Transcript Chapter 24

Chapter 24
Wave Optics
Conceptual questions: 3, 4, 13, 14, 17, 18
Quick Quizzes: 1, 2, 3, 4
Problems: 10, 17, 34
Interference
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Light waves interfere with each other
much like mechanical waves do
Two conditions which must be met
1. The sources must be coherent. They
must maintain a constant phase with
respect to each other
2. The waves must have identical
wavelengths
Producing Coherent Sources

Old method:
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
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New method: use a laser
Young’s Double Slit Experiment
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

Interference Patterns
Constructive
interference occurs
at the center point
 The two waves
travel the same
distance and they
arrive in phase

Interference Patterns, 2
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The upper wave
travels one
wavelength farther
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Therefore, the waves
arrive in phase
A bright fringe
occurs
Interference Patterns, 3
The upper wave
travels one-half 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
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A dark fringe occurs
Interference Equations
The path difference,
δ, is found from the
tan triangle
 δ = r2 – r1 = d sin θ
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This assumes the
paths are parallel
Interference Equations, 2
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For a bright fringe, produced by
constructive interference, the path
difference must be
δ=mλ
 m = 0, ±1, ±2, …
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δ = d sin θbright = m λ
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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
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Interference Equations, 3
When destructive interference occurs, a
dark fringe is observed
 This needs a path difference of an odd
half wavelength; δ = (m + ½) λ
 δ = d sin θdark = (m + ½) λ
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
m = 0, ±1, ±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
 Approximation
 θ is small and therefore tanθ ~ sin θ
 For bright fringes
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y bright
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L

m m  0,  1,  2 
d
For dark fringes
y dark
L 
1

 m   m  0,  1,  2 
d 
2
Quick quiz 24-1
In a two slit interference pattern
projected on a screen the fringes are
equally spaced on the screen
 A. everywhere
 B. only for large angles
 C. only for small angles
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Problem 24.10
A pair of slits, separated by 0.150 mm, is illuminated by
light having a wavelength of λ = 643 nm. An
interference pattern is observed on a screen 140 cm
from the slits. Consider a point on the screen located
at y = 1.80 cm from the central maximum of this
pattern.
(a) What is the path difference δ for the two slits at the
location y?
(b) Express this path difference in terms of the
wavelength.
(c) Will the interference correspond to a maximum, a
minimum, or an intermediate condition?
Phase Changes Due To
Reflection
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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
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Analogous to a reflected
pulse on a string
Phase Changes Due To
Reflection, cont
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There is no phase
change when the
wave is reflected
from a boundary
leading to a medium
of lower index of
refraction
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Analogous to a pulse
in a string reflecting
from a free support
Interference in Thin Films
Rules to remember
1. 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
2. There is no phase change in the reflected wave if
n 2 < n1
3. 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
Interference in Thin Films, 2
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
 Ray 2 also travels an
additional distance of
2t before the waves
recombine
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Interference in Thin Films, 3
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For constructive interference
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2 n t = (m + ½ ) λ m = 0, 1, 2 …
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This takes into account both the difference in
optical path length (2t) for the two rays and
the 180° phase change (1/2 λ)
For destruction interference
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2 n t = m λ m = 0, 1, 2 …
Interference in Thin Films, 4
An example of different
indices of refraction
 A coating on a solar
cell
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Quick quiz 24-2
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Supposed Young’s experiment is carried
out in air, and then, in a second
experiment, the apparatus is immersed
in water. In what way does the
distance between bright fringes
change?
A. they move further apart
 B. they move closer together
 C. no change
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Problem 24.17
A coating is applied to a lens to minimize
reflections. The index of refraction of the coating
is 1.55, and that of the lens is 1.48. If the coating
is 177.4 nm thick, what wavelength is minimally
reflected for normal incidence in the lowest
order?
Newton’s rings
Conceptual questions
3. Consider a dark fringe in an interference
pattern, at which almost no light energy is
arriving. Light from both slits is arriving at
this point, but the waves are canceling.
Where does the energy go?
4. If Young’s double slit experiment were
performed under water, how would the
observed interference pattern be affected?
13.Would it be possible to place a nonreflective
coating on an airplane to cancel radar waves
of wavelength 3 cm?
Reading a CD
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As the disk rotates, the
laser reflects off the
sequence of lands and pits
into a photodector
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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
land
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
Fraunhofer Diffraction
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Fraunhofer Diffraction
occurs when the rays
leave the diffracting
object in parallel
directions
A bright fringe is seen
along the axis (θ = 0)
with alternating bright
and dark fringes on
each side
Single Slit Diffraction
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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
Wave 1 travels farther
than wave 3 by an
amount equal to the
path difference (a/2)
sin θ
destructive interference
occurs when
sin θdark = mλ / a
Single Slit Diffraction, 2
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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
Problem 34
A screen is placed 50.0 cm from a single
slit, which is illuminated with light of
wavelength 680 nm. If the distance
between the first and third minima in the
diffraction pattern is 3.00 mm, what is
the width of the slit?
Quick quiz 24.3
In a single-slit diffraction experiment, as
the width of the slit is made smaller, the
width of the central maximum of the
diffraction pattern becomes
(a) smaller,
(b) larger,
(c) remains the same.
Diffraction Grating
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The condition for
maxima is
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d sin θbright = m λ
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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
QUICK QUIZ 24.4
If laser light is reflected from a
phonograph record or a compact disc, a
diffraction pattern appears. This occurs
because both devices contain parallel
tracks of information that act as a
reflection diffraction grating. Which
device, record or compact disc, results in
diffraction maxima that are farther apart?
Diffraction Grating in CD
Tracking
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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
Polarization of Light Waves
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Each atom
produces a wave
with its own
orientation of E
This is an
unpolarized wave
Polarization of Light, cont
A wave is said to be linearly
polarized if the resultant
electric field vibrates in the
same direction at all times at
a particular point
 Polarization can be obtained
from an unpolarized beam by
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selective absorption
reflection
scattering
Polarization by Selective Absorption
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The most common technique for polarizing light
Uses a material that transmits waves whose electric
field vectors in the plane parallel to a certain
direction and absorbs waves whose electric field
vectors are perpendicular to that direction
Malus’ law: I = Io cos2 θ
Polarization by Reflection
The angle of incidence for
which the reflected beam is
completely polarized is called
the polarizing angle, θp
θp is also called Brewster’s
Angle
 Brewster’s Law relates the
polarizing angle to the index
of refraction for the material

n
sin p
cos p
 tan p
Polarization by
Scattering
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The horizontal part of the electric
field vector in the incident wave
causes the charges to vibrate
horizontally
The vertical part of the vector
simultaneously causes them to
vibrate vertically
Horizontally and vertically
polarized waves are emitted
Conceptual question
14. Certain sunglasses use a polarizing material
to reduce intensity of light reflected from
shiny surfaces, such as water or a hood of a
car. What orientation of the transmission axis
should the material have to be most
effective?
18. Can a sound wave be polarized?
17. When you receive a chest x-ray at a
hospital, the ex-ray passes through a series
of parallel ribs in your chest. Do the ribs act
as a diffraction grating for x-rays?
Optical Activity
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Certain materials display the property of
optical activity
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A substance is optically active if it rotates
the plane of polarization of transmitted
light
Liquid Crystals
Rotation of a polarized light beam by a liquid
crystal when the applied voltage is zero
 Light passes through the polarizer on the
right and is reflected back to the observer,
who sees the segment as being bright
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Liquid Crystals
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When a voltage is applied, the liquid crystal does not
rotate the plane of polarization
The light is absorbed by the polarizer on the right
and none is reflected back to the observer
The segment is dark
MCAD
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Two light sources produce light with
wavelength . The sources are placed
22.5  and 45  away from point P. When
both sources are turned on and their
intensities, I, at point P are equal, the
resultant intensity at point P will be
A. 0
 B. 0.5 I
 C. I
 D. 2I
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The process discussed in the previous question is called
a. Diffraction
b.
Refraction
c. Interference
d.
Dispersion
Which of the following would result in greatest diffraction?
a. Small wavelengths moving through a small opening
b. Large wavelengths moving through a small opening
c. Small wavelengths moving through a large opening
d. Large wavelengths moving through a large opening