Fundamentals of Physics II

PHYS 201 - Lecture 18 - Wave Theory of Light

Chapter 1: Revisions to Geometric Optics [00:00:00]

Professor Ramamurti Shankar: I dropped a bombshell end of last class telling you that all the geometric optics I taught you has to be revised, and before letting you go for lunch I said you’ve got to tell me why I would do that.

Student: Experiments.

Professor Ramamurti Shankar: Yes, somebody said it’s the experiments, and that’s what I want you to remember. That’s the only way things change in physics is experiments tell you something’s wrong with your theory, and in the end if your theory doesn’t agree with the experiment it’s over. And conversely, if you’re a newcomer unknown to anybody, make predictions and they agree with experiment then you’re a rock star. Everything is based on experiment. That’s the only way we change our mind. Now you might say, “Why do you keep doing this to us? We believe you. We write everything down. We do the problems and then you say, ‘Oh, these guys are wrong. Here’s a better theory,’ what’s going on? Are people really wrong?” So I have to be very careful when I say people are wrong because the news leaks to the press and the media will say, “Physicists think they’re always wrong.”

In fact, I’ve gone on record here saying, “We’re always wrong.” What I mean by that is no matter how many laws we find one day you will find some new experiments which are not explained by these laws. That’s not really bad news. That’s what keeps us in business because we want to find something that doesn’t fit anything we know. For example, Newtonian mechanics in some sense is wrong because it doesn’t work when velocities approach the speed of light, but it’s not wrong in the sense that the predictions it made in its proper domain do not work anymore. It was supposed to work in a limited range of experimental observations. If you push the limit, if you build accelerators that send particles at very high speeds you may find they don’t obey Newtonian mechanics. Then you get Einstein’s special theory. Now if you have a special theory like Einstein, or a new theory that overthrows the old theory, and it explains new phenomena, there is still one extra requirement. Can you guess what that might be with a new theory? One thing we demand of every new theory, yes?

Student: That part of it coincides with the old theory?

Professor Ramamurti Shankar: Very good. If you go back to the old experiments, which are explained by the old theory, well your new theory should continue to explain that. In fact the new theory, when it’s a good one, will tell you why for hundreds of years people fell for the old theory and then it’ll also do new stuff. If you do new stuff and it doesn’t fit the old stuff that’s not good. So relativity is just like that. It works for all velocities up to the speed of light, but if you let v/c go to zero then you will finally get back Newtonian mechanics. Similarly, quantum mechanics works for very, very tiny objects, the atomic scale, but when you apply it to big things you’ll find that the world begins to look Newtonian. Similarly, the end of geometric optics came from a series of experiments, and I will tell you why we didn’t realize there was something wrong with it for a while.

So here’s an experiment you could do. Here a screen and there’s some light coming from here, and there’s a hole, and behind this screen is another screen. So I’m not good at drawing, so you should imagine this is coming out of the blackboard, and the ray of light is coming in. It’ll form a shadow. I happen to have in mind a circular hole here. It’ll form a disc shape exactly like that. And another thing, but you will find that as you shrink this, make it smaller and smaller, suddenly that won’t be true anymore. You will find that the light spreads out like that to bigger and bigger areas. So it’s no longer forming the geometric image of this aperture. It’s fanning out, so it’s not going to come from geometrical optics. That’s one experiment where something goes wrong.

There’s a more dramatic experiment which really puts the nail on all kinds of ray theories, and that’s the experiment involving interference. So you take a partition with two holes in it, and you send light from here, and you put a screen behind. Are you guys with me? This is a cross section of a plate coming out of the blackboard. It’s got only two holes in it.

Or here’s a better way to think about it. This is the top view of an experiment. I’m looking down at my lab from the top. There’s a wall with two holes in it, and there’s the back wall, and the light is coming from here. If I close this hole what we will find typically — let me make them slightly bigger to make my point. You’ll find some light. I didn’t draw it properly. Most of the brightest areas will be in front of this hole. So let’s see. It’ll tend to be big here and kind of fall off. Now if you block the other one and you shine light here you get something like that. It’s like two windows in the room. You’ve got one window, you get some sunlight. You’ve got another one you get another sunlight. Now the question is if you open both of them our expectation is that there is some energy coming here, some energy coming here, and if you open both you should get the sum of the two energies, which I’m assuming this is the sum of these two graphs plotted here. By the way, do you know how I’m plotting this? I hope you understand that this graph really means that distance is proportional to how bright this point is. See, normally you’re used to drawing graphs like this as a function of distance, but I’ve turned it around so that that is the graph of the brightness at different points on the screen. That’s what you expect. You open two windows you get the light equal to the sum of the two.

For example, if you’re feeling warm it’ll be warmer in this region because you’re getting light from two sources. But this experiment also gives you different results under certain conditions. I’ll tell you what the conditions are. What you find then if one slid open it looks like that. If the other slid open it looks like that, but with both open it looks like this. That’s called interference. The main point to notice is there are some places, like right here, where you get less light with one more slid open, okay? And more light with only one open, less light with two open. In fact, you can arrange it so this is really zero. I mean, it can come all the way down here. In other words that’s the point on the screen which used to be bright with one slit open. You open the second one that point becomes dark. That means if you’re warm there, and you tell somebody, “Hey, this is good, open another window,” suddenly you get nothing. That happens. So when did people realize this is what happens, and why did people fall for the other graph that looks like this before? The answer is this. In order to see this phenomenon in which it instead of being a shadow like this you get a huge spread, or instead of being simply the sum of these two featureless graphs it starts oscillating, and a certain condition has to be satisfied.

Chapter 2: Young’s Double Slit Experiment [00:08:21]

The condition is the following. This light, if it has a wavelength λ, the λ, if it is much, much smaller than any distance in the problem then you get geometrical optics. So in this example if this slit is thousand times the wavelength of light then the shadow, in fact, will be geometrical, but as you make the slit smaller and smaller until it becomes comparable to the wavelength then you’ll find it begin to fan out. Similarly here, if this distance between the two slits, d, is much, much bigger than the wavelength and each slit is much, much bigger than the wavelength, then you will find that you don’t see any of this interference effects. Because what happens is interference effects are always there, but they oscillate like mad on that screen. Suppose it oscillates like this, and this could be 10,000 oscillations per centimeter, and your eye is able to look at it only one millimeter at a time, then you’re averaging over a region and you only find the average of this graph. That looks like what you got without interference. So in order to see interference you need to make your apparatus probe the physics at a distance scale comparable to the wavelength. So that’s the answer.

We will see all the details, but it’ll turnout that if you look at light using any equipment, maybe hole in the wall, two holes in the wall, whatever, whose dimensions are much bigger than the wavelength you will not see the wave theory of light. You’ll be able to get away with the geometric theory. But once you probe light at length comparable to wavelength all these funny things will happen. That’s why people didn’t realize all these things about light. For example, the wavelength of light, typical light, is 5,000 angstroms, or 500 nanometers, that is 500×10^-9, or 5×10^-7 meters. That means if you forget the 5 there are 10 million oscillations of the wave within 1 meter, so if you have an instrument that can pick up that kind of fine features then you will see the wave nature, but normally you won’t. So it’s only when people probed light with very, very tiny holes placed very close to each other they saw it.

The second important thing to remember is the following. Even if you made the holes very small, and if you shine white light on this, you will not see any of these patterns. Why is that? Can you guess? I’m going to tell you everything, but can you guess why? Yes? You have a guess?

Student: Because they’re white light.

Professor Ramamurti Shankar: That’s right. White light is made of many colors and many wavelengths, so each wavelength will form its own pattern, but the maximum and minimum of one guy won’t coincide with that of the other. So where this one wants a maximum that’ll want a minimum, it’ll get washed out. So in order to see this effect you need monochromatic light sent by a source which is sending it at a definite frequency and wavelength. And these are all things you have to get right before you will see these effects.

Okay, so now let’s ask ourselves, what is the nature of waves or what is the nature of light that produces these funny effects? Light, as you know, obeys a wave equation. Let me call ψ as whatever it is that’s doing the oscillations, generic name is ψ. For example, ψ can be the height of this string that’s vibrating. If the string is doing this from equilibrium ψ is the deviation from equilibrium. ψ can be E or B in the electromagnetic case. ψ can be the height of water in a lake above some reference level. That’s the reference level, and the water does some oscillations, ψ is the deviation from the normal. In the case of a sound wave, as you know sound wave is deviations in air pressure. Again, this graph would stand for the normal air pressure, ambient air pressure, and when you talk you make some oscillations in the density or in the pressure, and there’s the deviation in pressure. So I’m going to call ψ as any one of these things. It’s something that’s oscillating in the wave. And all these ψ’s obey the wave equation. The wave equation looks like d²ψ/dx² minus 1 over the velocity d²ψ/dt² = 0. The context is different but the equation’s the same. The velocity will vary from problem to problem.

So I’m going to give a short name for this. I’m going to all it box ψ = 0. Turns out it’s not a crazy notation. A lot of people actually use that. Do you know why they draw the box? This is for people who have done somewhat more advanced mathematics. There’s something called del squared which stands for the three spatial derivatives squared. This has got the fourth one so you use the box. Anyway, it just means I’m tired of writing this. This stands exactly for this. That’s the main point. Now when the wave obeys the wave equation there’s a very, very important property. The important property is it’s a linear equation. It’s a linear equation which means if you multiply ψ by a number it also satisfies the wave equation. Because if I multiply by number 3 I can take the 3 inside all the derivatives, and I find 3 times ψ is also a solution. Another important property is that if someone gives you a solution called ψ₁, another person gives you another solution to the wave equation which is some other function of x and t called ψ₂, then you can check that if you take any number times ψ₁, and any number times ψ₂, and add this, you will find box of aψ₁ + bψ₂ is also 0.

You understand why? When the derivatives come it doesn’t care about a. It’s a constant. Box of ψ₁ will vanish and the box of ψ₂ will vanish. The whole thing will vanish. But the important property is you can take two solutions, ψ₁ and ψ₂, multiply each one a number you like, a constant, and add them, that’s also a solution. That’s the principle of superposition, and its origin is in the fact that the underlying ψ obeys a linear equation. It’s very important that it’s linear. I’ve told you many times if you have something like that plus ψ² = 0 you can verify that if ψ₁ is a solution and ψ₂ is a solution if you add them you will get ψ₁² + ψ₂², but that is not the same as ψ₁ + ψ₂ the whole thing squared. So for nonlinear equations you cannot add the solutions. These are linear because they involve the first part of ψ. Okay, now what that means is first of all ψ can be positive or negative. What if someone tried to prove to you that ψ is always positive? How will you shoot that person down? Do you know? Yes?

Student: Well, if the constant is negative 1.

Professor Ramamurti Shankar: Very good. If the constant is negative 1 then minus ψ is also a solution so no one can ever convince you ψ should always be positive. Well, if ψ can be positive and negative, which means generally it can go up and down in sign, it cannot stand for things which are always positive such as the brightness of light. The brightness of light is always positive. The worst think you can have is no light. You cannot have negative brightness. Therefore, in the electric field the brightness is not measured by the electric field, the one that oscillates, but by something quadratic in the field, and that’s called the intensity. The intensity is proportional to square of whatever is oscillating. It is the measure of energy contained in the wave or brightness. Intensity being positive definite has some chance of being a description of something that’s positive definite. Not only does it have a chance that happens to be the case, okay?

See, in all these problems it’s the square of the function that stands for something like brightness, something like energy per unit second. They’re all given by the square. So the important thing to notice is that when you have one source producing some light, and a second source producing some light, and you turn them both on, together they’ll produce a field which is the sum of the two fields. What you can add is the ψ, which in this case happens to be E or B. You cannot add the intensities. No one told you the intensities are additive. The correct process is if ψ₁ has an intensity I₁, let me call it ψ₁², and ψ₂ has an intensity I₂ which is ψ₂², then if you put the two waves together by producing them due to the two sources, I₁₊₂ is (ψ₁ + ψ₂)². That, you see, has got a ψ₁² and a ψ₂² + 2ψ₁ψ₂ which is I₁ + I₂ + something which is indefinite in sign. This can be positive. This can be negative. This can be 0. So I₁₊₂ is not equal to I₁ + I₂. It’s got the extra stuff, and that’s going to be the origin of all the things we see here.

So it turns out for waves there’s two levels as which things happen. There’s a thing that actually oscillates and then you have to square it to get things that you normally associate with energy or brightness. And the additive law applies to the things that oscillates and not to its square. This is the reason why if you’ve got two sources each one of it produces a positive intensity I₁. When you turn them both on the answer is not I₁ + I₂, but this cross term, and the cross term can come and neutralize these and make it vanish. That’s why the experiment when you open a second hole some places can become darker than they are. Now later on, I mean, not later on, later this week when we start quantum mechanics we’ll deal with the ψ. That’s called the wave function in quantum mechanics. Now that’s a very bizarre object. I don’t want to even talk about it now. Let me just say that it’s something which is intrinsically complex. See, normally for electromagnetic waves or harmonic oscillator if you’ve got x = Acosωt we’re used to taking x = Ae^iωt then taking the real part at the end. That everyone knows x is real. You bring the complex exponential because it makes it easier to solve certain equations. But in quantum mechanics all of ψ, real part and imaginary part together are needed. It is, by nature, complex. In fact the equations of quantum mechanism, the analog of Newton’s Laws, or the analog of the wave equation will have an i in them. There’s an i in the equation for ψ, so you cannot get rid of it.

So in those cases what you want to think about the analog of intensity cannot be ψ², because ψ², if you really mean by ψ² the real part of ψ + i times the imaginary part of ψ whole thing squared, you see that is ψ_real² - ψ_imaginary² + 2iψ_realψ_imaginary. What’s wrong with this? If you want this to stand for something positive definite this is definitely not it. Can you see that? In the simple case where ψ is purely imaginary you see it’s negative, but neither positive, not even real. So you must know what I should do in the case where ψ is intrinsically complex to define something positive. Any guess on what it is? Yes?

Student: Take the ________.

Professor Ramamurti Shankar: Could take the absolute value of ψ, or another way to say that is you want to take ψ*ψ. But anyway, I’m doing some of this today because I want you to be on top of this before Wednesday. I’ve given you enough warning. I ruined your spring break by sending you notes on complex numbers so you can read them. If you know it already it’s fine, but you should be able to understand the meaning of real part, imaginary part, absolute value squared, things like E_Iθ, okay? That’s going to be used a lot. By anyway, if I take ψ*ψ that becomes ψ_real + i times ψ_imaginary times ψ_real - i times ψ_imaginary. And if you do the whole thing you will find it’s ψ_real² + ψ_imaginary². That’s, of course, never negative. So when you do complex waves they will also obey the superposition principle for ψ, but the analog of intensity, which is the positive definite number you extract from ψ, will be that.

You can also write this as absolute value ψ². When you do complex don’t forget absolute value. When you do real either you can square the ψ or you can take the absolute value. It doesn’t make any difference. For the real function the simple squared and the absolute value squared are both the same. All right, so now I’m going to start with trying to understand interference of waves by doing the following. Usually you get interference when two people are sending you a signal. One person is sending you a signal ψ₁ and the signal may be traveling in space, but I want you to sit at once place and just register what’s coming to where you are. For example, you could be in one part of a lake and someone is rocking the boat there and sending out ripples, the ripples come to where you are, and let’s say it’s periodic, and at the location where you are let it be of the form Acosωt. So A is the amplitude and ω is the angular frequency. If you like, it’s 2Π times f where f is the normal frequency. Then another wave is coming towards you. It is Acosωt, but it comes at a different phase Φ.

Now the phase Φ, if you’ve got only a single wave you can choose Φ to be 0. Because Φ is — what does Φ do? cosωt likes to have a maximum of t = 0, cos(ωt + Φ) has a maximum somewhat earlier. You can always reset your clock so that whenever it hits a maximum you press your clock and set it to 0 you can get rid of Φ. But you cannot get rid of Φ from two wave functions. You cannot choose them both to have their maximum at the same time because they’re not. In fact this fellow goes like this. This one has got an extra lead on it, so it’ll sort of start here. So it’ll go like this. If you went backward it’ll have its maximum at some negative time. So what we want to do is to add these two signals to get ψ₁ + ψ₂. Think of this as water waves. That’s the best example. Okay, someone shaking the water, ripples are coming out. They’re coming to where you are. With only the first vibrator sending the waves you get this. With the second one you get this. But now if both are doing it then this is what you will get. And the height of the water will actually be simply the sum of the two heights. It’s not obvious but true, okay?

If you agitate the water with one source and it produces the height here, and with the second one that produces another height, if you do both it’s not obvious, it’s not a logical necessity, but it’s a fact that the height of the water at any point will simply be the sum of these two. So you’ve got to add these guys. So you do that you get Acosωt + Acos(ωt+Φ). Now you’ve got to go back to your trig identities and remember the following: cosine α + cosine β is twice cosine α + β over 2 times cosine α - β over 2. If you’ve got a formula like this, of which there are numerous ones, and if you have any doubts you can check some special cases. For example, if α is equal to β I want to get cos α + cos α or twice cosine α. On the right hand side I get twice cosine α + α over 2, which is cosine α times cosine of 0 which is 1. Now it doesn’t prove it’s the right formula, but it will tell you that if it is wrong in some way then you can tell.

Anyway, this is the right formula applied to this problem. If I apply it to this problem, I get twice cosine ψ over 2, I’m sorry, 2A cosine ψ over 2 times cosine ωt + Φ/2. This is just plugging in. It is an example. This whole thing looks like a pre-factor, doesn’t depend on time and this is oscillating. So this is a problem which looks like some amplitude cosine ωt + Φ/2. The amplitude à is 2A cosine Φ/2. And if I want the amplitude to be always positive I will take it to mean the absolute value of cosine Φ/2. So here’s the answer, guys. When you add these two waves — if you draw the pictures it’s kind of confusing, but this is the simple answer. It says the total will be a new wave whose amplitude is Ã. à can be as big as 2A, or it can be as small as 0 depending on what the cosine does. And then it will also not be in step with this guy, not in step with this guy, it’ll be half way between them. Its phase will be Φ/2. Now let’s check the formula in a couple of special cases. If Φ is equal to 0 that means the two signals are identical. We don’t need any fancy stuff to know the answer is 2Acosωt. If Φ is equal to 0, that’s Acosωt, cos 0 is 1 and you get the 2A. The one other case you can do very easily in your head is when Φ is equal to Π. If Φ is equal to Π cosine of ωt + Π is just minus cosine ωt. That means this wave starts like this. Wherever this is going up this is going, I’m sorry, it starts like this one. So when that one is doing this, this will do that and it’ll always be opposite to it. If you add them you’ll get identically 0. Anyway, they are the two extreme cases where you can check your formula. So this is the only formula you will need, so you should get that into your head. There is some natural frequency ω and the phase is increasing at the rate ω so it’s ωt. The other guy has shifted by phase Φ. Their sum has a new amplitude and a new phase. I’m doing a special case where the two waves have the same amplitude. If they’re different amplitudes it becomes more messy, but that’s not necessary.

So I’m going to do this once more, but with complex numbers, because I want you guys to get used to complex numbers. And I think the only way you’re going to get used to anything is if you use it a lot. You can sit down and read it at bedtime. It may help you go to sleep, but it doesn’t do any good. So I’m going to keep using the complex numbers, because starting Wednesday we’re going to be doing a lot more. But you don’t need very fancy complex numbers. It’s at the level that I’m doing now. So let us now do exactly this addition using complex numbers. So let’s take one single wave first, ψ₁ = Acosωt. We agree there’s no complex number here, but let’s write it as the real part Re(Ae^iωt). Now you remember the complex number is z = mod ze^iθ looks like this. It’s got a length equal to mod z and there’s an angle equal to θ. That’s the complex number z. If you want you can think of it as that. It starts at the origin and terminates at the point z. Its real part is really this one, this mod z cosine θ. The imaginary part is mod z sine θ, but I just want the cosine. It’s the real part of this guy. So what does this look like? Well, let me draw you a picture here. This looks like this. It’s got length A and the angle is ωt, and as time goes by it goes round and round.

So at any future time it’ll be on this circle. At this instant its real part is just this one, A cosine ωt. So if you like, the complex number has fixed magnitude, goes round and round in a circle. The real part of it has got a maximum of A here, a 0 there, and -A, and that’ll be the cosine, because the horizontal part of this is A cosine ωt. So we can keep track of the thing we’re interested in by letting this vector rotate in a circle by looking at a shadow on the x-axis, okay? If you have some light shining from here you ask, “What image does it cast?” Well, it comes there. So for a single complex number it’s not very useful, but now let me bring in a second complex number. Let’s bring in a second complex number, so I’ll draw a new figure for the two of them. So you should understand this is a dynamic situation. This is not a fixed vector. As t increases this is constantly rotating, okay? And as it rotates if you follow the shadow on the x-axis you get the real part, which is our function.

Now if you’ve got a second function, so let me take a circle of radius A, and here’s the first guy who’s angle ωt. The second one is also on that circle, but you’ve got an extra angle Φ. You follow that? So this ψ₁ this is ψ₂, actual ψ is actually the real part of these two guys, but I’m going to work with the complex number first. We all remember in the end to take the real part. And I’m using the fact that the real part of ψ₁ + ψ₂ is the real part of ψ₁ + the real part of ψ₂. Therefore if you want to add two real waves you can add the complex waves and then take the real part. You can take it before or you can take it after. Our philosophy is going to be to add the vectors first then take the real part. I want you to think about, what is the sum of these two guys going to do? As usual I’ll give you a few seconds to think about it, okay? By vector addition you do this parallelogram. You draw one copy of this here and one copy of that here, and you join this to that. That line is ψ₁ + ψ₂, so their sum would rotate on that circle, but let’s look at the sum. What’s this angle here?

You can tell from the similarity of this parallelogram that that angle is Φ/2, because the angle Φ between them is split evenly between the diagonal of the parallelogram. So you already tell that the new wave, which is obtained by adding these two, when this guy rotates we’ll have a phase of Φ/2. And what’s its amplitude? That we’re asking, what’s the length of the diagonal of this parallelogram? Then you use whatever theorem, and to say ò is the square of one guy, square of the other guy + 2 times length of A length of A cosine of the angle between them. I mean, cosine of this angle, Φ. So this becomes 2A squared times 1 + cosine Φ. Now I remind you of another trigonometry identity that 1 + cosine Φ is 2 times cosine squared θ/2. That is ò. Then for à if you take the square root of this it’s 2A times cosine Φ/2.

So we just reproduced the answer, but I want you to understand that the sum of these two complex numbers is a number of length 2A cosine Φ/2. It goes around at the same ω, but it is shifted. Its phase is half way between the phase of the two waves. That half comes because in a parallelogram the diagonal will bisect the angle Φ. So this answer is the same as what we got from trigonometry. It’s not going to give you anything new, but it’s helpful to think of our actual ψ₁ + ψ₂, if you like, is the real part, which is the projection on the x-axis, but it’s a lot easier to add these arrows than to add the real parts.

Okay, now I’m going to do the last part of it one more way, so once more with feeling. So let me do that, ψ₁ + ψ₂. This is when I’m not going to draw any pictures. Some people like to draw pictures. Some people like to do the algebra. So I’m one of these algebra types because when they draw a picture I don’t understand anything. So for fellows like me ψ₁ + ψ₂ looks like this, ψ₁ is Ae^Iωt, ψ₂ is Ae^Iωt+Φ, which I can write like this.

So let’s write it as a common Ae^iωt times 1 + e^iΦ. This much, I think, is automatic. Your spinal cord can do this calculation, but the next move requires a little bit of planning and we do this. Ae^iωt I’m going to pull half the phase of that e^iΦ/2 times e^iΦ/2 + e^+iΦ/2. You can see if I open all the brackets this goes back to being 1. This goes back to being e^iΦ. The reason I do this is that I want everybody to know this is a famous combination, e^iθ + e^-iθ is twice cosine θ, so twice cosine Φ/2. So the sum of all these complex numbers became 2A cosine Φ/2 times e^iωt+Φ/2. You can see that this is the amplitude. This is the length of the vector, and this is angle by which it’s rotating. Can you see that? That’s like mod ψ₁ + ψ₂, and there’s the i times whatever the phase of ψ₁ + ψ₂ is. There are many ways to do this, but having done it three ways you should at least remember the answer. When you add two waves both at the same frequency ω, with a relative phase between them, their sum oscillates at the same frequency with the phase half way between the two with an amplitude which is 2A cosine Φ/2.

Now I’m going to start applying this result to a variety of problems. The problem I’m going to do is the following. I’m going to take — let me start fresh here. Now I’m going to do the interference problem. For interference you take these two slits and some wave come from here. It hits the two slits. Now here’s some part of physics that I’m just going to tell you without proving it. I hate doing that, but once in a while we have to do that. I can only appeal to you sense of logic. That is the assumption that if you look at this wall which is completely covered except for two tiny holes, let’s say, then any light on the back, on the other side, will be in the form of two glowing objects, and with the two holes in the wall will act like two light sources. That’s how it works, for example. You open the window, the window looks like a source of light. It’s coming from somewhere, but to the people inside the room that a source of light and that’s a source of light. And if the things are very close to each other, this hole, or very small it’ll look like a point source of light, and for the point source of light the waves will emanate like this with that as a center, and another set of waves with this at the center, and the two will be in sync because this wave front hits them both at the same time. And I’m going to sit somewhere here, and I’m going to measure what’s going on.

So you imagine this is water waves. This is a top view of the water wave. This is a crest and there’s a trough in between, another crest, another trough, another crest, and I’m sitting here and these wave are going by, so two waves will come to me. One will come from here, and one will come from there. Let that distance be r. Let that distance be r + δ. So I am sitting here. So the water here will be bobbing up and down by an amount equal to what this guy tells me to do plus what that guy tells me to do. So I’m only asking you, “What is the amplitude of vibrations of the water here with both of them open?” Well, you’ve done it many, many times. If the first one sends a signal Acos(kr - ωt), by the way, you have seen waves in one dimension whether it’s cos or sine of kx - ωt. That’s the wave in which the fronts look like this. When you advance the next the lines of constant phase or maximum of the cosine are lines like this. But this is not a plane wave it’s a circular wave. It spreads out in all directions, and the points of constant phase are circles, if you like. See, here if x is a constant you get a certain phase, and as you vary x at a given instant you’ll get the various maxima here. Here things depend on r. At a given r the phase is constant, so if the cosine is maximized at one value of r it’s maximized at all values of r. So all the crests will lie on that semicircle on this side, and all the troughs will be half way in between. And I remind you that k is related to the family of wavelength by this formula, 2Π/λ. You see that? If this k was 2Π/λ, kr looks like 2Πr/λ. What that tells you is if to any given r you add a λ it doesn’t make a difference because you’re adding a 2Π to that trigonometric function. That’s the relation between k and λ, it’s inverted. Okay, that is ψ₁. ψ₂ has got the same amplitude cosine kr + kδ - ωt because this path, I’m assuming is longer by an amount δ, so the r measured from this hole is going to be r + δ. So what happens when you add them? It’s the same trick we have done many times. All you have to do is identify kδ as this number Φ. That’s the phase difference between the two. And that’s the phase difference because even though these apertures are acting as synchronized sources of waves it takes this wave somewhat longer to get to my point than this one, so there is a delay and that’s why it arrives with some shift Φ.

Yes? So what’s the amplitude here? Well, we don’t have to do this over and over again. I already told you à is 2A cosine of Φ/2 becomes cosine of kδ/2. So when is the amplitude big and when is the amplitude zero? That’s what we’re asking. If you go along the perpendicular bisector here you can see by symmetry that two r’s are equal and δ is 0 then amplitude is 2 times A. So here the two waves are completely in step and they give you a wave of double the height. And double the height means four times the intensity because intensity is proportional to square of the amplitude. So this point will have an amplitude which is 2A. I’m just going to draw that part there, but now let’s go to another place here where kδ/2 is Π/2. Well, cosine Π/2 is 0. At that point the amplitude will vanish, and then you will get no vibration of water there. Let’s understand precisely why that is true. So kδ/2, k is 2Π/λ times δ/2 is Π/2 canceling the 2 with the 2 the Π with the Π I find δ is λ/2. That means this second wave has to travel a distance half a wavelength more than the first one. I think you can see what that means.

I mean, here’s a wave going along. If that guy hits there and you’re half a wavelength behind this guy will hit there. When you go a half a wavelength in a wave you’re doing exactly the opposite of what you’re doing here. At every instant you can take any wave you want, take a flash photograph, go to some point and go half a wavelength you’ll find you’re doing minus of that. Because half a wavelength adds the phase of Π to the cosine or sine, and sine when you add a Π goes to minus itself. So that’s why you get destructive interference here because the signal from this one is telling the water to go up. The signal from that one is telling the water to go down so it doesn’t do anything. A little later the first one says go down, second one says go up then also nothing happens. So you can get two signals that can actually cancel each other. That’s the property of waves.

It’s not the property of intensity. You might naively think, “Hey, this used to be bright without the second slit. It used to be bright without the first slit. Maybe with both it’ll be twice as bright,” and that’s wrong because what adds is not intensity. What adds is the amplitude, the wave is the function ψ, and ψ being positive or negative has the possibility of getting canceled by a second source. So one source can cancel another source in ψ and therefore kill the intensity. That’s why if you now plot this graph you’ll get a minimum here, you’ll get a maximum there, and minimum here, and maximum there. So, where are the places where you get a maximum, and where are the places where you get a minimum? So let’s write that down. You remember the à was 2A cosine kδ/2. Therefore, à is equal to 2A whenever this cosine, I mean, let’s take the absolute value of this for the amplitude. Whenever kδ/2 is equal to Π, equal to 3Π plus or minus, plus or minus 5Π and so on that means 2Πδ/λ is equal to Π — let me see. I’m doing something wrong here. I’m trying to ask, when is it a maximum? The maximum is when this angle is either 0 or Π, that’s right, so why am I getting half a wavelength is what I’m trying to understand.

Okay, delta over 2. So this is 2Π/λ times δ/2 = 0. Oh, I’m sorry. I know what I did wrong. This whole thing has to be 0. That’s good. That means δ is 0. What’s the next time it’ll be a maximum? When this whole thing is equal to Π, so let me see what I get for that. These two cancel and I get δ is equal to λ. Can you see that, 2 cancels 2, Π cancels Π, δ equals λ, so you can have δ equal to 0, λ plus or minus, plus or minus. They’re all going to be bright if you’re talking about light. They’re going to be huge waves if you’re talking about water. On the other hand if this is equal to Π/2, 3Π/2 etcetera, plus or minus, then there’s going to be dark, that means that δ is equal to λ/2, 3λ/2 etcetera. So there will be bright and dark, and bright and dark fringes on the screen. If you go to the screen here the midpoint will be bright, but it’ll be oscillating like that many, many times.

Chapter 3: Interference and Diffraction of Light [00:50:53]

This was the experiment that was done by Young. Young did the experiment with light and showed that it has an interference pattern. He did not know what light was. He did not produce electromagnetic waves, but you don’t need to know that. You shine the light, you get the dark and bright, and dark and bright, and I will tell you in a minute how we can find the wavelength of light from this experiment. You can probably guess, but I’ll do that in a minute. He got the wavelength of light, but he didn’t know what it is that was vibrating, but he got the wavelength.

So let’s ask ourselves exactly where on this screen I will get dark and bright, and dark and bright. So for that you’ve got to do a little more geometry now. So let’s do that geometry here. So here is the double slit. I want to go in some direction. I’m assuming the screen is so far away that the rays can be treated as parallel. And I want to go at an angle theta so that these lengths are equal. This is the extra length delta here, and this is the distance between the two slits.

So you can see delta is equal to dsinθ, because this angle θ between the horizontal and the ray is the same as the perpendicular of the horizontal and the perpendicular to the ray. It’s at that angle. So dsinθ is δ, and that had to be equal to either 0 or an integer multiple of λ for a maximum, or equal to plus or minus λ/2, plus or minus 3λ over 2 half wavelengths for a minimum.

Can you imagine doing this calculation? If someone says, “Here is light of some wavelength, 5,000 angstroms or 500 nanometers, the slits are separated by some microns or whatever,” then you can say, “What’s the angle θ at which I will get a maximum or a minimum? Well, you can already do this calculation. If you wanted the first maximum you said dsinθ equal to wavelength. You solve for sine θ and look at a calculator, it’ll tell you what theta is. Then you can go and see, “Maybe I can get two lambda,” then that can also give you another θ. But remember, it cannot go on forever because at some point sine theta, which will be some multiple of λ/d where m is an integer, if it’s bigger than 1 no luck. Sine theta cannot be bigger than 1. So you’ll get some number of this maxima then it’ll stop. That’s because if you’ve got two signals coming from here to here only if you go in this direction will you get the maximum phase, the maximum path difference if you go straight up, and if that’s not enough to be some multiple of λ there’s no use going any further because theta you can get is 90 degrees, and the biggest path length you can get is then the space between the slits d, and your multiple wavelength should fit into that. So I hope all of you understand this very simple experiment, the double slit experiment. So light comes from the left. These two points act like sources of light. They send out spherical or circular waves. You go to a screen far away and you add the contribution from the two. If this and that differ in the path length by an integer number of wavelengths it doesn’t matter. They will arrive in step. I mean, the ninety-sixth maximum will come when the ninety-fifth one from this one comes, but we don’t care. A maximum is a maximum. If they differ by half a wavelength or three halves wavelengths then they will cancel, and what I said on the top also goes on the bottom.

Now sometimes there is a little extra geometry you can do which is the following. This may be some homework problem. If you put a screen here at the distance L and the patterns look like this, this is the central maximum, and I ask you, “Where is the central minimum measured from here,” well, you call that distance as y. Then you see y/L is tanθ and θ should be such that dsinθ is λ/2. Are you with me? You find the directions of minima. Once you find the found the direction of a maximum or a minimum it is simple trigonometry to find that if a screen is L meters away how much do you have to move in the vertical direction. That’s very easy trig. You can calculate and you can predict. That’s what Young did because he knew the distance between the maximum and the minimum. He knew how far the screens were. He knew the slit separation. All he did not know was the wavelength and he got the wavelength without knowing what the waves are made of.

Now you can see why you got fooled in the old days. If y/L is — so let’s make an estimation of how far apart these maxima and minima are in this graph. Since they repeat let me just take the first minimum, and that’s a rough measure of the oscillations. How far apart are these maxima and minima? Well, the first one occurs when dsinθ is equal to λ/2, or sinθ, which is roughly going to be θ for small angles is λ/2d. Now if you pick any λ you like, visible light is 5,000 angstroms. That is 5000×10^-8 centimeters. And suppose your screen is 1 centimeter away? Then you get 5×10^-3. So it’s roughly one-thousandth of a radian is the difference between the maximum and the minimum. The interference pattern, even though I’ve shown it this way here, actually oscillates very, very rapidly. Okay, the angle of difference between the maximum and a minimum and a maximum and a minimum is one-thousandth of a radian. So even if you go a centimeter away you will find that your probes, like your eyeball, if your eyeball is this big, okay? It’s going to pick up so many of these waves at one time. It cannot tell the maximum and minimum. That’s why we think there is no interference because you only pick the average value. That’s why you have to do very careful measurements. Okay, so that is double slit.

Okay, here’s another thing which will be very useful for people who are doing biology, or maybe even chemistry. Let’s call it diffraction grading. In a diffraction grading you take a piece of opaque glass covered with black soot then you draw fine lines on it, evenly spaced lines so that the effect is the following. This whole thing is dark except you’ve made some tiny places where light can come through like that, and the light’s coming like this. Are you with me? This is my solid screen with a few places through which light can come, and every one of them becomes a source. It’s like a double slit experiment except you’ve got million of lines, and you want to ask, “What will they do?” In the forward direction if you put an object, if you put a screen very far away that’s roughly infinity they all take the same path distance. They all go to the same distance. Therefore they will all be in step. So in the forward direction you will get a big maximum, because for every color the path is the same in the forward direction. But think of another direction which is not the forward direction. It’s some other angle. Then this guy and this guy are out of step by an amount δ. This guy and this guy are out of step by an amount δ. This guy and this guy are out of step by 2δ, so all the sources add up at face difference of δ, 2δ, 3δ, 4δ and so on.

And suppose you want them all to be still coherent. What will you require? What will ensure that all those things are coherent in a certain direction? Any guess on what δ should be? Yep?

Student: It’s equal to the wavelength.

Professor Ramamurti Shankar: If δ is equal to the wavelength this guy and this guy are a wavelength apart. We already know they are in step from the double slit experiment. This guy and this guy are a wavelength apart. This guy and this guy are two wavelengths apart. That is just fine. Therefore if you pick a direction in which that delta is either a wavelength or an integer multiple of a wavelength all the things will add coherently. Therefore the favorable direction for a bright maximum satisfy, again, dsinθ = mλ where m can be 0 plus or minus 1 plus or minus 2. These are the maximum. Except here in the forward direction where θ is equal to 0 — okay, here’s the point. Let us find the first maximum away from the center. The center is not even counted, center being white. All the colors will come to the forward direction and you will get a big bright area here, but look at what happens to the different colors. This condition, let’s take m = 1, dsinθ = λ. Whose λ are we talking about, right? Red light is one λ, blue light’s another λ. So each one will pick a different λ, each one will pick a different direction for this first order maximum.

So that means the white light, when it comes in this direction, will be split into many colors. It is just that they all shrink to one line here in the center, but any other place the lines will split. That’s because λ depends on the color; d is, of course, the spacing between these little lines on the grating, so sine θ is a function of λ. So it acts like a prism. It splits the colors. So this is called the first order maximum. Between this and that will be a minimum. Then there’ll be another minimum. Then there’s a maximum. It’s harder to calculate the minimum, very easy to calculate the maximum. So it’s easy to calculate the maximum. That’s when everybody has to add. Minimum is more complicated because everybody has to cancel. You’ve got to add the little arrows in different directions and ask when they cancel. You can do it, but it’s a little more complicated. But we can sort of imagine that between the maxima would lie the minima, but the colors will get split. That’s important.

So this is how people find out. So if you send white light it will split into all the colors. If you look at white light coming from the sun you’ll find some colors are missing because what happens is hydrogen in the surface of the sun will absorb some of the colors because its atomic — its atom likes to absorb light at a certain color. That color is missing, missing in the spectrum. So in this whole area you’ll have a little cavity. Somebody is missing and you can easily tell from the wavelength that it’s some emission line of hydrogen. That tells you the sun contains hydrogen. That’s how people know what elements are sitting on different planets. How does anybody know what they’re made of? It was not at all clear in the ancient days that stars and planets are made of the same thing you are, right? People thought moon was made of cheese. Well, that’s pretty much the same thing that you are, but something else, okay? But they didn’t know. They thought that heavenly bodies and earthly bodies, but now we find it’s the same elements because you can find the spectra lines either by their presence or by their absence. So this is the kind of spectroscopy you do.

Okay, now I’m going to do one other thing which is extremely important. That is to explain why when light comes through a hole of some size d instead of forming a shadow like this it spreads out a lot into something much broader. So let’s understand that, okay? Here is light coming in. Now what you can do is this aperture is going to be bright. When you see it from this side it’ll be glowing. So let’s treat every point of it as a source of light. In the forward direction everybody’s in step, but if you go off at some angle they’re not in step because you can see this guy is behind that guy, and that guy’s behind by that amount. Each one is somewhat behind the leader. Now my claim that if this number here is λ, what do you think will happen? In that direction if that spacing is λ is that going to be a maximum or a minimum? Anybody want to guess? Pardon me? Yes?

Student: Minimum.

Professor Ramamurti Shankar: And on what do you base that?

Student: Because the λ/2 ________ the minimum, and…

Professor Ramamurti Shankar: But this thing is λ. Usually λ between two holes was considered good, but now it’s bad. The reason is that it’s just not the top guy and the bottom guy. It’s all these little guys in between. So if this and that differ by λ, by similar triangles, you see that’s λ over two? So this fellow will cancel that fellow, and the second guy will be canceled by the second guy from below. So when the whole phase shift is λ, or the path difference is λ, you can pair the points into two at a time each differing by λ/2 and we know they will separately cancel. So for a single slit the fraction dsinθ = λ is the first minimum, and after that some funny things happen which are negligible, but the light spreads out to that angle. And this is the diffraction of light by an opening. And let’s look at it. Sine θ is λ/d.

Suppose λ is 10^-4 and d is 1? Sine θ = theta 10^-4 that means the angular spread is 1 over ten-thousandth of a radian. That’s the extent to which light won’t go straight. It’ll spread out, but the spread is 1 over ten-thousandth of a radian. It’s very small. But take a case where λ and d become comparable, maybe d is equal to 2λ, then sine θ is one-half. That’s 30 degrees. It’s already beginning to spread out. That’s why I said earlier that if you want to see the effect of wave optics you need to have the length in the problem, which is the size of the hole, and the wavelength in the problem comparable to each other. If the wavelength is much smaller than the dimensions of your experiment you can get away with the ray optics, and you need wave optics only when the condition is not met. So I remind you one more time why people thought ray optics was good. They were not able to produce slits which are this narrow. Most slits you produce by drilling a hole with your electric drill will be many, many times the wavelength of light. That means its angular spread will be slightly non-zero, but instead of a shadow that big you’ll get a shadow that big. You won’t know. Similarly the interference pattern, the interference pattern for microscopic separations between the slits will be so tight that your eye cannot detect it even if you use light of one color, and if you use light of many colors it’ll wash out anyway because each lambda likes to peak at a different angle, so different colors would wipe it out. So I wanted to let you know why it is that people believed in geometric optics in the beginning because they didn’t do experiments. This is why every theory will fall. One day somebody has the technology to produce slits which are very, very near and to get light which is very pure of a definite color then you will see something new. Then you’ve got to figure out what it is. But when Young did the experiment and saw the interference they knew right away it’s waves because people had played with water waves, and they knew with water waves that you can have interference.

Okay, here’s another thing you can do. Here is a bunch of atoms at the surface of some metal. Metal’s down here, and I’m shining light. You know how reflection works? The light hits the atom. Atom absorbs the light and it reemits the light. That’s what reflection is. You don’t just bounce off. You get absorbed and get reemitted. Now the problem is if you reemit you reemit in all directions. You don’t just reemit in one direction. The question you can ask is, why is it that i = r for reflection? That’s what I want to explain to you. So this guy emits something. There’s a wave emanating from this. This guy emits something, wave emanating from that. So let’s take two rays going off in some direction. And let’s just ask when these two guys will be in step in terms of the emitted wave. So you see the light signal is coming from here. It hits this one first. Are you with me? So this emits right way. Its signal has started taking off in the final direction. This guy is delayed because it hasn’t hit this, but once they start moving toward the eye this is closer than this one, so you see this one and that one, if you want to make a real comparison — well, this is bad picture. Let me draw you another picture where there’s a good competition between them.

So let’s see. Suppose I had a really bad idea and I thought they reflect like this. I’ll show you that won’t work. That’s all I want to do. You see what you should do is draw perpendicular here and that’s the extra distance this has to travel, right? And you will draw a perpendicular here and let’s see, what’s the extra distance the other has to travel? It has to travel that extra distance. So if you draw those two angles you will find — let’s see which angle I want to take. If you call this alpha you’ll find d cosine α₁ is d cosine α₂ or α₁ = α₂. That means I = r. You see that? Maybe I should repeat the picture here so everybody can see. I just need two of them. So one let me draw like this. Let me draw the second at what is definitely not i = r. So here I’m saying draw that perpendicular and draw that perpendicular. Starting from infinity these two have the same length, those two have the same length. This is an extra length traveled by this guy. That’s the extra length traveled by that guy, okay?

So if you call that angle alpha, you call that angle β, and if this is d, you can see that d cosine α must be d cosine β or α should be β. That is a very interesting result. It tells you why i = r is true. See, if light is made up of particles that bounce off a wall and ricochet with i = r you can understand that from conservation momentum. But light is a wave and when the atoms absorb light they emit in all directions, but only in this direction will everybody be in step because once I made sure these two guys are in step you can repeat it for these two and those two. You’ll find that’s the direction in which every single guy is in step. So i = r is the direction in which the re-radiated light arrives in step in only one direction, and that’s the direction with the angle of incidence is the angle of the reflection. You understand why there is a difference?

Okay, one last thing I should mention, because I’ve given a problem for you, and I just want to tell you one trick you may not know. And you can read up. The book is very good in that. If you’ve got some oil slick on a rainy day, there’s water here. This is an oil slick. You see all kind of colors in the oil slick? This is to explain why it is. So white light comes or any light comes, bounces off the top surface. The light also bounces off the bottom surface. And the sum of those two hit your eye. Suppose you want to arrange it so that blue light cancels out in the process? What will it take to cancel? I claim first of all the wavelength in this medium is n times the wave. It’s λ/n. If this is the wavelength here the wavelength in a medium of refractive and n is λ/n. If this whole thing is equal to a full wavelength, this round trip, I claim you’ll get a dark spot. You may think that’s wrong because this guy’s gone an extra full wavelength. That should be additive, but it turns out that when light bounces off in going from a rare to a dense medium it changes by an extra Π. At the other end when it goes from dense to rare it doesn’t. So you have to add an extra Π to everything you thought. Then you will find out that when this extra round trip is the full wavelength you get destructive interference. So if that occurs for a wavelength for blue light you take blue light out of white light you get some yellowish greenish color. That’s what you see in the oil slick. And if the oil slick thickness varies different colors will be missing from white at different point. That’s why you see all kinds of different colors in the oil slick. So I agree I didn’t do a good job of this one, but this is the easiest thing to me. The only thing to do, when you do your homework, if you didn’t read the section don’t forget there’s an extra factor of Π in the phase shift when you hit the hard medium.

[end of transcript]

Back to Top