23
ELECTROMAGNETIC WAVES
Answers to Multiple-Choice Problems
1. D 2. E 3. A, B, E
4. D 5. C
6. C
7. A 8. B
9. C
10. A, B
11. B
12. D 13. B
14. C.
15. D
Solutions to Problems
23.1. Set Up: The distance of the earth from the sun is the orbital radius of the earth, r 5 1.50 3 1011 m. The speed
of light is c 5 3.00 3 108 m / s.
x
1.50 3 1011 m
Solve: t 5 5
5 500 s 5 8.33 min
c
3.00 3 108 m / s
23.2. Set Up: The speed of electromagnetic waves in air is c 5 3.00 3 108 m / s.
Solve: A total time difference of 0.60 ms corresponds to a difference in distance of
cDt 5 1 3.00 3 108 m / s 2 1 0.60 3 1026 s 2 5 180 m.
23.3. Set Up: The speed of light in vacuum is c 5 3.00 3 108 m / s. d 5 ct. 1 yr 5 3.156 3 107 s.
3.84 3 108 m
d
5
5 1.28 s
c
3.00 3 108 m / s
(b) d 5 ct 5 1 3.00 3 108 m / s 2 1 8.61 yr 2 1 3.156 3 107 s / yr 2 5 8.15 3 1016 m 5 8.15 3 1013 km
Reflect: The speed of light is very large, but it can still take light a long time to travel enormous astronomical distances.
Solve: (a) t 5
23.4. Set Up: E 5 cB. The magnetic field of the earth is about 1024 T.
Solve: B 5
3.85 3 1023 V / m
E
5 1.28 3 10211 T. The field is much smaller than the earth’s field.
5
c
3.00 3 108 m / s
23.5. Set Up: The wave speed is c 5 3.00 3 108 m / s. c 5 fl.
3.00 3 108 m / s
c
5
5 6.0 3 104 Hz
l
5.0 3 103 m
3.00 3 108 m / s
5 6.0 3 1013 Hz
(ii) f 5
5.0 3 1026 m
3.00 3 108 m / s
c
(iii) f 5 5
5 6.0 3 1016 Hz
l
5.0 3 1029 m
3.00 3 108 m / s
c
(b) (i) l 5 5
5 4.62 3 10214 m 5 4.62 3 1025 nm
f
6.50 3 1021 Hz
3.00 3 108 m / s
(ii) l 5
5 508 m 5 5.08 3 1011 nm
590 3 103 Hz
Solve: (a) (i) f 5
23-1
23-2
Chapter 23
23.6. Set Up: c 5 3.00 3 108 m / s. The wavelength ranges for each color are: violet: 400 nm to 440 nm; red: 630 nm
to 700 nm.
c
Solve: f 5 . Largest l corresponds to smallest f.
l
(a) violet: 6.8 3 1014 Hz to 7.5 3 1014 Hz
(b) red: 4.3 3 1014 Hz to 4.8 3 1014 Hz
23.7. Set Up: For an electromagnetic wave propagating in the negative x direction, E 5 2Emax sin 1 vt 1 kx 2 .
2p
1
. T 5 . Emax 5 cBmax.
l
f
Emax
5 1.25 mT.
Solve: (a) Emax 5 375 V / m so Bmax 5
c
v
5 9.52 3 1014 Hz. k 5 1.99 3 107 rad / m so
(b) v 5 5.98 3 1015 rad / s so f 5
2p
v 5 2pf and k 5
l5
2p
1
5 3.16 3 1027 m 5 316 nm. T 5 5 1.05 3 10215 s.
k
f
This wavelength is too short to be visible.
(c) c 5 fl 5 1 9.52 3 1014 Hz 2 1 3.16 3 1027 m 2 5 3.00 3 108 m / s. This is what the wave speed should be for an
electromagnetic wave propagating in vacuum.
v 2p
v
5 is an alternative expression for the wave speed.
Reflect: c 5 fl 5
2p k
k
1 21 2
23.8. Set Up: Since the wave is traveling in empty space, its wave speed is c 5 3.00 3 108 m / s. c 5 fl.
1
Emax 5 cBmax. T 5 .
f
3.00 3 108 m / s
c
Solve: (a) f 5 5
5 6.94 3 1014 Hz
l
432 3 1029 m
(b) Emax 5 cBmax 5 1 3.00 3 108 m / s 2 1 1.25 3 1026 T 2 5 375 V / m
1
(c) T 5 5 1.44 3 10215 s.
f
E 5 E max sin 2p
1
2
1
x
t
x
t
2
2
5 1 375 V / m 2 sin 2p
215
T
l
432 3 1029 m
1.44 3 10
s
E 5 1 375 V / m 2 sin 1 3 4.36 3 1015 rad / s 4 t 2 3 1.45 3 107 m / s 4 x 2
B 5 B max sin 2p
1
2
2
t
x
2
5 1 1.25 3 1026 T 2 sin 1 3 4.36 3 1015 rad / s 4 t 2 3 1.45 3 107 m / s 4 x 2
T
l
2p
.
l
c
Solve: (a) f 5 so f ranges from 4.29 3 1014 Hz to 7.50 3 1014 Hz.
l
(b) v 5 2pf so v ranges from 2.70 3 1015 rad / s to 4.71 3 1015 rad / s.
2p
(c) k 5
so k ranges from 8.98 3 106 rad / m to 1.57 3 107 rad / m.
l
23.9. Set Up: c 5 3.00 3 108 m / s. c 5 fl. 2pf 5 v. k 5
2p
. Larger l corresponds to smaller f and k.
l
c
14
14
Solve: (a) f 5 . UVA: 7.50 3 10 Hz to 9.38 3 10 Hz. UVB: 9.38 3 1014 Hz to 1.07 3 1015 Hz.
l
2p
. UVA: 1.57 3 107 rad / m to 1.96 3 107 rad / m. UVB: 1.96 3 107 rad / m to 2.24 3 107 rad / m.
(b) k 5
l
23.10. Set Up: c 5 3.00 3 108 m / s. c 5 fl and k 5
Electromagnetic Waves
1
f
23.11. Set Up: c 5 3.00 3 108 m / s. c 5 fl. T 5 . k 5
Solve: f 5
23-3
2p
.
l
c
1
5 3.0 3 1018 Hz, T 5 5 3.3 3 10219 s, k 5 6.3 3 1010 rad / m.
l
f
23.12. Set Up: For an electromagnetic wave in air the wave speed is c 5 3.00 3 108 m / s. c 5 fl. k 5 2p / l.
v 5 2pf. E 5 cB.
3.00 3 108 m / s
c
Solve: (a) l 5 5
5 361 m
f
8.30 3 105 Hz
2p
(b) k 5
5 0.0174 rad / m
l
(c) v 5 2pf 5 1 2p 2 1 8.30 3 105 Hz 2 5 5.22 3 106 rad / s
(d) Emax 5 cBmax 5 1 3.00 3 108 m / s 2 1 4.82 3 10211 T 2 5 1.45 3 1022 V / m
23.13. Set Up: The speed of the wave is c 5 3.00 3 108 m / s. c 5 fl. Emax 5 cBmax.
3.00 3 108 m / s
c
5
5 6.90 3 1014 Hz
l
435 3 1029 m
2.70 3 1023 V / m
Emax
(b) Bmax 5
5
5 9.00 3 10212 T
c
3.00 3 108 m / s
S
S
Reflect: The directions of E and B and of the of propagation of the wave are all mutually perpendicular. Waves for
visible light have very high frequencies.
Solve: (a) f 5
S
S
23.14. Set Up: The direction of propagation of an electromagnetic wave is related to theS directions
of E and B
S
according to the right-hand rule illustrated in Figure 23.2 in Section 23.2. The directions of E and B in each case are
S
S
shown in Figures 23.14a-d. The direction of propagation is perpendicular to the plane of E and B.
y
y
S
B
S
S
E
B
x
x
z
z
(a)
(b)
S
E
z
S
y
x
E
x
S
B
S
S
B
E
z
(c)
Figure 23.14
(d)
y
23-4
Chapter 23
Solve: (a) The wave is propagating in the 1z direction.
(b) 1z direction
(c) 2y direction
(d) 2x direction
23.15. Set Up: At a distance r from the station the power output of the antenna is spread over a spherical surface of
area A 5 4pr 2. P 5 IA, where I is the intensity at distance r and P is the power output of the antenna.
0.0900 V / m
Emax
5 3.00 3 10210 T
Solve: (a) Bmax 5
5
c
3.00 3 108 m / s
(b) I 5 12 P0cEmax2 5 12 1 8.854 3 10212 C / N # m2 2 1 3.00 3 108 m / s 2 1 0.0900 N / C 2 2 5 1.076 3 1025 W / m2.
1
2
P 5 IA 5 1 1.076 3 1025 W / m2 2 1 4p 2 1 50.0 3 103 m 2 2 5 3.38 3 105 W.
1
2
Emax, 1
(c) P 5 IA 5 constant, so I14pr12 5 I24pr22. I 5 12 P0cEmax2 so Emax, 12r12 5 Emax, 22r22 and r2 5 r1
5
Emax, 2
Emax, 1
r1
5 2 1 50.0 km 2 5 100.0 km.
Emax, 1 / 2
Reflect: The intensity decreases as 1 / r 2 and the electric field decreases as 1 / r, as the distance r from the source
increases.
23.16. Set Up: P 5 IA. A 5 pr 2. Emax 5 cBmax. The intensity I 5 Sav is related to the maximum electric field by
I 5 12 P0cEmax2, where P0 5 8.85 3 10212 C2 / N # m2. The average energy density uav is related to the intensity I by
I 5 uavc.
0.500 3 1023 W
P
Solve: (a) I 5 5
5 637 W / m2.
A
p 1 0.500 3 1023 m 2 2
2 1 637 W / m2 2
Emax
2I
(b) Emax 5
5 2.31 mT.
5
5 693 V / m. Bmax 5
212
2
2
8
#
P
c
c
Å 0
Å 1 8.85 3 10 C / N m 2 1 3.00 3 10 m / s 2
637 W / m2
I
(c) uav 5 5
5 2.12 3 1026 J / m3.
c
3.00 3 108 m / s
23.17. Set Up: The surface area of a sphere of radius r is A 5 4pr 2. I 5 P / A. I 5 12 P0cEmax2, with P0 5
8.85 3 10212 C2 / N # m2. Emax 5 cBmax.
1 0.05 2 1 75 W 2
P
5 330 W / m2.
Solve: (a) I 5 5
A
4p 1 3.0 3 1022 m 2 2
2 1 330 W / m2 2
2I
(b) Emax 5
5
5 500 V / m.
Å P0c Å 1 8.85 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2
Bmax 5
Emax
5 1.7 3 1026 T 5 1.7 mT.
c
Reflect: At the surface of the bulb the power radiated by the filament is spread over the surface of the bulb. Our calculation approximates the filament as a point source that radiates uniformly in all directions.
23.18. Set Up: Intensity is average power per unit area and power is energy per unit time. If the energy is absorbed
during one cycle of the wave, then the average power absorbed is the energy divided by the period T of the wave. The
1
c
area of one end of the rod is pr 2. f 5 and T 5 , so the wavelength determines the period.
l
f
505 3 1029 m
l
5 1.68 3 10215 s.
Solve: T 5 5
c
3.00 3 108 m / s
I5
Pav
3.94 3 10219 J
5
5 7.46 3 107 W / m2.
215
A
1 1.68 3 10 s 2 p 1 0.0010 3 1023 m 2 2
23.19. Set Up: Intensity is average power per unit area and power is energy per unit time. I 5 12 P0cEmax2 and
Emax 5 cBmax.
Electromagnetic Waves
23-5
Solve: (a) For the beam, energy 5 Pt 5 1 2.0 3 1012 W 2 1 4.0 3 1029 s 2 5 8.0 3 103 J 5 8.0 kJ. This is spread
uniformly over 100 cells, so the energy given to each cell is 80 J.
(b) The cross sectional area of each cell is A 5 pr 2, with r 5 2.5 3 1026 m.
I5
(c) Emax 5
P
2.0 3 1012 W
5
5 1.0 3 1021 W / m2
1 100 2 p 1 2.5 3 1026 m 2 2
A
2 1 1.0 3 1021 W / m2 2
2I
5
5 8.7 3 1011 V / m.
Å P0c Å 1 8.85 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2
Emax
5 2.9 3 103 T
Bmax 5
c
I
c
23.20. Set Up: The radiation pressure on a totally absorbing surface is .
Solve:
8.00 W / m2
I
5 2.67 3 1028 Pa
5
c
3.00 3 108 m / s
I
c
23.21. Set Up: The radiation pressure is for a totally absorbing surface and
Solve: (a)
2I
for a totally reflecting surface.
c
6.00 W / m2
I
5 2.00 3 1028 Pa
5
c
3.00 3 108 m / s
2I
5 2 1 2.00 3 1028 Pa 2 5 4.00 3 1028 Pa
c
Reflect: For the same intensity of light, the radiation pressure is twice as great for a totally reflecting surface versus
a totally absorbing surface. For ordinary light intensities, the radiation pressure is very small.
(b)
23.22. Set Up: I 5 12 P0cEmax2.
Solve: Emax, 2 5
2 1 3I 2
2I2
5
5 "3E
Å P0c Å P0c
P
and I 5 12 P0cEmax2.
A
Emax, 1502
Emax, 150
150 W
Solve: P 5 12 AP0cEmax2.
and
5
5 "2 .
2
75 W
Emax, 75
Emax, 75
Reflect: The power output of the source is proportional to the square of the electric field amplitude in the emitted
waves.
23.23. Set Up: I 5
23.24. Set Up: For a totally reflecting surface the average pressure is
1 2
2I
I
and for a totally reflecting surface it is .
c
c
The radiation is the same so has the same intensity in the two cases.
I
1 2I
5 P/2
Solve: P 5 5
c
2 c
2I
. Pressure is force per unit area. Fnet 5 ma. The mass of
c
the sail is its volume V times its density r. The area of the sail is pr 2, with r 5 4.5 m. Its volume is pr 2t, where
t 5 7.5 3 1026 m is its thickness.
2 1 1400 W / m2 2
2I
A5
p 1 4.5 m 2 2 5 5.9 3 1024 N
Solve: (a) F 5
c
3.00 3 108 m / s
(b) m 5 rV 5 1 1.74 3 103 kg / m3 2 1 p 2 1 4.5 m 2 2 1 7.5 3 1026 m 2 5 0.83 kg.
23.25. Set Up: For a reflecting surface the pressure is
1 2
a5
5.9 3 1024 N
F
5
5 7.1 3 1024 m / s2.
m
0.83 kg
23-6
Chapter 23
(c) With this acceleration it would take the sail 1.4 3 106 s 5 16 days to reach a speed of 1 km/s. This would be useful only in specialized applications. The acceleration could be increased by decreasing the mass of the sail, either by
reducing its density or its thickness.
Reflect: The calculation assumed the only force is that due to the radiation pressure. The sun would also exert a
gravitational force on the sail; both forces are considered in Problem 23.73.
23.26. Set Up: For reflection, ur 5 ua. The desired path of the ray is sketched in Figure 23.26.
Solve: tan f 5
14.0 cm
, so f 5 50.6°. ur 5 90° 2 f 5 39.4° and ur 5 ua 5 39.4°.
11.5 cm
Mirror
11.5 cm
f
ua
ur
14.0 cm
Mirror
Figure 23.26
23.27. Set Up: For reflection, ur 5 ua. The angles of incidence and reflection at each reflection are shown in Figure 23.27. For the rays to be perpendicular when they cross, a 5 90°.
u
u
a
f
f
b
b
Figure 23.27
Solve: (a) u 1 f 5 90° and b 1 f 5 90°, so b 5 u.
a
1 b 5 90° and a 5 180° 2 2u.
2
(b) u 5 12 1 180° 2 a 2 5 12 1 180° 2 90° 2 5 45°.
Reflect: As u S 0°, a S 180°. This corresponds to the incident and reflected rays traveling in nearly the same direction. As u S 90°, a S 0°. This corresponds to the incident and reflected rays traveling in nearly opposite directions.
Electromagnetic Waves
23-7
23.28. Set Up: For reflection, ur 5 ua.
Solve: The law of reflection is used to trace each ray in parts a, b, and c of Figure 23.28. Ray 1 strikes mirror B at an
angle of incidence of zero so reflects straight back and retraces its path. When it reaches point P it then continues to
travel parallel to the surface of mirror B and has no further reflections. Ray 2 reflects from B and then travels parallel
to the surface of A and has no further reflections. Ray 3 strikes mirror B at zero angle of incidence and reflects straight
back. It reflects from mirror A and then travels parallel to the surface of mirror B and has no further reflections.
A
45°
A
45°
45°
45° 45°
1
P
P
2
45°
45°
45°
B
B
(a)
(b)
A
45°
45°
45° 45°
P
3
B
(c)
Figure 23.28
23-8
Chapter 23
23.29. Set Up: The two rays are shown in Figure 23.29. The two angles labeled f are equal because of the law of
reflection. The angle between the two reflected rays is u.
A
2
a
f
u
2
f
b
u
2
g
Figure 23.29
A
A
and a 5 since the lines forming one angle are parallel to the lines forming the other angle.
2
2
A u
A
A
a 1 f 5 90° and b 1 f 5 90°, so a 5 b and b 5 . 5 b 1 g 5 1 5 A. u 5 2A, as was to be shown.
2 2
2
2
Reflect: As A S 0°, u S 0°. As A S 90°, u S 180°. When A S 180° (flat surface), u S 360°. This corresponds to
the reflected rays traveling in nearly the same direction. All of these limiting results are sensible.
Solve: g 5
23.30. Set Up: The mirror in its original position and after being rotated by an angle u are shown in Figure 23.30.
a is the angle through which the reflected ray rotates when the mirror rotates. The two angles labeled f are equal and
the two angles labeled fr are equal because of the law of reflection. The two angles labeled u are equal because the
lines forming one angle are perpendicular to the lines forming the other angle.
Incident
ray
Normal
to mirror
Ray reflected
from mirror
Normal to
rotated mirror
u
f9
f f f9
a
Mirror
Ray reflected
from rotated mirror
Rotated mirror
Figure 23.30
Solve: From the diagram, a 5 2fr 2 2f 5 2 1 fr 2 f 2 and u 5 fr 2 f. a 5 2u, as was to be shown.
Electromagnetic Waves
23-9
23.31. Set Up: The law of reflection says that ur 5 ua.
Solve: The point and mirror are shown in Figure 23.31. The law of reflection is used to trace two rays after they
reflect from the mirror. If each ray is traced backward after reflection, they both appear to come from point Pr, and
y
y
that is the image of P. s 5
and sr 5
, so sr 5 s.
tan f
tan f
ur 5 f
ua 5 f
y
2
f
P
1
f
s
s9
P9
Figure 23.31
23.32. Set Up: In air, c 5 fl0. In glass, l 5
l0
.
n
3.00 3 108 m / s
c
5
5 517 nm
f
5.80 3 1014 Hz
l0
517 nm
(b) l 5
5
5 340 nm
n
1.52
Solve: (a) l 0 5
c
v
3.00 3 108 m / s
23.33. Set Up: n 5
Solve: n 5
1.82 3 108 m / s
5 1.65
c
v
23.34. Set Up: n 5 . The frequency of light doesn’t change when it passes from one material into another.
v 5 fl. l 5
l0
.
n
l0, r
400 nm
700 nm
5
5 299 nm. lr 5
5 522 nm. Range is 299 nm to 522 nm.
n
1.34
1.34
3.00 3 108 m / s
c
(b) Calculate the frequency in air, where v 5 c 5 3.00 3 108 m / s. fr 5
5
5 4.29 3 1014 Hz.
lr
700 3 1029 m
3.00 3 108 m / s
c
fv 5
5
5 7.50 3 1014 Hz. Range is 4.29 3 1014 Hz to 7.50 3 1014 Hz.
lv
400 3 1029 m
3.00 3 108 m / s
c
(c) v 5 5
5 2.24 3 108 m / s
n
1.34
Solve: (a) l v 5
l0, v
n
5
23-10
Chapter 23
l0
. From Table 23.1, n water 5 1.333 and n benzene 5 1.501.
n
n water
1.333
5 1 438 nm 2
5 389 nm.
Solve: (a) lwaternwater 5 lbenzenen benzene 5 l0. l benzene 5 lwater
nbenzene
1.501
(b) l0 5 lwaternwater 5 1 438 nm 2 1 1.333 2 5 584 nm
Reflect: l is smallest in benzene, since n is largest for benzene.
1
23.35. Set Up: l 5
23.36. Set Up: v 5
c
n
Solve: nv 5 c, which is constant, so n1v1 5 n 2v2. v2 5 v1
23.37. Set Up: l 5
l0
c
and v 5 .
n
n
2
1
2
1 2 1 2
n1
n
5v
5 v / 2.
n2
2n
n1
l2
2l1
5
5
5 2.
n2
l1
l1
v1
n2
1
5 .
(b) vn 5 c, which is constant. v1n1 5 v2n2. 5
n1
v2
2
Reflect: As l increases, n decreases and v increases.
Solve: (a) ln 5 l 0, which is constant, so l1n1 5 l2n2.
23.38. Set Up: na sin ua 5 nb sin ub. The light refracts from the liquid into the glass, so na 5 1.70, ua 5 62.0°.
n b 5 1.58.
na
1.70
sin ua 5
sin 62.0° 5 0.950 and ub 5 71.8°.
Solve: sin ub 5
nb
1.58
1 2
1 2
23.39. Set Up: The angle of incidence is ua 5 90° 2 47.5° 5 42.5°. The light refracts from air into the glass, so
n a 5 1.00 and n b 5 1.66.
Solve: ur 5 ua 5 42.5° so the reflected ray makes an angle of 90° 2 42.5° 5 47.5° with the surface of the glass.
na
1.00
sin ua 5
sin 42.5° 5 0.407 and ub 5 24.0°. The refracted ray makes an
(b) n a sin ua 5 n b sin ub. sin ub 5
nb
1.66
angle of 90° 2 24.0° 5 66.0° with the surface of the glass.
1 2
23.40. Set Up: v 5
1 2
c
n
d
2.50 m
5 8.33 3 1029 s.
5
c
3.00 3 108 m / s
The light traveling in the block takes time t 5 8.33 3 1029 s 1 6.25 3 1029 s 5 1.46 3 1028 s. The speed of
2.50 m
d
light in the block is v 5 5
5 1.71 3 108 m / s. The refractive index of the block is
t
1.46 3 1028 s
3.00 3 108 m / s
c
n5 5
5 1.75.
v
1.71 3 108 m / s
Solve: The time for the beam traveling in air to reach the detector is t 5
1 2
1 2
1 2
1 2
23.41. Set Up: Snell’s law is na sin ua 5 nb sin ub. Light is incident in material a and refracts into material b.
na
1.25
sin ua 5
sin 51.0° 5 0.585 and ub 5 35.8°.
nb
1.66
na
1.66
sin ua 5
sin 35.8° 5 0.777 and ub 5 51.0°.
(b) Now n a 5 1.66, n b 5 1.25, and ua 5 35.8°. sin ub 5
nb
1.25
Reflect: Reflected rays are also reversible.
Solve: (a) n a 5 1.25, n b 5 1.66, and ua 5 51.0°. sin ub 5
Electromagnetic Waves
1 2
1
2
1
23-11
2
23.42. Set Up: Snell’s law says na sin ua 5 nb sin ub. Let n1 and n2 be the two refractive indicies.
sin ua
sin 63.0°
sin 63.0°
. n 1 5 1 1.00 2
5 1.14 and n 2 5 1 1.00 2
5 1.33.
sin ub
sin 51.3°
sin 42.1°
(b) A ray from part of one letter is shown in Figure 23.42. When it passes into the air it splits into two rays, where
each ray has an angle of refraction corresponding to one of the two indicies of refraction. When traced backwards,
these two rays appear to come from two different points on the page, corresponding to two images of the letters.
Solve: (a) n b 5 n a
1
2
2
1
Figure 23.42
23.43. Set Up: The light first refracts from air into the glass and then from the glass into the water. The path of
the ray is sketched in Figure 23.43. ua 5 32.0°. n glass 5 1.60 and n water 5 1.333. ura 5 ub. Snell’s law says
na sin ua 5 n b sin ub.
ua
Air
u9a
Glass
ub
Water
u9b
Figure 23.43
1 2
1
2
Solve: For the refraction air S glass, n air sin ua 5 n glass sin ub. For the refraction glass S water, nglass sin ura 5
nair
1.00
nwater sin urb . But ura 5 ub, so nair sin ua 5 n water sin urb . sin urb 5
sin ua 5
sin 1 32.0° 2 5 0.3975 and
nwater
1.333
urb 5 23.4°.
Reflect: The angle of the ray in the water is the same as if the glass plate wasn’t there.
23.44. Set Up: Snell’s law says na sin ua 5 nb sin ub. Apply Snell’s law to the refraction from material X into the
water and then from the water into the air.
Solve: (a) material X to water: n a 5 n X, n b 5 n w 5 1.333. ua 5 25° and ub 5 48°.
na 5 nb
1 2
1
2
sin ub
sin 48°
5 1 1.333 2
5 2.34
sin ua
sin 25°
23-12
Chapter 23
(b) water to air: As Figure 23.44 shows, ua 5 48°. n a 5 1.333 and n b 5 1.00.
sin ub 5
1 2
na
sin ua 5 1 1.333 2 sin 48° 5 82°.
nb
Air
48°
Water
48°
65°
X
25°
Figure 23.44
23.45. Set Up: na sin ua 5 nb sin ub. The light is in diamond and encounters an interface with air, so na 5 2.42 and
nb 5 1.00. The largest ua is when ub 5 90°.
Solve: 1 2.42 2 sin ua 5 1 1.00 2 sin 90°. sin ua 5
1
and ua 5 24.4°.
2.42
23.46. Set Up: The critical angle occurs when light in liquid encounters an interface with air. na 5 nliq, nb 5 nair.
When ua 5 42.5°, ub 5 90°. na sin ua 5 n b sin ub.
Solve: (a) First solve for nliq: n liq sin 42.5° 5 n air sin 90°.
nliq 5 n air
sin 90°
1
5
5 1.48.
sin 42.5°
sin 42.5°
1 2
Then n a sin ua 5 n b sin ub with n a 5 1.48, n b 5 1.00 and ua 5 35.0° gives
sin ub 5
1 1.48 2 sin 35.0°
na
sin ua 5
5 0.849 and ub 5 58.1°.
nb
1.00
(b) Now n a 5 1.00, n b 5 1.48 and ua 5 35.0°. sin ub 5
1 2
1 1.00 2 sin 35.0°
na
sin ua 5
5 0.3876 and ub 5 22.8°.
nb
1.48
Electromagnetic Waves
23-13
23.47. Set Up: The largest angle of incidence for which any light refracts into the air is the critical angle for
water S air. Figure 23.47 shows a ray incident at the critical angle and therefore at the edge of the ring of light. The
radius of this circle is r and d 5 10.0 m is the distance from the ring to the surface of the water.
r
ucrit
d
ucrit
Ray
Figure 23.47
Solve: From the figure, r 5 d tan ucrit. ucrit is calculated from na sin ua 5 n b sin ub with n a 5 1.333, ua 5 ucrit, n b 5
1 1.00 2 sin 90°
1.00 and ub 5 90°. sin ucrit 5
and ucrit 5 48.6°. r 5 1 10.0 m 2 tan 48.6° 5 11.3 m. A 5 pr 2 5
1.333
p 1 11.3 m 2 2 5 401 m2.
Reflect: When the incident angle in the water is larger than the critical angle, no light refracts into the air.
23.48. Set Up: 48.7° is the critical angle for glass S water. nwater 5 1.333.
Solve: na 5 n glass, n b 5 n water. ua 5 48.7° when ub 5 90°. n a sin ua 5 n b sin ub gives
n glass 5
1 1.333 2 1 1.00 2
n water sin 90°
5
5 1.77.
sin 48.7°
sin 48.7°
23.49. Set Up: The ray has an angle of incidence of 0° at the first surface of the glass, so enters the glass without
being bent, as shown in Figure 23.49. If no light refracts out of the glass at the glass to air interface, then the incident
angle at that interface is ucrit. The figure shows that a 1 ucrit 5 90°.
a
ucrit
a
Figure 23.49
Solve: (a) For the glass-air interface ua 5 ucrit, n a 5 1.52, n b 5 1.00 and ub 5 1.00. na sin ua 5 n b sin ub gives
1 1.00 2 1 sin 90° 2
sin ucrit 5
and ucrit 5 41.1°. a 5 90° 2 ucrit 5 48.9°.
1.52
(b) Now the second interface is glass S water and nb 5 1.333. na sin ua 5 n b sin ub gives sin ucrit 5
1 1.333 2 1 sin 90° 2
and ucrit 5 61.3°. a 5 90° 2 ucrit 5 28.7°.
1.52
Reflect: The critical angle increases when the air is replaced by water and rays are bent as they refract out of the
glass.
23-14
Chapter 23
23.50. Set Up: The critical angle for total internal reflection is ua that gives ub 5 90° in Snell’s law. In Figure 23.50 the angle of incidence ua is related to angle u by ua 1 u 5 90°.
Air
Plastic
u
ua
Figure 23.50
Solve: (a) Calculate ua that gives ub 5 90°. n a 5 1.60, n b 5 1.00 so na sin ua 5 n b sin ub gives 1 1.60 2 sin ua 5
1.00
1 1.00 2 sin 90°. sin ua 5
and ua 5 38.7°. u 5 90° 2 ua 5 51.3°.
1.60
1.333
(b) n a 5 1.60, nb 5 1.333. 1 1.60 2 sin ua 5 1 1.333 2 sin 90°. sin ua 5
and ua 5 56.4°. u 5 90° 2 ua 5 33.6°.
1.60
vair
. When ua 5 ucrit, ub 5 90°.
v
344 m / s
vair
vair
Solve: (a) For air, n 5
5 1.00. For water, n 5
5
5 0.261. Air has a larger index of refraction for
v
v
1320 m / s
sound waves.
(b) Total internal reflection requires that the waves be incident in the material of larger refractive index. n a 5 1.00,
0.261
nb 5 0.261, ua 5 ucrit, and ub 5 90°. n a sin ua 5 n b sin ub gives sin ucrit 5
sin 90° and ucrit 5 15.1°.
1.00
(c) The sound wave must be traveling in air.
(d) Sound waves can be totally reflected from the surface of the water.
Reflect: Light travels faster in vacuum than in any material and n is always greater than 1.00. Sound travels faster in
solids and liquids than in air and n for sound is less than 1.00
23.51. Set Up: n 5
1
2
c
n
1 1.00 2 sin 57.0°
23.52. Set Up: Snell’s law is na sin ua 5 nb sin ub. a 5 air, b 5 glass. v 5 .
1 1.00 2 sin 57.0°
n a sin ua
5
5 1.36. violet: n b 5
5 1.40.
sin ub
sin 38.1°
sin 36.7°
3.00 3 108 m / s
3.00 3 108 m / s
c
c
(b) red: v 5 5
5 2.21 3 108 m / s; violet: v 5 5
5 2.14 3 108 m / s.
n
n
1.36
1.40
Solve: (a) red: n b 5
c
l
23.53. Set Up: f 5 , so when l increases, f decreases.
3.00 3 108 m / s
5 7.50 3 1014 Hz.
400 3 1029 m
3.00 3 108 m / s
For l 5 700 nm, f 5
5 4.29 3 1014 Hz.
700 3 1029 m
Solve: For l 5 400 nm, f 5
Electromagnetic Waves
23-15
The graph of n versus f for silicate flint glass is sketched in Figure 23.53. The refractive index increases as f
increases.
n
1.7
1.65
1.60
7.5 3 1014 Hz
4.3 3 1014 Hz
f
Figure 23.53
23.54. Set Up: na sin ua 5 nb sin ub. From Figure 23.29 in the textbook, for l 5 400 nm, n 5 1.67 and for
l 5 700 nm, n 5 1.62. The path of a ray with a single l is sketched in Figure 23.54.
ua 5 35.0°
55.0°
ub
55.0°
Figure 23.54
Solve: For l 5 400 nm, sin ub 5
na
1.00
1.00
sin ua 5
sin 35.0° and ub 5 20.1°. For l 5 700 nm, sin ub 5
sin 35.0°
nb
1.67
1.62
and ub 5 20.7°. Du is about 0.6°.
l0
c
and v 5 . n a sin ua 5 n b sin ub.
n
n
Solve: (a) red: n 5 1.62; violet: n 5 1.67
700 nm
400 nm
5 432 nm; lviolet 5
5 240 nm.
(b) lred 5
1.62
1.67
n violet
vred
c n violet
c
c
1.67
5 1.03. The red light has a smaller n so
;v
5
5
5
5
(c) vred 5
so
nred violet nviolet
nred
c
nred
vviolet
1.62
travels faster.
na
1.00
sin 65.0° and ub 5 34.0°. For violet light,
(d) For red light, sin ub 5 sin ua 5
nb
1.62
23.55. Set Up: l 5
1 21 2
sin ub 5
na
1.00
sin ua 5
sin 65.0°
nb
1.67
and ub 5 32.9°. The angle between the two rays will be Du 5 34.0° 2 32.9° 5 1.1°.
Reflect: The violet light has a larger n, slows down more in glass and is bent through a greater angle when it enters
the glass.
23-16
Chapter 23
23.56. Set Up: na sin ua 5 nb sin ub. Figure 23.29 in the textbook shows that nblue . nred.
Solve: sin ub 5
na
sin ua. As n b increases, ub differs more from ua. The blue light is deviated more.
nb
23.57. Set Up: For unpolarized light incident on a filter, I 5 12 I0 and the light is linearly polarized along the filter
axis. For polarized light incident on a filter, I 5 Imax 1 cos f 2 2, where Imax is the intensity of the incident light, and the
emerging light is linearly polarized along the filter axis.
Solve: (a) After the first filter, I 5 12 I0 and the light is polarized. After the second filter I 5 1 12 I0 2 1 cos 41.0° 2 2 5
0.285I0.
(b) The light is linearly polarized along the axis of the second filter.
23.58. Set Up: After passing through the first filter the light is linearly polarized along the filter axis. After the
second filter, I 5 Imax 1 cos f 2 2, where f is the angle between the axes of the two filters.
Solve: (a) I 5 I0 1 cos 22.5° 2 2 5 0.854I0
(b) I 5 I0 1 cos 45.0° 2 2 5 0.500I0
(c) I 5 I0 1 cos 67.5° 2 2 5 0.146I0
23.59. Set Up: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 21 and the
transmitted light is polarized along the axis of the polarizer. When polarized light of intensity Imax is incident on a
polarizer, the transmitted intensity is I 5 Imax cos 2f, where f is the angle between the polarization direction of the
incident light and the axis of the filter.
Solve: (a) At point A the intensity is I0 / 2 and the light is polarized along the vertical direction. At point B the intensity is 1 I0 / 2 2 1 cos 60° 2 2 5 0.125I0, and the light is polarized along the axis of the second polarizer. At point C the
intenisty is 1 0.125I0 2 1 cos 30° 2 2 5 0.0938I0.
(b) Now for the last filter f 5 90° and I 5 0.
Reflect: Adding the middle filter increases the transmitted intensity.
23.60. Set Up: For unpolarized light incident on a filter, I 5 12 I0 and the light is linearly polarized along the filter
axis. For polarized light incident on a filter, I 5 Imax 1 cos f 2 2, where Imax is intensity of the incident light, and the
emerging light is linearly polarized along the filter axis.
Solve: With all three polarizers, if the incident intensity is I0 the transmitted intensity is
75.0 W / cm2
I
I 5 1 12 I0 2 1 cos 23.0° 2 2 1 cos 3 62.0° 2 23.0° 4 2 2 5 0.256I0. I0 5
5
5 293 W / cm2. With only the
0.256
0.256
first and third polarizers, I 5 1 12 I0 2 1 cos 62.0° 2 2 5 0.110I0 5 1 0.110 2 1 293 W / cm2 2 5 32.2 W / cm2.
23.61. Set Up: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 21 and the
transmitted light is polarized along the axis of the polarizer. When polarized light of intensity Imax is incident on a
polarizer, the transmitted intensity is I 5 Imax cos2f, where f is the angle between the polarization direction of the
incident light and the axis of the filter.
Solve: (a) After the first filter I 5 I0 / 2 and the light is polarized along the vertical direction. After the second filter
we want I 5 I0 / 10, so I0 / 10 5 1 I0 / 2 2 1 cos f 2 2. cos f 5 "2 / 10 and f 5 63.4°.
(b) Now the first filter passes the full intensity I0 of the incident light. For the second filter I0 / 10 5 I0 1 cos f 2 2.
cos f 5 "1 / 10 and f 5 71.6°.
Reflect: When the incident light is polarized along the axis of the first filter, f must be larger to achieve the same
overall reduction in intensity than when the incident light is unpolarized.
23.62. Set Up: The polarizing angle up is given by tan up 5
Solve: n glass 5 1 1.00 2 1 tan 57.6° 2 5 1.58
nb
. n 5 1.00, nb 5 n glass.
na a
Electromagnetic Waves
23-17
nb
. When ua 5 ucrit, ub 5 90°.
na
nb
nb
5 sin ucrit. tan up 5 sin 1 54.2° 2
Solve: na sin ua 5 n b sin ub gives n a sin ucrit 5 n b sin 90° and 5 sin ucrit. tan up 5
na
na
and up 5 39.0°.
Reflect: The critical angle and the polarizing angle both depend on the ratio of the refractive indexes of the two
materials.
23.63. Set Up: The polarizing angle up is given by tan up 5
23.64. Set Up: The reflected beam is completely polarized when ua 5 up, with tan up 5
nb
. n 5 1.00, nb 5 n glass.
na a
nb
gives n glass 5 n a tan up 5 1 1.00 2 tan 54.5° 5 1.40.
na
1 1.00 2 1 sin 54.5° 2
n a sin ua
5
(b) n a sin ua 5 n b sin ub with ua 5 54.5° gives sin ub 5
and ub 5 35.5°. Note that
nb
1.40
ua 1 ub 5 90°.
Solve: (a) tan up 5
23.65. Set Up: The reflected beam is completely polarized when ua 5 up, with tan up 5
nb
. n 5 1.00, nb 5 1.333.
na a
up is measured relative to the normal to the surface.
1.333
Solve: (a) tan up 5
and up 5 53.1°. The sunlight is incident at an angle of 90° 2 53.1° 5 36.9° above the
1.00
horizontal.
(b) Figure 23.38 in Section 23.10 shows that the plane of the electric field vector in the reflected light is horizontal.
Reflect: To reduce the glare (intensity of reflected light), sunglasses with polarizing filters should have the filter axis
vertical.
23.66. Set Up: I 5 12 P0 cEmax2. Emax 5 cBmax. At the earth the power radiated by the sun is spread over an area of
4pr 2, where r 5 1.50 3 1011 m is the distance from the earth to the sun. P 5 IA.
2 1 1.4 3 103 W / m2 2
2I
Solve: (a) Emax 5
5
5 1.03 3 103 N / C.
Å P0 c Å 1 8.854 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2
Bmax 5
1.03 3 103 N / C
Emax
5 3.43 3 1026 T.
5
c
3.00 3 108 m / s
(b) P 5 I 1 4pr 2 2 5 1 1.4 3 103 W / m2 2 1 4p 2 1 1.50 3 1011 m 2 2 5 4.0 3 1026 W
23.67. Set Up: The wave speed in air is c 5 3.00 3 108 m / s. c 5 fl. Emax 5 cBmax. I 5 12 P0cEmax2. For a totally
I
absorbing surface the radiation pressure is .
c
3.00 3 108 m / s
c
5 7.81 3 109 Hz
Solve: (a) f 5 5
l
3.84 3 1022 m
1.35 V / m
Emax
(b) Bmax 5
5
5 4.50 3 1029 T
c
3.00 3 108 m / s
(c) I 5 12 P0 cEmax2 5 12 1 8.854 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2 1 1.35 V / m 2 2 5 2.42 3 1023 W / m2
1 2.42 3 1023 W / m2 2 1 0.240 m2 2
IA
(d) F 5 1 pressure 2 A 5
5
5 1.94 3 10212 N
c
3.00 3 108 m / s
Reflect: The intensity depends only on the amplitudes of the electric and magnetic fields and is independent of the
wavelength of the light.
23-18
Chapter 23
I
c
23.68. Set Up: For a totally absorbing surface the radiation pressure is .
1 36.0 3 103 W / m2 2 1 0.500 m2 2
IA
5
5 6.0 3 1025 N. A 1 kg mass has a weight of
c
3.00 3 108 m / s
9.8 N. The force the light beam exerts is much too small to be felt by the man.
Solve: F 5 1 pressure 2 A 5
I
c
23.69. Set Up: Energy 5 Pt. For absorption the radiation pressure is , where I 5
l0
P
. l 5 . I 5 12 P0 cEmax2 and
n
A
Emax 5 cBmax.
Solve: (a) Energy 5 Pt 5 1 250 3 1023 W 2 1 1.50 3 1023 s 2 5 3.75 3 1024 J 5 0.375 mJ.
P
250 3 1023 W
5 1.22 3 106 W / m2. The average pressure is
(b) I 5 5
A
p 1 255 3 1026 m 2 2
1.22 3 106 W / m2
I
5
5 4.08 3 1023 Pa.
c
3.00 3 108 m / s
3.00 3 108 m / s
l0
810 nm
v
c
5
5 604 nm. f 5 5
5
5 3.70 3 1014 Hz; f is the same in the air and in
n
1.34
l
l0
810 3 1029 m
the vitreus humor.
2 1 1.22 3 106 W / m2 2
2I
5
5 3.03 3 104 V / m.
(d) E 5
Å P0 c Å 1 8.85 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2
(c) l 5
Bmax 5
Emax
5 1.01 3 1024 T.
c
23.70. Set Up: The general wave function for the electric field is E 5 Emax sin 2p 1 vt 2 kx 2 . f 5
2p
v
,l5
2p
k
c
and v 5 fl. v 5 .
n
Solve: (a) By comparing the equation for E to the general form, v 5 3.02 3 1015 rad / s and k 5 1.39 3 107 rad / m.
v
2p
f5
5 4.81 3 1014 Hz. l 5
5 4.52 3 1027 m 5 452 nm. v 5 fl 5 2.17 3 108 m / s.
2p
k
3.00 3 108 m / s
c
5 1.38
(b) n 5 5
v
2.17 3 108 m / s
(c) In air, v 5 3.02 3 1015 rad / s, the same as in the plastic. l0 5 ln 5 1 4.52 3 1027 m 2 1 1.38 2 5 6.24 3 1027 m
2p
5 1.01 3 107 rad / m. The equation for E in air is
so k 5
l
E 5 1 535 V / m 2 sin 3 1 3.02 3 1015 rad / s 2 t 2 1 1.01 3 107 rad / m 2 x 4 .
23.71. Set Up: I 5
Solve: I 5
P
. I 5 12 P0cEmax2.
A
P
2.80 3 103 W
5
5 77.8 W / m2.
A
36.0 m2
Emax 5
2 1 77.8 W / m2 2
2I
5
5 242 N / C.
Å P0c Å 1 8.854 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2
Reflect: This value of Emax is similar to the electric field amplitude in ordinary light sources.
Electromagnetic Waves
23-19
23.72. Set Up: Find the force on her due to the momentum carried off by the light. Express this force in terms of
the radiated power of the flashlight. Find the acceleration produced by this force.
IA
P
I
5 , were P is the power output of the
. Force is (pressure)A, so F 5
c
c
c
flashlight. The acceleration produced by this force is
Solve: (a) The radiation pressure is
a5
P
200 W
F
5
5
5 4.44 3 1029 m / s2.
m
mc
1 150 kg 2 1 3.00 3 108 m / s 2
2 1 16.0 m 2
2x
5
5 8.49 3 104 s 5 23.6 h.
Å 4.44 3 1029 m / s2
Åa
(b) She could throw the flashlight in the direction away from the ship. By conservation of linear momentum she
would travel toward the ship with the same magnitude of momentum as she gave the flashlight.
x 5 v0xt 1 12 axt 2 gives t 5
2I
. Find the force due to this pressure and
c
mMsun
express the force in terms of the power output P of the sun. The gravitational force of the sun is Fg 5 G 2 .
r
Solve: (a) The sail should be reflective, to produce the maximum radiation pressure.
P
2I
A, where A is the area of the sail. I 5
, where r is the distance of the sail from the sun.
(b) Frad 5
c
4pr 2
mMsun
2A
P
PA
PA
Frad 5
5
. Frad 5 Fg so
5G 2 .
c 4pr 2
2pr 2c
2pr 2c
r
23.73. Set Up: For a totally reflective surface the radiation pressure is
1 2
1 21 2
A5
2p 1 3.00 3 108 m / s 2 1 6.67 3 10211 N # m2 / kg2 2 1 10,000 kg 2 1 1.99 3 1030 kg 2
2pcGmMsun
.
5
P
3.9 3 1026 W
A 5 6.42 3 106 m2 5 6.42 km2.
(c) Both the gravitational force and the radiation pressure are inversely proportional to the square of the distance
from the sun, so this distance divides out when we set Frad 5 Fg.
Reflect: A very large sail is needed, just to overcome the gravitational pull of the sun.
23.74. Set Up: na sin ua 5 nb sin ub. na 5 noil. nb 5 1.333. ua 5 30.0° and ub 5 45.0°.
Solve: n a 5
1 1.333 2 sin 1 45.0° 2
n b sin ub
5
5 1.89
sin ua
sin 1 30.0° 2
23.75. Set Up: na sin ua 5 nb sin ub. na 5 1.00. nb 5 1.80. ub 5 ua / 2. sin a 5 2 sin 1 a / 2 2 cos 1 a / 2 2 .
Solve: Snell’s law gives 1 1.00 2 sin ua 5 1 1.80 2 sin 1 ua / 2 2 . sin ua 5 2 sin 1 ua / 2 2 cos 1 ua / 2 2 , so
2 sin 1 ua / 2 2 cos 1 ua / 2 2 5 1 1.80 2 sin 1 ua / 2 2 .
cos 1 ua / 2 2 5 0.900. ua / 2 5 25.84° and ua 5 51.7°.
Reflect: When the angle of incidence increases the angle of refraction increases. The angle of refraction is smaller
than the angle of incidence when nb . n a, which is the case here.
I
c
23.76. Set Up: The weight of the paper is mg. For a totally absorbing surface the radiation pressure is and for a
P
2I
. Force is pressure times area. Intensity is I 5 .
c
A
I
A 5 mg.
Solve: (a) The radiation force must equal the weight of the paper, so
c
totally reflecting surface it is
I5
12
1 1.50 3 1023 kg 2 1 9.80 m / s2 2 1 3.00 3 108 m / s 2
mgc
5
5 7.16 3 107 W / m2
1 0.220 m 2 1 0.280 m 2
A
23-20
Chapter 23
(b) I 5 12 P0cEmax2. Emax 5
2 1 7.16 3 107 W / m2 2
2I
5
5 2.32 3 105 V / m.
Å P0c Å 1 8.85 3 10212 C2 / N # m2 2 1 3.00 3 108 m / s 2
1 2
Bmax 5
2.32 3 105 V / m
Emax
5
5 7.74 3 1024 T.
c
3.00 3 108 m / s
mgc
2I
2I
so
A 5 mg. I 5
5 3.58 3 107 W / m2.
c
c
2A
0.500 3 1023 W
P
(d) I 5 5
5 637 W / m2. The intensity of this laser is much less than what is needed to supA
p 1 0.500 3 1023 m 2 2
port a sheet of paper. And to support the paper, not only must the intensity be large, it also must be over a large area.
(c) The pressure is
P
. I 5 12 P0cEmax2. Emax 5 cBmax.
A
Solve: (a) P 5 IA 5 1 1.0 3 102 W / m2 2 p 1 0.75 3 1023 m 2 2 5 1.8 3 1024 W 5 0.18 mW
2 1 1.0 3 102 W / m2 2
Emax
2I
(b) E 5
5 9.13 3 1027 T
5
5 274 V / m. Bmax 5
212
c
Å P0c Å 1 8.85 3 10 C2 / N # m2 2 1 3.00 3 108 m / s 2
(c) P 5 0.18 mW 5 0.18 mJ / s
1m 2
5 0.010 W / cm2
(d) I 5 1 1.0 3 102 W / m2 2
102 cm
23.77. Set Up: I 5
1
2
23.78. Set Up: The angle of incidence at A is to be the critical angle. The ray is sketched in Figure 23.78.
ua
ub
ucrit
A
Figure 23.78
Solve: For glass S air at point A, Snell’s law gives 1 1.38 2 sin ucrit 5 1 1.00 2 sin 90° and ucrit 5 46.4°. ub 5
90° 2 ucrit 5 43.6°. Snell’s law applied to the refraction from air to glass at the top of the block gives
1 1.00 2 sin ua 5 1 1.38 2 sin 1 43.6° 2 and ua 5 72.1°.
Electromagnetic Waves
23-21
23.79. Set Up: na sin ua 5 nb sin ub
Solve: n a sin ua is constant, so n b1 sin ub1 5 n b2 sin ub2. n b1 5 n and nb2 5 2n. sin ub2 5
n b1
n
sin ub1 5 sin 60.0° 5
nb2
2n
1
2 sin 60.0°.
ub2 5 25.7°.
Reflect: Increasing the refractive index decreases the angle of refraction and causes the ray to be bent more when it
refracts into the glass.
23.80. Set Up: na sin ua 5 nb sin ub. If ua is the critical angle then ub 5 90°. For air, nair 5 1.00. For heart muscle,
nmus 5
344 m / s
1480 m / s
5 0.2324.
Solve: (a) na sin ua 5 n b sin ub gives 1 1.00 2 sin 1 9.73° 2 5 1 0.2324 2 sin ub. sin ub 5
(b) 1 1.00 2 sin ucrit 5 1 0.2324 2 sin 90° gives ucrit 5 13.4°.
sin 1 9.73° 2
and ub 5 46.7°.
0.2324
23.81. Set Up: The rays are incident on the prism in the normal direction so they do not change direction as they
enter the prism. Apply Snell’s law to the refraction of the rays as they exit the prism. The rays are sketched in Figure 23.81. The angle between the two rays after they emerge from the prism is 2g.
A
ub
A
A
b
g
g
Figure 23.81
Solve: The angle of incidence is A 5 25.0°. na sin ua 5 n b sin ub gives 1 1.66 2 sin 1 25.0° 2 5 1 1.00 2 sin ub.
ub 5 44.55°. b 5 90° 2 ub 5 45.45°. g 1 b 1 A 5 90°, so g 5 90° 2 A 2 b 5 90° 2 25.0° 2 45.45° 5
19.55°. The angle between the emerging rays is 2g 5 39.1°.
Reflect: If A S 0°, b S 90° and g S 0°. The rays remain parallel if the prism is replaced by a rectangular slab of
glass.
23-22
Chapter 23
23.82. Set Up: The path of the ray is sketched in Figure 23.82. The ray enters the prism at normal incidence so is
not bent. The incident angle at the prism S water interface is to be the critical angle. For water, nwater 5 1.333.
45°
45°
ucrit
Figure 23.82
Solve: From the figure, ucrit 5 45°. n a sin ua 5 n b sin ub gives nglass sin 45° 5 1 1.333 2 sin 90°. nglass 5
1.333
5 1.89.
sin 45°
23.83. Set Up: The path of the ray is sketched in Figure 23.83. The problem asks us to calculate urb.
u9b
ua
ub
u9a
f
f
Figure 23.83
Solve: Apply Snell’s law to the air S liquid refraction. 1 1.00 2 sin 1 42.5° 2 5 1 1.63 2 sin ub and ub 5 24.5°. ub 5 f
and f 5 ura, so ura 5 ub 5 24.5°. Snell’s law applied to the liquid S air refraction gives 1 1.63 2 sin 1 24.5° 2 5
1 1.00 2 sin urb and urb 5 42.5°.
Reflect: The light emerges from the liquid at the same angle from the normal as it entered the liquid.
23.84. Set Up: The angle of incidence for the glass S oil interface must be the critical angle, so ub 5 90°.
Solve: na sin ua 5 n b sin ub gives 1 1.52 2 sin 57.2° 5 n oil sin 90°. n oil 5 1 1.52 2 sin 57.2° 5 1.28.
nb
. n 5 1.00. nb 5 n glass.
na a
Solve: (a) nglass 5 1 1.00 2 tan up. Red: n glass 5 tan 60.0° 5 1.73. Blue: n glass 5 tan 70.0° 5 2.75.
n a sin ua
. ub differs more from ua when n b is larger. The blue light is refracted more upon entering the
(b) sin ub 5
nb
glass.
23.85. Set Up: The polarizing angle up is given by tan up 5
Electromagnetic Waves
23-23
23.86. Set Up: For linearly polarized light the intensity passed by a filter is I 5 I0 1 cos f 2 2, where I0 is the incident
intensity and I is the transmitted intensity. After passing through the filter the light is linearly polarized along the
filter axis.
Solve: (a) Use two filters, with the axis of the second filter at 90° from the original direction of polarization of the
light.
(b) Let u be the angle between the original direction of polarization of the light and the axis of the first filter. The
angle between the axes of the two filters therefore is 90° 2 u. The intensity passed by the two filters is I 5
I0 1 cos u 2 2 2 cos 3 90° 2 u 4 2 2. cos 1 90° 2 u 2 5 sin u so I 5 I0 1 cos u sin u 2 2. cos u sin u 5 12 sin 2u so I 5 14 I0 sin 1 2u 2 .
The transmitted intensity has a maximum value of I0 / 4 when u 5 45°.
23.87. Set Up: Let the light initially be in the material with refractive index na and let the final slab have refractive
index n b. In part (a) let the middle slab have refractive index n 1. Apply Snell’s law to each refraction.
Solve: (a) 1st interface: na sin ua 5 n 1 sin u1. 2nd interface: n1 sin u1 5 n b sin ub. Combining the two equations gives
n a sin ua 5 n b sin ub. This is the equation that would apply if the middle slab were absent.
(b) For N slabs, na sin ua 5 n 1 sin u1, n 1 sin u1 5 n 2 sin u2, c, n N22 sin uN22 5 n b sin ub. Combining all these equations gives n a sin ua 5 n b sin ub. The final direction of travel depends on the angle of incidence in the first slab and the
refractive indices of the first and last slabs.
23.88. Set Up: The reflected light is completely polarized when the angle of incidence equals the polarizing angle
nb
. n 5 1.66.
na b
1.66
Solve: (a) n a 5 1.00. tan up 5
and up 5 58.9°.
1.00
1.66
(b) n a 5 1.333. tan up 5
and up 5 51.2°.
1.333
up, where tan up 5
23.89. Set Up: The paths of rays A and B are sketched in Figure 23.89. Let u be the angle of incidence for the
combined ray.
B
A
u
u
f
n1
a
a
a
a
n2
Figure 23.89
Solve: For ray A its final direction of travel is at an angle u with respect to the normal, by the law of reflection. Let
the final direction of travel for ray B be at angle f with respect to the normal. At the upper surface, Snell’s law gives
n 1 sin u 5 n 2 sin a. The lower surface reflects ray B at angle a. Ray B returns to the upper surface of the film at an
angle of incidence a. Snell’s law applied to the refraction as ray B leaves the film gives n2 sin a 5 n1 sin f. Combining the two equations gives n1 sin u 5 n 1 sin f and u 5 f; the two rays are parallel after they emerge from the film.
Reflect: Ray B is bent toward the normal as it enters the film and away from the normal as it refracts out of the film.
23.90. Set Up: Both l-leucine and d-glutamic acid exhibit linear relationships between concentration and rotation
angle.
Solve: For l-leucine: rotation angle 5 1 20.11 deg # mL 2 C. For d-glutamic acid:
rotation angle 5 1 10.124 deg # mL 2 C.
23-24
Chapter 23
23.91. Set Up: The path of the ray is sketched in Figure 23.91. Let L be the length of the path of the ray inside the
plate.
ua
n
u9b
n9
ua 2 u9b
t
L
u9b
n
u9a
d
Figure 23.91
Solve: (a) Apply Snell’s law to the air S plate refraction: na sin ua 5 nr sin urb. Apply Snell’s law to the plate S air
refraction: nr sin urb 5 n sin ura. Combining these two equations gives n sin ua 5 n sin ura and ua 5 ura.
t sin 1 ua 2 urb 2
,
(b) d 5 L sin 1 ua 2 urb 2 and t 5 L cos urb. Combining these two equations to eliminate L gives d 5
cos urb
as was to be shown.
1 1.00 2 sin 66.0°
n sin ua
5
(c) n sin ua 5 nr sin urb gives sin urb 5
and urb 5 30.5°.
nr
1.80
d5
1 2.40 cm 2 sin 1 66.0° 2 30.5° 2
t sin 1 ua 2 urb 2
5
5 1.62 cm.
cos urb
cos 30.5°
Reflect: The result derived in part (b) applies to a plate of any thickness. ua 5 ura means the incident and emerging
rays are parallel. In the limit of a thin plate (t S 0), the lateral displacement d S 0.