I'm going to ask you a few questions which have to do with
a rainbow. All of us have looked at
rainbows. But the question always comes
up, have you really ever seen it?
I can ask you the same question about art.
All of us have looked at art, which is very different from
seeing art. And so, here I have 15
questions. And let's look at them.
And ask yourself, how many of them can I answer?
We've all looked at rainbows. So, the first question is,
what is the color sequence? Is the red on the outside or on
the inside? And, what is the radius of the
bow? And what I mean by that is the
following. Say if here's the horizon and
you see a rainbow. Then there is somewhere a point
that you can call the center of the circle.
And then you can always measure the angle and terms of degrees,
not in inches but in terms of degrees.
What is that [radius?] typically?
And then, we can ask the same question about the length of the
bow, the width of the bow. There are colors here.
How wide is that in terms of angle, roughly?
I don't expect you to be precise, but roughly.
Have you ever noticed that there is an enormous difference
and the light intensity outside the bow and inside the bow?
And if so, where is it bright and where is it dark?
Have you ever noticed that? It will be staring you in the
face when you look at the rainbow.
What time of day do you expect the rainbow?
Do you expect to see the north, east, south,
or west? Are there more than one bow?
And, yes, I already answered that for you.
There really is a secondary goal.
And then, the question is, what is the color sequence of
that second bow? And where do you have to look
for it? What is its radius,
and what is its width? And then comes the question
that all of you can answer because of my problem set number
ten. There was a 10.4,
which addressed the issue, are the bows polarized?
And then, what is the direction of polarization?
And is the polarization week or is the polarization strong?
All right, Professor Tegmark, how many questions can you
answer? That's not bad.
That is not bad. Who can answer all 15?
14? 13?
12? 11?
10? Don't be ashamed.
Nine? Eight?
Now comes an area where you may have to become ashamed.
Seven? Six?
Who can answer six? That's not bad.
Five? Four?
Who can only answer four? Three?
Two? One?
Who can answer zero? I'm proud of you.
All of you can at least answer one.
All right, I'm going to make you see a rainbow the way you
have never seen it before because most of you have never
seen a rainbow. This is the picture that you
also had on your problem set number ten.
Light from the sun in this case comes straight from the left.
We are going to give the sun an angle in the sky very shortly.
We used to call this angle theta one.
I called it I now for today, angle of incidence.
And I call this R. When the light gets into the
water drop, I call that the angle of refraction.
I call that R. And so, with [Snell's?] law,
we would then have that N1 times the sine of I,
which is the angle of incidence with which the light strikes the
water drop, N1 times sine I equals N2 times the sine of R.
And for all practical purposes N1 is one.
That is in air. And N2 is then an average value
for water, N2. Average value is about 1.336
averaged over the colors. I asked you in problem 10.4 to
concentrate on the following trajectory.
Light strikes the water. Some of it is reflected at A.
I'm not interested in that. I want to know what goes into
the water. At B, some of it comes out.
I'm not interested in that. I want to know how much is
reflected. And then, at C,
some of it is reflected into the water drop.
Some of it is reflected into the water drop.
I'm not interested in that. I want to know what's coming
out. And what you see now is what
comes out and makes an angle phi with this horizontal line.
And, this angle, phi, is going to play an
important role in the rainbow. The first thing I asked you in
problem 10.4 oh form of the first things to calculate,
phi, in terms of I and R, well, here is I,
and the same angle I shows up here.
Here is an R. Here is an R.
It's only a little bit of high school algebra to show that phi
is 4R minus 2I. And, you've all done that
because you've already turned in your homework.
So, what is interesting now: if you take I is zero degrees,
which is the light that strikes the rainbow head on,
so to speak, then I is zero,
right, because I is the angle between the incident radiation
and the normal to the surface. So, that is zero.
Then, R is also zero, the angle of refraction,
and consequently phi is zero. In fact, what the latest doing
penetrates the raindrop and it comes straight back.
So this angle, phi, is then zero.
But now comes something that is very nonintuitive.
If you increase I, then we will all accept that R
goes up. And we will also all accept
that phi goes up. What is not so intuitive is
that there comes a certain angle of I for which phi has a maximum
value which is roughly near 60 degrees for I.
And when you hit the water drop higher, that value for phi,
which is the maximum value, will go down.
So there is no longer an angle, phi, larger than a certain
maximum, which I call phi maximum.
And, you can see that very easily.
It is a simple exercise that is no more difficult than applying
Snell's law. You see Snell's law there at
the top, and I remind you that phi is 4R minus 2I.
So, you pick an angle of I. You calculate R with Snell's
law, and you calculate phi. And you see,
indeed, that in the beginning as I goes up,
R goes up and phi goes up. But, when you reach I of about
60°, you see that phi reaches a maximum, which is somewhere near
41.6 degrees. And, look when I is 70° and
80° that phi is distinctly lower.
And this now is going to be the key point in our rainbow.
The first question that I can ask you in 8.03 may be a little
harder to ask high school students.
Can you calculate, really, at what angle I you
will get at maximum value for phi?
In other words, this is sort of trial and
error. Can you actually algebraically
derive that? And that was also part of the
homework assignment. If what you now do is you say,
[defy?] the I equals zero. And then you have a long way to
go. It's actually surprising.
It takes a long way algebraically to finally
conclude that indeed you can calculate that angle of I for
which phi is a maximum. The cosine square of that angle
is N squared minus one divided by three.
In other words, this only gives you the angle
of I for which phi is a maximum. And so, you substitute in here
the value for N. In this case,
you take 1.336. And then you know what I is.
Well, once you know what I is, you can calculate what R is
with Snell's law. And then you know what phi
maximum is because you know that phi is, I wrote it on the
blackboard here, 4R minus 2I.
So, let us write now on the blackboard here what happens for
the different colors. So we have here,
N red is 1.331. Notice that I'm now going to be
rather specific about colors because that's,
of course, behind the rainbow. And violet, which is sort of
the shortest wavelength that we can see for violet light,
N is 1.343 for clean water. Seawater: it is a little
different. So, it notice there is a 1%
difference. That means the speed of light
for violet, blue light is about 1% lower in water than the speed
of light of red light. And so, with this equation,
now, with these different numbers for N,
we can now calculate what I is for which phi is a maximum
value. And, you will find,
now, 58, 59.53. You can now calculate the value
for R, for which phi reaches that maximum.
You do that from using Snell's law.
You get 40.36 degrees. And so, now you have your goal,
which is your phi max value, which is 42.37.
And if you want to, you can round all these off,
of course. And if we do the same for
violet light, then you get 58.83.
And then for R, we get 39.58.
And for phi max, you get 40.65.
So, what does this mean, now?
This means that if sunlight strikes a water drop,
just one water drop, that's all it takes,
that a cone of light will be reflected into the direction of
the sun because all angles for I are present.
The sunlight comes from the left, so all angles for I are
present. And so, let me put here one
raindrop. And let's still assume for now
that the sunlight comes straight from the left.
It means, then, that the largest value for phi
is going to be a cone of light because the whole problem is,
of course, [axial?] symmetric. You're dealing here with
spheres. So, red is really the winner.
Red has the largest value for phi maximum that exists.
And [through?] that, we have made a special triangle
which has here, for me, this angle of 42°.
This was made a long time ago. And so, I can now draw here an
angle of 42°, and 42° in this direction
because the whole thing is axial symmetric.
And, that means a cone of light whereby this angle rounded off
is then 42.4°. That is that maximum value for
phi. There's no value for phi that
is larger. And then, I need my violet
light. Where did I leave it?
Oh, we have one here. And then, the other colors have
a smaller value for phi maximum. This is not to scale.
But this is, then, the violet light.
And so, inside here, you have that cone for the
violet light for which this angle is smaller.
It's about 40.65°. At angles smaller than the
maximum value for violet light, all colors are being reflected
back. There is no restriction.
There is only a restriction on a maximum value for phi,
but not on the minimum value. That means that all other
colors can come back inside this cone.
And therefore, since all colors come back
roughly with the same intensity, it will be white light because
all colors overlap. Your brains will tell you that
it is white light. But, if you only concentrate on
this journey, refraction at A,
reflection at B, and refraction at A,
there is no light that can come out outside this cone.
Keep that in mind. So, suppose I had a screen
here, and I had a small opening through the screen so that the
sunlight can fall through. And I project on that screen,
I intersect this cone with that screen.
What you'd see, then, is the following.
You would see a circle on the outside, which is red.
And then, you would see all the other colors that follow.
And then, the last color is the violet, blue violet.
The inside would be white light.
And outside, there would be no light.
Now, there's a key question that some of you must be burning
to ask. And that is,
how come the violet light in a rainbow is so clear?
The red is clear. It's obvious why the red is
clear, because the red is not polluted by any other colors.
Only the red can come out at an angle of 42.4°.
But the violet comes out at a smaller angle.
At that same smaller angle of 40.65°, to read can also come
out. And the green can also come
out. And the yellow can also come
out. So, why then is it that the
violet is still so prominent? That comes now down to light
intensities. If you calculate for the
various colors, and here you plot phi,
the light intensity, and remember intensity is
always watts per square meter in terms of units,
then you will see that the red light is the winner.
This is this angle, this 42.4° has a maximum here.
It's enhanced. It's not obvious at all.
But, there is an enhancement at that phi maximum.
And then, the green light, which is a little bit inverse,
because phi maximum for green is a little lower,
would also have a maximum. And then, ultimately the violet
light, which is the last one, will also have a maximum.
And so, it is because of the enhancements at that phi
maximum, which is by no means obvious, which doesn't follow
from Snell's law, but it follows from Fresnel's
equations, that is the reason why the blue here,
the violet here, still dominates.
And here, although colors overlap.
And so, that's why you see the white light there.
So, now comes the question, why do we see a rainbow?
Well, I'm going to put you now somewhere at a location.
This is the ground. And here you are.
And the sun is in this direction, the sunlight.
And you know that because you see your shadow.
This is your head. This is your body.
And these are your legs. You're looking straight at the
shadow at your head here 180° away from the sun.
The sun is there and you see your head there.
And let us assume now because that is a condition for a
rainbow that it is raining or that there is water here,
and that somehow, the sunlight is not obscured.
In other words, if it's also raining here,
that's tough luck. Then the sunlight cannot get
through. And so, you won't see a
rainbow. So, it is important that it is
clear in the direction to the sun, and that the sunlight can
strike the raindrops without any interference of clouds.
So, it just so happens that it's raining.
And you look at the direction of the sky.
You are looking up in the sky. And you pick any raindrop that
you want on this line. There are zillions of them.
Let's just pick this one. So, the sunlight strikes that
raindrop like so. And what does this one drop do?
Like all the other zillions of drops, they reflect into the
direction of the sun a cone which is like this.
Each one does that individually.
So, there we go: 42°.
Here is that cone. And I will not draw the violet
one. And you look at that raindrop,
and at all the others in this direction will you see any
light. The answer is no.
If you accept this picture that no light can be further out than
the maximum value of the cone angle.
So, you don't see any light reflected from the sun.
So, you see the dark sky, because when it is raining in
general, the sky is dark. Now, you look in this
direction. You pick any raindrops in this
direction. There are zillions of them.
Let's pick this one. So, here comes the sun.
And this one raindrop, like zillions of others,
is going to reflect into the direction of the sun,
a cone of light [whereby?] this angle, 42°, and this angle is
42°. There it is.
What will you see? You see light.
What color? White.
You are looking straight into the inner cone because look how
small the angle is. You're looking right into this
area somewhere here. And so, you see white light.
So, you look high in the sky. You see the dark sky,
the background. You look lower,
and then you say, [ow?], white light from the
sun. And you will see in pictures,
it's enormous. But now, you look in a very
special direction. You are going to look into this
direction, the direction which is 42° away from this line
So this angle, now, is the famous 42.
If you want to make it 42.4, that's fine,
degrees. Away from this line,
away from your shadow, 42.4°, and here is water,
zillions of water drops along the line.
And here is the sun. And the sunlight comes in like
this. What is this water drop going
to do? It's going to refract back into
the direction of the sun a cone. This happens to be in the
direction of the 42°. And then, of course,
this is the direction of the 42°.
So, here is that cone. What do you see?
What will you see? You will see red light.
You will see the light, this light.
So, you look high in the sky. No light is reflected from the
water. You look at the 42.4° and you
see only red. And then you look way lower and
you see white light. And we now understand why you
are going to see the other colors, because we understand
the enhancement due to Fresnel equations of the green and of
the violet. And so, if here is the [SOUND
OFF/THEN ON], this is away from you now,
since the whole thing is axial symmetric, you can turn this
around that line because a sphere is axial symmetric,
right? All the rain drops have axial
symmetry. So, what you can do in this
direction, 42°, you can also do in this
direction, 42°. The sphere doesn't know the
difference between up and down and left and right.
And so, what you're going to see now is you're going to see
an outside bow, which has a radius of about
42.5 degrees. Here would be your shadow on
the ground. And so, this angle would be,
very roughly, then, the 42° if I round that
off. And then, you would see the
other colors follow. And then ultimately you would
see the violet. And, you know which angle that
would be. That would be at the 46.5
there. I draw the bow specifically a
little lower than your horizon because there is nothing wrong
with the rain falling, of course, nearby.
If the rain is very close, you can actually extend this
even further. So you often see that,
but the rainbow extends below the horizon, of course.
And so, this angle, then, indicates the elevation
of the sun, how high the sun is above the horizon.
And if the sun is very low above the horizon,
this point will be here. And you will see a much larger
bow. In fact, the bow at sunrise and
sunset, this angle is 90°. But, if the sun rises in the
sky, then this point goes down. And with that,
the bow goes down, and therefore,
during midday, you almost never see rainbows.
That's the reason, because the whole bow is then
below the horizon unless you spray water around you.
Nature produces two bows. One is called the primary,
which is that one. And other is called the
secondary. And the secondary bow is the
result of light that makes one extra reflection inside the
water. And so, I have not shown here
every time when the light comes in that some of it gets back
into the air. I've only shown the trajectory
of the key light that I need to make the secondary.
Here, the light strikes high up on the water drop.
Here it strikes low on water. In fact, this angle is around
72°. It refracts into the rain,
reflection number one, reflection number two,
and here it comes out. And this is then that angle,
phi. Compare that with this angle,
phi, right? The radiation comes out like
this, and I define this as the angle with the horizontal.
So, that's really this angle. Sorry that it's a little bit on
the edge of the blackboard. If you do your homework,
the same way that I have done for the primary,
you will see that now there is not a maximum value.
But there is now a minimum value for phi.
So now, we have phi minimum. And the values for phi minimum
for the red light is 50.37°. And for the violet light,
it is 53.47°. So, this means that the
secondary bow is higher in the sky than the primary because,
look, the radii are larger. But it also means that the
colors are reversed. The red is inside the violet
bow. So, if I make an attempt,
this would be the outside of the bow, and this would be the
inside of the bow. And then, very roughly,
the radius, if I round that off a little bit,
the radius is about 50 because this radius is about 52°.
Now, the secondary bow is much fainter.
And that's why many of you would have been staring at
rainbows were so impressed by the primary, which is so
dominant that you don't pay attention to the secondary.
It's much feinter (sic). The reason why it's feinter is
that you people know because you worked with Fresnel equations.
You have an extra reflection inside the water drop.
That means, of course, the light intensity goes down
dramatically. The secondary is substantially
feinter. So now, I'd like to look at
some slides to see what we have derived here,
whether we can see that also in our slides.
So, we'll turn this off, make sure that I have light
enough. Yes, it's OK to start with the
first slide. Yeah, if you turn the light
off, thank you very much. So, Marcos?
Oh, I have to do that myself, right?
I've been promoted to advancer of the slides.
So here you see a situation that, oh, I wanted to,
yeah, OK, that's fine. This, the maestro himself,
his name was Isaac Newton. He understood the rainbow quite
well. In his book optics,
you see this picture. And you see here the viewer,
which is you. This is, then,
that direction away from the sun.
So this is where your shadow would fall.
And here you see the inner [ball?], which is the primary.
[SOUND OFF/THEN ON] the lights from the sun has one reflection
in the back. And then here,
you see the secondary bow by the light hits the water drop
below, and then comes in this way.
So, the secondary bow is outside the primary bow.
And then, here is a sketch that I made that indicates that even
if the sun is very high in the sky, if you want an ego trip,
and if you spray your lawn, then you can have a rainbow
encircling your legs, which gives you a feeling of
great power. Believe me, I've done it many
times. This is a painting from the
eighth century from Turkey. And I've always been intrigued
by this painting. It undoubtedly has a connection
with the Bible because you see a hand here.
And I think in the Bible it says I do set my bow in the
clouds. Now, you wonder,
the color sequence is not correct.
But then again, there is only blue and there is
only red. So, I think in this case,
we have a few options. One option is that the artist
really didn't look carefully at a rainbow.
That's a possibility. It's another possibility is
that physics was different in the eighth century.
I think we can dismiss that possibility.
And then the third one, which I think is the most
likely, that it was a conscious decision on the part of the
artist. He purposely or she purposely
reversed the colors. It was his or her opinion,
impression of a rainbow. And that's what art is all
about. And then, two colors is
perfectly fine to get the idea across.
Anything in this country is for sale, so, also rainbows.
And they are dirt cheap. For [SOUND OFF/THEN ON] bucks
you can get a rainbow. But as always,
you get what you pay for. This is a fake.
Who wants a fake? Because, look,
the colors are the wrong sequence.
So let's not even look at this. Rainbow maker,
yeah, sure, ha. When I started lecturing 8.03,
that's a long time ago. It was in 1973.
And, I wanted so badly to pictures of rainbows that I have
made myself using a water hose and spraying water around.
And I managed to do that. In fact, my daughter did.
And so, water is coming from the left.
And what you see here is clearly the primary,
the red on the outside. You see the blue violet on the
inside. You see all this white light?
You understand now, why there is this white light.
But, look outside the bow. You don't see the white light
reflected. So, that's why you see the sky
so dark. My poor daughter,
Emma, she suffered so much with a father who is a physicist.
[LAUGHTER] It was January because my
lecture started in February. And, it really was freezing
cold. And she was crying.
But look, you have to sacrifice sometimes for the sake of
science. Poor Emma.
I also wanted to make an attempt to get a picture of a
secondary. That is very difficult because
the secondary is so feint. So, I did that over my
driveway. I lived in Winchester at the
time. And so, you clearly see the
primary. I don't have to point out where
the primary is. And this is the secondary.
It is further out. And you see that the colors are
reversed. And it is much feinter.
A friend of mine, Michael Sorgey [SP?],
sent me this slide, very nice.
It was a waterfall somewhere in Austria.
You see very dramatic the white lights from inside the bow that
reflected light from inside the bow.
And then, it terminates right here at the red,
and of course there is water here as well.
But, it doesn't come back in your direction.
And that's why you see the dark forest behind it.
And now comes here a picture that was given to me by Doc
Johnson. This picture was taken in
Sapporo where there is the famous VLA, the Very Large
Array, radio telescope. And now you see truly all the
features that you want to see. Your primary bow follows
sequence as predicted. And notice how much brighter
this guy is inside the bow than it is outside the bow.
And then you see the secondary. You see that the colors are
reversed. Now, keep in mind,
the secondary only has a phi minimum.
So, that means light can go out at a larger angle because phi
minimum means only [NOISE OBSCURES] brighter again than
here because white light can make it from two reflections.
So, you see here quite dramatically why it is so white
here, why it is so dark here, but why some of the light
returns because of the fact that the secondary has the phi
minimum. All this is the result of what
we call geometric optics: Snell's law and Fresnel,
nothing else. However, this is all very nice
and dandy as long as the water drops have a diameter of a few
millimeters. When the water drops become
smaller, and this effect begins to be noticeable around 1 or 2
mm, then refraction starts to play a role.
And reflection has the ability for constructive and destructive
interference as we have seen in 8.03.
And without going through the mouth, which I cannot do this
time, there is some other lectures that I do go through
the math. I'm just telling you the result
that you can see now sometimes dark bands inside the bow.
It's always on the inside of the primary, and then of course,
other areas are a little brighter.
The dark bands are then destructive interference.
And when you look carefully at this one, these are called by
the way super [numero rebows?]. And, when you look at this one,
you can actually see such a dark band here with a little bit
of imagination. Destruction becomes very
important if the size of the water drops gets down to 100 °,
1/10 of a millimeter, and 50 °.
And when it is even smaller than 50 °, diffraction ruins
all the colors. All the colors begin to
overlap, and the bow also becomes much lighter.
And then when the colors go away, you see white lights.
And so, you see a white rainbow.
It's also called a fog bow. I have never seen a white
rainbow. But I was so fortunate that
when I lectured in 1973 for the first time in 8.03,
that there was a student in my audience to that summer after
the lecture went on an expedition to the North Pole.
And he was at [Fletcher?] Island, and he sent me a picture
of a white rainbow which you will see very shortly that he
took 340 miles from the North Pole at Fletcher Island.
He took it at 2 a.m. in the morning at midnight in
July when the sun is above the horizon.
And there is a striking example of a white rainbow.
And whenever you see a white rainbow, it must still be water.
It's not ice. Ice doesn't give you rainbows.
You need spherical symmetry. You need spherical objects to
get rainbows. You always see super numero
rebows because of the fact that it is diffraction that is doing
it. And so, you see here this dark
band here. And then it starts white again.
So, this is not so uncommon that in white rainbows you see
super numero rebows. I've never seen this one.
And, there is another thing, another kind of rainbow that I
have never seen that I would like to see.
And that is a rainbow that you can see very shortly,
very close to sunset, very shortly before sunset,
or very strongly after sunrise. We all know because we have
covered [rainy?] scattering in 8.03.
Why, then, the sun only lets red lights penetrate to you?
The whole sky becomes red. The clouds become red.
There is only red light. There's no white light anymore.
There's no blue light. There's no green light anymore.
So, what is nature going to do when it produces a rainbow?
Everything turns red, and that is what you see them
here. This is a red rainbow.
And I wish I would ever see that.
Notice that all that white light from inside the boat is
red now. But also notice that the bow is
still there. That clearly is an enhancement
of red light at that angle of 42.4°.
That's because of Mr. Fresnel.
That enhancement is still there at phi maximum.
That is not undone by the red light.
And so, it's not surprising, then, that the bow itself still
stands out very clearly, and then that the inside of the
bow becomes now also red. You have now reached the point
that you should be able to answer all the questions.
And to prepare you for that, I think this is a good moment
to have our four minute break, our traditional break.
So, we will reconvene in four minutes.
And then I expect all of you to be able to answer these
questions. So, we will start again in four
minutes. [SOUND OFF/THEN ON] All right,
so here is your chance. Most of them,
of course, are trivial for you now.
Clearly red is on the outside. The radius is about 42°.
You work through that quantitatively,
and the length of the bow, yeah, that depends on where
it's raining. I mean, if it's only raining
there, you would only see a small portion of the rainbow,
of course. And what is important is the
elevation of the sun. The higher the elevation of the
sun, the less rainbow you see. We just discussed that.
And then, the width of the bow, ah, that's a good one.
The width of the bow, you would think that the width
of the bow is the difference between the two angles phi
maximum. And, that means that the width
of the bow would then be 1.8°. But, that's not true.
If there are any astronomers in my audience, you can probably
tell me why you have to add to the 1.8°, why you have to add
half a degree. The actual width is closer to
2.3°. The actual width is closer to
2.3°. Why is that?
I see one hand. I want to see a few more hands.
Any astronomers? [LAUGHTER] Why is it?
You tried, and the others didn't.
I said, is there an astronomer in my audience?
So, I'll give you a hint. The sun is have a degree in the
sky. That means each point of the
sun makes a rainbow. And since in this direction you
have half a degree, and in this direction you have
half a degree, imagine that when you see these
rainbows there that you obviously get to spread in this
direction of half a degree, and also in this direction.
And so, you have to add half a degree.
And that's actually quite noticeable.
The finite size of the sun, have a degree,
is the diameter of our sun. And so, also the width of the
secondary is not just the difference between these
numbers, which would give you about 3.1°.
But it's closer to 3.6°. And so, we know there is a
secondary. The time of the day,
yeah, well, when does it rain? And when is the sun low in the
sky? Mid-day, you are not likely to
see rainbows because the sun is too high in the sky.
It depends, of course, on the season,
whether it's summer or winter. I've been told that it rains
more often in the afternoon than in the morning.
If that's true, then the afternoon is a more
likely time than the morning. And that is when the bow would
obviously appear, then, in the east when the sun
is in the west. And certainly,
in Boston, the sun during midday would never be in the
north. You have to go for that to the
Southern Hemisphere. So, you are not likely to see a
rainbow in the south. But you might see rainbows in
the north and the winter when the sun is lower in the horizon.
So, that covers the idea of time of the day,
and in what direction. So, the secondary is there.
The color sequence is reversed. We just discussed the width of
the bow, and the radius of the bow is about 52°.
And now comes the question that all of you have worked out.
Problem set number ten is the bow polarized,
and the answer is yes. And, is it weakly polarized?
No. It is enormously polarized.
And you all came up, I hope, with the right answer
problem, 10.4 H I think it was. 91%, the linear degree of
linear polarization is 91%, and I will, of course,
demonstrate that today. As far as the direction of
polarization is concerned, well, think about Fresnel's
equations, and think about the Brewster angle because the
Brewster angle is really responsible for the fact that
the rainbow is so highly polarized.
Remember, the tangent of the Brewster angle in this case
would be one divided by 1.336 if I take this as the average index
of refraction for light. That leads, then,
to a Brewster angle of about theta Brewster of roughly
36.8°. But, remember that R,
this angle of reflection, when you make the rainbow,
that R is 40°. That's where the colors come
from. Well, that is only a few
degrees away from the Brewster angle.
So, as this light comes in un polarized, here it is still
perhaps not completely unpolarized, but not very
strongly polarized. But it is here that the action
starts. This angle is within a few
degrees of the Brewster angle. So this light that comes back
is polarized perpendicular to the plane of incidence.
And so, whatever comes out here is, of course,
also perpendicular to the plane of incidence.
So the E vector is oscillating like this.
Remember, the plane of incidents is defined as the
direction of the light and then normal through the surface.
So that, in this case, the blackboard.
And so, the radiation is polarized.
You have calculated it for 91% linearly polarized.
And, because of the actual symmetry of the problem,
that means it is polarized like this in this direction.
And here it's polarized like this.
And here it's polarized like that, nearly 100%.
When I see a rainbow, I always check whether it's
really polarized because I always worry about it someday
maybe it will be polarized. And so when I go to the beach,
which I often do: Plum Island,
one hour north of here; wonderful reservation land.
And in the afternoon, the sun is there and the ocean
water is there. Then the waves come in and the
water splashes up. And then, 42° away from my
shadow, when the water splashes up, rainbow.
Rainbow here, rainbow there.
The water splashes up high, beautiful.
Rainbow just for me. Beautiful.
And I remember a few years ago I took a friend to Plum Island.
His name is Bill Predorski [SP?].
He's also a physicist. And we were looking at it and
splashing the water. And I said, Bill,
look at the rainbow. Bill looks, what rainbow?
It's not raining, he says.
I said, look at the water. Look at the waves.
Bill looks: nothing. I said, well,
let's just wait for a really good one.
And boy, it was a real wave coming in.
And the water splashed up, a gorgeous rainbow.
Bill did not see a thing. So I got a little annoyed.
And then, I looked at Bill. I looked at Bill.
And he looked like this. [LAUGHTER]
And then, I knew. And he knew too,
because polarized sunglasses are designed in such a way which
was covered in 8.03 that the direction of polarization is
like this. But the bow is like that.
And so, you killed the bow. So, I said, Bill,
would you please take your sunglasses off?
And he did. And I was waiting,
and boy, there came a wave so fantastic and Bill looked and he
says, nice rainbow water, very low key.
He comes from the south, you know?
Nice rainbow water. Yeah.
That was a nice experience. So, now I want to demonstrate
to you a rainbow. I asked all of you to bring an
umbrella. But most of you didn't.
Who brought an umbrella? See, the problem is that the
majority did not. And we were afraid of that.
So therefore, we had to change the
demonstration from real rain to something that's a little bit of
rain, in fact, it is so little rain that it's
only one raindrop. It's not very much rain,
is it? Here it is.
One raindrop, can I make you see a rainbow
with one raindrop? No, because you see here why
you need millions of them. Can I make you see this with
one raindrop? Yes, I can.
Here's my light gun. I'm going to blind you.
And then, out of this one drop comes this cone,
red on the outside, blue on the inside,
and then white. And, that's what I can show
you. And I can show you that this
light is highly polarized. I asked you to bring your
polarizes, but not for this demonstration because when the
light reflects off the screen, the polarization is lost.
It's only on the way to the screen that it's polarized.
One water drop, don't expect too much.
It's only one water drop. So, the rainbow will be feint
if we call it a rainbow for now. And, of course,
you don't want to be blinded. So, we are also going to absorb
the sunlight so that you would actually be able to see the bow
there. The light is so feint that your
eyes have to adjust first to the darkness.
So, if we turn the lights off, I'll give you 30 seconds to get
used to the darkness. And in the meantime,
it is so properly timed that the screen is coming down.
So let your eyes adjust.
And there it is. If that's not red on the
outside, what is? If that's not blue on the
inside, what is? And you see the white light
here? This is the primary bow.
This is not the secondary. This is some weird reflection
because of the glass. And this is highly polarized.
I have a polarimeter [SP?] here, and I will hold that in
the beam. This is the way that the light
can go through. Can you see that?
It goes through. And now, I rotate it 90°,
and I can kill that light. It is nearly 100% polarized.
If I do it here, the angle is different.
Polarization angle is like this.
So, this is the way that I can let the light through.
Of course, there's always absorption, right,
with a polarized? Always 50% is lost anyhow,
and then there's, in addition,
some absorption. But it's clear that the light
goes through there. And now, I rotate it 90°,
and then you kill it. But, of course,
the white light, which is very close to the bow,
is also the result of reflection at that point,
B, in the back of the water drop so that white light,
certainly the white lake here and there is,
of course, also very close to the Brewster angle.
So, that is also highly polarized.
And so, you can see that the white light goes through here.
And, if I rotate it 90°, that white light also goes
away. And, of course,
when you go further in, then the degree of linear
polarization becomes less. So, this is,
then, what you can do. Yeah, we can have lights again.
This is what you can do, then, with one water drop.
It has all the ingredients, all the physics that you need
to explain and understand the rainbow.
But it's really not the rainbow itself.
There are other phenomenon in the sky which are very common.
And, they are also remarkable. And many of you may never have
seen them. And yet, they are so common
that I see them every week. And so I want you to become
alert to them without going into the details of the physics.
Ice crystals high up in the atmosphere can cause stunning
halos. The most famous one is the 22°
halo that you can see around the sun, and you can see around the
moon. You can see it's summer and
winter because high up in the atmosphere, temperature is way
below freezing. Red is on the inside.
Colors are never truly spectacular.
The 22° halo is so common that it's fair to say I see it almost
every week. And the reason why you probably
don't, who wants to look in the direction of the sun?
That's a terrible thing to do. Well, what I do,
I first hold up my fist, and I block out the sun.
And then I look. And that's why I see them so
often. The moon is,
of course, easier because you don't have to block out the
lights from the moon. So, if you look at the moon,
very often several times a month you see this gorgeous 22°
halo. And then, there are many other
phenomenon of which I will show you also a picture which are all
the results of ice crystals. They have names.
They are called sun dogs, a 46° halo.
24 individual names are identified in books in these
various arcs tangent that you see [SOUND OFF/THEN ON] from the
ice crystals. So, if I can have the next
slide on this and make it dark, then I think we are coming up
to the 22° halo. You see the red is on the
inside. It's not as colorful as a
rainbow, but it's very distinct. There is no question when you
see it. You know you are looking at a
22° halo. And this is,
then, a phenomenon that is very rare.
I've never seen it so complete. And this is the 22° halo.
And then there is here one that you don't see so well,
which is the 46° halo. And these things are called sun
dogs. And this arc has a special
name. The sun dogs I see often
actually here in Boston. They're not so uncommon at all.
This picture was taken in the South Pole station.
So, this business is ice crystals, is quite common.
I lost my light, but I found it.
There is another phenomenon which probably all of you have
seen, which is called The Glory. And The Glory is not the result
of a refraction and reflection. But The Glory is the result of
diffraction. You only see Glories when you
have extremely fine water drops. For instance,
as you have them in clouds, sometimes, and most commonly,
you can see it when you fly over clouds, and you have a seat
away from the sun. In fact, that's the reason why
I always, when I make reservations I always want a
seat away from the sun. Then you look in the direction
where the shadow of your plane would be.
And if you fly very low, you can actually see the shadow
of your plane. And then you see around the
shadow of your plane these beautifully colorful circles.
They are complete circles. Their diameter is a strong
function of the size of the water drops.
Like with all diffractions, the smaller the water drops,
the larger the angle. And the larger the water drops,
the smaller the angle. So, when you fly over various
clouds, you may see that the Glory changes in size.
I've seen that many times. And so, let me show you,
then, a picture that I took several years ago of a glory.
Many of you have seen this. In fact, I get countless
pictures by e-mail from students who took my lectures,
and people who didn't take my lectures who sent me this kind
of stuff and say, Professor Lewin,
we've seen a complete rainbow. Well, this has nothing to do
with the rainbow. Of course, you can see the
angle is also way smaller. But that depends,
again, on the size of the water drops.
The angle can be very large. And you can also know from this
picture where my seat was. You're always at the center,
of course, where your camera is at the center of the Glory.
So I was sitting just behind the wing, which is another
reason why I not only want to see on the side away from the
sun, but I also want to have a clear view of the ground,
which is a little bit more complicated sometimes.
I always dreamed of sainthood. [LAUGHTER]
Chances are small, but never zero.
And so, I decided if I could somehow create a Glory around my
own head, can't do any harm. Many years ago I visited
Georgia, [the caucuses?]. It was still part of the Soviet
Union. And my host took me to the 6 m
telescope they are, which was, at the time,
the largest optical telescope on Earth.
And I was welcomed very gracefully, very nice food.
And then, I talked to the local astronomers, and they said,
you know that this telescope is a joke.
It just cannot produce any science.
And I said, well, what do you mean by that?
They said, well, it was put at the wrong
location. Every evening just before
sunset, the fog comes up from the valley and you can't
observe. You are in the clouds all
night. So, it's the most ridiculous
thing here to have a telescope. And then I said to myself,
this is my chance, because if the fog comes up
from the valley, if then I can photograph my
shadow on the wall of fog, because they said it really
comes up like a wall, then I will get a glory around
my head. And I thought they were joking.
I said, well, nice story, but you must be
joking. And they said,
well, just come out at 5:30. We'll show you.
So, I came out at 5:30, and fair enough,
like a wall it came up. And in a matter of one minute,
we were in the clouds. And we stayed in the clouds the
whole night. [LAUGHTER] The next day I had
my camera. And the sun was there.
And the fog came perfectly on time.
The timing was very critical, because the fog comes very
quickly. Let me first show you that the
fog indeed comes like a wall. There it is.
That's the 6 m telescope in Georgia.
And then, this wall comes. And in a matter of one minute,
it's over you. But I was quick.
[LAUGHTER] St. Walter, after all.
Isn't that a nice example of a Glory around my head?
OK, then I have the next slide. Oh, I think I have to do that.
And then, I would like a little bit more light.
Can you put the light on TV? I suppose that you recognize
the slide. It is the mystery picture of
8.03 on the webpage. It was the astronomy picture of
the day on September 13. I received about 3,000
responses from viewers all over the world, and I answered each
one of them. It took me two months.
I did about 50 per day. About 50 people have the right
idea about what causes this phenomenon, only two from MIT.
But of those 50, there were really only about
five who had a complete understanding of the physics.
About 400 of the 3000 believed that it is a Glory,
while clearly you know now enough that this is not a glory.
Many explanations were very interesting.
Some believed it was an atomic explosion.
[LAUGHTER] More than one.
Others suggested that I was photographing the total solar
eclipse. Imagine, the day was given.
It was June 20. It never occurred to them that
there was no total solar eclipse on June 20.
But that's a detail. Some people who knew that I was
an astronomer said, ah, you were photographing a
supernova explosion. There were three people
independently who said the rings were caused by the vibration of
a jackhammer. That would make a very nice
problem for the final. How do you get this from a
jackhammer? And then there was one who says
it's the sun shining through people like those you find in
sex parlors. [LAUGHTER] And then he said
between parentheses, I know.
[LAUGHTER] The very best solution came from a five year
old girl who wrote me a letter, handwriting,
and I really want to quote her verbatim.
She says, Professor Lewin, it's very simple.
You painted the picture with crayons on the ground and you
photographed it. [LAUGHTER] Isn't that
wonderful? I mean, this five year old kid:
I mean, that's the easiest solution.
You just drew it on the ground and take a photograph.
So I sent her a very nice letter back and I said that she
was very close. [LAUGHTER] All right,
let's look at this picture. Read us on the outside.
The pilot is on the inside, and a white light comes from
inside the bow. This can only be produced by
spherical transparent beats. There is no other way.
The radius, very small, is about 20°.
It cannot possibly be due to water.
For one thing, there is no water.
Also, the width of the bow, the width, which you can easily
measure with a ruler, the width of the bow is about
16% of the radius, whereas with water [bow?]
that's only about 5%. On June 20, I visited the
DeCordova Museum in Lincoln, sculpture garden and with my
son and my significant other, who is in the audience.
It was about 1 p.m., and we walked by an area where
new construction was going on because DeCordova was building a
new visiting center. And my son, Chuck,
all of a sudden said, dad, look.
And we all looked at it. And this is what we saw.
I had never seen anything like that before in my life.
But I knew immediately within seconds that it can only be made
by spherical transparent beats. That's the only way it could be
made. So, I immediately thought of
maybe glass, maybe plastic, whatever.
But it had to be spherical transparent beats.
So, I wondered, why would there be so many of
these beats in this construction site?
It was not until days later when I actually discussed it
with Marcos, who is also in the audience, these glass beats,
about a quarter millimeter diameter, are being used for
sandblasting. And indeed, a lot of
sandblasting had been going on. You can see in that area.
And so, they had spilled a lot on the ground luckily.
In problem 10.4, I made an effort to give it
away to you so that you could score your extra credit for
8.03. I really tried to give it away,
but only one person got the message.
I wrote in problem 10.4 and the last question,
and I quote myself verbatim, in a world far,
far away, rain comes down as small drops of glass.
And then, after you have done all the work,
all I asked you, what is now the radius of a
glass bow, and what was the answer?
22.8°. But none of you made the
connection with this picture except for one person who did.
So, let's look at the physics. It doesn't take very much.
If I can have all the lights, so if we take the index of
refraction for glass 1.5, and I use my,
where is my cosine squared? I use this equation.
Then I can calculate what the maximum angle is of phi.
And I'm not going to do it for different colors.
I'll just take a mean value of 1.5.
So, I simply substitute 1.5 in this equation.
And that gives me, then, an I, for which the phi,
the maximum angle is reached. That is 49.8°.
That, then, translates with Snell's law to a phi maximum of,
angle of refraction of 30.6°. So, that's simply a matter of
Snell's law. And then, phi is 4R minus 2I.
And so, we now get phi max is about 22.8°.
You can calculate for my picture with a ruler the linear
size of the bow. Of course, we always think in
terms of angular size, but you can actually dactylic
linear size if you can give me TV because you know my head.
It's about 20 cm. And so, you can calculate now
what the radius of that bow is in terms of linear size.
And that comes out to be about 65 cm.
Now, the angle is about 20°. So I will bend over a little.
So, my head was about 5 feet from the ground.
And then I took the picture, and that exactly gives you then
the angle of about 20°. Now, I made on the spot a very
quick calculation in my head about the Brewster angle because
I remember, since I've taken 8.03, that the tangents of the
Brewster angle is one over the index of refraction.
So, that is one over 1.5. Now, I didn't have a calculator
on me, but I came roughly that the Brewster angle was probably
around 34°. And so, I concluded that that
was probably very close to the R value, in other words to the
value here, which makes the rainbow.
So, the theta Brewster for the transition from glass into air
is about 33.7°. And it is within 3° of this
value. And even though I couldn't be
so precise there on the spot, I concluded that they had to be
highly polarized. I always carry on me,
as you will from now on, always a linear polarizer on
me. So I took my linear polarizer
and I impressed my significant other.
You can ask her. And I said, look through the
linear polarizer. And indeed, when we looked at
the bow: highly polarized. I took some beats home,
scooped them up. There were lots of them because
I immediately realized I was going to use this for 8.03.
And we ordered a few kilograms which is dirt cheap.
It only cost a few dollars. And Marcos glued them on black
paper here.
Here is the black paper. Here are the glass beats on the
black paper. The sun is in the back of
6-120. And, we hired someone,
an expert, to turn on the Sun. And we hired someone to turn
off the lights in the lecture hall.
And here, I am standing in front of the glass beats.
On June 20, the sun was high in the sky.
And the glass beats were there. Today, December 7,
the Sun is there. And the glass beats are here.
And you won't believe what I'm seeing.
I'm seeing that. I see a beautiful glass bow.
I see white from the inside. I have my linear polarizer with
me. And boy, is it polarized!
[SOUND OFF/THEN ON]
And I want all of you to see this.
I don't think this has ever been demonstrated in any
lecture. When you see it,
you will just never forget it. Not only will you never forget
this, but whenever in your life you see a rainbow,
you'll think of me. I want you to be very careful
as you come down the stairs. Bring your linear polarizers,
and take your time. We have at least ten minutes
left in this lecture. I planned it that way.
Be very careful, don't run into this stuff.
I hope to see all of you on Thursday when I will give my
last farewell lecture, and then you're on your own if
you think you can handle it. So, walk by.
The closer you are, the nicer the bow is.
And, if you close one eye, you even see a better bow.
And, this is a question for you.
Why would it help if you close one eye, which I did not have to
do on June 20? So, take your time.
Don't rush, and use your linear polarizers.
Don't rush.
We have nine more minutes, and at nine minutes,
you can slowly get out of here. And, all of you should have
plenty of time to stand there for at least ten seconds.
Go very close. Close one eye.
When you are close, you see it even more dramatic.
Of course, the angle will always be 23° no matter what
your distance is. But if you're very close,
then the linear size of the bow will change, not the angle.
And then, convince yourself about the degree of
polarization. Do you have a polarimeter,
Jeffrey? So do I.
Yeah, yeah. You're a physicist,
of course. I didn't mean to insult you.
The angular size is independent of your distance,
but the linear size of the bow, of course, becomes smaller if
you are closer.
1121 1:17:08,000 Well, you can always borrow one.