Physics for Scientists and Engineers, 6e
Chapter 38 - Diffraction Patterns and
Polarization
Suppose the slit width in the figure below is made
half as wide. The central bright fringe
1
1.
becomes wider
2.
remains the same
3.
becomes narrower
2
3
4
5
33%
1
33%
2
33%
3
Equation 38.1 shows that a decrease in a
results in an increase in the angles at which
the dark fringes appear.
If a classroom door is open slightly, you can hear
sounds coming from the hallway. Yet you cannot see
what is happening in the hallway. Why is there this
difference?
1.
1
Light waves do not diffract
through the single slit of the
open doorway.
2.
Sound waves can pass through
the walls, but light waves
cannot.
3.
The open door is a small slit for
sound waves, but a large slit
for light waves.
4.
The open door is a large slit for
sound waves, but a small slit
for light waves.
2
3
4
5
25% 25% 25% 25%
1
2
3
4
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.
Consider the central peak in the diffraction envelope in the
figure of question 3. Suppose the wavelength of the light is
changed to 450 nm. What happens to this central peak?
1.
The width of the peak decreases and the number of
interference fringes it encloses decreases.
2.
The width of the peak decreases and the number of
interference fringes it encloses increases.
3.
The width of the peak decreases and the number of
interference fringes it encloses remains the same.
4.
The width of the peak increases and the number of
interference fringes it encloses decreases.
5.
The width of the peak increases and the number of
interference fringes it encloses increases.
6.
The width of the peak increases and the number of
interference fringes it encloses remains the same.
1
2
3
4
5
17% 17% 17% 17% 17% 17%
1
2
3
4
5
6
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. Equation 38.1 tells us that a decrease in
wavelength causes a decrease in the width of the
central peak.
Cat’s eyes have pupils that can be modeled as
vertical slits. At night, cats would be more
successful in resolving
1.
2.
1
2
headlights on a distant
car
50%
50%
vertically-separated lights
on the mast of a distant
boat
3
4
5
1
2
The effective slit width in the vertical direction
of the cat’s eye is larger than that in the
horizontal direction. 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.
Suppose you are observing a binary star with a
telescope and are having difficulty resolving the two
stars. You decide to use a colored filter to maximize
the resolution. (A filter of a given color transmits
only that color of light.) What color filter should you
choose?
25% 25% 25% 25%
1.
blue
2.
green
3.
yellow
4.
red
1
2
3
4
5
1
2
3
4
We would like to reduce the minimum angular
separation 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 aperture larger, relative to the light
wavelength, increasing the resolving power.
Thus, we should choose a blue filter.
If laser light is reflected from a phonograph record or a
compact disc, a diffraction pattern appears. This is due to the
fact that both devices contain parallel tracks of information that
act as a reflection diffraction grating. Which device results in
diffraction maxima that are farther apart in angle?
50%
1.
record
2.
compact disc
1
2
3
4
5
1
50%
2
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.
A polarizer for microwaves can be made as a grid of
parallel metal wires about a centimeter apart. The
electric field vector for microwaves transmitted
through this polarizer is _____ to the metal wires.
50%
1
1.
parallel
2.
perpendicular
2
3
4
5
1
50%
2
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.
You are walking down a long hallway that has many light
fixtures in the ceiling and a very shiny, newly waxed floor. In
the floor, you see reflections of every light fixture. Now you put
on sunglasses that are polarized. Some of the reflections of
the light fixtures can no longer be seen. (Try this!) The
reflections that disappear are those
1
1.
nearest to you
2.
farthest from you
3.
at an intermediate
distance from you
2
3
4
5
33%
1
33%
2
33%
3
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 horizontally, it
will not pass through your polarized
sunglasses. Tilting your head to the side will
cause the reflections to reappear.