Transcript Power Point Slides for Wave Optics

Physics 1404, Section 1
Chapter 25:
Wave Optics
Contents
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Coherence and Conditions for Interference
The Michelson Interferometer
Thin-Film Interference
Light through a Single Slit: Qualitative
Behavior
Double-Slit Interference: Young’s Experiment
of Light from a Single Slit
Diffraction Gratings
Optical Resolution and the Rayleigh Criterion
Why Is the Sky Blue?
Coherence and Conditions for Interference
Wave Optics
• The field of wave optics studies the properties of light that
depend on its wave nature
• Originally light was thought to be a particle and that model
successfully explained the phenomena discussed in geometric
optics
• Maxwell’s theory of electromagnetism convinced physicists that
light was a wave
• When discussing image characteristics over distances much
greater than the wavelength, geometric optics is extremely
accurate
• When dealing with sizes comparable to or smaller than the
wavelength, wave optics is required
Interference
• One property unique to waves is
interference
• Interference of sound waves can be
produced by two speakers
1. Constructive Interference
• When the waves are in phase, their maxima
occur at the same time at a given point in
space
• The total wave displacement at the
listener’s location is the sum of the
displacements of the two individual waves
• If two waves are in phase, the sum of their
displacements is large
2. Destructive Interference
• The maximum of one wave can
coincide with the minimum of the
other wave
• These waves are out of phase
• The interference is destructive when
the waves are out of phase
• If the waves are 180° out of phase, the
sum of the displacements of the two
waves is zero
General conditions for interference.
• Two waves can interfere if all the following conditions
are met.
1. Two or more interfering waves travel through different
regions of space over at least part of their propagation
from source to destination.
2. The waves are brought together at a common point.
3. The waves must have the same frequency and must also
have a fixed phase relationship. Thus, over a given
distance or time interval the phase difference between
the waves remains constant. Such waves are called
coherent.
The Michelson Interferometer
• The Michelson interferometer is
based on the interference of reflected
waves
• Two reflecting mirrors are mounted
at right angles
• A third mirror is partially reflecting
– Called a beam splitter
• The incident light hits the beam splitter and is divided
into two waves
• The waves reflect from the mirrors at the top and right
and recombine at the beam splitter
• After being reflected again from the beam splitter,
portions of the waves combine at the detector
• The only difference between the two waves is that they
travel different distances between their respective mirrors
and the beam splitter
• The path length difference is ΔL = 2L2 – 2L1
• The path length difference is
related to the wavelength of the
light
DL
N=
l
– If N is an integer, the two waves are in
phase and produce constructive interference
– If N is a half-integer the waves will produce
destructive interference
• For constructive interference, ΔL = m λ
• For destructive interference, ΔL = (m + ½) λ
– m is an integer in both cases
• If the interference is constructive, the light intensity at
the detector is large
– Called a bright fringe
• If the interference is destructive, the light intensity at
the detector is zero
– Called a dark fringe
Example
• A helium–neon (He–Ne) laser emits light
with a wavelength of approximately λHe–Ne
=633 nm. Suppose this light source is used
in a Michelson interferometer and one of
the mirrors is moved a distance d such that
exactly N = 1,000,000 bright fringes are
counted, calculate d.
Solution:
When the mirror moves a distance d, the distance traveled by the light
changes by 2d because the light travels back and forth between the
beam splitter and the mirror.
ΔL=2L2 – 2L1=2(L2 – L1)=2d
d = 0.5×1000000×633×10-9m=0.317m
Thin-Film Interference
• Assume a thin soap film rests on a flat glass surface
• The upper surface of the soap film is similar to the
beam splitter in the interferometer
• It reflects part of the incoming light and allows the rest
to be transmitted into the soap layer after refraction at
the air-soap interface
• The transmitted ray is partially reflected at the bottom
surface
• The two outgoing rays meet the conditions for
interference
– They travel through different regions
– They recombine when they leave the film
– They are coherent because they originated from the same
source and initial ray
• From the speed of the wave inside the film
lfilm ƒfilm
c
=
nfilm
• The wavelength changes as the light wave travels from a
vacuum into the film
2d
N=
• The number of extra wavelengths is
lfilm
Thin-Film Interference, Case 1
• Waves reflected by a thin film undergo a phase change
• The number of extra cycles traveled by the ray inside the
film completely determines the nature of the interference
• If the number of extra cycles, N, is an integer, there is
constructive interference
• If the number of extra cycles is a half-integer, there is
destructive interference
2d =
ml
constructive interference
nfilm
1ö
æ
m
+
l
ç
÷
2ø
è
2d =
destructive interference
nfilm
Only if: nair < nfilm < n(substance below the film)
Thin-Film Interference, Case 2
• Assume the soap bubble is surrounded by air
• There is a phase change at the top of the bubble
• There is no phase change at the bottom of the bubble
• Since only one wave undergoes a phase change, the
interference conditions are
1ö
æ
m
+
l
ç
÷
2ø
2d = è
constructive interference
nfilm
2d =
ml
destructive interference
nfilm
Example 25 .3 Color and Thickness of a Soap Film
• Consider a bubble formed from a thin, soapy film that
looks blue when viewed at normal incidence. Estimate
the thickness of the film. Assume its index of refraction
is nfilm=1.35 and blue light has a wavelength λblue=400
nm. Also assume the film is so thin that thinner films are
not able to give constructive interference.
Solution:
The interference here is destructive and we consider m=0:
Light through a Single Slit: Qualitative Behavior
• Light passes through a slit or opening and then
illuminates a screen
• As the width of the slit becomes closer to the wavelength
of the light, the intensity pattern on the screen and
additional maxima become noticeable
Single-Slit Diffraction
• Water wave example of single-slit
diffraction
– All types of waves undergo single-slit
diffraction
• Diffraction is the bending or
spreading of a wave when it passes
through an opening
• It is useful to draw the wave fronts
and rays for the incident and
diffracting waves
• Huygen’s Principle can be stated as
all points on a wave front can be
thought of as new sources of
spherical waves
Double-Slit Interference: Young’s Experiment
of Light from a Single Slit
Double-Slit Interference
• Light passes through two very narrow slits
• When the two slits are both very narrow, each slit acts as
a simple point source of new waves
• The outgoing waves from each slit are like simple
spherical waves
• The double slit experiment showed conclusively that
light is a wave
• Experiment was first carried out by Thomas Young
around 1800
Young’s Double-Slit Experiment
• Light is incident onto two slits and
after passing through them strikes a
screen
• The light intensity on the screen
shows an interference pattern
• If the slits are separated by a
distance d, then the difference in
length between the paths of the two
rays is
ΔL = d sinθ
• For constructive interference, d sinθ = m λ
– m = 0, ±1, ±2, …
– Will observe a bright fringe
– The light intensity will be large
– The waves will be in phase when they strike the
screen
• For destructive interference, d sinθ = (m + ½) λ
– m = 0, ±1, ±2, …
– Will observe a dark fringe
– The light intensity will be zero
– The waves will not be in phase when they strike the
screen
Double-Slit Intensity Pattern
• The angle θ varies as you
move along the screen
• Each bright fringe corresponds
to a different value of m
• Negative values of m indicate
that the path to those points on
the screen from the lower slit
is shorter than the path from
the upper slit
l
• For m = 1,
sinq =
d is the distance between the slits
W is the distance between the slits
and the screen
h is the distance between the adjacent
bright fringes
d
• Since the angle is very small, sin θ ~ θ and θ ~ λ/d
• Between m = 0 and m = 1, h = W tanθ
Approximations
• For small angles, tan θ ~ θ and sin θ ~ θ
• Using the approximations, h = W θ = W λ / d
Interference with Monochromatic Light
• Light with a single frequency is called monochromatic
(one color)
• Light sources with a variety of wavelengths are generally
not useful for double-slit interference experiments
– The bright and dark fringes may overlap or the total pattern
may be a “washed out” sum of bright and dark regions
– No bright or dark fringes will be visible
Example 25 .5 Measuring the Wavelength of Light
• Young’s double-slit experiment shows that light is
indeed a wave and also gives a way to measure the
wavelength. Suppose the double-slit experiment in
Figure 25.20 is carried out with a slit spacing d=0.40
mm and the screen at a distance W=1.5 m. If the bright
fringes nearest the center of the screen are separated by
a distance h=1.5 mm, what is the wavelength of the
light?
Solution:
Single-Slit Diffraction: Interference
• Slits may be narrow enough to exhibit
diffraction but not so narrow that they
can be treated as a single point source of
waves
• Assume the single slit has a width, w
• Light is diffracted as it passes through the
slit and then propagates to the screen
• The key to the calculation of where the fringes
occur is Huygen’s principle
• All points across the slit act as wave sources
• These different waves interfere at the screen
• For analysis, divide the slit into two parts
• If one point in each part of the
slit satisfies the conditions for
destructive interference, the
waves from all similar sets of
points will also interfere
destructively
• Destructive interference will
produce a dark fringe
Single-Slit Analysis: Dark Fringers
• Conditions for destructive interference are
w sin θ = ±m λ (m = 1, 2, 3, … )
– The negative sign will correspond to a fringe below the center
of the screen
Single-Slit Analysis: Bright Fringes
• There is no simple formula for the angles at which the bright
fringes occur
• The intensity on the screen can be calculated by adding up all
the Huygens waves
• There is a central bright fringe:
– The central fringe is called the central maximum
– The central fringe is about 20 times more intense than the bright
fringes on either side
– The width of the central bright fringe is approximately the
angular separation of the first dark fringes on either sideThe full
angular width of the central bright fringe = 2 λ / w
– If the slit is much wider than the wavelength, the light beam
essentially passes straight through the slit with almost no effect
from diffraction
Diffraction Gratings
• An arrangement of many slits is
called a diffraction grating
• Assumptions
– The slits are narrow
• Each one produces a single outgoing
wave
– The screen is very far away
• If the slit-to-slit spacing is d, then the path length
difference for the rays from two adjacent slits is
ΔL = d sinθ
• If ΔL is equal to an integral number of wavelengths,
constructive interference occurs
• For a bright fringe, d sin θ = m λ (m = 0, ±1, ±2, …)
• The condition for bright
fringes from a diffraction
grating is identical to the
condition for constructive
interference from a double slit
• The overall intensity pattern
depends on the number of slits
• The larger the number of slits,
the narrower the peaks
Example 25 .6 Diffraction of Light by a Grating
• A diffraction experiment is carried out with a grating.
Using light from a red laser (λ = 630 nm), the
diffraction fringes are separated by h=0.15 m on a
screen that is W=2.0 m from the grating. Find the
spacing d between slits in the grating.
Solution:
Optical Resolution and the Rayleigh Criterion
• For a circular opening of diameter
D, the angle between the central
bright maximum and the first
minimum is
1.22l
q=
D
– The circular geometry leads to the
additional numerical factor of 1.22
Rayleigh Criterion
• If the two sources are sufficiently far apart, they can
be seen as two separate diffraction spots (A)
• If the sources are too close together, their diffraction
spots will overlap so much that they appear as a single
spot (C)
A
B
C
• Two sources will be resolved as two distinct sources of
light if their angular separation is greater than the
angular spread of a single diffraction spot.
• This result is called the Rayleigh criterion
• For a circular opening, the Rayleigh criterion for the
angular resolution is
qmin
1.22l
=
D
• Two objects will be resolved when viewed through an
opening of diameter D if the light rays from the two
objects are separated by an angle at least as large as θmin
Example 25 .7 Rayleigh Criterion for a Marksman
• A rifle used for target shooting has a small telescope so
that the shooter can accurately aim at his target.
Suppose the telescope has a diameter D=1.5 cm and the
shooter is looking at two small objects that are L=200
m away. If the shooter is just barely able to tell that
there are two objects (and not one), how far apart are
the objects? Assume red light with λ=600 nm.
Solution:
Why Is the Sky Blue?
• The light we see from the sky is sunlight
scattered by the molecules in the atmosphere
• The molecules are much smaller than the
wavelength of visible light
– They scatter blue light more strongly than red
– This gives the atmosphere its blue color
• Blue sky
–
–
–
–
Although violet scatters more than blue, the sky appears blue
The Sun emits more strongly in blue than violet
Our eyes are more sensitive to blue
The sky appears blue even though the violet light is scattered more
• Sun near horizon
– There are more molecules to scatter the light
– Most of the blue is scattered away, leaving the red