We have discussed earlier the boundary conditions for ideal
conductors, and today we are going to aim at dielectrics.
In conductors, we couldn't have electric
fields and we couldn't have a changing magnetic field.
The situation is very different in dielectrics.
The dielectric is an ideal insulator.
It has zero conductivity. And there are electric fields.
And changing magnetic fields are certainly allowed inside the
dielectrics. One adjustment that we have to
make, which we discussed actually earlier,
is that wherever we have epsilon zero in Maxwell's
equations, we have to replace it by kappa of E epsilon zero,
the dielectric constant. And wherever we have mu zero,
we have to replace this by kappa of m times nu zero.
That is the magnetic permeability.
And that has certain consequences for the speed of
propagation of electromagnetic radiation.
I want to stress, though, that kappa of m,
except for paramagnetic materials which are rare,
kappa of m is always very close to 1.00000, except for
paramagnetic materials. We know that the speed of light
of electromagnetic radiation in vacuum was this.
And so now that is going to change.
We have to take this into account.
Now we get v, which is now the speed of
light, speed of electromagnetic radiation, which is now c
divided by the square root of kappa E times kappa m.
And that is also c divided by n, as in Nancy.
And n is called the index of refraction of the dielectric.
The dielectric constant is a strong functional frequency.
I have mentioned that earlier. That is the whole basic idea
behind dispersion. If I stick for now to light,
it is about 5 times 10 to the 14 hertz, and I give you an
example which you have seen before for water.
Kappa of E for light, that means for that high
frequency, is approximately 1.77.
And so the index of refraction is about 1.33.
But it is different for the different colors.
A little bit different, but it is different.
And if you take glass, there are many different kinds
of glass with many different indices of refraction,
we will typically take today a number of about 1.5.
03:32 That stands for incident.
What is going to be reflected I am going to call r.
That stands for reflected. And what penetrates into the
medium two, I will give the letter t which stands for
transmitted into. We also call it refracted.
But now you get reflection and refracted, that is confusing,
so I give it the letter t. And so there is here a k
vector. And this then is k of the
incident radiation. I call this angle theta one.
And then there will be a reflection at this surface.
And so I am going to put in here k reflected.
And this angle I will call theta three.
I will show you shortly that theta one is the same as theta
three. And then here some of that
radiation goes into the second medium.
And I have here k transmitted. The magnitude of these two k
vectors is, of course, the same because lambda is 2 pi
divided by k. But the magnitude of the k
vector here is different because the two have different indices
of refraction. 06:00
And then, if you prefer, I can write it down in complex
notation, omega t minus k of i dot r.
And this is the vector. If you would like to write it
down as the cosine, that is fine,
too, of course. And so this is the incident
wave. I can now change the i to an r
here and to an r here. And I can change this i to an r
here. This r has nothing to do with
that r. That is the position vector.
This means reflected. I can put a t here and a t here
and a t here, and I would have three
equations then for the incident one, for the reflective one and
for the transmitted one. And now comes a key point that
at z equals zero, that means really at the
boundary these three waves have to be inphase with each other,
except for a possible 180 degree phase change.
And this should remind you of something that we have dealt
with many times before. You have a string which is
connected to another string. And at the junction they are
inphase. The incident one and the
transmitted one are inphase with each other.
And the incident one and the reflected one either inphase,
mountain comes back as a mountain, or out of phase 180
degrees. And the same situation you have
here. And that means that this dot
product for these three cases, ki dot r, kr dot r and kt dot r
at z equals zero must be the same, except for the 180
degrees. But the 180 degrees we can
always deal with in terms of a minus sign.
If you massage this a little bit further, which is really not
much more than high school algebra but has worked out in
Bekefi and Barrett. If you massage this a little
further, you will be able to demonstrate it is the
consequence of the fact that these three dot products are the
same at z equals zero. On page 472 this is worked out
in great detail. You will find then that theta
one is theta three and that n1 times the sine of theta one
becomes n2 times the sine of theta two.
We call theta one the angle of incidence.
We call theta three the angle of reflections.
The angle of incidence is always the angle of reflection.
This law you may remember from your high school days,
if you had any physics, is Snell's law.
And that is a law that does not follow for Maxwell's equations.
Many of you think that Snell's law is the consequence of
Maxwell's equations. That is not true.
This law also holds for water. It holds for sound.
It is simply the consequence of the fact that there is a
different velocity in the medium here than there is in the medium
there. Maxwell's equations will come
up later and have nothing to do with Snell's law.
Willebrord Snellius was a mathematician.
He was Dutch and he discovered this relationship empirically,
which is quite an amazing thing, by the way.
In 1621, he discovered it for light at the time that it was
not known that the velocity of light in water was actually
different from the velocity of light, the speed of light in
air. Thank you very much.
This is theta, too.
Thank you. Much obliged.
Now I am going to apply Snell's law for two different
situations. I will first go from air to
water. Here we have air.
And we will just assume that n is very close to 1.00.
And then here we have water. This is n1.
One is always where you are two is where you are going.
We have water. And so n2 is 1.33.
And so here we have light coming in, and this angle is
then theta one. And here we have light being
reflected. And we know that the angle is
also theta one. And then here we have light.
This is the incident one, the reflected one and the
transmitted one. And this angle then is theta
two. And what you see here and what
I will demonstrate is not the most exciting thing.
If you increase theta one then you will see that theta two also
increases. That follows immediately from
Snell's law. You make this angle larger,
and so this opens up and this angle also opens up.
What is interesting is that you cannot make theta one any larger
than 90 degrees. And when you do make theta one
90 degrees, theta two has a maximum.
And that maximum follows immediately, of course,
from Snell's law. And, in this case,
the angle theta two, which is the maximum value that
you can have, is then 48.7 degrees.
That is the consequence of the indices of refraction the way I
have chosen them. And I will show you that.
What you can also do, which makes it way more
interesting, is you can also send light from water to air.
Now you must be careful. If you are in water that is now
your n1. If you go to air that is now
your n2. I will be repetitive,
but I do that for a good reason.
N1 is 1.33. And now this is air.
And so now n2 is 1.00. Here is now light coming in.
This now, by my definition, is theta one.
That is the angle of incidence. And so this now,
by my definition, is the angle of reflection,
which is also theta one. And then here in this medium in
air, you will have light coming out at any angle which we now
call theta two. And you can apply Snell's law
to find these angles. And it is now immediately
obvious that if you make theta one larger than 48.7 degrees
that there is no longer any solution for Snell's law,
because the sine of theta two cannot be larger than one.
And so now there is a crisis for nature.
And what nature now does is reflects all the light off this
surface. Nothing comes out anymore in
the air. And we call it total internal
reflection. And that happens at a critical
angle. Theta critical which in this
case is, of course, the 48.7 degrees for this
transition. And you will find that value
the sine of the critical angle n2 over n1.
And it is only meaningful if n1 is larger than n2.
There is no critical angle in this situation.
There is only a critical angle that means total internal
reflection if n1 is larger than n2.
We say, as physicists, you have to go from an optical
dense medium to an optically less dense medium.
That is when you can have total reflection here at the surface.
And so I would like to demonstrate that.
We are going to do that here. We are going to make it a
little dark for you so that you see it better.
I have here a laser light coming out.
Shall we turn that off? Yeah.
Can you turn that off? It would be nice.
Thank you very much. You have here light coming from
above. And it strikes the water.
And you see that some of that light is reflected.
And one of my major goals today, which we have not reached
yet, is to actually calculate the light intensity of that
reflected light and see how little is reflected.
But for now it is enough that you notice that the angle of
incident is the angle of reflection.
And notice how much goes into that second medium of water.
That is that angle theta two which is clearly small than the
angle theta one. What I can do now is increase
the angle. I can increase the angle theta
one, provided that I get my hands out of the way.
There we go. And this is not so exciting.
You still see reflected light coming here.
And you see that the angle theta two increases.
Is this exciting? No.
But this is a consequence of Snell's law, the breaking of
light in water. Now I do something that is way
more interesting. I now have light coming from
below. We manipulated the light using
a prism. It comes in now from below.
It doesn't come in from above, but it comes in from the water
here. And that is the situation I had
on the blackboard there on the right side.
Now you see a reflection in the water and you see some of that
radiation coming out into the air.
The intensity of the light in the air is actually quite
strong. Notice that.
But if now I make the angle of incidents larger than 49 degrees
then this, of course, will approach 90 degrees and
will then suddenly disappear and 100% will be reflected.
And that is what is coming up now.
I am going to make the angle larger, larger.
Notice that the light intensity of the beam in air is already
decreasing. And now I reach that critical
angle of 49 degrees. And now we have 100% internal
reflection. All the light that came out of
here, 100% is now reflected at that surface because we have
exceeded the critical angle. And this, as you can imagine,
plays a major and interesting role.
In fiber optics, this allows you to transport
images over thousands of miles without any loss of light
intensity. But, in a nutshell,
it comes down to this. If I had here some kind of a
glass fiber or plastic fiber, index of refraction say 1.5,
which would have a critical angle of about 42 degrees.
And if we send here light in then it is very easy to arrange
it that the angle there is larger than the critical angle.
And, if that is the case, then it is for 100% reflected.
But if it hits here it is, again, larger than the critical
angle. It is again for 100% reflected.
And you can send it five times around the earth,
and it will come out unaffected.
This is a very powerful way. And there are a lot of
companies who try to make money out of this who use this for
image transfer. 18:32
And the light that comes out here, you see it on the wall
there, I can put a knock in here.
I can do anything with it I want and it will not change the
intensity. I can bend it over in this
direction. There it is.
I can do anything with this I want to.
Look how tightly this is wound. I will see it here at the
ceiling. That is the idea of fiber
optics. And to show you that you can
actually also send a message, that means an image.
And that is, of course, where the big
companies are becoming interested in.
I want to demonstrate that, too.
I have a special message for you.
We have here this special message which I wrote this
morning. And then we have here the fiber
optics, a few thousand of these fibers.
And you can put a knock in it. The message will go through
here. The fibers are arranged in the
following way. This is about two centimeters
by six millimeters, and here are all these fibers.
And they come out at the other end, of course,
in the same sequence. You don't scramble them.
If you scramble them then you cannot read it,
which the military people use, by the way, in the old days to
transmit messages that were decoded.
But, in any case, we will not decode it.
We have the same image that comes in here,
the fibers come out in the same way there, and here we have a
television camera. And then you can see very
shortly there the message that I wrote for you.
Let me see whether we can get that there.
There you see the individual fibers.
For each one of the little honeycombs is a fiber.
And here comes the message. I cannot read it because I have
to concentrate, but maybe you can read it and
let me know what it reads. Can I hear you?
Oh, boy. You can read it?
Well, we knew that anyhow. You see, it is not so bad,
the quality. And this is only a few thousand
fibers. This is actually a very nice
way of showing you that you can transport that image.
And, as I do this, I hope you realize that nothing
changes. You see no change there.
And I make a crazy knot in here.
It will all go through and there is no loss of light
intensity. Now comes the key question that
is really at the heart of this lecture, and that is light
intensities. I now am interested in light
intensities. We have an incident beam,
we have a reflected beam and we have a transmitted beam.
And now, in order to get the light intensities,
I now must use Maxwell's equations.
And the first thing we have to do is we have to derive the
boundary conditions at the surface of the dielectric.
We did that before for conductors, but now we have to
do that for dielectrics. Here we have the boundary.
This is medium one, index of refraction n1,
kappa of E1, kappa of m1.
22:32 This is E1.
That means in that medium one. That is the total E vector of
the electromagnetic wave coming in, highly time variable,
frequency omega, but that speaks for itself.
And I am going to decompose this now, as I did before,
into two components. One that is parallel to the
surface. We call that the tangential
component. I am going to call this E1
tangential. 23:30
This n has nothing to do with that n.
That means normal. I still feel bad about my two
c's last time, but there is not always a way
that you can bypass using the same symbols.
I could have done that last time, though.
And now, in medium two, when the waves get through,
there is an electric vector which is now E2.
And I am going to decompose that now also into one component
tangential. This is now E2 tangential.
And this is now E2 normal. Now, we have four Maxwell's
equations, two dips and two curls.
The dips mean pill boxes. The curls mean closed loops.
Last time I did two out of four.
Today, I am a little greedier and will do one out of four.
And you do the other three. None of them are difficult.
It is just a straight-forward 8.02 application of the dips and
the curls. I will do only one.
I will do the curl of E is minus dB/dt.
It is my favorite one. It is Faraday's law.
It runs our economy. Whenever I see a chance,
I use Faraday's law. That means the closed loop
integral of E dot dL, dL is a little vector along the
path that I take in the direction of my path.
It is a dot product. It has to be a closed loop.
That now equals minus d phi. Let me put a B there.
It is a magnetic flux change. This is the magnetic flux
change through a surface that is open and that is attached to
that loop. You can choose any loop you
want to. You can even choose any
surface, provided it is everywhere attached to that
loop. And then Faraday's law holds.
An amazing law, truly incredible and always
holds. Here is going to be my path,
no different from what I did earlier when we derived the
boundary conditions for conductors.
And I am going to march around like this.
And let this be here dz. I will avoid the letter L
because we already have a dL here.
I call it just z. And let this from here to here
be just capital L. And so now we are going to do
the closed loop integral. Follow me closely.
I start here and I go here. Little dL is in this direction.
This one here in medium one has no effect because it is a dot
product. But this one has an effect.
dL and this component are in the same direction,
but when I go down here it is exactly the same value with a
minus sign because dL is in the down direction.
This contribution here and here will exactly kill each other.
There is no need writing them down.
And the same is true for medium two.
When I go from here to here, I get one component of the
normal vector with the dL. And here I get another one.
The two will cancel out each other because it is a dot
product. All I have to worry about is
the path from here to here and from here to here.
And the path from here to here, this component has no effect
because it is at 90 degrees, so it is only the horizontal
component that I deal with. And so I get that E1 tangential
times L that is in this direction.
This is my positive direction of E.
But now I go in this direction. 28:01
But now I am going to make dz zero.
That is always the trick we do when we go to the boundary.
We must make this zero because we want to know exactly what
changes at the boundary. We make dz zero.
But, once you make dz zero, the surface here,
that is the surface through which I want to measure my
changing magnetic field goes to zero.
So this part goes to zero. And so now you see that this
equals zero. And now we have derived our
first boundary condition, and the first boundary
condition them for a dielectric is the following,
that E1 tangential equals E2 tangential.
It is not too different from a conductor.
The only difference is that E tangential was zero for a
conductor. And so it was zero on both
sides. This one and this one look
alike, provided you keep in mind that for a conductor it had to
be zero. Now there are three Maxwell's
equations left. They are in your court.
And I will give you the results.
But you can derive them. It is really not difficult,
no more difficult than what I just did.
29:58 However, there is a huge
difference. This rho S was time variable as
the electromagnetic wave came in.
It changed like crazy with the frequency of the incoming wave.
This one cannot change because it is an ideal insulator.
There is no conductivity so no charges can move on the surface.
In other words, this rho S is the surface
charge density that somehow ended up on that surface that
you may have put there yourself. And it will never change.
It can also be zero. There is a big difference.
But it is, of course, physically the same rho S,
namely Coulombs per square meter.
But the function is very different.
It is very easy to do a transition whereby this is zero.
But if you put a lot of charge at that surface then this is,
of course, a large number. And so that is the number one.
And then we have B1n equals B2n.
And then we have the last one which is B1 tangential divided
by kappa m1. It is the same as B2 tangential
divided by kappa m2. Well, let me use a more
consistent notation. This is the two and this is m
and this is the one and this is the m.
That is the notation I have there.
32:00 You have them on your handout.
They are also on the Web. And I will derive two the four.
And I will leave you with the other two.
For that, I have to make some room.
And I think this we really don't need any more.
Let's do it on the center board.
This is not done as thoroughly and as well as I would have
liked to see in Bekefi and Barrett.
They do not derive in all its glory all four Fresnel
equations, yet they are absolutely fundamental to an
understanding of what happens with the light when it strikes
from one dielectric to another. Try to follow me closely.
I will try to go slowly, but if it confuses you I can
understand that. Here we have again the
transition from one medium to another.
We have here medium one with index of refraction n1 and here
we have medium number two with index of refraction n2.
We have incident radiation coming in here,
and this angle is theta one. And we have reflected
radiation, and this angle is also theta one.
And then we have radiation that goes into this medium,
and this angle is theta one. Incident.
Reflected. Transmitted.
The radiation that comes in, the E vector must be
perpendicular to the direction of propagation,
which is this. It must be perpendicular.
It must be somewhere in this plane perpendicular to the
blackboard and perpendicular to this line.
That is nonnegotiable. Therefore, I can always
decompose that if I want to. And you will see shortly why we
do that. I can always decompose any
electromagnetic wave that comes into this direction in two
components. One component which is
perpendicular to the blackboard, which we will give the symbol E
of i perpendicular. And one component in the
blackboard, which we will call E of i parallel.
And we run out of symbols so we try to do all kinds of different
ways to give names. This is a component of an
incoming electromagnetic wave. It has an electric vector in
some direction. It could even be circularly
polarized light. I can always decompose that in
two components. And I can always decompose it
one that oscillates like this and one that oscillates like
this. I will now march through this
problem and I will now show you that I can predict what this
component will come out here and how this component will come out
here. And then I will let you do the
job on this component, how that comes out here and how
this component comes out here. And then the job is done,
because you can always then combine this component with this
component. And that will tell you then
what the transmitted wave looks like.
And you can do the same here. And so that is my next task.
I am going to do it for the perpendicular component.
Once we have done it, for these two components
independently, we are safe for all cases.
Whether elliptically polarized light comes in,
circularly polarized light or linearly polarized lights,
we can then tackle any possibility.
Follow me closely. Here is an enlargement of the
incident electromagnetic wave. This is the direction it comes
in. I will only do the component
perpendicular to the blackboard. I am only looking at this
component E of i perpendicular. That component must have an
associated B vector. It is married to a B vector.
The two go hand in hand. One does not exist without the
other. And the E cross B of that
component must go in this direction.
And the B vector there must lie like so.
There is no other possibility. This must be the B vector of
the incident B. But it just so happens that it
lies in the plane of the blackboard.
Therefore, we give it the symbol parallel.
Because, if the E vector is oscillating like this,
the only way that the wave can propagate in this direction is
if the B vector oscillates like this.
It is the B parallel component that is married to the E
perpendicular component. This angle is theta one.
I am now going to decompose this one.
I will do it in green. B of i parallel times the
cosine theta one, and the vertical component is B
of i parallel times the sine of theta one.
This is the one. When it reaches z equals zero,
that is this one. That is why I am doing it that
way. I want to get a component that
hits the surface that I can handle with Maxwell's equations.
And, when this component hits the surface here,
that is also the tangential component.
That is this one. You see now why I am doing it
this way. I need components that I can
handle with the boundary condition.
That is the incoming stuff. Now we will do the reflected
stuff. Again, thing angle is theta
one. And I am only dealing with the
perpendicular component. Here is that perpendicular
component. That is E reflected
perpendicular. Now I want to know the B vector
with which it is married. This is the B vector
perpendicular in the blackboard that is associated with that E
vector. It is being reflected but it is
parallel. It is in the plane of
incidence. The plane of incidence is
always defined as the incident light and the normal through the
surface. The blackboard is what we call
the plane of incident. Now I can decompose that into a
horizontal component and I can decompose that into a vertical
component. I am not interested in the
vertical component because I am only going to use this equation.
I will only do the component in this direction.
And so this one here is then B reflected perpendicular times
the cosine of theta one. Now we have to go into medium
two. There is one more to come.
There it is. And this angle now is theta
two. And the light goes like this,
electromagnetic radiation. I am only dealing with E
perpendicular, which is what you are to do.
This is the E perpendicular component of the transmitted
wave. This is E transmitted
perpendicular. It is married to a B vector,
and that B vector must be in this direction.
To make sure that E cross B is in the direction of propagation,
this is B transmitted parallel. And the horizontal component is
then B transmitted parallel times the cosine of theta two.
All of this we need. But we are close.
I will give you a minute to digest it.
I will erase this so that we stay on the center board.
Now I go to the boundary conditions.
Before I go to the boundary conditions, there is one thing
that I want you to remember. That is there is a relationship
between B and E. They are married to each other.
And the ratio of their amplitude is c.
Well, not quite. It used to be c,
but now it is v because we are now working in dielectrics.
Since B is married to E, we can always write down,
and we are going to need that, that the magnitude of B
parallel must always be the same as the magnitude of E
perpendicular divided by v, which is also the magnitude of
E perpendicular divided by c times n.
That is the index of refraction.
And we are going to need this. This is completely kosher.
This is what we knew from Maxwell's equations way before
we dealt with this. We are close.
At the boundary condition, I now must make the E vector,
the tangential component on the left side the same as the
tangential component, not on the left side,
on the top side the same as on the medium two.
Medium one and medium two must have the same.
But medium one has two. Medium one has this component
when it reaches here, and it has this component when
it comes back. I always must add the incident
and the reflected E field, of course.
We get E incidence perpendicular plus E reflected
perpendicular, must now be E transmitted
perpendicular. Of course, transmitted
perpendicular component means in this plane.
That is that tangential thing. All these components are
tangential. You see now my problem with t,
a, n and t. It is awkward.
But this is the perpendicular component.
That is a given. That is the boundary condition.
Now I go to this boundary condition.
And, just for simplicity, I will leave the kappas out
because the kappas are one, as far as I am concerned.
If you want to put them in, be my guest.
What are those components? Well, I already eluded you to
that. This is one of them,
a medium one. This is another one,
a medium one. And this is one in medium two.
And the two in medium one together must be the same as the
one in medium two to meet the boundary condition.
I am going to write this one down.
And so I am going to get that B of i parallel times the cosine
of theta one. That is in this direction.
Then I have to subtract this one, because it is in the
opposite direction, minus B of r parallel time the
cosine of theta one. That is on one side.
Plus and minus must now be the same as B transmitted parallel
times the cosine of theta two. And now I am going to use this.
I want to convert all my b's into E perpendicular.
And, once I have done this, I am spitting distance away
from knowing what the ratios of those E vectors are.
And that is my goal. If you know the ratio of the E
vectors over the reflected one and the incident one,
you know the ratio of intensities, because that is the
pointing vector and you get squares and things like that.
I am going to replace now the parallel component by the E
perpendicular divided by v. 45:30
And look at this equation. If you know n1,
n2 and theta one, Snell's law gives you theta
two. Now you have two equations with
three unknowns. This intensity,
this electric field, this electric field and that
electric field. What you can do now is find the
ratios of these electric fields, and that is the only thing that
is meaningful. Because, clearly,
you can make the intensity of the incident one as strong as
you want to. By combining this equation with
this equation, it is clearly a high school
exercise to now find what is E transmitted divided by E
incident, of course, for the perpendicular
component? I admit that I have only done
the perpendicular component. You are going to do the
parallel component. And I can also find now what
the E vector of the reflected on is divided by the E vector of
the incident one, but only for the perpendicular
component. And that is my goal.
Once I have ratios of E vector, I can calculate light intensity
ratio. And that is when we were
looking at the demonstration there, when I wanted to show you
Snell's law, there is no way that we had to calculate why
there was so little light being reflected in that first
demonstration. And now you take your hands out
and look at page one. It is also on the Web.
There you see, in all glorious detail,
the four Fresnel equations, two of which I was in spitting
distance. To convert this to these two
ratios is no more than another three minute job,
but it is so trivial that I don't want to do it.
At the top you see Snell's law. That means if you tell me what
n1 is and what n2 is, and you give me theta one,
I will immediately tell you what theta two is.
That is nonnegotiable. That is easy.
Once you know n1, theta one, n2,
theta two, these four equations will tell you what the ratios
are of the incident electric vector over the reflected one.
But there is a parallel component and there is a
perpendicular component. And so the only two that I
derived, or I came very close to deriving, that was this one and
that was this one. I was within spitting distance
of these two. Had I made it one step further,
you would have seen this equation coming out.
The fact that I put a zero here means that I made it the
magnitude of the E vector. And I give it a shorthand
notation, a little parallel. Little r is the ratio of r over
i and the t is the ratio of t over i.
This is shorthand notation. I give you two columns.
They are the same, except that one is sometimes
easier to use than the other. This is extremely
consumer-friendly, though you may not think that.
It is extremely consumer-friendly.
You cannot go wrong if you simply apply these equations.
No one in the world could just in five minutes derive these
equations. It is a lot of work.
I spent 20 minutes on one, and I only came within spitting
distance. I didn't even finish it.
It is a lot of time. As long as you know what they
are, you just use them. I gave you theta one,
I gave you n1, I give you n2,
you calculate theta two, and out come these ratios.
There are things in here which are amazing which are very
opaque. They are not very transparent.
And I am going to work with you on that after the break.
I just want you to know you have them.
They are worth gold. They are not in Bekefi and
Barrett. Bekefi and Barrett only mention
one and a half or so for a very special case.
This is the complete set. And so I would have thought,
but let me check my notes, that this is an ideal moment
for a break. And then, after the break,
we will try to digest some remarkable consequences of these
equations. And so, if you can help me
handing out the mini-quiz number nine.
The idea was, but I could be wrong,
that you can answer this in 30 seconds.
That was my goal. And, if it takes you any more
than 30 seconds, you might as well not do it,
believe me. You will see why.
If you can help me handing it out.
Wait until the whistle blows. Can you help me handing it out?
And you? I hope you all have a handout.
There are handouts at all three entrances.
As I promised, I now want to give you a little
bit of insight, but I will only stretch the
surface of the consequences of Fresnel equations.
And let's first do a case whereby theta one is theta there
and is therefore zero. Snell's law tells you if theta
one is zero then theta two is also zero.
We give that a name. We call that normal incidence.
It then so happens, which is by no means obvious
when you look at these equations, that the first two
equations get you exactly the same result.
And so you will find then -- 52:30
And now I take a special case which is on your page number
two. That is the transition from air
to glass. That is the upper panel going
from air to glass. Where you are is always n1.
Never forget that. N1 is then very close to 1.00.
And n2 is very close to 1.5, if we take 1.5 for glass.
We will find then that for this case, this is minus 0.2.
What is the meaning of the minus?
Come on. We have seen that with strings.
It means that there is 180 phase flip.
That is all it means. It means that the E vector,
if it is in a positive direction as it arrives,
it reflects, it flips over.
That minus sign is what we see very often when we hit
boundaries where there is a change.
This is a 180 phase flip. And this is plus 0.8.
There is no phase flip in going from medium one to medium two?
Of course not. That was the same with the
string. If you have a string which is
connected to another string, if the junction goes up,
then the wave that goes into the second medium also goes up.
There is no way that it could down, remember?
It is the reflected one that can change phase,
but not the transmitted one. That is not too surprising,
by the way. If now you look at that upper
panel, which is a plot of Fresnel equations,
four Fresnel equations have been plotted here for the
transition from air to glass horizontally theta one,
and vertically is the ratio of the E vector in those four
cases. Reflected parallel,
reflected perpendicular, transmitted parallel and
transmitted perpendicular. And look.
Theta one is zero. What do you see there?
You see minus 0.2, which I just calculated for
you, and plus 0.8 at an angle of zero degrees.
Now, clearly we can also calculate the light intensity
now. That was my goal.
I don't give a damn whether that light changes phase by 180
degrees or not. That is a later issue.
I want to know light intensities.
And light intensities means I deal with pointing vectors,
E cross B. Well, B, in magnitude,
is always E divided by v. Whenever you compare light
intensities, all you have to do is take E squared and then you
divide it by v. But since the reflected one and
the incident one are both in the same medium, I don't even have
to worry about v because v is the same in both cases,
which is not true when I go from medium one to medium two.
Therefore, I can now calculate for you the intensity,
i stands for intensity, of the reflected wave divided
by the intensity of the incident wave.
And I claim that it is the E vector reflected divided by the
E vector incident squared. That is all it is because the
velocity in that medium is the same for reflected and incident
wave. This takes everything into
account. Mu zero is taken into account.
B is taken into account. The fact that we have a square,
that is the result of the B. Remember, it is E cross B?
And so this is 0.04. That is a famous number.
Every astronomer knows that number.
That means when light strikes a glass surface at normal
incidence 4% comes back. There is nothing you can do
about that. 4% of that radiation is lost.
If you want to get it through a lens there is 4% that comes
back. Therefore, in order to conserve
energy, i of t divided by i of i must be 0.96.
Well, you do that. The answer is yes,
it is 0.96. But I want you to calculate it
because now you must take into account that the velocity in
medium two is different from the velocity in medium one.
If you overlook that, you will never find 0.96.
If you take this number and you square it, you don't get 0.96,
but you cannot do that because in medium two the B vector is E
divided by its own v. And that own velocity is
different from the velocity in medium one.
I have set you off on the right track.
This is still correct but you cannot just take this ratio.
There is something so incredibly bizarre about Fresnel
equations. Look at equation number one,
and look at the right column. Suppose you make theta one plus
theta two 90 degrees. I will show you that it is very
easy. Then that thing goes to zero.
Because the tangent of 90 degrees is infinitely high.
R parallel, the way I have called that there,
goes to zero. That is amazing.
That means there must be one angle theta one,
if I hit it just right, that none of the parallel
component is reflected. The only thing that is
reflected is then the perpendicular component.
Now, it may not be reflected for 100%.
But some of it is reflected. Therefore, the reflected light
is now 100% polarized. There is only one angle for
which that is happens, and that angle has a name.
We call that the Brewster angle.
And the tangent of theta Brewster, that is that theta
one, is n2 divided by n1. I want to remind you,
one is always where you are and two is where you are going.
And I can easily show you that this is an immediate consequence
of that. I will prove it to you.
For one thing, if theta one plus theta two is
90 degrees then the cosine of theta one is the sine of theta
two. That is high school.
But n1 times sine theta one is n2 times the sine of theta two.
That is my countryman Snell. And, if you combine these two,
in 30 seconds you can prove this.
And so this is the angle that then will automatically give you
that theta one plus theta two is 90 degrees.
And that, of course, is so wonderful to demonstrate.
And I can demonstrate that in various ways.
And look at that upper panel that you have on your handout on
page two. Look at that Brewster angle.
If you go from air to glass, so now we go from air to glass,
and we know what n2 is, n2 is 1.5 and n1 is one,
then you will find that theta Brewster, which is this
equation, is 56.3 degrees. Of course, that is only correct
if n2 is exactly 1.5, which is almost never is.
And, if you use Snell's law, then you will find that theta
two is 33.7 degrees. Of course the sum must be 90,
and it is 90 if I did not make a mistake.
And then you get the crazy situation that r parallel is
zero. And so now you can calculate
what is now our perpendicular for which you have to go to your
Fresnel equations. There is no way that anyone can
immediately see what that is. Now you have to go to this
equation and substitute in here the values for theta one and
theta two and for n1 and n2. And you will find then that our
perpendicular, and I will just make sure that
I have it right, is minus 0.385.
That is what it is. If we want to know intensity,
all we have to do is square it because we are still in the same
medium. The reflected medium and the
incident medium is one in the same medium.
So 0.385 squared is about 0.15. What this means is that if you
have an E component perpendicular that comes in then
15% of that will be reflected. And zero percent will be
reflected of the parallel component.
That means if unpolarized light comes in, unpolarized light
means that half of it is perpendicular and half of it is
parallel, 7.5% of the incoming unpolarized light will come out
100% polarized in this direction.
Imagine. You now have a wave.
You start with unpolarized light.
You shine it onto a glass panel.
And, if the angle of incidence is close to 56 degrees,
7.5% reflects of the total intensity.
And it is 100% polarized in the direction perpendicular to the
plane of incidence. And that is what you see there.
On this plot you see the Brewster angle of 56.3%.
You notice that our parallel goes to zero?
Therefore, everything that comes back --
Of course, some of it penetrates into medium,
too. I didn't calculate how much
that is. But here it is the minus 0.385
that I just calculated that is this component.
That is refracted. And you can imagine that this
is, of course, great stuff to demonstrate.
Before I demonstrate that, I want you to appreciate that
if you go from glass to air that there is also a Brewster angle.
There is a different one. Of course, because then n1 and
n2 flip roles. And so, if you look at the
bottom panel of your page two, you will see that there is a
Brewster angle at 33.7 degrees. That means if you are now
inside glass and you bounce the light off the medium with air
that then, at an incident angle of 33.7 degrees,
then you can create also 100% polarized light.
Notice also that in this case, but only in this case,
there is such a thing as a critical angle.
Remember, earlier I mentioned you only have a critical angle
of total internal reflection if you go from optical dense medium
to a less dense medium. That is the case where you go
from glass to air. There was no such thing when
you go from air to glass. And that is what you see in
this panel. You see the moment you hit this
critical angle that 100% of the radiation is totally reflected.
You see, this is 1.0 and the r parallel is 1.0.
Nothing penetrates into the second medium.
The Fresnel equations, I want you to appreciate,
are unbelievably powerful. They are extremely
consumer-friendly because you can now calculate light
intensities of reflected and transmitted light for any angle
of incidence theta one. All you have to know is what n1
and what n2 is. Whatever the incoming light is,
if it is elliptically polarized you decompose it into an E
perpendicular and an E parallel component.
And then you turn the crank four equations.
And then you get the reflected component and you get the
transmitted component. You can do that for any
incoming light whether it is circularly polarized or
elliptically polarized or linearly polarized.
And you can now also evaluate to what degree the reflected
light is polarized. So far we always said 100%
polarized or it is unpolarized. Life is not that way.
There is also something, of course, of being partially
polarized. And I am going to give you now
one possible way of defining the degree of polarization,
the degree of linear, I stress, polarization.
And I will define that for you. I give it the capital letter v.
I don't think that is universal.
I just gave it the letter E. It is a degree of polarization.
And I define that as the intensity of the parallel
component minus the intensity of the perpendicular component
divided by the intensity of the parallel component plus the
intensity of the perpendicular component.
And I take the absolute value. This number can be anywhere
from zero to one. If it is zero,
it is totally unpolarized. And, if it is one,
then it is 100% linearly polarized.
You can have a situation that the answer is 0.5,
that it is partially polarized. It is not completely
unpolarized. That means one of these
components then dominates. And then you have partially
polarized light. And when you have your linearly
polarizer, you can actually tell that, as you rotate it around,
there is a change in light intensity.
Only if this is one is it 100% linearly polarized.
A small test. This test may come too early
for you, but just see how you react to this test.
You know that there are sunglasses which are made of
polarized polarizers. And so here is one of those
sunglasses and here is another one and here is another one.
And you are the manufacturer. You have to decide on the
direction of polarization of the linearly polarizers.
Would you do it this way or that way or this way or any
other way? And, if so, why?
Now, think about it. What would be the goal of
having sunglasses that are polarized?
What would be the goal? Why would you want that?
Just because it looks sexy and it is in?
No. There really is a reason for
that. What is the reason?
That is the point. The goal of polarizers is that
light that has been reflected, for instance of water,
when you drive your car and there is water,
or ice, that the reflected light is linearly polarized to
some degree. Now, if the light comes in at
the Brewster angle, it would be 100% polarized.
That would be very special. But suppose the sun is there
and I am driving my car here and the sunlight hits water and it
comes in this direction then it is highly polarized.
At the very minimum partially polarized in the perpendicular
direction, perpendicular to the plane of incidence.
This is the plane of incidence. It is polarized like this,
to a large degree. So how do I want my sunglasses?
Like this. Because now I kill this
component. And that is exactly the way you
make them. This is the correct answer.
And that, of course, calls for a demonstration.
I have here a light beam and here I have a pane of glass.
And the light reflects off that glass.
And we have aimed it roughly at the Brewster angle.
And you cannot do that, of course, exactly the beam is
not so well defined. And so when I look here at the
light that reflects off that plane, it is painful.
I am not even joking. It is enormously annoying.
There is a tremendous amount of light that reflects from it.
Who has seen the Matrix? You have seen the Matrix?
Remember me? [LAUGHTER] Now I look,
and I have no problem. I can just see there is an
astronaut walking on the moon. Now, you may say,
sure, you can say that, but what do we know?
I accept that. The light that comes back is
polarized in this direction. This is the plane of incidence.
It is almost 100% polarized in this direction.
I am holding this in front of my face.
This is the direction of polarization of this sheet.
Look at me. Now look at me.
Big difference, right?
What I am doing now, this is a super pair of
sunglasses, very expensive and specially made for me.
This one is super. It cuts out this component.
And, when I wear them like this, the sunglasses is not
doing anything. I mean, you can look at the
wall there, you see the same phenomenon.
That is what sunglasses are doing for you.
And this is an amazing thing that you cannot even read what
is under that glass plate. You cannot even see what it is,
but the moment you do this you have no problems.
That is the power of the Brewster angle.
And then the linear polarizers take care of that.
I would like you take out of your envelope,
if you have it with you, one linear polarizer.
And all I want you to do is relax for one minute and look
around in this room. And we have objects that are
made of metal and we have objects that are made of glass.
If unpolarized light strikes metal, it will not become
polarized. If unpolarized light strikes a
dielectric, it can become polarized to some degree.
In an extreme case for 100%. If you take your linear
polarizer and you rotate it in front of your eyes,
I want you to see that when you look at dielectrics like the
glass or varnished surfaces or leather that you may actually
see it change in light intensity, which you will not
see when you look at metal. If you do see it at metal,
it means that the light that struck the metal was already
partially polarized. And then, of course,
it remains polarized. Metal, an ideal conductor,
doesn't change the degree of polarization.
It keeps it as it was. But it is the dielectric that
can change it.
You have almost the Brewster angle?
But it is only for you. Not for other people.
It depends on where you are located.
You are saying if you look into the direction of the projector
there. It doesn't do much for me
because the angle of view is different from me.
You can see the effect. I can see it now very well.
Absolutely. One of those lamps goes out.
Boy, it is getting dark. That is what you had in mind.
Thank you very much. But look at some objects here.
Look at this, for instance.
I bet if you rotate this that certain lights change their
intensity. And it also looks at varnished
surfaces. Anything that is dielectric has
potentially this possibility. And now comes the ultimate
demonstration of Brewster angle. What I am going to do now is
create, out of 100% unpolarized light, 7.5% linearly polarized
light. The famous 7.5% that I
mentioned to you earlier. And the way I am going to do
that is as follows. We have here a light beam.
Actually, it goes in this direction but big deal.
And this is unpolarized light. And I am going to put in here
one and only one pane of glass. Here it is.
One pane of glass. And I bounce that light
unpolarized off this surface, and I am going to project it
here. And, when I do that,
then the light will go like this.
This angle is near 56 degrees. Then the light is polarized for
100% in the direction perpendicular to the plane of
incidence. That means, in this case,
since this is the plane of incidence, that is the incident
light and the normal, it will be polarized in this
direction. And, as I rotate that pane of
glass, the reflected beam will move on the board here.
And, when I approach the Brewster angle,
I will show you that I have it 100% polarized in this
direction. When I am over the Brewster
angle it is partially polarized. When I am under the Brewster
angle it is partially polarized, but I will come very close to
that. Once I have done that you are
going to get even more for your money.
I will show you something even more remarkable.
Most of the light will go through.
For one thing, the parallel component doesn't
come out at all, so that means it goes through.
And, of course, most of the perpendicular
component will still go through because only 7.5% of the total
radiation comes out here. Most of it goes through.
Well, I put another plane here. And then I put another one and
another one and another one and another one.
And Markos and I don't even know how many,
but I think there are 200 in here.
And every time that the light strikes that surface,
at the Brewster angle, whatever comes out,
if anything comes out, it is polarized in that
direction. Every time it is polarized in
this direction. What do you think then happens
with the light that ultimately makes it through?
100% polarized. I have sucked out every
perpendicular component. And so what remains then is the
parallel component. Imagine what I have done now.
I start with 100% unpolarized light.
With one reflection, I can get 7.5% linearly
polarized. With hundreds of reflections,
without being too specific, I can 100% polarized light
here, which you are going to see there, and I get 100% polarized
light here. The light that comes out here
is polarized in this direction. That is the perpendicular
component. The other component is in the
plane of incidence. Remember, one is perpendicular
the other is incidence. In the plane of incidence means
this is the normal to the glass, this is the incident beam so
this is the plane of incidence. It is in the plane of incidence
so it is polarized like this. When I use those 300 or 400
glass plates in a row, the light that you see here
will be polarized like this. And I can show you that because
I have a big polarizer. If we can lower the screen.
Markos, would you do that? Then I will turn this on.
There it is. We will make it very dark.
Not completely dark. We are going to set it at TV
level. First of all,
this only goes through one pane of glass.
You cannot use your polarizers because whenever the light
reflects off a screen it loses its degree of polarization,
at least to a high degree. This is the light that goes the
plane. And 7.5%, if I am near the
Brewster angle, comes out here.
93% go through and about 7.5% comes out here,
if I am close to the Brewster angle.
And I may actually be close here.
Actually, let me purposely not be close.
Now I am not very close. Here you see it is already
polarized to some degree, but you see clearly light
behind it. Now I am going to go through
the Brewster angle. And now I kill it.
I am very close to the Brewster angle now.
This light is almost 100% polarized.
This is the direction. The direction of polarization
is like this. That is the direction of my
polarizer. And here I kill it.
And that, of course, is hardly polarized at all.
I can show it by rotating it. It is hardly polarized at all.
It has lost some of its perpendicular component,
but not so much that you can really see the difference when I
rotate this. Now I am going to make the
light go through my hundreds of panes, and that is going to be
even more interesting. I think that is this one.
Now I do the same thing, except I do not have one pane
of glass but I have many. A few hundred,
Markos. We don't want to open it.
You see, when you open it you break it, so we have no idea.
But the idea is we agree that it will be close the Brewster
angle here. This now is polarized in this
direction. That means if it is polarized
in this direction, I should be able to kill it
like this. And I can.
This is when I let it through and this is when I kill it.
And I claim that this is polarized now in this direction,
100% polarized. That means I can let it through
like this, but I can kill it like this.
There you go. Now I have made 50% of my
incident light go out there, 50% comes out here,
roughly, 100% polarized, 100% polarized.
This is an absolutely remarkable thing.
We now have a method to convert unpolarized light into 100%
polarized radiation. A component like this and a
component like this. And one of the amazing things
of physics is that it makes the impossible possible.
And that is the power and the beauty of physics and is why I
love physics, to make the impossible
possible. It is like the impossible
dream. It is also to make difficult
things easy. Physics makes difficult things
easy. See you Thursday.