Today we are going to talk about interference of
electromagnetic radiation. And I will start,
as a warm-up, with the famous historical
experiment which was first done by young in 1801.
By that time the issue of whether or not light was waves
or whether it was particles was still unresolved.
Newton always wanted light to be particles but the Dutch
physicist Huygens wanted them to be waves.
And the issue was unresolved. Let us agree that if light are
particles and you have a screen that, say, has two openings and
you throw particles through their like tomatoes,
the tomatoes are particles, and you collect them here,
then those tomatoes that don't get stuck on the screen but that
make it would form a pile here of tomatoes and they would form
a pile there of tomatoes. That is very typical for
particles. But the situation is different
when we deal with waves because the moment you have waves coming
in, say, for instance, we have plan waves coming in
like this, then the wave can go through both openings
simultaneously. And that changes the picture
quite dramatically. It was known,
of course, already in the 17th century that if you have water
waves going through a small opening --
Here are water waves moving in like so.
And here is also water. That what you see coming out
there are circular waves. The waves then look like this.
And they are propagating out in a circular fashion.
And, if the velocity of the wave here is the same as the
velocity there then the wave length here would be the same as
the wavelength there. But if the velocities are
different, of course, you will see a difference in
the wavelength. Huygens, my countryman,
suggested in the 17th century the idea, which is now known as
Huygens' Principle that we can think of this in a very
different way. Huygens, by the way,
is difficult to pronounce. Any one of you who knows how to
say Huygens correctly is also Dutch.
Because you miss the uh in your language and you miss the hua in
your language. And the combination of uh and
hua is a complete killer for you guys.
You don't get extra 8.03 course credit if you can come to my
office as say Huygens, but it would certainly put you
in the Dutch category. In the 17th century then
Huygens came with an idea, which was later amended by
Fresnel in the 19th century and is now known as,
I will pronounced it your way, the Huygens-Fresnel Principle,
which works as follows. If we have a plane
monochromatic wave and it is incident on a screen with an
aperture, this would be an aperture, an opening.
Here you would have two apertures, two openings.
Then the Huygens-Fresnel Principle states the following.
All points in the aperture plane may be thought of as
secondary point sources of spherical waves.
And the point sources replace the real source,
which is floating the screen. And the screen itself is a
perfect absorber of the radiation falling upon it.
And so, I will repeat the Huygens-Fresnel Principle which
I will need today. All points in the aperture
plane may be thought of as secondary point sources of
spherical waves. In the case that we had water,
this is a two-dimensional surface, there would be circles.
But, when you deal with light, you can think of them as
three-dimensional. That means spherical waves.
Now, the Huygens-Fresnel Principle is very powerful,
though, there is a wide range of opinion to its scientific
merit. And, in a very famous book,
the principle of electrodynamics by Melvin
Schwartz, I read the following. And I quote verbatim from his
book. He says, Huygens' Principle
tells us to consider each point on a wave front as a new source
of radiation and add the radiation from all the new
sources together. Physically, this makes no sense
at all. Light does not emit light.
Only accelerating charges emit light.
Thus, we will begin by throwing out Huygens' Principle
completely. Later we will see that it
actually does give the right answer for the wrong reasons.
I will proceed with my lecture today to get the right answer,
but perhaps for the wrong reason.
Suppose now we have an electromagnetic wave.
I am particularly thinking of light.
And I have here an opening and I have here an opening,
and plane waves are coming in from the left moving with the
speed of light. And let this opening be A and
let this opening be B. And, here, the center of the
two is O. This could be a circular
opening. It could also be a slit.
Most of the experiments I will do today are with slits,
which are perpendicular to the blackboard.
06:03 Then what you get is that the
mountain of the E vector here will coincide with the valley of
the E vector from the other one. And so you would get the
situation that light plus light will give darkness.
We call that destructive interference.
The wave from here and the wave from there would be 180 degrees
out of phase. And, if you take all the points
for which the difference is one-half lambda,
that is a high paraboloidal surface.
It is like a bowl. It is not only in the
blackboard, but it also comes out of the blackboard.
It is like a bowl. Remember from your high school
days that if the sum of the distance AP and BP is a constant
then you get an ellipse, but if the difference is your
constant you get a hyperbola. You will get a hyperboloidal
surface and everywhere on that surface BP minus AP would then
be one-half lambda. And then you would get the
amazing consequence that light plus light will give darkness,
destructive interference. Tomatoes don't do that.
One tomato on top of another tomato does not give no tomato.
That is distinctly different from particles.
Now, of course, there are also hyperboloidal
surfaces for which there is constructive interference.
In other words, if I make this difference n
times lambda, whereby n is either zero or
plus or minus one or plus or minus two then I would get
surfaces for which we get construct interference.
Then the phase difference between the two waves is either
zero or 2pi or 4pi. We have surfaces with
constructive interference and then we have surfaces with
destructive interference. And that is going to be at the
heart of what I want to discuss with you now.
I will start a new drawing. I think we do not need this
anymore. And I will now have a screen
with two narrow openings. They could be slits,
as I said before. Here is one opening and here is
the other opening. And let the separation between
them be d. And the upper one we call
source number one, and the other one source number
two. Suppose now that we are looking
at this at a distance that is very far away.
Very, very far away. And where you are located in
space seen from this point is an angle theta.
The wave from this source will reach you.
And, since that point is very far away, these lines here are,
of course, parallel to each other, very close to parallel
because it is very far away. The wave that comes from this
one has to travel over the larger distance than the wave
that comes from this one. And that difference can easily
be expressed algebraically. 10:02
And that translates into a phase angle difference between
the E vectors. Theta.
First of all, d sine theta,
which is the past difference, this is how many wavelengths
you can fit on that past difference, so I divided by
lambda. And for each wavelength that I
can fit on there, the phase difference is going
to be 2pi. And so, the phase angle,
the difference between the wave from here and there is,
therefore, 2pi times d divided by lambda times sine theta.
And so, now I can set the condition for constructive
interference. What I want is delta to be a
multiple times 2pi. I already wrote that down
earlier, but I will do that again.
N can be zero, it can be plus or minus one,
it can be plus or minus two, etc.
If n is zero, by the way, you are right at
this plane that comes perpendicular out of the
blackboard. Clearly, when you are on this
plane, when n is zero, the distance to any point on
the plane form source number one and source number two is
obviously the same. 12:00
And now we can write down the condition for destructive
interference. Now you want that delta is pi,
3pi, 5pi. Now you have delta equals 2n
minus one times pi. And, if you want to write it
down in terms of the d sine theta then you get that d sine
theta of n, which is the n that you see there,
is now 2n minus one divided by two times lambda.
And so, this is at the heart of the idea of the interference,
in this case, of two of these sources,
two slits or two circular openings.
Let me check that we have that right.
D sine theta, the n makes connection with the
n, equals n lambda. Delta is a multiple of 2pi.
Destructive interference, we get 2n minus one times pi.
Yes, I can live with that. Now I want to be quantitative
because I am going to do a demonstration very much with the
numbers that you are going to see now on this blackboard.
I will give this back to you, although you may not need it.
We now have the two openings all the way on the left here.
And they are so close together that you cannot even separate
them anymore. This is the point where the two
openings are. And we are going to look on the
screen that is far away. Here is a screen,
which is going to be that screen.
And we want to see how the constructive interference,
that means light, shows up on the screen,
and how the destructive interference,
that means darkness, shows up on that screen.
And so here are these two openings.
Light comes in from the left. And let this distance be l.
And so, if I look under an angle theta, as seen from the
slits, I want to know what kind of pattern I will see.
Let's call this x equals zero. And we call this positive x.
And that is negative x. You can also say that the sine
of theta, if you want that, is approximately x divided by
L, but that is only true for small angles of theta.
And, with this kind of interference,
we sometimes have angles of theta which are not small at
all. I want to warn you that this is
not something you can use universally, but today you will
see that we can use it. I start now with red light.
Lambda 600 nanometers. That is the light I will use.
I use monochromatic light. And we will have a distance to
the screen, which is about five meters.
The setup is here. This is about five meters.
And we have light from a red laser which is a little larger
than 600 nanometers. But I just want some wrong
numbers. And the separation here,
d, between the two slits is one-quarter of a millimeter.
Now, you may think that it is rather small.
Well, you will see that that is what you need.
You need very small values of d to even see this phenomenon.
Now, let's evaluate what n is, let's evaluate what delta is
and let's evaluate what the sine is of theta of n.
And then we will evaluate what x of n is.
And we will do that for constructive interference.
The numbers that you are going to see here is for constructive.
And, of course, if you want to,
you can do the same exercise for destructive interference.
I need constructive interference.
So I have to turn to delta. And I make them n times 2pi.
I start with n equals zero. Well, that is easy.
N is zero, delta is zero, the sine theta n is zero and x
of n is zero. That is where you will see
maximum light, constructive interference.
Obviously, that is the plane that goes right through the
middle of those two openings. And the distance to each one of
those sources everywhere on that plane is, of course,
the same. And so, you expect here to see
light maximum. Now I go to plus or minus one.
And I calculate now what delta is.
Then I get plus or minus 2pi. And so the sine is now 2.4
times 10 to the minus 3. You can check that.
Because you know that d sine theta is n times lambda.
And I told you what lambda is. D sine theta is now n times
lambda. And we take n equals one,
and you know what lambda is, you know what d is,
so you can calculate the angle sine theta.
And that translates then into an x value of about plus or
minus plus n minus 1.2 centimeters.
Now I know what the sine of theta is, and I use the small
angle approximation which is more than adequate.
And I know now where on the this screen I get again maxima.
And that is then here, 1.2 centimeters on this side
and 1.2 centimeters on the other side because you get,
of course, plus as well as minus.
And then I can continue. For n equals one,
n equals two, I will just do plus or minus
ten. Then you get here plus or minus
20pi. Then you get here 24 times 10
to the minus 3. And so you get plus or minus 12
centimeters. It is almost linear because the
sine of theta, for these small angles,
is still the same as theta in radius.
Now, before I show the demonstration,
I want a little bit more information on the shape of the
dark and light areas that I am going to see.
What I plot here now is -- I will lower this and I will
raise it later because I want to work above my head so that you
can see what I am doing. I am plotting here the sine of
theta. And keep in mind it is linearly
proportional with x. But I always prefer to plot the
sine of theta. And so here is zero.
Let here be lambda divided by d.
That is where my first maximum will come, and you can see that.
And then here, at the same distance,
I have two lambda divided by d. And, on the other side,
I have minus lambda divided by d.
And all of these are maxima constructive interference.
And the destructive interference clearly falls smack
in the middle. I did not calculate them.
And so, the destructive interferences are here.
And so, you get light curves, light intensity just like this.
Light intensity, remember, is always the
pointing vector. This is light intensity.
And that is in watts per square meter.
Now, the first thing that I want you to notice,
and you will see that today in various demonstrations,
is that the location of the maximum depends on the
wavelength. So that means if red light,
650 nanometers, fall here as a maximum and it
falls here as a maximum and it has here a maximum and it has
there a maximum that blue light, which would have a wavelength,
say, I pick 400 nanometers, would not have its maximum at
the same location. Therefore, not the minimum
either. Because, look at the
relationship, d sine theta is n times lambda.
And so, the angles of theta are difference.
And so, you would get here, roughly, it is about two-thirds
down the way, the maximum for blue.
You would get here the maximum for blue.
Notice that for n equals zero, the red and the blue have the
maximum at the same location. But you would get the first
maximum, the blue here, you would get the second
maximum here and you would get the third maximum almost exactly
at the location where you get the second maximum for red.
The reason being that two times 650 is very close to three times
400. That is the reason.
Each lives a life of its own. The light intensity that you
see is a cosine square function of delta divided by two.
And this is so fundamental that first I thought I would ask you
to derive it, but then I decided to spend one
minute on it and divide it for you.
Because it is extremely important as we use that later
in our predictions. Let us assume that source
number one, and the other one I call source number two --
At source number one, E one has an amplitude E zero
one when it reaches the screen. And then, of course,
you have cosine omega t. And this is the frequency of
the light. That is the electromagnetic
radiation. E two, that is the other
source, has E zero two times cosine omega t.
And here is that crucial delta, which is the difference between
the two in phase. That is why you get maxima and
that is why you get minima. That is the whole idea of the
interference pattern. Now, if you were very far away
from the two openings then these E vectors are,
of course, very closely the same.
Remember E is proportional to one over r, but if the distance
r is very closely the same then we can assume to good
approximation that the E vector, the amplitudes are the same.
That means the total E vector that you get,
the vectorial sum of the two is then going to be 2E zero.
I simply call it E zero now. And then I get the cosine of
half the sum. I get the cosine of omega t
minus delta divided by two times the cosine of half the
difference, so that becomes the cosine of delta divided by two.
And now you see why the light intensity is proportional to
cosine square delta over two. Because, when you calculate the
mean value of the pointing vector, you have to take E cross
b. And so, b also has this term
and also has this term. And so, you get the square of
this term and the square of this term.
But the average of the square of this term is one-half.
Forget that for now. And so you see it is going to
be proportional to the square of delta over two.
And so this function that you see, delta is linearly
proportional to sine theta. This curve that you see is a
cosine square curve.
The experiment that I am going to show you has a separation of
the slits, which is one-quarter millimeter.
My wavelength is 633 nanometers, which is very much
the same. And the distance to the screen
is about five meters. And so the way that we have
this arranged is that the two slits, they are vertical,
they are like this. And so this separation is 1.4
millimeters. And then we put over there a
laser beam. And then the result is what you
are going to see there. And so I now have to think
about this is where it is. You are going to see it there.
And, in order for you to see it, I have to make it completely
dark.
When you look here on the screen, you will see areas which
are distinctly dark, and you will see areas whereby
you see the light. In other words,
you see the pattern that we just calculated.
You see constructive interference and you see
destructive interference. And you are looking here at an
experiment that is historically of great importance.
It is one of the most mysterious and puzzling issues
in all of physics. This interference pattern,
it was first shown by Young in 1801, is clearly convincing
evidence that light is a wave phenomenon.
However, 20th century physics has shown that light comes in
the form of individual photons with well-defined energies.
And photons can behave like bullets.
They have momentum, the produce radiation pressure,
and they can be localized in space.
We can detect individual photons in a way that we detect
individual tomatoes. Photons behave like particles.
However, the interference pattern that you were staring at
on that screen can only be explained if we assume that each
photon went through both slits. But how on earth can one
particle go through both slits? It will have to be one slit or
the other. And here lies the great
mystery. Light is both.
It is a wave, if you want it to be a wave,
and it is a particle if you want it to be a particle.
If you manage to determine for every photon through which of
the two slits it went, and that is possible if you do
it at very low light intensity, you can really determine
through which of the two slits each photon went.
Then you will not see any interference anymore.
The pattern on the screen will be like that of the tomato.
You will get two piles. The moment that you establish
the particle character of the light by determining through
which slit it went, you have destroyed its wave
character. And it will not behave like a
wave. It will behave like a particle.
However, as long as you do not determine through which slit the
photon went, as long as it remains your guess through which
one it went, light will reveal to you its wave character.
And you will see interference. The choice is yours,
but you cannot have it both ways.
And so you are looking now at something that had a huge impact
on physics. Clearly, at the time that this
experiment was done, it was accepted that light are
waves. But we now look at it in a very
different way. I want to show you a slide of
this what we call double-slit interference in white light.
We did it in red light. That is the easiest.
And you get very sharply defined maxima and minima.
And the reason why you get them so sharply defined is because
you do not have the interference of the other color.
But, of course, if you do it with white light
then you will the blue light has its own spacing and the red
light has its own spacing. And the net result,
of course, is that you see a somewhat different pattern.
And that I will show you here. I will make it a little darker
for that. The upper panel is then the
double-slit interference in blue light.
And you see the clearly dark lines, which are the destructive
inference locations. And you see,
as I already indicated on the blackboard, that the separation
of the red is larger because it is wavelength dependent.
And, if you then do it with white light, then you see
something like this. It is more difficult to
actually see the maxima and the minima.
Now, what you can do with light, if you can turn light
plus light into darkness then obviously you should also be
able to turn sound plus sound into silence.
It is just a matter of scaling up the whole experiment.
And that is what we have set up for you here.
We have two loud speakers. And so all the equations that
you have here apply verbatim, except that we scale it a
little bit. So our d is no longer a quarter
millimeter, but our d is now 1.5 meters.
Here is a source of sound and here is a source of sound,
and they are fed by the same electronics.
And so, the d, the separation is 1.5 meters.
And the frequency is 3000 hertz.
We do this experiment now for sound.
And so, lambda is about 11.3 centimeters if you take 340
meters per second for the speed of sound.
We have, in a way, monochromatic lights,
monochromatic sounds, more or less one wavelength.
And so you can calculate what lambda divided by d is.
It is about 0.075. And so you can calculate where
now those surfaces are. They are really hyperboloidal
surfaces, they are coming out from the center here and they
bend in your direction. Where all those hyperboloidal
surfaces are, where there is no sound.
And then you can calculate where all those high
hyperboloidal surfaces are where there is sound.
Let's first calculate how many hyperboloidal surfaces there
are. In order to do that,
we simply have to stick in our equation d sine theta equals n
lambda. We just put in theta is the
maximum value possible, and that is 90 degrees.
You put in theta max, and that is 90 degrees pi over
two radiance. And so, when you do that,
you will find that n max, that is the highest number of n
in that series 0, 1, 2, 3, 4, 5,
is then divided by lambda. Which, in our case,
is about 13. It is one over this number.
That means there are 13 of those surfaces going in this
direction, but there are also 13 going in this direction.
Remember, we have plus and minus?
We really have to multiply this by two.
And so, there are 26 surfaces, no more, and probably no less
in this audience of constructive interference.
And then there are probably also around 26 destructive
interferences. Now I want to be as qualitative
as I can be. And I would like to know the
students who are sitting about five meters away from me what
the distance is between one maximum, the zero order maximum
right at the center and the next maximum.
We get an idea where to search for those maxima and where for
those minima. 34:00
34:29 That separation x1,
I call it x1, is roughly L times that 0.075.
And that is about 38 centimeters.
Right there those guys I am looking at now,
you will have a maximum sound here.
And, when you move 38 centimeters, you will have
another maximum. And there is a minimum in
between. The minima and the maxima right
there is about 19 centimeters. Now, for you there in the back
it is larger. And for you here in front it is
smaller. Of course, remember those
surfaces go like this. And so here the separation is
smaller. I just give you this as an
example that you know roughly an order of magnitude of what you
are looking for. I am going to turn on this
sound now, and then I want you to wiggle back and forth,
as you did when you were in nursery school.
I want you to find the locations of maxima and minima.
And I can assure that the ones with minima, the ones with sound
silence are extremely well-defined because,
of course, we have done this several times ourselves.
Don't move too fast. Move very slowly.
And there will be points where you hear no sound and there will
be points where you hear a lot of sound.
Are we ready for this?
Keep moving. Most of you move too fast.
Who was able to clearly distinguish the maxima from the
minima? Not too many.
The reason is that all of you are lousy scientists.
[LAUGHTER] If the difference in space between maxima and minima
is 19 centimeters, that is this much,
what is the separation between your two ears?
It is about 20 centimeters. It is excellent that I chose
that. If you have a maximum here,
you would have a minimum there. What would I have to do?
You have to close one ear, and then you will be real
scientists. I am going to turn it back on
again. I want you to close one ear.
And then you will be able to find those minima and those
maxima. Move again back and forth.
Now, there may be some reflections from the wall which
may interfere with his experiment, no pun implied.
Go slowly. Go very slowly.
It is remarkable. Unbelievable.
I hear nothing here and here I hear a lot of noise.
So who heard it now? There you go.
All right. This is the great moment for
the mini-break. There is the mini-quiz.
If someone is going to help me --
I owe you something regarding Exam 2.
Here you see a histogram off Exam 2.
And, what is interesting and in a way helpful to me,
it is almost bimodal. Now, we have 47% to go in the
course of which 40% is for the final alone.
And so, clearly, I cannot say much about
dividing lines between a, b, c, d and f.
I want to make only a global statement.
And, I only warn those of you I put in the danger zone.
The danger zone for me are students whose cumulative,
assuming that they have taken all the exams,
is less than 30 or near 30. And those are really the
students, almost exclusively, who have less than 45 for Exam
2, or approximately 45. And this is the danger zone.
Now, that does not mean that you will fail the course.
It is just the matter of probability, I would say,
is high. And then those of you who are
dying to know whether they are going to get an A,
all I can say is if your cumulative now is more than 45,
assuming you have taken all exams, then you will probably
end up with an A. Will you end up with an A?
I do not know. It is all a matter of
probability, but the chances are you may.
That is as far as I can go regarding the dividing lines of
course grades. Now I will continue the idea of
interference which has very far-reaching consequences.
And I will turn to what we call thin film interference.
I will cover normal incidence. It really doesn't change very
much if you change to a different angle.
The concept is the same. I have here a thin film of oil.
And we will shortly see what I mean by thin.
This is air. We have plane waves of light
coming in this way, so it is normal incidence.
And this is n1 which is air, so that is about one.
And this is n2, which for oil is about 1.45.
Let's make it 1.5, just so we round it off.
And this is n3, which is again air in this
case, so let's make it one. It does not have to be air,
but I will make it air. And let the separation or the
thickness, I should say, of the oil be d.
And I call this surface A and this surface between oil and air
B. Light comes in,
normal incidence comes in from above over a large surface.
I just put in one arrow here. And let that intensity be i
zero. And now I am going to calculate
how much is reflected back into the air.
And I put the arrow here, so I offset it just for the
purpose of clarity. Some of it is reflected and
some of it goes straight through.
All of you are more than capable of calculating how much
is reflected. At that point A,
the r, the reflectivity of the E vector, is n1 minus n2 divided
by n1 plus n2. And that is minus 0.2.
The minus sign tells you that there is a flip here in the E
vector by 180 degrees. The intensity of the light that
is reflected is then 4%, the square of this number.
And, if 4% is reflected, clearly 96% go through.
If this is 100 then this 4.0. And then this is 96.
This light here hits the surface at b,
the boundary between oil and air.
And some of that light goes through, which I am not
interested in, and some of it comes back.
And so, I can calculate again the r value for that surface
near b, which is now n2 minus n3 divided by n2 plus n3.
This is n2. This is n3.
And so, that is plus 0.2. There is no flip in the E
vector as it returns. And the fraction of the light
that is reflected is, again, 4%.
However, it is 4% of the 96% because only 96% go through.
And 4% of 96% is 3.84. When it reaches this surface
here, again, 4% will be reflected and 96% will go
through. What comes out here then,
ultimately, is 96% of this number, which is 3.7.
I now want to ask the question, what happens when you are
looking here from above and you see this light coming back
straight from a and this light coming back at you,
which makes this journey through the oil,
whether or not the combination of those two can constructively
interfere and whether they can destructively interfere.
Now, as far as destructive interference is concerned,
let me make a qualitative statement.
The intensity of this is 4.0 and the intensity of this is
3.7. It can never be completely
darkness. But you can never kill 3.7
intensity with 4.0. But, if they are 180 degrees
out of phase, which we will be able to do,
there is very little light left.
There will only be the difference, which is 0.3 left,
so it will be quite dark. But I will still call that
constructive and destructive interference.
The question now is what is the phase difference between this
light here and this light there? Phase difference,
that is what matters. Well, let's first look at the
difference in path. The difference in path is
clearly 2d because this light has to go in this direction.
That is it has to come back. So that is 2d.
Now is the question, how many wavelengths fit on
that separation on that distance 2d?
Wavelengths in oil, of course.
That is what matters. That is this many.
What now is the phase difference?
Now you have to multiply this by 2pi.
Because, if this were three, then the phase difference is
three times 2pi. For each wavelength you get a
phase difference 2pi. Now you have to multiply this
by 2pi. Now you may think,
and that would not be unreasonable,
that this is the phase difference between this
radiation and this radiation. But what have you overlooked
when you do that?
That is right. You have overlooked the fact
that there is here a flip of 180 degrees which does not occur
here. There is an additional pi in
the phase difference, so I am going to write down
here now the total delta. Delta is now 4pi times d.
Now, I can write this as lambda oil.
It is perfectly fine. I can do that.
I would prefer to write it as lambda air.
And then all I have to do, of course, is define the lambda
air by the index of refraction of oil.
And so, yes, I could have written lambda oil
here, but I would prefer to write here lambda air.
47:00 But, if someone wanted to be
nasty to you on an exam, they could make n3 larger than
n2. What would change in this
equation? The pi would disappear.
That is all. The rest would remain the same.
Don't think of this as a universal equation.
There is no such thing. This pi is the result of the
fact that here n2 is larger than n1, but that here n3 is smaller
than n2. That is why you get only pi.
Now, we are in business. We now have here the criterion
for constructive interference.
Thank you very much. How could I?
It is d, right? Uh-oh.
My Greek heritage somehow. All right.
We agree that, in this case,
it is the phase angle. Now we are going to ask
ourselves the question what thickness are required to get
minima and to get maxima? The first very interesting case
is that if you make d equals zero that you kill this whole
term and you only have pi. That is destructive
interference. There is no question,
destructive. Now, you may say zero film,
what kind of nonsense is that? Is that meaningful?
Yes, it is very meaningful. I will demonstrate it to you.
I cannot make an oil film zero thickness, but that is not
necessary. All that is necessary,
that this d overlap the oil. It becomes exceedingly small.
For instant, 100.
If I make d 100 of the wavelengths of light in oil then
I am close enough. And you will indeed see then,
I will do this with soap, that in reflection you will see
no light. You get completely destructive
interference. Apart from the fact which I
mentioned that there is an imbalance between the 4.0 and
the 3.7. That is a different issue.
We will even be able to enjoy the d equals zero.
But now I want to be more general.
I want to give you a feeling quantitatively for what you are
going to look at, and also a feeling for what
thicknesses are really necessary to see these colors.
And so I am going to address now only the case of destructive
interference. And I am going to choose a
lambda in air of 400 nanometers. That is my choice.
That means that my lambda in oil is 400 divided by 1.5,
which is about 267 nanometers. And now, I want to calculate
for what thickness is d, I am going to get destructive
inference. I am going to set this equal to
2n minus one times pi. That is all I do.
And I choose n equals one, and what do I find?
I find d equals zero. Of course.
We knew that already. We just discussed that.
Because, if n equals one, then that requires that delta
is pi. And, if delta is pi,
then this term is zero. You see that d is zero.
You can immediately see that. Now we take n equals two.
And, if you substitute for n equals two in this equation,
you get 3pi. And this is already pi so this
has to be 2pi. And so, if now you calculate
what d is, you will find that d is 133 nanometers.
Well, that is obvious. It obviously must be half of
the wavelength of the light in oil.
Because, if the wavelength of light in oil is 267,
you have to travel this distance twice.
And, if you travel this distance twice,
that is exactly one wavelength. And one wavelength corresponds
to a phase angle of 2pi. And so, here you get 2pi.
And it is this pi that kills you.
That gives you the darkness. And then n equals three,
you will find that d is approximately 267 nanometers.
Now 2d translates into two wavelengths.
Now I am going to choose this thickness.
And, with that thickness now, I am going to work.
And I am going to radiate white light onto depth thickness.
I select that one for which I already know that in blue light
I kill it, destructive interference.
We now have white light. And we have d 133 nanometers,
so that is nonnegotiable. D is 133 nanometers.
I am going to write here lambda air in nanometers,
I am going to write here delta, and I am going to write here
the cosine squared of delta over two.
That is a measure for the light intensity.
It is proportional to that. 54:00
We have green. Now I am going to also shine on
it green light 500 nanometers. I know what d is.
That is nonnegotiable. And now my lambda air is now
500 nanometers. And so I can calculate now what
delta is. Untouched by human hands,
very simple. And I find now 2.6pi.
And, if now I take cosine square of delta over two,
I find 0.35. But now I go to red and I do
650 nanometers. Now I calculate what delta is,
and I find 2.2pi. Now I get 0.90.
Now I am very close to a maximum.
And so this is very convincing that if you make your film this
thin that it will look distinctly red.
Maybe not perfect red, but it will look distinctly
red. The red color dominates and the
blue, practically absent at all, apart then from the imbalance
that you may have between the 4.0 and the 3.7.
The question that I have asked over the years at the final exam
for 8.03 is whether thin film interference is the result of
the difference between the index of refraction between the
different colors? And the answer is absolutely
not. The index of refraction,
I have taken the same for all colors, for all practical
purposes. That difference does not
explain the colors. The color is explained by the
different path length divided by lambda.
That is why you see colors. It has nothing to do with the
index of refraction. You can see thin film
interference in soap, and I will try to demonstrate
that. You can see it in oil spills on
the road. I will show you an example.
An only thin film gives you colors.
Thick film does not give you colors.
If you go through the exercise, which I want you to do,
to make d 0.1 millimeter, which is, by practical
standards, still very thin, you will not see any colors.
And the reason is that if you go to this equation and you
calculate for which colors, you will see constructive and
destructive interference. There are so many colors in the
visible spectrum for which you get constructive interference,
so many, you will see, that your brains will tell you
that you see white lights. There is not one color that
dominates. The n values that you will
need, by the way, are going to be very high.
They are in the range 500 to 700.
Huge values of n are necessary in order to get the destructive
and the constructive interference.
Go through this exercise on your own, and you will see very
quickly that so many colors constructively interfere that
the film will look white. The first thing that I want to
do, I want to try, it is not that easy,
is to make you see these colors with soap, which all of you must
have seen. In fact, if you just take a
shower and you soap yourself, the bubbles themselves in
reflection already have these colors.
I will try to make one that is slightly larger in size.
But we do not always succeed. Let's try that.
The idea is I want you to see colors.
Well, maybe you did but that was a little quick.
Now you see colors. Did you?
That was all I wanted you to see.
Did you see them? If you had said no,
I would have done it again, but now I won't.
Do you want me to do it once more?
It is not easy, though.
Now you say it, right?
Yeah. Thank you.
Soap bubbles give these colors because they are exceedingly
thin. It also gives you a feeling of
how thin they have to be. Really, they can be much
thicker than a few times the wavelengths of light.
If they become much thicker, you lose the colors,
for the reason that I just mentioned to you.
There are too many colors that constructively interfere,
and then you don't see it anymore.
Now I have a demonstration which is very tricky.
It works most of the time but not all the time.
I am going to make a soap film on a metal frame.
It is going to be done in this box.
We have a metal frame. And we dip it in soap,
and so you get a soap film there.
And then gravity will make the film thinner at the top and
thicker at the bottom. As you make a cross-section
through here then the film for this is then the soap.
I can make this so thin that it is completely dark in
reflection. That is what I promised you
that d is not zero. But d divided by lambda is so
small, maybe 120 or 130 of the wavelength of light,
that it will turn completely black.
And, as this proceeds, in the beginning the top will
show you colors. It goes through these phases of
colors, then it gets thinner and thinner and thinner and then it
turns completely black. I will show it to you upside
down. But don't get confused.
It has to do with the way that we project it.
The thinnest part of the film will be low on the screen,
and then the part of the screen that is the thickest of the film
will be the upper part. As you look at it,
appreciate the fact that when we go into the dark phase,
if we succeed, that you are talking about the
thickness of the film, which is many times smaller
than the wavelengths of light, which is typically a half a
micron. Let me first get the film up.
Actually, let's not get the film up.
Let's just get the light up. Let's turn it off.
It has to be completely dark. Didn't we agree it has to be
completely dark? I am going to dip it in here.
The bottom is the thinnest. And you see the colors.
You see them? And you see that the thickness
changes over the film, and that is why there are areas
where the red dominates and then other areas where the red
dominates again. And so, the bottom gets thinner
and thinner and thinner. The white light that you see
now here is the result of Markos' trick.
In order to avoid that the film bursts, he has to put glycerin
in the soap. And what you see here is a
reflection of the layer of glycerin that forms on top of
the soap. We will have to wait a little
bit for that glycerin to also go down.
And then you will see the pure soap.
And that is then the area. And, boy, look at it,
black. That area that you are looking
here now is much thinner than the wavelength of light.
And so, the reflection is, well, when I say zero,
of course, that depends again on the asymmetry that you have
that we discussed here on the blackboard between the 4.0 and
the 3.7. It is a little different for
water than it is for oil. You see?
Black. You can just look at it for a
while. It may burst.
You also see some very interesting thin film
interference in the glycerin itself, which is the oil layer
that is on top of the soap. It is not uncommon.
You may remember that when you see oil spills on the road,
that you sometimes see these very nice structures,
brownish, reddish, bluish.
But look at this here. Isn't that incredible?
Isn't that fabulous? Boy, the film is holding up,
Markos. We cannot complain.
You see these nice oil structures here in the glycerin?
The nightmare that we have always is that it pops too fast,
and then you have to do again and again and again.
And our experience is if somehow the soap is not right
then, in general, no matter how often you try it,
it keeps popping. Look how beautiful,
look at how incredibly nice the [CNT?] film interference in the
glycerin. And then, here,
the total darkness because the thickness of the film,
of the soap is many times smaller than the wavelength of
the light in the soap.
I never get bored when I look at it.
Isn't it gorgeous? Think about the physics that is
going on in there.
All right. Let's look at some other
phenomenon of interference. Try to remember this.
This is a gorgeous picture.
There is a whole family of ways that you can see [CNT?] film
interference. And, in your problem sets,
we give you two problems to work on that quantitatively.
And there is also this take-home experiment that
supports the problem. And I want to concentrate on
those, too, because they are very nice to demonstrate.
I can demonstrate them to you in monochromatic light.
Whenever you have monochromatic light, that means you do not
have white light. Of course, the fringes,
we call these fringes, the dark and the bright areas,
we call them fringes, show up so much better than
when you have white light. Because then you get the
superposition of all the colors. And the maxima and the minima
depend on the color. Remember, from one and the same
thickness, you may have destructive interference for one
color, but you have constructive interference for the other.
If you do it in monochromatic light, you can see this very
dramatically. And then one of the classic
examples is, and you can do that at home, and you should do it at
home, is the two microscope slides.
You have a microscope slide which is way too thick to give
you thin film interference. Don't even think about the fact
that there is glass there. It will do nothing.
Way too thick. Probably tens of a millimeter.
It is ridiculous. And then there is another
microscope glass here. And I exaggerate highly this
area in between because they are lying flat on top of each other,
but there is an air gap between them.
And this is that air gap. And it is this air gap now that
acts like a thin film. You are going to get
interference now of light that bounces off this layer and
bounces off this layer. And so, it is this thickness d
now of the air layer that determines the darkness and the
brightness, destructive and the constructive interference.
And you're being asked, in your problem set,
to calculate where those maxima and minima are.
And you can also do that then with the take- home experiment.
And I will demonstrate it very shortly.
There is another one, a classic that I will also
demonstration and which is also part of your problem set.
This is a glass lens. The curvature is highly
exaggerated. And then here is a glass plate.
And so here is now an air gap. And now, if you shine
monochromatic light on this then, of course,
seen from above there is symmetry about this line.
You will see rings of darkness and rings of brightness.
And these rings of darkness and brightness will be very far
apart here and they will be very closely together there.
This is part of your homework assignment to show that.
It has to do with the fact that the change here in thickness is
very little because of the curvature and the change here is
much faster. This is called the Newton
rings. And then, when you just walk
out on the street, preferably on a day when it is
not sunny but is cloudy, you look on the ground and you
see oil spills. And they have phenomenal
colors. I would like to show you one of
my own pictures that I took some time ago of an oil spill.
Let me first turn it on, and then we will make it a
little darker.
I will make it a little darker so that you can see it better.
I photograph countless oil spills.
And, actually, the reason why I photograph so
many is that always I try to reconstruct thickness of the oil
on the road. And I was never very
successful. For one thing,
you can only get a spot like this if a car loses some oil.
So there is one drop of oil and then it starts to run out.
You would always think that it is thickest here and then
gradually tapers off near the edges.
And I have always wondered why then here you see so much white
light. You would expect that perhaps
it would be much darker. And so, many attempts that I
have made to convince myself that it is really thicker here
than there, many have failed. But there is no question what
you are looking at, of course, is a striking
example of thin film interference.
And, not only that, but you know that this film has
to be extremely thin, clearly very close in that
range of 100 to 200 nanometers. Otherwise, you would not see
over the entire area such beautifully equal blue.
And here you see this reddish that is not pure,
it never is, and very well-defined.
Clearly, it has a spherical symmetry so it is probably a
drop that then spread out. And so, now I want to
demonstrate then your microscope.
The microscopes we have here, the microscope glass.
Turn on the CD-TV there, if everything works.
And we do it with [neo?] monochromatic light,
this light from a mercury lamp which is largely in the green.
So you see very well-established maxima and
minima. That is the beauty,
if you do it with monochromatic light.
And so these two microscope slides are just on top of each
other. But, yeah, there is always a
little bit of air in between. And so, you can sort of make up
your own mind that they may be touching each other,
really touching where they have practically no air,
for reasons that are not so clear, which must be somewhere
here. And that is where the
separation between the dark lines is the largest.
And then, somehow here, you see these fringes,
the areas of darkness and maximum constructive
interference. When I push on one side of
these slides and, of course, I change this very
subtle structure of light in between these microscopes,
you will then see immediate change in the separation of the
lines. I am just pushing on it now,
changing the configuration.
Now this is the point of contact, more or less.
And you can really see that. What I am looking at right now,
I see it just as well as you can see it there.
And so, you should really do that experiment at home because
it is part of your take-home experiment.
I mentioned in the write-up that it is not necessary that
you turn in a written summary of your experiment,
so it won't count for your problem set as such,
but I will hold you responsible on the final.
So I advise you to do the experiments because I may ask a
few questions which you can only answer if you actually did this
experiment. Here I have two flats,
like the microscopes, but thicker so they are a
little sturdier. And this is the way opticians
actually test their optics. There is here an air gap that
grows in size. And when I push now on it,
you see now I make them flat. Now I see almost no fringes.
Now I push so hard that I open them again like this.
I make one go up like this. And now you see many fringes.
You've got a huge angle now. And then I want to show you
this, which is called the Newton rings.
We have it arranged in such a way that with some set screws we
can actually push down on this, so we can also make the air gap
larger and smaller. And that we are going to show
you there.
I can actually make it even darker.
That may be nicer for you. You already see the center
portion of the contact. Where the contact is here,
the air gap doesn't change very much.
And, therefore, you will see that the spacing
of the rings is large. But, when you go further out,
then the spacing gets smaller and it becomes difficult to see
perhaps. But, if you are close here,
I can really see the rings here, much closer than the rings
here. And I can put some pressure on
there, even though it has changed the geometry a little
bit. Now I squeeze a little harder.
I hope I don't break it. And when you squeeze harder
they touch each other over a larger surface in the center.
You see? Therefore, it opens up.
Over this very large area now the gap is the same.
And, when I loosen it up, then you expect those rings to
become smaller. And this is something,
I believe, that is also part of your take-home experiment.
I cannot wait to see the result of the mini-quiz.
I will be proud of you if you did well, let me tell you.
And I see then, I hope, on Thursday.