Anderson Video - Double-Slit Interference Example

Professor Anderson
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...Bored and don't want to do that. That's okay, yeah? They're hungry. They're hungry for knowledge, that's what i'm going to go with. Hungry for physics knowledge. That's my takeaway message here. Okay, so let's look at uh- number five, and number five says in a double slit experiment the slit separate separation is two millimeters, two wavelengths 750 and 900 nanometers illuminate the slits. A screen is placed two meters from the slits, at what distance from the set- from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other? Interesting question. Okay so this is a double slit experiment but we have two different colors. So we're gonna draw this. Hopefully carefully. Here's our double slit, here is our distant screen. And now we're going to send in two different colors. And we're going to pick red and blue, even though those colors are red and near infrared. So, blue light coming in, red light coming in. Those are both plane waves, they illuminate both slits. And there is going to be a diffraction pattern over here. And when we think about the two wavelengths, and we can label them, blue is going to be 750 nanometers, red is 900 nanometers. Again, these don't correspond to the actual colors, right. 750 is at the red end of the spectrum and 900 is at the near infrared end of the spectrum. But what do we know? Well, if we go in with shorter wavelength light, blue light, we know that it's going to make some diffraction pattern over here. And we kind of know what that pattern looks like. It looks like this. You have these bright regions and you have these dark regions. And now if I go in with red light, am I going to see the exact same thing or am I going to see something different? Well, Let's go back to the equation for a second for two slits. The equation for finding the two slit maxima is d sine theta equals m lambda. d is this distance between the slits. Theta is this. It's the angle from the optic axis to a particular maxima. Right, that would be the m equals 1, the next one would be the m equals 2, next one will be the m equals 3 and so forth. So if I go to a higher wavelength does theta get bigger or smaller? Bigger, right? Bigger over here means bigger over here because sine of theta increases all the way up to theta of 90 degrees. So red is, in fact, going to have an overlap with the central maximum, But then the next one is going to be pushed out a little bit. And so the red pattern is, in fact, going to look like this. Alright, so why don't we label these angles as the following. This one we're going to call theta for the blue pen, and this one we will call theta for the red pen. Now, clearly they don't line up right there the way we drew them. Clearly the second one doesn't look like it lines up either, so we need to figure out where they are going to line up. And if they're going to line up then that means that we have the following. They line up when theta r equals theta blue. When those two angles, in fact, line up somewhere, then they are going to overlap the bright fringes up there. Alright, if theta r equals theta b then sine of theta r has to equal sine of theta b. But if I look at this equation, I know exactly what sine of theta is. Sine of theta is going to be m lambda divided by d. This one is going to be m lambda divided by d. But now I have to be a little bit careful. Right, I have to be careful because this m is not necessarily the same as that m. This is m for red, this is m for blue. Lambda on this side is lambda for red, lambda on this side is lambda for blue. So that equation, in fact, becomes this relationship between the two of them. And now we can probably figure out how to do this. m B over m red is going to be what? Well, I divided by the m red, the d's are going to cancel out on either side, and so I get lambda red over lambda blue. And lambda red over lambda blue is 900 over 750, and if you punch in 900 over 750 into your calculator, tell me what you get. What'd you get? 1.2 Okay, so the ratio of these m's is going to be 1.2. But m has to be an integer. Right? That has to be some integer. So if I start with m equals 1, I could have m equals 1 or 2 on top, that's not going to do it. If I go m equals 2, then i could do 2 or 3 on top, that's still not going to do it. And so you just need to express this in terms of a fraction of integers. And you probably know what the answer is, m B over m R has to be what? 6 over 5. Okay, the sixth fringe of the blue pattern is going to overlap with the fifth fringe of the red pattern. So those are the relevant numbers. And the question was, at what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other? And so now that you know one of these, you can just pick the height here, y m. And now we just need to do a little bit of trig to figure out what that corresponds to. Alright so, let's pick the theta R. Which corresponds to the m equals 5 pattern. Okay, in the problem they told us d, they also told us l. And so what we can say is d sine theta equals m lambda, but sine of theta in the small angle approximation is approximately y m over l. And so that equation becomes d times y m over l equals m lambda. And now we can solve for y m. y sub m is equal to m lambda l all over d. And now we have all those numbers. We know that m for the red one is five, we know that lambda for the red one is 900 nanometers. We know l, they gave that to us. l was 2 meters. And then d, they also gave us d. d was 2 millimeters. Which is two times ten to the minus three meters. And so we can put that down here two times ten to the minus three. And now I don't even think we need a calculator for this. Right, we've got 5 times 9 times 10 to the minus 7 and then we've got a 2 and then we've got a 2 times a 10 to the minus 3. the twos cancel out, 5 times 9 is 45. And then we have 10 to the minus 7 and we have a 10 to the minus 3 so I need to add 3 so I get 10 to the minus 4. So that is, of course, in meters, but all the answers are in millimeters, and 45 times 10 to the minus 4 is the same as 4.5 times 10 to the minus 3, which is 4.5 millimeters. And if you look at the answers, answer d is, in fact, 4.5 millimeters Questions on that one? You guys like the multi-color approach here? Give it a little flavor.
...Bored and don't want to do that. That's okay, yeah? They're hungry. They're hungry for knowledge, that's what i'm going to go with. Hungry for physics knowledge. That's my takeaway message here. Okay, so let's look at uh- number five, and number five says in a double slit experiment the slit separate separation is two millimeters, two wavelengths 750 and 900 nanometers illuminate the slits. A screen is placed two meters from the slits, at what distance from the set- from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other? Interesting question. Okay so this is a double slit experiment but we have two different colors. So we're gonna draw this. Hopefully carefully. Here's our double slit, here is our distant screen. And now we're going to send in two different colors. And we're going to pick red and blue, even though those colors are red and near infrared. So, blue light coming in, red light coming in. Those are both plane waves, they illuminate both slits. And there is going to be a diffraction pattern over here. And when we think about the two wavelengths, and we can label them, blue is going to be 750 nanometers, red is 900 nanometers. Again, these don't correspond to the actual colors, right. 750 is at the red end of the spectrum and 900 is at the near infrared end of the spectrum. But what do we know? Well, if we go in with shorter wavelength light, blue light, we know that it's going to make some diffraction pattern over here. And we kind of know what that pattern looks like. It looks like this. You have these bright regions and you have these dark regions. And now if I go in with red light, am I going to see the exact same thing or am I going to see something different? Well, Let's go back to the equation for a second for two slits. The equation for finding the two slit maxima is d sine theta equals m lambda. d is this distance between the slits. Theta is this. It's the angle from the optic axis to a particular maxima. Right, that would be the m equals 1, the next one would be the m equals 2, next one will be the m equals 3 and so forth. So if I go to a higher wavelength does theta get bigger or smaller? Bigger, right? Bigger over here means bigger over here because sine of theta increases all the way up to theta of 90 degrees. So red is, in fact, going to have an overlap with the central maximum, But then the next one is going to be pushed out a little bit. And so the red pattern is, in fact, going to look like this. Alright, so why don't we label these angles as the following. This one we're going to call theta for the blue pen, and this one we will call theta for the red pen. Now, clearly they don't line up right there the way we drew them. Clearly the second one doesn't look like it lines up either, so we need to figure out where they are going to line up. And if they're going to line up then that means that we have the following. They line up when theta r equals theta blue. When those two angles, in fact, line up somewhere, then they are going to overlap the bright fringes up there. Alright, if theta r equals theta b then sine of theta r has to equal sine of theta b. But if I look at this equation, I know exactly what sine of theta is. Sine of theta is going to be m lambda divided by d. This one is going to be m lambda divided by d. But now I have to be a little bit careful. Right, I have to be careful because this m is not necessarily the same as that m. This is m for red, this is m for blue. Lambda on this side is lambda for red, lambda on this side is lambda for blue. So that equation, in fact, becomes this relationship between the two of them. And now we can probably figure out how to do this. m B over m red is going to be what? Well, I divided by the m red, the d's are going to cancel out on either side, and so I get lambda red over lambda blue. And lambda red over lambda blue is 900 over 750, and if you punch in 900 over 750 into your calculator, tell me what you get. What'd you get? 1.2 Okay, so the ratio of these m's is going to be 1.2. But m has to be an integer. Right? That has to be some integer. So if I start with m equals 1, I could have m equals 1 or 2 on top, that's not going to do it. If I go m equals 2, then i could do 2 or 3 on top, that's still not going to do it. And so you just need to express this in terms of a fraction of integers. And you probably know what the answer is, m B over m R has to be what? 6 over 5. Okay, the sixth fringe of the blue pattern is going to overlap with the fifth fringe of the red pattern. So those are the relevant numbers. And the question was, at what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with a bright fringe from the other? And so now that you know one of these, you can just pick the height here, y m. And now we just need to do a little bit of trig to figure out what that corresponds to. Alright so, let's pick the theta R. Which corresponds to the m equals 5 pattern. Okay, in the problem they told us d, they also told us l. And so what we can say is d sine theta equals m lambda, but sine of theta in the small angle approximation is approximately y m over l. And so that equation becomes d times y m over l equals m lambda. And now we can solve for y m. y sub m is equal to m lambda l all over d. And now we have all those numbers. We know that m for the red one is five, we know that lambda for the red one is 900 nanometers. We know l, they gave that to us. l was 2 meters. And then d, they also gave us d. d was 2 millimeters. Which is two times ten to the minus three meters. And so we can put that down here two times ten to the minus three. And now I don't even think we need a calculator for this. Right, we've got 5 times 9 times 10 to the minus 7 and then we've got a 2 and then we've got a 2 times a 10 to the minus 3. the twos cancel out, 5 times 9 is 45. And then we have 10 to the minus 7 and we have a 10 to the minus 3 so I need to add 3 so I get 10 to the minus 4. So that is, of course, in meters, but all the answers are in millimeters, and 45 times 10 to the minus 4 is the same as 4.5 times 10 to the minus 3, which is 4.5 millimeters. And if you look at the answers, answer d is, in fact, 4.5 millimeters Questions on that one? You guys like the multi-color approach here? Give it a little flavor.