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I'm very proud of you.
You did very well on the last exam.
Class average is a little bit above 70.
Congratulations.
There were 22 students who scored 100.
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 135 to 138.
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 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 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 R1 from source number one and at a distance R2 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 180 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, 1 / R.
Because you're talking about 3-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 1 / 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 R1 and R2 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 180 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 R2 - R1 is for instance plus or minus one-half lambda, but it could also be plus or minus 3/2 lambda, 5/2 lambda, and so on.
And so in general you would have destructive interference if the difference between R2 and R1 is 2N + 1 times lambda divided by 2 whereby N is an integer, could be 0, or plus or minus 1, or plus or minus 2, and so on.
That's when you would have destructive interference.
We would have constructive interference if R2 - R1 is simply N times lambda.
So then the waves at point P are in phase and N is again, could be 0, plus or minus 1, plus or minus 2, 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 R2 - R1 is 0 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 0, 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 different R2 - R1 equals lambda would again give me constructive interference.
That would be a hyperbola then, R2 - R1 equals lambda, that would again be a maximum, and you can draw the same line on this side, and then R2 - R1 being 2 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 R2 - R1 is 3/2 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 2D 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 2D 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 10 centimeters, so we're talking about water here.
Uh, we will tap with a frequency of about 7 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 1 / the square root of R.
In a three-dimensional wave must fall of as 1 / 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 180 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 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 1 / 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 8.02, 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 R1 and we called this earlier R2, it is the distance to that point very far away.
If I want to know what R2 - R1 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 R2 - R1.
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 R2 - R1 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 0, that is the center line, N equals 1, N equals 2, N equals 3, and so on.
And then I have the whole family of destructive interference.
Which would require that lambda R2 - R1 which is D sine theta must now be 2N +1 times lambda/2.
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 2N+1 times lambda / 2D.
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 0 here, that the tangent of theta is X/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 0, 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 2N+ 1 times lambda / 2D.
That is simple geometry.
