (d) 1/2

30. In Young’s experiment, the monochromatic light is used to illuminate two slits A and B as shown in Fig. 10.41. Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed normally in the path of beam coming from the slit A, then(a) the fringes will disappear(b) the fringe width will increase(c) the fringe width will decrease(d) there will be no change in fringe width

Screen

A

B

G

Figure 10.41



31. Which is the similarity between the sound waves and light waves?(a) Both are electromagnetic waves(b) Both are longitudinal waves(c) Both have the same speed in a medium(d) They can produce interference

32. Two coherent sources of intensity ratio 1 : 4 produce an interference pattern. The fringe visibility will be(a) 1 (b) 0.8 (c) 0.4 (d) 0.6

33. In a Young’s double-slit experiment, let b be the fringe width, and let I0 be the intensity at the central bright fringe. At a distance x from the central bright fringe, the intensity will be

(a) ⎛ ⎞⎜ ⎟⎝ ⎠b0 cos xI (b) ⎛ ⎞

⎜ ⎟⎝ ⎠b2

0 cos xI

(c) ⎛ ⎞⎜ ⎟⎝ ⎠

20 cos xI p

b (d) ( ) ⎛ ⎞⎜ ⎟⎝ ⎠

0 2cos4I xp

b

34. In Young’s double-slit experiment, we get 15 fringes per cm on the screen, while using light of wavelength 5896 Å (sodium lamp). How many fringes per cm we get with light of wavelength 7370 Å?(a) 18 (b) 15 (c) 12 (d) 10

35. What happens to the interference pattern if the two silts S1 and S2 in Young’s double-slit experiment are illuminated by two independent but identical sources?(a) The intensity of the bright fringes is doubled(b) The intensity of the bright fringes becomes four times(c) Two sets of interference fringes overlap(d) No interference pattern is observed

36. The idea of secondary wavelets for the propagation of a wave was first given by(a) Newton (b) Huygens (c) Maxwell (d) Fresnel

37. A wavefront AB passing through a system C emerges as DE as shown in Fig. 10.42. The system C could be(a) a slit (b) a biprism (c) a prism (d) a glass slab

A D

E

C

?

B

Figure 10.42

38. Figure 10.43 shows a wavefront P passing through two systems A and B, and emerging as Q and then as R. The systems A and B could, respectively, be(a) a prism and a convergent lens (b) a convergent lens and a prism(c) a divergent lens and a prism(d) a convergent lens and a divergent lens

R

BQ

AP

Figure 10.43

39. Which one of the following does not show polarization?(a) Transverse waves in gas (b) Longitudinal wave in gas(c) Both (a) and (b)(d) None of the above

40. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are(a) 5I and I (b) 5I and 3I(c) 9I and I (d) 9I and 3I

41. When a polaroid is rotated, the intensity of light varies but never reduces to zero. It shows that the incident light is(a) unpolarized(b) completely plane polarized(c) partially plane polarized(d) none of the above

42. The velocity of light emitted by a source S observed by an observer O, who is at rest with respect to S is c. If the observer moves toward S with velocity v, then the velocity of light as observed will be(a) c + v (b) c − v

(c) c (d) 2

21 vc

−

43. In a Young’s double-slit experiment, if the slits are of unequal width, the(a) fringes will not be formed(b) the positions of minimum intensity will not be

completely dark(c) bright fringe will not be formed at the center of the

screen(d) distance between two consecutive bright fringes will

not be equal to the distance between two consecutive dark fringes

44. An unpolarized beam of intensity I0 is incident on a pair of Nicols making an angle of 60° with each other. The intensity of light emerging from the pair is(a) I0 (b) I0/2 (c) I0/4 (d) I0/8

45. Two sources of light are said to be coherent when both give out light waves of the same(a) amplitude and phase(b) intensity and wavelength(c) speed(d) wavelength and a constant phase difference



46. Though quantum theory of light can explain a number of phenomena observed with light, it is necessary to retain the wave nature of light to explain the phenomenon of(a) photoelectric effect (b) diffraction(c) compton effect (d) blackbody radiation

47. Two coherent sources of light can be obtained by(a) two different lamps(b) two different lamps but of the same power(c) two different lamps of same power and having the same

color(d) none of the above

48. The angle of polarization for any medium is 60°, what will be critical angle for this

(a) −1sin 3 (b) −1tan 3

(c) −1cos 3 (d) −1 1sin3

49. The interference phenomenon can take place(a) in all waves(b) in transverse waves only(c) in longitudinal waves only(d) in standing waves only

50. Which of the following is conserved when light waves interfere?(a) Phase (b) Intensity (c) Amplitude (d) None of these

51. Diffraction pattern is obtained using a beam of blue light. What happens if red light is used in place of blue light?(a) Bands become broader and farther apart(b) Banks become narrower and crowded(c) No change(d) Bands disappear

52. White light is used to illuminate the two slits in Young’s experi-ment. The separation between the slits is b and the screen is at a distance >>( )d b from the slits. At a point directly in front of one of the slits, certain wavelength is missing. The missing wavelength(a) =l 2 /b d (b) =l 2 /5b d

(c) =l 2 /3b d (d) All of these

53. Huygens conception of secondary waves(a) allow us to find the focal length of a thick lens(b) is a geometrical method to find a wavefront(c) is used to determine the velocity of light(d) is used to explain polarization

54. Two beams of light will not give rise to an interference pat-tern, if(a) they are coherent(b) they have the same wavelength(c) they are linearly polarized perpendicular to each other(d) they are not monochromatic

55. Light passes successively through two polarimeter tubes each of length 0.29 m. The first tube contains dextro-rotatory solution of concentration 60 kg/m3 and specific rotation 0.01 rad m2/kg. The second tube contains levo-rotatory

solution of concentration 30 kg/m3 and specific rotation 0.02 rad m2/kg. The net rotation produced is

(a) 15° (b) 0°(c) 20° (d) 10°

56. A thin slice is cut out of a glass cylinder along a plane paral-lel to its axis. The slice is placed on a flat glass plate as shown in Fig. 10.44.

The observed interference fringes from this combination shall be(a) straight(b) circular(c) equally spaced(d) having fringe spacing, which increases as we go

outward

Figure 10.44

57. Two Nicols are oriented with their principal planes making an angle of 60°. The percentage of incident unpolarized light which passes through the system is(a) 50% (b) 100% (c) 12.5% (d) 37.5%

58. In Young’s double-slit experiment, the separation between the slits is halved and the distance between the slits and screen is doubled. The new fringe width is(a) unchanged (b) halved(c) doubled (d) quadruped

59. Consider Fraunhoffer diffraction pattern obtained with a single slit at normal incidence. At the angular position of first diffraction minimum, the phase difference between the wavelets from the opposite edges of the slit is(a) p/4 (b) p/2(c) p (d) 2p

60. Two crossed polaroids are placed in the path of a light beam. In between these a third polaroid is placed whose polariza-tion axis makes an angle f with the polarization axis of the first polaroid. Then, intensity of light emerging from the third polaroid, in term of f , is

(a) f21 sin 24

(b) f0 2sin4I

(c) f0 2cos4I f0 2cos

4I (d) f0 2cos 2

2I f0 2cos 2

2I

61. Two powerful 100 W bulbs are used to study the interference of light. How will interference pattern by affected?(a) Fringes will become narrow(b) Fringes will become broader(c) It would vanish(d) Pattern will be colored green–yellow

62. C.V. Raman was awarded the Nobel prize for his work associated with which of the following phenomenon of radiations?



(a) Scattering (b) Diffraction(c) Interference (d) Polarization

63. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffrac-tion pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is(a) 1.2 cm (b) 1.2 mm (c) 2.4 cm (d) 2.4 mm

64. In a biprism experiment, when the setup is shifted from air to the inside of a still, clear lake water, the fringe pattern(a) disappears (b) gets shrunk(c) gets enlarged (d) remains unchanged

65. A polaroid produces a strong beam of light which is(a) circularly polarized(b) elliptically polarized(c) plane polarized(d) unpolarized

66. The ratio of intensities of successive maxima in the diffrac-tion pattern due to single slit is(a) 1 : 4 : 9 (b) 1 : 2 : 3

(c) 2 24 41: :

9 25p p (d) 2 2

4 91: :p p

67. Two coherent beams of light of same wavelength superpose in a certain region of space. If the intensity of one beam is 4 times that of the other, then the ratio of intensity at a bright point to that at a dark point is(a) 16 : 1 (b) 2 : 1(c) 5 : 3 (d) 9 : 1

68. In the visible region of the spectrum the rotation of the plane of polarization is given by = +q l2( / )a B . The optical rotation produced by a particular material is found to be 30° per mm at l = 5000 Å and 50° per mm at l = 4000 Å. The value of constant a will be

(a) °+ 509

per mm (b) °− 509

per mm

(c) °+ 950

per mm (d) °− 950

per mm

69. A plane wavefront (l = 6 × 10−7 m) falls on a slit 0.4 mm wide, A convex lens of focal length 0.8 m placed behind the slit focuses the light on a screen. What is the linear diameter of second maximum(a) 6 mm (b) 12 mm (c) 3 mm (d) 9 mm

70. Polaroid glass is used in sun glasses because(a) it is cheaper(b) it is fashionable(c) it has good color(d) it reduces the light intensity to almost half on account

of polarization

71. If torch is used in place of monochromatic light in Young’s experiment, what will happen?(a) Fringe will occur as from monochromatic source(b) Fringe will appear for a moment and then it will

disappear(c) No fringes will appear(d) Only bright fringe will appear

72. When a thin film of thickness t is placed in the path of light wave emerging out of S1, then increases in the length of optical path will be (a) −m( 1)t (b) +m( 1)t(c) mt (d) m/t

73. The waves y1 = A1 sin (wt – b1) and y2 = A2 sin (wt – b2) super-impose to form a resultant wave whose amplitude is

(a) 2 21 2 1 2 1 22 cos( )A A A A+ + −b b

(b) 2 21 2 1 2 1 22 sin( )A A A A+ + −b b

(c) A1 + A2

(d) |A1 + A2|

More Than One Option Correct Type

1. If white light is used in a Young’s double-slit experiment, the(a) bright white fringe is formed at the center of the

screen(b) fringes of different colors are observed clearly only in the

first order(c) the first-order violet fringes are closer to the center

of the screen than the first-order red fringes(d) the first-order red fringes are closer to the center of the

screen than the first-order violet fringes

2. In a Young’s double-slit experiment, let A and B be the two slits. A thin film of thickness t and refractive index m is placed in front of A. Let b be the fringe width. The central maximum will shift(a) toward A (b) toward B

(c) by t(m – l) bl (d) by t

bm l

3. In the Young’s double-slit experiment, the ratio of intensities of bright and dark fringes is 9. This means that

(a) the intensities of individual sources are 5 units and 4 units, respectively

(b) the intensities of individual sources are 4 units and 1 unit, respectively

(c) the ratio of their amplitudes is 3(d) the ratio of their amplitudes is 2

4. In a Young’s double-slit experiment, the separation between the two slits is d and the wavelength of the light is l. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s)(a) If d = l, the screen will contain only one maximum(b) If l < d < 2l, at least one more maximum (besides the

central maximum) will be observed on the screen(c) If the intensity of light failing on slit 1 is reduced so that

it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase

(d) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase



5. White light is used to illuminate the two slits in a Young’s double-slit experiment. The separation between the slits is d and the screen is at a distance D (�d) from the slits. At a point on the screen directly in front of one of the slits, cer-tain wavelengths are missing. Some of the missing wave-lengths are(a) l = d2/D (b) l = 2d2/D(c) l = d2/3D (d) l = 2d2/3D

6. In a Young’s double-slit experiment, the separation between the two slits is d and the wavelength of the light is A. The intensity of light falling on the slit 1 is four times the inten-sity of light falling on slit 2. Choose the correct choice(s)(a) If d = l, the screen will contain only one maximum(b) If l < d < 2 l, at least one more maximum (besides the

central maximum) will be observed on the screen(c) If the intensity of light falling on the slit 1 is reduced,

so that it becomes equal to that of the slit 2, the inten-sities of the observed dark and bright fringes will increase

(d) If the intensity of light falling on the slit 2 is increased, so that it becomes equal to that of the slit 1, the inten-sities of the observed dark and bright fringes will increase

7. 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 slits are 4 units and

1 unit respectively(c) the amplitude ratio is 3(d) the amplitude ratio is 2

8. In Young’s double-slit experiment, white light is used. The separation between the slits is b. The screen is at a dis-tance d (d�b) from the slits. Some wavelengths are missing exactly in front of one slit. These wavelengths are

(a) =l2b

d (b) =l

22bd

(c) 2

3b

d=l (d)

223bd

=l

9. If white light is used in a Young’s double-slit experiment, the(a) Bright white fringe is formed at the center of the

screen(b) Colored fringes will be observed(c) The first-order violet fringes are closer to the center

of the screen than the first-order red fringes(d) The first-order red fringes are closer to the center of the

screen than the first-order violet fringes

10. If the first minima in a Young’s double-slit experiment occurs directly in front of one of the slits (distance between slit and screen l = 12 cm and distance between slits d = 5 cm), then the wavelength of the radiation used can be(a) 2 cm (b) 4 cm(c) 2/3 cm (d) 4/3 cm

11. An interference pattern is formed on the screen, when light from two different monochromatic sources are allowed to interfere. Then, it is true that the

(a) frequencies of light from the two sources are equal to each other

(b) the sources are coherent(c) the sources should be located in the same medium(d) the path difference should either be an even or, an odd

multiple of l/2, where l is the wavelength of light

12. The intensity of a plane progressive wave in loss free medium is(a) directly proportional to the square of amplitude of the

wave(b) directly proportional to the velocity of the wave(c) directly proportional to the square of frequency of the

wave(d) inversely proportional to the density of the medium

13. If one of the slits of a standard Young’s double-slit experi-ment is covered by a thin parallel sided glass slab so that it transmits only one half the light intensity of the other, then(a) the fringe pattern will get shifted toward the covered

slit(b) the fringe pattern will get shifted away from the cov-

ered slit(c) the bright fringes will become less bright and the dark

ones will become more bright(d) the fringe width will remain unchanged

14. If instead of monochromatic light, white light is used in Young’s double-slit experiment, then (a) a bright white fringe is formed at center of screen(b) fringes of different colors are clearly observed only in

the first-order place(c) the first-order red fringes are relatively close to the

screen than the first-order violet fringes(d) the first-order violet fringes are relatively close to the

center of the screen than the first-order red fringes

15. Huygens principle of secondary wavelets may be used to(a) find the velocity of light in vacuum(b) explain the particle behavior of light(c) find the new position of a wavefront(d) explain Snell’s law

16. A parallel beam of light ( 5000 Å)=l is incident at an angle a = 30° with the normal to the slit plane in a Young’s double-slit experiment. Assume that the intensity due to each slit at any point on the screen is 0 .I Point O is equidis-tant from 1S and 2S (Fig. 10.45). The distance between slits is 1 mm(a) the intensity at O is 4l0

(b) the intensity at O is zero(c) the intensity at a point on the screen 1 m below O is 4l0

(d) the intensity at a point on the screen 1 m below O is zero

S1

S2O

a

3 m

Figure 10.45



17. White light is used to illuminate the two slits in Young’s double-slit experiment. The separation between the slits is b and the screen is at a distance d (�b) from the slits. At a point on the screen directly in front of any one of slits, certain wavelengths are missing. Some of these missing wavelengths are (a) 2/b d=l (b) 22 /b d=l(c) 2 /3b d=l (d) 22 /3b d=l

18. Which of the following can give sustained interference?(a) Two independent laser sources(b) Two independent light bulbs(c) Two independent sound sources(d) Two independent microwave sources

19. A Young’s double-slit experiment is performed with white light. Then(a) the central fringe will be white(b) there will not be a completely dark fringe(c) the fringe next to the central will be red(d) the fringe next to the central will be violet

20. Fraunhofer diffraction can be observed with(a) one wide slit (b) one narrow slit(c) two narrow slits (d) large number of narrow

slits

21. A person views the interference pattern, produced by two slits illuminated by white light, on placing a green filter in front of his eyes. Then(a) the will see sharply distinguishable dark and bright

fringes(b) the fringes will not be sharp but are differentiable(c) on replacing the green filter by a blue filter, the fringes will

be again sharp but closer than those by the green filter

(d) on using both the filters simultaneously, the central bands will be maximum bright

22. In a Young’s double-slit experiment apparatus, we use white light. Then(a) the fringe next to the central will be red(b) the central fringe will be white(c) the fringe next to the central will be violet(d) there will not be a completely dark fringe

23. Two sources are called coherent if they produce waves(a) of equal wavelength(b) of equal velocity(c) having same shape of wavefront(d) having a constant phase difference

24. To observe a stationary interference pattern formed by two light waves, it is not necessary that they must have(a) the same frequency(b) same amplitude(c) a constant phase difference(d) the same intensity

25. If one of the slits of a standard Young’s double-slit experi-ment apparatus is covered by a thin parallel sided glass slab so that it transmit only one half of the light intensity of the other, then(a) the fringe pattern will get shifted toward the covered slit(b) the fringe pattern will get shifted away from the cov-

ered slit(c) the bright fringes will be less bright and the dark ones

will be more bright(d) the fringe width will remain unchanged

ASSERTION–REASONING TYPEIn the following set of questions, two statements “Assertion” and “Reason” are given. Choose the correct answer from the following options:

(a) Assertion and Reason are true and the Reason is the correct explanation of the Assertion.

(b) Assertion and Reason are true but the Reason is not a cor-rect explanation of the Assertion.

(c) Assertion is true but the Reason is false.(d) Assertion and Reason both are false.

1. Assertion: Thin films such as soap bubble or a thin layer of oil on water show beautiful colors, when illuminated by white light.

Reason: It happens due to the interference of light reflected from the upper surface of the thin film.

2. Assertion: The phase difference between any two points on a wavefront is zero.

Reason: Corresponding to a beam of parallel rays of light, the wavefronts are planes parallel to one another.

3. Assertion: Radio waves can be polarized.

Reason: Sound waves in air are longitudinal in nature.

4. Assertion: Light waves can be polarized.

Reason: Light waves are transverse in nature.

5. Assertion: The law of conservation of energy is violated during interference.

Reason: For sustained interference the phase difference between the two waves must change with time.

6. Assertion: All bright interference bands have same intensity.

Reason: Because all bands receive same light from two sources.

7. Assertion: Colored spectrum is seen, when we look through a muslin cloth.

Reason: It is due to the diffraction of white light on passing through fine slits.

8. Assertion: When the apparatus of Young’s double-slit experi-ment is brought in a liquid from air, the fringe width decreases.

Reason: The wavelength of light decreases in the liquid.

9. Assertion: In Young’s double-slit experiment the two slits are at distance d apart. Interference pattern is observed on



a screen at distance D from the slits. At a point on the screen when it is directly opposite to one of the slits, a dark fringe is observed. Then, the wavelength of wave is proportional to square of distance of two slits.

Reason: For a dark fringe intensity is zero.

10. Assertion: Light waves are transverse.

Reason: Because they can be diffracted.

11. Assertion: No interference pattern is observed, when two coherent sources are infinitely close to each other.

Reason: The fringe width is inversely proportional to the distance between the two slits.

12. Assertion: Newton’s rings are formed in the reflected sys-tem. When the space between the lens and the glass plate is filled with a liquid of refractive index greater than that of glass, the central spot of the pattern is dark.

Reason: The reflection in Newton’s ring case will be from a denser to a rarer medium and the two interfering rays are reflected under similar conditions.

13. Assertion: The film which appears bright in reflected sys-tem will appear dark in the transmitted light and vice versa.

Reason: The conditions for film to appear bright or dark in reflected light are just reverse to those in the transmitted light.

14. Assertion: In Young’s double-slit experiment, the fringe width for dark fringes is different from that for white fringes.

Reason: In Young’s double-slit experiment performed with a source of white light, only black and bright fringes are observed.

15. Assertion: For best contrast between maxima and minima in the interference pattern of Young’s double-slit experi-ment, the intensity of light emerging out of the two slits should be equal.

Reason: The intensity of interference pattern is proportional to square of amplitude.

16. Assertion: In Young’s double-slit experiment, the fringes become indistinct if one of the slits is covered with cello-phane paper.

Reason: The cellophane paper decreases the wavelength of light.

17. Assertion: In Young’s double-slit experiment, the two slits are at distance d apart. Interference pattern is observed on the screen. When it is directly opposite to one of the slits, a dark fringe is observed. Then, the wavelength of the wave is proportional to the square of the distance of the slits.

Reason: For a dark fringe, the intensity is zero.

18. Assertion: The unpolarized light and polarized light can be distinguished from each other by using polaroid.

Reason: A polaroid is capable of producing plane polarized beams of light.

19. Assertion: Nicol prism is used to produce and analyze plane polarized light.

Reason: Nicol prism reduces the intensity of light to zero.

20. Assertion: Standard optical diffraction gratings cannot be used for discriminating between different X-ray wavelengths.

Reason: The grating spacing is not of the order of X-ray wavelengths.

21. Assertion: In everyday life the Doppler’s effect is observed readily for sound waves than light waves.

Reason: Velocity of light is greater than that of sound.

22. Assertion: In Young’s experiment, the fringe width for dark fringes is different from that for white fringes.

Reason: In Young’s double-slit experiment the fringes are performed with a source of white light, then only black and bright fringes are observed.

23. Assertion: Radio waves can be polarized.

Reason: Sound waves in air are longitudinal in nature.

24. Assertion: When a tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the center of shadow of the obstacle.

Reason: Destructive interference occurs at the center of the shadow.

25. Assertion: A solution of sucrose in water is dextro-rotatory. But oh hydrolysis in the presence of a little hydrochloric acid, it becomes levo-rotatory.

Reason: Sucrose on hydrolysis gives unequal amounts of glucose and fructose. As a result of this, change in sign of rotation is observed.

26. Assertion: Thin films such as soap bubble or a thin layer of oil on water show beautiful colors when illuminated by white light.

Reason: It happens due to the interference of light reflected from the upper surface of the thin film.

27. Assertion: Microwave communication is preferred over optical communication.

Reason: Microwaves provide large number of channels and band width compared to optical signals.

28. Assertion: Interference pattern is made by using blue light instead of red light, the fringes becomes narrower.

Reason: In Young’s double-slit experiment, fringe width is given by relation /D d.=b l

LINKED COMPREHENSION TYPEParagraph for questions 1 and 2: Angular width of central maxima in single-slit diffraction pattern for light of wavelength 600 nm is measured. For light of wavelength 1 the angular with decreases by 30%. The same decrease in angular width is measured when entire apparatus is immersed in a liquid of refractive index m.

1. Value of l is(a) 4200 nm (b) 420 nm(c) 800 nm (d) 320 nm



2. Value of m is

(a) 43

(b) 1.5

(c) 1.43 (d) 1.86

Paragraph for questions 3–5: An initially parallel cylindrical beam travels in a medium of refractive index m(I) = m1 + m2I, where, m0 and m2 are positive constants and I is the intensity of the light beam. The intensity of the light beam is decreasing with increas-ing radius.

3. The initial shape of the wavefront is(a) convex(b) concave(c) convex near the axis and concave near the periphery(d) planar

4. The speed of the light in the medium is(a) maximum on the axis of the beam(b) minimum on the axis of the beam(c) the same everywhere in the beam(d) directly proportional to the intensity I

5. As the beam enters the medium, it will(a) diverge(b) converge(c) diverge near the axis and converge near the periphery(d) travel as a cylindrical beam

Paragraph for questions 6–10: When waves from two coherent sources, having amplitudes a and b superimpose, the amplitude R of the resultant wave is given by 2 2 2 cos ,R a b ab= + + fwhere f is the constant phase angle between the two waves. The resultant intensity I is directly proportional to the square of the amplitude of the resultant wave, that is, I ∝ R2, that is, I ∝ (a2 + b2 + 2ab cos f). For constructive interference, f = 2np and Imax = (a + b)2. For destructive interference, f = (2n − 1)p and Imin = (a − b)2. If I1, I2 are intensities of light from two slits of widths w1 and w2, then 2 2

1 2 1 2/ / / .I I a b= =w w Light waves from two coherent sources of intensity ratio 81: 1 produce interference.

6. The ratio of amplitudes of two sources is(a) 9 : 1 (b) 81 : 1(c) 1 : 9 (d) 1 : 81

7. The ratio of slit widths of the two sources is(a) 9 : 1 (b) 81 : 1(c) 1 : 9 (d) 1 : 81

8. The ratio of maxima and minima in the interference pattern is(a) 9 : 1 (b) 81 : 1(c) 25 : 16 (d) 16 : 25

9. If two slits in Young’s experiment have width ratio 1 : 4, the ratio of maximum and minimum intensity in the interference pattern would be(a) 1 : 4 (b) 1 : 16(c) 9 : 1 (d) 9 : 16

10. The ratio of amplitudes of light waves from two sources is(a) 1 : 4 (b) 4 : 1(c) 2 : 1 (d) 1 : 2

Paragraph for questions 11–13: Consider the situation shown in Fig. 10.46. The two slits S1 and S2 placed symmetrically around the central line are illuminated by monochromatic light of wave-length l. The separation between the slits is d. The light transmit-ted by the slits falls on a screen S0 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance Z from S3. Another screen Sc is placed a further distance D away from Sc. Find the ratio of the maximum to minimum intensity observed on Sc.

S1

d

D D

S2

S4

S3

Sc

Z

P0

x1

Figure 10.46

11. If /2z D d= l(a) 1 (b) 1/2(c) 3/2 (d) 2

12. If /z D d= l(a) 4 (b) 2(c) ∞ (d) 1

13. If /4z D d= l

(a) 2[3 2 2 ]− (b) 2[3 2 ]+

(c) 2[3 2 ]− (d) 2[3 2 2 ]+

Paragraph for questions 14 and 15: A lens of focal length f = 40 cm is cut along the diameter into two identical halves. In this pro-cess, a layer of the lens t = 1 mm in thickness is lost, then the halves are put together to form a composite lens. In between the focal plane and the composite lens, a narrow slit is placed, near the focal plane. The slit is emitting monochromatic light with wavelength l = 0.6 mm. Behind the lens a screen is located at a distance L = 0.5 m from it (Fig. 10.47).

S1

S2

S

Lu < f

V

D

d

Figure 10.47

14. Find the fringe width for the pattern obtained under given arrangement on the screen.(a) 0.19 m (b) 0.24 m(c) 0.38 m (d) 0.43 m

15. The expression for the number of visible maxima that which are obtained through above said arrangement will turn out to be

(a) 2

2Lt

fl (b)

2

22Lt

fl



(c) 22Lt

fl (d)

2

22Lt

fl

Paragraph for questions 16 and 17: In the arrangement shown in Fig. 10.48, light of wavelength 6000 Å is incident on slits S1 and S2. Slits S3 and S4 have been opened such that S3 is the position of first maximum above the central maximum and S4 is the clos-est position where intensity is same as that of the light used, below the central maximum. The point O is equidistant from S1 and S2 and O’ is equidistant from S3 and S4. The intensity of inci-dent light is I0. Find

S1

S2

S3

O O¢

1 m 1 m

S4

3 mm

Figure 10.48

16. The intensity at O’ (on the screen)(a) 4.5I0 (b) 3I0

(c) 2I0 (d) 5I0

17. The intensity of the brightest fringe.(a) 9I0 (b) 5I0

(c) 3I0 (d) 7I0

Paragraph for questions 18–20: In Fig. 10.49 a screen is placed nor-mal to the line joining the two point coherent sources S1 and S2. The interference pattern consists of concentric circles.

P

O

d D

S1

y

S2

Figure 10.49

18. Find the radius of the nth bright ring.

(a) 1 1 nDd

⎛ ⎞−⎝ ⎠l (b) 2 1 nD

d⎛ ⎞−⎝ ⎠

l

(c) 2 2 1 nDd

⎛ ⎞−⎝ ⎠l (d) 2 1

2nD

d⎛ ⎞−⎝ ⎠

l

19. If d = 0.5 mm, l = 5000 Å and D = 100 cm, then find the radius of the closest second bright ring.(a) 888 (b) 830(c) 914 (d) 998

20. Find the value of n for this ring.(a) 6.32 cm (b) 5.52 cm(c) 4.7 cm (d) 3.25 cm

Paragraph for questions 21–23: Two coherent sources emit light of wavelength, l. Distance between the two coherent sources is d = 4l (Fig. 10.50).

d = 4l

S2

S1

D >> D

d sin q =

Ñ

xD

q

qx

y

P

Figure 10.50

21. A detector moves along the y-axis. What is the maximum number of maxima the director moves along the y-axis?(a) 6 (b) 4(c) 8 (d) 2

22. If the detector moves along the x-axis the maximum num-ber of maxima observed(a) 2 (b) 3(c) 0 (d) 5

23. In a Young’s double-slit experiment, the upper slit is cov-ered by a thin glass plate of refractive index 1.4 while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wave-length 5400 Å. It is found that the point P on the screen, where the central maxima (n = 0) lay before the glass plates were inserted, now has 3/4 the original intensity. It is fur-ther observed that what used to be the fifth maxima ear-lier lies below point P while the sixth minima lies above P (Fig. 10.51). Calculate the thickness of the glass plate (absorption of light by the glass plate may be neglected).(a) 9.3 mm (b) 19.3 mm(c) 6.3 mm (d) 12.3 mm

m1

m2

S1

S2

Sixth minima

Fifth minima

Figure 10.51

Paragraph for questions 24–26: A Young’s double-slit experi-ment is performed in a medium of refractive index 4/3. A light of 600 nm wavelength is falling on the slits having 0.45 nm sepa-ration. The lower slit S2 is covered by a thin glass plate of thick-ness 10.4 mm and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 m from the slits as shown Fig. 10.52. (All the wavelength in this problem are for the given medium of refractive index 4/3, ignore absorption.)



S

y

O

S1

S2

Figure 10.52

24. Find the location of the central maximum (bright fringe with zero path difference) on the y-axis.(a) 2.33 mm (b) 4.33 mm(c) 6.33 mm (d) 4.43 mm

25. Find the light intensity at point O relative to maximum fringe intensity.

(a) max14

I (b) max54

I

(c) max12

I (d) max34

I

26. Now, if 600 nm light is replaced by white light of range 400–700 nm, find the wavelength of the light that forms maxima exactly at point O.

(a) 1300 1300nm, nm,2 2

… (b) 1300 1300nm, nm,4 4

…

(c) 1500 1500nm, nm,2 2

… (d) 1300 1300nm, nm,6 6

…

Paragraph for questions 27 and 28: In a modified Young’s double-slit experiment, source S is kept in front of slit S1 as shown in Fig. 10.53. Find the phase difference at a point O that is equidis-tant from slits S1 and S2, and point P that is in front of slit S1 in the following situations:

x0

S P

O

S1

S2

Figure 10.53

27. A liquid of refractive index m is filled between the screen and slits.

(a) 2

2 20 0

22

dd x x

D⎡ ⎤⎡ ⎤+ + +⎢ ⎥⎣ ⎦⎣ ⎦

mpl

(b) 2

2 20 0

22

dd x x

D⎡ ⎤⎡ ⎤+ − +⎢ ⎥⎣ ⎦⎣ ⎦

mpl

(c) 2

2 20 0

22

dd x x

D⎡ ⎤⎡ ⎤− + +⎢ ⎥⎣ ⎦⎣ ⎦

mpl

(d) 2

2 20 0

22

dd x x

D⎡ ⎤⎡ ⎤− − +⎢ ⎥⎣ ⎦⎣ ⎦

mpl

28. Liquid is filled between the slit and the source S.

(a) 2

2 20 0

22dd x x

D⎡ ⎤+ − −⎢ ⎥⎣ ⎦

p ml

(b) 2

2 20 0

22dd x x

D⎡ ⎤+ − +⎢ ⎥⎣ ⎦

p ml

(c) 2

2 20 0

22dd x x

D⎡ ⎤− + +⎢ ⎥⎣ ⎦

p ml

(d) 2

2 20 0

22dd x x

D⎡ ⎤− − −⎢ ⎥⎣ ⎦

p ml

MATRIX-MATCH TYPEMatch the statements in Column I labeled as (A), (B), (C), and (D) with those in Column II labeled as (P), (Q), (R), and (S). Any given statement in Column I can have correct matching with one or more statements in Column II.

1. Column I Column II

(A) Interference (P) Corpuscular theory(B) Compton effect (Q) Longitudinal waves(C) Diffraction (R) Transverse waves(D) Polarization (S) Dual theory

2. In Fig. 10.54, A, B, and C are the three slits each of which individually produces the same intensity 0 ,l at point P0 when they are illuminated by parallel beam of light of wavelength l. (Take 2

0 0 0BP AP /2, , 4 W/m ;d D l− = << =l Amplitude of each wave = 2 units; 6000 Å.=l )

C

B

A P0

2d

d

Figure 10.54

Column I Column II

(A) Resultant intensity at P0 in (W/m2)

(P) 4

(B) Resultant amplitude at P0 (in unit)

(Q) zero

(C) If slit C is closed then resultant intensity at P0

(R) 2

(D) If slit B is closed resulting amplitude at P0 (in SI unit)

(S) 2 2

3. Column I Column II

(A) Interference (P) Size of aperture is much larger than wavelength of light used

(B) Diffraction (Q) Size of aperture is comparable to wavelength of light used

(C) Rectilinear propa-gation of light

(R) Disapproval of existence of either

(D) Lasers (S) Coherent sources

4. Match the observation given in Column I with the experiments given in Column II where these observation occur.



Column I Column II

(A) Central fringes are bright and all fringes are equally spaced

(P) Interference in wedge-shaped thin

(B) Central fringes are dark and all fringes are equally spaced

(Q) Young’s double-slit experiment

(C) Central fringes are bright and fringes are not equally spaced

(R) Fraunhofer single-slit diffraction experiment

(D) All fringes are equally spaced

(S) Lloyd’s mirror experiment

5. In Young’s double-slit experiment, the distance between slits is d, distance between slit and screen is D, and the wavelength of light used is l.

Column I Column II

(A) Condition for bright fringe (P) 2D

dl

(B) Condition for dark fringe (Q) ( 1)D td

−m

(C) Displacement of fringe when glass plate of thickness t is placed

(R) Path difference = nl

(D) Distance between central maxima and first dark fringe when glass plate of thickness t is used

(S) Path difference = (2n − 1)l/2

6. In a Young’s double-slit experiment d = 1 mm, D = 2000 mm,510 mm−=l positions are measured from central maxima

on screen.

Column I Column II

(A) Optical path difference at 1/ 2 mmx =

(P) 1 mm

(B) Optical path difference at x = 2000 mm

(Q) Just a distance below central maxima (0.02 mm)

(C) Path difference (l) is almost

(R) 3(1/2 2 ) 10 mm−×

(D) Maximum path difference

(S) Far away from cen-tral maxima

7. The arrangement of Lloyd’s mirror experiment is shown in Fig. 10.55. S is a point source of frequency 146 10 Hz.× A and B are the two ends of a mirror placed horizontally, and LOM represents the screen.

1 mm

5 cm 5 cm

L

O

M

2 m

S

A mirror B

Figure 10.55

Column I Column II

(A) The minimum height, above point O, fringes will form is (in mm)

(P) 39

(B) The maximum height, above point O, fringes will be formed is (in mm)

(Q) 19

(C) Total number of fringes are (R) 500

(D) Fringe width (in μm) is (S) 40

INTEGER TYPE The answer to each question is a non-negative integer.

1. In the arrangement shown in Fig. 10.56, the wavelength of light used is l. The distance between slits S1 and S2 is d (�D) The distance between S3 and S4 is /3 .u D d= l If the ratio of maximum to minimum intensity observed on screen is k. Find k.

D

S1

d

S3

S2

S4

u P

Screen

D

Figure 10.56

2. Light used in a young’s double-slit experiment consists of two wavelengths 450 nm and 720 nm. In the interference pattern, eighth maximum of the first coincides with the nth maximum of the second. What is n?

3. In Young’s double-slit experiment, d = 2, D = 2 m, and l = 500 nm. If intensity of two slits are I0 and 9I0, then the intensity at (1/6) mmy = is KI0. Find K.

4. Two identical coherent sources of wavelength l are placed at (100l, 0) and (−50l, 0), respectively. A detector moves slowly from origin to (2l, 0) along x-axis. What is the number of maxima detected [including origin and (2l, 0)]?

5. For the situation depicted in Fig. 10.57, /3BP AP− = l and D�d. The slits are of equal widths, having intensity I0. If the intensity of P is KI0, find K.

Parallel beam oflight of wavelength l

Screen

PA

B

C

D

d/2

d/2

Figure 10.57



More Than One Option Correct Type

6. In a Young’s double-slit experiment, the point source of light is placed as shown in Fig. 10.58 that follows. The source starts moving at t = 0 from the shown position in upward direction along AB with constant velocity 3 m/s the wave-length is l and D�d and d�l. Find the velocity of central maxima (in m/s).

A

d

2DD

S1

S2

B

Source

Screen

Figure 10.58

7. A plane monochromatic light wave falls normally on a single slit. At a distance D = 4 m from it, there is a screen showing a certain diffraction pattern. The slit width is decreased n = 2 times. Calculate the new distance D′ at which the screen should be positioned to obtain the diffraction pattern simi-lar to previous one but diminished n times.

8. One slit of a Young’s experiment is covered by a glass plate 1( 1.4)=m and the other by another glass plate 2( 1.7)=m

of the same thickness. The point of central maxima on the screen, before the plates were introduced is now occupied

by the third bright fringe. Find the thickness of the plate, the wavelength of light used is 4000 Å.

9. In a modified Young’s double-slit experiment, a monochro-matic uniform and parallel beam of light of wavelength 6000 Å and intensity 2(10/ ) W/mp is incident normally on two apertures A and B of radii 0.001 m and 0.002 m, respec-tively. A perfectly transparent film of thickness 2000 Å and refractive index 1.5 for the wavelength of 6000 Å is placed in front of aperture A (see Fig. 10.59). Calculate the power (in W) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.

A

B

F

Figure 10.59

A N S W E R K E Y

One Option Correct Type

1. (d)

2. (a)

3. (b)

4. (c)

5. (a)

6. (c)

7. (c)

8 (b)

9. (b)

10. (a)

11. (b)

12. (c)

13. (d)

14. (b)

15. (a)

16. (b)

17. (c)

18. (a)

19. (b)

20. (d)

21. (a)

22. (d)

23. (a)

24. (b)

25. (c)

26. (d)

27. (c)

28. (b)

29. (a)

30. (d)

31. (d)

32. (b)

33. (c)

34. (c)

35. (d)

36. (b)

37. (c)

38. (b)

39. (b)

40. (c)

41. (c)

42. (c)

43. (b)

44. (c)

45. (d)

46. (b)

47. (d)

48. (d)

49. (a)

50. (d)

51. (a)

52. (d)

53. (b)

54. (c)

55. (b)

56. (a)

57. (c)

58. (d)

59. (d)

60. (a)

61. (c)

62. (a)

63. (b)

64. (b)

65. (c)

66. (c)

67. (b)

68. (b)

69. (a)

70. (d)

71. (b)

72. (a)

73. (a)

1. (a), (b), (c)

2. (a), (c)

3. (b), (d)

4. (a), (b)

5. (a), (c)

6. (a), (b)

7. (b), (d)

8. (a), (c)

9. (a), (b), (c)

10. (a), (c)

11. (a), (b)

12. (a), (b), (c)

13. (a), (c), (d)

14. (a), (b), (d)

15. (c), (d)

16. (a), (c)

17. (a), (c)

18. (a), (c), (d)

19. (a), (b), (d)

20. (b), (c), (d)

21. (a), (c)

22. (b), (c), (d)

23. (a), (b), (c), (d)

24. (b), (d)

25. (a), (c), (d)


Answer Key 669

Assertion–Reasoning Type

1. (a)

2. (b)

3. (b)

4. (a)

5. (d)

6. (c)

7. (a)

8. (a)

9. (b)

10. (b)

11. (a)

12. (d)

13. (a)

14. (d)

15. (b)

16. (c)

17. (b)

18. (a)

19. (c)

20. (a)

21. (b)

22. (d)

23. (b)

24. (c)

25. (b)

26. (c)

27. (a)

28. (a)

Linked Comprehension Type

1. (b)

2. (c)

3. (d)

4. (b)

5. (b)

6. (a)

7. (b)

8. (c)

9. (c)

10. (d)

11. (a)

12. (c)

13. (d)

14. (b)

15. (d)

16. (b)

17. (a)

18. (b)

19. (d)

20. (a)

21. (b)

22. (c)

23. (a)

24. (b)

25. (d)

26. (a)

27. (d)

28. (b)

Matrix-Match Type

1. (A) → (Q), (R), (S); (B) → (P), (S); (C) → (Q), (R), (S); (D) → (R), (S)

2. (A) → (R); (B) → (R); (C) → (Q); (D) → (Q)

3. (A) → (R), (S); (B) → (Q); (C) → (P); (D) → (S)

4. (A) → (Q); (B) → (S); (C) → (R); (D) → (P)

5. (A) → (R); (B) → (R); (C) → (Q); (D) → (P)

6. (A) → (R); (B) → (R); (C) → (Q); (D) → (P), (S)

7. (A) → (Q); (B) → (P); (C) → (S); (D) → (R)

Integer Type

1. 9

2. 5

3. 7

4. 1

5. 3

6. 6

7. 1 mm

8. 4 μm

9. 7 μW



Wiley Chap 10

Documents

electric field

challenge

negative phase

a1 a2

entire experimental

electric field

destructive

film appears