Physics For Scientists And Engineers 6E - part 311
Answers to Quick Quizzes
1241
aperture larger, relative to the light wavelength, increasing
the resolving power. Thus, we should choose a blue filter.
38.7 (b). The tracks of information on a compact disc are
much closer together than on a phonograph record. As
a result, the diffraction maxima from the compact disc
will be farther apart than those from the record.
38.8 (c). With the doubled wavelength, the pattern will be
wider. Choices (a) and (d) make the pattern even wider.
From Equation 38.10, we see that choice (b) causes sin !
to be twice as large. Because we cannot use the small
angle approximation, however, a doubling of sin ! is not
the same as a doubling of !, which would translate to a
doubling of the position of a maximum along the screen.
If we only consider small-angle maxima, choice (b) would
work, but it does not work in the large-angle case.
38.9 (b). Electric field vectors parallel to the metal wires cause
electrons in the metal to oscillate parallel to the wires.
Thus, the energy from the waves with these electric field
vectors is transferred to the metal by accelerating these
electrons and is eventually transformed to internal energy
through the resistance of the metal. Waves with electric-
field vectors perpendicular to the metal wires pass
through because they are not able to accelerate electrons
in the wires.
38.10 (c). At some intermediate distance, the light rays from
the fixtures will strike the floor at Brewster’s angle and
reflect to your eyes. Because this light is polarized hori-
zontally, it will not pass through your polarized sun-
glasses. Tilting your head to the side will cause the
reflections to reappear.
0.012
14.5
0.016
15.3
0.015
16.1
0.010
16.9
0.004 4
17.7
0.000 6
18.5
0.000 3
19.3
0.003
20.2
Answers to Quick Quizzes
38.1 (a). Equation 38.1 shows that a decrease in a results in an
increase in the angles at which the dark fringes appear.
38.2 (c). The space between the slightly open door and the
doorframe acts as a single slit. Sound waves have
wavelengths that are larger than the opening and so
are diffracted and spread throughout the room you are
in. Because light wavelengths are much smaller than the
slit width, they experience negligible diffraction. As a
result, you must have a direct line of sight to detect the
light waves.
38.3 The situation is like that depicted in Figure 38.11 except
that now the slits are only half as far apart. The diffrac-
tion pattern is the same, but the interference pattern is
stretched out because d is smaller. Because d/a " 3, the
m " 3 interference maximum coincides with the first
diffraction minimum. Your sketch should look like the
figure below.
π
I
/2
β
π
–
38.4 (c). In Equation 38.7, the ratio d/a is independent of
wavelength, so the number of interference fringes in the
central diffraction pattern peak remains the same. Equa-
tion 38.1 tells us that a decrease in wavelength causes a
decrease in the width of the central peak.
38.5 (b). The effective slit width in the vertical direction of
the cat’s eye is larger than that in the horizontal direc-
tion. Thus, the eye has more resolving power for lights
separated in the vertical direction and would be more
effective at resolving the mast lights on the boat.
38.6 (a). We would like to reduce the minimum angular sepa-
ration for two objects below the angle subtended by the
two stars in the binary system. We can do that by reducing
the wavelength of the light—this in essence makes the
©
2003 by Sidney Harris