Chapter 32 Physical Optics
In making some experiments on the fringes of colors
accompanying shadows, I have found so simple and so
demonstrative a proof of the general law of the
interference of two portions of light, which I have
already endeavored to establish, that I think it right
to lay before the Royal Society a short statement of
the facts, which appear to me to be thus decisive.”
Thomas Young
32.1 Introduction
In chapter 29, we saw that light is an electromagnetic wave, and the equation for
the electric portion of the plane wave was given by equation 29.38 as
E = E0 sin(kx − ωt)
(29.38)
In chapter 30, using the fact that light is a wave, we used Huygens’ principle to
derive the law of reflection. In chapter 31, we again used the wave nature of light to
derive the law of refraction by another application of Huygens’ principle. However,
once these two laws were developed, we treated light as a ray and explained a large
number of optical phenomena by geometrical optics. However, we need to ask
ourselves, is there any limit to the use of geometrical optics? Is it ever necessary to
treat light strictly from its wave nature? The answer to each of these questions is
yes. To discuss the experimentally observed phenomena of interference and
diffraction, we need to treat light as a wave. The study of light from its wave nature
is called physical optics, or sometimes wave optics.
To see at what point light must be treated as a wave, let us consider a series
of plane waves and their associated light rays, as they impinge on the openings in
figure 32.1. The light rays are shown perpendicular to the wave fronts, as discussed
Figure 32.1 Transition from geometrical optics to physical optics.
32-1
Chapter 32 Physical Optics
in chapter 30. When the opening is very large compared to the wavelength of light,
that is, when a ! λ, figure 32.1(a), the light passes through the opening as a series of
geometrical rays and a sharp image of the opening is found on the screen. This is the
region in which light can be treated from the point of view of geometrical optics.
Because the wavelength of red light, the longest of the visible spectrum, is
approximately 720.0 nm or 7.2 × 10−5 cm, the size of the opening a does not have to
be very large, from a practical viewpoint, to be very much larger than the
wavelength λ. This is why geometrical optics gives such a good description of so
many optical problems.
When the size of the opening is made smaller and smaller, until the slit
opening a becomes the same order of magnitude as the wavelength of light λ, a
breakdown in geometrical optics occurs. Instead of the image on the screen becoming
smaller and sharper as a approaches the size of the wavelength λ, the image on the
screen starts to become larger, and the edges fuzzier, as shown in figure 32.1(b).
Instead of light propagating in the straight line motion of geometrical optics, the
rays of light bend into the region that would normally be considered as a region of
shadow. This bending of light around an obstacle, into the region that should be a
shadow area, is called diffraction. The obstacle in this case is the edge of the
opening. We should note here that diffraction is very much different from refraction,
which is the bending of light as it goes from one medium into another. In the case of
diffraction there is no different medium. The bending occurs in the same medium.
If the slit opening is made smaller still, until the opening a is less than the
wavelength λ, then the diffraction of the light wave becomes even greater, as seen in
figure 32.1(c). In fact, in figure 32.1(c), the opening is so small that it now appears
as a point source of light, with light emanating radially from the slit.
Diffraction is not limited to light waves but occurs in all wave phenomena
such as water waves and sound waves. The greater the wavelength of the wave the
greater the amount of diffraction. Since the wavelength of light is very small the
bending of the light ray is very small. The wavelength for sound waves, varies from
16.5 m, for a sound of frequency 20.0 Hz, to 1.65 cm, for a frequency of 20,000 Hz.
Hence, the wavelength of sound is very much greater than the wavelength of light,
which is no larger than 7.2 × 10−5 cm. For large wavelengths such as sound,
diffraction is very pronounced. Thus, when someone speaks in another room, the
sound waves bend around the doorway so the person can be heard, even though the
light from that person cannot bend around the doorway and the person cannot be
seen. In fact, the diffraction of sound waves is such a common phenomenon, most
people are unaware of it, unless someone points it out to them.
The wave theory of light was first proposed by Christian Huygens and
diffraction was first observed by Francesco Maria Grimaldi (1618-1683), a Professor
of Mathematics at the University of Bologna. In essence, Grimaldi measured the
bright image, such as in figure 32.1(b), and found it to be greater than the size of
the opening of the slit. He also placed an obstacle in the path of the beam of light
and found that its projected image on a wall was smaller than the obstacle itself.
32-2
Chapter 32 Physical Optics
Both of these examples of diffraction can be explained by the application of
Huygens’ principle to the wave fronts and is shown in figure 32.2. A plane wave
Figure 32.2 The bending of light around an obstacle.
front is seen approaching an obstacle. The new position of the wave front is found by
drawing new secondary wavelets as described in chapter 26. When the wave fronts
hit the obstacle, these waves cannot send out new wavelets in the forward direction.
However, the waves above the top of the obstacle continue to send out secondary
wavelets and, as seen in the figure, these secondary wavelets now extend into the
normally dark region. Because the new wave front is found by drawing a line tangent
to all these wavelets, we see that the wave front bends around the obstacle, into the
shadow region. Thus, Grimaldi’s observation that the measured bright image was
greater than the size of the opening of the slit can be explained by the light bending
at both sides of the slit into both shadow areas. The fact that an obstacle in the path
of the beam of light formed an image on the wall that was smaller than the obstacle
itself, can be explained by the light bending around both sides of the obstacle into
the normal shadow area of the obstacle.
The interference of waves is caused by a superposition of waves and was
described in chapter 12 on wave motion. The interference of light waves is a direct
extension of the superposition of waves and we will treat it in detail in sections 32.2
and 32.3.
32.2 The Interference of Light -- Young’s Double-Slit
Experiment
The first proof that light had wave characteristics was furnished in 1800 by the
crucial experiment on the interference of light from a double slit by Thomas Young
32-3
Chapter 32 Physical Optics
(1773-1829), an English physicist. The experiment is now referred to as Young’s
double-slit experiment. The experimental set-up is shown in figure 32.3(a). A
monochromatic source of light (light of a single wavelength) is passed through a
small hole in the first opaque screen and falls on the second opaque screen that
contains two small slits. If light is not a wave then the light should follow the
geometrical paths shown in figure 32.3(a), and thus give the intensity distribution
shown in figure 32.3(b). That is, only two light images should appear on the third
screen with a large dark shadow between the bright fringes. Instead, the
Figure 32.3 Young’s double-slit experiment.
experiment showed that there was a series of bright and dark fringes on the screen,
as shown in figure 32.3(c). There is no way that this strange result can be explained
by geometrical optics. However, the result can be easily explained if light is a wave.
Young’s double-slit experiment is shown again in figure 32.4. Light from the
source passes through the narrow slit in the first screen and is diffracted in the
Figure 32.4 The waves in Young’s double-slit experiment.
same way as shown in figure 32.1(c). When light from the first slit reaches the
second set of slits, it is again diffracted as if there was a point source at each of the
32-4
Chapter 32 Physical Optics
second slits. The waves from each slit now propagate toward the screen where they
interfere, or superimpose, with each other.
The double-slit experiment is redrawn in figure 32.5. The double slits are
labeled s1 and s2 and are located a distance d apart. If we consider the arbitrary
Figure 32.5 Geometry of Young’s double-slit experiment.
point P, then the light intensity at that point is the result of a superposition of a
wave of light from slit 1 and a wave from slit 2. Notice that the wave from slit 2 has
to travel a greater distance r2 than the wave from slit 1, the distance r1. Hence,
there is a difference in the optical path between path 1 and path 2. It is this path
difference that is responsible for the bright and dark fringes. We can write the path
difference as
(32.1)
Path difference = PD = r2 − r1
A perpendicular is dropped from the center of slit 1 to the path r2, at the point A. In
the experimental setup the distance R from the plane of the slits to the screen is
very large, compared to the slit separation d. Therefore, the length AP is
approximately the same length as r1. This is equivalent to rotating the distance r1
about the point P until r1 comes into coincidence with the length AP. The arc
associated with that rotation is approximately equal to the chord S1A. Hence, S1A is
perpendicular to BP and S2P. Two of the triangles of figure 32.5 are redrawn in
figure 32.6 in order to show the relationship between the angles in these triangles
more clearly. The angle θ is the angle that defines the location of the fringe at the
point P and is angle PBO of triangle I. We will call angle BPO angle α, whereas
angle POB is a right angle. Thus, in triangle I,
θ + α + 900 = 1800
In triangle II, angle S1BC is equal to the same angle, α, of triangle I because
alternate, interior angles of two parallel lines are equal. Angle S1CB is a right
angle. Let us determine the angle β. In triangle II we have
32-5
Chapter 32 Physical Optics
Figure 32.6 A closer look at the angles.
β + α + 900 = 1800
Comparing these two equations, we see that
β=θ
Thus, angle S2S1A is equal to the angle θ, and hence, the side opposite angle θ S2A is
equal to d sin θ, as shown in figure 32.5. Therefore, we can write the path length r2
as
r2 = r1 + d sin θ
(32.2)
The path difference between waves 1 and 2, equation 32.1, becomes
PD = r2 − r1 = r1 + d sin θ − r1
PD = d sin θ
(32.3)
We will now analyze the experiment in two ways. First, we will examine a simple
intuitive geometrical solution showing where the bright and dark fringes are
located, and then we will explore a more detailed analytical solution showing the
actual intensity distribution of the light.
Geometrical Solution
As seen in equation 32.3, there is a path difference between wave 1 and wave 2.
Wherever the waves are in phase when they superimpose, there is constructive
interference and a bright image or bright fringe on the screen. If the path difference
between the two waves is a whole number of wavelengths, then the two waves are
in phase at the screen, and a bright fringe appears. Thus the condition for a bright
fringe to appear on the screen is
32-6
Chapter 32 Physical Optics
Bright Fringe
PD = d sin θ = mλ
(32.4)
where m is an integer.
Wherever the waves are out of phase by 1800, or a half of a wavelength (λ/2),
there is destructive interference and a dark region or dark fringe appears on the
screen. If the path difference between the two waves is an odd multiple of half
wavelengths, then the two waves are out of phase at the screen, and a dark fringe
appears. Thus the condition for a dark fringe to appear on the screen is
Dark Fringe
PD = d sin θ = (2m − 1) λ
2
(32.5)
Notice that the term (2m − 1) always gives an odd number for any integer value of
m.
Hence, the observed distribution of light, that is, bright and dark fringes, can
be explained by the processes of diffraction and interference, both wave
manifestations of light.
Analytical Solution
Let us now determine the light intensity distribution analytically. We can represent
the wave from slit 1 as the plane wave
E1 = E0 sin(kr1 − ωt)
(32.6)
where E0 is the amplitude of the wave, k is the wave number, and ω is the angular
frequency of the wave. These quantities have the same relations as previously
defined in chapter 12 on wave motion. We can represent the wave from slit 2 as
E2 = E0 sin(kr2 − ωt)
(32.7)
Since the light comes from the same source, E0, k, and ω are the same for each
wave. If the value of r2 from equation 32.2 is placed into equation 32.7, we get
or
E2 = E0 sin[k(r1 + d sin θ) − ωt]
E2 = E0 sin(kr1 − ωt + kd sin θ)
(32.8)
If we compare E2, equation 32.8, with E1, equation 32.6, we see that they look alike
except for the term, kd sin θ, in E2. Recall, from equation 14.39 of chapter 14, that a
wave
y2 = A sin(kx − ωt + φ)
(14.39)
represents a wave, just like
(14.37)
y1 = A sin(kx − ωt)
32-7
Chapter 32 Physical Optics
except for the term φ. Recall that φ is a phase angle and measures how far wave 2 is
displaced from wave 1, or how much wave 2 is out of phase with wave 1. Therefore,
let the term
kd sin θ = φ
and since the wave number k = 2π/λ, φ becomes
φ = 2πd sin θ
λ
(32.9)
Hence, φ is the phase difference between wave 1 and wave 2. We can now write
wave 2 as
E2 = E0 sin(kr1 − ωt + φ)
(32.10)
The resultant wave at the point P can now be determined by the interference of the
wave from slit 1 with the wave from slit 2. Thus, the resultant wave at P is given by
E = E1 + E2
(32.11)
Substituting the waves from equations 32.6 and 32.10 into equation 32.11, gives
E = E0 sin(kr1 − ωt) + E0 sin(kr1 − ωt + φ)
Factoring out the amplitude E0 from each term, gives
E = E0 [sin(kr1 − ωt) + sin(kr1 − ωt + φ)]
But this sum looks like the trigonometric identity for the sum of two sine waves
given in appendix B and used in chapter 14,
sin B + sin C = 2 sin
with
(B + C)
cos B − C
2
2
B = kr1 − ωt
and
C = kr1 − ωt + φ
Hence, the resultant wave at P becomes
E = E 0 2 sin
kr 1 − ✬t + kr 1 − ✬t + "
kr 1 − ✬t − kr 1 + ✬t − "
cos
2
2
Simplifying,
32-8
Chapter 32 Physical Optics
E = 2E 0 sin kr 1 − ✬t +
−"
"
cos
2
2
Because the cosine is an even function, we have
cos
Thus,
−"
"
= cos
2
2
E = 2E 0 cos
"
"
sin kr 1 − ✬t +
2
2
(32.12)
Equation 32.12 is the resultant wave at the arbitrary point P.
Since bright and dark fringes are observed on the screen, the intensity
distribution of this light as it hits the screen must be determined. Recall from
chapter 29 on electromagnetic waves that the intensity I is given by
I = ε0cE2
(29.63)
where ε0 is the permittivity of free space, c is the speed of light, and E is the value of
the electric field. We can now obtain the intensity distribution by substituting E
from equation 32.12 into equation 29.63. Thus,
I = ✒ 0 c4E 20 cos 2
"
"
sin 2 kr 1 − ✬t +
2
2
(32.13)
Because the frequency of visible light is so high (about 5 × 1014 cycles/s), the human
eye cannot see the effect of each wave as it hits the screen but instead can see only
their average value. We saw in chapter 28 that the average value of the sin2θ is
(sin2θ)avg = 1
2
Hence, the average intensity on the screen becomes
I avg = ✒ 0 c4E 20 cos 2
or
"
2
I avg = 2✒ 0 cE 20 cos 2
1
2
"
2
(32.14)
This equation can be further simplified by letting
I 0 = 2✒ 0 cE 20
Thus, the intensity distribution of the light on the screen is
32-9
(32.15)
Chapter 32 Physical Optics
I avg = I 0 cos 2
"
2
(32.16)
Equation 32.16 says that the intensity of the light on the screen varies with the
phase angle φ. But since φ is given by equation 32.9 as
φ = 2πd sin θ
λ
we see that the intensity varies with the angle θ in figure 32.5.
The location of a bright fringe on the screen can be determined by realizing
that a bright fringe corresponds to maximum light intensity. The intensity I in
equation 32.16 is a maximum whenever the cosine term is a maximum. This occurs
whenever the angle φ/2 is a multiple of π. That is, a bright fringe is obtained when
φ = mπ
2
Bright Fringe
(32.17)
where m is a whole number such as
m = 0, ± 1, ± 2, ± 3, …
For example, if we substitute equation 32.17 into equation 32.16, we get
Iavg = I0 cos2(mπ)
But cos(mπ) = ±1, and cos2(mπ) = 1. Therefore,
Iavg = I0
That is, the maximum intensity of light I0 occurs for φ/2 = mπ. Substituting equation
32.9 into equation 32.17, gives
φ = 2πd sin θ = mπ
2
2λ
Therefore,
Bright Fringe
d sin θ = mλ
for m = 0, ± 1, ± 2, ± 3, … (32.18)
Equation 32.18 is the condition for a bright fringe and is the same as equation 32.4,
which was derived geometrically. Notice from figure 32.5 that d sin θ is the
difference in the path length between the wave from slit 1 and the wave from slit 2.
When this path difference is equal to a whole multiple of the wavelength (mλ), then
the two waves are in phase at the point P, and there is constructive interference or a
bright fringe. The value m = 0 corresponds to the central maximum located at O in
32-10
Chapter 32 Physical Optics
figure 32.5. The first bright fringe after the central maximum occurs for m = 1, the
second fringe for m = 2, and so on. The location of the mth fringe on the screen is
found from the geometry of figure 32.5 as
ym = R tan θ
However, since R is so much larger than d in the experimental set-up, figure 32.5,
the angle θ is a very small angle. When the angle θ is small, we can use the small
angle approximation, and
tan θ ≈ sin θ
Hence, the mth fringe is located at
But from equation 32.18,
ym = R sin θ
(32.19)
sin θ = mλ
d
(32.20)
Substituting equation 32.20 into equation 32.19 yields the location of the mth bright
fringe on the screen at the position given by
Bright Fringe
ym = Rmλ
d
(32.21)
The dark fringes observed on the screen are at the positions of minimum light
intensity. The intensity, equation 32.16, is a minimum whenever φ/2 is equal to an
odd multiple of π/2, because then the cosine of a multiple of π/2 is zero. Hence, the
dark fringes occur when
Dark Fringe
φ = (2m − 1) π
2
2
for m = 1, 2, 3, …
(32.22)
Note that the term (2m − 1) is an odd number for m = 1, 2, 3, .… Substituting
equation 32.22 into the intensity equation 32.16 gives
I = I 0 cos 2 (2m − 1 ) ✜
2
But, cos(π/2) = 0 and cos[(2m − 1)π/2] = 0. Therefore, the intensity I is equal to zero
whenever the phase difference over two, φ/2, is an odd multiple of π/2, and this gives
a dark fringe. Substituting the phase angle φ from equation 32.9 into equation
32.22, gives
φ = 2πd sin θ = (2m − 1) π
2
2λ
2
32-11
Chapter 32 Physical Optics
and the condition for a dark fringe becomes
Dark Fringe
d sin θ = (2m − 1) λ
2
for m = 1, 2, 3, …
(32.23)
As we can see in figure 32.5, d sin θ corresponds to the path difference between the
wave from slit 1 and the wave from slit 2. Equation 32.23 says that whenever this
path difference is an odd multiple of half-wavelengths, the wave from slit 1 is 1800
out of phase with the wave from slit 2, and destructive interference occurs at point P,
thus producing zero light intensity and a dark fringe there. Notice that equation
32.23 is the same as equation 32.5 derived earlier in the geometrical solution.
We can also obtain the location of the dark fringe on the screen from the geometry
of figure 32.5 as
ym = R tan θ = R sin θ
Location of Dark Fringe
ym = R(2m − 1) λ
d
2
(32.24)
The overall intensity distribution for the double-slit interference can be found
by plotting the intensity I versus the angle φ/2 for equation 32.16, as is shown in
figure 32.7. The maximum peaks correspond to the bright fringes, and the zeros
Figure 32.7 The intensity distribution for Young’s double-slit experiment.
of the intensity correspond to the dark fringes. Hence, it is easy to explain, by the
wave theory of light, the bright and dark fringe pattern found on the screen.
If one of the slits in figure 32.5 were covered up, the interference pattern
would disappear. For example, if slit 2 were covered up, then the only wave arriving
at P would be wave 1, and the intensity associated with wave 1 would be
I1 = ε0cE12 = ε0cE02 sin2(kr1 − ωt)
(32.25)
The average value of sin2(kr1 − ωt) is again equal to 1/2, so the average value of I1
would be
I1 = ε0cE02
(32.26)
2
32-12
Chapter 32 Physical Optics
which is a constant and does not vary with the angle θ. If we compare this intensity
with the value of I0 from equation 32.15, we see that
I1 = 1 I0
4
This intensity distribution is shown as the dashed line in figure 32.7. The screen
would be uniformly illuminated.
If the two waves came from different sources they would not interfere with
each other. To explain the reason for this, we need to describe the light that comes
from a distributed light source. Light from a distributed light source consists of a
very large number of atoms acting as independent light sources. The phase
difference between the light waves from different atoms is random. Hence, the term
containing the phase angle φ in equations 32.12 and 32.9 would not be constant in
time but would change in a random way. Thus, there would not be any positions on
the screen where the light would always be either interfering constructively or
destructively to produce the bright or dark fringes.
If the phase difference between the two light waves of a source of light is a
constant in time, we say the light source is coherent. If the phase difference varies
with time in a random way, we say the source is incoherent. Young was able to
produce coherent waves by using a single point source of light, which then fell on
the two slits as shown in figure 32.4. If light comes from two different sources, the
waves do not have a constant phase angle between them, and there is no
interference. The intensity at the screen then is simply the sum of each individual
intensity, that is,
I = I1 + I2
= ε0cE02 + ε0cE02
2
2
= ε0cE02 = I0
2
which is also shown in figure 32.7. That is, there is an even illumination of the
entire screen, but there are no interference fringes.
Example 32.1
A double slit. In a double-slit experiment, light of 500.0-nm wavelength impinges on
a double slit that has a separation of 0.350 mm. If the screen is placed 5.00 m from
the double slit, find (a) the value of θ corresponding to the first four bright fringes,
(b) the value of y locating the four bright fringes on the screen, (c) the value of θ
corresponding to the first three dark fringes, and (d) the value of y locating the
three dark fringes.
32-13
Chapter 32 Physical Optics
Solution
a. The angle θ corresponding to the bright fringes, found from equation 32.18, is
d sin θ = mλ
Therefore
θ = sin−1 mλ
d
The first fringe corresponds to the central maximum and occurs for m = 0. Hence,
For m = 0,
θ0 = 0
For m = 1,
θ1 = sin−1 λ
d
−7
500.0
nm
10
cm 10 mm
= sin −1
0.350 mm 1.00 nm
1 cm
−1
−3
= sin (1.4286 × 10 )
= 0.08180
For m = 2,
θ2 = sin−1 2λ
d
−6
mm
−1 (2)(500.0 nm) 10
= sin
1.00 nm
0.350 mm
= 0.16400
For m = 3,
θ3 = sin−1 3λ
d
(3)(500.0
nm)
10 −6 mm
= sin −1
0.350 mm
1.00 nm
= 0.2460
b. The value of ym locating the bright fringes on the screen, found from equation
32.21, is
ym = mRλ
d
For m = 0,
y0 = 0
For m = 1,
y1 = Rλ
d
(5.00 m)(500.0 nm) 10 −6 mm 10 3 mm
=
0.350 mm
1.00 nm
1m
= 7.14 mm
32-14
Chapter 32 Physical Optics
For m = 2,
y2 = 2Rλ = 2y1
d
= 14.3 mm
For m = 3,
y3 = 3Rλ = 3y1
d
= 21.4 mm
c. The value of θ corresponding to the dark fringes, found from equation 32.23, is
d sin θ = (2m − 1) λ
2
Hence,
For m = 1,
θ = sin−1 (2m − 1) ✘
2d
θ1 = sin−1 (2 − 1)
(500.0 nm)
10 −6 mm
(2)(0.350 mm)
1 nm
= 0.04090
θ2 = sin−1 (4 − 1)
(500.0 nm)
10 −6 mm
(2)(0.350 mm)
1 nm
0
= 0.123
θ3 = sin−1 (6 − 1)
(500.0 nm)
10 −6 mm
(2)(0.350 mm)
1 nm
0
= 0.205
For m = 2,
For m = 3,
d. The location ym of the dark fringe on the screen, found from equation 32.24, is
ym = (2m − 1)Rλ
2d
For m = 1,
y1 = (2 − 1)Rλ = Rλ
2d
2d
(5.00 m)(500.0 nm) 10 −6 mm
=
1.00 nm
(2)(0.350 mm)
= 3.57 mm
For m = 2,
For m = 3,
10 3 mm
1m
y 2 = (4 − 1 ) R✘ = 3 R✘ = 3y 1
2d
2d
= 10.7 mm
y 3 = (6 − 1 ) R✘ = 5 R✘ = 5y 1
2d
2d
32-15
Chapter 32 Physical Optics
= 17.9 mm
Note, in this example, the very small values of the angle θ, which justifies the
assumption of the small angle approximation in the derivation.
To go to this Interactive Example click on this sentence.
Example 32.2
Using the double-slit experiment to determine the wavelength of light. A double-slit
experiment is performed in order to determine the wavelength of the source of light.
If R = 5.00 m, d = 0.350 mm, and the location of the first bright fringe on the screen
is observed to be at 8.25 mm, find the value of the wavelength λ.
Solution
The wavelength, determined from equation 32.21, is
ym = mRλ
d
λ = d ym
mR
(0.350 mm)(8.25 mm)
1m
=
10 3 mm
1(5.00 m)
= 578 nm
1.00 nm
10 −6 mm
The double-slit experiment is, therefore, a simple experiment that can be used to
determine the wavelength of a particular light source.
To go to this Interactive Example click on this sentence.
32.3 The Interference
Interferometer
of
Light
--
The
Michelson
A very accurate device to measure the wavelength of light was designed by Albert
Michelson (1852-1931). The device, now called the Michelson interferometer, is
shown schematically in figure 32.8(a), and a typical laboratory interferometer is
shown in figure 32.8(b). A diffuse source of monochromatic light impinges on a
half-silvered mirror, called a beam-splitter. Because the beam-splitter mirror is not
completely silvered, some of the incident light is transmitted through the
32-16
Chapter 32 Physical Optics
Figure 32.8 The Michelson interferometer.
beam-splitter and passes on to the fixed mirror M1. This ray of light is reflected from
mirror M1 and returns to the beam-splitter where it is reflected into the telescope
for the eye to observe. The rest of the original incident light falling on the
beam-splitter is reflected from the half-silvered mirror and travels to the movable
mirror M2, where it is reflected back to the beam-splitter. Here it is transmitted
through the beam-splitter and passes through the telescope where it can be
observed by the eye. Thus, the incident ray is split into two separate rays, each of
which follows a different optical path, as seen in the figure. When these two rays
come together at the eye they interfere with each other, thereby producing
interference fringes. Note that ray 2 passes through the beam-splitter three times,
whereas ray 1 only passes through the splitter once.
In order for both rays to pass through the same length of glass a compensator
plate is placed between the beam-splitter and the fixed mirror M1. The compensator
plate is a piece of plane glass having the same index of refraction and the same
thickness as the glass of the beam-splitter. Thus, ray 1 now goes through three
thicknesses of glass, the same as ray 2. Hence, the only difference in the optical
path between ray 1 and ray 2 depends on the difference in length between the path
L1 and the path L2.
Because of the diffuse nature of the light source and the fact that, in practice,
mirrors M1 and M2 are not exactly perpendicular to each other, the interference
fringes appear as a series of straight lines when viewed through the telescope, as
shown in figure 32.9. A hairline in the telescope is placed directly over the central
bright fringe. Mirror M2 in figure 32.8 is capable of being moved forward or
backward by a very accurate micrometer screw. If M2 is moved backward by a
distance of λ/4, then ray 2 travels an additional distance of λ/4 on the way to the
mirror and an additional distance of λ/4 after it is reflected from the mirror. Hence,
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Chapter 32 Physical Optics
Figure 32.9 Interference fringes in the Michelson interferometer.
in all, the path length of ray 2 changes by λ/2 from the path length of ray 1. Thus,
ray 1 and ray 2 are now out of phase by a half-wavelength as they pass through the
telescope and cause destructive interference. Therefore, a dark fringe appears where
originally there was a bright fringe. If the mirror is moved an additional distance of
λ/4, then there is another half-wavelength change in the optical path and the rays
cause constructive interference or a bright fringe. Hence, a total movement of M2 by
a distance of λ/2 causes one bright fringe to move across the cross hair of the
telescope. If the observer counts m fringes passing the cross hair of the telescope
when the mirror is moved a distance x, as measured by the micrometer screw, then
that distance x corresponds to a distance of m half-wavelengths or
x=m λ
2
(32.27)
If the wavelength λ of the light source is known then the interferometer, through
the use of equation 32.27, can be used to measure a length x very accurately. On the
other hand, if we wish to determine the wavelength λ of a light source, we can
rearrange equation 32.27 to solve for λ when x is a measured reading of the
micrometer screw. Thus,
λ = 2x
(32.28)
m
Example 32.3
Using the Michelson interferometer. A Michelson interferometer is used to determine
the D spectral line in sodium. If the movable mirror moves a distance of 0.2650 mm,
as read from the micrometer screw, when 900.0 fringes are counted passing the
telescope cross hair, find the wavelength of the D line.
32-18
Chapter 32 Physical Optics
Solution
The wavelength, found from equation 32.28, is
λ = 2x
m
= 2(0.2650 mm)
900.0
= (5.889 × 10−4 mm) 16nm
10 mm
= 588.9 nm
To go to this Interactive Example click on this sentence.
32.4 Interference -- Thin Films
Interference fringes can also be caused by multiple reflections from a thin film, a
very thin piece of transparent material. The interference can be caused by a phase
change that occurs on reflection and/or a phase change that occurs because of the
difference in the optical path of the two interfering light waves. Let us first consider
what happens when a light wave is reflected from a boundary.
Reflection at a Boundary between a Less Dense Medium and a
More Dense Medium
Recall from our study of waves on a string in chapter 12 that when the wave went
from the less dense string to the more dense string, the reflected wave was inverted.
That is, the reflected wave was 1800 out of phase with the incident wave (see figure
12.10), while the transmitted wave did not change its phase. Although that result
was obtained for waves on a string, it is a general characteristic of wave motion.
When a ray of light in air is incident on a piece of glass, the light wave is reflected at
the boundary between the less optically dense air and the more optically dense glass.
Hence, the reflected wave is found to be 1800 out of phase with the incident wave,
figure 32.10(a). The light wave transmitted through the glass does not change its
phase with respect to the incident wave but its wavelength decreases in the more
dense glass, as shown in chapter 27, equation 27.7:
λ2 = λ1
n21
Applying equation 27.7 to the current problem, the wavelength in the glass is
32-19
Chapter 32 Physical Optics
Figure 32.10 Reflection and phase changes of light at a boundary.
λg = λair = λair
n21 ng/nair
= nairλair
ng
But since nair = 1,
λg = λair
ng
(32.29)
Thus, the wavelength of the transmitted wave in the glass is decreased.
Reflection at a Boundary between a More Dense Medium and a
Less Dense Medium
It was shown in figure 12.11 that when a wave goes from a more dense medium to a
less dense medium, the reflected wave is not inverted. That is the reflected wave is
not out of phase with the incident wave. Although the result was established for
waves on a string it is a characteristic of wave motion in general. Hence, when a
light wave is reflected at a boundary between a more dense medium and a less dense
medium, the reflected wave is not inverted but has the same phase as the incident
wave. Thus, in figure 32.10(b) the incident ray in the more dense glass is reflected
at the boundary between the glass and the less optically dense air, and there is no
phase change on reflection. The transmitted wave is always in phase with the
incident wave. The effect of the phase change, or lack of it, in reflection becomes
important in the interference of light waves from thin films.
Interference of Light from a Soap Bubble or Thin Film of Oil
Consider an incident ray of light in air that impinges on a thin film of water, such
as a soap bubble found in a child’s toy for making bubbles, and shown schematically
in figure 32.11. We assume that the incident ray is almost normal to the thin film of
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Chapter 32 Physical Optics
Figure 32.11 Reflection from a thin layer of water.
water, although it is shown at an exaggerated angle in figure 32.11 so we can more
easily see what is happening. As the incident ray hits the boundary between the air
and the water, part of the incident light wave is reflected and part is transmitted
into the film of water. The reflected ray is labeled as ray 1 in figure 32.11. Since this
ray has been reflected from a boundary between a less optically dense medium, the
air, and a more optically dense medium, the water, the reflected light wave
undergoes a phase change of 1800 on reflection. The part of the wave that is
transmitted through the water, ray 2, undergoes a reflection at point A at the
boundary between the water and the air, as shown in the figure. This reflection
occurs at a boundary between a more optically dense medium, the water, and a less
optically dense medium, the air, so there is no phase change on reflection for ray 2.
If the thickness of the film d were zero, then ray 1 would be 1800 out of phase
with ray 2, and then ray 1 and ray 2 would superimpose to give destructive
interference. However, the thickness d is not zero, and ray 2 moves through the
distance OAB before it emerges into the air, and this additional distance moved by
ray 2 causes a path difference between ray 1 and ray 2. Because the incident ray is
almost normal, the distance OAB is approximately twice the thickness of the film.
That is, the path difference between ray 1 and ray 2 is
Path difference = PD = 2d
If this path difference is equal to a whole multiple of wavelengths, then ray 2 has
the same phase at position B as ray 1 at position O, if ray 1 did not have a phase
change of 1800. But ray 1 has experienced a phase change of 1800 on reflection, so
when ray 2 reaches point B it is 1800 out of phase with ray 1 and causes destructive
interference when viewed by the eye in figure 32.11. Hence, the condition for a dark
fringe or destructive interference is
Dark Fringe
Path difference = PD = 2d = mλn
32-21
m = 1, 2, 3, … (32.30)
Chapter 32 Physical Optics
In equation 32.30 the subscript n on the wavelength indicates that this is the
wavelength in the medium with the index of refraction n. Hence, λn represents the
wavelength of light within the thin film of water. The wavelength of light within the
water is related to the wavelength of light in air by equation 32.29 with a change in
subscript from g to w representing the water medium rather than a glass medium.
Hence,
λn = λw = λair
(32.31)
nw
Substituting equation 32.31 back into equation 32.30, we get
Dark Fringe
2d = mλair
nw
m = 1, 2, 3, …
(32.32)
Thus, the equation for destructive interference, or a dark fringe in the reflected light
from a soap bubble, is
Dark Fringe
2nwd = mλair
m = 1, 2, 3, …
(32.33)
We use the index of refraction of water for the index of refraction of the soap bubble
because it is made up largely of water. (If the interference were from a thin film of
oil, then n would, of course, be the index of refraction of oil.)
If the path length of ray 2 in the water is equal to an odd multiple of λ/2, then
ray 2 should be out of phase with ray 1. However, because ray 1 suffers a phase
change of 1800 by reflection, ray 1 is now in phase with ray 2. Therefore, the
superposition of ray 1 and ray 2 now causes constructive interference or a bright
fringe when viewed by the eye. Thus, the condition for a bright fringe is
PD = (2m − 1) λn
2
m = 1, 2, 3, …
where the term (2m − 1) always gives an odd number for any value of m = 1, 2, 3, .…
But we have already seen that the path difference for nearly normal incidence is 2d.
Thus,
Bright Fringe
2d = (2m − 1)λn
2
Substituting for λn from equation 32.31 gives the condition for constructive
interference or a bright fringe as
Bright Fringe
2d = (2m − 1)λair
2nw
32-22
(32.34)
Chapter 32 Physical Optics
Example 32.4
Minimum reflection from a thin film. A soap bubble is 220.0 nm thick. For what
wavelength of the incident light will the intensity of the reflected light be zero?
Solution
The intensity of the reflected light is zero when destructive interference occurs
between the wave reflected from the air-water boundary and the wave reflected
from the water-air boundary. The condition for this dark fringe is found from
equation 32.33. Solving that equation for the wavelength in air we get
λair = 2nwd
m
= 2(1.33)(220.0 nm)
1
= 585 nm
Thus, if a ray of light of wavelength 585 nm fell on the soap bubble there would be
no reflected light and hence all the light would be transmitted through the soap
bubble.
To go to this Interactive Example click on this sentence.
Example 32.5
Maximum reflection from a thin film. For the same soap bubble of 220.0-nm
thickness, what wavelength of the incident light would give the maximum
reflection?
Solution
The maximum reflection occurs when there is constructive interference between the
reflected waves. The condition for this bright fringe is given by equation 32.34.
Solving for the wavelength, we get
λair = 4nwd
(2m − 1)
For m = 1,
λ1air = 4(1.33)(220.0 nm)
1
= 1,170 nm
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Chapter 32 Physical Optics
But this wavelength falls out of the visible spectrum (380.0 − 720.0 nm) and could
not be seen by the eye.
For m = 2,
λ2air = λ1air = 1170 nm
3
3
= 390 nm
Therefore, a wavelength of 390 nm, a blue light just barely in the visible spectrum,
gives a maximum intensity of reflected light. For values of m = 3 or greater, the
wavelength occurs in the ultraviolet portion of the spectrum, which is of course, not
visible.
To go to this Interactive Example click on this sentence.
If we shine white light, a mixture of all the wavelengths from 380.0 to 720.0
nm, the only bright reflection occurs for the blue wavelength of 390 nm. If we shine
monochromatic light of any wavelength other than the 390 nm, we will not get a
bright reflection.
In practice, the soap bubble is usually not of a constant thickness d.
Therefore, many different wavelengths can be reflected for the different values of d,
thus accounting for the beautiful array of colors usually associated with the soap
bubble film.
Nonreflecting Glass
In many sophisticated optical systems, there may be as many as four to ten lenses
in the optical system. The amount of light reflected from a single lens is actually
quite small, of the order of 4%. However with four lenses the light lost by reflection
is about 15%, whereas for 10 lenses it amounts to over 33%. Hence, for a multilens
system, the amount of reflected light can become quite significant. Our analysis of
the reflected light from the thin film of a soap bubble suggests that a thin film of
transparent material, placed on the surface of a lens, can be used to reduce the
amount of reflected light from a lens. Glass, so coated with a thin film, is called
nonreflecting glass.
A thin film of magnesium fluoride (MgF2) is, therefore, placed over a piece of
glass as indicated in figure 32.12. The index of refraction of MgF2 is 1.38, which is
greater than the index of refraction of air, but less than the index of refraction of
glass. When the incident ray impinges on the MgF2 surface at the point O, part of
the light is reflected as ray 1 and part is transmitted as ray 2. Ray 1 is 1800 out of
phase with the incident ray because of the reflection from a boundary between the
less dense and more dense medium. When the transmitted ray, ray 2, hits the glass
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Chapter 32 Physical Optics
Figure 32.12 Nonreflecting glass.
at point A of the figure, the reflected ray is also 1800 out of phase with the incident
ray because the reflection at the magnesium fluoride-glass boundary is a reflection
at the boundary between the less dense medium, MgF2 (n = 1.38) and the more
dense glass (n = 1.50). Thus, if the thickness d of the magnesium fluoride were
effectively zero, the two reflected waves would be in phase. However, d is not equal
to zero and there will, therefore, be a difference in the optical paths between ray 1
and ray 2. If this difference in the optical path is an odd multiple of a half
wavelength λn/2, ray 1 and ray 2 are out of phase. Their superposition produces
destructive interference, and there is no reflected light. Stated mathematically the
condition for destructive interference is
Path difference = 2d = (2m − 1) λn
2
(32.35)
where the term (2m − 1) always gives an odd number for m = 1, 2, 3 .… The
wavelength within the magnesium fluoride medium, λn, is given by equation 32.31
as
λn = λair
(32.36)
nMgF2
Substituting equation 32.36 into equation 32.35 gives the condition for zero
reflection as
Minimum reflection
2d = (2m − 1) λair
2nMgF2
(32.37)
Note that equation 32.37 corresponds to destructive interference while equation
32.34 looks almost the same, yet equation 32.34 represents a constructive
32-25
Chapter 32 Physical Optics
interference from a thin film of water. The difference between the two equations is
based on the phase change at reflection. For the thin soap bubble film only the first
reflected wave is out of phase by 1800. For the thin film on the glass lens, both
reflected rays are 1800 out of phase. Hence the problems are different and we should
expect to get different results.
Example 32.6
Minimum reflection from nonreflecting glass. What thickness of magnesium fluoride
should be deposited on a glass lens of ng = 1.50, such that the greatest portion of
incident white light should not be reflected?
Solution
Visible light lies in the range of 380.0 to 720.0 nm. It is desired that the thin film
give minimum reflection at the center of the visible spectrum. That is, the center of
the visible spectrum is equal to
380.0 nm + 720.0 nm
2
= 550.0 nm
Hence, a thickness d is picked that gives destructive interference for λ = 550.0 nm,
the center of the visible spectrum. Solving equation 32.37 for the thickness d, gives
d = (2m − 1) λair
4nMgF2
For m = 1,
d = (2 − 1) 550.0 nm
4(1.38)
= 99.6 nm
Therefore, a thickness of 99.6 nm of magnesium fluoride placed on a glass lens
causes destructive interference for light of 550.0 nm and none of this light is
reflected.
To go to this Interactive Example click on this sentence.
Example 32.7
Maximum reflection from nonreflecting glass. For the thickness of magnesium
fluoride in example 32.6, is there any wavelength within the visible spectrum that
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Chapter 32 Physical Optics
will give maximum reflection of light, that is, give constructive interference?
Solution
For constructive interference the path difference between ray 1 and ray 2 must be a
whole multiple of the wavelength λ of the light. Hence the path difference must be
PD = 2d = mλn
(32.38)
Substituting the value of λn from equation 32.36 into equation 32.38 yields for the
condition of maximum reflection
Maximum reflection
2d = mλair
nMgF2
(32.39)
Solving for the wavelength that gives this maximum reflection we have
λair = 2nMgF2 d
m
= 2(1.38)(99.6 nm)
1
= 275 nm
That is, for the thin film thickness of 99.6 nm, maximum reflection occurs at 275
nm. But 275 nm lies outside the visible portion of the spectrum and cannot be seen.
To go to this Interactive Example click on this sentence.
At 550.0 nm complete constructive interference occurs. At wavelengths
greater than 275 nm and less than 550.0 nm, superposition of rays 1 and 2 still
occurs with some intermediate value of intensity. Thus, at the lower portion of the
visible spectrum, 380.0 nm, some light is reflected. The closer we get to the 550.0
nm wavelength, the smaller is the reflection. There is also some reflection for
wavelengths greater than 550.0 nm just as there were some in the lower region.
Interference from an Air Wedge
If two plates of glass are separated by a very small amount at one edge, as in figure
32.13(a), an interference pattern of dark and bright fringes occurs. Ray 1 is reflected
from a glass-air interface (more dense to less dense) and no phase change occurs.
Ray 2, on the other hand, is reflected from an air-glass interface (less dense to more
dense) and suffers a 1800 phase change on reflection. Thus, ray 1 and ray 2 are out
of phase by 1800. However, there is an additional phase shift caused by the
32-27
Chapter 32 Physical Optics
Figure 32.13 An air wedge.
difference in path of 2d traversed by ray 2 as it passes through the air gap, between
the upper and lower plate, after reflection. The sizes of h and d in figure 32.13(a)
are greatly exaggerated so we more easily see the reflections and path differences.
Hence, the total phase difference between ray 1 and ray 2 is equal to the phase
difference caused by the difference in optical path between the two rays plus the
phase change that occurs to ray 2 on reflection. If the path difference is equal to a
multiple of the wavelength λ, then the only phase difference is the 1800 one caused by
reflection and ray 1 is then out of phase with ray 2 and destructive interference
occurs. Hence, a dark fringe occurs when
Dark Fringe
2d = mλ
m = 0, 1, 2, 3, …
(32.40)
We can find the location of this dark fringe from the geometry of figure
32.13(a):
tan θ = d
x
and also
tan θ = h
l
Therefore,
d = h
x
l
Solving for x,
x=dl
(32.41)
h
The value of d for the mth fringe, found from equation 32.40, is
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Chapter 32 Physical Optics
d = mλ
2
(32.42)
Substituting equation 32.42 into equation 32.41 gives for the location of the mth
fringe
Location of Dark Fringe
xm = mλl
m = 0, 1, 2, 3, …
(32.43)
2h
The separation between fringes is found by subtracting the location of the mth
fringe from the (m + 1)th fringe. That is,
∆x = xm+ 1 − xm
= (m + 1)λl − mλl
2h
2h
∆x = λl
2h
(32.44)
Since λ, l, and h are all constant for the geometry of figure 32.13, the separation
between dark fringes is uniform.
We should note that the special case of m = 0 in equation 32.43 corresponds
to d = 0, or approximately zero, at x = 0, and is the location of a dark fringe. The
reason for this is that the only phase change that occurs there, one of 1800, occurs
from the reflection from the lower air-glass interface.
Between the dark fringes, where destructive interference occurs, there are
bright fringes, where constructive interference occurs. The fringe pattern is shown
in figure 32.13(b). We find the condition for the bright fringe from the geometry of
figure 32.13(a), where the optical path difference must now be an odd multiple of a
half a wavelength. That is, for a bright fringe we must have
Bright Fringe
2d = (2m − 1) λ
2
m = 1, 2, 3, …
(32.45)
And the location of the mth bright fringe, found from equations 32.41 and 32.45, is
Location of Bright Fringe
xm = (2m − 1)λl
4h
m = 1, 2, 3, …
(32.46)
Example 32.8
The air wedge. A hair, 8.50 × 10−5 m in diameter, is placed between two 10.0-cm
plates of glass (n = 1.52) at their edge. Light of 589.0-nm wavelength shines on the
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Chapter 32 Physical Optics
glasses and an interference pattern is formed. Find (a) the location of the fifth dark
fringe, and (b) the location of the third bright fringe.
Solution
a. The location of the fifth dark fringe, found from equation 32.43, is
xm = mλl
2h
Because the first dark fringe occurs for m = 0, the fifth dark fringe occurs for m = 4.
Therefore,
x5 = 4λl
2h
= 4(589.0 × 10−9 m)(0.100 m)
2(8.50 × 10−5 m)
= 1.39 × 10−3 m = 1.39 mm
b. The location of the third bright fringe, found from equation 32.46, is
xm = (2m − 1)λl
4h
x3 = [2(3) − 1](589.0 × 10−9 m)(100 mm)
4(8.50 × 10−5 m)
= 0.866 mm
To go to this Interactive Example click on this sentence.
Newton’s Rings -- A Variable Air Wedge
An interesting interference pattern can be observed if light shines on a planoconvex
lens that is placed on a flat piece of glass, as shown in figure 32.14(a). The
interference pattern consists of a family of concentric dark and bright circles, known
as Newton’s rings and shown in figure 32.14(b).
The interference between the reflected light waves is the same as for the air
wedge studied above, but because of the curvature of the lens, the angle of the air
wedge is continuously varying. A dark fringe, or destructive interference, again
occurs when the optical path difference is equal to a whole number of wavelengths,
that is,
2d = mλ
(32.47)
where d, as shown in figure 32.14(a), is a variable. Since a fixed value of d occurs all
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Chapter 32 Physical Optics
Figure 32.14 Newton’s rings.
around the lens, the fringes are circular. The radius of the fringe is related to the air
gap d by the geometry of figure 32.14(a). Here R is the radius of curvature of the
convex lens, while r is the location of the air gap d. From the figure we see that
R2 = (R − d)2 + r2
= R2 − 2Rd + d2 + r2
However, since R ! d, d2 = 0. Therefore,
Solving for d, we get
R2 = R2 − 2Rd + r2
0 = −2Rd + r2
d = r2
2R
Substituting this value of d into equation 32.47, the condition for a dark fringe, we
get
2(r2) = mλ
2R
Solving for rm, the radius of the mth fringe, we get
rm = m✘R
32-31
(32.48)
Chapter 32 Physical Optics
Example 32.9
Newton’s rings. Light of 589.0-nm wavelength falls normally on a planoconvex lens
whose radius of curvature is 10.0 m, and Newton’s rings are observed. Find the
radius of the 15th fringe.
Solution
The radius of the 15th fringe, found from equation 32.48, is
rm = m✘R
= 15(589.0 % 10 −9 m)(10.0 m )
= 9.40 mm
To go to this Interactive Example click on this sentence.
32.5 Diffraction from a Single Slit
As seen in the introduction to this chapter, diffraction is the bending of light waves
around an obstacle. Let us now consider the diffraction of light from a single slit,
where each side of the slit represents the obstacle that the light bends around.
There are two distinct ways to observe diffraction from a single slit. The first
technique is called Fresnel diffraction, after its discoverer, Augustin Jean Fresnel
(1788-1827), and is diagramed in figure 32.15. Monochromatic light from a point
Figure 32.15 Fresnel diffraction.
32-32
Chapter 32 Physical Optics
source impinges on the slit. The waves from the point source are spherical and the
wave fronts are circular as they fall on the slit opening. The waves are diffracted at
the slit opening and spread out into the shadow region and a diffraction pattern is
found on the screen, figure 32.15(b). The pattern resembles the interference pattern
of a double slit, with alternate dark and bright fringes, but with decreasing
intensity. The screen is relatively close to the slit in this arrangement.
The second technique for observing the effect of diffraction from a single slit
is due to Joseph von Fraunhofer (1787-1826) and is shown in figure 32.16(a). The
Figure 32.16 Fraunhofer diffraction.
source is very far away from the slit so that the incident light can be treated as a
series of plane waves. The screen is so far away from the source that the light
waves, after the diffraction, can again be assumed to be plane. Again a diffraction
pattern occurs on the screen as in figure 32.15(b). This second technique is called
Fraunhofer diffraction and can be approximated in the laboratory by the use of
two converging lenses, as shown in figure 32.16(b). A converging lens is placed in
front of the monochromatic light source such that the source is at the principal
focus. Hence, light rays emanate from the lens as parallel rays, which means that
the waves are, of course, plane light waves. The plane waves are diffracted by the
slit and the light rays that head toward the slit are all parallel. A converging lens is
32-33
Chapter 32 Physical Optics
placed a distance f in front of the screen and these parallel rays are then converged
by the lens to the point P on the screen.
Our analysis of diffraction from the single slit will center on Fraunhofer
diffraction because the plane waves are easier to handle than spherical waves.
Figure 32.17(a) shows a single slit of width a. Points 1, 2, and 3 are points on the
Figure 32.17 Single slit-diffraction.
same wave front. Secondary Huygens’ wavelets from each of these waves are in
phase with each other before they arrive at the slit. These straight ahead waves are
also in phase after the slit and are converged by the lens to the point P0 to form the
central bright fringe. Figure 32.17(b) shows rays 1, 2, 3, 4, and 5 associated with
waves that are all in phase as the plane waves approach the slit. After the
diffraction the rays are still parallel when they enter the lens, and are then
converged to the point P. However, because the point P is no longer symmetrically
located with respect to the slit, the light rays 1 through 5 travel different path
lengths in arriving at the point P. We can see in figure 32.17(b) that the path
difference between ray 1 and ray 3 is
PD3 = a sin θ
2
32-34
Chapter 32 Physical Optics
If this path difference between ray 1 and ray 3 is exactly a half a wavelength, then
ray 1 and ray 3 interfere destructively when they combine at point P. Hence, the
condition for obtaining a dark fringe at P is that
PD = a sin θ = λ
2
2
(32.49)
Also note that all the rays between ray 1 and ray 2 are completely out of phase with
all the corresponding rays between ray 3 and ray 4 and all interfere destructively at
the point P.
The path difference between ray 2 and ray 1, from the figure, is
PD2 = a sin θ
4
while the path difference between ray 4 and ray 1 is
PD4 = 3a sin θ
4
Hence, the path difference between ray 2 and ray 4 is
PD4 − PD2 = 3a sin θ − a sin θ = a sin θ
4
4
2
But as just seen in equation 32.49, (a/2) sin θ = λ/2, and therefore ray 2 and ray 4
are also 1800 out of phase when they combine at point P, also producing a dark
fringe. All the rays between ray 2 and ray 3 are also completely out of phase with all
the corresponding rays between ray 4 and ray 5. Hence, all the rays from the single
slit interfere destructively at the point P and form a dark fringe there. Therefore,
the condition for the first dark fringe, obtained from equation 32.49, is
First Dark Fringe
a sin θ = λ
(32.50)
In general, there is more than one dark fringe and destructive interference
occurs whenever the quantity (a sin θ) is a whole multiple of the wavelength λ.
Thus, the general condition for a dark fringe is
Dark Fringe
a sin θ = mλ
m = 1, 2, 3, …
(32.51)
Secondary bright fringes are found to exist approximately half way between
the dark fringes. Although too long to derive here, the intensity distribution for the
single-slit diffraction pattern is shown in figure 32.18 as a function of the angle θ.
32-35
Chapter 32 Physical Optics
Figure 32.18 Intensity distribution for single-slit diffraction.
The intensity of the secondary maxima shown in the figure are exaggerated for
clarity in the diagram. We should note a very interesting result of this intensity
distribution. There is a central bright fringe occurring at θ = 0, and secondary
bright fringes occurring on both sides of the central maximum. However, the
intensity of these secondary maxima diminish more rapidly than is shown in the
figure.
Example 32.10
Diffraction from a single slit. Find the location for the first three dark fringes for
diffraction from a single slit of width a = 10λ.
Solution
The locations of the dark fringes are found from equation 32.51 as
a sin θ = mλ
sin θ = mλ
a
θ = sin−1 mλ
a
−1
= sin mλ
10λ
−1
= sin m(0.1)
For m = 1
For m = 2
For m = 3
θ = sin−1(0.1) = 5.740
θ = sin−1(0.2) = 11.50
θ = sin−1(0.3) = 17.50
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Chapter 32 Physical Optics
It is interesting and informative to return to Young’s double-slit experiment
and note that each slit in that experiment is a single slit and must have a
diffraction pattern associated with it. Thus, the interference fringes of Young’s
double-slit experiment must be modified to show the effects of diffraction. The
combined effects of interference and diffraction are shown in figure 32.19. The effect
of the diffraction is to reduce the intensity of the interference fringes farther away
from the central bright fringe.
Figure 32.19 Interference and diffraction.
32.6 The Diffraction Grating
The interference of light from a double slit causes a characteristic interference
pattern of bright and dark fringes. However, what is so special about two slits?
Suppose three slits, or four slits, or even N slits, were used in the experiment.
Would the characteristic interference pattern still be observed? The answer is yes.
Interference of light from any number of slits does give the characteristic pattern of
bright and dark fringes.
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Chapter 32 Physical Optics
Several parallel slits of equal width a, equally spaced a distance d apart, is
called a diffraction grating. Diffraction gratings are made by marking equally
spaced grooves on a glass plate. The spaces between the grooves represent the slits.
Typical gratings have anywhere between 400 slits per mm to 1200 slits per mm. A
schematic of a diffraction grating is shown in figure 32.20. For light rays emerging
Figure 32.20 The diffraction grating.
from each slit at an angle θ, there is a path difference of d sin θ between each ray
and the one directly above it. If this path difference is an exact multiple of
wavelengths then the rays from each slit would interfere constructively when they
reach the screen and would therefore produce a bright fringe there. Thus, the
condition for the formation of a bright fringe is
Bright Fringe
d sin θ = mλ
m = 0, 1, 2, …
(32.52)
Note that this is the same condition found for double-slit interference in equation
32.18. The effect of the increased number of slits is to form a sharper and narrower
interference pattern, as shown in figure 32.21.
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Chapter 32 Physical Optics
Figure 32.21 Comparison of interference pattern from a double slit to a diffraction
grating.
Example 32.11
The diffraction grating. Monochromatic light of 500.0-nm wavelength shines on a
diffraction grating of 1200 lines per mm. Find the location of the first three bright
fringes.
Solution
The distance d between slits is found by taking the reciprocal of the number of lines
per mm, thus
6
d = 1 mm = (8.33 % 10 −4 mm ) 10 nm = 833.0 nm
1200
1 mm
The location of the bright fringes, found from equation 32.52, is
and
d sin θ = mλ
θ = sin−1 mλ
d
The first bright fringe occurs for m = 0 and therefore
θ = sin−1 0 = 00
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Chapter 32 Physical Optics
That is, the central maximum occurs at θ = 00, as expected. For m = 1, the
first-order bright fringe is located at
θ1 = sin−1 (1)(500.0 nm) = 36.90
833.0 nm
For m = 2,
θ2 = sin−1 (2)(500.0 nm) = sin−1 1.20
833.0 nm
But the sine of θ cannot exceed the value of 1. Yet for m = 2, the sin θ = 1.20, which
is impossible. Therefore the second-order bright fringe (corresponding to m = 2) does
not exist.
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The Language of Physics
Diffraction
The bending of light around an obstacle, into the region that should be a shadow
area (p. ).
Monochromatic light
Light that consists of a single wavelength. In contrast, white light consists of light
of very many wavelengths. A laser is a good source of monochromatic light (p. ).
Young’s double-slit experiment
An experiment that superimposes light from two different slits to form a series of
bright and dark fringes upon a screen. The fringes can only be explained if light has
the characteristics associated with waves (p. ).
Michelson interferometer
An optical device that measures distances or wavelengths very accurately by
superimposing light from two different paths to give a series of dark and bright
fringes (p. ).
Thin film
A very thin piece of transparent material. When monochromatic light shines on the
film, interference effects can be observed due to a phase change that occurs on
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Chapter 32 Physical Optics
reflection and/or a phase change that occurs because of the difference in the optical
path of the two interfering light waves (p. ).
Nonreflecting glass
A piece of glass on which a thin film of transparent material has been placed to
cause reflected rays to be out of phase and hence, interfere destructively (p. ).
Newton’s rings
An interference pattern that occurs when light shines on a planoconvex lens that is
placed on a flat piece of glass. The pattern consists of a family of concentric dark
and bright fringes (p. ).
Fraunhofer diffraction
Diffraction from a slit in which the source is very far away from the slit, and the slit
is very far away from the screen, so that the light waves can be assumed to be plane
waves. When the effects of diffraction are combined with the interference of light
from a double slit, the resulting intensity distribution is similar to the interference
pattern except that the intensity of the fringes decreases farther away from the
central bright fringe (p. ).
Diffraction grating
Several parallel slits of equal width, equally spaced, that gives a characteristic
intensity distribution of sharper and narrower fringes (p. ).
Summary of Important Equations
E = E0 sin(kx − ωt)
Equation of plane light wave
φ = 2πd sin θ
λ
"
"
E = 2E 0 cos
sin kr 1 − ✬t +
2
2
Phase difference in double slit
Resultant wave in double slit
Iavg = ε0cE2
Intensity of an electromagnetic wave
Intensity distribution for a double slit I = ✒ 0 c4E 20 cos 2
Average intensity distribution for a double slit
where
(29.38)
I0 = 2ε0cE02
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(32.12)
(29.63)
"
"
sin 2 kr 1 − ✬t +
2
2
I avg = I 0 cos 2
(32.9)
"
2
(32.13)
(32.16)
(32.15)
Chapter 32 Physical Optics
Maximum intensity occurs for bright fringe for double slit
Condition for bright fringe for double slit
φ = mπ
2
d sin θ = mλ
Location of mth bright fringe for double slit
(32.17)
(32.18)
ym = Rmλ
d
Minimum intensity for double slit
φ = (2m − 1) π
2
2
Condition for dark fringe for double slit
d sin θ = (2m − 1) λ
2
Location of mth dark fringe for double slit
ym = (2m − 1)Rλ
2d
Michelson interferometer
x=m λ
2
λ = 2x
m
Decreased wavelength in a glass medium
λg = λair
ng
Destructive interference for thin film
2d = mλair
nw
Constructive interference for thin film
2d = (2m − 1)λair
2nw
Destructive interference for nonreflecting glass
2d = (2m − 1) λair
2nMgF2
Constructive interference for nonreflecting glass
2d = mλair
nMgF2
Condition for dark fringe for an air wedge
2d = mλ
(32.21)
(32.40)
Location of dark fringe for an air wedge
(32.43)
(32.22)
(32.23)
(32.24)
(32.27)
(32.28)
(32.29)
(32.32)
(32.34)
(32.37)
(32.39)
xm = mλl
2h
∆x = λl
2h
2d = (2m − 1) λ
2
xm = (2m − 1)λl
4h
rm = mλR
(32.45)
Condition for dark fringe for single slit diffraction
a sin θ = mλ
(32.51)
Condition for bright fringe for diffraction grating
d sin θ = mλ
(32.52)
Separation between dark fringes
Condition for bright fringe for an air wedge
Location of bright fringe for an air wedge
Radius of mth fringe for Newton’s rings
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(32.44)
(32.46)
(32.48)
Chapter 32 Physical Optics
Questions for Chapter 32
1. Why do we use monochromatic light rather than white light when doing
experiments with interference and diffraction?
2. How can you hear a radio station that is on the other side of the mountain
from you?
3. Two separate light sources, such as the headlights of your car, do not cause
interference. Why not?
4. What effect does the medium have on interference fringes for (a) light and
(b) sound?
5. When destructive interference occurs, what happens to the energy in the
light waves?
6. What effect does changing the slit width and the slit separation have on
the diffraction pattern?
7. Why don’t you observe interference from a thick film, such as an ordinary
piece of window glass?
8. Using an audio oscillator, can you set up a double-slit experiment with
sound waves? What would the approximate size of the slits have to be? What would
you hear on the far side of the double slit?
9. Why are diffraction gratings used more than prisms in spectroscopy?
*10. How is the resolving power of a telescope, the smallest resolution of an
object that can be determined by the telescope, affected by diffraction?
Problems for Chapter 32
32.2 The Interference of Light-- Young’s Double-Slit Experiment
1. In a Young’s double-slit experiment, a screen is placed 7.00 m behind a
double slit of 0.200-mm separation. If light of 589.0 nm shines on the slit, find the
value of the angle, θ, corresponding to the first 3 bright fringes.
2. In a Young’s double-slit experiment, a screen is placed 5.00 m behind a
double slit of 0.250-mm separation. If light of 589.0-nm wavelength shines on the
slit, find the value of θ corresponding to the first three dark fringes.
3. In a Young’s double-slit experiment, a screen is placed 7.00 m behind a
double slit of 0.400-mm separation. If light of 589.0-nm wavelength shines on the
slit, find the value of y on the screen corresponding to the first three bright fringes.
4. In a Young’s double-slit experiment, a screen is placed 6.00 m behind a
double slit of 0.387-mm separation. If light of 589.0-nm wavelength shines on the
slit, find the value of y on the screen corresponding to the first three dark fringes.
5. In a Young’s double-slit experiment, a screen is placed 7.00 m behind a
double slit of 0.200-mm separation. If light of 589.0-nm wavelength shines on the
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Chapter 32 Physical Optics
slit, find (a) the value of the angle θ corresponding to the first three bright fringes,
(b) the value of θ corresponding to the first three dark fringes, and (c) the value of y
on the screen corresponding to the first three bright and dark fringes.
Diagram for problem 5.
6. In a double-slit experiment the location of the second bright fringe on the
screen is found to be at 12.5 cm. If the distance to the screen is 7.00 m and the
separation between slits is 0.176 mm, find the wavelength of the light used.
7. Light of wavelength 600 nm is incident on a double slit whose separation is
0.00480 m. A bright fringe is observed at an angular displacement θ = 0.02000.
What is its order (i.e., the value of the integer m)?
8. Adjacent bright fringes from a double slit separated by 0.700 mm are
measured to be 1.71 mm apart on a screen 1.00 m away. What is the wavelength of
light used?
9. A Young’s double-slit experiment is performed with a slit separation of
0.0230 m. A second-order bright fringe is observed at 0.200 mm from the center of
the bright central fringe. If the wavelength of the light is 490 nm, find the distance
between the double slit and the screen on which the fringes were observed.
10. Show that the distance between bright fringes in a double-slit experiment
is equal to the distance between dark fringes.
11. What is the distance between fringes in a double-slit experiment when
the slit separation is (a) doubled and (b) halved?
32.3 The Interference of Light -- The Michelson Interferometer
12. A Michelson interferometer is used to determine the wavelength of a
source of light. If the movable mirror moves a distance of 0.250 mm when 800
fringes move across the telescope cross hair, find the wavelength of the light source.
13. A length of 5.00 cm is to be measured very accurately. If a Michelson
interferometer is to be used to measure this length, how many fringes of
monochromatic light of 589.0-nm wavelengths must pass the cross hair of the
telescope to correspond to this length?
14. How far must the movable mirror of the Michelson interferometer be
moved in order to observe 900 fringes of light of 589.0-nm wavelength?
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Chapter 32 Physical Optics
Diagram for problem 14.
*15. A glass tube, 10.0 cm long, is used as one arm of the Michelson
interferometer and the standard interference pattern is obtained using a
wavelength of light of 589 nm. Water is now slowly introduced into the glass tube.
When the tube is full, how many fringes will have passed through the cross hair of
the eyepiece?
32.4 Interference -- Thin Films
16. A light wave of 700 nm passes from air into glass. Find the wavelength of
the light in the glass.
17. A light wave has a wavelength of 450 nm in water. Find its wavelength
when it enters air.
18. A light wave of 589-nm wavelength in air passes from air into water and
then into glass. Find the wavelength of the light in the water and the glass.
19. A wave of light will be reflected and transmitted at the interfaces of the
air-glass-water-air surfaces, shown in the diagram. If the thickness of each surface
is negligible, find the phase relations between waves 1, 2, 3, and 4.
Diagram for problem 19.
20. Repeat problem 19 but let the thickness of the water surface equal the
thickness of the glass surface, which is equal to half of the incident wavelength of
the light in air.
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Chapter 32 Physical Optics
21. Repeat problem 19 but let the thickness of the water surface equal the
thickness of the glass surface, which is equal to the incident wavelength of the light
in air.
22. Repeat problem 19 but let the thickness of the water surface equal half of
the incident wavelength of the light in air and the thickness of the glass surface be
equal to the incident wavelength of the light in air.
23. A soap bubble is 200.0 nm thick. For what wavelength of light will (a) the
intensity of the reflected light be zero and (b) the intensity of the reflected light be
maximum?
24. A soap bubble is 300.0 nm thick. If white light shines on it, what color will
it appear in reflected light?
25. The wavelength of the maximum reflected light from a soap bubble is 710
nm. Find the thickness of the soap bubble.
26. What thickness of an oil film (n = 1.40) will give complete destructive
interference for the reflected light of 589.0-nm wavelength? For this thickness is
there any wavelength that will be completely reflected?
27. A thin film of oil of 100-nm thickness lies on a surface. Maximum
reflection from the oil occurs for a wavelength in air of 650 nm. Find the index of
refraction of the oil.
28. We want to prevent any light of 700.0-nm wavelength from being
reflected from a lens. What thickness of magnesium fluoride should be deposited on
this lens? For this thickness is there any wavelength of light that gives maximum
reflection?
29. If 4% of the incident light is lost by reflection at each lens of a multiple
lens optical system, how much light is lost if the system contains six lenses?
30. An air wedge is formed by placing a sheet of paper between the edges of
two glass plates 10.0 cm long. If 15 bright fringes are observed per cm, when light of
589.0-nm wavelength shines on the glasses, what is the thickness of the paper?
31. A hair is placed at the edge between two 10.0-cm glass plates of n = 1.52.
Light of 400-nm wavelength shines on the glass and an interference pattern is
formed. If the separation between the second and fourth dark fringe is measured to
be 0.800 mm, find the diameter of the hair.
32. Repeat problem 31 but use fused quartz glass (n = 1.46) and the glass is
immersed in glycerine (n = 1.47).
33. Show that the separation distance between bright fringes in an air wedge
is equal to the separation distance between dark fringes.
34. Find the equation for the width of a fringe in the interference of light from
an air wedge.
35. Newton’s rings are used to determine the wavelength of light. If the
radius of the fifth fringe is 7.00 mm and the radius of curvature of the lens is
measured by a spherometer to be 10.0 m, find the wavelength of light.
36. The radius of the tenth fringe in Newton’s rings is 10.0 mm. If the
wavelength of light used is 589.0 nm, what is the radius of curvature of the lens?
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Chapter 32 Physical Optics
32.5 Diffraction from a Single Slit
37. In the diffraction from a single slit of width, a = 15λ, find the angle θ at
which the second dark fringe occurs.
38. The first dark fringe for a single slit is found at an angle θ of 5.740 for
light of 589.0-nm wavelength. Find the value of the angle θ if the slit width is
halved.
39. In a single-slit experiment, the first dark fringe occurs at θ = 0.0100. If the
slit is 0.0035 m wide, find the wavelength of the light.
40. A source of light of λ = 633 nm shines on a single slit and a diffraction
pattern is found on a wall 1.00 m away. The first dark fringe is found at x = 5.00 cm
from the central maximum. Find the width of the slit.
41. In a single-slit experiment, a slit 3.05 × 10−4 m wide receives light of
wavelength 640 nm. Find the order of the dark fringe at the angular displacement θ
= 0.6000.
32.6 The Diffraction Grating
42. A diffraction grating has 800 lines per mm. Find the distance between
each slit.
43. A diffraction grating has 800 lines per mm. If monochromatic light of
589.0-nm wavelength shines on the grating, where will the first three bright fringes
be found?
44. How many bright fringes will occur on either side of the central maximum
for a diffraction grating of 600 lines per mm if the wavelength of light is 589.0 nm?
45. The first bright fringe of a diffraction grating of 600 lines/mm, is found at
an angle of 21.00. Find the wavelength of light of the source.
46. A first dark fringe is observed at angular displacement θ = 15.00 due to
light of frequency 6.00 × 1014 Hz passing through a diffraction grating. Find the
separation of the slits in the grating.
47. Find the longest wavelength that can be observed in second order for a
diffraction grating of 800 lines/mm.
48. Using light of 589.0-nm wavelength, a diffraction grating produces a
first-order fringe at 44.90. Find the number of lines/mm of this grating.
Additional Problems
*49. A double-slit experiment is performed with the light source, slits, and
screen placed under water. If the separation of the double slit is 0.350 mm, the
wavelength of light in air is 589.0 nm, and the screen is 2.00 m away from the slit,
find the value of y on the screen corresponding to the third bright fringe.
*50. In a double-slit experiment the intensity of the light on the screen varies
from zero at a dark fringe to I0 at the center of a bright fringe. Find the width of a
bright fringe assuming that a bright fringe exists for an intensity of I ≥ 0.
*51. A piece of flint glass (n = 1.57) is being ground into a planoconvex lens.
The focal length of the lens is to be 25.0 cm. The radius of curvature of the plane
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Chapter 32 Physical Optics
side of the glass is infinite. (a) Find the radius R1 of the convex side. Newton’s rings
will be used to verify that the lens has been ground to the correct radius. The lens is
placed on a flat piece of glass, and light of 589-nm wavelength illuminates the lens
and the characteristic pattern of Newton’s rings is obtained. (b) What must be the
radius of the fifth fringe such that the lens has the correct focal length?
*52. When two objects are too close together the diffraction pattern of each
object can overlap and hence the two objects cannot be seen distinctly. The smallest
angular distance between two objects that can be seen distinctly with an optical
system is called the resolution of the optical system. The smallest angular
resolution is determined by the distance between the central maximum and the first
intensity minimum (dark fringe) of the diffraction pattern of the single slit. (Then
the two diffraction patterns will not overlap and each object is distinct.) For the
single slit discussed in this chapter, the angular separation between the central
bright fringe and the dark fringe was determined from sin θ = λ/a. For a circular
aperture the result can be shown to be sin θ = 1.22 λ/a. What is the range of the
angular resolution of a human eye if the pupil diameter can vary between 2 and 6
mm, and the wavelength of light used is 550 nm?
53. If the human eye can resolve two objects when the angle between them is
1 minute of arc, how far in advance can you detect two headlights of a car if they are
separated by a distance of 1.32 m?
54. A diffraction grating of 1000 lines per cm is to be used for the entire
spectrum of visible light. (a) Find the angle of diffraction for the first-order red light
(720.0 nm) and first-order blue light (380.0 nm). (b) Find the angular separation.
(c) If the screen is placed 1.00 m away, find the linear separation.
55. Calculate the angular separation between the first-order fringes of the
two yellow lines in the sodium spectrum, having wavelengths of 589.0 nm and 589.6
nm, for a diffraction grating of 1200 lines/mm.
Interactive Tutorials
56. Young’s double-slit intensity. In a double-slit experiment, light of
wavelength λ = 589.0 nm impinges on a double slit that has a separation d = 0.285
mm. The screen is placed at a distance R = 3.50 m from the double slit. Find the
intensity distribution on the screen as a function of the angle θ in figure 32.5.
57. A double slit. In a double-slit experiment, light of wavelength λ = 589.0
nm impinges on a double slit that has a separation d = 0.285 mm. The screen is
placed at the distance R = 3.50 m from the double slit. Find (a) the value of θ
corresponding to the m = 3 bright fringe; (b) the value of y locating the bright fringe
on the screen, as seen in figure 32.5; (c) the value of θ corresponding to the m = 3
dark fringe; and (d) the value of y locating the dark fringe on the screen.
58. A soap bubble. A soap bubble of thickness d = 300 nm is illuminated by
white light. If the index of refraction of the soap film is n = 1.28, calculate (a) the
first three wavelengths (λ) of the incident light that will give a dark fringe and
(b) the first three wavelengths of the incident light that will give a bright fringe.
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Chapter 32 Physical Optics
59. A diffraction grating. Monochromatic light of wavelength λ = 589.0 nm
shines on a diffraction grating that has 400 lines per mm. Find (a) the distance d
between slits and (b) the value of θ corresponding to the m = 3 bright fringe.
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32-49