Number of Dark Fringes on a Screen

by Patrick Ford
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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.