Unformatted text preview: elative to the earth by solving the wavelength
version of Equation 24.6 for vrel: λ 434.1 nm vrel = s = 8 m/s) 1 −
c 1 –
(3.0 ×10
3.1× 6 =10 m/s λ 438.6 nm o 36. REASONING There are two Doppler frequency changes in the emitted wave in this case.
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 (source) frequency fs. The wave then
reflects and returns to the police car, where it is observed to have a frequency f o′ that is
different than its frequency f o at the instant of reflection. Although the police car is now
moving, the relative motion of the two vehicles is one of approach. In Example 6, it is
shown that, when the source and the observer of the radar are approaching each other, the
magnitude of the difference between frequency of the emitted wave and the wave that
returns to the police car after reflecting from the speeder's car is v f o′ – fs ≈ 2 fs rel c
where vrel is the relative speed between the speeding car and the police car.
SOLUTION Since the police car is moving to the right at 27 m/s, while the speeder is
coming from behind at 39 m/s, the relative speed vrel is 39 m/s – 27 m/s = 12 m/s . The total
Doppler change in frequency is, therefore, 12 m/s f o′ – f s =
2(8.0 × 109 Hz) 640 Hz
=
8 3.0 × 10 m/s ___________________________________________________________________________ Chapter 24 Problems 37. SSM REASONING 1299 The Doppler effect for electromagnetic radiation is given by fo
Equation 24.6, = fs (1 ± vrel / c ) , where f o is the observed frequency, fs is the frequency
emitted by the source, and vrel is the speed of the source relative to the observer. As
discussed in the text, 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.
SOLUTION
a. At location A, the galaxy is moving away from the earth with a relative speed of
vrel = (1.6 ×106 m/s) − (0.4 ×106 m/s) = 1.2 ×106 m/s Therefore, the minus sign in Equation 24.6 applies and the observed frequency for the light
from region A is 1.2 ×106 m/s vrel 14
14
f o = fs 1 – = 6.175 ×10 Hz = (6.200 ×10 Hz) 1 −
8
c
3.0 ×10 m/s b. Similarly, at location B, the galaxy is moving away from the earth with a relative speed of
6 6 6 vrel = (1.6 ×10 m/s) + (0.4 ×10 m/s) = 2.0 ×10 m/s The observed frequency for the light from region B is vrel 2.0 ×106 m/s 14
14
fo = fs 1 – = 6.159 ×10 Hz = (6.200 ×10 Hz) 1 −
8
c
3.0 ×10 m/s ______________________________________________________________________________ v
fo
38. REASONING The observed frequency is= fs 1 ± rel according to Equation 24.6. The
c frequency fs emitted by the source is the same in each case, so that only the direction of the
relative motion and the relative speed vrel determine the observed frequency. In situations A
and B the observer and the source move away from each other, and the minus sign in
Equation 24.6 applies. In situation C the observer and the source move toward each other,
and the plus sign applies. Thus, the observed frequency is largest in C. To distinguish
between A and B, we note that the relative speed in A is 2v − v = v, whereas in B the relative
speed is 2v + v = 3v. The greater relative speed means that the term vrel/c is greater in B than
in A, and since the minus sign applies, the observed frequency is more reduced in B than in
A. We conclude, then, that the situations are ranked in descending order according to the
observed frequencies as follows: C (largest), A, B
SOLUTION To apply Equation 24.6 to calculate the observed frequencies, we need the
relative speed in each situation. The relative speeds and the observed frequencies are: 1300 ELECTROMAGNETIC WAVES [Situation A, minus sign in Equation 24.6] vrel = 2v − v = v v v
f o = fs 1 − rel = fs 1 − =
c c ( 4.57 ×1014 Hz ) 1 − 1.50 ×108 m/s =
3.00 ×10 m/s
6 4.55 ×1014 Hz [Situation B, minus sign in Equation 24.6]
vrel = 2v + v = 3v ( ) = 4.50 ×1014 Hz ) = 4.62 ×1014 Hz 3 1.50 ×106 m/s vrel 3v 14
1 −
f o = fs 1 − = f 1 − = 4.57 ×10 Hz c s
c
3.00 × 108 m/s ( ) [Situation C, plus sign in Equation 24.6]
vrel = v + v = 2v ( 2 1.50 × 106 m/s vrel 2v 14
f o = fs 1 + = f 1 + = 4.57 ×10 Hz 1 + c s
c
3.00 ×108 m/s ( ) 39. REASONING AND SOLUTION
a. The polarizer reduces the intensity of the light by a factor of two or to 0.55 W/m 2 .
b. The intensity of the light leaving the analyzer is given by Malus’ law.
S = (0.55 W/m2) cos2 75° = 3.7 × 10 –2 W/m 2 ______________________________________________________________________________
40. REASONING If the incident light is unpolarized, the average intensity of the transmitted
light is onehalf the average intensity of the incident light, independent of the angle of the
transmission axis. Thus, the average intensity of the transmitted light remains the same as
the polarizing material is rotated.
If the incident light is polarized along the z axis, the direction of polarization and...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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