electricitymagnetism-34

{Professor Lewin 8.02 Lecture 34} (0:00 - 9:40)

Last time, we discussed the interference pattern due to two coherent light sources. Today I will expand on this by exploring many many light sources

Suppose that instead of having two slits, to which I allow the light to go, I have many, I have N, capital N. And that the separation between two adjacent ones be “d,” so plane parallel waves come in, and each one of these sources is going to be a Huygens source: is going to produce spherical waves. So we can ask ourselves the same question that we did before.

And that is, look at a long distance, far away at a certain angle theta, where will we see maxima and where will we see minima? And then we can put up here a screen at a distance L, and we will call this x = 0. And then we can even ask the question, “Where exactly on this screen will we see these maxima?”

You will have constructive interference in exactly the same situation that we had with the double slits interference pattern: when the sine of theta of n equals n lambda divided by d. And, if you are dealing with very small angles theta you should all remember that the sine of an angle is the same as the angle itself provided that you work in radians. So, for small angles you can always use this approximation if you remember that it is radians and that is only in the small angle approximation.

So, the conclusion is that if we work in radians for now, then theta of n for the maxima is n lambda divided by d. And being zero here and being one right here and being two right there. And if you want to express that in terms of a linear displacement from zero, then x of n (again for small angles) is L times that number.

And so, now you get the displacement here in terms of centimeter or in terms of millimeters. So you will say, “Big deal, that’s the same result that we had for the double slit interferometer. We had exactly the same equation there was no difference. And “d” now is the separation between two sources.” It is obvious that it is the same. Because if these two are constructively interfering, then these two will too and these two will too and these two will too. So, all of them will. So it’s not too surprising that you get exactly the same results.

But now comes the big surprise, we haven’t discussed yet the issue where the locations are: where light plus light gives darkness. We haven’t discuss the destructive interference. And, to derive that properly is very tricky. In fact, if you take 8.03 you will see a perfect derivation, but I will give you the results.

What is not so intuitive is that if you have N sources, that between two major maxima, I mean like between this maximum at n = zero and the maximum at n = one, there are now N, capital N, minus one minima. And minima means complete destructive interference, so if capital N is two (which we did last time), then two minus one is one. Exactly, that was correct, we had only one zero between the two maxima. But, that is not the case anymore for N much larger than two. So let me now make you a sketch whereby I plot the intensity of the light as a function of angle theta. And this is the intensity, so that’s in watts per square meter (Remember that’s the Poynting vector.). And, let this be zero. And let the angle theta one be here and for small angles that’s lambda divided by d. And here you have theta two which is two lambda divided by d. And so on. I take the small angle approximation so this angle is now in radians.

What you are going to see now is the following intensity as a function of theta. You see here a peak. You are going to see here a peak and you are going to see here one, and so on. And the same (of course) is true on the other side.

And here in between you are going to see now N minus one locations whereby you have total destructive interference. And the same is the case here. And this can be used: N can be a few hundred. So you have many locations where you have one hundred percent destructive interference.

And so this point (this first location where we hit the zero, that now is at the position lambda divided by d divided by capital N. And I will call that angle, from the maximum to that zero, from this maximum to this zero. I will call that angle (for now) delta theta. Because that delta theta is a measure for the width of the line. Here it is maximum, here it is zero. And so that angle delta theta in terms of radians is lambda divided by d times n. Which then is approximately theta one divided by n, for theta one itself is lambda divided by d. and so you see that it is N times smaller than this distance, and so if n is large, these lines become extremely narrow. That’s the big difference between two slit interference and multiple slit interference and the larger N is the higher these peaks will be. The height of these peaks (the intensity here) is proportional to N squared.

You may say, “Jee, why is it not linearly proportional to N?” Well that’s easy to see. Suppose I increase capital N, the number of sources, by a factor of three, then the electric field vector, where there are maxima, is three times larger. But if the electric field vector is three times larger, then the Poynting vector is nine times larger. So you get nine times more light. Now you might say, “That’s a violation of the conservation of energy: three times more sources, nine times more light. How can that be?” Well you overlooked then that if you make N go up by a factor of three, that the lines get narrower by a factor of three because of this N here. So, they get higher by a factor of nine, they get narrower by a factor of three, so you get a gain of a factor of three in light. Of course there’s a gain of three since if you have three times more sources, you get three times more light, so there’s no violation of the conservation of energy here.

And I want to demonstrate this to you, using a red laser which we have used before, and I will use what we call a grating: a grating is a transparent plate which is specially prepared, which has grooves in it. And the one which I will use has 2500 grooves (we call them lines) per inch. That means that the separation d between two adjacent groups (in my case) is about 10.16 microns. A micron is ten to the minus six meters [10^-6m]. And the wavelength that I am going to use is our red laser which is about 6.3 times ten to the minus seven meters. And I’m going to put the whole thing there and I’m going to make you see there at a distance L which is about ten meters. So this allows me now to calculate where the zero-order will fall, where the first-order and where the second-order will fall. When n is zero we call that zero-order. So this is zero order.

When n is one we call that first-order, and when n is two we call that second-order. And you have (of course) the first-order also on this side, and the second-order also on this side. Everything that you have here, you also have to think of it as being on the other side. So, I can predict now where the zero-order will be when n is zero, that is zero degrees. That is immediately obvious I can use that equation that when n is zero, the zero-order is always right at the center, provided that all these sources are in phase. And, they will be in phase because I used plane waves, so Hyugens will tell you that they are going to oscillate exactly at the same time.

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From Johnne [80.250.32.13] - 7/20/06 9:51 AM