AP Physics – Light
Let there be light.
Huh? Where’s that from?
Well, it’s from the Bible and
it leads off the Book of
Genesis, which relates how
the world began. According
to the good book, it began
with light. What is
interesting about that is that
it is consistent with the Big
Bang Theory for the origin of
the universe.
Anyway, this unit is called
“Light” but really it should be called “electromagnetic waves”. Light (by this the Physics Kahuna
means visible light) is a type of electromagnetic wave – it’s just the one that we are most familiar
with.
Until the 18th century light was the only electromagnetic wave that was known. Even at that, it was
poorly understood. The first really good scientific treatment of light was done by the good Sir Isaac
Newton (well, we say that, but in truth he wasn’t so good, i.e., he wasn’t a very nice man, but, we
hasten to point out, a terrific physicist). Newton believed that light was a stream of particles that
traveled through space. This nicely explained how light could reach the earth from the sun through
the vacuum of space. Little teeny fast particles could easily travel through space, couldn’t they?
This was Newton’s corpuscular theory of light – he called the little particles corpuscles.
(“Corpuscles” is one of those words that were immensely popular in the old timey days, but which
have gone out of favor in these modern times. Corpuscles is still the name used for blood cells. It
was also the name the J.J. Thompson gave to electrons when he initially discovered them, but that
name, obviously, is no longer used.)
A competing theory was also in vogue at about the same time. This theory viewed light as a wave
rather than as a stream of particles. Christian Huygens, a Dutch astronomer, was the leader of this
faction. The wave idea had, unfortunately, a big problem - what was the medium for this wave? At
that time, it was a well-established fact that all waves had to have a medium that they propagated
through. So light needed to have one as well. But what was the medium? There wasn’t one! At
least one that anybody could discover.
Eventually physicists, resourceful devils that they were, invented an invisible, undetectable
substance that would provide just such a medium for the propagation of light. They called the stuff
luminiferous ether (which turned out to not exist, but that’s another story best told later in the
course).
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Both theories were able to explain the behavior of light, at least as it was known at the time. But in
1801 a British physicist and physician named Thomas Young conclusively demonstrated that light
was a wave. This was accomplished in a very famous experiment - Young’s double slit experiment.
What Young did was to show that light exhibited interference patterns (interference patterns result
from constructive and destructive interference, which, the Physics Kahuna is sure, you will recall is
a result of the superposition law of waves). It was well known at the time that waves could
interfere with each other causing interference patterns. In fact, at the time, it was believed that only
waves could do this (This turns out not to be true – but this discovery came about much later). This
put the particle theory to rest and it was gracefully retired. Throughout the 19th century physicists
were cheered by the certain knowledge that light was a wave and not a particle. They were,
however, stuck with the ether business.
In 1865 James Clerk Maxwell, considered by many
(including the Physics Kahuna) to be the greatest
physicist in the 19th century, unified electricity and
magnetism and was able to show that a moving
charge would produce a changing electric and
magnetic field. Said fields would then propagate
through space. He was able to calculate the speed
of these waves using his equations (collectively
known as Maxwell’s laws) and came up with a
number that was equal to the speed of light. He
boldly stated that light had to be one of his
electromagnetic waves. He also predicted that
different ones would be discovered in the fullness
of time.
In 1900 Max Planck was trying to explain a thing
called the black body problem (more on that later
when we study quantum mechanics). This had to
do with the energy of light as a function of its
frequency. Experiments had yielded results that
classical physics could not explain. In order to
come up with something that would match up with the experimental results, Planck had to imagine
that a beam of light was somehow arriving as a stream of energy bits. He called the energy bundles
“quanta” (this led to the quantum theory of light, which later led to the quantum atomic theory).
Max came up with a formula that would calculate the energy of one of these quanta.
This is one of the most famous equations in all of physics. Physics before Planck’s equation is
called classical physics. All physics after the equation is called modern physics. Here is the
equation.
Ehf
E is the energy of a quanta of light, h is called Planck’s constant, and f is the
frequency of the light.
Planck’s constant is: h = 6.63 x 10-34 J s or
4.14 x 10-15 eVs
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It should be pointed out that Planck did not believe these quanta were real. He thought of the
equation as being a sort of mathematical artificiality that allowed him to closely approximate the
energy provided by light as a function of its frequency, but that it was not a true description of the
nature of light. We won’t do much with this equation . . . . . yet.
Color Film: A Public Radio Commentary by Bill Hammack
I know of no better illustration of science being a creative process, and not just a list of facts, than
the invention of color film by two friends who loved first and foremost, not photography, or
chemistry, but music.
It all began when they saw a color movie. Dissatisfied with the blurry image they decided, with the
hubris of youth, to make a better color film.
At that time color photographs and movies were made by using three pieces of film, one for each
primary color. These separate pieces were combined in a projector to create a full color image, yet
always blurry because the three pieces of film never aligned correctly.
Godowsky and Mannes, realized that to get sharp color images they needed all three layers to be
part of the same piece of film. In their lab they found it very tricky to create a thin layer for each
color, called an emulsion, then to separate these with an even thinner layer of clear gelatin.
Although dedicated to making color film, their nineteen year quest was often interrupted by their
love for music. At one point Godowsky studied violin at the University of California, and Mannes
studied piano at Harvard. Then Mannes won Pulitzer and Guggenheim scholarships to study music
composition in Italy.
By 1930 they found their experiments so complex that they could no longer fund them from their
musical performance fees. Luckily, the Research Director for Kodak had learned of their work. He
hired Godowsky and Mannes to work in the Kodak laboratories. This arrangement fit them ideally:
They could spent days in the lab, then evenings performing at the nearby Eastman School of Music,
although the music wasn't completely separate from their lab work.
Godowky and Mannes would sing as they worked in their labs, not for fun, but as an essential part
of developing their color film. In their darkened lab they were unable to see a watch, so they timed
the reactions by singing passages from their favorite musical pieces, whose length they knew by
heart.
By 1935 they perfected their color film. In an odd press conference, the inventors announced their
discovery, showed sample photos, then played a violin and piano sonata for the reporters. After this
they developed no more film, but instead returned full time to musical careers.
Now to some it seems odd that these two would return to music, but to me there is no discontinuity.
Creating color film, or playing a sonata are both creative acts. So, to continue playing music for
them, was one and the same, at a fundamental level, with developing another type of film.
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In 1887 Heinrich Hertz (a Deutsche physicist) discovered the photoelectric effect. This was a very
mysterious deal. What happened (what he found) was that light shining on certain metals would
cause electrons to leave the surface of the metal. Why? How?
It remained a mystery until 1905 when a 3rd class technical inspector for the Swiss Patent Office
published a paper that explained the effect. This was Albert Einstein before he became famous.
Actually the paper helped make him famous – at least within the world of physics. Einstein won
the Noble prize for his discovery. The main thing about this was that Einstein explained the
photoelectric effect by assuming that the light was made up of a stream of actual, physical particles
that were banging into the metals physically knocking electrons loose from the atoms.
According to Einstein light wasn’t a wave at all, it was a stream of particles just like Newton had
said. The light particles are now called photons. So we end up coming back to Newton after all the
trouble with the waves and particles.
So who is right? Is light a particle or is it a wave?
Well, this is where physics gets a little sticky. Many people are very unhappy with what follows,
but it is the best physics we have for now. Hopefully someone will make a few odd discoveries and
come up with a theory that is more elegant.
But don’t hold your breath.
Here it is. Light acts a both particle and wave. This is called wave/particle duality.
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How simple! How elegant! How confusing! How can light be a particle and a wave both? The
two are very different things. But that is the way it is. Electromagnetic waves are particles and
waves at the same time and there is nothing we can do about it.
It is obvious that we do not properly understand the nature of light (or matter as well). We have a
theory that works, but it is a clumsy thing and no one is happy with it.
It turns out for some things, it is easier to imagine that light is a wave – optics and interference for
example. For other concepts it is better to think of it as a particle – the photoelectric effect comes
to mind.
Electromagnetic waves are transverse waves (when we think of light as a wave). We have a
changing electric field that has a spatial orientation that is ninety degrees from a changing magnetic
wave. Here is a rough drawing of the thing.
And yet another drawing.
E
B
c
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Speed of Light:
Light speed, that is the speed of light in a vacuum, is a universal constant. It
has been given its very own symbol, c – this is a true measure of how important it is.
c speed of light
The latest and greatest value for the speed of light (at least the last time the Physics Kahuna
checked on the thing) is 2.997 924 574 x 108 m/s. Note, this is the speed of light in a vacuum.
When light travels through other mediums it slows down.
For our humble purposes, we will use:
c = 3.00 x 108 m/s
All electromagnetic waves travel at c.
In astronomy a common unit of distance is the light year, which is defined as the distance that light
will travel in one year (which to us is a heck of a lot of distance, but to the universe is pathetically
small).
What is a light year in meters?
v
x
t
x vt
365 day
m
x 3.00 x 108 1 yr
s
1 yr
24 h
1 day
3.6 x 103 s
h
1
9.46 x 1015 m
The Electromagnetic Spectrum:
The electromagnetic spectrum encompasses all the
various types of electromagnetic waves. The sorting of these waves into specific groupings is, with
the exception of visible light, arbitrary. Anyhow, the specturm is the width and breadth of all the
electromagnetic waves. These waves are all the same – a changing magnetic and electric field, but
as the frequency varies, the energy changes and this changes the way that they interact with the
universe. The spectrum has been arbitrarily chopped up into named groups of waves that have
similar characteristics.
Frequency
Wavelength
Radio Micro Infrared Visible Ultraviolet X- rays - rays
waves waves light
light light
Above is a representation showing the various types of electromagnetic waves in the spectrum.
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Note that there are no numbers on the graphic. You will not be expected to remember frequencies
or wavelengths for these waves. What you will be expected to know is the relative positions of the
different types of waves. You should also be cognizant of the fact that the greater the frequency,
the greater the energy carried by a light photon (remember E = hf). Thus the greatest energy
photons are gamma rays. Also notice that as the frequency gets bigger, the wavelength gets
smaller. You do need to remember the wavelength minimum and maximum for visible light; 400
nm to 700 nm.
Gamma rays have the smallest wavelengths and radio waves have the longest.
Since light is a wave, we know that its speed must be equal to its frequency multiplied by its
wavelength. In other words:
c f
Here are the different types of waves with a general description of their uses and characteristics.
Radio Waves: Radio waves were discovered in 1888 by Hertz.
It didn’t take long to come up
with a practical use for them – radio! Today radio waves are used for, well, radio – AM radio and
short-wave radio both use radio waves. Other communication systems use them as well - cordless
telephones, radio controlled toys, cab radios, etc.
Microwaves: Microwaves were discovered in the early 1900’s.
Microwaves are used for FM
radio, broadcast TV, radar, cooking food, telephone communication systems, and numerous other
communications systems (different types of radio). The cooking food thing is interesting. It turns
out that microwaves excite molecules that are about the size of the ones that make up water. Thus,
anything that contains a significant amount of water can be heated by microwaves.
Infrared: Infrared electromagnetic waves are radiant heat.
They are often times referred to as
IR. We get a great deal of energy from the sun in the form of infrared waves. When you go out on
the first warm spring day and bask in the sun, absorbing all that wonderful heat, you are taking in
infrared electromagnetic waves. The wavelength of IR is just right to penetrate your outer skin and
excite the molecules in your epidermis, thus heating you up.
Visible Light: Visible light is important to we humans because we make great use of it to
create mind pictures of our surroundings. We arbitrarily arrange light into colors. The main colors,
when we think of colors, are red, orange, yellow, green, blue, and violet. The colors are listed in
increasing order of frequency. So red light photons have less energy than blue light photons.
Commit to memory the energy order of visible light. So in order of increasing energy: red, orange,
yellow, green, blue, and violet. But these are arbitrary names and the number of colors the human
eye can see is vast. There are a wide range of colors that we call “red” for example. Don’t believe
the Physics Kahuna? Okay, go to the makeup section of a store and look at all the shades of
lipstick. There are thousands to chose from.
The wavelength of red photons begins at 700 nm (700 x 10-9 m or 7 x 10-7 m). The highest energy
violet light photons have a wavelength of 400 nm (4 x 10-7 m).
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Ultraviolet Light: Ultraviolet light or UV is the general group of electromagnetic waves that
follows directly after visible light. We can’t see UV, but other critters can. Insect eyes can pick up
UV. It turns out that many plain looking flowers reflect lovely patterns when viewed in the UV –
no doubt to attract the esthetic senses of discriminating insects.
UV represents the point on the spectrum where the photons have enough energy to be harmful to
earthling hominid types. Besides the IR and visible light the sun radiates, it also radiates UV.
Fortunately for us, most of this UV is absorbed high in the atmosphere by the ozone layer (ozone is
a fancy oxygen molecule that has three oxygen atoms bonded together, O3). But some of it gets
through. When you go out and expose your outer surface to the sunlight, the UV, especially the
shorter wavelengths, has enough energy to damage skin cells. We call this damage sunburn. Have
you ever gotten a bad case of sunburn?
Sunburn is damaging. The skin ages and gets premature wrinkles. Even worse, too much sun has
been linked to skin cancer. People can die from skin cancer.
X-Rays: x-rays are very high-energy waves that are produced by the decay of unstable
radioactive elements or by exciting elements with high voltages. x-rays have a lot more energy
than UV. They have so much energy that they can pass right through a person’s soft tissue, but the
bones and denser organs absorb them. This is how x-rays at the doctor's office work. A large
piece of unexposed photographic film in a protective wrapper is placed on a table. Then the part of
the body the doctor is interested in examining is placed on top of the film. x -rays are then zapped
into the area. They travel through the soft tissue and expose the film (they affect it just as visible
light would). The bones absorb the x-rays so that the film under the bones is not exposed. The
film is developed and a picture of the interior of the body is revealed - shadows where the bones
blocked the x-rays.
x-rays have a great deal of energy, so doctors are very careful to prescribe them only when they
are really needed. Over exposure to x-rays is very dangerous. The x-rays penetrate the body and
have collisions with atoms in the various tissues. They can knock atoms loose and break chemical
bonds. When they mess up DNA molecules a body can end up with mutations. One can also get a
really bad form of skin cancer called melanoma.
Gamma Rays: Above the x -rays are the gamma rays (-rays). -rays are also produced by
the decay of radioactive elements. They are the most energetic of all the electromagnetic waves
and can be enormously destructive. They have tremendous penetrating power. Gammas are
produced naturally in the earth and atmosphere by the decay of radioactive elements. We also get
a lot of them from space – they are one of the constituents of what we call “cosmic rays”. So we
are exposed to them all the time - they are part of what is called the earth’s background radiation.
It is a most fortunate fact that our bodies can deal with the damage caused by this form of
radiation. Exposure to artificial -rays above the natural background, however, is extremely
dangerous. People whose work could expose them to such radioactivity have strict limits on the
amount of exposure they are allowed. Their exposure is monitored with personal film badges as
well as installed radiation detectors.
Vision: We see things because light rays have somehow traveled from the things we see into our
eyes. This can happen in several ways. Many objects give off their own light rays – candles, light
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bulbs, LED’s, etc. We say that these objects are luminous. The other way we can see things is
when the object reflects light rays. For example, light rays from the sun reflect off the leaves of a
tree and then come into our eye. So we see the tree.
Red objects appear red because they reflect red light. The other wavelengths are absorbed.
A green object is green because it reflects green light. And so on.
When we see all the colors at once, we see “white”. “Black” is the absence of light. Black objects
absorb all wavelengths of light. White objects reflect all wavelengths of light.
What happens to the light that is absorbed? Well, it basically heats the object up a slight amount.
Light travels in straight line path until it reaches a boundary. Once it reaches a boundary it can be
reflected, absorbed, or transmitted.
Transparency: Transparent objects allow light to travel through them. The way this happens
is kind of interesting. Photons enter the transparent medium and are absorbed by atoms. The
photon is absorbed by an electron within the atom. This increases its energy and it jumps to a
higher energy level. This is called a quantum leap (did you ever see the TV program of the same
name?). The electron is not stable at the higher energy level and falls back down to a lower energy
state. The energy lost by the electron is converted into a photon of the same energy as the one that
was absorbed. So light travels through a transparent medium in a series of photon absorptions and
emissions. These absorptions and emissions take time, so light is effectively slowed down as it
travels through the new medium. Thus light travels as a slower speed when it enters a medium
from a vacuum.
Reflection: There are two types of reflection that light can undergo.
Oh, by the way, when we
are looking at the path of light, we talk about light rays. A light ray is the path that a single photon
will follow – in general, a straight line. (At least on earth. Einstein found that light follows curved
paths in space. This curvature is caused by the bending of spacetime by objects with a great deal of
gravity.)
Here are the two types of reflection:
Diffuse reflection - the rays are reflected in random directions.
Specular reflection - parallel rays are reflected parallel to each other.
Objects that exhibit diffuse reflection are said to be opaque. Most things are opaque – clothing
(usually), paper, wood, leaves, concrete, people, hair, dirt, rugs, etc. are all examples of opaque
objects.
Specular reflection is often called mirror or regular reflection. Mirrors, polished metal surfaces,
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Diffuse reflection
Specular reflection
the surface of calm water, and other really smooth surfaces exhibit mirror reflection.
Specular Reflection:
Light rays obey the law of reflection during specular reflection. In the
rest of the course, when we discuss the reflection of light we will be talking about specular
reflection.
The reflection of light must obey the law of reflection, this means that the angle of reflection is
equal to the angle of incidence.
i r
Angle i is the angle of incidence – this is the angle made by an incoming light ray to
a normal to the surface. A normal to the surface is a line that is perpendicular to the
surface. Angle r is the angle of reflection. This is the angle the reflected ray makes
with a normal to the surface.
Refraction: Refraction occurs when a wave travels from one medium to another.
The light rays
are said to be “bent”. You can see this in a glass of water. Place a pencil in the glass – the pencil
looks like it is broken, shifted to the side, and larger in the water.
This occurs because light rays from the pencil are bent as they travel through the
water.
Refraction of light change in the path direction in a different
medium.
The refraction happens because of the velocity difference for light in the two mediums.
This brings us to a very important concept. The index of refraction. This is the ratio of the speed
of light in a vacuum to the speed of light in the new medium.
The index of refraction ratio of the speed of light in a vacuum to the
speed of light in a different medium.
n
speed of light in vacuum c
speed of light in medium v
n
c
v
n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of
light in the medium.
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You will be provided with this equation on the AP Physics Test.
Directly to the right, the Physics Kahuna has placed a table of
values for the index of refraction for different substances. Using
this table, one can easily calculate the speed of light in any of the
listed substances.
What is the speed of light in ice?
We use the index of refraction equation to find the speed in the
ice.
n
c
v
v
c
n
3.00 x 108
1.309
m
s
2.29 x 108
m
s
Important Thingee: Observe the table for the index of
Substance
Index of
Refraction
2.419
1.458
1.52
1.66
1.309
1.49
Diamond
Fused quartz
Crown glass
Flint glass
Ice
Polystyrene
Zircon
1.923
Benzene
1.501
Ethyl alcohol
1.361
Water
1.333
Air
1.000 293
Carbon dioxide 1.000 45
refraction for air. It is essentially one. So on all problems, let the
index of refraction be one for all calculations.
Snell’s Law:
The amount of refraction is of great interest to we physicists – you are interested
are you not?
N ormal
If the light rays travel from one medium into another, they
will be bent. The bigger the difference in the light speed in
the two media, the bigger the refraction. If the speed of
light is the same in both media then there will be no
refraction.
1
You should remind the Physics Kahuna to show you some
of his nifty refraction demonstrations.
2
In the drawing to the right, the angle of incidence is I, the
angle of refraction is 2. These angles are measured
relative to a normal to the surface of the boundary between the two substances.
Another important point is that the frequency of light does not change as it goes from one medium
to another. The wavelength can change and the speed of light can change, but the frequency does
not. v f must hold true.
The amount of refraction, that is, the angle of refraction, is
related to the angle of incidence by Snell’s law. The
drawing to the right shows a ray of light passing from one
medium into another. In the first medium it has an index of
refraction of n1 in the second medium its index of refraction
is n2. Here is Snell’s law:
n1 substance 1
1
n2
2
substance 2
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n1 sin 1 n2 sin 2
n1 is the index of refraction in the first medium, n2 is the index of refraction in the
second medium, 1 is the angle of incidence, and 2 is the angle of refraction.
You will be provided with Snell’s law on the AP Physics Test.
A beam of light that has a wavelength of 651 nm traveling in air is incident on a slab of
transparent material. The angle of incidence is 35.0. The angle of refraction is 23.4. Find
index of refraction for slab.
n1 sin 1 n2 sin 2
n2 n1
sin 1
sin 2
For all practical purposes, the index of refraction in air is 1.
sin 35.1o
n2 1
sin 23.4o
1.45
Wavelength and Frequency:
What happens to the wavelength and frequency as light
travels from one medium to another? We could, actually you could, develop any number of
equations relating wavelength and frequency or index of refraction. Unfortunately, you won’t have
access to any of those equations. So Physics Kahuna is going to introduce you to solving the
various problems using brute force and the equations that you are given. It’s not very elegant, but it
works. Feel free, however, to derive relationships if you like.
Let’s look at a typical problem.
A Beam of light with a wavelength of 565 nm is traveling in air and is incident on slab of
transparent material. The angle of incidence is 32.0. The refracted beam makes an angle of
20.5. (a) Find the index of refraction for the slab and (b) find the wavelength of the light in the
second medium.
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(a) n1 sin 1 n2 sin 2
n2 n1
sin 1
sin 2
sin 32.0o
n2
1.51
sin 20.5o
(b)
We know the index of refraction in the second medium. We can use this to find the speed of
light in this medium:
n
c
v
v
c
n
3.00 x 108
m
s
1.51
1.987 x 108
m
s
Using this and the equation relating wavelength and velocity, we can find the wavelength in the
new medium
v f But what is the frequency? Well we can solve for the frequency of the light in air (recall
that the frequency of the light will be the same in both mediums):
v f
f
m
s
9
567 x 10 m
3.00 x 108
v
0.005291 x 1017 Hz
5.291 x 1014 Hz
Okay, we can plug this into the velocity equation, which we solve for wavelength.
v f
v
v
f
m
1
6
9
1.987 x 108
0.376 x 10 m 376 x 10 m
1
s 5.291 x 1014
s
376 nm
A beam of laser light, wavelength 633.8 nm in air is incident on a block of polystyrene at an
angle of 32.0. Its wavelength in the new medium is 345 nm. Find (a) the index of refraction of
the light in the polystyrene and (b) the angle of refraction in the polystyrene.
(a) We can find the frequency of the light in air and that’s the same as it is in the polystyrene.
v f
f
v
3.00 x 108
f 4.733 x 1014 Hz
v f
m
1
17
0.004733 x 10 Hz
9
s 633.8 x 10 m
Next we find the speed of the light in the new medium:
1
4.733 x 1014 345 x 109 m
s
1633 x 105
m
s
1.633 x 108
m
s
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Now we can find the index of refraction:
n
c
v
m
s
m
1.633 x 108
s
3.00 x 108
1.84
(b) This is simple, just use Snell’s law.
n1 sin 1 n2 sin 2
sin 32.0o
0.2880
sin 2
1.84
sin 2
2
n1
sin 1
n2
16.7o
Dear Doctor Science,
Before electric lighting, did we have acoustic lighting?
-- Joe Futrelle from Urbana, IL
Dr. Science responds:
We did, but it was so loud that it strained the ears, as well as the eyes so it
was quickly prohibited by law, causing people to use dim but quiet forms of
illumination such as candles and the oil lamps. Acoustic lighting was even
more orange than sodium vapor lights, and anything red appeared black
under its garish glow. For some reason, a certain portion of the populace
suffered from seizures when exposed to this light, and it's thought that the
Cleveland Nonsense Riots of 1889 were
caused by the Cleveland Indians using acoustic lighting to illuminate the first
night game.
Total Internal Reflection:
When a light ray encounters a new medium, some of the wave
is reflected and some of the energy is transmitted. So at a boundary when the incident wave is not
perpendicular to the surface we would see both reflection and refraction.
normal
Air
Water
normal
transmitted
light ray
reflected
light ray
transmitted
light ray
reflected
light ray
normal
transmitted
light ray
critical
angle
reflected
light ray
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As the angle of incidence increases, the angle of refraction also increases. Eventually the angle of
refraction is big enough so that it is bent into the surface of the boundary. This means that none of
the energy goes into the second medium, all of it is trapped in the first medium. We call this total
internal reflection. This is really big in the field of fiber optics.
Total internal reflection will occur at the angle where the refracted ray stays in the boundary. This
is called the critical angle. The critical angle is given by the formula:
sin C
n1
n2
The nice thing is that you will be given this equation on the AP Physics Test.
air
What is the critical angle for water and air?
sin C
n1
n2
1
1.333
C
48.6o
c
water
Polarization:
We have already learned that light is a transverse wave. It is created by electrons when they change
energy levels in atoms. As they do this, the energy that they give up is converted into photons of
light. This happens in all directions and orientations – there is no “up” or “down” to an atom,
electron, or photon. So the electromagnetic waves that are formed are oriented in all possible
planes.
Light can undergo processes that orient all the waves so that they are in the same plane. Such light
is said to be polarized.
Please note that just about all sources of light – the sun, a candle, lightbulbs, etc. produce light that
is unpolarized. In fact the idea of polarization was not even dreamed of until the 1940’s when a
brilliant inventor, Edmund Land, discovered it. He also learned how to make a filter that would
produce polarized light. These are called polarizing filters. When normal light passes through a
polarizing filter, only light in one plane gets through, the rest is absorbed by the filter.
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E
B
c
What happens to the filter when it absorbs this light? (It heats up, right?)
Here is a poor drawing that the Physics Kahuna made to try to give you an idea of this polarizing
filter thing. Does the drawing make any sense at all?
horizontal wave can
pass through horizontal
filter
vertical wave
cannot pass through
the filter
The other way that light can get polarized is during reflection off planer surfaces (i.e., flat surfaces).
This is called polarization by reflection and is quite common.
Edmund Land invented polarizing filters in the 1940's. One of their first applications - and still
popular today - is in sunglasses. Most of the surfaces that cause glare (reflected sunlight) are
horizontal: bumpers, car windows, water puddles, the polished hood of a car, etc. Land made
sunglasses that had the filters vertically polarized. Normal light which contains light oriented in all
planes can come through the lenses, but only the light that is vertically polarized gets through. So
the light reaching the eyes is diminished (just what you want in sunglasses). Light reflected off
horizontal surfaces is polarized horizontally. Normally this reflected light causes glare. But the
vertically oriented polarized filters won’t let this light come through. So they do a bang up job of
reducing glare.
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Polarized light is also used in 3D movies. We see in three dimensions (3D stands for three
dimensions). This is because we have two eyes that are set slightly apart. Because of this, we have
parallax, each eye sees a slightly different picture. The brain receives two slightly different
images. Objects that are very close are situated very differently in the two images. Objects that are
very far away look about the same. Hold a finger a few inches from you nose and look at it with
one eye closed. Then close the other eye. The finger will look very different. Next look at
something farther away in the same manner. You will see that there isn’t much of a change
between what the right and left eye sees when the object is distant. Anyway, your brain puts the
two images from each eye together and gives you a sense of depth. You can tell just by looking if a
thing is close or distant. In a 3D movie, the film that is projected on the screen is made up of two
such images. One image is polarized horizontally, the other is polarized vertically. Each member
of the audience is given special glasses. One of the lenses is polarized horizontally and the other
vertically. Only one of the pictures is allowed through each of the filters so the brain has two
images to put together, just as in real life. So you see a 3D image.
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Dear Doctor Science,
What is polarized light?
-- Melissa Kitchen from Emporia, VA
Dr. Science responds,
Polarized light is light that has passed over both the North and South poles. Thanks to the
atmosphere's refractive nature, light from the sun bends when it hits the stratosphere, and begins
to circle the planet. Heavy light, an oxymoron, sinks quickly into the oceans where it forms algae
and plankton. Zippy, peppy, vivacious light, or lite light, zooms around the earth, sometimes
passing over the entire planet in less than a few minutes. This is the rare and exotic "polarized
light" which you can only see with special glasses. Sure, they're more expensive than ordinary
sunglasses, but they're worth it. Sometimes you'll see and angry polar bear or a lovesick penguin,
but those are virtual images, remnants from polarized light's amazing journey.
Dear Doctor Science,
What colors attract a child's attention most? Do some colors encourage learning better than
others?
-- Annie O'Keefe, Webster Groves, MO
Dr. Science responds:
Color attraction is the key to personality type. The happy, extroverted child will naturally be
attracted to joyous, primary colors. Children who prefer mauve and burnt umber are suffering
from some malady, usually a biochemical imbalance that can be cured with exposure to loud rock
music and a steady diet of chewing gum. Learning is best stimulated by an effective reward system.
Money works very well, as does candy and access to cute animals. Punishing children by making
them watch so-called educational television rarely accomplishes anything.
Dear Doctor Science,
No matter whether I put white, blue, red or green clothing in my dryer, the lint always comes
out gray. Why?
-- Marsha Wegman from Minneapolis, MN
Dr. Science responds,
You know that one black sock the dryer keeps eating? Well, it stays deep within the bowels of the
drier, providing a necessary service. Not only does its reluctant fluff lubricate the spinning drum,
but it allows you to know that your dryer is indeed, working. Sure, your clothes are dry when they
come out, but they were dry when you put them into your washing machine. Dryer lint is the only
absolute indicator that something has happened. Gray lint was found to be the most reassuring,
providing this tangible gauge of progress, without signifying destruction of your clothing. Very
few clothes are gray.
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Dear Cecil:
I'm sitting here placing cold compresses on my eyes after being suckered into another
3-D movie. After being tortured as a child with those ridiculous candy-wrapper
glasses, you'd think I'd learn, but nooooo. Though the movie, Space Hunter, was the
most natural, i.e., it wasn't hitting you in the face at every turn with special effects, it
still fell victim to the low state of 3-D technology. Is there a reason, save Foster
Grant's special interest, that we must suffer through the present form of projection?
Why can't a coherent 3-D image just appear on the screen?
--John F., Evanston, Illinois
Cecil replies:
John, you knucklehead, think about it. The movie screen is flat! Reality is 3-D! There is a
basic problem trying to get a 3-D image to "just appear" on a 2-D surface. What you have
to do is fool the eye by presenting each eyeball with a slightly different image, which the
brain then fuses into the illusion of 3-D. (Remember those old Viewmaster 3-D image
viewers? They worked the same way.)
To create a 3-D movie, two images are projected simultaneously, one for each eye. You
wear special glasses with different-colored lenses so that one eye sees one image, the other
eye sees the other. Voila, 3-D action--plus untold misery as the glasses dig into your nose.
But hey, what's your complaint? Haven't you heard of suffering for your art?
For years one heard rumors of an experimental 3-D screen that made glasses unnecessary.
(Supposedly it worked along the lines of those novelty photos where the image changed as
you looked at the thing from different angles.) However, as far as I can tell, this process
never became commercially practical.
The main problem with Space Hunter, in any event, was not the glasses but the print. Years
ago 3-D required two separate projectors, which were hard to keep properly aligned and in
sync. To eliminate such problems, Space Hunter used a new process in which both images
were printed on the same piece of film. Unfortunately, since you were still using the same
old 35mm stock, each image got to be only half as big as before, and consequently was a
lot fuzzier when it got enlarged to big-screen size. In addition, since you only had one light
source (as opposed to two with the old 3-D system), the screen image was a lot dimmer.
The result was eyestrain and migraine headaches for viewing audiences around the
country.
--CECIL ADAMS
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AP Physics – Geometric Optics
We’ll start off our exploration
of optics with the topic of
mirrors. The type of mirror you
are most familiar with is called
a plane mirror. Your first
thought is probably, “What do
we care about mirrors on
airplanes for?” Well, not that
kind of plane mirror. Although
this stuff would certainly apply
to mirrors on planes. Anyway,
plane mirrors are flat and have a
metal surface. Some times they
are made of metal – the first
mirrors found by archeologists
were made of polished brass.
Jason (owner of the good ship
Argo) used a polished metal
shield as a mirror when he
chopped off the head of the
Gorgon Medusa (he had to use a
mirror because if you look at a
Gorgon, they like turn you into
stone). Modern mirrors are made of glass or clear plastic and have a thin, very smooth layer of
metal on one side. Nowadays the metal is usually aluminum but it mostly used to be silver, so the
metal surface is called the silvered surface. Most mirrors have the silvering on the far side under
glass so it won’t get scratched. Mirrors for very precise applications (like in projection TV’s) are
front silvered.
Plane mirrors form images. When you look into the mirror you see all sorts of wondrous things.
Remember that the only way you see anything is for light rays to enter your eyes. When you look
at a mirror, everything that you see in the mirror is being reflected off the surface of the mirror. Our
eyes, primitive as they are, do not understand that light rays can be reflected. Our eye/brain system
is convinced that light rays always travel in straight lines. When we look into a mirror, light rays
seem to be coming from objects that are on the other side of the mirror. In reality the objects are on
the same side of the mirror as we are and the light rays are reflected into our eyes. The drawing
below shows how an image of an arrow is formed. Two rays are shown coming off the object (the
arrow). A ray from the top and a ray from the bottom. The ray from the top reflects off the mirror
into our eye and the ray from the bottom does the same. Our brain “ray traces” the reflected rays
back behind the mirror. We call these the back rays. Your eye and brain are back ray tracing.
The image is formed in our brain on the opposite side of the mirror. It is the same size as the object
and the same distance from the mirror as the object.
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The image is called a virtual image. Think of virtual image as meaning that the image is made by
phantom rays. No real light rays form the virtual image.
Plane mirrors form images that are erect, non magnified, and virtual.
Object
Image
Here is a baby, the baby does not understand about virtual
images and is foolishly trying to grab its image (which it
thinks is behind the mirror) with its free hand.
Spherical Mirrors:
Spherical mirrors produce a very different type of image. So first things
first, what is a spherical mirror?
A spherical mirror has a curved surface. The surface has a constant radius of curvature. You could
think of it as sort of like a circular part of a mirror that is a true sphere. Sort of a cookie cut out of
the thing. The little round cookie has a curved surface. So the mirrored surface is part of sphere.
There are two types of spherical mirrors, convex mirrors and concave mirrors. The type a mirror
depends on where the silvering is placed.
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On the drawing above, you can see that on a convex mirror, the silvering is on the side that bulges
out. A concave mirror has the silvering on the side that caves in. In fact the cute little way to
remember which is which is that a “concave mirror makes a cave” or some such nonsense.
Here is a drawing showing the geometry of a concave mirror.
Principle axis
C
V
f
R
C is the center of curvature. Its distance from the center of the mirror is
equal to the radius of curvature R for the mirror.
The principle axis is a line that goes through the center of curvature to the center of the mirror.
f is the principle focus. It is also called the focal point or sometimes simply the focus.
The focal length is the distance from the principle focus to the center of the mirror. It is equal to
one half of the radius.
f
R C
2 2
Rules For Ray Diagrams:
The reflection of light rays that enter spherical concave mirrors
follow certain specific rules. Here are those very rules.
1. Rays that enter the mirror parallel to the principle axis are reflected off the mirror
through the principle focus.
2. Rays that pass through (or appear to originate from) the principle focus are reflected
parallel to the principle axis.
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3. Rays that pass through (or appear to originate from) the center of curvature enter the
mirror and are reflected back through the center of curvature.
1
2
3
f
C
Ray 1 -- Comes in parallel to the principal axis, is reflected through f.
Ray 2 -- Enters by passing through the focal point, is reflected off the mirror parallel to the
principle axis.
Ray 3 -- Goes through center of curvature, is reflected off the mirror through the center of
curvature. It retraces itself.
Image Formation:
Let’s look at a typical image formed by a concave spherical mirror.
The object is located between the center of curvature
and the principle focus.
The object has a humungeous number of rays reflecting
off it in all directions. We are only interested in the
rays that follow our rules.
C
f
So the first ray we look at is one coming off the top of
the object parallel to the principle axis. It will be
reflected off the mirror through the focal point.
C
f
C
f
The next ray that we draw is the one that goes through
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the principle focus. It is reflected off the mirror parallel to the principle axis.
These two rays fix the image, the top of the object (the arrowhead) is located where the two rays
intersect. The bottom of the object is fixed by the
principle axis. Rays from the principle axis that travel
through the focal point are parallel to the principle
axis, they also appear to originate from the center of
f
curvature, so the base of the object stays attached to
C
the principle axis.
This is a real image because it is made up of real light
rays that actually intersect each other. It is inverted.
Also note that for this specific example that the
distance from the image to the mirror center is greater
than the object distance. Also the image is magnified.
Be sure to remind the Physics Kahuna to show you his real image demonstrations.
As the object is moved about in front of the mirror, the image
size and image distance will vary. The image can be smaller or
larger than the object.
It is interesting to look at what happens when the object is placed
at the focal point:
f
C
Note that the two rays are parallel. They will never intersect, so
there can be no image. Thus no image is formed of an object that
is at the principle focus of the concave mirror.
What happens when the object is placed inside the principle
focus? Let’s look at the rays and see what happens.
When we trace out the three rays, they don’t cross in front of the
mirror. So what is going on?
Did you see it? Yes, your brain appraises the situation and says
to itself, “Hey these rays all look like they’re coming from
behind the mirror!” Your brain does a bit of back ray tracing is
what happens. So we end up with a virtual image that is erect
and magnified.
C
f
451
C
f
People buy these mirrors so that they can get a virtual image of their faces in the thing. The
manufacturers call them makeup mirrors (if the market is women) or shaving mirrors (if the market
is men).
Concave mirrors are also known as converging mirrors because reflected light rays can converge to
form real images.
Convex Mirror Images: Convex mirrors also form images. But the image that forms is
always virtual and smaller. Convex mirrors are also known as diverging mirrors because the
reflected rays always diverge and can never form a real image.
Here we see a convex mirror with an object placed in front of
it. The main difference here is that the principle focus is on
the opposite side of the mirror from the object.
mirror surface
f
The rules for the rays are pretty much the same, but there are
a couple of little differences. Notice also that we don’t really
care about the center of curvature.
Rays that are parallel to the principle axis are reflected as if
they originated from the principle focus. The ray is reflected
off the mirror as if it had started out at the principle focus.
Rays that look as if they are going to go through the
principle focus are reflected off the mirror parallel to the
principle axis. Let’s look at this ray.
f
Note that the rays will never cross in front of the mirror. So,
the old brain has to go in and do some major back ray tracing.
f
452
You get something that looks like this:
For a convex mirror, the image will always be virtual and will
always be smaller than the object.
f
Convex mirrors are used to give a big field of view. Because
objects in front of the mirror are smaller as an image, you get
a bigger view in the mirror. So the mirrors are used on cars
and trucks as rear view mirrors. They are also used in stores as security mirrors. A couple of
mirrors mounted in the corners give a view of the entire interior of a store.
Lenses:
Enough already on mirrors. Let’s talk about lenses. You know what they are – you
may have a pair of them in your eyeglasses, or do you wear contact lenses?
Lenses refract light in a sort of organized way. The type of lenses that we’ll be looking at are called
double spherical lenses – both sides of the lens have the same radius of curvature. There are two
types of these lenses: convex lenses and concave lenses. Concave lenses cave inward and convex
lenses bulge outward.
Convex lenses are also known as converging lenses because the refracted rays that pass through
can converge. Convex lenses can form real or virtual images.
Concave lenses are called diverging lenses because the light rays always diverge. Concave lenses
can only form virtual images.
Lens Stuff:
Here is the geometry of the double convex lens.
Q
f
f
Q is the center of the lens, f and f’ are the focal points on either side of the lens. Also note that
lens has a principle axis.
Here are the rules for image formation:
1. Rays that are parallel to the principle axis are refracted through the focal point on the
opposite side of the lens.
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2. Rays the travel through the center of the lens are not refracted and travel in a straight
line.
3. Rays that travel through the focal point before they reach the lens are refracted out of the
lens parallel to the principle axis.
Note: the Physics Kahuna likes to draw a vertical line on the lens, that is a line through the center
of the thing. No doubt it will be clear how this vertical line is used.
1
2
f
3
f
Ray 1 is the parallel ray that is refracted through focal point on the opposite side of
the lens.
Ray 2 is the ray that goes through the center of the lens.
Ray 3 is the ray that goes through the focal point and is then refracted parallel to the
principle axis.
The image is formed where the rays intersect. Clearly you can see that this will be a real
image.
f
f
Converging lenses form real images, they can also form virtual images as well. This is what
happens when you use one as a magnifying glass.
454
f
f
Here is a virtual image formed by a converging lens. Notice that the object is inside the focal
point. When an object is placed at the focal point no image forms. This is shown in the next
drawing. This is because all the rays are refracted in such a way that they are parallel to one
another. They never cross, so they form no image.
f
f
What about a diverging lens? Let’s look at how diverging lenses form images. Keep in mind
that a diverging lens always produces a virtual image.
1. The parallel ray is refracted as if it had started out at the focal point on the same
side of the lens as it is.
2. The ray that goes through the center of the lens is not refracted.
The rays look like this:
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f
f
Notice that the rays all diverge. This means that you have to do the old back ray tracing
thing to find the virtual image.
f
f
You should be prepared to use ray tracing to locate images for spherical mirrors and lenses.
Multiple Lens Deals:
Many instruments that use lenses use several of them in series with
one another - telescopes, microscopes, camera lenses, film projectors, etc. The way this works is
that the first lens forms an image of an object. This image serves as the object for the next lens and
so on.
Here is a typical simple set up. There are two lenses and a single object.
object
First we find the image formed by the first lens.
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1st image
The first image will then act as an object to the second lens, which will form a final image.
Final image
Thin Lens Equation:
The relationship between the image distance and the object distance
is given by the thin lens equation.
1 1 1
si so f
so is the object distance, si is the image distance, and f is the focal length for the lens.
Lens sign conventions:
so - Positive if object is in front of the lens.
so - Negative if the object is behind the lens.
si - Positive if the image is behind the lens.
si - Negative if the image is in front of the lens.
f is positive for a converging lens.
A lens has a focal length of 25 cm. An object is placed 32 cm from the lens. What is the object
distance?
1 1 1
si so f
1 1 1
si f so
457
1
1
1
si 25 cm 32 cm
si
110 cm
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Magnification:
One of the reasons that lenses are used is that they can provide some
magnification – make something look bigger.
If you divide the size of the image by the size of the object you get a number that represents the
magnification. If the image is 10 cm tall and the object is 5 cm tall then the magnification is two.
The magnification is given by:
M
hi
s
i
ho
so
M is the magnification, hi is the image height, ho is the object height, si is the image
distance, and so is the object distance.
A double convex thin lens has a focal length of 36.0 cm. A 2.30 cm tall object is placed 12.0
cm from the lens, find (a) the type of image, (b) the image distance, (c) the magnification, (d)
the image height.
(a) Virtual Image. We know this because the object distance is less than the focal length.
(b)
1 1 1
si so f
1 1 1
si f so
1
1
36.0 cm 12.0 cm
si
18.0 cm
18.0 cm
si
Note: no units for magnification. Also the
1.50
so
12.0 cm
Physics Kahuna dropped the negative sign. It just tells us that it is a virtual image.
(c) M
(d)
M
hi
ho
hi Mho
1.50 2.30 cm
3.45 cm
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We have us a two lens system. The focal length for the 1st lens is 10.0 cm, the focal length
of the second lens is 20.0 cm. (a) Find the final image distance. (b) find the magnification of
the final lens.
35.0 cm
(a)
1 1 1
si so f
22.0 cm
1 1 1
si f so
1
1
1
si 10.0 cm 35.0 cm
si 14.0 cm
The image is 14.0 cm from the first lens. Since the second lens is 22.0 cm from the first lens, the
distance from the first image to the second lens, which will be its object distance, is the difference
between the two distances:
22.0 cm 14.0 cm 8.0 cm
Now we can find the image distance for the final image formed by the second lens:
1 1 1
si f so
1
1
20.0 cm 8.0 cm
si
13.0 cm
(b) Finding the magnification.
M
si
so
13.0 cm
14.0 cm
0.928
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Changes in the Lens: What happens to the focal length if the radius of curvature for a lens
is changed? What happens if the index of refraction for the mirror is changed? What happens if the
lens is immersed into a different medium that has a different index of refraction than does air?
The basic thing to think of is this: what happens to the angle of refraction? Anything that increases
the angle of refraction will produce a smaller focal length as the light is bent a greater amount.
First let’s look at changes to the radius of curvature, R. In general, this is what happens. As R
increases, the focal length increases. Thick fat lenses (which have a small radius of curvature) have
a small focal length. Thin lenses (which have a large radius of curvature) have a longer
wavelength. The bigger the radius, the bigger the amount of refraction, the smaller the focal length.
If the index of refraction is changed? If the index of refraction increases the focal length decreases.
I f the index of refraction decreases the focal length increases. This is because the light is bent
more with a larger index of refraction.
large radius
of curvature
small radius
of curvature
If the lens is placed into a new medium, what happens is that the focal length increases. This is
because air has, essentially, an index of refraction of 1. Any other medium you put the thing into
will have a larger index of refraction. The bigger the difference in the index of refraction, the more
the light will be refracted. Right? Makes sense, don’t it? Anyway. Since the difference in the
index of refraction has to be smaller (since the new medium has to be bigger than 1) the light is bent
a smaller amount so the focal length has to increase.
Dear Dr. Science,
I wear glasses and contact lenses. Why does squinting help me see better?
---- Keith Lawler, Sunderland, England
Dr. Science responds:
I have bad news for you, Keith. You're supposed to wear either glasses or contact lenses,
not both at the same time. No wonder you're spending much of your time squinting. Keep
this up and you'll be begging for radial keratotomy as well. Whenever I visit any kind of
doctor, I always try to pay attention to the part at the end of the visit, where they tell me
how to use the new artificial brain, or whatever, they've just installed. Sure, I ignore most
of what they tell me, but if I can remain teachable for just a few moments, I won't end up in
your shoes, squinting and walking into walls.
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Dear Cecil:
How do "night" rear view mirrors work? One flick of the button and it seemingly dims all.
--Chris Gaffney, Toronto
Cecil replies:
Here at Straight Dope University, we have explanations suited to all levels of intellectual
attainment. We offer the intro-level course first.
In a dimming rear view mirror you've got two reflecting surfaces--one with high reflectance, one
with low. During the day you use the high reflector. At night the dimmer button swings the low
reflector into place, dimming glare from headlights behind you. Satisfied? Then cut to the funnies,
wimp. Still thirsting for more? Coming right up.
The trick is that the two reflecting surfaces are the front and back of the same piece of glass. Said
glass is specially ground so that the back surface is slightly tilted relative to the front one. In other
words, the glass looks wedge-shaped from the side. The back surface is coated with silver like a
bathroom mirror, making it highly reflective. The front surface isn't coated, but it's still slightly
reflective, like all glass.
Because the two surfaces are out of parallel, any time you look at the rear-view mirror, you're
seeing two different reflections simultaneously. During the day with the mirror tilted into the
normal position, the silvered surface shows you the road behind you. The non-silvered surface,
meanwhile, shows you the back seat of the car--but it's so dim you don't notice it.
At night the situation is reversed. When you flip the dim button, the silvered surface tilts so it's
showing you the car's ceiling, which is so dark you don't notice it. But now the non-silvered
surface is showing you the road.
Because the headlights of the cars behind you are so bright, the non-silvered surface reflects
enough light to let you see what's behind you. But it's not so bright that you're blinded.
The folks at GM tell me that on Cadillacs you can now get a high-tech "electrochromic" mirror that
dims at night automatically, without having to flip a switch. The Caddy mirror has only one
reflective surface, but there's a special film in front of it that gradually darkens at night through the
magic of electronics.
Very impressive. But for sheer low-tech genius the tilting surfaces are hard to beat.
--CECIL ADAMS
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Dear Cecil:
I was looking in the mirror the other day without wearing my glasses, which I
occasionally use because I'm nearsighted. I noticed that things that were far away, even
when reflected in the mirror, were blurry. When I put my glasses on and looked in the
mirror again, everything was in focus. I found this strange. I thought everything should
have been in focus without my glasses, because the mirror was close to my eyes and so (I
thought) were the reflections. I guess that's why people don't use mirrors for vision
correction, huh?
--Kirsten Munson, Santa Barbara, California
Cecil replies:
You got it, babe. The reflection is out of focus, even though you're close to the mirror, because
you're not looking at the mirror. You're looking at the image in the mirror, a different matter
entirely.
You can prove this by a simple experiment. Look at a mirror from a distance of 6-12 inches.
With your glasses off, focus as best you can on some distant object reflected in the mirror--say,
a bathroom towel on the wall behind you. No doubt the image of the towel is pretty fuzzy, and
not just because you haven't cleaned the lint screen on the dryer. Now look at something on the
surface of the mirror, such as a dust speck. You'll observe that (1) it requires a noticeable effort
to adjust your eyes--in other words, you're refocusing--but that when all is said and done (2) the
speck, unlike the towel, is in reasonably sharp focus. This clearly demonstrates (to me, anyway)
that when you look at a reflection in the mirror, you're not looking at the mirror's surface.
So what are you looking at? For purposes of focusing, at the object itself (in this example, the
towel). Without going into the the technical details, the image of the towel in the mirror is out
of focus for the same reason that the towel is out of focus when you look at it directly. In both
cases the light travels more or less the same distance from the object to your eyes; the fact that
in one instance it bounces off the mirror en route is irrelevant. Unless you want me to get out
my giant model of the exposed human eye--and it IS looking a little bloodshot--I say we leave
it at that.
--CECIL ADAMS
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AP Physics – Interference
In the 18-century, physicists discovered that waves displayed interference patterns. One of the
reasons that Newton thought that light was a stream of particles was that light did not, apparently,
display interference patterns.
The reason for this was that they weren’t looking at the proper scale. The wavelength of light is
very small and the interference patterns are on a very small scale as well.
Actually they aren’t hard to observe at all. Take one of your hands and hold your fingers straight
and together. Place the hand between your eyes and a strong light and move the fingers towards
your eyes until you find a little gap between the fingers. If you bring this close to your eyes, you
will see little lines between the fingers. These are interference patterns.
Thomas Young was the first to show the wave nature of light with his
double slit experiment. The experiment made use of a wave
phenomenon called diffraction. Diffraction is when a wave moves
through a small opening. To diffract light, you need a really narrow
opening – we call these things slits. Young used two slits that were
close together. He made the slits by coating a piece of glass with
carbon and then inscribing two scratches close together onto the glass
through the carbon.
The little demonstration with the fingers is an example of the
diffraction of light. We’ll look at single slit interference patterns a
little later. But for now, back to the double slit deal.
Diffraction of wave
through a slit
To get a double slit interference pattern, what is
needed is a single coherent light source. In coherent
light all the rays are going in the same direction,
they have the same wavelength, and they are in
phase. In these enlightened modern times the easy
way to get coherent light is to use a laser beam.
Young didn’t have that luxury, so he had to fake it.
He passed a single light source – a bright “bulls
eye” lantern – through an opening. The lantern
acted like a flashlight, forming a light beam that
would spread out as it traveled through the air.
After it went through the opening, it would be even more directional. It wasn’t pure coherent light,
but was close enough for Young’s experiment.
For the experiment to work, a single light source is required. Two sources won’t produce the right
sort of interference patterns.
After the light goes through the single opening, it is incident on the two narrow slits. The light
diffracts through the slits and the expanding wavefronts interfere with each other. We get
constructive and destructive interference. A white card is placed at some distance from the slits and
464
the interference patters can be seen on the card. Where constructive interference has taken place we
see a bright line. A dark line indicates destructive interference (no light energy). These lines, both
the bright and dark ones, are called fringes.
Let’s look at the wave diffracting through one of the slits and then both of the slits:
Diffraction through
one slit
Diffraction through
both slits
Constructive interference takes place where the wave crests intersect. Destructive interference
takes place sort of in between. If you place a card in the interference pattern, you can see the
fringes.
Max
The destructive fringes are called minima and the
constructive fringes are called maxima.
This is basically what the thing looks like:
The radial lines simply mark the intersection of the
wave crests where the constructive interference is
taking place.
Reason for Interference Patterns:
The basic reason that the interference patterns show
up on the screen is that the light rays from each slit
travel a different distance. Because of this, when
they finally arrive at the screen they may or may
not be in phase.
If the waves arrive in phase from each slit, we get
constructive interference and a maxima (bright
fringe) appears on the screen. If they arrive out of
phase, destructive interference takes place and we
get a minima (dark fringe).
Min
Max
Min
Max
Min
Max
Min
Max
Min
Max
Min
Max
Min
Max
Min
Max
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You can see this in the drawing below. At point P on the screen the waves arrive in phase and we
get a bright fringe. Below P
we see destructive
interference in the drawing
to the right.
p
p
These drawings are very
inaccurate, however – the
Bright area
scale is way off. Major,
Dark area
big-time off to be exact.
Anyway, the wavelength is
so small, the slits are really
tiny, and the distance to the screen is enormous when compared with the wavelength and the slit
width. The rays are actually almost parallel to each other when they reach the screen and are not
separated by a big angle like the picture shows.
The arrival of the waves at the screen looks more like this:
In phase
Out of phase
The critical thing for the waves coming through the slits and meeting at the screen is this: in phase
or out of phase? If they traveled the same distance, they would have to be in phase, but they don’t.
They travel a different distance. So if we look at the path difference between the rays coming out of
the slits we can figure this thing out. Is the path difference enough to make everything be in phase
or out of phase?
Path
difference
In phase
Path
difference
Out of phase
466
If the path difference is zero or an integer multiple of the wavelength, we would get constructive
interference. This would mean that the path difference could be the wavelength . Or the path
difference could be 2 , 3 , 4 , 5 , etc.
If the wave arrive out of phase we get destructive interference. This would happen if the path
difference is
1
3
5
7
or or or , etc. Here we would have an odd integer multiple of half
2
2
2
2
the wavelength. Do you see why this is so? Stop and think about it until you do.
Here is a really fancy drawing showing the geometry of the path lengths. We have two slits, S1 and
S2. Two rays are shown - the straight lines that go from the slits to the point P.
P
r1
S1
d
Q
S2
y
r2
d sin
O
L
P is the point on the screen where the two waves arrive. d is the spacing between the slits. Q is the
midpoint between the two slits and O is the point on the screen straight across from the slits.
The angle is the angle between a line that connects Q to P.
The path difference for light rays traveling through the slits is r2 r1 . Using a bit of geometry,
you can see that a line dropped from the center of the first slit drawn perpendicular to the ray from
the center of the second slit ( r2 ) makes a right triangle. This line has the angle . One side of the
right triangle is d (the hypotenuse). The opposite side to the angle is d sin .
This is a very key thing.
The symbol for the path difference is .
So the path difference is:
r2 r1 d sin
Clearly then, if the path difference is an integer multiple of the wavelength, we will get constructive
interference.
467
d sin
m
where
m 0, 1, 2, 3, &tc
Thus:
d sin
m
m fringe number
3
2
1
This is the equation you will have available on the
AP Physics Test. Use it wisely.
You end up with a central bright fringe. This is the
zeroth-order maximum (m = 0). You can see that the
angle would be zero for this fringe. The bright
central fringe is bracketed by a series of smaller
bright fringes for the different integer values of m.
The next set of fringes is the first-order maximum,
then we have the second-order maximum, and so on.
0
1
2
3
Because the slits are so small, the wavelength is really incredibly tiny, and the distance to the screen
is humungous, the rays forming a fringe are essentially parallel and the angle is the same. Like
this ridiculous drawing:
r1
r2
S1
d
S2
Example Problem:
468
A double slit setup has a slit spacing of 1.50 mm. A screen is set up 3.50 m from the double
slit. Monochromatic light of wavelength 565 nm is incident on the slits. So find the angle for
the first order bright fringe.
d sin m
m 1 sin
d
565 x 109 m
1.50 x 103 m
376.7 x 106
0.0003767
0.0216o
Destructive Interference: Destructive interference will take place if the path difference is
half of the wavelength or an odd multiple of half the wavelength. This way the waves will arrive
180 out of phase and we will get a minima.
The equation for constructive interference is:
d sin
m
So for the first order dark fringe, the first minima where m is 1 we get:
d sin
1
2
For the second minima we would get
d sin
3
2
And so on.
Finding the Spacing Between Fringes:
How do we calculate the spacing between the bright fringes?
We start out with a drawing of the geometry of the double slit
system. You can see it off to the right. We assume that L is
much greater than d. The distance y is the spacing between
the zeroith-order fringe and the first-order fringe.
r1
y
d
We also assume that d is much greater than .
r2
d sin
L
The angle is very small, so small, that the sine of is
essentially the same value as the tangent of . (This is only true for very small angles.)
Therefore we can say:
sin tan
y
L
The reason we do this is that we can measure y and L.
Look at the equation for the path difference:
469
d sin
sin
m
We solve it for sin
m
d
We plug in this value for sin in the path difference equation.
sin
ym
m
d
y m
L
d
L
ym m
d
Solve the thing for y:
L
d
Here ym is the spacing between the central fringe and the mth order fringe (does that make sense?).
m is the integer, is the wavelength of the light, L is the distance from the slits to the screen, and d
is the spacing between the two slits.
Similarly:
yDark
L
1
m
d
2
On the AP Physics Test the equation has a slightly different form:
xm m
L
d
You can see that it is the same thing as the one we developed, right?
Red light ( = 664 nm) is used in Young’s experiment with slits separated by a distance of 1.20
x 10-4 m. The screen is located 2.75 m from the slits. Find the distance y on the screen between
the central bright fringe and the third order bright fringe.
xm m
y 3
L
d
664 x 109 m 2.75 m
4
1.20 x 10 m
4560 x 105 m
4.56 x 102 m or
4.56 cm
A screen is separated from a double slit setup by a distance of 2.00 m. The slit spacing is 0.025
mm. Light passing through the slits has a second-order maximum that is 6.55 cm from the
centerline. Find (a) the wavelength, (b) the distance between the adjacent fringes.
470
(a) The phrase “second-order maximum” means that m is 2. So there is a central bright fringe, a
first order fringe, and a second order fringe. We have the spacing for the second order fringe from
the center bright fringe.
103 m
5
2.5 x 102 mm
2.5 x 10 m
1 mm
L
dx
xm m
mL
d
2.5 x 10
5
m 6.55 x 102 m
2 2.00 m
1 nm
4.09 x 107 m 9
10 m
409 nm
(b) Distance between fringes: What we must do is find the distance between the second order fringe
and the first order fringe. This is the difference between the spacings of the fringes.
y ym1 ym
4.09 x 10
x
7
m 1
m 2.00 m
2.5 x 10
5
m
L
d
m
L
d
m
L
d
L
d
m
L
d
L
d
3.27 x 102 m or 3.27 cm
Reflection & Phase: Light reflecting from a boundary can do so in phase (sort of a free end
reflection) or out of phase (a fixed end reflection). The thing that determines whether the reflected
wave is in or out of phase is the difference in speed for light in the two media. The wave will
180 phase
change
N o phase
change
n1 n 2
n1
n2
n1
n1 n 2
n2
undergo a 180 phase change when it is reflected from a medium that has a higher index of
refraction than the one it came through.
471
There will be no phase change if the wave is reflected from a medium that has a lower refractive
index.
You would get the phase change for light traveling through air and reflecting off glass. Glass has a
higher index of refraction than air.
You would not get a phase change for light traveling through glass and being reflected off water,
since water has a lower index of refraction than glass.
This lead us to:
Thin Film Interference: This occurs when light travels through a very thin layer of
transparent material. Thin film interference occurs with oil films, soap bubbles, etc.
Light that is incident on the film has several things happen to it. Some of the light is reflected off
the top of the film. These waves have a 180 phase change since the index of refraction for the film
is greater than for air. Next, the light that goes into the
180 phase No phase
film is refracted as it travels from air into the film. Some
change
of the light goes into the air on the other side of the film.
change
This light is refracted (back the other way). Finally,
some of the light is reflected off the air/film surface. This
light does not undergo any phase change.
1
We have a lovely drawing showing how all this works.
(See drawing to the right.)
2
Air
The film has a thickness of t.
We let n be the index of refraction for the film. The
index of refraction for air is, of course, 1.
Ray 1 reflecting off the surface of the film has a 180
phase change.
Film
t
Air
Ray 2 reflecting off the opposite film surface has no phase change.
The two rays are out of phase.
The two waves will recombine when you look into the film and the rays enter your eyes. If the path
difference is half of the wavelength, or an odd multiple of the wavelength, then the waves will end
up in phase and you will see constructive interference – a bright fringe.
The basic kind of problem involves finding the minimum thickness that will cause constructive or
destructive interference. This minimum would be when the wave came straight down onto the film.
This means that the angle of incidence is zero.
The surface reflected wave undergoes a 180 phase change. The wave that reflects off the bottom
surface does not undergo a phase change. In order to get the bottom reflected wave to match up
472
with the first one, the path difference must be different by a half wavelength. The totals distance
that the wave travels is twice the thickness of the film, 2 t. This must equal half a wavelength.
Now the wavelength we’re talking about here is the wavelength of the light in the film. This is
different than the wavelength in air (or the first medium if it ain’t air). So we have to carefully
specify which wavelength we’re talking about.
We’ll call the wavelength in the film f. This means that the minimum thickness is given by:
1
2t f
2
Unfortunately, we’re usually given the wavelength of the light in a vacuum, which is the same as
the wavelength in air for our purposes, right? But we aren’t given the wavelength of the light in the
film. So what do we do? Oh, poor poor pitiful us, whatever will we do?
Well, we can figure this out, can’t we? We can solve for the wavelength in the film! We start with
the equation for the index of refraction.
n
c
v
We solve for the speed of light.
c nv
This is true for any medium.
So:
c nv
for air
c nf vf
and
for the
film set them equal
nv n f v f
v f
The speed of a wave is given by:
We can plug this into our equation for v.
nv n f v f
n f n f f f
n n f f
The frequency is the same in each medium:
f
so the wavelength for the film is:
n
nf
So we can plug that into the equation:
1
2t f
2
t
1 n
2t
2 n f
2t
n
2n f
n
4n f
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Index of Refraction and Wavelength:
We’ve developed a very useful relationship in solving our little problem: n n f f .
n11 n22
In general, we see that:
Sadly, you won’t be given this equation on the AP Physics Test. So be prepared to figure it out.
Light with a wavelength of 555 nm is incident on a soap bubble. What is the minimum
thickness for thin film interference to take place for this wavelength of light? The index of
refraction for the bubble goop is 1.35.
2t
The minimum thickness is
f
2
The path difference has to be half the wavelength of the light in the film and, of course, the path
difference is twice the thickness of the film.
But we don’t know what the wavelength of the light is in the film – we can figure it out however.
n11 n22
2
1 n
2t 1 1
2 n2
t
n11
n2
n11
4n2
555 nm
4 1.35
103 nm
Destructive Interference on Thin Film: Destructive interference will take place if the
path difference is equal to a single wavelength or an integer multiple of the wavelength. To find the
minimum thickness, we go through the same deal. The minimum path length is 2 t, and it must be
an integer multiple.
2t m f
m 1, 2, 3, ...
For the case where m is 1, we get
2t f
Find minimum film thickness for destructive interference in reflected light if a thin film (n =
1.35) is illuminated by light that has wavelength of 585 nm.
The minimum thickness is
2t f
474
n11 n22
2t f
2
2t
n11
n2
n11
n2
t
n11
2n2
585 nm
2 1.35
217 nm
180 phase 180 phase
change
change
Thin Coatings: If the thin film rests on a surface that has
a different index of refraction than what is on the on the other
side, it is a thin coating. Thin coatings are used on glass lenses.
Another example of a thin coating would be an oil slick, where
a thin layer of oil rests on the water surface.
Air
If the coating has a lower index of refraction than the surface it
rests on, light rays will undergo a 180 phase change from the
bottom surface of the film as well as the upper surface.
This makes everything opposite to what we saw for a thin film.
Film
Glass
If the surface the coating rests on has a smaller index of
refraction than the coating, then it is the same as for a thin film and we don’t need to go anywhere
else with the thing.
So for your standard thin coating, both reflected waves have the same phase. The interference again
will depend on the path difference. You can see that the minimum thickness for constructive
interference will be given by:
2t f
The minimum thickness for destructive interference is given by:
2t
f
2
Find the minimum film thickness for constructive interference in
reflected light for a coating (nf = 1.30 ) on glass (ng = 1.60) is
illuminated by light that has wavelength of 555 nm.
2t
n
t
Air
2n
Film
Glass
475
We know that n11 n f f
t
2n
f
so
1 5.55 nm
1 n11
t
2 n nf
2 1.42
n11
nf
195 nm
Non Reflective Coatings:
Non reflective coatings are used on camera lenses to get rid of unwanted glare and light highlights.
To get rid of the flares, we want destructive interference to take place. This gives us the least
amount of reflection.
180 phase 180 phase
change
Both rays undergo a 180 phase change.
change
Air
Net change in phase from reflection is zero.
To get destructive interference:
2t
Film
2
A camera lens needs coating that will minimize reflection.
nf = 1.25. Figure wavelength of light at 545 nm. What
thickness?
2t
2t
2
2
n11 n22
so
1 545 nm
1 n
t 1 1
4 n2
4 1.25
f
Glass
n11
n2
109 nm
476
Dear Doctor Science,
Since there is a speed of light and a speed of sound, is there a speed of smell?
-- Andrea M. Allan Ph.D. from Dept of Pharmacology, Univ. of New Mexico,
Albuquerque, NM
Dr. Science responds:
There is, but it's very slow. It's speed depends on several factors, most notably the
stinkability index of the smellee. Other factors include, wind velocity, humidity, and the
sensitivity of the smeller's
receptor, or in layman's terms, nose. So the smell of, say, Parmesan cheese on a winter
day, traveling indoors, would be around 6 miles per hour. An Iowa hog lot, on a still
summer morning, would propagate at around 550 miles per hour, close to the speed of
sound. And if you're wondering if hogs would smell as bad if there were no one near to
smell them, any Iowan would reply, "you bet".
Dear Cecil:
Do you have any info on the so-called "green flash"? It's not a superhero, but rather an
optical phenomenon involving a burst of pure green light that occurs just as the sun rises or
sets over the ocean. I've seen it several times but my friends won't believe me, saying it's
just delayed mescaline aftereffects. Set these unbelievers straight.
--E.N., Hollywood, California
Cecil replies:
Adds a sort of postmodern element to the process of scientific discovery, doesn't it? "I believe
I've discovered a new perturbation in the space-time continuum! However, it could just be the
drugs." Don't worry, though--there really is such a thing as green flash. Usually it's a thin green
band or splotch visible for a split second at or near the top edge of the sun as it sinks beyond the
horizon. You can see it at sunrise too. Sometimes it lasts longer; sometimes it's blue or violet or
turns from green to blue. To see it you need a clearly delineated horizon and a haze-free sky. The
ocean (or any large body of water) will do fine, as will a desert or mountain.
Most people have never seen green flash and think it's a myth, ascribing it to retinal fatigue on
the part of the observer or other causes. One reason they're so adamant is that green flash is
impossible to photograph with an ordinary camera--the image is too small too register. But
researchers managed it in the 1950's using telescopes. (For some of their handiwork, see the
January, 1960 Scientific American.)
Green flash is caused by atmospheric refraction--that is, the bending of sunlight as it passes
through the air so that it splits into a rainbow of colors. Refraction causes the solar disk to be
surrounded by ghost images like a cheap TV, with a violet/blue/green "shadow" above and a
red/orange/yellow one below. None of this is visible except at sunrise and sunset, when
477
refraction hits the max and the sun's light is so reduced that the ghosts don't wash out. The red
ghost disappears below the horizon, the orange and yellow ones are absorbed by the atmosphere,
the blue and violet ones scatter (usually), and what's left is green. Count yourself lucky if you've
seen it; you're one of a privileged few.
--CECIL ADAMS
Dear Cecil:
Are cats and dogs really color-blind? How do they know?
--Jim L., Chicago
Dear Jim:
You ever see a cat who could pick out a tie? Believe me, cats'll wear things you wouldn't put on a
dog. Scientists, however, are not content with anecdotal evidence. They often test animal color
sensitivity by trying to link color with food. One such experiment was conducted in 1915 by two
scientists at the University of Colorado, J.C. DeVoss and Rose Ganson. They put fish in two jelly
jars and then lined both with paper, one gray and one colored. If a cat picked the colored jar, it
got to eat the fish. Nine cats, 18 months, and 100,000 tries later, the researchers established that
cats picked the right jar only half the time--the level of pure chance. On the other hand, cats
could readily distinguish between different shades of gray. Ergo, cats are color-blind.
Doubts about this conclusion arose some years later, however. Cats have cones as well as rodtype vision receptors in their retinas, and cones have long been associated with color vision in
humans. Neurologists who wired up feline brains with electrodes discovered that, on laboratory
instruments at least, cats responded to light of different wavelengths--which is to say, color. So
researchers went back for another round of fish experiments. Finally, in the 1960s, they managed
to teach the cats to discriminate between colors. But it took some doing--one group found it took
their cats between 1,350 and 1,750 tries before they got the hang of it.
From this one might deduce one of two things: either cats are exceptionally dense, a proposition
Cecil has no trouble buying, or else they just don't give a hoot about color. Most cat scholars
have opted for choice #2, saying that the ability to distinguish colors is obviously of no
importance to cats and hence not something they learn readily.
Less work has been done on dogs than on cats, but what there is suggests canine color sensitivity
isn't very good either. Much the same can be said for mammals in general, with the exception of
primates. In contrast, some of your supposedly lower order creatures, such as fish, turtles, and
especially birds, can distinguish color with ease. The fact that these primitive beasts should have
more advanced visual abilities than their mammalian betters has always struck observers as a
little odd; clearly the evolutionary progress of color vision has been more erratic than one might
expect.
LATE NEWS!
To the Teeming Millions:
OK, so we know cats can see in color. Now comes new research indicating that dogs can see in
478
color, too. Three scientists at the University of California at Santa Barbara adopted the
traditional strategy of trying to tempt the dogs with food. The menu, frankly, could have stood
some improvement: would YOU cooperate with people whose idea of a reward was a cheeseand-beef-flavored pellet? Nonetheless, the researchers found three mutts who were sufficiently
desperate to play along. They showed the dogs three screens lit up from behind with colored
lights--two of one color, the third of a different color. The mutts got the pellet if they poked the
odd-colored screen with their noses.
The dogs had no difficulty distinguishing colors at the opposite ends of the visible spectrum,
such as red and blue, and they proved to be demons with blues in general, quickly learning to
differentiate blue from violet. But they bombed at other colors, confusing greenish-yellow,
orange, and red.
The researchers concluded that dogs suffer from a type of colorblindness that in humans is called
deuteranopia. Normal humans have three types of color receptors for red, green, and blue.
Deuteranopes lack the green receptor, and thus (apparently) can't tell a lemon from a lime--or, for
that matter, a red traffic light from a green one. One more reason to put your foot down next time
the pooch says he wants to drive.
--CECIL ADAMS
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AP Physics - More Dadgum Interference
Single Slit Diffraction:
A single slit will also form an interference pattern when light
passes through it.
Each part of the slit acts as source of waves. This is described in Huygen’s principle.
Huygen’s principle Every point on a wave front acts as a source of tiny wavelets that
move forward with the same speed as the wave. The wave front at a later instant of time is
the surface that is tangent to the wavelets.
You can imagine that across the width of the
slit, little wavelets originate and travel
through the slit. These waves pass through
the slit and form a bright central fringe on the
screen, which is at a far distance from the slit.
d
This distance is so far that all the waves are
essentially parallel to one other. All the
wavelets travel the same distance and arrive
at the screen in phase with each other and we
incident
get constructive interference. This creates a
wave
bright central fringe at the center of the
screen directly opposite the slit.
Central Maxima
midpoint
of center
maxima
Very distant
screen
The wavelets that originate in the slit can also
interfere destructively. Here’s a drawing that shows the very thing.
First
dark
fringe
d
incident
wave
Wavelet
fromation for
dark fringe
midpoint
of center
maxima
How interference
takes place
480
Light from one part of slit interferes with light from another part of slit, forming the patterns. Again
the cause of the interference is the path difference for the waves (wavelets in this case).
The patterns that form can be described in this way:
There will be a bright central fringe surrounded by two dark fringes, then a set of weaker bright
fringes, a set of dark fringes, a set of weaker (than the central fringe) bright fringes, and so on.
Let’s look at the geometry of the thing.
d
The same equation is used with single slit diffraction as with the double slit diffraction, except that
the angle we get, , is the angle from the center of the slit to the center of the dark fringe.
d sin m
This equation describes destructive interference.
d is the width of the slit, is the angle to the center of the dark fringe, m is the
integer order number, and is the wavelength of the light.
We can analyze it the same way we did the double slit deal to find the spacings between the central
fringe and the dark fringes.
sin
d
sin tan
y
L
481
y
L d
y
L
d
This general case is:
xm
m L
d
While this is the same equation as for a double slit deal, for a single slit it gives you the distance
from the center of the bright central fringe to the desired dark fringe.
y
2
y
d
1
Sin
d
Sin
d
Sin
0
y
1
Sin
y
2
Sin
575 nm light passes through a slit of width 0.250 mm. An
observing screen is set up 3.00 m away. (a) Find the
position of the first dark fringe. (b) What is the width of the
central maxima?
(a) This is the first minima, so m = 1. The spacing is given by:
L
m L
xm
d
y1
1 575 x 109 m 3.00 m
3
0.250 x 10 m
6900 x 106 m
d
d
y
1
y
1
6.90 mm
Diffraction Grating:
Diffraction gratings are a recent invention (well, a lot more recent than the old double slit deal).
Basically, a diffraction grating is a piece of transparent material that has parallel cuts scribed in it.
The scribings are so small that you can’t really see them. At any rate, the grating has a very large
482
number of equally spaced parallel slits cut into it. This would be on
the order of hundreds to several thousand lines per centimeter.
The grating acts like a double slit setup. It produces a large number
of very bright, sharp fringes separated from one another by fairly
wide minima.
Maxima are given by the same equation as we have seen before:
d sin m
m is the order number, d is the spacing between the slits,
is the wavelength of the light, and is the angle formed
by a normal to the grating to a line at the center of the fringe.
Light from a distant star enters a telescope and then passes through a diffraction grating onto a
screen. A first order red line appears on the screen at an angle of 25.93. The lines of the
grating are separated by 1.50 x 10 –6 m. What is the wavelength of the light?
d sin m
d sin
1.50 x 106 m sin 25.93o 0.656 x 106 m 656 x 109 m 656 nm
632.8 nm laser beam passes through a diffraction grating that has 6 000.0 lines per centimeter.
An observing screen is set up 3.00 m away. Find separation of the maxima.
The slit separation is the inverse of slit density.
d
1
cm 1.667 x 104 cm 1.667 x 106 m
6 000
d sin m
sin
d
The angle is not small and we cannot make the
assumption that the sine of is equal to the tangent.
Y
We can find from the equation and then use the
to find y.
sin
d
632.8 x 109 m
1.667 x 106 m
379.6 x 103
tangent
L
0.3796
22.3o
483
tan
y
L
y tan L tan 22.3o 3.00 m
1.23 m
The pattern of maxima from a diffraction grating looks like this:
M
2
d
1
d
0
0
-1
d
-2
d
The maxima are very bright and sharp, they are also widely separated from one another. For this
reason, the diffraction grating is preferred to double slits. So diffraction gratings are better because:
Get very sharp maxima
Get very wide dark areas
2 1
With a few easy to make measurements, one can easily
calculate the wavelength of the light.
Newton’s Rings:
The last thing we want to talk
about in our lovely discussion of light and optics are
Newton’s rings. These are not the ugly lines he had under
his eyeballs, nor are they the fabulous finger wear he
sported. This is an optic effect thingee.
r
P
O
To the right you can see a drawing of the thing. A rectangular glass plate lies on a table top. Lying
atop the plate is a circular lens that is flat on top and has a constant radius curvature on its lower
surface. It’s your basic half of a lens.
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Ray 1 is reflected from the surface of the plate. It undergoes a 180 phase change.
Ray 2 is reflected from bottom of lens. Because air has a smaller index of refraction than glass
there is no phase change on the reflected ray. Because of this, there exists a path difference
between the two rays. This will give us interference patterns – minima and maxima. These will
appear in the lens when you look down into it. You will see dark and bright concentric circles –
they look like rings, hence the name. Newton was the dude who discovered them.
Dispersion:
The index of refraction, n, is a function of wavelength. The angle of refraction is varies for
different wavelengths of light. The index of refraction generally decreases with increasing
wavelength. The means that red light ( = 600 nm) bends less than does blue light ( = 470 nm).
red
blue
dispersion in a prism
This phenomenon is called dispersion. It was discovered by Isaac Newton. He built a prism and
used it to separate out or disperse all the visible colors from white light. Prisms are especially
valuable in devices called spectrometers. The prism separates light into a spectrum of colors which
can then be focused onto a screen with a system of lenses. Light from distant stars can be analyzed
in this way to determine the star’s chemical composition.
prism
increasing
wavelength
white
light
Dispersion is responsible for rainbows. Little droplets of water act like prisms hanging in the sky
and disperse the sun’s light, forming a spectrum, which we call a rainbow.
Here’s an interesting thing – how do rainbows work? Ever wonder about that?
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Okay, here’s the deal with rainbows. You see rainbow right before, after, during rain, right? You
also see them around sprinklers, waterfalls. etc.
The basic idea is that they are around water sprays. This is where you get water drops floating in
the air. The rainbow is the result of light undergoing dispersion in these tiny drops of water
suspended in the atmosphere.
The water drops act like prisms. Light enters near the top of
the drop on one side, is reflected (total internal reflection),
which means it undergoes two refractions. (Refraction as it
enters the water drop and refraction when it leaves the water
drop.) The colors are therefore dispersed . Violet light is
refracted the most (since it goes through the water at a slower
speed, recall the higher the frequency of light, the greater the
refraction) and red light is bent the least. The angle between
the sunlight entering the drop and the red light leaving the drop is forty two degrees (maybe that's
the real question for the ultimate answer). The angle for the violet light is 40. The other colors are
somewhere in between. This
dispersion does not happen with just
one raindrop. It happens with
gazillions of the little devils.
Therefore one sees colors
everywhere along a 40 - 42 arc.
We call this effect a rainbow.
To see the rainbow, you must have
your back to the sun. It appears as a
circle, but roughly half the circle is in the ground so you don't notice the lower half of the circle.
You only see the upper half. If you see a rainbow from the air, like when you are flying in a plane,
you will see a raincircle instead of a
rainbow.
You can see only one color from each
raindrop. So if violet light enters your
eye from a drop, the red light travels in
a path from a drop that is actually a bit
higher than the drop that provided you
with the red color. So when you see a
rainbow, red light is the color that is at
the top of the rainbow and violet light
is below. All the other colors fall in
between - red, orange, yellow, green,
blue, and violet.
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You see the colors directly overhead (at 42 ), but you
also see them in an arc that is 42 on either side of you.
Sometimes you not only see one rainbow but a bonus,
extra one as well. The first rainbow is a regular one red light on top; violet light on bottom. The second
rainbow is above this one, but the colors are reversed.
Violet is on top and red is on the bottom.
This second rainbow is caused by light being reflected
twice in the water drops. It enters the bottom of a drop,
is reflected upward, reflected yet again, and then leaves
the drop. The second reflection of the light is not at the
critical angle, so some of the light escapes, so the second rainbow is dimmer and not as bright.
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Rainbows appears as cones that extends from your eyes. Your view of the rainbow is unique - no
one else can see it in the same way. It's your own personal,
private rainbow! As you move, the rainbow moves with
you. This is why chasing after the end of a rainbow is a no
win deal - you can never get there because as you move
towards it, it moves with you. So chasing after the gold pot
at the end of a rainbow is a foolish task. Sort of the ultimate
wild goose chase. Our ancestors knew this, hence the stories
of gold at the end of a rainbow - they were cruel jokers on
occasion.
Another related effect that one can see is the classic "ring around the moon". This is caused by the
refraction of light through ice crystals. There is no internal reflection within the ice crystals, so
there is no dispersion, so no rainbow.
Dear Cecil:
I've been told that thousands of years ago mankind recognized the existence of only
three colors. For obvious reasons I'm rather skeptical about this piece of historical
wisdom. When were additional colors recognized? Did someone "discover" them--e.g.,
William of Orange? Eric the Red? You just can't find this out in the World Book.
--Cary C., Santa Monica, California
Cecil replies:
In the restaurant of knowledge, Cary, remember this: the World Book is a snack, the Straight
Dope is a meal. But enough of this folderol. Anthropologists and classicists have been
arguing about color terminology since the nineteenth century, and it's only recently that the
situation has begun to clarify. The opening shot was fired by William Gladstone, the British
politician and Homeric scholar. He pointed out that abstract color terminology was virtually
absent from Homer's work, and claimed that the Greeks had no sense of color at all, having
only the ability to distinguish light from dark. He believed "... that the organ of color and its
488
impressions were but partially developed among the Greeks of the heroic age"--i.e., the Greeks
were physiologically incapable of perceiving color.
Lazarus Geiger, a naturalist, expanded on Gladstone's ideas. Based on his examination of Greek
literature, the Vedic hymns, and other ancient writings, he claimed that man's color sense had
developed only gradually, in fairly recent times. He thought that man had become aware of colors
in the order that they appear in the spectrum, starting with the longest wavelengths. First, he said,
people dimly realized that something was "colored," then they could distinguish black and red,
then black and red plus yellow, then white, then green, and finally blue. (I realize white is not a
"color" of the spectrum, but this is Geiger's theory, not mine.) He pointed out that "Democritus and
the Pythagoreans assumed four fundamental colours, black, white, red, and yellow.... Nay, ancient
writers (Cicero, Pliny, and Quintilian) state it as a positive fact that the Greek painters, down to the
time of Alexander, employed only these four colours."
Later writers conceded the relative poverty of color terminology among ancient peoples, but
denied that it reflected a physical inability to distinguish color. A major breakthrough on the
question occurred in 1880 when an opthalmologist named Hugo Magnus organized a study
involving missionaries working with primitive tribes around the world. Using standardized color
samples and a rigorous testing procedure devised by Magnus, the missionaries found that primitive
peoples with a limited color vocabulary nonetheless could distinguish colors every bit as well as
persons from highly developed cultures--they just didn't have _names_ for all the colors.
Not everybody bought this conclusion. As late as 1901 one researcher was arguing that the
members of one primitive culture literally could not see any difference between blue and green
because their retinas were more strongly pigmented than those of Europeans. But in general
anthropologists came to accept the view that physiological differences did not explain the
variations in color vocabulary among cultures.
So what does explain the variations? That's still a matter of dispute. The majority view, I would
venture to say, is that the designation of colors in different cultures is totally arbitrary. For
instance, H.A. Gleason notes, "There is a continuous gradation of color from one end of the
spectrum to the other. Yet an American describing it will list the hues as red, orange, yellow,
green, blue, purple, or something of the kind. There is nothing inherent either in the spectrum or
the human perception of it which would compel its division in this way" (An Introduction to
Descriptive Linguistics, 1961). Similarly, Verne Ray says "there is no such thing as a natural
division of the spectrum. Each culture has taken the spectral continuum and has divided it up on a
basis which is quite arbitrary" ("Techniques and Problems in the Study of Human Color
Perception," Southwestern Journal of Anthropology, 1952).
More recent research, however, suggests that color terminology may not be so arbitrary after all.
Brent Berlin and Paul Kay (Basic Color Terms: Their Universality and Evolution, 1969), to whom
Cecil is indebted for much of the preceding discussion, suggest that there is a remarkable degree of
uniformity in the way different cultures assign color names. In a study of 98 languages from a
variety of linguistic families, they found the following "rules" seem to apply:
489
1. All languages contain terms for white and black.
2. If a language contains three terms, then it contains a term for red.
3. If a language contains four terms, then it contains a term for either green or yellow (but not
both).
4. If a language contains five terms, then it contains terms for both green and yellow.
5. If a language contains six terms, then it contains a term for blue.
6. If a language contains seven terms, then it contains a term for brown.
7. If a language contains eight or more terms, then it contains a term for purple, pink, orange,
grey, or some combination of these.
Berlin and Kay also found that the number of basic color terms tends to increase with the
complexity of the civilization. They speculated that this explains the relative poverty of color
terminology among the ancients--e.g., the Greeks had terms only for black, white, yellow, and red
because theirs was a relatively uncomplicated culture, at least from a technological standpoint. But
Berlin and Kay admit they don't know why the "rules" should operate as they do. For more detail,
check out their book.
--CECIL ADAMS
490
AP Physics - Light Wrap Up
Here beith the equations for the light/optics deal. There are several of them, but not nearly enough.
v f
Here we have the equation for the velocity of a wave as a function of frequency and
wavelength.
n
c
v
This is the equation for the index of refraction.
n1 sin 1 n2 sin 2
This is Snell’s law. Use the thing to find the angle of incidence or refraction.
sin C
n1
n2
This is the equation for the critical angle – you know, the whole total internal
reflection deal.
1 1 1
si so f
This is the equation for a thin lens. It relates image distance ( si ), object distance
( so ) and focal length.
M
hi
s
i
ho
so
Magnification anyone? This calculates the magnification provided by a lens as a
function of image height, object height, image distance, and object distance.
f
R
2
This equation calculates the focal length of a converging (concave) spherical mirror
as a function of the radius of curvature.
d sin m
491
This equation is used for interference problems with double slits, single slits, and
diffraction gratings. It equates the path difference ( d sin ) to the wavelength of the
light. The m term has to do with order number of the fringe. m is 0, 1, 2, 3, etc.
xm
m L
d
This equation is used to find the spacing between fringes for the double slit, single
slit, and diffraction grating deal.
Here’s the stuff we’re supposed to be able to do.
B. Physical Optics
1. You should understand the interference and diffraction of waves so you can:
a. Apply the principles of interference to coherent sources oscillating in phase in order to:
(1) Describe 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.
Can you do this? Do you know why waves interfere constructively and destructively? The
Physics Kahuna hopes you do. Basically its just the old law of superposition. The waves
add algebraically. d sin m can be used to determine the spacings and etc.
(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.
Basically two sources produce waves that spread out in circles (if we’re talking about two
dimensions) or spheres (for three dimensions). Where the crests meet we get a maxima –
the waves are in phase and add up. Where a crest meets a trough we get destructive
interference and a minima – the out of phase waves cancel each other out.
(3) Relate amplitude and intensity produced by two or more sources that interfere
constructively to the amplitude and intensity produced by a single source.
For a single source, the incident waves arrive one after the other. Think of a crest
followed by a trough. For waves from two or more sources there are bunches of waves
that arrive either in phase, out of phase, or somewhere in between. If they arrive in
phase we get constructive interference and a large amplitude is produced (much larger
than what we saw for a single source). If they arrive out of phase, we get destructive
interference. So for multiple sources we get minima and maxima.
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:
492
(1) Sketch or identify the intensity patterns that result when monochromatic waves pass
through a single slit and fall on a distant screen, and describe how this pattern will
change if slit width or the wavelength of the waves is changed.
Not difficult. Consult the handout to see what a single slit diffraction pattern ought to
look like. As for changing the slit width, look at the equation.
xm
m L
d in the
d
equation is the slit width. As it gets larger the distance between the dark fringes
decreases, as the slit width gets smaller, the dark fringe spacing would increase.
(2) Calculate for a single-slit pattern, the angles or the positions on a distant screen
where the intensity is zero.
Easy as pie. Simply use the equation:
xm
m L
. This equation calculates the
d
various positions 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 singleslit diffraction and which from two-slit interference.
Consult the handout to learn how to do this.
(4) Calculate, for a two-slit interference pattern, the angles or the positions on a distant
screen at which intensity maxima or minima occur.
Two equations help you do this: d sin m and xm
m L
. Simply apply the
d
equations.
(5) Describe or identify the interference pattern formed by a grating of many equally
spaced narrow slits, 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.
This is the grating deal where you use d sin m and xm
m L
. Once again,
d
simply apply the equations.
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.
Okay, the phase reversal occurs when light is reflected by a medium that has a higher index
of refraction than the medium the light was going though originally. When the new medium
has a smaller index of refraction, no phase change takes place.
493
(1) Determine whether rays of monochromatic light reflected 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 the
reflection of visible light.
Just apply the rules for light reflection above with the explanation in the handout on
Newton’s rings. For the coating thing, you need nonreflective coatings which means you
want destructive interference. This means that the minimum thickness for the film is
2t
2
. The index of refraction for the film must be such that we get a phase change on the
top and at the film/glass interface.
2. You should understand dispersion and the electromagnetic spectrum so you can:
a. Relate a variation of index of refraction with frequency to a variation in refraction.
Dispersion is the deal where short wavelength light has a bigger index of refraction than does
light with a longer wavelength in a transparent media. So blue light is bent more than red
light. Examples are the prism and rainbows.
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.
This is a simple bit of memorization for a bright AP Physics student such as yourself.
3. You should understand the transverse nature of light waves so they can explain qualitatively
why light can exhibit polarization.
See the notes. The Physics Kahuna did a fine job of explaining the whole thing in the
handout. Please consult it.
C. Geometrical Optics
1. You should understand the principles of reflection and refraction so you can:
a. Determine how the speed and wavelength of light change when light passes from one
medium to another.
This is the old index of refraction deal.
b. Show on a diagram the directions of reflected and refracted rays.
The Physics Kahuna showed you how to make these drawings. Consult your notes or the
handout.
c. Use Snell’s laws to relate the directions of the incident ray and the reflected ray, and the
indices of refraction of the media.
494
This basically is using Snell’s law to find the various angles of incidence and refraction, et al.
d. Identify conditions under which total internal reflection will occur.
The conditions is when you get the critical angle.
2. You should understand image formation by plane or spherical mirrors so you can:
a. Relate the focal point of a spherical mirror to its center of curvature.
Focal length is half the radius of the mirror. The center of curvature is positioned a distance
of one radius from the center of the mirror.
b. Given a diagram of a mirror with the focal point shown, locate by ray tracing the image
of a real object and determine whether the image is real or virtual, upright or inverted,
enlarged or reduced in size.
We did a bunch of these using the various rules for the different rays reflecting off the
mirror.
3. You should understand image formation by converging or diverging lenses so you can:
a. Determine whether the focal length of a lens is increased or decreased as a result of a
change in the curvature of its surface or in the index of refraction of the material of
which the lens is made or the medium in which it is immersed.
If the curvature changes, the focal length changes. In general, the greater the curvature the
shorter the focal length. As the index of refraction is changed; the greater the index of
refraction the smaller will be the focal length (bigger angle of refraction so smaller focal
length, right?)
b. Determine by ray tracing the location of real object located inside or outside of the focal
point of the lens, and state whether the resulting image is upright or inverted, real or
virtual.
We did a bunch of these using the various rules for the different rays being refracted by the
lens.
c. Use the thin lens equation to relate the object distance, image distance, and focal length
for a lens, and determine the image size in terms of the object size.
Just use the equation. We did several problems.
d. Analyze simple situations in which the image formed by one lens serves as the object for
another lens.
The idea here is that the first lens forms an image which is then the object for the next lens,
which can then form another, final image. We did some of these problems and your handout
shows you an example.
495
From 2001:
In an experiment a beam of red light of wavelength
675 nm in air passes from glass into air, as shown
above. The incident and refracted angles are 1
and 2, respectively. In the experiment, angle 2
is measured for various angles of incidence 1,
and the sines of the angles are used to obtain the
line shown in the following graph.
(a) Assuming an index of refraction of 1.00 for air, use the graph to determine a value for the index
of refraction of the glass for the red light. Explain how you obtained this value.
n1 sin 1 n2 sin 2
sin 2
n1
sin 1 The slope is n1/n2 since n2 is 1, the index of
n2
refraction for the glass, n1 is the slope.
slope
sin 2
sin 1
0.8 0
0.5 0
1.6
(b) For this red light, determine the following.
i. The frequency in air
c f
f
c
3 x 108
m
1
s 675 x 109 m
4.44 x 1014 Hz
ii. The speed in glass
496
n
c
v
v
c
n
3 x 108
m 1
s 1.6
1.88 x 108
m
s
iii. The wavelength in glass
v f
v
f
m
1
1.88 x 108
s 4.44 x 1014 1
s
423 nm
(c) The index of refraction of this glass is 1.66 for violet light, which has wavelength 425 nm in air.
i. Given the same incident angle 1, show on the ray diagram on the previous page how the
refracted ray for the violet light would vary from the refracted ray already drawn for the red
light.
ii. Sketch the graph of sin 2 versus sin 1 for the violet light on the figure on the previous page that
shows the same graph already drawn for the red light.
Pick a point on the graph, say where sin 1 = 0.5, we can use Snell’s law to find the value
for sin 2 :
n1 sin 1 n2 sin 2
sin 2
n1
sin 1
n2
1.66 0.5
1
0.83
We plot the point (0.5, 0.83) and connect it to the origin. This gives us the new curve. Note
that the slope is slightly different..
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(d) Determine the critical angle of incidence C, for the violet light in the glass in order for total
internal reflection to occur.
sin C
n2
n1
1
1.66
c 37.0o
From 1998:
A transmission diffraction grating with 600 lines/mm is used to study the line spectrum of the
light produced by a hydrogen discharge tube with the setup shown below. The grating is 1.0 m
from the source (a hole at the center of the meter stick). An observer sees the first-order red line
at a distance yr = 428 mm from the hole.
a.
600
Calculate the wavelength of the red line in the hydrogen spectrum.
lines
mm
Number of mm/line
1
line
600
mm
mm
0.001667
line
mm 103 m
6
1.667 x 10
1.667 x 10 m
line 1 mm
3
428 103 m
o
23.17
1.0 m
d sin m
tan 1
1.667 x 10 m sin 23.17
6
1
o
0.656 x 106 m 656 x 109 m
656 nm
498
From 1997:
An object is placed 30 mm in front of a lens. An image of the object is located 90 mm behind
the lens.
a. Is the lens converging or diverging? Explain your reasoning.
Converging. Only converging lenses can project real images behind the lens.
Diverging lens always produce a virtual image on the same side as the object.
b. What is the focal length of the lens?
1 1 1
si so f
c.
1
1
1
90 mm 30 mm f
f
22.5 mm
On the axis below, draw the lens at position x = 0. Draw at least two rays and locate the
image to show the situation described below.
f
d.
e.
Based on your diagram in (c), describe the image by answering the following questions in
the blank spaces provided.
Is the image real or virtual? Real
i.
Is the image smaller than, larger than, or same size as the object? Larger
ii.
Is the image inverted or upright compared to the object? Inverted
iii.
The lens is replaced by a concave mirror of focal length 20 mm. On the axis below, draw
the mirror at position x = 0 so that a real image is formed. Draw at least two rays and
locate the image to show this situation
499
From 1996:
Coherent monochromatic light of wavelength in air is incident on two narrow slits, the centers
of which are 2.0 mm apart, as shown. The interference pattern observed on a screen 5.0 m away
is represented in the figure by the graph of light intensity I as a function of position x on the
screen.
a. What property of light does this interference experiment demonstrate?
Diffraction, Wave Property of Light, or Constructive & Destructive interference
b. At point P in the diagram, there is a minimum in the interference pattern. Determine the
path difference between the light arriving at this point from the two slits.
m = 1 (first bright fringe is at m = 0) so the dark fringe must happen at m + ½, the
path difference is d sin . So we can use the equation given for the thing.
d sin m
c.
1
1
2
3
2
Determine the wavelength, , of the light.
xm
m L
1 L
m
d
2 d
2 x 10 m 1.8 x 10
3
1 5.0 m
d.
1
d sin m
2
3
3 L
x
2 d
dx 2
mL 3
m 2
6
7
0.48 x 10 m 4.8 x 10 m
3
480 nm
Briefly and qualitatively describe how the interference pattern would change under each
of the following separate modifications and explain your reasoning.
The experiment is performed in water, which has an index of refraction greater
i.
than 1.
xm
m L
For m = 0 Wavelength is less under water. So xm is less, therefore
d
the interference pattern will be compressed toward the center.
500
One of the slits is covered.
New pattern is a single slit pattern. The pattern will spread with a larger central
maximum.
The slits are moved farther apart.
iii.
ii.
xm
mL
d
as d increases x decreases. Interference pattern will be compressed
toward the center.
501
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