Chapter 24
Wave Optics
Waves and Optics
Physical Optics (1.5 weeks)
Learning Objectives:
1. Interference and diffraction
Students should understand the interference and diffraction of waves, so they can:
a) Apply the principles of interference to coherent sources in order to:
(1) Describe the conditions under which the waves reaching an observation point
from two or more sources will all interfere constructively, or under which the
waves from two sources will interfere destructively.
(2) Determine locations of interference maxima or minima for two sources or
determine the frequencies or wavelengths that can lead to constructive or
destructive interference at a certain point.
(3) Relate the amplitude produced by two or more sources that interfere
constructively to the amplitude and intensity produced by a single source.
b) Apply the principles of interference and diffraction to waves that pass through a
single or double slit or through a diffraction grating, so they can:
(1) Sketch or identify the intensity pattern that results when monochromatic waves
pass through a single slit and fall on a distant screen, and describe how this
pattern will change if the slit width or the wavelength of the waves is changed.
(2) Calculate, for a single-slit pattern, the angles or the positions on a distant screen
where the intensity is zero.
(3) Sketch or identify the intensity pattern that results when monochromatic waves
pass through a double slit, and identify which features of the pattern result from
single-slit diffraction and which from two-slit interference.
(4) Calculate, for a two-slit interference pattern, the angles or the positions on a
distant screen at which intensity maxima or minima occur.
(5) Describe or identify the interference pattern formed by a diffraction grating,
calculate the location of intensity maxima, and explain qualitatively why a
multiple-slit grating is better than a two-slit grating for making accurate
determinations of wavelength.
c) Apply the principles of interference to light reflected by thin films, so they can:
(1) State under what conditions a phase reversal occurs when light is reflected from
the interface between two media of different indices of refraction.
(2) Determine whether rays of monochromatic light reflected perpendicularly from
two such interfaces will interfere constructively or destructively, and thereby
account for Newton’s rings and similar phenomena, and explain how glass may
be coated to minimize reflection of visible light.
2. Dispersion of light and the electromagnetic spectrum
Students should understand dispersion and the electromagnetic spectrum, so they can:
a) Relate a variation of index of refraction with frequency to a variation in refraction.
b) Know the names associated with electromagnetic radiation and be able to arrange
in order of increasing wavelength the following: visible light of various colors,
ultraviolet light, infrared light, radio waves, x-rays, and gamma rays.
Read and take notes on
pgs. 480-490
in Conceptual Physics Text
Read and take notes on
pgs. 790-794
in College Physics Text
Wave Optics
• The wave nature of light is needed to explain
various phenomena.
– Interference
– Diffraction
– Polarization
• The particle nature of light was the basis for
ray (geometric) optics.
Introduction
Interference
• Light waves interfere with each other much
like mechanical waves do.
• All interference associated with light waves
arises when the electromagnetic fields that
constitute the individual waves combine.
Section 24.1
Conditions for Interference
• For sustained interference between two
sources of light to be observed, there are two
conditions which must be met.
– The sources must be coherent.
• The waves they emit must maintain a constant phase
with respect to each other.
– The waves must have identical wavelengths.
Section 24.1
Producing Coherent Sources
• Light from a monochromatic source is allowed to
pass through a narrow slit.
• The light from the single slit is allowed to fall on a
screen containing two narrow slits.
• The first slit is needed to insure the light comes from
a tiny region of the source which is coherent.
• Old method
Section 24.1
Producing Coherent Sources, Cont.
• Currently, it is much more common to use a
laser as a coherent source.
• The laser produces an intense, coherent,
monochromatic beam over a width of several
millimeters.
• The laser light can be used to illuminate
multiple slits directly.
Section 24.1
Link to a cool YouTube
video on Double Slit Experiment
Link to Bright Storm on
Double Slit Experiment
Young’s Double Slit Experiment
• Thomas Young first demonstrated interference in
light waves from two sources in 1801.
• Light is incident on a screen with a narrow slit, So
• The light waves emerging from this slit arrive at a
second screen that contains two narrow, parallel
slits, S1 and S2
Section 24.2
Young’s Double Slit Experiment, Diagram
• The narrow slits, S1 and
S2 act as sources of
waves.
• The waves emerging
from the slits originate
from the same wave
front and therefore are
always in phase.
Section 24.2
Resulting Interference Pattern
• The light from the two slits form a visible pattern on
a screen.
• The pattern consists of a series of bright and dark
parallel bands called fringes.
• Constructive interference occurs where a bright
fringe appears.
• Destructive interference results in a dark fringe.
Section 24.2
Fringe Pattern
• The fringe pattern formed
from a Young’s Double Slit
Experiment would look like
this.
• The bright areas represent
constructive interference.
• The dark areas represent
destructive interference.
Section 24.2
Interference Patterns
• Constructive
interference occurs at
the center point.
• The two waves travel
the same distance.
– Therefore, they arrive in
phase.
Section 24.2
Interference Patterns, 2
• The upper wave has to
travel farther than the
lower wave.
• The upper wave travels one
wavelength farther.
– Therefore, the waves arrive in
phase.
• A bright fringe occurs.
Section 24.2
Interference Patterns, 3
• The upper wave travels onehalf of a wavelength farther
than the lower wave.
• The trough of the bottom
wave overlaps the crest of
the upper wave.
• This is destructive
interference.
– A dark fringe occurs.
Section 24.2
Geometry of Young’s Double Slit Experiment
Section 24.2
Interference Equations
• The path difference, δ, is
found from the small
triangle.
• δ = r2 – r1 = d sin θ
– This assumes the paths are
parallel.
– Not exactly parallel, but a
very good approximation
since L is much greater than d
Section 24.2
Interference Equations, 2
• For a bright fringe, produced by constructive
interference, the path difference must be
either zero or some integral multiple of the
wavelength.
• δ = d sin θbright = m λ
– m = 0, ±1, ±2, …
– m is called the order number.
• When m = 0, it is the zeroth order maximum.
• When m = ±1, it is called the first order maximum.
Section 24.2
Interference Equations, 3
• When destructive interference occurs, a dark
fringe is observed.
• This needs a path difference of an odd half
wavelength.
• δ = d sin θdark = (m + ½) λ
– m = 0, ±1, ±2, …
Section 24.2
Interference Equations, 4
• The positions of the fringes can be measured
vertically from the zeroth order maximum.
• y = L tan θ  L sin θ
• Assumptions
– L >> d
– d >> λ
• Approximation
– θ is small and therefore the approximation tan θ  sin θ
can be used.
• The approximation is true to three-digit precision only for angles
less than about 4°
Section 24.2
Interference Equations, Final
• For bright fringes
• For dark fringes
Section 24.2
Uses for Young’s Double Slit
Experiment
• Young’s Double Slit Experiment provides a
method for measuring wavelength of the light.
• This experiment gave the wave model of light
a great deal of credibility.
– It is inconceivable that particles of light could
cancel each other.
Section 24.2
Since there is no Khan academy videos on wave optics you will watch
Walter Lewin from MIT. These videos are appropriate for AP Physics
B students. This one lesson is broken down into two parts totaling
one hour and twenty minutes. These are long but a must watch!
Link to Walter Lewin on double slit and interference
(start to minute 30:36)
Link to sound interference Optional
(30:36-40:20)
EXAMPLE 24.1 Measuring the Wavelength of a Light Source
Goal Show how Young's experiment can be used to measure the wavelength of
coherent light.
Problem A screen is separated from a double-slit source by 1.20 m. The distance
between the two slits is 0.0300 mm. The second-order bright fringe (m = 2) is
measured to be 4.50 cm from the centerline. Determine (a) the wavelength of the
light and (b) the distance between adjacent bright fringes.
Strategy The bright fringe equation relates the positions of the bright fringes to the
other variables, including the wavelength of the light. Substitute into this equation
and solve for λ. Taking the difference between ym+1 and ym results in a general
expression for the distance between bright fringes.
SOLUTION
(a) Determine the wavelength of the light.
Solve the bright fringe equation for the wavelength and substitute the values m = 2,
y2 = 4.50 10-2 m, L = 1.20 m, and d = 3.00 10-5 m.
y2d (4.50 10-2 m)(3.00 10-5 m)
=
λ=
2(1.20 m)
mL
= 5.63 10-7 m = 563 nm
(b) Determine the distance between adjacent bright fringes.
Use the bright fringe equation again to find the distance between any adjacent bright
fringes (here, those characterized by m and m + 1).
Δy = ym +1 - ym =
=
λL
d
(m + 1) -
λL
λL
m=
d
d
(5.63 10-7 m)(1.20 m) = 2.25
cm
3.00 10-5 m
LEARN MORE
Remarks This calculation depends on the angle θ being small because the small
angle approximation was implicitly used. The measurement of the position of the
bright fringes yields the wavelength of light, which in turn is a signature of atomic
processes, as is discussed in the chapters on modern physics. This kind of
measurement therefore helped open the world of the atom.
Question Which of the following make the separation between fringes greater in the
two slit interference experiment? (Select all that apply.)
Narrower slits.
Smaller separation of the two slits.
Larger separation of the two slits.
Wider slits.
Read and take notes on
pgs. 795-798
in College Physics Text
Lloyd’s Mirror
• An arrangement for
producing an interference
pattern with a single light
source
• Waves reach point P either
by a direct path or by
reflection.
• The reflected ray can be
treated as a ray from the
source S’ behind the mirror.
Section 24.3
Interference Pattern from the Lloyd’s Mirror
• An interference pattern is formed.
• The positions of the dark and bright fringes
are reversed relative to pattern of two real
sources.
• This is because there is a 180° phase change
produced by the reflection.
Section 24.3
Phase Changes Due To Reflection
• An electromagnetic wave undergoes a phase change of
180° upon reflection from a medium of higher index of
refraction than the one in which it was traveling.
– Analogous to a reflected pulse on a string
Section 24.3
Phase Changes Due To Reflection, Cont.
• There is no phase change when the wave is reflected from a
boundary leading to a medium of lower index of refraction.
– Analogous to a pulse in a string reflecting from a free support
Section 24.3
Read and take notes on
pgs. 490-494
in Conceptual Physics Text
Interference in Thin Films
• Interference effects are commonly
observed in thin films.
– Examples are soap bubbles and oil on water
• The interference is due to the interaction of
the waves reflected from both surfaces of
the film.
Section 24.4
Interference in Thin Films, 2
• Facts to remember
– An electromagnetic wave traveling from a medium
of index of refraction n1 toward a medium of index
of refraction n2 undergoes a 180° phase change
on reflection when n2 > n1
• There is no phase change in the reflected wave if n2 <
n1
– The wavelength of light λn in a medium with index
of refraction n is λn = λ/n where λ is the
wavelength of light in vacuum.
Section 24.4
Interference in Thin Films, 3
• Ray 1 undergoes a phase
change of 180° with
respect to the incident ray.
• Ray 2, which is reflected
from the lower surface,
undergoes no phase change
with respect to the incident
wave.
Section 24.4
Interference in Thin Films, 4
• Ray 2 also travels an additional distance of 2t before
the waves recombine.
• For constructive interference
– 2 n t = (m + ½ ) λ m = 0, 1, 2 …
• This takes into account both the difference in optical path length
for the two rays and the 180° phase change
• For destructive interference
– 2 n t = m λ m = 0, 1, 2 …
Section 24.4
Interference in Thin Films, 5
• Two factors influence interference.
– Possible phase reversals on reflection
– Differences in travel distance
• The conditions are valid if the medium above the top
surface is the same as the medium below the bottom
surface.
• If the thin film is between two different media, one
of lower index than the film and one of higher index,
the conditions for constructive and destructive
interference are reversed.
Section 24.4
Interference in Thin Films, Final
• Be sure to include two effects when analyzing
the interference pattern from a thin film.
– Path length
– Phase change
Section 24.4
Newton’s Rings
• Another method for viewing interference is to
place a planoconvex lens on top of a flat glass
surface.
• The air film between the glass surfaces varies in
thickness from zero at the point of contact to
some thickness t.
• A pattern of light and dark rings is observed.
– These rings are called Newton’s Rings.
– The particle model of light could not explain the origin of the rings.
• Newton’s Rings can be used to test optical lenses.
Newton’s Rings, Diagram
Section 24.4
Problem Solving Strategy with Thin
Films, 1
• Identify the thin film causing the interference.
• Determine the indices of refraction in the film
and the media on either side of it.
• Determine the number of phase reversals:
zero, one or two.
Section 24.4
Problem Solving with Thin Films, 2
• The interference is constructive if the path difference
is an integral multiple of λ and destructive if the path
difference is an odd half multiple of λ.
– The conditions are reversed if one of the waves undergoes
a phase change on reflection.
• Substitute values in the appropriate equation.
• Solve and check.
Section 24.4
Problem Solving with Thin Films, 3
Equation
m = 0, 1, 2, …
1 phase
reversal
2nt = (m + ½) l constructive
2nt = m l
destructive
Section 24.4
0 or 2 phase
reversals
destructive
constructive
Interference in Thin Films, Example
• An example of different
indices of refraction
• A coating on a solar cell
• There are two phase
changes
Section 24.4
Link to Walter Lewin on Thin Film
(From minute 40:20 - 53:00 then 56:00 – 117:00 )
EXAMPLE 24.2 Interference in a Soap Film
Goal Study constructive interference effects in a thin film.
Problem (a) Calculate the minimum thickness of a soap-bubble film (n = 1.33) that
will result in constructive interference in the reflected light if the film is illuminated
by light with wavelength 602 nm in free space. (b) Recalculate the minimum
thickness for constructive interference when the soap-bubble film is on top of a glass
slide with n = 1.50.
Strategy In part (a) there is only one inversion, so the condition for constructive
interference is 2nt = (m + 1/2)λ. The minimum film thickness for constructive
interference corresponds to m = 0 in this equation. Part (b) involves two inversions,
so 2nt = mλ is required
SOLUTION
(a) Calculate the minimum thickness of the soap bubble film that will result in
constructive interference.
Solve 2nt = λ/2 for the thickness t and substitute.
λ 602 nm
= 113 nm
t= =
4n 4(1.33)
(b) Find the minimum soap film thickness when the film is on top of a glass slide
with n = 1.33.
Write the condition for constructive interference, when two inversions take place.
2nt = mλ
Solve for t and substitute.
mλ 1·(602 nm)
=
= 226 nm
t=
2(1.33)
2n
LEARN MORE
Remarks The swirling colors in a soap bubble result from the thickness of the soap
layer varying from one place to another.
Question A soap film looks red in one area and violet in a nearby area. In which
area is the soap film thicker?
In the red region, because red light has a shorter wavelength than violet light.
In the red region, because red light has a longer wavelength than violet light.
In the violet region, because red light has a shorter wavelength than violet light.
In the violet region, because red light has a longer wavelength than violet light.
EXAMPLE 24.3 Nonreflective Coatings for Solar Cells and Optical Lenses
Reflective losses from a silicon solar cell are minimized by coating it with a
thin film of silicon monoxide (SiO).
Goal Study destructive interference effects in a thin
film when there are two inversions.
Problem Semiconductors such as silicon are used to
fabricate solar cells, devices that generate electric
energy when exposed to sunlight. Solar cells are
often coated with a transparent thin film, such as
silicon monoxide (SiO; n = 1.45), to minimize
reflective losses. A silicon solar cell (n = 3.50) is
coated with a thin film of silicon monoxide for this
purpose. Assuming normal incidence, determine the minimum thickness of the film
that will produce the least reflection at a wavelength of 552 nm.
Strategy Reflection is least when rays 1 and 2 in the figure meet the condition for
destructive interference. Note that both rays undergo 180° phase changes on
reflection. The condition for a reflection minimum is therefore 2nt = λ/2.
SOLUTION
Solve 2nt = λ/2 for t, the required thickness.
λ 552 nm
= 95.2 nm
t= =
4n 4(1.45)
LEARN MORE
Remarks Typically, such coatings reduce the reflective loss from 30% (with no
coating) to 10% (with a coating), thereby increasing the cells efficiency because
more light is available to create charge carriers in the cell. In reality the coating is
never perfectly nonreflecting because the required thickness is wavelength
dependent and the incident light covers a wide range of wavelengths.
Question To minimize reflection of light with a smaller wavelength, should the
thickness of the coating be thicker or thinner?
thicker
thinner
EXAMPLE 24.4 Interference in a Wedge-Shaped Film
Interference bands in reflected light can be observed by illuminating a wedgeshaped film with monochromatic light. The dark areas in the interference pattern
correspond to positions of destructive interference.
Goal Calculate interference effects when the film has
variable thickness.
Problem A pair of glass slides 10.0 cm long and with n = 1.52 are separated on one
end by a hair, forming a triangular wedge of air, as illustrated in the figure. When
coherent light from a helium—neon laser with wavelength 633 nm is incident on the
film from above, 15.0 dark fringes per centimeter are observed. How thick is the
hair?
Strategy The interference pattern is created by the thin film of air having variable
thickness. The pattern is a series of alternating bright and dark parallel bands. A dark
band corresponds to destructive interference, and there is one phase reversal, so 2nt
= mλ should be used. We can also use the similar triangles in the figure to obtain the
relation t/x = D/L. We can find the thickness for any m, and if the position x can also
be found, this last equation gives the diameter of the hair, D.
SOLUTION
Solve the destructive-interference equation for the thickness of the film, t, with n = 1
for air.
t = mλ/2
If d is the distance from one dark band to the next, then the x-coordinate of the mth
band is a multiple of d.
x = md
By dimensional analysis, d is just the inverse of the number of bands per centimeter.
d = (15.0 bands/cm)-1 = 6.67 10-2 cm/band
Now use similar triangles and substitute all the information.
t mλ/2 λ D
=
=
=
x md 2d L
Solve for D and substitute given values.
λL (6.33 10-7 m)(0.100 m)
-5
=
4.75
10
m
D= =
2(6.67 10-4 m)
2d
LEARN MORE
Remarks Some may be concerned about interference caused by light bouncing off
the top and bottom of, say, the upper glass slide. It's unlikely, however, that the
thickness of the slide will be half an integer multiple of the wavelength of the helium
neon laser (for some very large value of m). In addition, in contrast to the air wedge,
the thickness of the glass doesn't vary.
Question If the air wedge is filled with water, how are the wavelength and the
distance between dark bands affected? (Select all that apply.)
The wavelength decreases.
together.
The interference bands get closer
The interference bands get farther apart.
The wavelength increases.
Read and take notes on
pgs. 800-801
in College Physics Text
CD’s and DVD’s
• Data is stored digitally.
– A series of ones and zeros read by laser light reflected
from the disk
• Strong reflections correspond to constructive
interference.
– These reflections are chosen to represent zeros.
• Weak reflections correspond to destructive
interference.
– These reflections are chosen to represent ones.
Section 24.5
CD’s and Thin Film Interference
• A CD has multiple tracks.
– The tracks consist of a sequence of pits of varying
length formed in a reflecting information layer.
• The pits appear as bumps to the laser beam.
– The laser beam shines on the metallic layer
through a clear plastic coating.
Section 24.5
Reading a CD
• As the disk rotates, the laser
reflects off the sequence of
bumps and lower areas into
a photodector.
– The photodector converts the
fluctuating reflected light
intensity into an electrical
string of zeros and ones.
• The pit depth is made equal
to one-quarter of the
wavelength of the light.
Section 24.5
Reading a CD, Cont.
• When the laser beam hits a rising or falling bump
edge, part of the beam reflects from the top of the
bump and part from the lower adjacent area.
– This ensures destructive interference and very low
intensity when the reflected beams combine at the
detector.
• The bump edges are read as ones.
• The flat bump tops and intervening flat plains are
read as zeros.
Section 24.5
DVD’s
• DVD’s use shorter wavelength lasers.
– The track separation, pit depth and minimum pit
length are all smaller.
– Therefore, the DVD can store about 30 times more
information than a CD.
Section 24.5
EXAMPLE 24.5 Pit Depth in a CD
Cross section of a CD showing metallic pits of depth t and a laser beam detecting the edge of a pit.
Goal Apply interference principles to a CD.
Problem Find the pit depth in a CD that has a plastic transparent layer with index of
refraction of 1.60 and is designed for use in a CD player using a laser with a
wavelength of 7.80 102 nm in air.
Strategy Rays 1 and 2 (see figure) both reflect from the metal layer, which acts like
a mirror, so there is no phase difference due to reflection between those rays. There
is, however, the usual phase difference caused by the extra distance 2t traveled by
ray 2. The wavelength is λ/n, where n is the index of refraction in the substance.
SOLUTION
Use the appropriate condition for destructive interference in a thin film.
2t = λ/2n
Solve for the thickness t and substitute.
λ 7.80 102 nm
= 1.22 102 nm
t= =
(4)(1.60)
4n
LEARN MORE
Remarks Different CD systems have different tolerances for scratches. Anything
that changes the reflective properties of the disk can affect the readability of the disk.
Question Given two plastics with different indices of refraction, the material with
the larger index of refraction will have:
a larger pit depth.
a smaller pit depth.
the same pit depth.
Read and take notes on
pgs. 802-804
in College Physics Text
Diffraction
• Huygen’s principle requires
that the waves spread out
after they pass through slits.
• This spreading out of light
from its initial line of travel is
called diffraction.
– In general, diffraction occurs
when waves pass through
small openings, around
obstacles or by sharp edges.
Section 24.6
Diffraction, 2
• A single slit placed between a distant light source
and a screen produces a diffraction pattern.
– It will have a broad, intense central band.
– The central band will be flanked by a series of narrower,
less intense secondary bands.
• Called secondary maxima
– The central band will also be flanked by a series of dark
bands.
• Called minima
Section 24.6
Diffraction, 3
• The results of the single slit
cannot be explained by
geometric optics.
– Geometric optics would say
that light rays traveling in
straight lines should cast a
sharp image of the slit on the
screen.
Section 24.6
Fraunhofer Diffraction
• Fraunhofer Diffraction
occurs when the rays leave
the diffracting object in
parallel directions.
– Screen very far from the slit
– Converging lens (shown)
• A bright fringe is seen along
the axis (θ = 0) with
alternating bright and dark
fringes on each side.
Section 24.6
Single Slit Diffraction
• According to Huygen’s
principle, each portion of
the slit acts as a source of
waves.
• The light from one portion
of the slit can interfere with
light from another portion.
• The resultant intensity on
the screen depends on the
direction θ
Section 24.7
Single Slit Diffraction, 2
• All the waves that originate at the slit are in phase.
• Wave 1 travels farther than wave 3 by an amount
equal to the path difference (a/2) sin θ
– a is the width of the slit
• If this path difference is exactly half of a wavelength,
the two waves cancel each other and destructive
interference results.
Section 24.7
Single Slit Diffraction, 3
• In general, destructive interference occurs for
a single slit of width a when sin θdark = mλ / a
– m = 1, 2, 3, …
• Doesn’t give any information about the
variations in intensity along the screen
Section 24.7
Single Slit Diffraction, 4
• The general features of the
intensity distribution are
shown.
• A broad central bright fringe
is flanked by much weaker
bright fringes alternating
with dark fringes.
• The points of constructive
interference lie
approximately halfway
between the dark fringes.
Section 24.7
Diffraction Grating
• The diffracting grating consists of many
equally spaced parallel slits.
– A typical grating contains several thousand lines
per centimeter.
• The intensity of the pattern on the screen is
the result of the combined effects of
interference and diffraction.
Section 24.8
EXAMPLE 24.6 A Single-Slit Experiment
Goal Find the positions of the dark fringes in single-slit diffraction.
Problem Light of wavelength 5.80 102 nm is incident on a slit of width 0.300 mm.
The observing screen is placed 2.00 m from the slit. Find the positions of the first
dark fringes and the width of the central bright fringe.
Strategy This problem requires substitution to find the sines of the angles of the first
dark fringes. The positions can then be found with the tangent function because for
small angles sin θ ≈ tan θ. The extent of the central maximum is defined by these
two dark fringes.
SOLUTION
The first dark fringes that flank the central bright fringe correspond to m = ±1.
λ 5.80 10-7 m
-3
sin θ = ± =
=
±1.93
10
a 3.00 10-4 m
Relate the position of the fringe to the tangent function.
tan θ = y1/L
Because θ is very small, we can use the approximation sin θ ≈ tan θ and then solve
for y1.
sin θ ≈ tan θ ≈ y1/ L
y1 ≈ L sin θ = (2.00 m)(±1.93 10-3) = ±3.86 10-3 m
Compute the distance between the positive and negative first—order maxima, which
is the width w of the central maximum.
w = +3.86 10-3 m - (-3.86 10-3 m) = 7.72 10-3 m
LEARN MORE
Remarks Note that this value of w is much greater than the width of the slit. As the
width of the slit is increased, however, the diffraction pattern narrows,
corresponding to smaller values of θ. In fact, for large values of a, the maxima and
minima are so closely spaced that the only observable pattern is a large central bright
area resembling the geometric image of the slit. Because the width of the geometric
image increases as the slit width increases, the narrowest image occurs when the
geometric and diffraction widths are equal.
Question Suppose the entire apparatus is immersed in water. If the same wavelength
of light (in air) is incident on the slit immersed in water, is the resulting central
maximum larger or smaller? Explain. (Select all that apply.)
The wavelength has become smaller.
The central maximum width is
smaller.
The wavelength has become larger.
The wavelength has remained
the same.
same.
The central maximum is wider.
The central maximum width is the
Read and take notes on
pgs. 805-807
in College Physics Text
Diffraction Grating, Cont.
• The condition for maxima is
– d sin θbright = m λ
• m = 0, ±1, ±2, …
• The integer m is the order
number of the diffraction
pattern.
• If the incident radiation
contains several
wavelengths, each
wavelength deviates
through a specific angle.
Section 24.8
Diffraction Grating, Final
• All the wavelengths are
focused at m = 0
– This is called the zeroth order
maximum
• The first order maximum
corresponds to m = 1
• Note the sharpness of the
principle maxima and the
broad range of the dark area.
– This is in contrast to the
broad, bright fringes
characteristic of the two-slit
interference pattern.
Section 24.8
Diffraction Grating in CD Tracking
• A diffraction grating can be
used in a three-beam
method to keep the beam
on a CD on track.
• The central maximum of the
diffraction pattern is used
to read the information on
the CD.
• The two first-order maxima
are used for steering.
Section 24.8
EXAMPLE 24.7 A Diffraction Grating
Goal Calculate different—order principal maxima for a diffraction grating.
Problem Monochromatic light from a helium—neon laser (λ = 632.8 nm) is incident
normally on a diffraction grating containing 6.00 103 lines/cm. Find the angles at
which one would observe the first order maximum, the second-order maximum, and
so forth.
Strategy Find the slit separation by inverting the number of lines per centimeter,
then substitute values into the interference pattern maxima equation.
SOLUTION
Invert the number of lines per centimeter to obtain the slit separation.
1
= 1.67 10-4 cm = 1.67 103 nm
d=
3
-1
6.00 10 cm
Substitute m = 1 to find the sine of the angle corresponding to the first order
maximum.
632.8 nm
λ
sin θ1 = =
= 0.379
3
d 1.67 10 nm
Take the inverse sine of the preceding result to find θ1
θ1 = sin-1 0.379 = 22.3°
Repeat the calculation for m = 2.
2λ 2(632.8 nm)
sin θ2 = =
= 0.758
d 1.67 103 nm
θ2 = 49.3°
Repeat the calculation for m = 3.
3λ 3(632.8 nm)
sin θ3 = =
= 1.14
d 1.67 103 nm
Because sin θ can't exceed 1, there is no solution for θ3.
LEARN MORE
Remarks The foregoing calculation shows that there can only be a finite number of
principal maxima. In this case only zeroth-, first-, and second-order maxima would
be observed.
Question If a diffraction grating has more lines per centimeter, how does that affect
the spacing of the lines on the grating and the separation between adjacent principal
maxima? (Select all that apply.)
The separation between adjacent maxima increases.
lines on the grating increases.
The spacing of the
The spacing of the lines on the grating
decreases.
The separation between adjacent maxima decreases.
of the lines on the grating remains the same.
The spacing
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Discussion Questions
Unit 4C Physics /Wave Optics
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Unit 4 C Physical / Wave
Optics Homework
Chapter 21
Electromagnetic Waves
Read and take notes on
pgs. 714-717
in College Physics Text
Link to Bright Storm on
Electromagnetic Waves
Electromagnetic Waves Produced by
an Antenna
• When a charged particle undergoes an acceleration,
it must radiate energy.
– If currents in an ac circuit change rapidly, some energy is
lost in the form of EM waves.
– EM waves are radiated by any circuit carrying alternating
current.
• An alternating voltage applied to the wires of an
antenna forces the electric charges in the antenna to
oscillate.
Section 21.10
EM Waves by an Antenna, Cont.
• Two rods are connected to an ac source, charges oscillate between the
rods (a).
• As oscillations continue, the rods become less charged, the field near the
charges decreases and the field produced at t = 0 moves away from the
rod (b).
• The charges and field reverse (c).
• The oscillations continue (d).
Section 21.10
EM Waves by an Antenna, Final
• Because the oscillating
charges in the rod produce
a current, there is also a
magnetic field generated.
• As the current changes, the
magnetic field spreads out
from the antenna.
• The magnetic field is
perpendicular to the
electric field.
Section 21.10
Charges and Fields, Summary
• Stationary charges produce only electric fields.
• Charges in uniform motion (constant velocity)
produce electric and magnetic fields.
• Charges that are accelerated produce electric and
magnetic fields and electromagnetic waves.
• An accelerating charge also radiates energy.
Section 21.10
Electromagnetic Waves, Summary
• A changing magnetic field produces an electric
field.
• A changing electric field produces a magnetic
field.
• These fields are in phase.
– At any point, both fields reach their maximum
value at the same time.
Section 21.10
Electromagnetic Waves are Transverse
Waves
• The and fields are
perpendicular to each other.
• Both fields are
perpendicular to the
direction of motion.
– Therefore, EM waves are
transverse waves.
Section 21.10
Properties of EM Waves
• Electromagnetic waves are transverse waves.
• Electromagnetic waves travel at the speed of light.
– Because EM waves travel at a speed that is precisely the
speed of light, light is an electromagnetic wave.
Section 21.11
Properties of EM Waves, 2
• The ratio of the electric field to the magnetic field is
equal to the speed of light.
• Electromagnetic waves carry energy as they travel
through space, and this energy can be transferred to
objects placed in their path.
Section 21.11
Properties of EM Waves, 3
• Energy carried by EM waves is shared equally by the
electric and magnetic fields.
– Intensity (I) is average power per unit area.
Section 21.11
Properties of EM Waves, Final
• Electromagnetic waves transport linear
momentum as well as energy.
– For complete absorption of energy U, p=U/c
– For complete reflection of energy U, p=(2U)/c
• Radiation pressures can be determined
experimentally.
Section 21.11
Determining Radiation Pressure
• This is an apparatus for
measuring radiation
pressure.
• In practice, the system is
contained in a vacuum.
• The pressure is determined
by the angle at which
equilibrium occurs.
Section 21.11
Properties of EM Waves, Summary
• EM waves travel at the speed of light.
• EM waves are transverse waves because the electric
and magnetic fields are perpendicular to the direction
of propagation of the wave and to each other.
• The ratio of the electric field to the magnetic field in an
EM wave equals the speed of light.
• EM waves carry both energy and momentum, which
can be delivered to a surface.
Section 21.11
Read and take notes on
pgs. 720-723
in College Physics Text
Link to Bright Storm on
Electromagnetic Spectrum
The Spectrum of EM Waves
• Forms of electromagnetic waves exist that are
distinguished by their frequencies and
wavelengths.
– c = ƒλ
• Wavelengths for visible light range from 400
nm to 700 nm.
• There is no sharp division between one kind of
EM wave and the next.
Section 21.12
The EM Spectrum
• Note the overlap between
types of waves.
• Visible light is a small
portion of the spectrum.
• Types are distinguished by
frequency or wavelength.
Section 21.12
Notes on The EM Spectrum
• Radio Waves
– Used in radio and television communication
systems
• Microwaves
– Wavelengths from about 1 mm to 30 cm
– Well suited for radar systems
– Microwave ovens are an application
Section 21.12
Notes on the EM Spectrum, 2
• Infrared waves
–
–
–
–
Incorrectly called “heat waves”
Produced by hot objects and molecules
Wavelengths range from about 1 mm to 700 nm
Readily absorbed by most materials
• Visible light
– Part of the spectrum detected by the human eye
– Wavelengths range from 400 nm to 700 nm
– Most sensitive at about 560 nm (yellow-green)
Section 21.12
Notes on the EM Spectrum, 3
• Ultraviolet light
– Covers about 400 nm to 0.6 nm
– The Sun is an important source of uv light.
– Most uv light from the sun is absorbed in the stratosphere
by ozone.
• X-rays
– Wavelengths range from about 10 nm to 10-4 nm
– Most common source is acceleration of high-energy
electrons striking a metal target
– Used as a diagnostic tool in medicine
Section 21.12
Notes on the EM Spectrum, Final
• Gamma rays
– Wavelengths from about 10-10 m to 10-14 m
– Emitted by radioactive nuclei
– Highly penetrating and cause serious damage
when absorbed by living tissue
• Looking at objects in different portions of the
spectrum can produce different information.
Section 21.12
Crab Nebula in Various
Wavelengths
Section 21.12
Grading Rubric for Unit 4C Physical and Wave Optics
Name: ______________________
Conceptual Physics Text
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Pgs 490-494---------------------------------------------------_____
Advanced notes from text book:
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Pgs 800-801 --------------------------------------------------_____
Pgs 802-804 --------------------------------------------------_____
Pgs 805-807 --------------------------------------------------_____
Pgs 714-717 --------------------------------------------------_____
Pgs 720-723 --------------------------------------------------_____
Example Problems:
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24.2 a -------------------------------------------------------------_____
24.3 a -------------------------------------------------------------_____
24.4 a -------------------------------------------------------------_____
24.5 a -------------------------------------------------------------_____
24.6 a -------------------------------------------------------------_____
24.7 a -------------------------------------------------------------_____
Web Assign 1 (a), 2 (a-c), 3 (a-b), 4 (a-b), 5 (a), 6 (a4), 7 (a), 8 (a), 9 (a3-b3) 10 (a) ------____