I'm very proud of you. You did very well on the last
exam. Class average is a little bit
above seventy. Congratulations.
There were twenty-two students who scored a hundred.
Many of you are interested in where the dividing line is
between C and D. If I take only the three exams
into account, forget the quizzes,
forget the homework, forget the
motor, and you add up the three grades of your three exams,
the dividing line between C and D will be somewhere in the
region a hundred thirty-five to a hundred thirty-eight.
So you can use that for your calibration where you stand.
The controversy between Newton and Huygens about the nature of
light was settled in eighteen oh one when Young demonstrated
convincingly that light shows all the
characteristic of waves. Now in the early twentieth
century, the particle character of light surfaced again and this
um mysterious and very fascinating duality of being
waves and particles at the same time is now beautifully merged
in quantum mechanics. But today I will focus on the
wave character only.
Very characteristic for waves are interference patterns which
are produced by two sources, which simultaneously produce
traveling waves at exactly the same frequency.
Let this be source number one and let this be source number
two. And they each produce waves
with the same frequency, therefore the same wavelength,
and they go out let's say in all directions.
They could be spherical, in the case of water surface,
going out like rings. And suppose you were here at
position P in space at a distance R one from source
number one and at a distance R two from source number two.
Then it is possible that at the point P the two waves that
arrive are in phase with each other.
That means the mountain from two arrives at the same time as
a mountain from one, and the valley from two arrives
at the same time as the valley from one.
So the mountains become higher and the valleys become lower.
We call that constructive interference.
It is also possible that the waves as they arrive at point P
are exactly a hundred eighty degrees out of phase,
so that means that the mountain from two arrives at
the same time as the valley from one.
In which case they can kill each other, and that we call
destructive interference. You can have this with water
waves, so it's on a two-dimensional surface.
You can also have it with sound, which would be
three-dimensional. So the waves go out on a
sphere. And you can have it with
electromagnetic radiation as we will
also see today, which is of course also three
dimensions. If particles oscillate then
their energy is proportional to the square of their amplitudes.
So therefore since energy must be conserved,
the amplitude of sound oscillations and also of the
electric vector in the case of electromagnetic radiation,
the amplitude must fall off as one over the distance,
one over R. Because you're talking
about three-D waves. You're talking about spherical
waves. And the surface area of a
sphere grows with R squared. And so the amplitude must fall
off as one over R. Now if we look at the
superposition of two waves in this case at point P and we make
the distance large so that R one and R two are much much larger
than the separation between these two points,
then this fact that the amplitude of
the wave from two is slightly smaller than the amplitude from
the wave from one can then be pretty much ignored.
Imagine that the path from here to here is one-half of a
wavelength longer than the path from here to here.
That means that this wave from here to here will have traveled
half a period of an oscillation longer than this one.
And that means they are exactly a hundred eighty
degrees out of phase and so the two can kill each other.
And we call that destructive interference.
And so we're going to have destructive interference when R
two minus R one is for instance plus or minus one-half lambda
but it could also be plus or minus three-halves lambda,
five-halves lambda, and so on.
And so in general you would have destructive interference if
the difference between R two and R one is two N plus one times
lambda divided by two whereby N is an integer,
could be zero, or plus or minus one,
or plus or minus two, and so on.
That's when you would have destructive interference.
We would have constructive interference if R two minus R
one is simply N times lambda. So then the waves at point P
are in phase and N is again could be zero,
plus or minus one, plus or minus two,
and so on. If the sum of the distance to
two points is a constant you get an ellipse in mathematics.
If the difference is a constant, which is the case
here, the difference to two points is a constant value,
for instance one-half lambda, then the curve is a hyperbola.
It would be a hyperbola if we deal with a two-dimensional
surface. But if we think of this as
three-dimensional, so you can rotate the whole
thing about this axis, then you get hyperboloids,
you get bowl-shaped surfaces. And so if I'm now trying to
tighten the nuts a little bit, suppose I have here two of
these sources that produce waves and the separation between them
is D, then it is obvious that the line right through the
middle of them and perpendicular to them is always a maximum if
the two sources are oscillating in phase.
So this line is immediately clear that R two minus R one is
zero here. If the two are in phase.
And they always have to generate the same frequency of
course. So this line would be always a
maximum. Constructive interference.
It's this zero substitute there.
And in case that we're talking about three-dimensional,
this is of course a plane. Going perpendicular to the
blackboard right through the middle.
The difference R two minus R one equals lambda would again
give me constructive interference.
That would be a hyperbola then, R two minus R one equals
lambda, that would again be a maximum, and you can draw the
same line on this side, and then R two minus R one
being two lambda again would be a maximum.
And again if this is three-dimensional you can rotate
it about this line and you get bowls.
And so in between you're obviously going to get the
minima, the destructive interference,
lambda divided by two, and then here you would have R
two minus R one is three-halves lambda.
We call these lines where you kill each other destructive
interference, we call them nodal lines or in
case you have a surface it's a nodal surface.
And the maxima are sometimes also called antinodes,
but I may also refer to them simply as maxima.
And so this is what we call an interference pattern.
If you look right here between on the line between the two
points then you should be able to convince
yourself that the linear separation here between two
lines of maxima is one-half lambda.
Figure that out at home. That's very easy.
Also the distance between these two yellow lines here right in
between is one-half lambda. And so that tells you then that
the number of lines or surfaces which are maxima is very roughly
two D divided by one-half lambda.
So this is the number of maxima, which is also the same
roughly as the number of minima, is then approximately two D
divided by lambda. And so if you want more maxima,
if you want more of these surfaces, you have a choice,
you can make D larger or you can make the wavelength shorter.
And if you make the wavelength shorter you can do that by
increasing the frequency if you had that control.
The first thing that I'm going to do is to make you see these
nodal lines with a demonstration of water.
We have here two sources that we can tap on the water and the
distance between those two tappers, D, is ten centimeters,
so we're talking about water here.
Uh we will tap with a frequency of about seven hertz and what
you're going to see are very clear nodal lines,
this is a two-dimensional surface, where the water doesn't
move at all. The mountains and the valleys
arrive at the same time. The water is never moving at
all. So let me make sure that you
can see that well. And so I have to change my --
my lights. I'll first turn it on,
that may be the easiest. Starts tapping already.
I can see the nodal lines very well.
So here you see the two tappers and here you see a line whereby
the water is not moving at all. At all moments in time it's
standing still. Here's one.
Here is one. And you even with a little bit
of imagination can see that they
are really not straight lines but they are hyperbolas.
If you're very close to one tapper, the zero can never be
exactly zero, because the amplitude of the
wave from this one then will always be larger than the
amplitude from that one, because as you go away from the
source the amplitude must fall off on a two-dimensional surface
as one over the square root of R.
In a three-dimensional wave must fall of as one over R.
But if you're far enough away then the distance is
approximately the same and so the amplitudes of the individual
waves are very closely the same and you can then like you see
here the water is absolutely standing still.
And here are then the areas whereby you see traveling waves,
they are traveling waves, they're not standing waves,
that here you see if you were sitting here in space the water
would be up and down, bobbing up and down,
and the amplitude that you would have is twice the
amplitude that you get from one, because the mountains add
to the mountains and the valleys add to the valleys.
But if you were here in space you would be sitting still.
You would not be bobbing up and down at all.
And that is very characteristic for waves.
If I were to tap them a hundred eighty degrees out of phase,
which I didn't -- they were in phase -- then all nodal lines
would become maxima and all maximum lines
would become nodes, that goes without saying of
course. It is essential that you --
that the frequencies are the same, that is an absolute must.
They don't have to be in phase, the two tappers,
if they're not in phase then the positions in space where you
have maxima and minima will change but a must is that the
frequency is the same. Now I was hiking last year in
Utah when I noticed a butterfly in the
water of a pond which was fighting for its life.
And you see that butterfly here.
Tom, perhaps you can turn off that uh overhead.
You see the butterfly here, and you see here projected on
the bottom the beautiful rings dark and bright,
because these rings on the water act like lenses,
and what you see very dramatically is indeed what I
said, that the amplitude of the wave must go down with distance,
because energy must be conserved of
course in the wave, and since the circumference
grows linearly with R, the amplitude must go down as
one over the square root of R because the energy in the wave
is proportional to the amplitude squared.
So when I saw this it occurred to me that it would be a good
idea to catch another butterfly, put it next to it,
and then photograph -- make a fantastic photograph of an
interference pattern. But I realized of course
immediately having taken eight oh two that the
frequencies of the two butterflies would have to be
exactly the same and so I gave up the idea and I decided not to
be cruel. So no other butterfly was
sacrificed. If we look at the directions
where we expect the maxima as seen from the location of the
sources, then I want to remind you of what a
hyperbola looks like. If here are these two sources
and here is the center I can draw a line here,
then a hyperbola would look like this.
Let me re- remove the part on the left, doesn't look too good,
but it's the same on the left of course.
And what you remember from your high school math,
that it approaches that line. And therefore you can define
angle theta as seen from the center between these two which
are the directions where you have maxima and where you have
minima. And that's what I am going to
work out for you now on this blackboard here.
So here are now the two sources that oscillate,
there's one here and there's one
here and here is the center in between them,
and let this separation be D. And I am looking very far away
so that I'm approaching this line where the hyperbolas merge
so to speak with the straight line.
And so I look very far away without being -- committing
myself how far, I'm looking in the direction
theta away. This is theta.
And so this is theta. And I want to know in which
directions of theta I expect to see maxima, and in which
direction I expect to see minima.
So this is what we called earlier R one and we called this
earlier R two, it is the distance to that
point very far away. If I want to know what R two
minus R one is that's very easy now.
I draw a line from here perpendicular to this line and
you see immediately that this distance here is R two minus R
one. But that distance is also you
realize that this angle is theta it's the same one as that one,
so that distance here is also D sine theta.
And so now I'm in business, I can
predict in what directions we will see constructive
interference. Because all we are demanding
now, requesting, that R two minus R one is N
times lambda. And so we need that D sine
theta and I'll give it a subindex N, as in Nancy,
equals N times lambda. In others words that the sine
of theta N is simply N lambda divided by D.
And that uniquely defines all those directions,
the whole zoo of directions N equals zero, that is the center
line, N equals one, N equals two,
N equals three, and so on.
And then I have the whole family of destructive
interference. Which would require that lambda
R two minus R one which is D sine theta must now be two N
plus one times lambda divided by two.
Just as we had it on the blackboard there.
We discussed that earlier. And so that requires then that
the sine of theta N for the destructive interference is
going to be two N plus one times lambda divided by two D.
So this indicates the directions where
we expect maxima and where we expect minima as seen from the
center between the two sources. But now I would like to know
what the linear distance is if I project this onto a screen which
is very far away. And so let us have a screen at
a distance capital L which has to be very far away,
so here are now the two sources.
It's a different scale. And here is a screen.
And the distance b- from the two sources to the screen is
capital L. And here is one of those
direction theta. And you see immediately that if
I call this the direction X, X being zero here,
that the tangent of theta is X divided by L.
If but only if I deal with small angles,
the tangent of theta is the same as the sine of theta.
And therefore I can now tell you where the maxima will lie on
that screen away from the center line, which I call zero,
that is now when X of N is L times the sine of theta,
in small angle approximation. So this is approximately L
times N lambda divided by D, and for the same reason you
will get here c- destructive
interference when X of N is going to be L times two N plus
one times lambda divided by two D.
That is simple geometry. So now we have all the
ingredients here on the blackboard and I'm going to
leave it there for the rest of the lecture.
Whenever we're going to do an experiment with two sources
which are in phase at the same frequency, you can predict the
directions of maxima and minima and you can
even predict the separation, the linear separation,
if you know how far away you are from these sources.
And the first demonstration that I'm going to do is with
sound. We have here two loudspeakers.
And the distance between those two loudspeakers,
we're going to do it with sound, D, is one point five
meters. That's
a given. And the frequency is three
thousand hertz. The wavelength therefore,
lambda, equals V divided by the frequency, the speed of sound is
about three hundred forty meters per second, divided by three
thousand, is about oh point one one three meters.
So the wavelength is about eleven point three centimeters.
I can now calculate everyone who is sitting here
fweet right in the middle through this whole plane will
have a maximum of sound, and then when we go away at
angle theta, some will again have maxima, and we go further
away theta, again maxima, and in between will be the
minima. And I'm going to calculate
where they fall in the lecture hall.
The first thing that I'm going to do is I'm going to give you N
as in Nancy and calculate that angle theta
of N and I will do it for the maxima.
In other words, I'm going to use constructive
interference and you see that the sine of theta N is lambda
divided by D. That's the equation I use.
When N is zero the angle is zero.
That is zero angle. Everyone here will hear a
maximum. When N is one,
and you may want to check that at home, I find an angle of four
point three degrees, and when N is two,
the angle is about double that, is about eight point seven
degrees, and when N is three, it should be close to thirteen
degrees, thirteen point one. In case you take N is ten,
so I skip a few, you get about forty-nine
degrees. This is where the maximum fall.
And so there's going to be a maximum here and then four point
three degrees away is again a maximum.
But surely we would like to know how far you in the
audience will have to move in order to go from a maximum to a
minimum. And so the way you have to
think of this is that if I make here a picture of the lecture
hall, if here are these two sources, you are at a distance L
away from here. Some of you are five meters
away. Some are ten meters away.
Some are fifteen meters away, all the way in the back of the
audience. And you want to know where
you're going to hear the maxima. I call this X one,
I call this X two, and I call this X three,
and this is zero. So this is the meaning of theta
one. And this is the meaning of
theta three, and this angle here would be theta two.
That's the meaning of these angles.
And so I can calculate now how far
you have to move depending upon what capital L is to hear to go
from one maximum in sound to another maximum.
And we raise the a little more. And so I will show you now s-
some of the results for maxima. So I only go now for
constructive interference. And I have done this for three
different distances. Those of you who are five
meters away from me, ten meters away from me,
and fifteen meters away from me.
And what you see on the left side is going to be X,
that is the linear separation, and these, so these were in
meters, forgive me but I will do these in centimeters.
And this is X one, if you are five meters away
from me, you will have -- I will put X one a
little lower than I have it now, you will see shortly why I
put it a little lower -- X one this is about thirty-eight
centimeters. So the linear separation from
one to the next is thirty-eight centimeters.
And you're ten meters away, it's double that,
that's no surprise, seventy-six centimeters.
And if you're fifteen meters away it is a hundred and
thirteen centimeters. And then X two which is the
position where you have another maximum would be at seventy-six
centimeters and it would be at a hundred and fifty-two
centimeters and it would be at two hundred and twenty-eight
centimeters if you're fifteen meters away from me.
So the minima will fall almost exactly in between and so the
minima where in an ideal case there is no sound at all,
sound plus sound gives silence, think about it,
sound plus sound will give silence, will be when you are
roughly at nineteen centimeters, half of this,
this will be thirty-eight centimeters.
Half of this, and here will be something like
fifty-seven centimeters. And you can calculate what
these values are, they are exactly in between.
And so the conclusion is that if you're five meters away from
me and you're near the center line,
but you can also be a little bit in this direction,
that the separation between bright sound,
loud sound, which is always at zero of course in the middle,
to silence is nineteen centimeters.
And then you move another nineteen centimeters and then
you hear loud sound. If you are however ten meters
away from me, just past the cameras,
then you have to move thirty-eight centimeters to go
from loud sound to silence. And if you're all the way in
the audience, in the back of the
audience, it's more like sixty centimeters.
And this is what we're going to do now together.
I want you all to stand up and I'm going to make you listen to
three thousand hertz. And what I want you to do when
I turn on the two loudspeakers, I want you to move your head
very slowly and try to find locations where you hear
silence. The position of silence is
extremely well-defined, so don't go too fast,
you miss it, also keep in mind that there
are reflections of the sound from the walls,
and from the blackboard, and so the pattern that I have
calculated here is not perfect. But you will see that there
will be locations where sound plus sound will give you
silence. [tone]
[tone] You are a couple of lousy scientists.
If the separation between a lot of sound and silence is nineteen
centimeters, that's about the separation of your ears,
you dummies, so one ear could be at a
maximum, the other ear could be at a minimum,
so at least close one ear. [laughter] [tone] Go very
slowly. [tone]
Who has found clear location where [tone] the sound is nearly
zero [tone] or practically zero? [tone] Most of you.
And you certainly can hear if you move that there's an
enormous difference [tone] in sound intensity.
[tone] [tone] So again, who has found locations [tone]
whereby you clearly say this is practically silence?
[tone] Ah, you see them all the way in the back,
[tone] [tone] and the separation, how far you have to
move, depends on how far you are away from me.
Sit down again. [audience noise] Young was a
sound engineer and as a sound engineer he was very familiar
with the interference of sound. He knew that sound and sound
can make silence. And so in eighteen oh one he
demonstrated in a convincing way that light plus light can create
darkness. That would be the nail in the
coffin that would demonstrate uniquely
that light are indeed waves, and there was still this
controversy between Huygens and Newton as you perhaps remember.
Newton wanted light to be particles but Huygens wanted
them to be waves. And the way that Young did his
experiment is as follows. He had a screen,
don't think of it as this big, you're talking now about
extraordinarily small dimensions, you will understand
shortly how small, and in this screen are
two openings, two pinholes,
and light is coming from the left and think of light as being
plain waves. They reach these two openings
and these two openings according to Huygens will produce circular
waves, spherical waves of course, three-dimensionally.
These openings become Huygens sources and spherical waves will
propagate out in this direction. And so now we have exactly the
situation that we had with our sound.
Now if all works well there should be directions theta away
from this line where you see darkness and other directions
where you see bright light. And we are going to do it in a
way, we have the luxury of laser beams, so we have very strong
light sources, which Young did not have.
The way we are going to do it, we have a -- a slide,
which is completely black, but with a razor blade two
lines have been drawn on it. And so I will draw these lines
as white lines. But they're really openings.
And there is another one here. And the separation between
these lines D is oh point oh eight eight millimeters,
less than a tenth of a millimeter.
When you look at them you cannot even see that they are
two lines. Our laser beam has a diameter
of about three millimeters, which is thirty times larger
than this distance, thirty times larger.
So what I'm going to show you now that this is our laser beam,
is not to scale, the laser beam is much larger
than that. And so the light will go
through some parts of these slots, as far as our laser beam
reaches, and we are now capable of predicting when we're going
to project it there, here are the two slots which
are like so, and so you're going to get
interference patterns in these directions theta,
and we can calculate what the position X is going to be there
between the maxima. And so if that is the screen
and if this is X equals zero, and if this is X one and this
is X two, and of course the whole thing is symmetric,
you can always go in the opposite direction,
you can now calculate, and you have all the tools,
I did it for you in great detail using sound,
but you have all the tools to do it now, you know D,
I'm going to tell you what lambda is, it's six thousand
three hundred and twenty-eight Angstroms, and one Angstrom is
ten to the minus ten meters, so you can calculate all the
direction theta for which there are maxima and for which there
are minima. Minima means light plus light
gives darkness, an amazing concept.
And you can then if you know the distance from
here to the screen, which is capital L,
you can calculate what the separation is as we see it on
the screen, and L is roughly ten meters, maybe eleven,
but that's not so important. And so I calculated,
and you can confirm that -- and you should confirm that that the
angle theta one, I will only calculate theta
one, which is the angle then to this point, theta zero is of
course always zero, right, that's the easiest,
I find that theta one is oh point four one degrees,
that is for maximum, and that means that X one given
the distance of ten meters then becomes seven point two
centimeters. So from here to here on the
screen, from maximum to maximum, will be about seven point two
centimeters, and from here to here will then of course also be
about seven two centimeters, and in between you will see
darkness. The light from the two sources,
a hundred eighty degrees out of phase, and that will give you
darkness. Let me turn on the laser.
And turn off the lights. Make sure I have my -- OK.
And there you see it. There you see a maximum,
darkness, a maximum, darkness,
a maximum, darkness, and so on.
And the separation if I didn't make a mistake between the
maxima is indeed about seven centimeters.
Imagine what an incredible moment this is in your life,
that you actually see that light plus light can make
darkness. So the waves go simultaneously
through both openings, and each opening acts like a
Huygens source, and the net result is that
these two waves arrive there on the screen a hundred eighty
degrees out of phase at the locations of darkness.
The censor is of course that they have exactly the same
frequency which is what they do, because we have one laser gun
going in and so the wave goes through both slots.
So we're guaranteed, and that was the secret,
that Young understood you're guaranteed that the waves are
not only the same frequency but they're even in phase
because they both go through to both slots.
Now if you look very carefully here you will see of course that
these maxima don't have the same strength.
We will understand next lecture why that's the case.
They would have very closely the same strength if the opening
where we scratched out the black on the slide was much much
smaller than the separation between the two slots,
so to speak. And that separation is
point oh eight eight millimeter.
If we make the openings much narrower, indeed,
the light intensities would be more uniform,
each maximum would be approximately the same strength,
but then very little light will go through.
And so it's a tradeoff. And the moment you make these
two openings, these two slots,
larger and larger, you will understand Friday why
then the light intensities are not the same,
why the light intensity is a maximum at the center and then
falls off near the edge.
As you see. It's a maximum here,
and then the light intensities become smaller.
I've shown you now the interference pattern with sound
and for red laser light, but imagine now that I did the
same with white light. The situation would be very
different, and maybe even disappointing for you.
Let this be the location on the screen.
So we -- we have X here, and here X is zero.
And I want to know where the maxima are in the red,
well that's very easy, there will be a maximum here
when this position is L times lambda divided by D.
This is when Nancy is one. And there will also be a
maximum here when we have two L times lambda divided by D.
And of course there will be one on this side same distance.
And there will be one here, this is when N is zero.
N is one. N is two.
The red light will have maxima. How about the blue light?
The blue light will have maxima here, where L lambda
divided by D, but lambda is different,
lambda for blue light is smaller.
Substantially smaller than red light, so the maximum of the
blue will fall here, the maximum of the blue will
always fall at N equals zero together with red,
and then Nancy equals two the blue will fall here,
so this is Nancy two, Nancy one, Nancy zero.
And here Nancy zero, Nancy one, Nancy two.
And so the red and the blue and therefore all the other colors
live a life of their own. They don't talk to each other.
They come in with their own separation in terms of angles
and in terms of locations X. That's the reason why I chose
one and only one frequency with the sound.
Because if I had exposed you to many different frequencies,
many different wavelengths, then the location of silence
for one wavelength is not the location of silence for the
other wavelength. And so the experiment would not
have worked. And that's why it worked so
well with the laser, the red laser,
which is practically one wavelength, and so the minima
and the maxima are extremely well-defined.
If we had done the experiment with white light,
it wouldn't have been so impressive, and on the next
slide I show you what you w- would have seen then.
This is what white light would have
done, this is a two-slit interference pattern.
This is what red light would have done.
Red light is a narrow bandwidth of wavelengths,
well-defined black lines, light plus light give darkness,
well-defined maxima, and the blue notice that the
separation between the dark lines and therefore also the
separation between the bright lines is substantially smaller.
Because blue light has a wavelength of about forty-five
hundred Angstroms and red light roughly
sixty-five hundred. So there's a big difference.
And so white light would then give you the superposition of
all these colors, and so you don't really get a
very nice interference pattern of dark areas and bright areas,
because all the colors begin to overlap and each live a life of
their own. What I can do with sound and
what I did with water and what I have done with laser light
I can also do with radio electromagnetic waves.
With radar we have a ten-gigahertz transmitter here
that we have used earlier in this course.
And so I will now show you that with radar you can also show
interference patterns and the calculation that you see there
are absolutely identical. The only thing I want to remind
you of, that the approximation when you
know capital L that the tangent theta is roughly the same as the
sine of theta is only true for small angles.
Five degrees is fine, ten degrees is fine,
but by the time that you reach fifty, sixty or seventy degrees
that approximation is not true. So then you really have to take
the tangent of theta. That's no problem because you
first calculate what theta is, because that equation is
correct, and then you can calculate always where X is,
but then you use the tangent and not the sine.
So these are approximations which hold for small angles.
And so if now we look at a ten-gigahertz transmitter,
that means we have two transmitters,
one here and one here. And their separation D is
twenty-three centimeters. You see them here.
This is where they are. Here's one and here's the
other, twenty-three centimeters apart.
At ten gigahertz the wavelength is three centimeters.
You can confirm that. The speed is speed of light.
Lambda is the speed of light divided by frequency.
That gives you the wavelength. And we have here at a distance
L which is a hundred and twenty centimeters, we have here a
receiver and a track, so this is X equals zero and
here we can move it along X and so you can calculate now at what
angles seen from this point there will be a maximum there.
Theta zero is obvious. Right here there will be a
maximum. The two waves,
the distance between them is zero, R two minus R one is zero.
So they will constructively interfere.
But there is another angle, theta one, for which again
there will be constructive
interference. And you can confirm that I
found for these numbers that theta one is about
seven-and-a-half degrees. This is now for maxima.
And so roughly at an angle which is half that value you
will find silence. Silence means that the two
radio waves will kill each other.
Essential for the maximum to be here is of course that the two
transmitters are in phase. We could have rigged it up so
that they were a hundred eighty degrees out of phase in which
case there would be silence here.
Silence in this case means that the radar would kill the radar.
But I do use the word silence for a good reason,
because the way we rigged this up is the same way we did it
before. We modulate this ten-gigahertz
signal with a thousand-hertz audio signal.
We call that amplitude modulation.
And the receiver which is here receives the
ten-gigahertz radiation which is modulated at a thousand hertz.
We feed it to an amplifier. We demodulate it and you will
hear the thousand hertz. And so we can also move it
along this track here and find the location X one whereby we
have our first maximum apart from the zero.
And I found that that is very roughly at fifteen point five
centimeters. And you should confirm that
using those equations.
Equations are the same. Whether you deal with sound or
with red laser light or with gigahertz makes no difference.
And so let's turn now to this demonstration.
I will turn on the -- the two transmitters and here is the
receiver which is exactly at angle theta zero.
So there's a maximum. I'm now going to close one
transmitter, [tone] put my hand in front.
[tone] And you think about what will be the reduction of
intensity of the sound here. [tone] If I close one,
it's substantially down. You may think that it is down
by a factor of two. Because we have only one
transmitter instead of two. You're wrong.
If you think that, you do not understand
interference. It is four times lower when I
hold my hands over one. Figure it out for yourself.
I'll test you on the final to see
whether you really understood that.
So now the sound is four times larger than when I cover one up.
I'll cover the other one up. It's down by a factor of four.
[tone] I can cover this one up and then you hear nothing of
course. [tone] Now I will move this one
[tone] to the location [tone] where there is destructive
interference, [tone] which should be about
half of fifteen point six centimeters.
[tone] [cough] Maybe you have good ears but I
hear nothing anymore. Now I go through it and find
the maximum, which is about fifteen point six centimeters.
[tone] Here it is. [tone] And the other side,
[tone] here's the maximum at center, so here should be a
minimum, there it is, and I go to the other side,
there should be a maximum. [tone] And there it is.
[tone] So what I have shown you today is I've shown you the
interference pattern of sound, of water, of red laser light,
of radar, and at the very least I hope I've convinced you as
Young convinced the world in eighteen oh one that light are
waves. And that means that Huygens was
right and Newton was wrong. Now that should perhaps not
surprise you because Huygens was Dutch.