The controversy between Newton and Huygens
about the nature of light was settled in 1801 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 are 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 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
3D 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-- it's 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 802 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 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 betw-- from the two sources
to the screen is capital L. And here is one of those directions
theta. Then 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 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 [fsswwt] 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 n 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
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 then 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. [sound of people standing up, chairs flipping] 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. [high tone] You are a couple of lousy scientists. You are a couple of lousy scientists.
[tone stops] 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 continues] Go very slowly. [tone] Who has found clear locations
where the sound is nearly zero? Or practically zero? Most of you. And you certainly can hear,
if you move, that there's an enormous difference
in sound intensity. So again,
who has found locations whereby you clearly say
this is practically silence? Ah, you see them all the way in the back and the separation,
how far you have to move, depends on how far
you are away from me. [tone stops] 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 1801 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. Young's exact setup was somewhat different
from my drawing on the blackboard. But my drawing simplifies Young's setup
and is therefore easier to understand. 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 for 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 radar 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 the're
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 science [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 six 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... [tone] ...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, put my hand in front. And you think about what will be
the reduction of intensity of the sound here. 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. I can cover this one up
and then you hear nothing of course. Now I will move this one to the location
where there is destructive interference, which should be about half of fifteen point six centimeters. [tone fades away] 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 volume increases again] Here it is. And the other side,
here's the maximum at center, [tone fades away] so here should be a minimum, [no tone] there it is and I go to the other side,
there should be a maximum. [tone volume increases] And there it is. [tone] So what I have shown you today-- is I have shown you the interference pattern. Of sound, of water, of red laser light, of radar, and-- at the very least
I hope I have convinced you as Young convinced the world in 1801
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.
