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Unformatted text preview: rel ≈ –2 fs rel c c c Chapter 24 Problems 1309 where we have assumed that fs and f o differ only by a negligibly small amount, so that
f o ≈ fs . Rearranging, we have v fs – f o′ ≈ 2 fs rel c
SOLUTION Solving for the relative speed vrel gives fs – f o′ 320 Hz
8
vrel ≈ = (3.0 × 10 m/s) = 6.9 m/s
9 2 f c 2(7.0 ×10 Hz) s The relative speed vrel is related to the speeds of the vehicles with respect to the ground by = vspeeder − vpolice . Therefore, the speeder's speed with respect to the ground is
vrel
vspeeder =rel + vpolice = m/s + 25 m/s = 32 m/s
v
6.9
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55. SSM REASONING Since the incident beam is unpolarized, the intensity of the light
transmitted by the first sheet of polarizing material is onehalf the intensity of the incident
beam. The beams striking the second and third sheets of polarizing material are polarized,
so the average intensity S of the light transmitted by each sheet is given by Malus’ law,
S = S0 cos 2 θ , where S0 is the average intensity of the light incident on each sheet.
SOLUTION The average intensity S1 of the light leaving the first sheet is onehalf the =
intensity of the incident beam, so S1 ( ) = 630.0 W/m 2 . The intensity S 2 of
1260.0 W/m 2 1
2 the light leaving the second sheet of polarizing material is given by Malus’ law, Equation
2
24.7, S 2 = S1 cos θ , where θ is the angle between the polarization of the incident beam and
the transmission axis of the second sheet: =
S2 ( 630.0 W/m=
) cos (55.0° − 19.0°)
2 2 412 W/m 2 The intensity S3 of the light leaving the third sheet of polarizing material is S3 = S 2 cos 2 θ ,
where θ is the angle between the polarization of the incident beam and the transmission
axis of the third sheet: =
S3 cos (100.0° − 55.0° )
( 412 W/m )=
2 2 206 W/m 2 ______________________________________________________________________________ 1310 ELECTROMAGNETIC WAVES 56. REASONING The average power P delivered to the wall is equal to the energy delivered
divided by the time t. Thus, the time is equal to the energy divided by the average power, or
t = Energy / P . The energy is given. The average power is equal to the average intensity S
times the area A, so P = S A . The area is known, and the average intensity is related to the
2
rmsvalue Brms of the wave’s magnetic field by S = cBrms / µ0 . SOLUTION The time required to deliver the energy to the wall is
t= Energy
P (6.10b) Since P = S A (Equation 16.8), the expression for the time can be written as
=
t Energy Energy
=
P
SA (1) 2
The average intensity is related to Brms by S = cBrms / µ0 (Equation 24.5c), where c is the speed of light in a vacuum and µ0 is the permeability of free space. Substituting this
expression into Equation (1) and noting that A = 1.30 cm2 = 1.30 × 10−4 m2 gives
t
= Energy
Energy
=
2
SA cBrms µ A 0 1850 J
= 0.129 s
( 3.00 ×108 m/s )( 6.80 ×10−4 T )2 (
1.30 ×10−4 m 2 )
−7
4π ×10 T ⋅ m/A
______________________________________________________________________________ = 57. REASONING According to Equation 16.3, the displacement y of a wave that travels in the
+x direction and has amplitude A, frequency f, and wavelength λ is given by
2π x y = A sin 2π f t –
λ This equation, with y = E, applies to the traveling electromagnetic wave in the problem,
which is represented mathematically as ( )( ) E = sin 1.5 ×1010 s –1 t – 5.0 ×101 m –1 x E0 Chapter 24 Problems 1311 As E0 is the maximum field strength, it represents the amplitude A of the wave. We can
find the frequency and wavelength of this electromagnetic wave by comparing the
mathematical form of the electric field with Equation 16.3.
SOLUTION
a. By inspection, we see that 2π = 1.5 ×1010 s –1 . Therefore, the frequency of the wave is
f
=
f 1.5 ×1010 s –1
=
2π 2.4 ×109 Hz b. As shown in Figure 17.15, the separation between adjacent nodes in any standing wave is
onehalf of a wavelength. By inspection of the mathematical form of the electric field and
comparison with Equation 16.3, we infer that 2π /= 5.0 ×101 m –1 . Therefore,
λ λ
= 2π
= 0.126 m
5.0 ×101 m –1 Therefore, the nodes in the standing waves formed by this electromagnetic wave are
separated by λ / 2 = 0.063 m .
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58. REASONING The polarizer, the insert and the analyzer in the setup in Figure 24.24a all
reduce the intensity of the light that reaches the photocell. The polarizer reduces the
intensity by a factor of onehalf, as described in Section 24.6 of the text; if the average
intensity of the incident unpolarized light is I , the average intensity of the polarized light
that leaves the polarizer and strikes the insert is S 0 = I / 2 . According to Malus' law (see
Equation 24.7), the average intensity S i nsert of the light leaving the insert is
Sinsert = S0 cos 2 θ , where θ is the relative angle between the transmission axes of the
polarizer and the insert. The intensity of the light...
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 Spring '13
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 Physics, The Lottery

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