Earlier in this course, we discussed linear
polarization of electromagnetic radiation, and I demonstrate
this at seventy-five megaHertz and at ten gigaHertz.
Today, I will concentrate exclusively on the polarization
of light, which is at a much higher frequency.
The light from the sun or light from light bulbs is not
polarized. So I can ask myself the
question, now, what does it mean when light is
not polarized? Let's think of individual light
photons as plane waves, with a well-defined direction
of polarization. So each one is linearly
polarized. A beam is coming straight out
of the blackboard. The first photon arrives,
it's linearly polarized in this direction.
This second photon arrives, linearly polarized in this
direction, so the electric field vector is oscillating like that.
Another photon, another photon,
and another photon. And what you see here,
very clearly, that there is no preferred
direction which you average over time, and that's what we call --
call unpolarized light. It was Edwin Land who,
in nineteen thirty eight, developed a material that can
turn this into linearly polarized light,
for which he became very famous, in addition to this
demonstration that I showed you last time.
If I take one of Edwin Land's sheets, which will turn light
into polarization in this direction, and I first take one
photon, for instance, this one.
That one comes in from the blackboard towards you,
and so here it is. Oscillating the E vector like
this, E zero is the maximum value of the electric field
strength in that plane electromagnetic wave.
And this is the direction of the polarizer that I have
through which this photon goes. I can now make a simple
calculation, by projecting this E-vector onto the preferred
direction of polarization, and this new E-vector is now
down by the cosine of theta, if this angle is theta,
this E-vector is now E -- E zero times the cosine of theta.
If you ask me now, whether the light is reduced in
intensity, I would have to say, "Yes, of course," because light
intensity depends on the pointing vector,
and the pointing vector is always proportional to E zero
squared, because the pointing vector is the cross-product
between E and B. And if E is reduced,
B is also reduced. And so we get a cosine square
reduction. If, now, I average over all
incoming photons -- so I take all of these,
which represent an unpolarized beam -- so I get not only one
like so, but I get one like so, and one like so,
and one like so -- then
clearly, I have to calculate, now, the mean value of cosine
square theta. And the mean value of cosine
square theta is one-half, and so if the intensity of the
unpolarized beam, unpolarized light was
originally I zero, once it comes through
this polarizer that Edwin Land gave me, then I get one-half I
zero, but that is now hundred percent polarized.
And it is hundred percent polarized in this direction.
And the one-half is the result of the average value of cosine
square theta. If this were the case,
it would be an extremely ideal polarizer, we would call this an
H N fifty polarizer -- they don't exist,
it's only in your head -- and the fifty refers to the fact
that fifty percent get through polarized.
In the optic skits that we hand out today that we will need
throughout this course, you don't have H N fifty
polarizers, they don't exist. I don't quite know what yours
is, I didn't measure it, yours may be an H N twenty
five, or maybe an H N thirty, which would then mean that the
I zero strength of an unpolarized light of beam would
not be half of I zero, but maybe only point two five,
or point three. But in any case,
the light that will come through your linear polarizers
will be very closely two hundred percent polarized.
So what I will do now, I will take unpolarized light,
and I will have this light coming straight out of the
blackboard perpendicular to you, with strength I zero,
and here is one of my polarizers, and the light that
comes through here is linearly polarized in
this direction. And so we already know that
one-half I zero will come through if it is an ideal
polarizer, and it is polarized in this direction.
I take a second sheet, an identical one,
I put it also in the plane of the blackboard,
but I rotate it over an angle theta.
So here is now a second sheet, which has a preferred direction
of polarization in -- in this direction, and the angle is
rotated over an angle theta.
So between this one and this one is an angle theta.
And so you can now immediately tell what the intensity of the
light is that comes through this second polarizer.
It must, of course, be polarized in this direction,
because that is the allowed direction polarization for that
second sheet -- and the intensity must now be one-half I
zero, because that's what comes in, and then I have to multiply
it by the cosine square of theta.
I don't have to average it now over all angles,
because there is only one value of theta between this sheet and
this sheet, so this is now the new intensity,
and it's all polarized in this direction.
And this law, whereby the light intensity is
reduced by the factor cosine square theta,
is known as Malus' Law. Malus' Law.
If theta were thirty degrees,
the light intensity here would be one-half I zero ties the
cosine square of thirty degrees, which is oh point seven five.
If theta were zero degrees, that means that this sheet is
in the same direction as this one, if everything were ideal,
one-half I zero would get through the second sheet.
If theta is ninety degrees, then nothing will get through,
because the cosine of ninety degrees is zero.
We call that crossed polarizers.
If you cross them like this, no light will get through.
Now, before I demonstrate this, I have to be honest with you,
because the idea of reducing the energy of individual photons
by reducing their electric field strength, as I did,
is a cheat. A light photon has a
well-defined energy which depends uniquely on the
frequency of the light. Blue light has a higher
frequency than red light, so blue light has a higher
energy than red light.
And when you send blue light through a polarizer,
the way I did here, it either comes through or it
doesn't come through. But if it does come through,
it is still blue light, there is no such thing as a
reduction in energy. Whereas this reduction,
by cosine theta, would imply that the energy
goes down, and that moo- would imply, then, that there would be
a color change, that it would no longer be
blue. And that's not the case.
If you want to treat this properly, you have to do it in a
quantum mechanical way.
The interesting thing is that if you use quantum mechanics,
you find exactly the same law, you find also Malus' Law.
So the law is OK, even though the derivation is
not kosher. Now, I want you to get out of
your envelope one of your green plates, which is a linear
polarizer. This is the kind of plate that
you have, you have three in there. Only take one out. If this light that comes in is Ah, ready for this? I want you to see two things.
These two lights shining on me, unpolarized light.
So the light that comes to you now is unpolarized.
I'm now going to hold in front of my face this polarizer.
So the light that comes through is linearly polarized in this
direction. And you are going to play the
role of the second polarizer. Close one eye,
put the polarizer in front of your eye, and rotate it.
And you will see a huge difference in light
intensity. If you cross-polarize with me,
then you can't see me. That may make you very happy.
But keep in mind, if you can't see me,
then I can't see you, either.
So rotate it around and convince that this light that
reaches you is now, indeed, linearly polarized,
and when you rotate around your polarimeters -- we call the
polarimeters, we call them polarizers -- you
can see me either,
or you cannot see me at all, and there is anything and
everything in between. Very well.
There is a second way that we can produce hundred percent
linearly polarized light, and we can do that by
reflecting unpolarized light off a dielectric.
For instance, water or glass.
None of this follows from Snell's Law, Snell's Law was two
hundred fifty years before Maxwell, polarization wasn't
even known in the days of Snell.
But Maxwell's equations allow you to properly deal with
refraction and reflection, including the polarization.
And I will make no attempt to derive this for you in detail,
that's really part of eight oh three if you ever take it,
but I will present you with some results so that you can at
least appreciate the far-reaching consequences of the
reflection in which we can produce a hundred percent
polarized light. Suppose I have unpolarized
light coming in here, medium one, index of refraction
N one, medium two, index of refraction N two.
It's coming in at an incident angle of theta one,
it's unpolarized. Some of it is reflected -- and
this angle is also theta one as we discussed earlier -- and some
of it is refracted into this medium, and this angle we'll
call theta two. So this light,
unpolarized comes in, reflects, and refracts.
If you want to use Maxwell's equations, at the action,
where everything is happening right here at the surface
between the two, you will have to decompose the
incoming light, the electric field vector in
two directions. And one direction is
perpendicular to the blackboard, so this is the E-vector,
and the other direction is in the blackboard.
You will have to do the same here, and you have to do the
same here. And if you look at this
decomposition, then notice that both
components -- this component, as well as this component,
is perpendicular to the direction of propagation.
That is always a must with traveling electric -- magnetic
waves. You see the same here.
This component and this are both perpendicular to the
reflected beam, this component and this are
both perpendicular to the refracted beam.
This one, the one perpendicular to
the blackboard, we normally give a symbol
perpendicular, we call this the incidence
plane, the plane through the incident light and then normal
to the surface, we call that the incident
plane, in this place -- in this case, that is the blackboard.
We call this the perpendicular component, and we call this the
parallel component. And this is,
of course, the incident beam. So this one we call the
perpendicular component, this one we call the parallel
component, and this is now of the refracted beam.
This is the parallel component, this is the perpendicular
component of the reflected beam. The incident light is not
polarized in this sense. So if you average,
there is no preferred direction of the electric field.
That means, then, that in this representation,
the strength of this component and the strength of this
component must be exactly equal, because if one were stronger
than the other, then it wouldn't be
unpolarized, then there would be, on average,
a preferred direction. So this component has the same
strength a that component for the incoming light.
What Maxwell's equations now can do for us -- it's a lot of
work, but you may see it in eight oh three -- it can relate
the parallel component in reflection with the parallel
component of incidence, the parallel component of
refraction with the parallel component of
incidence, it gives you two relations, two -- two equations.
It can also relate the perpendicular component of
reflection with this perpendicular component,
and the perpendicular component in refraction with this
component. So you get four equations. If this light that comes in is Ah, ready for this? I want you to see two things.
unpolarized, the strength of this component is the same as
this one. In general, you will see,
if you apply these four equations,
that that's no longer the case here.
They're no longer the same intensity, and they're no longer
the same intensity here. That means this reflected light
and the refracted light has now become partially polarized.
I will only give you the relation, one of those four
equations. I will only address today the
parallel component of the incident beam,
and the parallel component of the reflected beam.
And I don't imagine that any one of you will try to remember
that equation.
I don't either, I have to look it up every time
that I deal with this. So E zero, which always
represents, then, the maximum possible value of
the electric field, of the parallel component,
of the reflected beam, that's the one that I'm after,
I'm going to make polarized light by reflecting,
is E zero, the parallel component of the incident beam
times -- and this depends now on the angle of
incidence and on the indices of refraction -- is going to be N
one times the cosine of theta two minus N two times the cosine
of theta one. Believe it or not,
all of this follows from Maxwell -- divided by N one
times the cosine of theta two plus N two times the cosine of
theta one. And if you apply Snell's Law,
you can simplify this equation --
in this case it will help us -- and so you get minus E zero
parallel incidence, and now you get here the
tangent of theta one, minus theta two,
divided by the tangent of theta one plus theta two.
So these two equations are identical.
Though that may not be obvious to you, certainly not obvious to
me, either, but if you substitute
Snell's Law in here, you can show that these two are
the same. Keep in mind if you're ever
interested in the intensity of this light, then you must always
remember that the pointing vector is proportional to E zero
squared, so you always have to square these numbers when you're
interested in light intensity. This is just the strength of
the E-vector. There is something very special
hidden in this equation, and that is,
when theta one plus theta two is ninety degrees,
then the downstairs here is infinitely large.
And so that means that the parallel component in reflection
is zero. So E parallel in refection goes
to zero. And if that one goes to zero,
there is only this one left which is not zero,
and that means the reflected light is now hundred
percent polarized in this direction, because I have killed
this component completely. But it only works if this is
met, this condition. If this condition is met,
that theta one plus theta two is ninety degrees,
then it follows from high school math that the sine of
theta two is then the cosine of theta one.
That's immediately obvious, right?
You remember the triangle, theta one plus theta two is
ninety degrees, the sine of one angle is the
cosine of the other.
And if now, I remember Snell's Law, which says that the sine of
theta one divided by the sine of theta two, if N two divided by N
one, I can replace now, only for this special case,
the sine of -- sine theta I can replace by the cosine of theta
one. And so I get here,
now, the tangent of theta one. And so if this is the tangent
of theta one, that is under these
conditions, then we have met the condition that I was looking
for, that I end up with hundred percent linearly polarized
light. And so this is the secret to
getting hundred percent polarized light.
And this angle is called the Brewster angle.
And so if we, for instance,
look at the transition from air to glass, glass has a index of
refraction approximately one point
five -- depends on the kind of glass that you have -- if I go
from air to glass, which is what I will do in my
demonstration, then the tangent of this angle
is N two divided by N one, this is glass,
so this is one point five, and one is one,
then you will find that the Brewster angle,
theta Brewster turns out to be about fifty six degrees.
I can also make linearly polarized
light by going from glass to air, by bouncing it off this
way. Then, of course,
I have to invert this, and then you will get a
Brewster angle which is smaller, which is thirty-four degrees.
But since I will do it with -- from air to glass,
I want you to concentrate on the fifty-six degree angle.
The way I'm going to do this demonstration,
it's right set up here, we have light,
a light beam that strikes a piece of plane parallel glass.
That's all it is, there's nothing special about
this piece of glass. So the light comes in like so,
and here, I have a piece of glass.
This is the angle of incidence, theta one.
And so it's going to be reflected in this direction
whereby this angle is also theta one, and something will go in
here, that is that angle theta two, which I don't worry about,
because I want you to see that this
can become hundred percent polarized.
As this light comes in, it is unpolarized,
this component and this component have equal strength,
if theta one is fifty-six degrees or somewhere in that
vicinity, this light is now hundred percent polarized.
And I'm going to project that onto the screen,
and I'm going to convince you that it is, indeed,
polarized. You cannot use your own
polarizers to see that, because the light in this beam
is going to be hundred percent polarized.
However, once it reflects off the screen, it no longer is,
so you cannot use your polarimeter, so I have to use my
own polarimeter to show you this.
So if you can turn the light off there, off the overhead --
thank you very much -- I will turn on the -- the light of my
light beam, there it is, and I'm going to make it very
dark for you so that we can see that very well.
So the light comes in this direction, hits the glass,
and the angle of incidence is now about
forty-five degrees. I purposely didn't make it
fifty-six yet. I have here a large set of
polarizer, one of Edwin Land's polar- polarizers,
and I will rotate that in this beam.
You will see that it is partially polarized,
not yet hundred percent, but it's already partially
polarized. So there is already an
imbalance between the perpendicular and the parallel
component. So if I hold it in the beam,
and I rotate it, you will clearly see now that
it is much fainter than -- than it is now.
And now I will go for the fifty-six angle,
fifty-six degree angle, roughly, and so now,
I rotate my polarizer, and now, notice,
I can kill that light completely.
Hundred percent linearly polarized.
I may not have the angle perfect, but that's OK,
you get the idea. It is very close to totally
dark. Now you see the light,
it's polarized in this direction, and now I can kill
it. So that's quite a remarkable
thing, that if we reflect light off a
dielectric, that we can -- at one angle, and one angle only,
which is the Brewster angle, that we can turn it into
hundred percent linearly polarized light.
This does not apply to conductors.
The behavior of conductors is very different from dielectrics
such as -- such as water and glass.
You can use Maxwell's equations,
of course, to study the reflection of electromagnetic
waves off metals, but you get a very different
result. And so never expect to get
linearly polarized light which bounces off metals.
I have here, for your pleasure,
a metal sphere, and I have a glass sphere,
and if you still have your linear polarizers at hand,
uh, you can now -- or a little later -- just hold them in front
of your eyes, and rotate them around -- of
course, you're not seeing light at the Brewster angle,
the chances are that some of the light that's reflected off
this glass that you can clearly see that it is partially
polarized. You can see a difference in
light intensity as you rotate it.
You're not going to see that off this metal.
Now, I come to a third way of polarization.
There is a third way that we can make hundred percent
polarized light by the scattering of unpolarized light,
and we have to scatter it off very fine particles,
preferably a tenth of a micron. Dust particles would work very
well. Now, the theory of light
scattering is extremely complicated, but I will be able
to convince you that if I scatter the light over an angle
of ninety degrees -- so it comes in like this,
and it scatters over an angle of ninety degrees -- that it
becomes hundred percent linearly polarized.
I will stay on the center board.
Suppose I have light coming in like
so. And I have one light photon,
I concentrate on one to start with, and it happens to be that
that light photon is linearly polarized in this direction,
and so the E-vector is oscillating like this.
Later, we're going to add all directions that we want.
I just picked one now. And here are my find dust
particles, and these dust particles have electrons,
and the electric field passes by,
and these electrons, which are charged,
are going to oscillate in this direction.
They're going to experience an acceleration,
which is the force that they experience, divided by their
mass, and therefore, that is the charge that they
have, times the electric field, divided by their mass.
And so as this electric field vector, this electric field
component, oscillating with frequencies omega,
is passing by these electrons, they themselves are going to
oscillate with frequency omega, and this is the force that they
will experience. Uh, this is the acceleration
they will experience. Notice that the electrons will
experience a way higher acceleration than the protons,
because the protons have a mass which is more than eighteen
hundred times larger than the electron.
So whatever follows, it's really the electrons that
do the job and not the protons. So, we have charges that move
up and down, and now comes the question which we have discussed
earlier, so this is just simply to
refresh your memory, if you're here at point P,
in what direction will you now see electromagnetic radiation
that is produced by charges that are being accelerated?
We discussed that earlier, and we even had a movie about
that. And perhaps you will remember
that the electric field, at point P, is now oscillating
in this direction. A spherical wave goes out,
if I accelerate charges here. And the rules about this
direction of the electric field are very simple.
E is always perpendicular to the direction of propagation,
which is the position vector -- I call this the position vector
R -- and the second rule is that A R and E are in one plane,
and that happens to be, in this case,
the plane of my blackboard. But of course,
that doesn't have to be the plane of the blackboard,
because I can choose my point P in space here and these rules
still apply. The incident photon,
which in this case, I picked only one,
is completely destroyed. It is absorbed by the dust.
And the electrons which are going to radiate,
reradiate a photon at exactly the same frequency,
because if this is oscillating with angular frequency omega,
then the acceleration would be with angular frequency omega,
and so this E field will have angular frequency omega.
So it really is as if the photon comes in and takes off in
a different direction, that's why we call this
scattering. So the frequency remains the
same, but the direction changes. And the probably that that
photon will go in this direction or this direction is zero,
because you remember that no electromagnetic wave is
propagated in the direction of the acceleration,
there's a high probability that it
goes out in this plane, perpendicular to A,
and at this angle theta, the probability is a little
lower. We discussed that earlier.
Now, I'm going to convince you why light that scatters over
ninety degrees that comes in like this, and then [wssshhht]
comes to you, why that is hundred percent
linearly polarized. Here is a beam of light.
And this beam of light is unpolarized.
That means, if I look down on this beam, you have individual
photons, and I described through those -- maybe a little bit
artificially, but I will do that,
it was successful with Malus' Law -- I describe through them,
individual directions of polarization.
So each photon, on its own, has a uniquely
defined direction of polarization.
It is a picture that is not kosher, you really need quantum
mechanics to do this right, but on the other hand,
the result that you find is probably
correct. You will find the same result
if you did it in a quantum mechanical way.
So here, you have these dust particles, and so what is going
to happen is, this light comes in,
and then [wssshhht], one may be scattered in this
direction, [wssshhht], another one in this direction,
another one in forward direction.
And you can go in all directions as you please.
I'm now going to look in the plane
perpendicular to the blackboard, because in the plane
perpendicular to the blackboard, every photon that ends up there
is at ninety degrees angles. It comes in like this,
that is ninety degrees, this is ninety degrees,
this is also ninety degrees. And here is that plane.
I just draw a circle, but there is nothing special
about the circle. And so the photons come now
straight to you, perpendicular to the
blackboard. That's the picture that I have
now in mind. And let's say you were
looking here. So you were sitting there,
so you have to lift this up, so you're really sitting there.
So this is where you are. And let's take one photon that
comes in, which happens to be linearly polarized in this
direction. We pick one photon,
but it's unpolarized light because we're going to add so
many photons that, on average, there is no
preferred direction. But I pick one to start with.
And now I ask myself the question, if that photon is
scattered in your direction, in this direction,
how is the E-vector oscillating here?
So this is now the position vector R, and this is the
direction A in which the electrons are going to shake,
because the photon comes in, with an E field shaking like
this, so the electrons are going to shake like this.
And you will immediately conclude that the electric
vector here must be oscillating like this.
Why? Because it has to be
perpendicular to R, which it is,
and it has to be in the plane of A and R.
There's only one solution, and then this is the correct
solution. Now there is a next
photon coming in. And the next photon -- I will
give it color, just to distinguish the two --
happens to be oscillating in this direction.
And I ask myself the question, if that photon is scattered in
your direction, in what direction is the
E-field oscillating? And you'll come to exactly the
same conclusion, in this direction.
Why? Because it has to be
perpendicular to R, which it is,
and it has to be in the plane A and R, that's the only solution.
A little later, there is another photon that
comes in. How is that electric field
being observed here? Of course, in this direction.
And so, no matter how they come in, unpolarized light,
you will always see that if the photon is scattered over ninety
degrees, you will always see it polarized in this direction,
and therefore, you have created linearly
polarized light. So if you watch here,
that is at ninety degrees angle -- or if you watch here at
ninety degrees angles -- but of course,
it's a whole plane -- then you end up with hundred percent
linearly polarized light. If you go through the same
exercise, say, at an angle here,
of forty five degrees, and you look here,
it's only partially polarized. Indeed, if you rotate in front
of your eye the polarimeter, you will see clearly light
intensity changes. But not hundred percent
polarized. You couldn't turn it into
darkness. And if you look from above --
and you may want to go through that
exercise for yourself, you will see that the light
remains completely unpolarized. So it is only the ninety-degree
angle that is very special. And that's what I'm going to
demonstrate to you. But before I demonstrate this,
there is something that I have to tell you, that I cannot hide
from you, I wish I could, but I can't.
It's not something that is my goal during this lecture,
it has nothing to do with polarization.
But that's the fact, that the probability,
when you scatter light off very small
particles, a tenth of a micron or so, dust particles,
that the probability of scattering is way higher for
blue light than it is for blue light.
The shorter the wavelength, the higher the probability.
If you ever take eight of three, you're going to see a
derivation of that, a quantitative derivation.
Blue light has a ten times higher probability to be
scattered than red light. And so whenever I'm going to do
my scatter experiments on very fine particles,
you will see that that light is going to be bluish.
You can't miss that. Now, if the particles off which
I scatter are larger than a tenth of a micron,
say one micron, this effect of color on
probability of scattering is highly reduced,
and if I scatter off very large particles -- ten microns or so
-- then there is no dependence any more at all.
And in this lies the secret why the sky is blue -- I will get
back to that later today -- the reason why cigarette smoke can
be blue, if the smoke particles are very small,
and it's the reason why clouds are white.
Because the sunlight hits the clouds, the light scatters,
but the water drops in the clouds are not oh point one
microns, but they are much larger, they are more like ten
microns and up, and so there is no preferred
wavelength that scatters, and so the clouds look white.
And so the first demonstration that I want to do is very much
like what you see here. I'm going to send unpolarized
light up here, straight up.
We have bright spotlights, and the light goes straight up.
In here, I'm going to put very small dust particles.
And I have decided to do that smoke, simply cigarette smoke.
So I'm going to hold cigarette smoke in these beams,
and the light that will come to you, no matter where you sit,
must have scattered closely over ninety degrees,
right? It comes up like this,
but if you see it, almost for everyone in the
audience, ninety degree angle scattering.
So with your linear polarizers, you will be able to see that
that light is polarized, and it's going to be polarized
in this direction, which is the direction that I
have here. And that's the first thing I'm
going to do with this demonstration.
And so I need cigarettes, and I need smoke. As much as I hate this. If this light that comes in is Ah, ready for this? I want you to see two things.
[laughter]. OK. That should do. If this light that comes in is Ah, ready for this? I want you to see two things.
[laughter]. OK.
So, you need a lot of light, and as light comes,
presumably from here yes, there it is. If this light that comes in is Ah, ready for this? I want you to see two things.
OK, so have your polarizers ready. If this light that comes in is Ah, ready for this? I want you to see two things.
Number one, that the light is bluish, and number two,
that it is polarized. Take your time for that.
If you don't see it as blue, then the reason for that is
that at low-light intensities, your eyes are not very
sensitive for color any more. It looks quite bluish to me, though. If this light that comes in is Ah, ready for this? I want you to see two things.
Now I want to do something in addition.
I mentioned that if the particles grow in size,
that the scattering is no longer preferred in the blue.
And I can demonstrate that. I can kill two birds with one
stone. What I can do is I can hold the
smoke in my lungs for a while, and when I do that,
the the water vapor in my lungs will precipitate on these
dust particles, and they will form small water
drops. And when I puff that out,
you will see a distinct difference in color between what
you see now, and the smoke that comes out of my
lungs when the particles are ten microns and even larger.
You will see, then, that the light is
whitish. So this is -- this comes extra,
over and above, it comes for free.
In order to make you see the difference, shortly before I
puff out the smoke in here, I will again show you this
smoke as it is now, so you can compare the colors,
and you will see that there is a difference.
Even though this doesn't -- this may not look very bluish to
you, for reasons that I mentioned,
in darkness, you don't have a very good
sensitive for color. So I'm going to hold this smoke
now -- as much as I hate it, this is one of the worst
demonstrations that I have to do -- I hold it in my lungs for a while. If this light that comes in is Ah, ready for this? I want you to see two things.
I see a huge difference, but I'm very close.
The second puff [huh] [whew] was very wide compared to the
first one. So we were able to catch two
birds with one stone here. The sky is blue because of this
phenomenon. Here, you are standing
innocently on the Earth. And sunlight is coming in,
onto the Earth's atmosphere. The sun is there.
Sunlight comes in, and the light scatters.
And the light that reaches you, that scatters off these
extremely fine dust particles, and it also scatters off the
air molecules themselves. There are thermal fluctuations
that go on all the time, which causes density
fluctuations in the air, and they are sufficient to act
as scatterers. And so if light from here comes
to you, the chances are that it is blue, because it has a higher
probability than red. And this is also likely to be
blue. And so when you look at the
sky, the sky looks blue, that's the reason,
it has to do with this strong preference for color to be
scattered when it is blue light. If you look in the direction of
the sky at an angle of ninety degrees to the direction of the
sun, the sky is also linearly polarized for the reasons that
we now understand, because there is
scattering over ninety degrees. And so when the sun is out
there, there is always a whole plane, great circle in the sky,
which is ninety degrees away from the sun.
So when you come out with your linear polarizes,
as soon as the weather clears, look at the sky,
at the blue sky, and look ninety degrees away
from the sun, and you will see that the sky
is very strongly polarized. If you look at angles different
from ninety degrees, it's partially polarized.
But not as strongly polarized as it is at ninety degrees.
So this explains, in a very natural way,
why the sun, when it rises,
and why the sun, when it sets,
why the sun is red. Because if the sun rises or the
sun sets, the sun is near the horizon.
So now the sunlight comes in like this.
Imagine how much atmosphere it has to travel through,
how many scattering particles it will encounter on the way.
And so, right here there is scattering.
That is blue light, that is blue light,
that is blue light,
that has a higher probability, that is blue light.
So what do you think is left over for you?
There isn't very much left over.
What is left over, the blue is gone,
the green is gone, so if there's anything left
over, it is red. And it is that red light that
you see. That's why the sun looks red
when it sets, and it looks red when it rises,
the same is true for planets, and bright stars,
and the moon. When they're just above the
horizon, they look very reddish. And if
there happens to be a cloud here in the sky,
well, the cloud will also see that red light,
so you get a red cloud. And that's exactly what you see
at sunset, all these clouds turn red.
And so that is, again, the consequence of the
fact that the probability for light scattering of blue light
is ten times larger, roughly, that the scattering
for red light. So it explains both the blue
skies and the reason why the sky -- why the sunrise and the
sunsets are -- are red. I can show you two slides,
whereby you do see this phenomenon, that light that
scatters to you becomes bluish. We can see it in astronomy,
and the first slide is a picture of the Pleiades,
called the Seven Sisters, very hot stars,
they are surrounded by very small, fine dust,
and the light that reaches you is not only polarized -- which
you cannot see, of course, on the slide -- but
it's also bluish. And so let's take a look at
that first. So here you see the Pleiades,
it's called the Seven Sisters -- some people think there are
only seven stars in there, but there are hundreds -- and
you see here these very bright stars, and here you see the dust
surrounding these stars, and this is distinctly blue.
That's the effect, the fact that short wavelengths
have a higher probability to be scattered than long wavelengths.
So the blue has a higher probability than the red.
And the next slide shows you a man on the moon.
And this person is walking on the moon, but as he walks on the
moon, he produces dust, by just walking,
very fine dust particles that come from the soil that he
brings up. And what he is doing,
he is creating around himself sort of a blue atmosphere,
a blue sky, because the sunlight comes from the right,
and so the light that you see that comes in your direction is
being scattered, of course.
Because there is no air on the moon,
so that can only be the dust that he has produced,
and you see that that dust is blue.
It would probably also be strongly polarized,
because if it changes ninety degrees angle,
then, of course, it would also be strongly
polarized. Now I want to do a
demonstration, the last one,
which catches more than two birds with one stone,
it's going to kill three. I have here a bucket with
thiosulfate. And we have lights coming from
this side which we s- shine through that bucket.
And we're going to put a little bit of sulfuric acid in there,
and when we do that, sulfur will precipitate,
small particles of sulfur. These small particles of sulfur
will become the scatterers, and so this light,
which is unpolarized white light, will begin to scatter.
And you're going to see it. If you're sitting at right
angles, then you will see that that light is linearly
polarized, you can enjoy that. If you're sitting there,
you're not so well off, because then the light is not
at ninety degrees. But you will see it partially
polarized. If you're sitting there,
it's not at ninety degrees either, you will see it
partially polarized. But you will also see it blue,
because the probability that blue light scattered is higher
than red light. So you're going to see,
slowly in front of your eye, a blue sky is going to develop.
It's going to be polarized in the vertical direction for those
people who are sitting at right angles, partially polarized for
the rest of the audience,
and then we're going to look at the light that remains after the
light had penetrated the atmosphere.
After the blue light is slowly exhausted.
All right. I will also take a polarizer
with me, so I can also show you, then, the polarization.
Let me first put that sulfuric acid in.
I will do that now, so that I know that I OK.
So very slowly am I creating, now, my own atmosphere.
And I'm going to make it completely dark.
And so you see white light, keep an eye on the -- on the
bucket, the bucket cannot be seen by all of us so well,
depending upon where you sit in the audience,
but I can already begin to see that it turned slightly bluish.
Of course, I have -- I'm very close.
I admit I'm very close, which is very good.
It's slightly bluish, more and more sulfur is going
to precipitate as time goes on, we have to be patient.
Oh, boy, it's bluish. I will show you that it's
polarized. I will rotate in front of it a
polarimeter, and you see, there's a distinct polarization
of this light, if it comes out at ninety
degrees. More and more sulfur is
precipitating, and look at the sun -- if you
think of this as the sun. The sun is, um,
is getting a little yellowish, it's not so white any more.
And you wonder why. Well, you should be able to
answer that, now. Because the light that is
scattered in the atmosphere -- I call this the atmosphere -- is
blue. It has a higher probability
than red. And so what is left over in the
direction that you see on the screen there is the remaining
light. And the more blue leaves,
and the more green leaves, the redder the sun is going to
be. And if we're going to be a
little bit more patient -- and we certainly have the time for
that -- you're going to see the sun getting pretty -- pretty
bloody red. So keep an eye on it,
also have your linear polarizers, and try to see that
the light that comes to you from the atmosphere -- if you are at
ninety degree angles, I will once more do it with my
polarizers -- that that's strongly polarized.
Oh, boy, look at the sun. We're getting close,
we're getting close. Whew.
Imagine you're now on the beach, very romantic,
and there is the sun, you're with a friend,
you don't have your polari- well, from now on,
you should always have your polarizer with you,
of course. You can really impress your
friend, believe me. For one thing,
you can point out that the sky at ninety-degree angles from the
sun is nearly a hundred percent
polarized. You can even tell your friend
why the sky is blue. And as you experience this
romantic sunset, you can also explain why the
sun is getting red, because all that blue light --
and the green light -- gets out first, and this is,
then, what is left over. And the sun is really beautiful
now, I already feel these butterflies in my stomach,
and ants in my pants, this is sunset all right?
[sniffles] This is very,
very nice sunset. Oh, man.
What a beautiful sunset. Yes, indeed,
indeed! We are approaching sunset!
OK, see you Wednesday. [applause]