Interference of electromagnetic waves
More on single slit interference
- So I got to be honest with you. Single slit interference is confusing. In fact, this argument I gave earlier, first time I heard about this, I thought this was just mathematical mumbo-jumbo. I was, like, "What are you talking about? "This makes no sense." It seems like you could argue anything like this, But you can't. And I'm going to try to show you that in this video. Specifically, I'm going to try to show you that this same argument that we made for the destructive points won't work for the constructive points. And I think that will help a little bit. In other words, this half wavelength relationship won't give the constructive points, not exactly. It'll give it approximately, but it won't work exactly. Why? All right, well, let's look at this. So let's get rid of all of that. So let's say we were going to try to derive the formula for the constructive points. The first thing I'd do is, I'd say, well, all right, again, each point on here defracts when it gets to the hole. I'll have infinitely many sources, but I can't draw infinitely many so let's just consider eight again. One, two, three, four, eight, I get my interference pattern on the wall, and this is the graphical representation. I mean, they're really, like, smudgy lines over here, but the graphical representation looks something like this. You'd have a big old bright spot in the middle. These points on the end, you keep getting them, but I can't keep drawing them over here. So it keeps on going. And I'd pick a constructive point. Right here, this is a bright spot. So this, if I had to give a guess for what point would be totally constructive for all the waves, I'd say it's that point. So I'd take my topmost wave, I'd say it travels a certain distance to this bring spot. I'd take my wave in the middle, travels a certain distance to get here. I'd imagine my line straight down here, to try to figure out the path length difference. Now remember, this here is the path length difference. What should that be in order for these two purple waves to interfere constructively over here? It's got to be an integer wavelength. So one wavelength, two wavelengths ... since this is the first one from the center, we'll just say that's one wavelength. What's the relationship? We know that. Remember, the relationship for the path length difference and the theta, the angle that it's at, was just d sine theta. D is, well, the whole width of the hole -- it's a w. So what width is this? This width between this source of light and this source would be w over two. And the relationship I'd get? All right. D is w over two times sine of theta, would equal ... For this first point, I'd say it's equal to lambda. So let's just assume these two are interfering constructively at this point. And that would give me: w times sine theta equals two lambda gives me a constructive point. Now, I'm already confused. What? W sine theta equals two lambda? Constructive? We already proved this is a destructive point. Remember our relationship for the destructive points that we derived was: w sine theta equals m lambda as long as m is not zero, but one, two, three, four, five. These are giving destructive for any m equals one, two, it could even be negative if you want to consider down here, any integer. Looks like we just proved these are constructive. How are these constructive? Well, they're not, really. They kind of are, but watch what happens. Just, here we go. So if I follow this argument through, the thing that fails ... Our previous argument's fine. The argument that fails is this current one with constructive, because yes, these two are constructive there, but watch. This point here, now I ... just remember the game we played. We said if these two are constructive, then the rest of them should all be constructive. Is that so? Well, let's go down one, let's go down one. I imagine these two waves getting here. So far it's looking good. They're at the same angle. They're the same distance between them. I mean, this length here is still w over two. So I'd still get w over 2, sine of the same angle, because it's the same point on the wall. So if w over two is the same, sine of the theta's the same, then that's got to also be path length difference of lambda, which means these two blue waves also interfere constructively. So this is looking pretty good, which is kind of bad. I'll show you why. These two would also be constructive. Well, is this a constructive point? Is m equals two a constructive point or a destructive point in that end? It's a destructive point. This argument's failing, and it fails because ... watch this ... Even though these two purple ones interfere constructively over here -- here's a wave cycle. Even though the two purple ones meet up constructively, let's say the top one was there, that means the one in the middle, this one here, was also at the peak. So those two interfere constructively. How about the next two? Well, those two are going to interfere ... Now maybe those two are, like, at this point. They're both constructive, but they're not necessarily the same as the two purple ones. And how about the orange ones? Orange ones might be constructive, because they're both at the same point in the phase, but they're not at the same point as all the rest of them. You can have more. What about these down here? Oh, these might be down here. Those two together are also constructive, but you see the problem. Even though these two are constructive, this one's not constructive with this guy, and these all add up. In fact, for the most part, they cancel. That's why these are so little. You get these weak, you get these really weak fringes on the sides of the single slit, because you're not going to get points where they all add up really well necessarily. You get points where a lot of them sort of cancel out. And it doesn't completely cancel those. Here's where I lied, for the diffraction grating. Remember, for the diffraction grating ... Let me get rid of this. For the diffraction grating, we had a single line and we made a ton of holes in it. And I said that diffraction gratings are great because, if you come over to here, you make a ton of holes in here, instead of getting a smudgy pattern on the wall, you get a big bright spot right in the middle and then a well-defined, well-defined, well-defined on each side of them, evenly spaced, basically just zero and then extremely sharp. And then zero and then extremely sharp. And then zero and extremely sharp. And the whole argument I made for defraction gradings was that the reason it's zero in between, the reason these are giving zero everywhere except these constructive points was precisely because ... we come back over to here ... was precisely because of this effect right here. This effect where they, for the most part, cancel. Now I'm saying, "Eh, they don't actually completely cancel necessarily." So these wiggles here are actually in a defraction grading pattern. They're just so small and unpronounced, compared to these, you don't really notice them. What I'm saying is, if I wanted to draw this more realistically, I would definitely have this bright spot right here, but I'd have this in between small variations, small points where it becomes a little more, a little less, constructive or destructive. What you have for a single slit is this: just one center bright spot. It's not going to be as well-defined because it's a defraction grading, it's a single slit. But you still get these. You get these weird wiggles that, for the most part, you ignore for a defraction grading, but they're there. And for a single slit, that's kind of all you got. So, can't really ignore it so much. Those are going to be there. It's because these don't completely cancel. Our argument does not work. It works in the sense that two of these might be constructive. That means, you can pair these off in constructive, but they won't all be at the same point on their phase, which would give you a completely constructive point there. So that's why we don't ... it's hard to find an exact formula. What's the formula for the constructive points? Well, getting this formula -- not quite as simple. You need to know a little bit more physics to do that. And so, typically, in introductory physics classes, you aren't asked to find the exact locations of the most constructive points over here. Even these most constructive points partially cancel. You do know how to find the exact locations of the destructive points, though. And, if you wanted an approximate location of a constructive point, well, you can find the exact location of two neighboring destructive points, which, if you really wanted to, I mean, the constructive's in there, approximately in the middle, if you wanted to get a rough idea. I can still see some of you being upset, though. You might say, "Wait. "Hold on a minute. "So we're saying this formula's good "for the destructive points, but is this problem "we ran into for constructive points "also a problem for destructive points?" And it's not. I doesn't matter if they're at different points in their phase for the destructive, because each pair cancels. In other words, when we ran through this argument for the destructive points, look at -- if these two purple ones cancel, then they cancel. I mean, if one was at the peak, and then the other's at the trough or the valley, those add up to zero. They're gone. Any effect they might have had on light hitting this point on the screen is gone, completely negated. And so, what about the next two, These two blue ones? Well, those two, if these two purple ones cancel, remember the argument went that these two blue ones would have to cancel. So no matter where they're at, they're at some different point on this cycle, let's say one's here and the other's ... well, that looked like the same. So let's say one's here and the other's at this corresponding 180 degree out-of-phase point. Well, they still cancel. That adds up to zero. And so it doesn't even matter that they're at different points in their phase. It doesn't matter. No matter where they're at, one's 180 degrees out of phase with the other, every contribution cancels out. You add up a bunch of zeros, you get zero. So destructive works fine. You don't run into the same problem with constructive. It's a problem for the constructive points because these might add up to some big number, and then the blue ones add up to a different number, and the orange one adds up to a different number, and then the red ones might add up to a negative number, and you keep getting these different numbers. You try to add them all up, well, what do you get? That's why this formula's not so easy to find. Adding up zeros, that's easy -- just gives you zero. So, I hope I showed you that this crazy mumbo-jumbo argument can't say anything whatsoever. And hopefully, that gives you a little more justification, hopefully it makes you believe a little more into this formula that we derived for the destructive points. Those it does work for. And so, we can find destructive points just fine. One more thing we can find is the width of this center bright fringe here, the center bright spot. It's going to be wide. And since this goes to m equals one, now that first destructive is over here, this is wide. This is, in fact, twice as wide as all of these between these destructive points. And how wide is this? Well, you can find the angle to this first destructive point up here, m equals one. You can find it to the m equals negative one. You do a little trigonometry, you can actually get this length. That's another thing you can find exactly, is the width of this center bright spot, and the location is right in the center. But the location of these constructive points up here, the exact location, that's a little harder. You can find their width again, because you can find the locations where they terminate. But finding where it actually peaks in here, don't have an exact formula. We do have an exact formula for the single slit destructive points. And that's typically what you're going to have to find in these problems.