9. (a) The rest length L0 = 130 m of the spaceship and its length L as measured by the
timing station are related by Eq. 37-13. Therefore,
L L0 1 (v / c) 2 130 m 1 0.740 87.4 m.
2
(b) The time interval for the passage of the spaceship is
t
87.4 m
L
3.94
4. Due to the time-dilation effect, the time between initial and final ages for the daughter
is longer than the four years experienced by her father:
tf daughter ti daughter = (4.000 y)
where is Lorentz factor (Eq. 37-8). Letting T denote the age of the f
19. (a) Using the notation of Sample Problem Pointillistic paintings use the diffraction
of your eye,
D
2(50 106 m)(1.5 103 m)
L
019 m .
.
122 d
.
122(650 109 m)
.
(b) The wavelength of the blue light is shorter so Lmax 1 will be larger.
20. Using the no
4. (a) Eq. 36-3 and Eq. 36-12 imply smaller angles for diffraction for smaller wavelengths.
This suggests that diffraction effects in general would decrease.
(b) Using Eq. 36-3 with m = 1 and solving for 2 (the angular width of the central
diffraction max
36. (a) On both sides of the soap is a medium with lower index (air) and we are
examining the reflected light, so the condition for strong reflection is Eq. 35-36. With
lengths in nm,
3360 for m = 0
1120 for m = 1
2n2L
for m = 2
672
=
= 480
1
for m = 3
14. (a) For the maximum adjacent to the central one, we set m = 1 in Eq. 35-14 and obtain
1
sin 1
0.010 rad.
m 1
100
m
d
1 sin 1
(b) Since y1 = D tan 1 (see Fig. 35-10(a), we obtain
y1 = (500 mm) tan (0.010 rad) = 5.0 mm.
The separation is y = y
90. (a) Now, the lens-film distance is
F 1 1 I F 1 1 IJ
iG J G
H f p K H 5.0 cm 100 cmK
1
1
5.3 cm.
(b) The change in the lens-film distance is 5.3 cm 5.0 cm = 0.30 cm.
1305
40. We use Eq. 34-8 (and Fig. 34-11(d) is useful), with n1 = 1.6 and n2 = 1 (using the
rounded-off value for air):
1.6 1 1 16
.
pi
r
Using the sign convention for r stated in the paragraph following Eq. 34-8 (so that
r 5.0 cm ), we obtain i = 2.4 cm for o
47. The law of refraction states
n1 sin 1 n2 sin 2 .
We take medium 1 to be the vacuum, with n1 = 1 and 1 = 32.0. Medium 2 is the glass,
with 2 = 21.0. We solve for n2:
n2 n1
FG
H
IJ
K
sin 32.0
sin 1
.
(100)
.
148.
sin 210
.
sin 2
51. (a) Approximating
33. Let I0 be the intensity of the unpolarized light that is incident on the first polarizing
1
sheet. The transmitted intensity is I1 2 I 0 , and the direction of polarization of the
transmitted light is 1 = 40 counterclockwise from the y axis in the dia
55. We use vS = r (with r = 0.600 m and = 15.0 rad/s) for the linear speed during
circular motion, and Eq. 17-47 for the Doppler effect (where f = 540 Hz, and v = 343 m/s
for the speed of sound).
(a) The lowest frequency is
v0
f f
526 Hz .
v vS
(b)
38. The frequency is f = 686 Hz and the speed of sound is vsound = 343 m/s. If L is the
length of the air-column, then using Eq. 1741, the water height is (in unit of meters)
h 1.00 L 1.00
nv
n(343)
1.00
(1.00 0.125n) m
4f
4(686)
where n = 1, 3, 5, wi
6. Let be the length of the rod. Then the time of travel for sound in air (speed vs) will
be ts / vs . And the time of travel for compressional waves in the rod (speed vr) will be
tr / vr . In these terms, the problem tells us that
1 1
ts tr 0.12s .
v s
18. The volume of a cylinder of height is V = r2 = d2 /4. The strings are long,
narrow cylinders, one of diameter d1 and the other of diameter d2 (and corresponding
linear densities 1 and 2). The mass is the (regular) density multiplied by the volume: m
=
4. The distance d between the beetle and the scorpion is related to the transverse speed vt
and longitudinal speed v as
d vt tt v t
where tt and t are the arrival times of the wave in the transverse and longitudinal
directions, respectively. With vt 50 m/