chapter 37 Wave Optics
with index of refraction 1.38, calculate the fringe sepa-
ration for this same arrangement.
48. In the What If? section of Example 37.2, it was claimed
that overlapping fringes in a two-slit interference pat-
tern for two different wavelengths obey the following
relationship even for large values of the angle u:
(a) Prove this assertion. (b) Using the data in Example
37.2, find the nonzero value of y on the screen at which
the fringes from the two wavelengths first coincide.
49. An investigator finds a fiber at a crime scene that he
wishes to use as evidence against a suspect. He gives
the fiber to a technician to test the properties of the
fiber. To measure the diameter d of the fiber, the tech-
nician places it between two flat glass plates at their
ends as in Figure P37.37. When the plates, of length
14.0 cm, are illuminated from above with light of wave-
length 650 nm, she observes interference bands sepa-
rated by 0.580 mm. What is the diameter of the fiber?
50. Raise your hand and hold it flat. Think of the space
between your index finger and your middle finger as
one slit and think of the space between middle finger
and ring finger as a second slit. (a) Consider the inter-
ference resulting from sending coherent visible light
perpendicularly through this pair of openings. Com-
pute an order-of-magnitude estimate for the angle
between adjacent zones of constructive interference.
(b) To make the angles in the interference pattern easy
to measure with a plastic protractor, you should use an
electromagnetic wave with frequency of what order of
magnitude? (c) How is this wave classified on the elec-
51. Two coherent waves, coming from sources at different
locations, move along the x axis. Their wave functions
5860 sin c
5860 sin c
are in volts per meter, x
in nanometers, and t is in picoseconds. When the
two waves are superposed, determine the relation-
ship between x
that produces constructive
52. In a Young’s interference experiment, the two slits are
separated by 0.150 mm and the incident light includes
two wavelengths: l
5 540 nm (green) and l
5 450 nm
(blue). The overlapping interference patterns are
observed on a screen 1.40 m from the slits. Calculate
the minimum distance from the center of the screen
to a point where a bright fringe of the green light coin-
cides with a bright fringe of the blue light.
53. In a Young’s double-slit experiment using light of
wavelength l, a thin piece of Plexiglas having index of
refraction n covers one of the slits. If the center point
reflected light of wavelength 540 nm (in air) be miss-
ing? (b) Are there other values of t that will minimize
the reflected light at this wavelength? Explain.
41. Two glass plates 10.0 cm long are in contact at one
end and separated at the other end by a thread with a
diameter d 5 0.050 0 mm (Fig. P37.37). Light contain-
ing the two wavelengths 400 nm and 600 nm is inci-
dent perpendicularly and viewed by reflection. At what
distance from the contact point is the next dark fringe?
Section 37.6 the Michelson Interferometer
42. Mirror M
in Figure 37.13 is moved through a displace-
ment DL. During this displacement, 250 fringe rever-
sals (formation of successive dark or bright bands)
are counted. The light being used has a wavelength of
632.8nm. Calculate the displacement DL.
43. The Michelson interferometer can be used to mea-
sure the index of refraction of a gas by placing an
evacuated transparent tube in the light path along
one arm of the device. Fringe shifts occur as the gas
is slowly added to the tube. Assume 600-nm light is
used, the tube is 5.00 cm long, and 160 bright fringes
pass on the screen as the pressure of the gas in the
tube increases to atmospheric pressure. What is the
index of refraction of the gas? Hint: The fringe shifts
occur because the wavelength of the light changes
inside the gas-filled tube.
44. One leg of a Michelson interferometer contains an
evacuated cylinder of length L, having glass plates on
each end. A gas is slowly leaked into the cylinder until
a pressure of 1 atm is reached. If N bright fringes pass
on the screen during this process when light of wave-
length l is used, what is the index of refraction of the
gas? Hint: The fringe shifts occur because the wave-
length of the light changes inside the gas-filled tube.
45. Radio transmitter A operating at 60.0 MHz is 10.0 m
from another similar transmitter B that is 180° out of
phase with A. How far must an observer move from A
toward B along the line connecting the two transmit-
ters to reach the nearest point where the two beams
are in phase?
46. A room is 6.0 m long and 3.0 m wide. At the front of
the room, along one of the 3.0-m-wide walls, two loud-
speakers are set 1.0 m apart, with the center point
between them coinciding with the center point of the
wall. The speakers emit a sound wave of a single fre-
quency and a maximum in sound intensity is heard at
the center of the back wall, 6.0 m from the speakers.
What is the highest possible frequency of the sound
from the speakers if no other maxima are heard any-
where along the back wall?
47. In an experiment similar to that of Example 37.1,
green light with wavelength 560 nm, sent through
a pair of slits 30.0 mm apart, produces bright fringes
2.24 cm apart on a screen 1.20 m away. If the apparatus
is now submerged in a tank containing a sugar solution
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scope reveals a difference in index of refraction as a shift
in interference fringes. The idea is exemplified in the
following problem. An air wedge is formed between two
glass plates in contact along one edge and slightly sepa-
rated at the opposite edge as in Figure P37.37. When
the plates are illuminated with monochromatic light
from above, the reflected light has 85 dark fringes. Cal-
culate the number of dark fringes that appear if water
(n 5 1.33) replaces the air between the plates.
60. Consider the double-slit arrangement shown in Figure
P37.60, where the slit separation is d and the distance
from the slit to the screen is L. A sheet of transparent
plastic having an index of refraction n and thickness
t is placed over the upper slit. As a result, the central
maximum of the interference pattern moves upward a
distance y9. Find y9.
61. Figure P37.61 shows a radio-wave transmitter and a
receiver separated by a distance d 5 50.0 m and both a
distance h5 35.0 m above the ground. The receiver
can receive signals both directly from the transmitter
and indirectly from signals that reflect from the
ground. Assume the ground is level between the trans-
mitter and receiver and a 180° phase shift occurs upon
reflection. Determine the longest wavelengths that
interfere (a) constructively and (b)destructively.
Problems 61 and 62.
62. Figure P37.61 shows a radio-wave transmitter and a
receiver separated by a distance d and both a distance
h above the ground. The receiver can receive signals
both directly from the transmitter and indirectly from
signals that reflect from the ground. Assume the
ground is level between the transmitter and receiver
and a 180° phase shift occurs upon reflection. Deter-
mine the longest wavelengths that interfere (a) con-
structively and (b) destructively.
63. In a Newton’s-rings experiment, a plano-convex glass
(n 5 1.52) lens having radius r 5 5.00 cm is placed on a
flat plate as shown in Figure P37.63 (page 1158). When
on the screen is a dark spot instead of a bright spot,
what is the minimum thickness of the Plexiglas?
54. Review. A flat piece of glass is held stationary and
horizontal above the highly polished, flat top end of a
10.0-cm-long vertical metal rod that has its lower end
rigidly fixed. The thin film of air between the rod and
glass is observed to be bright by reflected light when
it is illuminated by light of wavelength 500 nm. As the
temperature is slowly increased by 25.0°C, the film
changes from bright to dark and back to bright 200
times. What is the coefficient of linear expansion of
55. A certain grade of crude oil has an index of refrac-
tion of 1.25. A ship accidentally spills 1.00 m3 of this
oil into the ocean, and the oil spreads into a thin, uni-
form slick. If the film produces a first-order maximum
of light of wavelength 500 nm normally incident on it,
how much surface area of the ocean does the oil slick
cover? Assume the index of refraction of the ocean
water is 1.34.
56. The waves from a radio station can reach a home
receiver by two paths. One is a straight-line path from
transmitter to home, a distance of 30.0 km. The sec-
ond is by reflection from the ionosphere (a layer of
ionized air molecules high in the atmosphere). Assume
this reflection takes place at a point midway between
receiver and transmitter, the wavelength broadcast by
the radio station is 350 m, and no phase change occurs
on reflection. Find the minimum height of the iono-
spheric layer that could produce destructive interfer-
ence between the direct and reflected beams.
57. Interference effects are produced at point P on a
screen as a result of direct rays from a 500-nm source
and reflected rays from the mirror as shown in Figure
P37.57. Assume the source is 100 m to the left of the
screen and 1.00 cm above the mirror. Find the distance
y to the first dark band above the mirror.
58. Measurements are made of the intensity distribution
within the central bright fringe in a Young’s interfer-
ence pattern (see Fig. 37.6). At a particular value of y, it
is found that I/I
5 0.810 when 600-nm light is used.
What wavelength of light should be used to reduce the
relative intensity at the same location to 64.0% of the
59. Many cells are transparent and colorless. Structures of
great interest in biology and medicine can be practi-
cally invisible to ordinary microscopy. To indicate the
size and shape of cell structures, an interference micro-
chapter 37 Wave Optics
at the center that is surrounded by 50 dark rings, the
largest of which is at the outer edge of the lens.
(a) What is the thickness of the air layer at the center
of the interference pattern? (b) Calculate the radius
of the outermost dark ring. (c) Find the focal length of
66. A plano-convex lens has index of refraction n. The
curved side of the lens has radius of curvature R and
rests on a flat glass surface of the same index of refrac-
tion, with a film of index n
between them, as shown
in Figure 37.66. The lens is illuminated from above
by light of wavelength l. Show that the dark Newton’s
rings have radii given approximately by
where r ,, R and m is an integer.
67. Interference fringes are produced using Lloyd’s mirror
and a source S of wavelength l 5 606 nm as shown in
Figure P37.67. Fringes separated by Dy 5 1.20 mm are
formed on a screen a distance L 5 2.00 m from the
source. Find the vertical distance h of the source above
the reflecting surface.
68. The quantity nt in Equations 37.17 and 37.18 is called
the optical path length corresponding to the geometrical
distance t and is analogous to the quantity d in Equa-
tion 37.1, the path difference. The optical path length
is proportional to n because a larger index of refrac-
tion shortens the wavelength, so more cycles of a wave
fit into a particular geometrical distance. (a) Assume a
mixture of corn syrup and water is prepared in a tank,
with its index of refraction n increasing uniformly
from 1.33 at y 5 20.0 cm at the top to 1.90 at y 5 0.
Write the index of refraction n(y) as a function of y.
light of wavelength l 5 650 nm is incident normally,
55 bright rings are observed, with the last one precisely
on the edge of the lens. (a) What is the radius R of cur-
vature of the convex surface of the lens? (b) What is
the focal length of the lens?
64. Why is the following situation impossible? A piece of trans-
parent material having an index of refraction n 5 1.50
is cut into the shape of a wedge as shown in Figure
P37.64. Both the top and bottom surfaces of the wedge
are in contact with air. Monochromatic light of wave-
length l 5 632.8 nm is normally incident from above,
and the wedge is viewed from above. Let h 5 1.00 mm
represent the height of the wedge and , 5 0.500 m its
length. A thin-film interference pattern appears in the
wedge due to reflection from the top and bottom sur-
faces. You have been given the task of counting the
number of bright fringes that appear in the entire
length , of the wedge. You find this task tedious, and
your concentration is broken by a noisy distraction
after accurately counting 5 000 bright fringes.
65. A plano-concave lens having index of refraction 1.50 is
placed on a flat glass plate as shown in Figure P37.65.
Its curved surface, with radius of curvature 8.00m, is
on the bottom. The lens is illuminated from above
with yellow sodium light of wavelength 589 nm, and a
series of concentric bright and dark rings is observed
by reflection. The interference pattern has a dark spot
73. Both sides of a uniform film that has index of refraction
n and thickness d are in contact with air. For normal
incidence of light, an intensity minimum is observed in
the reflected light at l
and an intensity maximum is
observed at l
, where l
. (a) Assuming no inten-
sity minima are observed between l
, find an
expression for the integer m in Equations 37.17 and
37.18 in terms of the wavelengths l
. (b) Assum-
ing n 5 1.40, l
5 500nm, and l
5 370 nm, determine
the best estimate for the thickness of the film.
74. Slit 1 of a double slit is wider than slit 2 so that the light
from slit 1 has an amplitude 3.00 times that of the light
from slit 2. Show that Equation 37.13 is replaced by the
equation I 5 I
(1 1 3 cos2 f/2) for this situation.
75. Monochromatic light of wavelength 620 nm passes
through a very narrow slit S and then strikes a screen
in which are two parallel slits, S
, as shown in
Figure P37.75. Slit S
is directly in line with S and at a
distance of L 5 1.20 m away from S, whereas S
placed a distance d to one side. The light is detected at
point P on a second screen, equidistant from S
When either slit S
is open, equal light intensities
are measured at point P. When both slits are open, the
intensity is three times larger. Find the minimum pos-
sible value for the slit separation d.
76. A plano-convex lens having a radius of curvature of r 5
4.00 m is placed on a concave glass surface whose
radius of curvature is R 5 12.0 m as shown in Figure
P37.76. Assuming 500-nm light is incident normal to
the flat surface of the lens, determine the radius of the
100th bright ring.
(b) Compute the optical path length corresponding to
the 20.0-cm height of the tank by calculating
(c) Suppose a narrow beam of light is directed into
the mixture at a nonzero angle with respect to the
normal to the surface of the mixture. Qualitatively
describe its path.
69. Astronomers observe a 60.0-MHz radio source both
directly and by reflection from the sea as shown in
Figure P37.17. If the receiving dish is 20.0 m above sea
level, what is the angle of the radio source above the
horizon at first maximum?
70. Figure CQ37.2 shows an unbroken soap film in a cir-
cular frame. The film thickness increases from top to
bottom, slowly at first and then rapidly. As a simpler
model, consider a soap film (n 5 1.33) contained
within a rectangular wire frame. The frame is held ver-
tically so that the film drains downward and forms a
wedge with flat faces. The thickness of the film at the
top is essentially zero. The film is viewed in reflected
white light with near-normal incidence, and the first
violet (l 5 420 nm) interference band is observed
3.00 cm from the top edge of the film. (a) Locate the
first red (l 5 680 nm) interference band. (b) Deter-
mine the film thickness at the positions of the violet
and red bands. (c) What is the wedge angle of the film?
71. Our discussion of the techniques for determining
constructive and destructive interference by reflec-
tion from a thin film in air has been confined to rays
striking the film at nearly normal incidence. What If?
Assume a ray is incident at an angle of 30.0° (relative
to the normal) on a film with index of refraction 1.38
surrounded by vacuum. Calculate the minimum thick-
ness for constructive interference of sodium light with
a wavelength of 590 nm.
72. The condition for constructive interference by reflec-
tion from a thin film in air as developed in Section
37.5 assumes nearly normal incidence. What If? Sup-
pose the light is incident on the film at a nonzero angle
(relative to the normal). The index of refraction of
the film is n, and the film is surrounded by vacuum.
Find the condition for constructive interference that
relates the thickness t of the film, the index of refrac-
tion n of the film, the wavelength l of the light, and
the angle of incidence u
When plane light waves pass through a small aperture in an opaque barrier, the aper-
ture acts as if it were a point source of light, with waves entering the shadow region behind
the barrier. This phenomenon, known as diffraction, was first mentioned in Section 35.3,
and can be described only with a wave model for light. In this chapter, we investigate the
features of the diffraction pattern that occurs when the light from the aperture is allowed
to fall upon a screen.
In Chapter 34, we learned that electromagnetic waves are transverse. That is, the elec-
tric and magnetic field vectors associated with electromagnetic waves are perpendicular to
the direction of wave propagation. In this chapter, we show that under certain conditions
these transverse waves with electric field vectors in all possible transverse directions can be
polarized in various ways. In other words, only certain directions of the electric field vectors
are present in the polarized wave.
38.1 Introduction to Diffraction Patterns
In Sections 35.3 and 37.1, we discussed that light of wavelength comparable to or
larger than the width of a slit spreads out in all forward directions upon passing
through the slit. This phenomenon is called diffraction. When light passes through
a narrow slit, it spreads beyond the narrow path defined by the slit into regions that
would be in shadow if light traveled in straight lines. Other waves, such as sound
waves and water waves, also have this property of spreading when passing through
apertures or by sharp edges.
38.1 Introduction to Diffraction
38.2 Diffraction Patterns from
38.3 Resolution of Single-Slit
and Circular Apertures
38.4 The Diffraction Grating
38.5 Diffraction of X-Rays
38.6 Polarization of Light Waves
c h a p p t t e r
The Hubble Space Telescope does
its viewing above the atmosphere
and does not suffer from the
atmospheric blurring, caused by air
turbulence, that plagues ground-
based telescopes. Despite this
advantage, it does have limitations
due to diffraction effects. In this
chapter, we show how the wave
nature of light limits the ability of
any optical system to distinguish
between closely spaced objects.
(NASA Hubble Space Telescope Collection)
38.2 Diffraction patterns from Narrow Slits
You might expect that the light passing through a small opening would simply
result in a broad region of light on a screen due to the spreading of the light as it
passes through the opening. We find something more interesting, however. A dif-
fraction pattern consisting of light and dark areas is observed, somewhat similar
to the interference patterns discussed earlier. For example, when a narrow slit is
placed between a distant light source (or a laser beam) and a screen, the light pro-
duces a diffraction pattern like that shown in Figure 38.1. The pattern consists of
a broad, intense central band (called the central maximum) flanked by a series of
narrower, less intense additional bands (called side maxima or secondary maxima)
and a series of intervening dark bands (or minima). Figure 38.2 shows a diffraction
pattern associated with light passing by the edge of an object. Again we see bright
and dark fringes, which is reminiscent of an interference pattern.
Figure 38.3 shows a diffraction pattern associated with the shadow of a penny.
A bright spot occurs at the center, and circular fringes extend outward from the
shadow’s edge. We can explain the central bright spot by using the wave theory of
light, which predicts constructive interference at this point. From the viewpoint
of ray optics (in which light is viewed as rays traveling in straight lines), we expect
the center of the shadow to be dark because that part of the viewing screen is com-
pletely shielded by the penny.
Shortly before the central bright spot was first observed, one of the supporters
of ray optics, Simeon Poisson, argued that if Augustin Fresnel’s wave theory of light
were valid, a central bright spot should be observed in the shadow of a circular
object illuminated by a point source of light. To Poisson’s astonishment, the spot
was observed by Dominique Arago shortly thereafter. Therefore, Poisson’s predic-
tion reinforced the wave theory rather than disproving it.
38.2 Diffraction Patterns from Narrow Slits
Let’s consider a common situation, that of light passing through a narrow open-
ing modeled as a slit and projected onto a screen. To simplify our analysis, we
assume the observing screen is far from the slit and the rays reaching the screen
are approximately parallel. (This situation can also be achieved experimentally by
using a converging lens to focus the parallel rays on a nearby screen.) In this model,
the pattern on the screen is called a Fraunhofer diffraction pattern.1
Figure 38.4a (page 1162) shows light entering a single slit from the left and dif-
fracting as it propagates toward a screen. Figure 38.4b shows the fringe structure of
The diffraction pat-
tern that appears on a screen when
light passes through a narrow vertical
slit. The pattern consists of a broad
central fringe and a series of less
intense and narrower side fringes.
created by the illumination of a
penny, with the penny positioned
midway between the screen and
Notice the bright spot at
Light from a small source passes by the edge of an
opaque object and continues on to a screen. A diffraction pattern
consisting of bright and dark fringes appears on the screen in the
region above the edge of the object.
1If the screen is brought close to the slit (and no lens is used), the pattern is a Fresnel diffraction pattern. The Fresnel
pattern is more difficult to analyze, so we shall restrict our discussion to Fraunhofer diffraction.
chapter 38 Diffraction patterns and polarization
a Fraunhofer diffraction pattern. A bright fringe is observed along the axis at u 5 0,
with alternating dark and bright fringes on each side of the central bright fringe.
Until now, we have assumed slits are point sources of light. In this section, we
abandon that assumption and see how the finite width of slits is the basis for under-
standing Fraunhofer diffraction. We can explain some important features of this
phenomenon by examining waves coming from various portions of the slit as shown
in Figure 38.5. According to Huygens’s principle, each portion of the slit acts as a
source of light waves. Hence, light from one portion of the slit can interfere with
light from another portion, and the resultant light intensity on a viewing screen
depends on the direction u. Based on this analysis, we recognize that a diffraction
pattern is actually an interference pattern in which the different sources of light are
different portions of the single slit! Therefore, the diffraction patterns we discuss
in this chapter are applications of the waves in interference analysis model.
To analyze the diffraction pattern, let’s divide the slit into two halves as shown in
Figure 38.5. Keeping in mind that all the waves are in phase as they leave the slit,
consider rays 1 and 3. As these two rays travel toward a viewing screen far to the
right of the figure, ray 1 travels farther than ray 3 by an amount equal to the path
difference (a/2) sin u, where a is the width of the slit. Similarly, the path difference
between rays 2 and 4 is also (a/2) sin u, as is that between rays 3 and 5. If this path
difference is exactly half a wavelength (corresponding to a phase difference of 180°),
the pairs of waves cancel each other and destructive interference results. This cancel-
lation occurs for any two rays that originate at points separated by half the slit width
because the phase difference between two such points is 180°. Therefore, waves from
the upper half of the slit interfere destructively with waves from the lower half when
or, if we consider waves at angle u both above the dashed line in Figure 38.5 and
sin u5 6
Dividing the slit into four equal parts and using similar reasoning, we find that
the viewing screen is also dark when
Likewise, dividing the slit into six equal parts shows that darkness occurs on the
Pitfall Prevention 38.1
Diffraction Versus Diffraction
Pattern Diffraction refers to the
general behavior of waves spread-
ing out as they pass through a slit.
We used diffraction in explaining
the existence of an interference
pattern in Chapter 37. A diffraction
pattern is actually a misnomer, but
is deeply entrenched in the lan-
guage of physics. The diffraction
pattern seen on a screen when a
single slit is illuminated is actually
another interference pattern. The
interference is between parts of
the incident light illuminating dif-
ferent regions of the slit.
(a) Geometry for
analyzing the Fraunhofer diffrac-
tion pattern of a single slit. (Draw-
ing not to scale.) (b) Simulation
of a single-slit Fraunhofer diffrac-
The pattern consists of a
central bright fringe flanked
by much weaker maxima
alternating with dark fringes.
Each portion of the slit acts as
a point source of light waves.
The path difference between
rays 1 and 3, rays 2 and 4, or
rays 3 and 5 is (a/2)
Paths of light rays
that encounter a narrow slit of
width a and diffract toward a
screen in the direction described
by angle u (not to scale).
38.2 Diffraction patterns from Narrow Slits
Therefore, the general condition for destructive interference is
m561, 62, 63,c
This equation gives the values of u
for which the diffraction pattern has zero
light intensity, that is, when a dark fringe is formed. It tells us nothing, however,
about the variation in light intensity along the screen. The general features of the
intensity distribution are shown in Figure 38.4. A broad, central bright fringe is
observed; this fringe is flanked by much weaker bright fringes alternating with
dark fringes. The various dark fringes occur at the values of u
that satisfy Equa-
tion 38.1. Each bright-fringe peak lies approximately halfway between its bordering
dark-fringe minima. Notice that the central bright maximum is twice as wide as the
secondary maxima. There is no central dark fringe, represented by the absence of
m 5 0 in Equation 38.1.
Q uick Quiz 38.1 Suppose the slit width in Figure 38.4 is made half as wide. Does
the central bright fringe (a) become wider, (b) remain the same, or (c) become
WW Condition for destructive
interference for a single slit
Pitfall Prevention 38.2
Similar Equation Warning! Equa-
tion 38.1 has exactly the same
form as Equation 37.2, with d,
the slit separation, used in Equa-
tion 37.2 and a, the slit width,
used in Equation 38.1. Equation
37.2, however, describes the bright
regions in a two-slit interference
pattern, whereas Equation 38.1
describes the dark regions in a
single-slit diffraction pattern.
Example 38.1 Where Are the Dark Fringes?
Light of wavelength 580 nm is incident on a slit having a width of 0.300 mm. The viewing screen is 2.00 m from the slit.
Find the width of the central bright fringe.
Conceptualize Based on the problem statement, we imagine a single-slit diffraction pattern similar to that in Figure 38.4.
Categorize We categorize this example as a straightforward application of our discussion of single-slit diffraction pat-
terns, which comes from the waves in interference analysis model.
Analyze Evaluate Equation 38.1 for the two dark
fringes that flank the central bright fringe, which
correspond to m 5 61:
Let y represent the vertical position along the viewing screen in Figure 38.4a, measured from the point on the screen
directly behind the slit. Then, tan u
/L, where the subscript 1 refers to the first dark fringe. Because u
small, we can use the approximation sinu
< tan u
; therefore, y
5 L sin u
The width of the central bright fringe is
twice the absolute value of y
L sin u
5 7.73 3 1023 m 5
Finalize Notice that this value is much greater than the width of the slit. Let’s explore below what happens if we
change the slit width.
What if the slit width is increased by an order of magnitude to 3.00 mm? What happens to the diffraction
Answer Based on Equation 38.1, we expect that the angles at which the dark bands appear will decrease as a increases.
Therefore, the diffraction pattern narrows.
Repeat the calculation with
the larger slit width:
Notice that this result is smaller than the width of the slit. In general, for large values of a, the various maxima and min-
ima are so closely spaced that only a large, central bright area resembling the geometric image of the slit is observed.
This concept is very important in the performance of optical instruments such as telescopes.
chapter 38 Diffraction patterns and polarization
Intensity of Single-Slit Diffraction Patterns
Analysis of the intensity variation in a diffraction pattern from a single slit of width
a shows that the intensity is given by
pa sin u/l
pa sin u/l
is the intensity at u 5 0 (the central maximum) and l is the wavelength
of light used to illuminate the slit. This expression shows that minima occur when
pa sin u
m561, 62, 63,c
in agreement with Equation 38.1.
Figure 38.6a represents a plot of the intensity in the single-slit pattern as given by
Equation 38.2, and Figure 38.6b is a simulation of a single-slit Fraunhofer diffrac-
tion pattern. Notice that most of the light intensity is concentrated in the central
Intensity of Two-Slit Diffraction Patterns
When more than one slit is present, we must consider not only diffraction patterns
due to the individual slits but also the interference patterns due to the waves com-
ing from different slits. Notice the curved dashed lines in Figure 37.7 in Chapter
37, which indicate a decrease in intensity of the interference maxima as u increases.
This decrease is due to a diffraction pattern. The interference patterns in that fig-
ure are located entirely within the central bright fringe of the diffraction pattern,
so the only hint of the diffraction pattern we see is the falloff in intensity toward
the outside of the pattern. To determine the effects of both two-slit interference
and a single-slit diffraction pattern from each slit from a wider viewpoint than that
in Figure 37.7, we combine Equations 37.14 and 38.2:
pd sin u
pa sin u/l
pa sin u/l
Although this expression looks complicated, it merely represents the single-slit dif-
fraction pattern (the factor in square brackets) acting as an “envelope” for a two-slit
interference pattern (the cosine-squared factor) as shown in Figure 38.7. The broken
intensity of a single-slit
Condition for intensity
minima for a single slit
(a) A plot of light
intensity I versus (p/l)a sin u for
the single-slit Fraunhofer diffrac-
tion pattern. (b) Simulation of a
single-slit Fraunhofer diffraction
A minimum in the curve in a
corresponds to a dark fringe in b .
Documents you may be interested
Documents you may be interested