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7/29/2019 sol_s_34

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Halliday♦ Resnick ♦Walker

FUNDAMENTALS OF PHYSICS

SIXTH EDITION

Selected Solutions

Chapter 34

34.1934.25

34.39

34.49

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19. (a) The average rate of energy flow per unit area, or intensity, is related to the electric field amplitudeE m by I = E 2

m/2µ0c, so

E m =

2µ0cI =

2(4π × 10−7 H/m)(2.998× 108 m/s)(10× 10−6 W/m

2)

= 8.7× 10−2 V/m .

(b) The amplitude of the magnetic field is given by

Bm =E mc

=8.7× 10−2 V/m

2.998× 108 m/s= 2.9× 10−10 T .

(c) At a distance r from the transmitter, the intensity is I = P/4πr2, where P is the power of thetransmitter. Thus

P = 4πr2

I = 4π(10× 103

m)2

(10× 10−6

W/m2

) = 1.3× 104

W .

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25. (a) Since c = λf , where λ is the wavelength and f is the frequency of the wave,

f = cλ

= 2.998× 108

m/s3.0 m

= 1.0× 108 Hz .

(b) The magnetic field amplitude is

Bm =E mc

=300V/m

2.998× 108 m/s= 1.00× 10−6 T .

B must be in the positive z direction when E is in the positive y direction in order for E × B to bein the positive x direction (the direction of propagation).

(c) The angular wave number is

k =2π

λ=

2π

3.0 m= 2.1rad/m .

The angular frequency is

ω = 2πf = 2π(1.0× 108 Hz) = 6.3× 108 rad/s .

(d) The intensity of the wave is

I =E 2m

2µ0c=

(300 V/m)2

2(4π × 10−7 H/m)(2.998× 108 m/s)= 119 W/m

2 .

(e) Since the sheet is perfectly absorbing, the rate per unit area with which momentum is delivered toit is I/c, so

dp

dt=

IA

c=

(119W/m2

)(2.0 m2)

2.998× 108 m/s= 8.0× 10−7 N .

The radiation pressure is

pr =dp/dt

A=

8.0× 10−7 N

2.0 m2= 4.0× 10−7 Pa .

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39. Let I 0 be the intensity of the incident beam and f be the fraction that is polarized. Thus, the intensityof the polarized portion is fI 0. After transmission, this portion contributes fI 0 cos2 θ to the intensity

of the transmitted beam. Here θ is the angle between the direction of polarization of the radiationand the polarizing direction of the filter. The intensity of the unpolarized portion of the incident beamis (1 − f )I 0 and after transmission, this portion contributes (1 − f )I 0/2 to the transmitted intensity.Consequently, the transmitted intensity is

I = fI 0 cos2 θ +1

2(1− f )I 0 .

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

I min =1

2(1− f )I 0

to a maximum of

I max = fI 0 +

1

2 (1−

f )I 0 =

1

2 (1 + f )I 0 .

The ratio of I max to I min isI max

I min

=1 + f

1− f .

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

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49. Let θ be the angle of incidence and θ2 be the angle of refraction at the left face of the plate. Let n be theindex of refraction of the glass. Then, the law of refraction yields sin θ = n sin θ2. The angle of incidence

at the right face is also θ2. If θ3 is the angle of emergence there, then n sin θ2 = sin θ3. Thus sin θ3 = sin θand θ3 = θ. The emerging ray is parallel to the incident ray. We wish to derive an expression for x interms of θ. If D is the length of the ray in the glass, then D cos θ2 = t and D = t/ cos θ2. The angle αin the diagram equals θ − θ2 and x = D sinα = D sin(θ − θ2). Thus

x =t sin(θ − θ2)

cos θ2.

If all the angles θ, θ2, θ3, and θ − θ2 are small and measured in radians, then sin θ ≈ θ, sin θ2 ≈ θ2,sin(θ − θ2) ≈ θ − θ2, and cos θ2 ≈ 1. Thus x ≈ t(θ − θ2). The law of refraction applied to the point of incidence at the left face of the plate is now θ ≈ nθ2, so θ2 ≈ θ/n and

x ≈ tθ −θ

n =

(n− 1)tθ

n

.

←−−−−−− t −−−−−−→

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