College Physics: Explore and Apply
Second Edition
Chapter 24
Wave Optics
Prepared by
Kendra L Wallis
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What’s new in this chapter
• In our chapter on reflection and refraction, we developed
two models of light:
– Particle-bullet model in which a light beam is modeled as a
stream of tiny “light bullets”
– Wave model in which light is modeled as a propagating
vibration—a wave
 Hippolyte Fizeau and Léon Foucault established light travels
more slowly in water, which disproved the particle model in
1850.
• This chapter explores wave properties of light.
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Testing experiment Table 24.1
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Young’s double-slit experiment
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Qualitative wave-based explanation of
Young’s experiment
The interference pattern of dark and bright bands produced
by light passing through two closely spaced narrow slits can
be explained by the addition or superposition of circular
wavelets originating from the two slits. The length of the path
that each wave travels to a particular spot on the screen
determines the phase of the wave at that location. Locations
where two waves arrive in phase are the brightest; locations
where two waves arrive with different phases are darker; and
locations where the waves arrive completely out of phase
are darkest.
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Interference
• An interference pattern can be explained using the idea of
interference of wavelets.
– Constructive interference occurs when the crests from
the wavelets overlap, resulting in double-sized crests.
– Destructive interference means no light is present: the
wavelets are out of phase and cancel each other out.
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Quantitative analysis of the double-slit
experiment
• Waves in phase give constructive interference.
• Waves completely out of phase give destructive
interference.
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Mathematical location of the mth bright
band (1 of 2)
• Using trigonometry, we find:
mλ
sin  m 
d
(24.1)
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Mathematical location of the mth bright
band (2 of 2)
• We can find the distance ym between the 0th order
maximum and the mth maximum:
ym
tan  m 
L
(24.2)
• If the angle θm is very small, we can approximate and find:
mλL
ym 
d
(24.3)
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Testing experiment Table 24.2
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Double-slit interference (1 of 2)
When light of wavelength λ passes through two narrow slits
separated by distance d, it forms a series of bright and dark
bands on a screen beyond the slits. The angular deflection
θm of the center of the mth bright band to the side of the
central m = 0 bright band is determined using the equation
d sin  m  mλ
(24.1)
where m = 0, ±1, ± 2, ± 3, etc. The distance ym on a screen
from the center of the central bright band to the center of the
mth bright band depends on the angular deflection θm and
on the distance L of the screen from the slits:
L tan  m  ym
(24.2)
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Double-slit interference (2 of 2)
For small angles:
mλL
ym 
d
(24.3)
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TIP
We could have done the derivation for the minima (the
locations where two waves arrive out of phase and cancel
each other) instead of for the maxima and obtained a
relation for the angles at which the dark bands occur. This
happens when the path length difference from the two slits
to the dark band equals an odd number of half wavelengths:
Δ = (2m + 1)(λ/2), where m = 0, ±1, ±2, ±3, etc., and
consequently
(2m  1)λ / 2
sin  
d
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Young’s interference with white light
• When white light is used in the double-slit experiment, we
see the following:
• Because the angular deflection of red light appears greater
than that of blue light, we can conclude that red light must
have a longer wavelength than blue light.
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Relating the refractive index and the
speed of light in a substance
• The wave model of light
explains:
– Why light bends at the
boundary of two media
– Snell’s law by connecting the
medium’s index of refraction to
the speed of light in that
medium
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Refractive index
The refractive index n of a medium equals the ratio of the
speed of light c in vacuum (or in air) to the speed of light v in
the medium:
c
n
v
(24.4)
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Refractive index and the color of light
• Different indexes of refraction
for different colors mean that
light of different colors or light
waves of different frequencies
travel at different speeds in
the same medium.
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TIP
Note that when light travels into a medium with a different
index of refraction, its frequency does not change, but its
wavelength does.
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Chromatic aberration in lenses: A
practical problem in optical instruments
• The image locations for each wavelength of light are
slightly different, leading to distortions.
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Monochromatic and coherent waves (1 of 2)
• Two light bulbs do not produce an interference pattern.
• The waves are not coherent; they add together randomly
and produce no interference pattern on the wall.
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Monochromatic and coherent waves (2 of 2)
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Coherent monochromatic waves
Only waves of constant frequency (monochromatic) and
having constant phase difference (coherent) can add to
produce an interference pattern. In Young’s double-slit
experiment, the light from the slits is coherent because the
disturbance at each slit is caused by the same passing wave
front.
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Gratings: An application of interference
• Gratings allow us to analyze
wavelengths of light emitted
by many sources.
• Bright bands sharpen with
increasing number of slits.
• Interference maxima are
observed at angular
deflections given by:
sin  m 
mλ
, where m  0,  1,  2,  3, 
d
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TIP
Gratings are usually labeled in terms of the number of slits
per millimeter or per centimeter, for example, 100 slits/cm. In
such a grating, the distance between the slits is one
hundredth of a centimeter. If a grating has 1000 slits/mm, the
distance between the slits is one thousandth of a millimeter.
You can find the distance between the slits by dividing 1 by
the number of slits per unit length. The distance comes out
in the respective units (cm or mm).
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White light incident on grating
• A spectrum produced by a grating is a result of the light of
different wavelengths interfering constructively at different
locations.
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CDs and DVDs: Reflection gratings
• The grooves in a CD play the role of the slits: the reflected
white light forms interference maxima for different colors at
different angles.
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Spectrometer
• A spectrometer is used to
analyze wavelengths of
light from different sources.
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Thin-film interference
• The beautiful, swirling colors on soap bubbles, oil slicks,
butterfly wings, and a peacock’s tail feathers are the result
of thin-film interference.
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Bright and dark bands due to reflected
monochromatic light (1 of 2)
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Bright and dark bands due to reflected
monochromatic light (2 of 2)
• Light reflection occurs at both the front surface and the
back surface. Two factors affect the superposition of these
two reflected waves:
– Phase change upon reflection
– Phase difference due to path length difference
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Path length difference and wavelength
• If the refractive index n of the thin film is greater than 1.0,
the wavelength of the light in the film is:
λ medium
vmedium c /nmedium
λ air
c /f




f
f
nmedium nmedium
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TIP
When light reflected off a soap bubble interferes
constructively, incident light is reflected. Soap bubbles are
transparent, which means that most of the incident light
passes through, even when constructive interference occurs.
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Thin film on glass surface
• Glass surfaces are often covered with a thin film. Waves
reflecting from the film interfere destructively, minimizing
reflected light.
• All incident light passes through.
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Examples of thin-film interference for
monochromatic incident light (1 of 2)
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Examples of thin-film interference for
monochromatic incident light (2 of 2)
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Reflection patterns on a soap bubble in
white light
• White light (wavelengths from 400 to 700 nm) is usually
incident on a thin film, rather than monochromatic light.
– Due to different wavelengths, different thicknesses, and
different angles, light of only a small wavelength range
is destructively reduced in intensity at any particular
location on a soap bubble.
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Complementary colors
• Complementary colors are the colors that are left when
light of a small wavelength range is subtracted from white
light.
• Different from the spectrum produced by a grating or
prism.
– These devices separate in space the primary colors
that are combined inside a beam of white light.
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TIP
Remember that it is the frequency of light that determines
its color. The frequency does not change when light
travels from one medium to another. The wavelength does
change. Thus when you hear “the wavelength of green
light is 550 nm,” check whether this statement assumes
that the light is propagating in a vacuum.
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Lens coatings
• Film of a particular thickness is used to reduce the
reflected light of a particular wavelength.
• The thickness of the coating on glass lenses for cameras,
microscopes, and eyeglasses is usually chosen to reduce
light at a wavelength of 550 nm, the center of the visible
spectrum.
• A lens with a thin-film coating has a purple hue because it
reflects red and violet light more than other colors.
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Bird and butterfly colors
• Many colors in the natural world, such as those of flower
petals and leaves, are caused by organic pigments that
absorb certain colors and reflect others.
• Some feathers and insect bodies consist of microscopic
translucent structures that act like thin films to produce
destructive and constructive interference of light.
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TIP
Remember that interference patterns for light that are stable
in time and thus observable appear when two coherent
waves arrive at the same location at the same time. These
coherent waves can be created in two ways:
1. by combining waves from two sources that produce
coherent, monochromatic waves (as in double-slit
interference), and
2. by dividing a monochromatic wave into two waves and
combining them (as in thin-film interference).
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Diffraction of light (1 of 2)
• In the pattern produced by light passing through two slits
on the screen, notice that in addition to alternating bright
and dark bands, there is an overall periodic modulation of
the brightness in the pattern.
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Diffraction of light (2 of 2)
• Shine laser light through a narrow slit with variable width.
• Width of central maximum increases as width of slit
decreases.
• Spreading of light combined with additional bright and dark
regions is diffraction.
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Quantitative analysis of single-slit
diffraction
• The slit is not infinitely narrow.
• Model slit as consisting of
multiple tiny mini-slit regions
that become sources.
• Light emitted by all mini-slits
can interfere because they
travel different distances.
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TIP
Notice that w sin θm = mλ is similar to Eq. (24.1) for doubleslit interference. However, for a double slit, Eq. (24.1)
describes the angles at which the maxima (bright bands) are
observed, and m = 0 is allowed. For a single slit, Eq. (24.6),
w sin θm = mλ, describes angles at which the minima (dark
bands) are observed, and m = 0 is not allowed since there is
no dark band at θ = 0°.
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Single-slit diffraction
When monochromatic light is incident on a slit whose width
is approximately 1000 wavelengths of light or less, we
observe a series of bright and dark bands of light on a
screen beyond the slit. The bands are caused by the
interference of light from different mini-slit regions within the
slit. The angle between lines drawn from the slit to the
minima, the dark bands, and a line drawn from the slit to the
central maximum is determined using the equation
w sin  m  mλ
(24.6)
where w is the slit width, λ is the wavelength of the light, and
m = ±1, ±2, ±3, … (not zero). The dark bands on the screen
are located at positions ym = L tan θm relative to the center of
the pattern. L is the slit-screen distance.
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The Poisson spot—testing the wave
model of light
• In 1818, Poisson set out to
disprove the wave model of light.
• Narrow beam of light shined at
small round obstacle should
illuminate edges, which become
sources.
• Bright spot should appear in
middle of shadow.
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Diffraction and everyday experience
• Wavelengths of visible light are small, so we seldom
observe diffraction in everyday life.
• All waves exhibit diffraction, including sound waves, which
have much longer wavelengths.
• Diffraction explains why we can hear a person talking
around a corner in another room when the door is open.
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Resolving power
• Diffraction pattern due to
light passing through a
small hole resembles
narrow slit pattern.
• Angle between line drawn
from hole to center of
pattern and line drawn
toward first dark ring is
found by:
1.22λ
sin  
D
(24.7)
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Resolving ability of a lens (1 of 3)
• Diffraction produces
angular spread.
• Instead of passing
through focal point at
distance f from lens,
parallel light rays form
central disk.
• Wave properties of light
make perfectly sharp
image impossible.
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Resolving ability of a lens (2 of 3)
• Telescope lenses of
increasing diameter
resolve two stars.
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Resolving ability of a lens (3 of 3)
• Angular separation of two sources affects ability to resolve
them.
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Rayleigh criterion
The minimal angular separation of two objects that can be
resolved (perceived as separate) is the limit of resolution αres
of the instrument:
 res
1.22λ

D
(24.8)
where D is the diameter of the opening through which light
enters and λ is the wavelength of the light. In this equation,
αres is measured in radians.
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TIP
When the limit of resolution of the instrument increases, it
means that its resolving power decreases. The larger αres,
the more difficult it is to resolve closely separated objects.
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Problem-Solving Strategy 24.1
• Analyzing processes using the wave model of light
–
–
–
–
Sketch and translate
Simplify and diagram
Represent mathematically
Solve and evaluate
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Skills for analyzing processes using the
wave model of light (1 of 2)
• Sketch and translate:
– Visualize the situation and then sketch it.
– Identify given physical quantities and unknowns.
• Simplify and diagram:
– Decide if the sources in the problem produce coherent
waves.
– Decide if the small-angle approximation is valid.
– Decide if the slit widths for multiple slits are wide enough
that you have to consider single-slit diffraction as well as
multiple-slit interference.
– If useful, represent the situation with a wave front diagram
showing the overlapping crests and troughs of the light
waves from different sources.
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Skills for analyzing processes using the
wave model of light (2 of 2)
• Represent mathematically:
– Describe the situation mathematically.
– Use geometry if needed.
• Solve and evaluate:
– Use the mathematical description of the process to solve for
the desired unknown quantity.
– Evaluate the result. Does it have the correct units? Is its
magnitude reasonable? Do the limiting cases make sense?
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A final note about light
• We have applied the wave model of light to explain
situations in which light passes through small openings or
around small objects. The wave model also explains the
interaction of light with large objects (reflection and
refraction).
• We have not yet established a mechanism for the wave
model. What is actually vibrating when light waves
propagate?
• This question will be explored in the next chapter.
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Summary (1 of 6)
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Summary (2 of 6)
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Summary (3 of 6)
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Summary (4 of 6)
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Summary (5 of 6)
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Summary (6 of 6)
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