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Unformatted text preview: ± rel according to Equation 24.6. The frequency
c fs emitted by the source is the same in each case, so that only the relative speed vrel and the
direction of the relative motion determine the observed frequency. In each case either the
source or the observer is moving, so the relative speed is just the magnitude of the velocity
vector shown in the drawing. Since the velocity vector has the same magnitude in each case,
the relative speed is the same in each case. Thus, it is only the direction of the relative
motion that needs to be considered here. In A and C the source and observer are moving
apart at the same relative speed, the minus sign applies in Equation 24.6, and the observed
frequencies are the same. In B and D they are coming together at the same relative speed, the
plus sign applies in Equation 24.6, and the observed frequencies are the same, but greater
than that in A and C. 1278 ELECTROMAGNETIC WAVES 7. 3.0 × 106 m/s
8. (c) When the incident light is completely unpolarized, half of its intensity is absorbed by the
polarizer on the left, and half passes through. The half that passes through is completely
polarized along the vertical direction, which is the same as the transmission axis of the
second polarizer. Thus, the second polarizer absorbs none of the light, and the intensity of
the exiting light is half that of the incident light. When the incident light is completely
polarized along the vertical direction to begin with, it passes through both sheets of material
with none of its intensity being absorbed. In this case the exiting light has the same intensity
as the incident light. Thus, the exiting light has a greater intensity when the incident light is
polarized.
9. (d) When the unpolarized light strikes the first polarizer, the light that passes through it is
polarized in the vertical direction. When this polarized light strikes the second polarizer, all
of it is absorbed, since the two polarizers are crossed. When the polarized light strikes the
first polarizer, all of it passes through, since the direction of polarization and the transmission
axis are both vertical. When this polarized light strikes the second polarizer, all of it is
absorbed, since the two polarizers are crossed. Thus, no light exits the polarizer on the right
in either case.
10. (e) When the light is incident from the left, Malus’ law (Equation 24.7) indicates that the
transmitted light has an average intensity that is reduced relative to the incident intensity by a
factor of cos 2 45° = 1 . When the light is incident from the right, Malus’ law also applies.
2
Now, however, polarizer 2 has its transmission axis at an angle of 45º with respect to the
polarization direction of the incident light. Malus’ law indicates that the intensity is reduced
by a factor of 1 . But the light leaving polarizer 2 is polarized at an angle of 45º with respect
2
to the transmission axis of polarizer 1. Malus’ law again applies and indicates that the
intensity is reduced by a second factor of 1 . The transmitted light, therefore, has an intensity
2
that is reduced by a factor of 1×1
22 1
= relative to the initial intensity.
4 Chapter 24 Problems 1279 CHAPTER 24 ELECTROMAGNETIC WAVES
PROBLEMS
______________________________________________________________________________
1. REASONING The distance traveled by the Xrays is equal to their speed multiplied by the
elapsed time. Therefore, the elapsed time is equal to the distance divided by the speed. The
distance is known. Since Xrays are electromagnetic waves, and all electromagnetic waves
move through a vacuum at the speed of light c, the speed is known.
SOLUTION The time it takes for the Xrays to travel from the sun to the earth is =
t Distance d 1.50 ×1011 m
==
= 5.00 × 102 s
8
c 3.00 × 10 m/s
Speed Since 1 min = 60 s, we have ( 5.00 ×10 1 min =
t =2 s ) 8.33 min 60 s ____________________________________________________________________________________________ 2. REASONING The distance d between earth and the probe is determined by d = ct
(Equation 2.1), where c is the speed of light in a vacuum and t is the time for the radio signal
to reach earth.
SOLUTION The elapsed time t is given in hours, so it must be converted to seconds: 3600 s t = 2.53 h = 9110 s 1h ( ) (1) The distance, then, between earth and the probe is ( ) d = m/s ( 9110 s ) =
ct =
3.00 ×108
2.73 ×1012 m
______________________________________________________________________________
3. SSM REASONING AND SOLUTION This is a standard exercise in units conversion.
We first determine the number of meters in one lightyear. The distance that light travels in
one year is 365.25 days 24 hours 3600 s 15
d = ct = (3.00 ×108 m/s)(1.00 year) = 9.47×10 m 1 year 1.0 day 1 hour 1280 ELECTROMAGNETIC WAVES Thus, 1 light year = 9.47 × 1015 m . Then, the distance to Alpha Centauri is 9.47 ×1015 m (4.3 lightyears) = 4.1×1016 m 1 lightyear ______________________________________________________________________________ 4. REASONING In order to pick up radio waves,...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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