35: Chapter Interference
Problem 1 The speed of yellow light (from a sodium lamp) in a certain liquid is measured to be 1.92 108 m/s. What is the index of refraction of this liquid for the light? Answer 1.56 Solution The index of refraction is equal to the speed of light in vacuum divided by the speed of light in the liquid:
m c s = 1.56 n= = vl 1.92 10 8 m s 3 10 8
Problem 3 How much faster, in meters per second, does light travel in sapphire than in diamond? See Table 33-1. Answer 4.55 107 m/s Solution The difference in the speed of light in sapphire and diamond is
vs - vd = c c 1 1 - = 3 10 8 m/s - = 4.55 10 7 m/s 1.77 2.42 ns nd
(
)
Problem 4 The wavelength of yellow sodium light in air is 589 nm. (a) What is its frequency? (b) What is its wavelength in glass whose index of refraction is 1.52? (c) From the results of (a) and (b) find its speed in this glass. Answer (a) 5.09 1014 Hz. (b) 388 nm. (c) 1.97 108 m/s.
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Chapter 35: Interference
Solution (a) Since the speed of light in air is about the same as c, its speed in a vacuum, the frequency is m 3 10 8 c s = 5.09 1014 Hz f = = 5.89 10 -7 m (b) The frequency of the light does not change. If v is the speed of the light in the glass then the wavelength in glass is
g =
but
n= c v
v f
so
m 3 10 8 c s g = = = 388nm nf 1.52 5.09 1014 Hz
(c)
The speed of the light in the glass is
v = g f = 388nm 5.09 1014 Hz = 1.97 10 8 m s
Problem 9 Suppose that the two waves in Fig. 35-4 have wavelength 500 nm in air. In wavelengths, what is their phase difference after traversing media 1 and 2 if (a) n1 = 1.50, n2 = 1.60, and L = 8.50 m; (b) n1 = 1.62, n2 = 1.72, and L = 8.50 m; and (c) n1 = 1.59, n2 = 1.79, and L = 3.25 m? (d) Suppose that in each of these three situations the waves arrive at a common point after emerging. Rank the situations according to the brightness the wave produces at the common point. Answer (a) 1.70; (b) 1.70; (c) 1.30; (d) all tie. Solution (a) The number of wavelengths difference in the two paths is
n1
n2 L Figure 35-4
=
L L n2 L n1 L L 8.50 m - = - = ( n2 - n1 ) = (1.6 - 1.5 ) = 1.7 2 1 a a a 500nm
Chapter 35: Interference (b) The difference is
35-3
=
(c)
L 8.50 m (1.72 - 1.62 ) = 1.7 ( n2 - n1 ) = a 500nm
The difference is
=
L 3.25 m (1.79 - 1.59 ) = 1.3 ( n2 - n1 ) = a 500nm
(d) The brightness will be the same for all three since the intensity is proportional to the square of the cosine of half the phase difference where the phase difference is = 2.
cos2 1 1 ( 2 1.7 ) = cos2 ( 2 1.3) 2 2
Problem 13 Two waves of light in air, of wavelength 600.0 nm, are initially in phase. They then travel through plastic layers as shown in Fig. 35-37, with L1 = 400 m, L2 = 3.50 *m, n1 = 1.40, and n2 = 1.60. (a) In wavelengths, what is their phase difference after they both have emerged from the layers? (b) If the waves later arrive at some common point, what type of interference do they undergo? Answer (a) 0.833; (b) intermediate interference Solution (a) The phase difference in wavelengths after exiting the layers is
=
1 1 (1.60 3.5 m - 1.40 4.0 m ) = 0 ( n2 L2 - n1L1 ) = a 600nm
but after the wave that passes through layer 2 travels the additional 0.50 mm needed to meet the wave that passes through layer 1 there is a phase difference of
=
0.5 m = 0.833 600nm
(b) If the phase difference were 0.5 wavelengths then there would be destructive interference and if the phase difference were 1.0 wavelengths then there would be constructive interference therefore we have intermediate interference which is close to constructive than to destructive.
35-4
Chapter 35: Interference
Problem 16 In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and the adjacent maximum? What is the distance between these maxima on a screen 50.0 cm from the slits? Answer (a) 0.01 rad; (b) Solution (a) The location of the mth maximum measured from the central maximum is given by
sin ( ) = m d
For small angles sin() in radians so the angular separation is
=
m 1 = = 0.01rad d 100
(b) The distance, ym, from the central maximum to the mth maximum on the screen a distance D from the slits is given by ym = D*tan() D* = 50 cm * 0.01 rad = 0.50 cm Problem 19 Suppose that 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? Answer 2.25 mm. Solution The fringe separation is given by
y =
D 500nm 5.04m = = 2.25mm d 1.20mm
Chapter 35: Interference
35-5
Problem 27 A thin flake of mica (n = 1.58) is used to cover one slit of a double-slit arrangement. The central point on the screen is now occupied by what had been the seventh bright side fringe (m = 7) before the mica was used. If = 550 nm, what is the thickness of the mica? Answer 6.64 m. Solution The phase difference for the seventh bright fringe is 14. This phase difference is due to difference between the number of wavelengths of the light which occur in the mica and in the air layer which the mica replaces. The number of wavelengths in thickness t is t/. Therefore the phase difference is
t t 14 = 2 - mica air
or, since mica =
air nmica
n t t 2 t 14 = 2 mica - = ( nmica - 1) air air air
or
t=
7 air 7 550nm = = 6.64 m nmica - 1 1.58 - 1
Problem 30 Find the sum y of of the following quantities: y1 = 10 sin(t) and y2 = 8.0 sin(t + 30). Answer 17.4 sin(t + 13.3) Solution Referring to Diagram 35-30 in which we have set t equal to zero, we can sum the horizontal components of the phasors yh = 10 cos(0) + 8 cos(30) = 16.9 and the vertical components y1 Diagram 35-30 y 30 y2
35-6 yv = 10 sin(0) + 8 sin(30) = 4.00 Therefore the magnitude of the sum is
Chapter 35: Interference
( y ) + ( y )
2 h v
2
= 16.9 2 + 4.00 2 = 17.4
and the angle between the phasor sum and the y1 phasor is
4.00 = arctan = 13.3 16.9
Therefore the sum is y = 17.4 sin(t + 13.3)
Problem 33 Add the quantities y1 = 10 sin(t) y2 = 15 sin(t + 30) y3 = 5 sin(t - 45) using the phasor method: Answer 26.8 sin(t + 8.5). Solution Referring to Diagram 35-33 in which we have set t equal to zero, we can sum the horizontal components of the phasors y2 y 30 y1 Diagram 35-33 yh = 10 cos(0) + 15 cos(30) + 5 cos(-45) = 26.5 and the vertical components y3 45
Chapter 35: Interference
35-7
yv = 10 sin(0) + 15 sin(30) + 5 sin(-45) = 3.96 Therefore the magnitude of the sum is
( y ) + ( y )
2 h v
2
= 26.5 2 + 3.96 2 = 26.8
and the angle between the phasor sum and the y1 phasor is
3.96 = arctan = 8.5 26.5
Therefore the sum is y = 26.8 sin(t + 8.5)
Problem 37 Bright light of wavelength 624 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference? Answer (a) 117 nm; 352 nm. Solution The wavelength of the light when in the soap film is
s =
air 624nm = = 469nm n 1.33
There is also a phase shift of upon reflection from the first surface of the film. Therefore we will have fully constructive interference if the difference in path length in the film is equal to odd integer multiples of s/2. Since the extra path length traveled by the ray which reflects from the second surface of the film is twice the film thickness, 2t, we will have fully constructive interference when
2t = m
s ; m = 1, 3, 5, 7... 2
The least thickness will be when m = 1 and
t= m s 1 469nm = = 117nm 4 4
35-8 The second least thickness will be when m = 3 and
t= m s 3 469nm = = 352nm 4 4
Chapter 35: Interference
Problem 40 A thin film of acetone (index of refraction = 1.25) is coating a thick glass plate (index of refraction = 1.50). White light is incident normal to the film. In the reflections, fully destructive interference occurs at 600 nm and fully constructive interference at 700 nm. Calculate the thickness of the acetone film. Answer 840 nm Solution If the thickness of the film is t and n is the index of refraction of acetone then for fully interference constructive with wavelength c in air
2t = m c n
and for fully destructive interference with wavelength d in air
1 2t = m + c 2 n
Eliminating m between these two equations yields
t=
c d 700nm 600nm == = 840nm . 4n ( c - d ) 4 1.25 ( 700nm - 600nm )
Problem 55 The reflection of perpendicularly incident white light by a soap film in air has an interference maximum at 600 nm and a minimum at 450 nm, with no minimum in between. If n = 1.33 for the film, what is the film thickness, assumed uniform? Answer 338 nm
Chapter 35: Interference
35-9
Solution If the thickness of the film is L then there is an interference maximum for wavelength max when
2L = m + 1 max 2 n
If we decrease the wavelength until we encounter an interference minimum at wavelength min then this will be the (m +1)th minimum
2L = ( m + 1)
min n
Eliminating m between these two equations yields
L= 1 max min 1 600nm 450nm = 338nm 4n max - min 4 1.33 600nm - 450nm
Problem 71 In Fig. 35-46, a broad source of light (of wavelength 680 nm) illuminates at normal incidence two glass plates 120 mm long that touch at one end and are separated by a wire 48.0 m in diameter at the other end. How many bright fringes appear over the 120-mm distance? Incident Light
Figure 35-46 Answer 141. Solution There will be complete destructive interference at the left end where the two plates touch because although there will be no phase difference due to path difference there will be a phase shift of at the bottom plate. As we move toward the right we will see complete constructive interference when the path difference becomes half a wavelength (air film thickness of one quarter of a wavelength). There will then be another bright fringe each time the path difference increases by a wavelength (air film thickness increases by half a
35-10
Chapter 35: Interference
wavelength). At the right end of the plates the air film is 0.048 mm/680 nm = 70.6 wavelengths thick. Subtract 0.25 wavelengths for the first bright fringe and this leaves 70.3 wavelengths which permits another 140 bright fringes (one for each half wavelength change in thickness) for a total of 141 bright fringes. Problem 83 Ocean waves moving at a speed of 4.0 m/s are approaching a beach at an angle 1 = 30 to the normal, as shown from above in Fig. 35-50. Suppose the water depth changes abruptly at a certain distance from the beach and the wave speed there drops to 3.0 m/s. (a) Close to the beach, what is the angle 2 between the direction of wave motion and the normal? (Assume the same law of refraction as for light.) (b) Explain why most waves come in normal to a shore even though at large distances they approach at a variety of angles Answer (a) 22. Solution (a) The waves are refracted according to
sin ( 2 ) sin (1 ) = v1 v2
Shoreline
2
Shallow water
Deep water
1 Figure 35-50
or
m 3 v2 sin ( 2 ) = sin (1 ) = s sin ( 30 ) = 0.375 m v1 4 s therefore
2 = 22. Problem 99 If the distance between the first and tenth minima of a double-slit pattern is 18 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, what is the wavelength of the light used? Answer 600 nm
Chapter 35: Interference Solution Since sin() is approximately equal to tan() which is equal to ym/D
dym = ( 2m + 1) D 2
35-11
or
ym = ( 2m + 1)
D 2d
Therefore the distance to the first minimum is
y1 = ( 2 + 1)
D 3 D = 2d 2d
and the distance to the 10th minimum is
y10 = ( 20 + 1)
D 21 D = 2d 2d
The distance between the first and tenth minima is
y10 - y1 = 21 D 3 D 9 D - = 2d 2d d
Therefore the wavelength is
= ( y10 - y1 )
d 18mm = 0.15mm = 600nm 9D 9 50cm
Problem xx What is the phase difference between the waves from the two slits arriving at the mth dark fringe in a Young's double-slit experiment? Answer (2m + 1). Solution The phase difference is
= 2 2 ( path difference ) = d sin ( )
But d sin ( ) = ( 2m + 1)
for the mth minimum. Therefore 2
35-12 = (2m + 1)
Chapter 35: Interference
Problem xx A and B in Fig. 36-29 are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power.
Figure 36-29 (a) If a detector is moved to the right along the x axis from point A, at what distances from A are the first three interference maxima detected? (b) Is the intensity of the nearest minimum exactly zero? (Hint: Does the intensity of a wave from a point source remain constant with an increase in distance from the source?) Answer (a) 15/28 m, 39/20 m, 55/12 m. (b) No. Solution (a) First, we note that d = 4. Referring to Diagram 27,
Diagram 27 the phase difference between the two waves received at point P on the x axis is = (2/)(path difference) = (2/)(BP - AP) or, letting AP = x,
Chapter 35: Interference = (2/)[(x2 + d2)1/2 - x] The point P will be the location of a minimum if the phase difference is (2m + 1). Therefore (2m + 1) = (2/)[(x2 + d2)1/2 - x] or (2m + 1) = 2[(x2 + d2)1/2 - x] or, solving for x x = d2/[(2m + 1)] - (2m + 1)/4 Calculating the value of x for various values of m yields m 0 1 2 3 4 x 63/4 m 49/12 m 39/20 m 15/28 m -17/36 m
35-13
For any value of m > 3 x is negative which is unacceptable since x is a distance. Therefore the three smallest values of x are 15/28 m, 39/20 m, 55/12 m. (b) The intensity of the wave at the minima is not zero since the two sources are different distances from the minima and the intensities from the two sources are inversely proportional to the square of the distance from the source. Problem xx A lens with index of refraction greater than 1.30 is coated with a thin transparent film of index of refraction 1.30 to eliminate by interference the reflection of red light at wavelength 680 nm that is incident perpendicularly on the lens. What minimum film thickness is needed? Answer 131 nm. Solution There will be a phase shift of from both the first and second surfaces of the thin film since both surfaces are transitions from small index of refraction to larger index of refraction. These two phase shifts "cancel" each other and there will be cancellation of the reflected light if the thickness causes a phase difference of or the total distance traveled in the film is f/2 where f is the wavelength of the light in the film. The total distance traveled in the film is twice the thickness, t. Therefore
35-14 2t = f/2 = a/2nf or t = a/4nf = (680 nm)/(4 * 1.30) = 131 nm.
Chapter 35: Interference
Problem xx Bright light of wavelength 585 nm is incident perpendicularly on a soap film (n = 1.33) of thickness 1.21 m, suspended in air. Is the light reflected by the two surfaces of the film closer to interfering fully destructively or fully constructively? Answer Fully constructive interference. Solution The wavelength of the light when in the soap film is s = /n = 585 nm/1.33 = 439 nm The light which is reflected back from the second surface of the soap film travels through the film twice for a total distance of 2.42 m which is 2.42 m/439 nm = 17.4 wavelengths which is a phase difference of 34.8. There is also a phase shift of upon reflection from the first surface of the film for a total phase difference of 33.8. The phase difference is closer to an even integer multiple of pi, 34, than an odd integer multiple of pi, 33, therefore the reflected light is closer to fully constructive interference. Problem xx A lens with index of refraction greater than 1.30 is coated with a thin transparent film of index of refraction 1.30 to eliminate by interference the reflection of red light at wavelength 680 nm that is incident perpendicularly on the lens. What minimum film thickness is needed? Answer 131 nm. Solution There will be a phase shift of from both the first and second surfaces of the thin film since both surfaces are transitions from small index of refraction to larger index of refraction. These two phase shifts "cancel" each other and there will be cancellation of the reflected light if the thickness causes a phase difference of or the total distance traveled in the film is f/2 where f is the wavelength of the light in the film. The total distance traveled in the film is twice the thickness, t. Therefore 2t = f/2 = a/2nf or t = a/4nf = (680 nm)/(4 * 1.30) = 131 nm.