Solucionario Purcell

38
Diffraction Patterns and Polarization
CHAPTER OUTLINE
38.1
38.2
38.3
38.4
38.5
38.6

Introduction to Diffraction
Patterns
Diffraction Patterns from
Narrow Slits
Resolution of Single-Slit and
Circular Apertures
The Diffraction Grating
Diffraction of X-Rays by
Crystals
Polarization of Light Waves

ANSWERS TO QUESTIONS
Q38.1

Audible sound has wavelengths on the order ofmeters or
centimeters, while visible light has a wavelength on the order
of half a micrometer. In this world of breadbox-sized objects,
is large for sound, and sound diffracts around behind walls

λ
a

λ
is a tiny fraction for visible light passing
a
ordinary-size objects or apertures, so light changes its direction
by only very small angles when it diffracts.
Another way of phrasingthe answer: We can see by a
small angle around a small obstacle or around the edge of a
small opening. The side fringes in Figure 38.1 and the Arago
spot in the center of Figure 38.3 show this diffraction. We
cannot always hear around corners. Out-of-doors, away from
reflecting surfaces, have someone a few meters distant face
away from you and whisper. The high-frequency, shortwavelength,information-carrying components of the sound do
not diffract around his head enough for you to understand his
words.
with doorways. But

Q38.2

The wavelength of light is extremely small in comparison to the dimensions of your hand, so the
diffraction of light around an obstacle the size of your hand is totally negligible. However, sound
waves have wavelengths that are comparable to thedimensions of the hand or even larger.
Therefore, significant diffraction of sound waves occurs around hand-sized obstacles.

Q38.3

If you are using an extended light source, the gray
area at the edge of the shadow is the penumbra. A
bug looking up from there would see the light
source partly but not entirely blocked by the book.
If you use a point source of light, hold it and the
bookmotionless, and look at very small angles out
from the geometrical edge of the shadow, you may
see a series of bright and dark bands produced by
diffraction of light at the straight edge, as shown in
the diagram.
FIG. Q38.3

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Diffraction Patterns and Polarization

3 × 10 8 m s

Q38.4

An AM radio wave has wavelength on the order of

Q38.5

The intensity of the lightcoming through the slit decreases, as you would expect. The central

~ 300 m . This is large compared to
1 × 10 6 s −1
the width of the mouth of a tunnel, so the AM radio waves can reflect from the surrounding ground
as if the hole were not there. (In the same way, a metal screen forming the dish of a radio telescope
can reflect radio waves as if it were solid, and a hole-riddled screen in thedoor of a microwave oven
keeps the microwaves inside.) The wave does not “see” the hole. Very little of the radio wave energy
enters the tunnel, and the AM radio signal fades. An FM radio wave has wavelength a hundred
times smaller, on the order of a few meters. This is smaller than the size of the tunnel opening, so
the wave can readily enter the opening. (On the other hand, the longwavelength of AM radio
waves lets them diffract more around obstacles. Long-wavelength waves can change direction more
in passing hills or large buildings, so in some experiments FM fades more than AM.)

λ
for
a
destructive interference on each side of the central maximum, θ increases as a decreases.
maximum increases in width as the width of the slit decreases. In the condition sin θ =

Q38.6It is shown in the correct orientation. If the horizontal width of the opening is equal to or less than
the wavelength of the sound, then the equation a sin θ = 1 λ has the solution θ = 90° , or has no
solution. The central diffraction maximum covers the whole seaward side. If the vertical height of
the opening is large compared to the wavelength, then the angle in a sin θ = 1 λ will be...