HW #9 Solutions
21.41
(a) The magnitudes of the fields associated with an electromagnetic wave are related by
E
= c.
B
Therefore,
E = cB = (3.00 x 108 m/s)(1.5 x 10-7 T) = 45 N/C
(b) The average power per unit area may be computed as
EmBm 45 N/C(1.5 x 10-7 T)
Average power per unit area =
=
= 2.7 W/m2
2 µo
2(4π x 10-7 N/A2)
21.45
(a)
(b)
21.57
21.58
3.00 x 108 m/s
= 556 m,
540 x 103 Hz
min
c
3.00 x 108 m/s
λmin = f
=
= 188 m.
1600 x 103 Hz
max
c
3.00 x 108 m/s
For the FM band: λmax =
=
= 3.4 m,
fmin
88 x 106 Hz
λmax = f
c
λmin = f
c
max
=
=
3.00 x 108 m/s
= 2.78 m.
108 x 106 Hz
Emax 2.0 x 10-7 V/m
=
= 6.7 x 10-16 T.
c
3.0 x 108 m/s
(a)
Bmax =
(b)
I=
(c)
P = IA = (5.3 x 10-17 W/m2)(100π m2) = 1.7 x 10-14 W.
(a)
B=
(b)
E2max
(2.0 x 10-7 V/m)2
=
= 5.3 x 10-17 W/m2.
2 µo c
2(4π x 10-7 Ns2/C2)(3.0 x 108 m/s)
E
6.0 V/m
=
= 2.0 x 10-8 T.
c 3.0 x 108 m/s
EmaxBmax (6.0 V/m)(2.0 x 10-8 T)
I=
=
= 4.8 x 10-2 W/m2, and
2 µo
2(4π x 10-7 Ns2/C2)
P = IA = 4πr2I = 4π(103 m)2(4.8 x 10-2 W/m2) = 6.0 x 105 W = 600 kW.
22.5
(a)
so
(b)
From geometry, 1.25 m = d sin40°
d = 1.94 m.
At 50° above horizontal, or parallel to the incident ray.
50°
50°
40°
d
i = 40°
40°
50°
1.25 m
50°
22.17
The air is medium 1, and the ice is medium 2.
Snell's law gives the angle of refraction as:
n2 sinθ2 = n1 sinθ1, or 1.309 sinθ2 = sin40.0°.
θ2 = 29.4 °.
Thus, sinθ2 = 0.491, and
Also, from the law of reflection, Φ = 40.0 °.
The angle between the reflected and refracted ray
(see sketch) is found as: θ2 + α + Φ = 180 °.
Thus, α = 110.6 °.
Φ
40 °
α
air
ice
θ
2
The angle of refraction, θ2, equals 19.5° (see problem 16). Let h represent the distance from
point a to c (i.e., the hypotenuse of triangle abc). Then, h is found by use of triangle abc as:
θ1
a
θ1
θ2
α
d
2.00 cm
= cos19.5°, from which, h = 2.12 cm.
h
We also see that: θ2 + α = θ1, or,
α = θ1 - θ2 = 30.0° - 19.5° = 10.5°
Finally, d = h sinα = (2.12 cm)sin10.5° = 0.386 cm.
2 cm
22.11
b
c
22.28
The angle of refraction for the blue light is found from Snell's law as:
1
θB = 17.64°.
sinθB =
= 0.3030, and
2(1.650)
1
Likewise, for red, we have: sinθR =
= 0.3096, and θR = 18.03°.
2(1.615)
Thus, the angle between the two rays is: ∆θ = 0.39°.
22.31
From Snell's law, we find the index of refraction of the fluid:
n2sinθ2 = n1sinθ1 _ n2 sin22° = 1.0 sin30.0°, or
n2 = 1.335.
n2
1
θc = 48.5°.
: sinθc =
= 0.749, and
Then, from sinθc =
n1
1.335
22.37
The critical angle for a water-air boundary is:
n2
1
3
sinθc =
=
= , and θc = 48.6°.
n1 4/3 4
The circular raft must cover the area of the surface through which light from the diamond could
emerge. Thus, it must form the base of a cone (with apex at the diamond) whose half angle is θ,
where θ is greater than or equal to the critical angle. We have:
rmin
= tanθc, or rmin = (2.00 m)tan48.6° = 2.27 m, and the diameter is dmin = 2 rmin =
h
4.54 m.