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Unformatted text preview: he intensity in either situation. When A is removed, however, D is a third
sheet that absorbs light intensity. In contrast, when D is removed, A is present as a third
sheet, but it absorbs none of the light intensity. This is because the transmission axis of A is
vertical and matches the direction in which the incident light is polarized. We conclude,
therefore, that the greater light intensity is transmitted when D is removed.
SOLUTION When light with an average intensity S0 is polarized at an angle θ with respect
to the polarization axis of a polarizer, the average intensity S that is transmitted through the
polarizer is given by Malus’ law as S = S0 cos 2 θ (Equation 24.7). The light that passes
through is polarized in the direction of the transmission axis. In this problem, each of the
polarizers, therefore, transmits a light intensity that is smaller than the incident light by a
factor of cos2θ. We use this insight now to determine the transmitted light intensity in the
two situations that result when A is removed and when D is removed.
[A is removed; B, C, and D remain]
S ( 27 W/m2 ) cos2 ° cos2 ° cos2 ° 30.0 60.0 30.0 = due to B due to C [B is removed; A, C, and D remain]
S = 0 W/m 2 [C is removed; A, B, and D remain]
S = 0 W/m 2 [D is removed; A B, and C remain] due to D 3.8 W/m 2 Chapter 24 Problems =
( 27 W/m2 ) cos° cos= 0.0 ° cos ° 30.0 60.0 due to A due to B 1307 5.1 W/m 2 due to C ______________________________________________________________________________
49. REASONING AND SOLUTION Using Equation 16.1, we obtain c
2.9979 ×108 m/s
= 11.118 m
26.965 × 106 Hz
λ 50. REASONING The magnitude E of the electric field in an electromagnetic wave is related to
the magnitude B of the magnetic field according to E = cB (Equation 24.3), where c is the
speed of light.
SOLUTION Using Equation 24.3, we find that ( )( ) E = 3.00 ×108 m/s 3.3 ×10−6 T = N/C
990 51. SSM WWW REASONING The electromagnetic wave will be picked up by the radio when the resonant frequency f0 of the circuit in Figure 24.4 is equal to the frequency of the
broadcast wave, or f0 = 1400 kHz. This frequency, in turn, is related to the capacitance C ( and inductance L of the circuit via f 0 = 1/ 2π LC ) (Equation 23.10). Since C is known, we can use this relation to find the inductance. ( ) SOLUTION Solving the relation f 0 = 1/ 2π LC for the inductance L, we find that
4π f 0 C 4π 2 1400 × 103 Hz
2 ( )(
2 = 1.5 × 10−4 H
8.4 × 10−11 F ) ______________________________________________________________________________
52. REASONING AND SOLUTION According to Equation 16.1, the wavelength of these
waves is λ = c/f. Therefore, λMRI c / f MRI
λPET c / f PET f PET 1.23 × 1020 Hz
= 1.93 × 1012
f MRI 6.38 × 10 Hz
______________________________________________________________________________ 1308 ELECTROMAGNETIC WAVES 53. SSM REASONING The rms value Erms of the electric field is related to the average
energy density u of the microwave radiation according to u = ε 0 Erms (Equation 24.2b). SOLUTION Solving for Erms gives
4 × 10−14 J/m3
= 0.07 N/C
8.85 × 10−12 C2 /(N ⋅ m 2 )
= 54. REASONING Using the Doppler effect, we will find the relative speed between the
speeding car and the police car. Since we know the speed of the police car relative to the
ground, we can determine the speed of the car relative to the ground, once the relative speed
vrel is found.
There are two Doppler frequency changes in this situation. First, the speeder's car observes
the wave frequency coming from the radar gun to have a frequency f o that is different from
the emitted frequency fs . The second Doppler shift occurs after the wave reflects from the
speeder's car and returns to the police car.
The Doppler frequency for electromagnetic radiation is given by = fs [1 ± (vrel / c)]
(Equation 24.6), where vrel is the relative speed between the source and the observer of the
radiation, and the plus sign applies when the source and the observer are moving toward one
another, while the minus sign applies when they are moving apart. Since the distance
between the police car and the speeder's car is increasing, they are moving apart, and
according to Equation 24.6, the first Doppler frequency change is given by
f o – fs = – fs (vrel / c) . After the wave reflects from the speeder's car it returns to the police car where it is observed to have a frequency f o′ that is different from its frequency f o at the
instant of reflection. Equation 24.6 may again be used, this time to determine the second
Doppler frequency shift: f o′ – f o = – f o (vrel / c) . We can use these two equations for the
frequency shifts to determine an expression for the total Doppler change in frequency.
Adding the two equations, we have v v ( f o′ – f o ) + ( f o – f s ) = – f s rel – f o rel
c v v v f o′ – fs = – f o rel + fs...
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