Young's Double Slit Experiment

Hey, guys, In this video, we're going to talk about Young's double slit experiment, which is just analysis of how light rays we're gonna interfere with one another after initially column mated like passes through a double slit. Okay, let's get to it. A beam of light that's shown onto a double slip. That just means it's aimed at a double slip. Okay, was thought initially before two fractions 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 cultivated light through a double slit, we expect to get just uniformed brightness at a single point on the screen the width of the slip. Okay, but what actually happens? What was demonstrated is that there are multiple bright spots on the screen that this initially culminated light when it passes through. The double slit does not produce a uniformed brightness, a constant brightness at one spot. It actually produces a varying brightness that goes 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 inter fear incense that as thes 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 lightwaves interfering with one another when they reach the screen. Okay. And this is entirely due to diffraction because without diffraction Sorry, that column mated light encountering the slit goes through undisturbed. It goes through column mated, but with diffraction. When that column made light encounters the slit, it spreads apart and comes out ice a tropic. So you get all those different angles for the light coming through. Okay, Now each slit produces array in every direction. All right is a tropic, 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, Ray, we have a minimum and When you have a maximum and a minimum or a peak in a trough, you get destructive interference right down here. You can see that both waves are at a maximum when both waves or a maximum or both waves or at a minimum, you get this destructive interference. And this depends upon where along the screen You wanna look at the light. If the light happens toe 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 gonna also destructively interfere and they're going to produce Ah, light with less amplitude or a dark fringe. And we're going to get alternating spots are 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, where the bright fringes and where the dark fringes are? Okay, The bright fringes air located angles given by this one formula, that sign of the theta is M Lambda over D. Okay, now what's M? And this is what we would call an indexing number. And it's allowed to be 0123 etcetera. All positive integers and zero. The reason we have this so called indexing number is because there is not a single angle where a brightness occurs a bright fringe. But there are a multitude of angles where this occurs. OK, I have on this figure here the diagram that you're always gonna draw when analyzing a double slit problem. Okay, here we have the double slit where D is the width. 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 equals zero peak. And if we choose an arbitrary bright spot which is given by an arbitrary index, fate A M 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 in Okay, it really doesn't matter that we use em. And in maybe your book uses I and J. Maybe it uses him 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 in plus one half Lambda over D. Okay, Where d once again is the separation between the two slits. En is the indexing number for the dark fringes. And land is just the wavelength. Lambda has always been okay. And you want to memorize this figure because you're always gonna draw this figure when trying to solve double slit problems. All right, let's do a quick example. A 650 nanometer laser is shone through a double slit of 10 millimeter separation. What angle is the fourth brightest fringe located at? If this double slit is to 8 m from the screen, how far from the brightest fringe is? The fourth, 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 gonna have a central, bright fringe, and then you're gonna 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 am does the fourth brightest fringe? A correct? Well, M can be etcetera. Zero is that first brightest fringe, that central fringe. The second brightest is M equals one. The third brightest is M equals two. And that fourth brightest is actually M equals three. Okay, so this problem is 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 four in this case, because it starts at zero. I am actually equals three. Okay, so we wanna find that angle fate of three that the n equals three fringe occurs at or the fourth brightest fringe occurs at. So sign Fada three is three Lambda over D. Right, Which is gonna be three Lambda the Wavelength. We're told us 650 nanometer laser. Okay, so 650 nana list times 10 to the negative nine 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 three. And so, plugging this into a calculator, you get 0. or you get that lamb to three equals degrees. Okay, Something to note here is that this is an incredibly small angle, the angles air always really going to be very, very small. All right. And that's because actually, Thio 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 you're professor points out. But that's the reason why the angles should be small. Okay, They're always gonna be this this 0.11 degrees. All right, It's not even It's barely over 100 of a degree. It's a very, very small angle. Okay, Now, the next part of the question is, if the double slit is to 8 m 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. Why? Okay, look at this. This is just a triangle, so I'm just gonna draw this triangle. This angle is our state of three angle, which is 0011 degrees. The base of the triangle is 2.8 m. Right? That's how far the screen is from the double slit and the vertical distance. Sorry. Let me actually call that Y sub three. That's just that vertical distance off that third that chemicals threepeat. Okay, in the height of this triangle is gonna be why so clearly I can use trigonometry. I have the opposite edge in the adjacent edge so I can use tangent and say tangent of our angle. 0.11 degrees equals the opposite edge. Why three Sorry. Over the adjacent edge, which is 2.8 m. Or we have that y three. If I just multiply to 8 m up, we have, like three equals 2.8 m times tangent 0.11 degrees which plugging into a calculator is just zero point 00054 m or 054 millimeters. So you see how small that angle is, even though this horizontal distances 2.8 m. Because that angles so small because it's this tiny angle. The height. Why three is on Lee. Half a millimeter roughly. Alright, guys, that wraps up this talk about Young's double slit experiment. Thanks for watching.

Hey, guys, let's do an example about Young's double slit experiment. Ah, laser of unknown wavelength shines monochromatic light through a double slit of 0.2 millimeter separation. If a screen is 5.5 m behind the double slit, you find the angular separation of each bright fringe to be 0.15 degrees. What is the wavelength of the laser? Okay, first I wanna approach. I just want to discuss this big word right here. Monochromatic. Okay, this is fancy for single colored. Okay. Mono is latin for one chrome. Oh, is Latin for color. Monochromatic. Single colored. This just means light at a single wavelength. Okay? And lasers are most typically monochromatic. Okay, but there are multi chromatic lasers. Okay, so it's specifically monochromatic. The light is on. Lee admitted at a single wavelength, and that wavelength is unknown. That's what we want to find. So the first thing we're gonna do is we're gonna draw the situation because that's what we always do with these double slit problems. We're told that the screen is 5.5 millimeters behind the double slit. Sorry. 5.5 m behind the double slip. And what we're told is that the angular separation of each bright fringe is 15 degrees. What does this mean? Well, we have our central bright fringe, right, and then the second bright fringe and third bright fringe, etcetera. What this is saying is the angle separating each bright fringe is 0. degrees. If I were to draw a line through the next bright fringe, that angle would also be 0.15 degrees. If I were to draw a line through this next bright fringe right here, this would also be 0.15 degrees of separation between every single bright fringe. The angular separation is 0.15 degrees. So notice that first raid that I drew this blue one right here. This has to be our fada zero angle because that's the angle. Sorry, Fatal one. My bed for M equals one. We're talking about the angle between the M equals zero fringe and the M equals one fringe. So that is fatal One. And remember, for the angular location of bright fringes, our equation is sign of Fada M equals M lambda D. We know that the angle that we're looking for is of that second bright fringe, the one just after the central bright fringe which occurs when m equals one. So we have m equals one. And so sign of Fada one that angle that we want to find is one lambda over tea. Lambda is our unknown. We know what data one is right is 10.15 degrees. Just out of curiosity, what do you guys think fatal to would be the angle at the location of the M equals to do you think it be 0.15 degrees? No, it's the entire sweep of this angle. So it's 0.15 plus 0.15 which is 0.3. And you could also solve this problem by doing that by finding the fate of two angle that would also tell you the same wavelength. All right. And you guys can double check what I'm saying at the end of the problem, if you want for the fate of two or the fate of three or the fate of four. Okay, but for theta one, we know what D is the separations. 10.2 millimeters. We know what fate of one is. 15 degrees wavelength is our unknown. So I'm gonna multiply the D up to the other side. And this lambda is just d sign fate of one which is 0.2 millimeters. So times 10 to the negative 3 m and fada is 0.15 degrees. So our wavelength is 5 to times 10 to the negative 7 m. Now you can leave it like this and be done. That's the answer. But I'm gonna rearrange this slightly. When I'm gonna do is I'm gonna increase the order of magnitude by two. I'm moving the decimal place over two points, which is gonna increase my order of magnitude by two. So I need to decrease my ex phone and by two Alright. If I'm gaining two in the decimal place, I have to lose two in the power. Alright, This times 10 to the negative nine is nanometer. So this is 524 nanometers and you're gonna see most of your problems are gonna describe wavelength in nanometers because on the hundreds of nanometers about nanometers to 750 nanometers. If I remember correctly, that is visible light. That's likely. You can see 450 nanometers is purple light. The lowest wavelength light in 750 nanometers. Is red light the highest wavelength light. All right, guys. Sorry That wraps up this problem. Thanks for watching.