Download Reflection and Refraction Video Script

Physics 1402 – Reflection and Refraction
Instructor
At the end of our last program, I showed you this little pink pig, and you saw that my fingers go right
through it. So it must be an image. Well, here’s the real pig. It sits on the bottom of a concave dish
that is mirrored on the inside. And the top is also a concave mirror. So we get reflection off both
mirrors. Put your pencils down and watch this.
(diagram on screen)
VO
Let’s find the image of the pig’s nose. An incident ray goes from the object, hits the top mirror, and is
reflected like this. When the reflected ray hits the bottom mirror, it reflects again, like this.
Now we need one more ray from the pig’s nose to the top mirror, then to the bottom, and back up
here. Where the two reflected rays meet, we see the image of the pig’s nose, right here, where the
hole is in the top mirror. It is a real image. We also get real images of the pig’s tail and everything in
between.
Instructor
If you’ll notice, the mirrors in the toy and satellite dishes are shaped more like dishes than parts of a
large sphere. When spherical mirrors are very large, the image they form is fuzzy rather than sharp.
Watch this to see what the problem is and how it is corrected.
(diagram on screen)
VO
The rule about incident rays parallel to the principal axis reflecting through F only works for small
mirrors like the one we used in the lab, or for rays close to the principal axis in large mirrors.
As you can see here, parallel rays far from P reflect at different points on the principal axis. This
produces a blurred image. This effect is called spherical aberration. One way to solve the problem
would be to use smaller mirrors or shield the outer parts of larger mirrors. But mirrors used in
astronomical telescopes and satellite dishes need to collect a large amount of light. So the solution is
to reshape the mirror into a parabolic mirror, such as this one.
Instructor
Because of a mistake in grinding, the Hubble telescope had a problem with spherical aberration,
focusing only 15 percent of the light at F instead of the 70 percent it was designed to focus. It was
found that the giant mirror was too flat on one edge by 1 / 50th of the width of a single human hair. In
December, 1993, the Space Shuttle Endeavor captured and modified the Hubble Space Telescope by
adding a camera to correct the problem.
OK. It’s time to turn things around and look into convex mirrors. Let’s look at some common uses
for these mirrors first. Then we’ll go straight to the lab to investigate how they form images.
(convex mirror on screen)
VO
The second type of curved mirror is a convex mirror. Convex mirrors curve outward and are often
called wide-angle mirrors. They are a common sight on side mirrors of large trucks and in department
and convenience stores. Convex mirrors provide drivers and clerks with a very wide field of vision.
(students on screen)
VO
In this lab, the concave mirror on the meter stick has been replaced by a convex mirror. Our students
place a lighted candle in front of the mirror. Then they try to find a real image by moving the screen
back and forth in front of the mirror, but no image forms on the screen.
When they move the candle farther away from the mirror, they still can’t find an image with the
screen. But when he looks into the mirror, our student sees an image.
Tell your teacher the type of image a convex mirror must form.
When our student looks into the mirror, here’s what he sees. All images formed by convex mirrors
appear to be behind the mirror, so they are virtual. The images are reduced and erect.
Convex mirrors are called diverging mirrors because the reflected rays seem to be coming from a
point behind the mirror.
Instructor
Earlier, you saw some uses for a convex mirror. Because of its shape and the erect image it forms,
this type of mirror can help you see around corners. So you see them in stores, libraries, and at drivethrough windows.
Now here’s a physics challenge for you. Convex mirrors are often used as side-view mirrors on cars
to eliminate the blind spot at the back corner of the car.
But why is the warning that “Objects in mirror are closer than they appear.” printed on the mirror?
Tell your teacher.
Did you remember that the image is reduced? And when we see a small image of a car, it’s easy to
make the conclusion that the car is far away. Of course, that could be dangerous.
Now it’s time for you to learn to draw ray diagrams to locate convex mirror images. Grab a ruler and
a sharp pencil, and we’ll get started.
(diagram on screen)
VO
The parts of a convex mirror are the same as those for a concave mirror, except for the fact that the
center of curvature and focal point are behind the surface of the mirror. The three special rays we use
to draw diagrams are also the same.
Let’s start with the one we always draw: In from the object to the mirror parallel to the principal axis.
When the ray of light hits the mirror, it reflects as if it were coming from F, which is here.
So turn the ruler around and line it up with F and the mirror at the point of incidence. And draw the
reflected ray.
Now let’s skip number two because it’s difficult to draw. We’ll go to ray three since it is so easy.
And, by the way, it always works. Line up your ruler with the object and C. The ray hits the mirror
and reflects straight back along the same line.
Look at the two reflected rays. They diverge from the mirror and will never meet in front of the
mirror. So our eyes and minds extend the lines behind the mirror. They meet here. When we
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complete the image, you can see that it is virtual, erect, and reduced, just like we saw in the lab.
You’ll get a chance to practice these diagrams later.
Instructor
You know, during the last few programs, I’ve had a feeling that something’s been missing.
But I can’t quite put my fingers on it. Oh yeah. It’s math. Well, we’re going to solve that problem
today. Let’s say you’re told that an object is placed in front of a concave mirror, between C and F.
And you’re asked to locate and describe the image. You can do one of three things. You can try to
remember what you saw in the lab, or you could draw a ray diagram…. or you could use some mirror
math.
It’s really simple. Just plug some numbers into two equations and chug out the answer.
(green chalkboard on screen)
VO
The first equation helps you locate the image. In this equation, “d sub i” is the distance the image is
from the mirror. And “d sub o” is the distance the object is from the mirror. You already know that
“f” is the focal length of the mirror. You will always be given this equation, but you must know what
each symbol represents. And you must know whether each value is positive or negative.
A positive value of “d sub i” or “f” means in front of the mirror. And a negative value means behind
the mirror. We’ll show you how to use these facts in a minute.
The second mirror equation is the magnification equation. Capitol “M” represents magnification,
which is the ratio of the height of the image, “h sub i” to the height of the object, “h sub o.” And this
ratio equals negative “d sub i” over “d sub o.”..
After you plug in the numbers and chug out the answer, a positive value for “h sub i” will mean that
the image is erect. And you know that erect images are virtual. A negative value will mean that the
image is inverted, so the image must be real.
VO
Let’s try an example problem together. The object is placed 60 centimeters in front of a concave
mirror with a focal length of 20 centimeters. First, we want to calculate the distance of the image
from the mirror, “d sub i.”
We’ll use this equation and rearrange to solve for “d sub i.” Now we plug in the numbers and
calculate. Use the reciprocal “one over x” button on your calculator, and don’t forget to punch
reciprocal button one more time after you subtract. The answer is 30 centimeters, which is between C
and F. In fact, if you look at your lab results from the last program, you’ll see that when the candle
was placed 60 centimeters from the mirror, the image formed 30 centimeters from the mirror.
Now we want to find the height of the image, so we’ll use this equation and rearrange to solve for “h
sub o.” When we plug and chug, we get a height of negative two centimeters, which is inverted, real,
and reduced, just like you found in the lab.
And a ray diagram shows the same thing.
Now it’s time for you to try some problems. Your teacher will give you these problems and time to
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work the problems. Come back when you’ve finished, and have checked your answers with your lab
results from the last program.
Local Teachers, turn off the tape and give students problem set number one from the facilitator's
guide.
(Pause Tape Now graphic)
(problem on screen)
VO
In “a,” the object is at C, and “d sub i” is 40 centimeters, which is the same distance from the mirror
as the object. The height of the image is negative four centimeters, which means that the image is
real, inverted, and the same size as the object. Your lab results and this ray diagram agree.
In “b” the object is between C and F, and “d sub i” is 60 centimeters, which beyond C. The height of
the image is negative eight centimeters, which means that the image is real, inverted, and enlarged.
Again, your lab results and this ray diagram agree.
In “c,” the object is placed at F. You should have gotten an error message on your calculator, which
means that there is no image. The ray diagram shows the same thing.
In “d” the object is between F and the mirror, and “d sub i” is negative 20 centimeters, which behind
the mirror. The height of the image is positive eight centimeters, which means that the image is erect,
virtual, and enlarged. Again, your lab results and this ray diagram agree.
Now, let’s try a problem using a convex mirror. This time the object is 60 centimeters in front of a
convex mirror. The only difference between this one and a concave mirror is that the focal point is
behind the mirror, so “f” is negative 20 centimeters. When you plug into the mirror equation, you get
a negative value for
“d sub i”, which equals negative 15 centimeters. This means that the image is behind the mirror.
Now, to calculate the height of the image, we plug the negative 15 centimeters into this equation.
And the double negative gives us a height of positive 1.0 centimeters. This tells us that the image is
erect, so it is virtual. And the image is reduced, which we know happens with convex mirrors.
Instructor
Well, that completes our study of mirrors, but this program is not over. Your teacher may want to
stop the tape now and let you practice what you’ve learned about mirrors, ray diagrams, and mirror
math. And then you’ll come back for Part Two.
(Pause Tape Now graphic)
VO
In Part Two of this program you will define optical density and index of refraction and relate this to
the reflection and refraction of light. You will use the concept of refraction of light to explain how
mirages and other refracted images are formed. Finally, you will describe total internal reflection and
use this concept to explain how optical fibers work.
Instructor
In the first program of this unit, I showed you the disappearing beaker trick. And you got clue
number one. I hope you have your clue sheet handy because here comes number two.
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(clue #2 on screen)
VO
When we say that one medium is different from another, as far as waves are concerned, what do we
mean? Which wave property does a medium affect? Go back to your notes on waves if you need
help.
Instructor
In this part of our program, we will be talking about what happens when light waves go from one
medium to another and either speed up or slow down. We say that a medium is more optically dense
if light slows down in it. And the best measure of a medium’s optical density is called the index of
refraction. Watch this, and then we’ll give you some notes on index of refraction.
(trees on screen)
VO
When light waves pass from one medium into another, their velocity changes. This change in
velocity causes the rays to be bent or refracted. The greater the change in velocity, the more the light
is bent. The degree to which a substance slows down light is called the index of refraction of the
substance. The index of refraction of space is 1 and of water is 1.33. This means that light travels
through water about 33% slower than through space.
(green chalkboard on screen)
VO
The index of refraction of a medium is a measure of its optical density, its ability to slow down light.
The index of refraction, represented by the symbol, “n” is the ratio of the speed of light in a vacuum
to the speed of light in the medium.
The greater the index of refraction, the more slowly light moves.
(clue #3 on screen)
VO
Here’s clue number three, with some common media and their “n” values.
Instructor
Here’s your next clue. I’ve disappeared from the neck down. Now what could my body be made of
so that light would go straight through me instead of reflecting back to your eyes?
And let me ask you this. What hot liquid could I pour into both beakers to make the inner beaker
disappear? If I had any arms to pour with.
Now, what does all this have to do with corn oil? You tell me. I think it’s time for you to put all the
clues together and explain why the little beaker disappears. Write a paragraph, using correct physics
terms. After your teacher has collected your papers, come back to solve the mystery of the
disappearing beaker.
(Pause Tape Now graphic)
Instructor
Were you able to put all the clues together? Let me take you through it. We see an object because
light reflects off it and reaches our eyes. And the more different the new medium is from the old, the
more light reflects and the less is transmitted.
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The difference we’re talking about is how the medium affects the speed of light. The index of
refraction tells us this.
Now when light goes from air to glass to air to glass, and so on, the difference is great enough that
some of the light is reflected back to our eyes. And we see the boundaries.
But if we poured hot melted glass into both beakers, the light would go from glass to glass to glass,
and no boundaries could be seen.
The same thing would happen if my head were made of a substance with the same index of refraction
as air. I guess that would make me an air-head, huh.
But what about corn oil? It just happens that the index of refraction of corn oil is the same as that of
the glass used to make beakers. So when light goes from glass to corn oil and back to glass, there is
no optical boundary to make it reflect. So the light goes straight through. Did you get it?
For the rest of this unit, we’ll be studying what happens when light enters a different medium and
bends or refracts. Refracted images may not be as familiar to you as reflected ones, so let’s start with
something you can do in class or at home. Our students will show you what to do.
(students on screen)
VO
Place a penny in a dish, with the penny on the opposite side, farthest from your eyes. Lower your
head until the penny is just out of sight.
Hold your head still while your partner slowly adds water to the dish. What happens?
In case you weren’t able to try this, we’ll let the camera act as your eyes. You can see that as water is
added, the penny became visible.
Next, use a straw or pencil to act as a spear. With your head in the position as before, prop the spear
on the side of the dish and line it up with the penny. Quickly spear the penny.
You’ll always hit the side of the dish above the penny and never the penny itself. Let’s see why.
(diagram on screen)
VO
Light from the room reflects off the penny in all directions. To make things simple, we’ll concentrate
on a point at the center. When you lower your head, the rays that would have hit your eyes are
blocked by the dish, so you can’t see the penny.
Here’s what happens when water is added. This ray would normally go right over your head. But
when it goes from water to air, it’s going from slow to fast, which makes it bend or refract away from
the normal, like this. This refracted ray hits your eye.
Now, your eyes and minds don’t think of light bending. So you think of a straight ray of light coming
toward your eyes. When we extend the ray straight back, we see a sharp image of the penny above
where it actually is.
When you followed this ray to try to spear the penny, you hit the spot where the refracted image was,
above the actual penny.
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(Fact or Fiction on screen)
VO
Is this statement fact or fiction?
We can see the sun for several minutes each day before it rises and after it sets.
It’s a fact. Let’s see how this can be.
(diagram of earth on screen)
VO
In this picture, the earth has rotated so that straight rays from the sun will not reach this point on the
earth. We say that the sun has dropped below the horizon. However, when the light goes through the
atmosphere it is going from fast to slow, so it refracts downward. Since our minds don’t think of light
as bending, we extend the rays coming down toward us, and we see the refracted image of the sun,
which is higher in the sky. This happens before sunrise and after sunset, extending the time we see
the sun by about four minutes each day. And because light rays from the bottom of the sun bend more
than those from the top, you’ll often see a distorted image of the sun when it is close to the horizon.
Instructor
Let’s see. There is no end to the rainbow, a beautiful sunset might mean that the sky is polluted, and
the sun we see at sunrise isn’t even the real sun, but a refracted image. It’s enough to make a guy get
depressed.
Well, while we’re thinking about seeing things that aren’t really there, we might as well talk about
mirages. Watch this.
(truck on screen)
VO
Mirages are also the result of refraction. A layer of very warm air often exists just above hot sand or
pavement. Hot air has a lower index of refraction than cool air. Light from the sky is refracted as it
passes from cool to hot air. As a result, a refracted image of the sky that resembles water is seen on
the sand or pavement.
(diagram of tree on screen)
VO
Mirages are caused by a drastic change in air temperature between the object and the ground. In the
summer, the air above sand or pavement gets very hot. And light travels faster in this hot air.
To explain this mirage, let’s concentrate on this point on the palm tree. Light rays leave it, going in
all directions, and you will see the tree because of the ones that go straight to your eyes.
Normally, this light ray would go right to the ground, and you wouldn’t see it. But on a hot summer
day, this ray hits a layer of hot air and bends upward, away from the normal. Your mind extends the
rays in a straight line, and you see a refracted image of the bottom of the tree here.
The overall result would be an inverted image of the tree. In addition, you would also see a refracted
image of the sky on the ground. These refracted images can be photographed, so you’re not just
imagining them.
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Instructor
Now here’s where your imagination comes in. We interpret an image of the sky on the ground as
water.
And if you were stranded in the desert and saw a palm tree plus an inverted image in what looks like
water, you would jump to the conclusion that you had found an oasis, with a tree growing beside a
pool of water.
As you get closer, the image disappears, and reality sets in again. Well, I don’t know about you, but
I’ve enjoyed about as much of this reality as I can stand for one day. I’m going to take a break while
you SHOW WHAT YOU KNOW!!
Jot down your choice for each question. Your local teacher will go over the correct answers with you.
(Read Show What You Know questions on screen)
Instructor
I … I can’t believe its almost over between us. We only have one more program to go, and then our
time together will be over … forever. I heard that giggle in the back. So just go and I’ll muddle
through my depressed state alone. I need a tissue.
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