Guys, in this video, we're going to talk about Young's double slit experiment, which is just an analysis of how light rays are going to interfere with one another after initially collimated light passes through a double slit. Okay? Let's get to it. A beam of light that's shown onto a double slit, that just means it's aimed at a double slit. Okay? Was thought initially before diffraction was understood to produce a single spot of brightness. Okay? However, when the experiment was performed, that wasn't the case. So in this image here on the left, we have the expected outcome of the experiment. If we shine this initially collimated light through a double slit, we expect to get just uniform brightness at a single point on the screen, the width of the slit. Okay? But what actually happens, what was demonstrated, is that there are multiple bright spots on the screen. That this initially collimated light, when it passes through the double slit, does not produce a uniform brightness, a constant brightness at one spot. It actually produces a varying brightness that alternates between bright and dark along the screen, and it produces a diffraction pattern. That's what we call this. This diffraction pattern is due to the lights interfering, so this is due to interference. That as these waves pass through these slits, they come out at different directions. So maybe here, you get constructive interference, and you get a bright spot. Maybe down here, you get destructive interference, and you get a dark spot. But that pattern of alternating brightness and darkness, that diffraction pattern is produced entirely by those two light waves interfering with one another when they reach the screen. Okay? And this is entirely due to diffraction, because if without diffraction, sorry, that collimated light encountering the slit goes through undisturbed. It goes through collimated. But with diffraction, when that collimated light encounters the slit, it spreads apart and comes out isotropic. So you get all those different angles for the light coming through. Okay? Now each slit produces a ray in every direction. Alright? Isotropic, right? The same in all directions. This means that some rays are going to interact constructively. You can see in the figure that here we have a maximum, and here on the top right we have a minimum. And when you have a maximum and a minimum, or a peak and a trough, you get destructive interference. Right? Down here you can see that both waves are at a maximum. When both waves are at maximum or both waves are at a minimum, you get this constructively interfere, and this depends upon where along the screen you want to look at the lights. If the light happens to constructively interfere, that means that the amplitude is going to get larger, you get bright, what are called, fringes. Up until this point, I've been calling them spots, but they're technically called fringes. Okay? That's the physics term for this. Some of those light rays are going to also destructively interfere, and they're going to produce a light with less amplitude or a dark fringe. And we're going to get alternating spots or alternating dark and light fringes, alternating spots of brightness and darkness. Okay? Now that's all conceptually, but how do we actually apply math to figure out where these bright spots are or the bright fringes and where the dark fringes are? Okay? The bright fringes are located at angles given by this one formula, that sinθ=mλd. Okay? Now what's m? M is what we would call an indexing number, and it's allowed to be 0, 1, 2, 3, etcetera. All positive integers and 0. The reason we have this so-called indexing number is because there is not a single angle where brightness occurs, a bright fringe, but there are a multitude of angles where this occurs. Okay? I have on this figure here the diagram that you're always going to draw when analyzing a double slit problem. Okay? Here we have the double slit where d is the width oh, sorry, the separation between the two slits. And we're going to draw a dashed center line that lines up with the central brightness peak, the m=0 peak. And if we choose an arbitrary bright spot, which is given by an arbitrary index, θmm is the angle that the light has to travel at to reach that bright spot. Okay? Likewise, we can also find the angles that dark spots occur at or dark fringes occur at. This is indexed by a different letter, n. Okay? It really doesn't matter that we use m and n. Maybe your book uses I and j. Maybe it uses m twice. That's also popular. The indexes are different for dark fringes and bright fringes. Okay? Because they occur at different positions, and the equation is slightly different. This is n+12λ/d. Okay? Where d once again is the separation between the two slits, n is the indexing number for the dark fringes, and λ is just the wavelength, what lambda has always been. Okay? And you want to memorize this figure because you're always going to draw this figure when trying to solve double slit problems. Alright? Let's do a quick example. A 650 nanometer laser is shown through a double slit of 10 10 millimeter separation. What angle is the 4th brightest fringe located at? If this double slit is 2.8 meters from the screen, how far from the brightest fringe is the 4th brightest? Okay? So let's draw our little diagram of the double slit problem, which is always what you want to start with. Okay? Now you're going to have a central bright fringe, and then you're going to have a less bright one, a less bright one, and a less bright one. Okay? And this is the first brightest. This is the second brightest. This is the third brightest, and this is the fourth brightest. Now remember that the equation for the angle for bright fringes is indexed by m, we called it. So the question is, what m does the fourth brightest fringe occur at? Well, m can be 0, 1, 2, 3, 4, etc. 0 is that first brightest fringe, that central fringe. The second brightest is m equals 1. The third brightest is m equals 2, and that fourth brightest is actually m equals 3. Okay? So this problem was written in a very specific way to teach you that just because this is the fourth brightest fringe, doesn't mean you have to assume that m equals 4. In this case, because m starts at 0, m actually equals 3. Okay? So we want to find that angle θm3 that the n equals 3 fringe occurs at or the fourth brightest fringe occurs at. So sinθ3=3λd, right? Which is going to be 3 lambda, the wavelength, we're told to 650 nanometer laser. Okay? So 650 nanos times 10 to the negative 9, and d is that separation distance between the slits. We're told that there's 10 millimeters of separation or 10 times 10 to the negative 3. And so plugging this into a calculator, you get 0.000195. Okay? Or you get that λθ3 equals 0.011 degrees. Okay? Something to note here is that this is an incredibly small angle. The angles are always really going to be very, very small. Alright? And that's because actually to get this equation that we use, we actually have to assume that the angles are always small. This might be something that you read in your book. This might be something your professor points out, but that's the reason why the angle should be small. Okay. They're always going to be this point 011 degrees. Alright. It's not even it's barely over a hundredth of a degree. It's a very, very small angle. Okay? Now the next part of the question is, if the double slit is 2.8 meters away from the screen, how far from the brightest fringe is the fourth brightest? We want to know this distance. How far is it? And I'll call that distance y. Okay? Look at this. This is just a triangle. So I'm just going to draw this triangle. This angle is our θm3 angle, which is 0.011 degrees. The base of the triangle is 2.8 meters, right? That's how far the screen is from the double slit, and the vertical distance sorry, let me actually call that y sub 3. That's just that vertical distance of that third, that m equals 3 peak. Okay? And the height of this triangle is going to be y. So clearly I can use trigonometry. I have the opposite edge and the adjacent edge, so I can use tangent, and say tangent of our angle 0.011 degrees equals the opposite edge y 3, sorry, over the adjacent edge, which is 2.8 meters. Or we have that y 3 if I just multiply 2.8 meters up, we have y 3 equals 2.8 meters times tangent 0.011 degrees, which plugging into our calculator is just 0 point 00054 meters or 0.54 millimeters. Okay? So you see how small that angle is? Even though this horizontal distance is 2.8 meters, because that angle is so small, because it's this tiny angle, the height y 3 is only half a millimeter roughly. Alright, guys. That wraps up this talk about Young's double slit experiment. Thanks for watching.
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Young's Double Slit Experiment
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