Wave Optics
Unit 11
Light
• In the last unit, we studied several properties
of light, including refraction and reflection.
• However, these were geometric properties,
and had nothing to do with the fact that light
is a wave.
• Now we will no longer be able to ignore this
reality.
Interference
The Double-Slit Experiment
Interference
• One of the most conclusive examples of the
wave nature of light was the Double Slit
Experiment.
• Early physicists, such as Newton, had thought
of light as a particle.
• However, Thomas Young was able to show
light acts as a wave, and was even able to
measure the wavelength of visible light.
Interference
• Young created a setup that
allowed light from a single
source to fall on two
closely spaced slits.
• If the particle picture of
light were correct, we
would expect to see two
bright lines on a screen
placed behind the slits.
Interference
• However, instead a series
of bright lines were
observed on the screen.
• Young explained this effect
as a wave-interference
phenomenon.
Interference
• To see how this effect
occurs, let’s consider Young’s
setup.
• Plane waves of light with a
single wavelength are
striking the slits.
• Light of this type is called
monochromatic, meaning
“one color.”
Interference
• When the waves strike the
slits, they spread out in all
directions on the other
side.
• This is analogous to what
happens when two rocks
are thrown into a pond, or
when two speakers are
placed close together.
Interference
• Recall from the speaker
problem that the waves from
the two slits will have
traveled different distances
depending on where they
strike the screen.
• As a result, the waves may
arrive out of phase.
• This can lead to interference
when the waves add back
together.
Interference
• Let’s consider
monochromatic light with
wavelength λ striking slits
S1 and S2, which are
separated by a distance d.
• If we look at the center of
the screen, we see a bright
spot.
Interference
• This is not surprising since
the light from each source
travels the same distance.
• The two waves arrive at the
screen in-phase, resulting in
constructive interference.
• The brightness of the spot is
brighter than the light from
either slit.
Interference
• Likewise, if we look at
certain points on the
screen that are off the axis,
we also see a bright spot.
• These are also points
where constructive
interference is occurring.
Interference
• From our experience with
sound waves, we know
that this means the path
length difference must be
some multiple of a
wavelength.
• We will quantify this in a
moment.
Interference
• At other points, no light
can be seen.
• These points indicate
destructive interference is
occurring.
• The path length difference
must be a multiple of a
half wavelength.
Interference
• To develop a formula for
determining the location of the
bright (or dark) fringes, let’s
look at the geometry of the
system.
• We want to know how far each
ray will have to travel to reach a
point at an angle θ above the
center.
Interference
• Notice that ray 2 travels farther
than ray 1.
• This is the path length
difference.
• Notice also that if forms a side
of a right triangle with the
spacing between the slits.
Interference
• Based on this, we can use
trigonometry to write the path
length difference in terms of d
and θ, quantities we can easily
measure:
path difference = d sinq
Interference
• This gives us the formula
for constructive
interference:
d sinq = ml
m = 0, 1, 2,...
Interference
• This also gives us the
formula for destructive
interference:
d sinq = ( m +
m = 0, 1, 2,...
1
2
)l
Interference
• Notice also that there is a
second right triangle that
involves the distance from
the center on the screen.
• From this we can see that
y
sin q =
L
Interference
• This lets us rewrite our
formulas:
dy
= ml
L
dy
1
= (m + 2 ) l
L
m = 0, 1, 2,...
Notes
• These formulas are only valid when the
distance between the slits (d) is very small
compared to the distance to the screen (L).
• As a result, θ must be very small.
• The value of m is called the order of the
interference fringe.
Example
Two slits are 0.1 mm apart and are located 1.2
m from a viewing screen. Light from a He-Ne
laser beam (λ = 633 nm) is shined on the slits.
a)How far from the center is the second
order interference minimum located?
b)What angle above the horizontal is this?
You Try
The He-Ne laser from the last example is
replaced with a new laser. You measure the
location of the first order interference maximum
to be 0.6 cm above the axis. What is the
wavelength of the light being emitted by this
laser? Express your answer in nm.
Homework
• Read 24-3.
• Do problems 1, 2, 3, and 5 on page 692.
Announcements
• Paper final draft due next Monday.
• Presentations May 2 – May 5.
Problem Day
• Do problems 5, 10, and 11 on page 692.
• We will whiteboard at the end of class.
Homework
• Read 24-1 and 24-5.
• Do problem 12 on page 692.
Huygens’s Principle and
Diffraction
Huygens’s Principle
• Christian Huygens was a Dutch scientist who
lived about the same time as Newton.
• He proposed a wave model for light that is still
used today to describe many phenomena.
• Recall that for 2D and 3D waves, we describe
the position of the wave using the wave front.
Huygens’s Principle
Every point on a wave front can
be considered to be a source of
tiny wavelets that spread out in
the forward direction at the
speed of the wave itself.
The new wave front is the
envelope of the wavelets.
Huygens’s Principle
• Note that this is a model of
how a wave propagates.
• However, it is extremely useful
for explaining phenomena such
as diffraction.
Diffraction
• In 1819, the French scientist Augustin Fresnel
used Huygens’s principle to explain how light
could “bend” around the edges of solid
objects.
• This phenomenon is known as diffraction.
• As we will see in a moment, diffraction
produces results similar to interference. The
two effects are related, but are not the same.
Diffraction
• Fresnel considered a circular disk that was
illuminated by monochromatic light.
• Using Huygens’s principle, we can consider the
points around the edge of the disk to be point
sources of wavelets.
Diffraction
• These wavelets spread out in all directions.
• On the outside of the disk, the wavelets add
together with other parts of the beam.
Diffraction
• However, the wavelets that spread out toward
the inside of the disk, do not encounter each
other until they reach the screen.
• Thus, Fresnel predicted a bright spot at the
center of the shadow of a disk.
Diffraction
See also the pictures on p 673.
Question
• What effect do those bright/dark fringes
remind you of?
• In fact, these fringes are due to the
interference of the diffracted light waves with
themselves.
• The image as a whole is called a diffraction
pattern.
Single-Slit Diffraction
• To see how a diffraction pattern arises, we will
consider the simplest case of monochromatic
light passing through a single, narrow slit.
• We will assume the rays striking the slit are
parallel and the viewing screen is very far away.
Single-Slit Diffraction
• From Huygens’s Principle, we can model the
slit as being made up of a bunch of point
sources.
• These sources send out wavelets in all
directions.
Single-Slit Diffraction
• For the wavelets that continue on straight
ahead, all the waves travel the same distance
to the screen.
• This means they arrive in phase and interfere
constructively, producing a bright spot.
Single-Slit Diffraction
• However, if we move off axis by an angle θ,
then the rays from the top of the slit will have
to travel a longer distance than the rays from
the bottom.
• This leads to interference at the screen.
Single-Slit Diffraction
• Consider first an angle where the top ray
travels one λ farther than the bottom ray.
• This means the middle ray will travel λ/2
farther than the bottom ray.
Single-Slit Diffraction
• This means the middle ray will interfere
destructively with the bottom ray.
• The ray just above the bottom will interfere
destructively with the ray just above the
middle.
Single-Slit Diffraction
• Thus all the rays interfere with each other at
this angle.
• The result is a dark spot on the screen.
Single-Slit Diffraction
• If we increase the angle so that now the top
ray travels 3λ/2 farther than the bottom, we
still get some destructive interference.
• The bottom 2/3 of the light from the slit
cancel out as before.
Single-Slit Diffraction
• However, now there are no rays to cancel the
light coming from the top third of the slit.
• This light reaches the screen, resulting in a
bright spot.
Single-Slit Diffraction
• If we increase the angle further so the path
difference is 2λ, we get total destructive
interference again.
• The bottom fourth cancels with the second fourth
and the third fourth cancels with the top fourth.
Single-Slit Diffraction
• The result is a diffraction pattern with a
central bright spot and several dimmer spots
seen as you go off the axis in either direction.
Single-Slit Diffraction
• To locate the diffraction minima, we need to
look at the geometry again.
• Notice that, as with the double-slit, the path
difference is equal to Dsinθ, where D is the
width of the slit.
Single-Slit Diffraction
• Notice that, unlike with the double-slit, the
diffraction minima occur when the path length
difference is equal to a whole number
multiple of the wavelength (i.e. 1λ, 2λ, etc.).
Single-Slit Diffraction
• This allows us to write a formula for locating
the diffraction minima:
Dsinq = ml
m =1, 2, 3,...
Things to Notice
Dsinq = ml
m =1, 2, 3,...
• This formula looks extremely similar to the
formula for finding interference maxima.
• However, there are important differences.
• First, this formula gives the diffraction
minima.
Things to Notice
Dsinq = ml
m =1, 2, 3,...
• Second, the minima only occur for m ≥ 1. There
is a bright spot for m = 0.
• Third, here D is the width of a slit rather than the
distance between two infinitely narrow slits.
• Lastly, this formula cannot be used to find
diffraction maxima, which are more complicated.
Example: Single-Slit
Light of wavelength 750 nm masses through a
slit that is 1.0 x 10-3 mm wide.
a) What is the angular distance to the first
minimum?
b) If the pattern is viewed on a screen 20 cm
away, how far from the central axis is the
minimum?
In-class Work
• Do problems 17, 18, 20, 22, and 23 on page
693.
• Your homework is to finish these problems.
Problem Day
• Do problems 25 and 62 on pages 693-694.
• We will whiteboard these problems and the
homework problems in ~25 mins.
Homework
• Study your notes on interference and
diffraction.
• Do problems 24 and 67 on pages 693-695.
Dispersion
The Visible Spectrum: A Review
• There are two aspects of light that can be
explained by the wave theory.
• The brightness of the light corresponds to the
intensity of the electromagnetic wave.
• The color of the light corresponds to the
frequency (or wavelength) of the light.
The Visible Spectrum: A Review
• The human eye is sensitive to EM waves with
frequencies from 4 x 1014 Hz to 7.5 x 1014 Hz.
• In air, this corresponds to wavelengths of 400
nm to 700 nm.
The Visible Spectrum: A Review
• This region of the EM spectrum is referred to as
the visible spectrum.
• Light with wavelengths shorter than 400 nm is
called ultraviolet (UV).
• Light with wavelengths longer than 750 nm is
called infrared (IR).
Prism Demo
Prisms
• When light passes from
one medium to another, it
is refracted according to
Snell’s Law.
• However, the prism
seemed to separate the
white light into many
different colors.
• Why did this happen?
Prisms
• It turns out that the index
of refraction of a material
depends on the wavelength
of light that shines on it.
• The index is greatest for
light that has a shorter
wavelength (blue-purple).
Prisms
• As a result (see Snell’s Law),
the blue-purple light gets
refracted at a larger angle
than the red light.
• Thus, if we send white light
(which contains all colors)
through a prism, each color is
refracted at a different angle.
• This spreading of light is
called dispersion.
Rainbow
• We see dispersion on a regular basis (well, not
in Phoenix) through rainbows.
• A rainbow appears when you look at falling
water droplets with the sun behind you.
Rainbow
• The sunlight contains all wavelengths of light.
• When light strikes the water droplets, it is
refracted due to the change in medium.
• The angle of refraction depends on the
wavelength of the light.
Rainbow
• The light is reflected off of the back of the
droplet and travels toward your eye.
• The red light is bent the least, and so you see
red light from droplets higher in the sky.
Rainbow
• The purple light is bent the most, so you see
purple light from droplets that are lower in
the sky.
• The other colors are seen from droplets in
between, resulting in a rainbow.
Homework
• Read 24-4, 24-6 and 24-7.
• Do problems 14 and 15 on page 693.
Multiple Slit Interference and
Diffraction
Where We Are…
• We have already looked at interference effects
for a two-slit setup and diffraction effects for a
single slit.
• We also observed that a real-world two-slit
system has both interference and diffraction
effects present.
• Now we will look at what happens when light
shines on more than 1 slit.
Multiple Slit Interference
• When monochromatic light is
shined on a series of 3 slits, we
can still see an interference
pattern.
• However, the maxima are
narrower than for the two slit
case.
• There are also smaller maxima
that can be seen in between
the larger one.
Multiple Slit Interference
• As the number of slits is
increased, the large
maximums become even
narrower.
• There are more little
maximums, but these
become dimmer and dimmer
as more slits are added.
Multiple Slit Interference
• Why does this happen?
• For the two slit case, if we
move a little distance away
from a maximum, the waves
from the two slits will be only
slightly out of phase.
• As a result, the interference
will be almost constructive.
Multiple Slit Interference
• This means the interference
fringes will be fairly wide for
two slits.
• However, as more slits are
added, the fringes sharpen.
• This is because there are now
more waves interfering at
each point.
Multiple Slit Interference
• This means the interference
fringes will be fairly wide for
two slits.
• However, as more slits are
added, the fringes sharpen.
• This is because there are now
more waves interfering at
each point.
Multiple Slit Interference
• The waves from two adjacent
slits might be only slightly
out of phase.
• But they will be canceled out
by waves from other slits.
• As a result, bright fringes
only occur when all the
waves arrive in phase.
Multiple Slit Interference
• The smaller maximums are
where waves from 2 or 3 slits
arrive in phase.
• As you increase the number
of slits, the brightness of
these smaller maximums
decreases to zero.
Multiple Slit Interference – Real World
• When we looked at the
double-slit experiment in the
lab, we noticed that some of
the interference fringes were
brighter than others.
• This is because the slits in the
lab are not infinitely narrow.
They have a finite width.
Multiple Slit Interference – Real World
• As a result, both interference and
diffraction effects are present.
• The pattern we see is an
interference pattern where the
brightness is determined by the
diffraction effect.
• The spacing of the interference
fringes is determined by the
spacing of the slits. The brightness
is determined by the width of each
slit.
Multiple Slit Interference – Real World
• The same thing happens when
we view multiple slit
interference in the real world.
• We see an interference pattern
contained within a diffraction
pattern.
• This will help us analyze
diffraction gratings.
Diffraction Gratings
Diffraction Gratings
• A diffraction grating is a collection of a large
number of slits (usually around 104).
• When the number of slits is large, the primary
maximums of the interference pattern
become very narrow.
• At the same time, the intermediate
maximums become so dim they disappear.
Diffraction Gratings
• If we shine monochromatic light on a
diffraction grating, we will see series of
narrow fringes on the screen.
• Because the interference fringes are so
narrow, the diffraction grating is a much
better instrument for measuring the
wavelength of light.
Diffraction Gratings
• Since the position of the fringes result from
interference, we can find the location of each
fringe using the old formula
d sinq = ml
m = 0, 1, 2,...
• Here, d is the distance between two adjacent slits
on the grating.
Example
A diffraction grating contains 10,000 lines/cm.
Determine the location of the 2nd order maxima
for
a) 400 nm light.
b) 700 nm light.
Diffraction Grating – White Light
• A more interesting question is what happens
when white light is shined on a diffraction
grating?
• As we know, white light is composed of all the
wavelengths of the visible spectrum.
• To see what happens, let’s first consider a
simple case where the incident light is a
mixture of two wavelengths.
Diffraction Grating – White Light
• Suppose the incident light had only two
wavelengths: 400 nm (purple) and 700 nm
(red).
• We already know that the location of the
diffraction maximums depends on
wavelength.
d sinq = ml
Diffraction Grating – White Light
• Suppose the incident light had only two
wavelengths: 400 nm (purple) and 700 nm
(red).
• We already know that the location of the
diffraction maximums depends on wavelength
(except for the central fringe).
d sinq = ml
Diffraction Grating – White Light
• The location central fringe will be the same for
both wavelengths (it’s at 0°).
• But the higher order fringes for the red and
blue lines will be at different locations since
the two colors correspond to different
wavelengths.
Diffraction Grating – White Light
• What we are seeing is basically two different
interference patterns on the same screen.
• Question, what would happen to the pattern if
the incident beam also included green light
(say 525 nm)?
Diffraction Grating – White Light
• If the incident beam contains all wavelengths,
then each is diffracted at a slightly different
angle.
• As a result, the light is separated into rainbows
for each order.
Diffraction Grating – White Light
• Question: why is the light for the second order
fainter than the first?
• The answer is the diffraction effect limits the
brightness of the higher orders.
• Notice also that it is possible to have the
rainbows overlap depending on the spacing of
the slits.
Example: Spectra Overlap
White light strikes a diffraction grating
containing 4000 slits/cm. Show that the blue
(450 nm) end of the 3rd order spectrum
overlaps with the red (700 nm) of the 2nd order
spectrum.
Homework
• Reread 27-6.
• Do problems 27, 28, and 29 on page 693.
Thin Film Interference
Thin Film Interference
• Interference gives rise to many every-day
phenomena.
• These include the colors reflected off a soap
bubble or an oil slick.
Thin Film Interference
• To see how this effect
occurs, let’s look at a thin
lair of oil on the surface of
water (or some other
surface with higher n).
• For simplicity, let’s assume
the light is monochromatic
(for now).
Thin Film Interference
• When the beam strikes the
oil, part of the beam is
reflected off the surface,
and part is refracted into
the oil.
• The refracted beam
continues until it strikes the
water. Again, part is
reflected and part is
refracted.
Thin Film Interference
• The reflected beam passes
back out of the oil and into
the air, forming a parallel
beam with the first
reflected ray.
• However, the second ray
has traveled a longer path
length than the first.
Thin Film Interference
• When these two rays arrive
at your eye, there will be a
phase difference that
depends on the path length
difference ABC.
• If ABC is equal to a
wavelength (or 2λ, 3λ, etc),
the rays arrive in phase and
interfere constructively.
Thin Film Interference
• If the thickness of the film is
a multiple of a halfwavelength, the rays will be
out of phase when they
reach your eye and you will
see no light.
• However, we have to be
careful, since the
wavelength of the light
changes in the film.
Thin Film Interference
• Therefore the path length
difference must be a
multiple of the wavelength
in the medium for
constructive interference to
occur.
• In other words,
ABC = mln
m =1, 2,3...
Thin Film Interference
• To find λn, use the index of
refraction
c
n=
v
• Then use the relation
v= fl
Thin Film Interference
• Let’s look at the following situation.
• A curved piece of glass is placed in contact with a
flat surface.
• The system is illuminated with monochromatic
light.
Thin Film Interference
• The air gap between the two pieces of glass
acts like a thin film.
• The two reflected rays have a phase difference
due to the path length difference.
Thin Film Interference
• However, the width of the film changes as you
move away from the center.
• Thus, at some points ABC is a whole multiple
of a wavelength.
Thin Film Interference
• At other points it is a half multiple of a
wavelength.
• This leads to a series of bright and dark circles
resulting from interference.
Newton’s Rings
• These fringes are called Newton’s rings.
• Notice that the central fringe is actually a dark
fringe.
Newton’s Rings
• This is surprising, since the two surfaces are in contact and
the path difference is zero.
• Yet there is a dark spot, indicating the rays are out of phase.
• It turns out that this is an important statement about the
behavior of light.
Reflection and Phase Changes
If a beam of light is reflected off of a surface
with a greater index of refraction than the
medium it is traveling through, it changes phase
by 180° (in other words, the wave flips).
Thin Film Interference
• An interference pattern can
also be seen when two pieces
of glass are separated by a
wedge-shaped region of air.
• Again, the air acts like a thin
film, leading to interference
that depends on the path
difference.
Thin Film Interference
• Because the region of air varies
in thickness, the path
difference changes.
• As a result, constructive
interference occurs at some
points and destructive at
others.
Homework
• Read 24-8.
• Complete your lab sheets.
Thin Film Interference
• If a thin film is illuminated by
white light instead of
monochromatic light, a series
of different colors is seen.
• Why do you think this
happens?
Thin Film Interference
• When white light is used,
interference is still occurring.
• However, the type of
interference that occurs for a
given wavelength depends on
the thickness of the film.
• So, different colors will interfere
constructively at for different
thicknesses.
Soap Bubbles and Oil Slicks
• This explains the colors seen on soap bubbles
and oils slicks.
• The film is not the same thickness throughout.
Soap Bubbles and Oil Slicks
• At at one point the thickness leads to
constructive interference for red light.
• At another point, its yellow, (then green, blue,
etc.).
Soap Bubbles and Oil Slicks
• Eventually, the thickness leads to constructive
interference for red light again, and the pattern
repeats.
• As a result, you see a series of different colors at
different points.