Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

Table of contents

34. Wave Optics

1

concept

12m

Play a video:

Was this helpful?

Hey, guys, in this video, we're gonna talk about single slit diffraction. So what happens to the light as it passes through a single slip, as opposed to what we saw before with the double slit? All right, let's get to it now. Light shone through a double slit had unexpected results, as we talked about. If you don't consider diffraction okay, if you do not consider diffraction, then it's an un obvious result. And obviously, back before they understood diffraction, they had certain expectations for the experiment, and the experiment turned out differently. Likewise, like Shone through, a single Slip also displays this same sort of unexpected result, which we call a diffraction pattern, which is alternating peaks. Sorry, alternating spots of brightness and darkness, right? The big difference between the double slit experiment and the single slit experiment is concerned with the central bright spot. Okay, in a double slit, the central bright spot is just as wide as all of the others. It's the same with as all the others, so every single bright spot across the entire screen is gonna be of uniform with. But in a single slit. The central bright spot is actually twice as large as all of the other ones. It's also considerably brighter, so that central one is definitely going to be larger than any of the other bright ones. But all the other bright spots, all the other bright fringes are gonna have the same width. Okay, that only applies to the central bright fringe. All of the dark fringes have the same width in the single slip, just as they did in the double slit experiment. Okay, let me remind myself, so we can see this figure like in a double slit. The diffraction pattern is produced due to interference. Okay, The big difference between the double slit and the single slit is that in the double slit, you actually have two sources of light that air interfering in the single slip. You have one source of light. It's just that light leaving at the top part of the slit and light leaving at the bottom part of the slit does not leave at the same angle. Okay. Light leaving different parts of the slit leave at different angles. Okay, so you have all of these different angles that the light can travel at leaving both slits. Okay, sometimes two beams of light will arrange themselves so that when they arrive they're both at a peak, right when they arrive on the screen there, both at a peak. This produces constructive interference and like that constructively interferes produces bright fringes. The amplitude of the light increases under constructive interference. Other times you can have a wave arrive at a peak. One wave arrived at a peak and another wave arrived at a trough. And when you have a peak and a trough meeting, you have destructive interference. And like that, destructively interferes, produces a dark spot or a dark fringe. Okay, because with destructive interference comes a smaller amplitude for the interfered wave. Smaller amplitude means it's darker. Okay, just like we did for the double slit experiment. We talked about the single slip conceptually, and now we want to actually talk about the mathematics of solving single slit problems. Where are these fringe is actually located. OK, now, the key difference in the math between the single slit and the double slit experiment is that in the single slit experiment, you Onley have an equation for bright friend. Sorry, dark fringes. Okay. In the double slit, we had two equations, One for the bright fringes. One for the dark fringes. For the single slit. We Onley have one for the dark fringes. Okay, dark fringes air located at angles given by sign if they t m equals m lambda over de. Okay, where m is our indexing number this time. Okay, but because em index is the dark fringes and not the bright fringes, it turns out that a requirement is we cannot have m equals zero. Okay, this is crucial to remember there's no m equals zero index for dark fringes due to a single slip. Okay, so what we have here is we have the first dark fringe or the M equals one dark fringe. We have a corresponding M equals one dark fringe on the other side. And then we would have the M equals two dark fringe on the top side and the bottom side, right? And then we could say some arbitrary M. The Mpath dark fringe is given by Fada Sub m where Fada follows this equation. Okay, Now, solving problems with a single slip is going to be exactly the same as solving problems with a double slit. Okay, The first thing that we're gonna do is we're gonna draw the figure that I have above me in the green box for a single slip, and it's gonna look identical to the figure for a double slit. Okay, let me minimize myself so I could draw this off to the side. All right, so here's our single slit. Right. This drawing is going toe look absolutely identical to that of a double slit. The Onley differences that I physically drawn one slip instead of two. Now I'm gonna draw that central axis, okay? And let me read the problem before continuing. Ah, 450 nanometer laser is shone through a single slip of with 4500. millimeters. If the screen is at a distance of 140 centimeters away from the slit, how wide is the central bright spot? Okay, so the screen is a distance of 140 centimeters away, which is equivalent to 1.4 m. Okay. And what we're looking for is the width of the central bright spot. So the central bright spot looks like this, right, and it will continue and there'll be a second bright spot and a third bright spot, etcetera. The equation that we have for the single slip tells us the locations off the dark fringes. So we know this is the M equals one dark fringe. This is the equivalent M equals one dark French on the bottom side. And this is a fatal one, that first dark fringe angle. And this is fatal one as well. And notice this distance is just that distance that we're looking for. This is the width, which I'll call w. That's the width of the central bright spot. Okay, now, to make solving this problem easier, we can notice that the top triangle triangle above the horizontal axis and the triangle below the horizontal axis are identical. So this height, why one and this height, Why one are the same. Okay, now all we need to do is find the angle fate a one using our equation so that we confined. Why one? So first, remember that our equation for the location of dark fringes in a single slit experiment is M lambda over D, where M starts at one. Remember, this is different then the location for the bright spots on a single slit experiment where M equals sorry. M started at zero. We're looking for theta one, so it is just gonna be one. What's Lambda? It's a 450 nanometer laser. Nano's 10 to the negative nine. What is the single slit width D O K. D was also different in the the single slit than it is in the double slit in the double slit. D represented the separation between the two slits for a single slit. D represents the width of the single slit. Okay, we're told that the single slides a width of 0.1 nanometers. So this is 0.1. Sorry millimeters and millions 10 to the negative three. Okay. Plugging this into a calculator, you're gonna find that this is 45 or that fate a one 0.26 degrees minus itself again. Now we can see we've just found fate. A one with this triangle right here. We can then find why one Okay, so I'm gonna redraw that triangle. The angle is data one which we know is 0.26 degrees. The base is 14 m and the height we called. Why one Okay. And once again, the width is not gonna be. Why won the width is actually going to be two times. Why one? Okay, so let's solve this triangle. Notice that we have the opposite edge and the adjacent edge to that angle. So we can say that the tangent of the angle is equal to the opposite edge, divided by de adjacent edge. And so all I have to do is multiply 1.4 up to the other side of the equation, Plug this into a calculator and we get an answer of 0.64 m, which is equivalent to 6.4 millimeters. Don't forget, this is not the final answer. Okay, This is simply why one the width w, which is what we're looking for, is two times why one It's twice this height or the some of those two heights, but they're the same. So this is to time 6. millimeters, which is 12.8 millimeters. Okay, Now, don't forget that the central bright fringe is twice as wide as all the other bright fringes. So what would the width of all the other bright fringes be? It would just be Why one which was 6.4 millimeters. Alright, guys, that wraps up this video on the single slit diffraction. Thanks for watching

2

Problem

Light from a 600 nm laser is shown through a single slit of unknown width. If a screen is placed 4.5 m behind the slit captures a diffraction patter with a central bright fringe of width 20 mm, what is the width of the single slit?

A

0.13 mm

B

0.11 mm

C

0.44 mm

D

0.26 mm

3

example

9m

Play a video:

Was this helpful?

Hey, guys, let's do an example. Light from a 500 nanometer laser is shone through a single slip of width 5000.5 millimeters with the screen placed 3.5 m from the slit. If the screen is two centimeters wide, how many dark fringes can fit on the screen? Okay, Thio illustrate what's going on in this problem. I'm actually going to draw on extra big diagram, Okay? Because this problem is a little bit different than problems that we've seen before. Now I'm gonna draw my central axis. Okay? We're told that this is the screen is 3.5 m behind the Sorry, the single slit. And we're told that the screen itself is two centimeters wide. Okay? And we want to know how many dark fringes fit. Okay, So what we're gonna have is we're gonna have, like, a dark fringe and a bright fringe in the dark and the bright right, these air alternating the huge central, bright fringe. It said, or something like that. Okay, now there's going to be a maximum angle. That light can leave the single slip and still reach the screen. Right? That's this angle right here. to the edge of the screen, and I'll call this fate a max. Clearly, if light leaves at a larger angle than that, it will not reach the screen. It'll go past the screen and so you won't see a diffraction pattern. The diffraction pattern ends at the screens with Okay, we haven't really addressed in any problem, either a double slit or a single slip problem. What happens when you consider the finite width of the screen? Okay, so this is the first problem that we're actually talking about that. But if you exceed this maximum angle, then obviously the light just goes past the edge of the screen and it's not captured and you don't see anything. Okay, how do we relate this to the number of dark fringes? Well, notice that each dark fringe Okay, I actually skipped one here, let me reposition myself. There's one. There's two. There's three. There's four etcetera. However many fit right. This is the M equals one dark fringe. This is the M equals two. This is the M equals three. This is the M equals four. Something important to remember is that the dark fringes are evenly spaced, angular. Early Okay, so whatever angle is right here, which I'll call five. The angle between the central axis and the first dark fringe is the same angle between the first dark fringe and the second dark fringe. And that's gonna be the same as the angle between the second and the third dark fringe, which is gonna be the same as the angle between the third and the fourth Dark fringe. OK, they are evenly spaced, angular, early. All right. Now notice that the angle phi is equal to the angle between the central axis and the first dark fringe. This angle right here, well, that we know by definition has to be theta one. So the first bit of information that we know is that five equals fate a one and that's really important. Our equation for the angles for dark fringes in a single slit is that sign of fate. A sub m is m lambda over D. So that tells us that sign of Fada one has to equal just lambda over D right when him is one. Okay, we know what Lambda is. We know what D is. So we confined sign of fatal one and then fatal one. This is a 500 nanometer laser Nano's 10 to the negative nine. The width of the slit is half a millimeter. So let's just point 5,000,010 to the negative three. And if you plug this into your calculator, you get an answer of zero 0.1 which is equivalent two fatal one equals 0. degrees. Okay. And remember that that equals five. The angle between each successive dark fringe. OK, now, how do we find out how many dark fringes there are? Well, if each dark fringes spaced by fi and we have some total angle Fada Max in the number of dark fringes is just gonna be Fada max divided by five. Okay, so the number of dark I'm just gonna call it number Dark is the maximum angle that light can still reach the screen divided by thigh. This isn't quite right, though. The reason it's not quite right is because look at theta Max. Theta Max is actually the angle between the central access and the furthest ray. But that's not going to tell us how many dark fringes we have. That's only going to tell us how many dark fringes we have on this side. You need the total angle, this angle to figure out how many dark fringes you have across the entire screen. So, technically, this should actually be two times data Max over five. Okay, so this is gonna be the equation that we're gonna use to figure out what the number of dark fringes are. We already know, Fi. Now we need to find fate of Ax. We'll notice that Theta Max is just given by this triangle right here. We know what the base of this triangle is. 3.5 m, right. What's the height of this triangle? While the width of the entire screen is two centimeters. So this should just be half that or one centimeter. So now I could just redraw that triangle, all right? And that should give us fate of Max. The adjacent edge, right. The base is 3.5 m, and the height I said was one centimeter. Okay, so we know the opposite edge. We know the adjacent edge so we can use the tangent. So the tangent of Theta Max is going to be the opposite edge, which is one centimeter since he is 10 to the negative too. And the adjacent edges the base or 3.5 m. If you plug this into your calculator, you get Which means that fate a max is 0 16 degrees. Okay, now we also have Fada Max. So all we need to do is plug Betamax and Fi into the above equation to find the number of dark fringes. That's two theta max over five. So that's two times 0.16 degrees. It's fine to leave this in degrees. You don't need to convert it to raid IANS. And this is 0057 degrees the width, the angular width or the angular distance between each dark fringe and plugging this into your calculator. You get 56 okay, so you can fit, or you can find at least five dark fringes in those two centimeters. Okay. You would also partially have some bright fringes hanging off the edge off both of those ends. Okay? Because you actually have 5.6, right? If this was five on the nose, then you would have a dark fringe at the edge at each edge of the screen. But since it's slightly above five, you have some sort of brightness at the edge of each screen. Doesn't end at a dark fringe. Alright, guys, that wraps up this problem. Thanks for watching.

© 1996–2023 Pearson All rights reserved.