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Unformatted text preview: ting the calculation for the third order maximum ( m = 3 ), we find that
( m = 3)
for violet
for red θ = 24 °
θ = 41° d. Comparisons of the values for θ calculated in parts (a), (b) and (c) show that the
second and third orders overlap . 62. REASONING Initially, the field mice are too close together to be resolved as separate
objects. But as the eagle gets closer, the angle θ between the field mice increases. When it
reaches the value θmin = 1.22 λ (Equation 27.6), where λ is the wavelength of the light that
D
the eagle’s eye detects and D is the diameter of the eagle’s pupil, the eagle will be able to
resolve them. We can also express the angle θmin in terms of the distance s between the field
s
mice and the distance r between the eagle and the mice via θ min = (Equation 8.1). Let r0
r
denote the distance between the eagle and the mice at the instant when the eagle begins to
dive. Then the time t it takes the eagle to travel a distance r0 − r toward the mice at a
constant speed v is given by Equation 2.1: t= SOLUTION r0 − r (2.1) v Setting the right sides of θmin = 1.22 λ
D (Equation 27.6) and θ min = s
r (Equation 8.1) equal and solving for r yields θmin = 1.22 λ
D = s
r or 1.22rλ = sD Substituting Equation (1) into Equation 2.1, we obtain or r= sD
1.22λ (1) Chapter 27 Problems t= r0 − r
v ( 0.010 m ) ( 6.0 ×10−3 m )
sD
176 m −
1.22 ( 550 ×10−9 m ) = 5.1 s
1.22λ =
v
17 m/s r0 − = 1473 63. REASONING The angle θ that specifies the location of the m dark fringe is given by
sin θ = mλ / W (Equation 27.4), where λ is the wavelength of the light and W is the width of
the slit. When θ has its maximum value of 90.0°, the number of dark fringes that can be
produced is a maximum. We will use this information to obtain a value for this number.
th SOLUTION Solving Equation 27.4 for m, and setting θ = 90.0°, we have
m= W sin 90.0° ( 5.47 ×10−6 m ) sin 90.0° = 8.40
=
651×10−9 m λ Therefore, the number of dark fringes is 8 . 64. REASONING The number m of bright fringes in an air wedge is discussed in Example 5,
where it is shown that m is related to the thickness t of the film and the wavelength λfilm of
the light within the film by (see Figure 27.12) 2
3t Extra distance
traveled by
wave 2 + 1
2 λ film 123 = mλ film where m = 1, 2, 3, ... Half wavelength
shift due to reflection
of wave 1 For a given thickness, we can solve this equation for the number of bright fringes.
SOLUTION Since the light is traveling in a film of air, λfilm = λvacuum = 550 nm. The number
m of bright fringes is
m= 2t λ vacuum c h 2 1.37 × 10 −5 m
1
1
+=
+ = 50 .3
−9
2
2
550 × 10 m Thus, there are 50 bright fringes . 1474 INTERFERENCE AND THE WAVE NATURE OF LIGHT 65. REASONING AND SOLUTION
a. The angular positions of the specified orders are equal, so λ /dA = 2λ /dB, or
dB = dA 2 b. Similarly, we have for the mA order of grating A and the mB order of grating B that
mAλ /dA = mBλ /dB, so mA = mB/2.
The next highest orders which overlap are mB = 4 , mA = 2 mB = and 6 , mA = 3 66. REASONING AND SOLUTION
a. The condition for constructive interference is
2t = mλfilm = m(λvacuum/nwater) For λvacuum = 432 nm, F
G
H I
J
K F
G
H I
J
K t= m 432 nm
2
1.33 t= m ′ 648 nm
2
1.33 For λvacuum = 648 nm, Equating the two expressions for t, we find that m/m' = 1.50. For minimum thickness, this
means that m = 3 and m ′ = 2 . Then
t= F
G
H m 432 mn
2
1.33 432
I = 3 F nm I =
J 2 G1.33 J
KH K 487 nm b. The condition for destructive interference is 2t = (m + 1/2) (λvacuum/nwater). Thus, λ vacuum = 2 n water t
m+ 1
2 = bg
c 2 1.33 487 × 10 −9 m
m+ h 1
2 The wavelength lying within the specified range is for m = 2, so that λvacuum = 518 nm CHAPTER 28 SPECIAL RELATIVITY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (e) In each of the situations in answers ad, the person and the frame of reference is subject to
an acceleration. In an accelerated reference frame Newton’s law of inertia is not valid, so the
reference frame is not an inertial reference frame.
2. (a) The worker measures the proper time, because he is at rest with respect to the light and
views the flashes as occurring at the same place.
3. 1.89 s
4. (d) To see the proper length of an object, an observer must be at rest with respect to the two
points defining that length. Observers in either spacecraft see the other spacecraft as moving.
Therefore, neither the observers in spacecraft A nor those in spacecraft B see the proper
length of the other spacecraft.
5. 8.19 lightyears
6. (b) The runner sees home plate move away from his feet and first base arrive at his feet.
Thus, the runner sees both events occurring at the same place and measures the proper time.
The catcher is the one at rest with respect to home plate and first base. Therefore, he
measures the proper length between the two points.
7. (c) According to the theory of special relativity, the equations apply when both observers
have constant velocities with respect to an inertial reference...
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 Spring '13
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 Physics, The Lottery

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