CHAPTER 27 INTERFERENCE AND
THE WAVE NATURE OF LIGHT
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION A radio station broadcasts simultaneously from two transmitting
antennas at two different locations. The radio that receives the broadcast could have better reception
depending on the location of the receiving antenna. According to the principle of linear
superposition, when the electromagnetic waves from the transmitting antenna arrive at the same
point, the resultant wave is the sum of the individual waves. The amplitude of the resultant wave
depends on the relative phase between the two waves. The relative phase between the waves
depends on the path length difference between the two waves. If the two waves arrive at the
receiving antenna so that the path length difference between them is equal to an integer number of
wavelengths, the two waves will be in phase, they will reinforce each other, and constructive
interference will occur. The resulting amplitude of the radio wave will be larger than it would be
from either transmitting antenna alone; therefore, the radio reception will be better. If the two waves
arrive at the receiving antenna so that the path length difference between them is equal to an odd
multiple of half of a wavelength, the two waves will be exactly out of phase, they will mutually
cancel, and destructive interference will occur. The receiving antenna will receive little or no signal
in this situation. As a result, the radio reception will be very bad. Thus, having two transmitting
antennas does not necessarily lead to better reception.
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2.
REASONING AND SOLUTION
a. The light waves coming from both slits in a Young's double slit experiment have their phases
shifted by an amount equivalent to half a wavelength. Since the light from both slits is changed by
the same amount, the relative phase difference between the light from the two slits is zero when the
light leaves the slits. When the light reaches the screen, the relative phase difference between light
waves from the two slits will be the same as if the phase of the waves had not been shifted at the slits.
Therefore, the pattern will be exactly the same as that described in the text.
b. Light coming from only one of the slits in a Young's double slit experiment has its phase shifted by
an amount equivalent to half of a wavelength. Now, there is a relative phase difference of one half of
a wavelength between the light leaving the slits. When the light reaches the screen, there will be a
relative phase difference due to the fact that light from each slit, in general, traveled along different
paths. In addition, there will be the initial phase difference that is equivalent to half of a wavelength.
Therefore, the pattern will be similar to that described in the text; however, the points of constructive
interference will be points of destructive interference, and the points of destructive interference will
be points of constructive interference. In other words, the positions of the light and dark fringes are
interchanged.
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3.
REASONING AND SOLUTION According to Equation 27.1, the bright fringes of a double slit
experiment occur at values of  for which sin  = m / d where m = 0, 1, 2, 3, ... According to
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INTERFERENCE AND THE WAVE NATURE OF LIGHT
Equation 27.2, the dark fringes occur at values of  for which sin  = [m  (1/ 2)] / d where m = 0,
1, 2, 3, . . . The slits S1 and S2 in Figure 27.5 are replaced with identical loudspeakers and use the
same ac electrical signal to drive them. The two sound waves produced will then be identical, and we
have the audio equivalent of Young's double slit experiment. Sound waves that reach the center of
the screen have traveled along paths of equal length; therefore, the sound waves from each
loudspeaker arrive there in phase and constructive interference occurs. Consequently, at the center
point on the screen, the sound will have maximum loudness. As you proceed to walk away from the
center (in either direction), the sound will decrease in volume. When the angle  is such that
sin  = (1/ 2) / d , the sound will decrease to zero amplitude. If you continue to walk beyond this
point, the loudness of the sound will increase. It will reach a maximum (a point of constructive
interference) when the angle  is such that sin  = /d . The sound will again decrease in loudness
until the next point of destructive interference occurs. This point occurs when sin  = (3/ 2) / d . In
general, there will a pattern of alternating loud and soft sounds heard as you walk away from the
center of the screen. The intensity at the points of constructive interference varies in a manner that is
similar to that shown in Figure 27.7. The central maximum has the greatest intensity. The sound at
each region of constructive interference decreases in intensity as one proceeds away from the center
(in either direction).
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4.
REASONING AND SOLUTION The angle  for the interference maximum in Young's double-slit
experiment is given by Equation 27.1: sin   m / d where m = 0, 1, 2, 3, . . . When the wavelength
 of the light is greater than the distance d between the slits, the ratio  / d is greater than one;
however, sin  cannot be greater than one. Therefore, it is not possible to see interference fringes
when the wavelength of the light is greater than the distance between the slits.
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5.
REASONING AND SOLUTION A camera lens is covered with a nonreflective coating that
eliminates the reflection of perpendicularly incident green light. If the light were incident on the
nonreflective coating at 45°, rather than perpendicularly, it would not be eliminated by the coating
and an observer would see it.
Following the reasoning of Example 4, part (b), if the refractive index of the film is less than the
refractive index of the lens, the minimum nonzero thickness of the nonreflecting coating is given by
t  coating / 4 . This result follows from the fact that the extra distance traveled by the wave that
travels through the coating is 2t when the wave strikes the film at normal incidence.
Chapter 27 Conceptual Questions
159
If the wave strikes the thin film at 45°, then
air
  = 45°
the extra distance traveled by the wave that
travels through the coating is no longer 2t.
n1
Instead, it is 2d, where d is the distance
shown in the figure at the right. The value of
n2
d depends on the value for 2, the angle of
d d
refraction. This angle, in turn, depends on
t
the refractive index n1 for air and n2 for the

lens coating
lens coating, according to Snell's law
(Equation 26.2). To eliminate the light that
is incident at 45°, the coating would have to
have a minimum nonzero thickness such that
d  coating / 4 . Since the coating is designed such that t  coating / 4 , it does not eliminate the light
incident at 45°.
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6.
REASONING AND SOLUTION When sunlight reflects from a thin film of soapy water, the film
appears multicolored, in part because destructive interference removes different wavelengths from the
light reflected at different places, depending on the thickness of the film. As the film becomes
thinner and thinner, the path length difference between the light reflected from the top of the film and
the light reflected from the bottom of the film becomes closer to zero. The light reflected from the
top of the film undergoes a phase change equivalent to half of a wavelength. Therefore, as the
thickness of the film approaches zero, the light reflected from the top of the film is almost exactly out
of phase with the light reflected from the bottom of the film. Therefore, as the film becomes thinner
and thinner, it looks darker and darker in reflected light. It appears black just before it breaks,
because at that point destructive interference occurs between the light reflected from the top and the
light reflected from the bottom of the film.
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7.
REASONING AND SOLUTION Two pieces of the same glass are covered with thin films of
different materials. The thickness t of each film is the same. In reflected sunlight, however, the films
have different colors.
This could occur because, in general, different materials have different refractive indices.
Following the reasoning of Example 4, parts (a) and (b), the wavelength air that is removed from the
reflected light corresponds to a value of film in the film that satisfies either of the following
conditions:
2t  mfilm ,
and
m = 0, 1, 2, 3, . . .
 
2t  m 
1
2
film ,
m = 0, 1, 2, 3, . . .
nfilm  nglass ,
(1)
nfilm  nglass .
(2)
However, the value of the wavelength film of the light in the film depends on the index of refraction
of the film. According to Equation 27.3, film  vacuum / n , where n is the index of refraction of the
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INTERFERENCE AND THE WAVE NATURE OF LIGHT
film. If the index of refraction of the two materials are different, then conditions (1) and (2) will be
satisfied for two different values of film. These values of film correspond to different values of air,
and, therefore, different wavelengths will be removed from the reflected light.
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8.
REASONING AND SOLUTION When a spherical surface is put in contact with an optically flat
plate, circular interference patterns, called Newton's rings, can be observed in reflected light, as
shown in Figure 27.16b. There is a dark spot at the center of the pattern of Newton's rings. The dark
spot occurs for the following reason. At the center of the plate, the spherical surface is in contact
with the plate. Destructive interference occurs there, because the thickness of the "air wedge" is
essentially zero at this point. Therefore, the only difference in phase between the light reflected from
the bottom of the spherical surface and the light reflected from the top of the optical flat is the halfwavelength phase change due to reflection from the optical flat. This situation is similar to that
discussed in part (b) of Example 6 in the text.
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9.
REASONING AND SOLUTION A thin film of material is floating on water (n = 1.33). When the
material has a refractive index of n = 1.20, the film looks bright in reflected light as its thickness
approaches zero. But when the material has a refractive index of n = 1.45, the film looks black in
reflected light as its thickness approaches zero.
When the material has a refractive index of n = 1.20, both the light reflected from the top of the
film and the light reflected from the bottom of the film occur under conditions where light is traveling
through a material with a smaller refractive index toward a material with a larger refractive index.
Therefore, both the light reflected from the top and the bottom of the film undergoes a phase change
upon reflection. Both phase changes are equivalent to half of a wavelength. Therefore, the
reflections introduce no net phase difference between the light reflected from the top and bottom of
the film. As the thickness of the film approaches zero, the path difference between the light reflected
from the top and the bottom of the film approaches zero. Since the path difference is close to zero,
and there is no relative phase difference due to reflection, the light reflected from the top will be in
phase with the light reflected from the bottom of the film, and constructive interference will occur.
Therefore, the film looks bright in reflected light.
When the material has a refractive index of n = 1.45, only the light reflected from the top of the
film is traveling through a material with lower refractive index toward a material with a larger
refractive index. Therefore, only the light reflected from the top undergoes a phase change upon
reflection. This phase change is equivalent to half of a wavelength. Therefore, there is a net phase
difference equivalent to half of a wavelength between the light reflected from the top and the light
reflected from the bottom of the film. As the thickness of the film approaches zero, the path
difference between the light reflected from the top and the light reflected from the bottom of the film
approaches zero. Since the path difference is close to zero, and there is a relative phase difference
due to reflection that is equal to half of a wavelength, the light reflected from the top will be exactly
out of phase with the light reflected from the bottom of the film, and destructive interference will
occur. Therefore, the film looks black in reflected light.
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10. REASONING AND SOLUTION A transparent coating is deposited on a glass plate and has a
refractive index that is larger than that of the glass. For a certain wavelength within the coating, the
Chapter 27 Conceptual Questions
161
thickness of the coating is a quarter wavelength. The coating enhances the reflection of the light
corresponding to this wavelength.
As discussed in part (a) of Example 4, only the light reflected from the top of the coating is
traveling in through a material (air) with lower refractive index toward a material coating with a
larger refractive index. Therefore, only the light reflected from the top of the coating undergoes a
phase change upon reflection. This change in phase is equivalent to half of a wavelength. Therefore,
there is a net phase difference equivalent to half of a wavelength between the light reflected from the
top and light reflected from the bottom of the coating. Since the thickness of the coating is a quarter
wavelength, the light reflected from the bottom of the coating traverses a distance equal to two times
a quarter wavelength, or half a wavelength, as it travels through the coating. The overall relative
phase difference between the light reflected from the top and the light reflected from the bottom of
the coating is equivalent to half of a wavelength due to the reflection from the top of the coating and
another half of a wavelength due to the path difference between the light reflected from the top and
the light reflected from the bottom. Thus, the overall phase difference is equivalent to a whole
wavelength, and the light reflected from the top is in phase with the light reflected from the bottom.
Constructive interference occurs, and the coating enhances light corresponding to that wavelength.
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11. REASONING AND SOLUTION Rayleigh's criterion states that two point objects are just resolved
when the first dark fringe in the diffraction pattern of one falls directly on the central bright fringe in
the diffraction pattern of the other. Equation 27.6 approximates the minimum angular displacement
between the two objects so that they are just resolved by an optical instrument:  min 1.22  / D ,
where  is the wavelength of light used and D is the diameter of the aperture of the optical
instrument. According to Equation 27.6, optical instruments with the highest resolution should have
the largest possible diameter D and utilize light of the shortest possible wavelength.
The f-number setting on a camera gives the ratio of the focal length of the camera lens to the
diameter of the aperture through which the light enters the camera. Therefore, smaller f-numbers
mean larger aperture diameters. According to Equation 27.6, if we wish to resolve two closely
spaced objects in a picture, we should adjust the aperture to have the largest possible diameter. As a
result, we should use a small f-number setting.
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12. REASONING AND SOLUTION According to Equation 27.6, Rayleigh's criterion is satisfied for
two point sources viewed through a circular lens when the angular separation  min is given by
 min  1.22  / D , where D is the diameter of the lens and is the wavelength of light as it travels
from the lens to the film. Since the camera is watertight, the region between the lens and the film is
filled with air, regardless of whether the camera is in air or under water. Thus, the wavelength is the
same in both situations, so both give rise to the same resolution.
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13. REASONING AND SOLUTION As discussed in the text, the extent to which a wave bends or
diffracts around the edges of an obstacle or opening is determined by the ratio /W, where  is the
wavelength of the wave and W is the width of the obstacle or opening. As the ratio /W becomes
larger, diffraction effects become more pronounced.
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INTERFERENCE AND THE WAVE NATURE OF LIGHT
The wavelength of light is extremely small and is normally expressed in nanometers
(1 nm = 10 –9 m ). Therefore, the ratio /W is very small for light, and when light passes through a
typical doorway, diffraction effects are not noticeable. Many sound waves encountered in the
everyday world have wavelengths that are comparable to the dimensions of a typical doorway.
Therefore, the ratio /W is nearly unity for these sounds, and when sound passes through a doorway,
diffraction effects are noticeable; you can hear around corners, although you cannot see around
corners.
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14. REASONING AND SOLUTION Light enters the eye through the pupil. The inner side of the pupil
has a refractive index of 1.36. When green light of wavelength vacuum = 550 nm passes through the
opening in the pupil, its wavelength is shortened according to Equation 27.3:
 eye 
 vacuum
n

550 nm
 404 nm
1.36
Light also passes through a hole in a sheet of opaque material. The hole has the same diameter as the
pupil and is surrounded by air. When green light of wavelength vacuum = 550 nm passes through the
hole, its wavelength remains the same, since nair is nearly 1.
As discussed in the text, the extent to which a wave bends or diffracts around the edges of an
obstacle or opening is determined by the ratio /W, where  is the wavelength of the wave and W is
the width of the obstacle or opening. As the ratio /W becomes larger, diffraction effects become
more pronounced. Since  air > eye, and W is the same for both the hole and the pupil, the ratio /W
is largest for the light that passes through the hole in the opaque material. Therefore, green light will
diffract more when it passes through the hole than when it passes through the pupil.
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15. REASONING AND SOLUTION Four light bulbs are arranged at the corners of a rectangle that is
three times longer than it is wide. You look at this arrangement perpendicular to the plane of the
rectangle. From very far away, your eyes cannot resolve the individual bulbs and you see a single
rectangular "smear" of light. From close in, you see the individual bulbs. The figures below illustrate
what you would see at two positions between these two extremes.
In the first case, you are still far away, but you have moved close enough so that you can begin to
resolve the bulbs at opposite ends of the diagonal. Essentially, you see a rectangular "smear" of light
with a gap in the middle.
gap
In the second case, you have moved closer still and now can resolve the two bulbs on one of the short
sides of the rectangle from the two on the other short side. Essentially, you see two "smears" of light,
each smear consisting of the two bulbs at the ends of a short side of the rectangle.
Chapter 27 Conceptual Questions
163
two bulbs on
two bulbs on
short side
short side
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16. REASONING AND SOLUTION The resolving power, or the ability to resolve two closely spaced
images, through any circular aperture (such as the pupil of your eye) is limited by diffraction effects.
The Rayleigh criterion for resolution focuses on these effects and is given by  min 1.22  / D
(Equation 27.6), where  is the wavelength of light, D is the diameter of the aperture, and min is the
minimum angle that can be resolved between two point objects by a circular aperture. A smaller
value for min or /D means a greater resolving power.
a. Suppose the pupil of your eye were elliptical instead of circular in shape, with the long axis of the
ellipse oriented in the vertical direction. Since the resolving power depends on the diameter of the
aperture and since the ellipse is longer than it is wide, the resolving power of your eye would not be
the same in the horizontal and vertical directions.
b. If the pupil of your eye were elliptical instead of circular in shape, with the long axis of the ellipse
oriented in the vertical direction, the "diameter" or "width" would be larger in the vertical direction
and smaller in the horizontal direction. Consider two "planes" of light rays. Plane 1 is oriented so
that neighboring rays are horizontal, and plane 2 is oriented so that neighboring rays are stacked
vertically. Since the "diameter" of your eye is largest in the vertical direction, the ratio /D is smaller
for light in plane 2 than it is for light in plane 1. Thus, diffraction effects will be less noticeable for
light rays stacked vertically. Therefore, the resolving power will be greater in the vertical direction.
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17. REASONING AND SOLUTION Rayleigh's criterion states that two point objects are just resolved
when the first dark fringe in the diffraction pattern of one falls directly on the central bright fringe in
the diffraction pattern of the other. Equation 27.6 approximates the minimum angular displacement
min between the two objects so that they are just resolved by an optical instrument:
 min 1.22  / D , where  is the wavelength of light used and D is the diameter of the aperture of the
optical instrument. According to Equation 27.6, optical instruments with the highest resolution
should have the largest possible diameter D and utilize light of the shortest possible wavelength.
Suppose you were designing an eye and could select the size of the pupil and the wavelength of the
electromagnetic waves to which the eye is sensitive. As far as the limitation created by diffraction is
concerned, the following list ranks the design choices in order of decreasing resolving power (greatest
first): (a) large pupil (large D) and ultraviolet wavelengths (small ); (c) small pupil (small D) and
ultraviolet wavelengths (small ); (b) small pupil (small D) and infrared wavelength (larger ).
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18. REASONING AND SOLUTION In our discussion of single-slit diffraction, we ignored the height
of the slit, in effect assuming that the height was much larger than the width. Suppose that the height
and width were the same size, so that diffraction occurred in both dimensions. The resulting pattern
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INTERFERENCE AND THE WAVE NATURE OF LIGHT
would change in the following way. The height of the horizontal fringes would be reduced.
Furthermore, there would be two sets of fringes. One set would be along the horizontal axis, and the
other set would be along the vertical axis. The two sets would share a common central maximum.
The intensity of the central maximum will be greatest. The intensity will decrease on either side of
the central maximum, in both the horizontal and the vertical directions. The figure below shows the
central maximum and the less intense higher order on either side of the central maximum in both
dimensions.
(continues)
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19. REASONING AND SOLUTION According to Equation 27.7, the angles  that locate principal
maxima of a diffraction grating satisfy the condition sin   m / d , where m = 0, 1, 2, 3, . . .
If the entire interference apparatus of a diffraction grating (light source, grating, and screen) were
immersed in water, the wavelength of the light would decrease from the value air to the value water.
This is because these values are related by Equation 27.3, water  vacuum / nwater  air / nwater ,
where we have used the fact that vacuum and air are nearly the same, since nair is nearly 1. Thus,
angles  that locate principal maxima of a diffraction grating immersed in water will be smaller than
when the apparatus is immersed in air. Consequently, the principle maxima will be closer together
when the apparatus is immersed in water relative to their location when the entire apparatus is
immersed in air.
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