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Current time:0:00Total duration:15:00

Video transcript

let's talk about single slit interference now if I were you I'd already be upset and a little mad single slit interference interference wave interference is by definition multiple waves overlapping at a single point so how could a single slit ever produce multiple waves that could overlap I mean when we had a double slit if I put a barrier in here and we have a double slit at least then okay I send in my wave it gets over to here there's a small hole we know what waves do at a small hole they diffract which is to say they spread out at least with a double slit you would have two waves spreading out now they can overlap interference but for a single slit how are we ever going to get this well I never really told you what why do the wave spread out at a hole why does the fraction happen at all why when waves encounter a hole do they spread out and the answer to this question is the key to single slit interference and the the answer to why they spread out at a hole is something called Huygens principle and I can't say it this is a Dutch physicist scientist who figured this out Huygens principle and I apologize right now to all the Dutch people out there I'm butchering this name Huygens principle easier to spell than to say what he said he figured out something ingenious he figured out this if you've got a wave coming in these wave fronts number these wave fronts are like Peaks and in between them are the troughs or the valleys if you've got a wave front coming in propagating this way you could say yeah that wave front moves from here to there that's what it does or he realized with a wave you could treat every point on this wave as a source of another wave that spreads out spherically if in the forwards direction this wave spreads out spherically this point here he said that a wave front you can think of as an infinite source of waves each point is the source of another wave and you're thinking this is horribly complicated what what kind of messes is going to give you well if you add this up these are going to interfere with each other constructive destructive in a way that just gives you this same wavefront right back this is crazy but true if you let every point on this way be another wave source it will just add up to another wavefront here you'll just get the same thing back and this is the key to understanding why diffraction happens is because the wave spread the wave was already diffracting so to speak it was already doing diffraction every point on here was doing diffraction it's just it always added up with the other waves around it at every other point and gave you the same wave back but when there's a barrier when there's something in the way these here can't rejoin up with their buddies you just get this one here spreading you know and then this one down here spreads out all the rest of these get blocked now that these are blocked they're not going to get to interfere constructively and destructively with these points here and so what do you see when it hits the hole you just see this thing spreading out so it was always diffracting so to speak we just didn't notice it because it always added up when you've got a hole or a barrier that's when we actually notice it and this is the key to single slit interference because if I get rid of all that if we imagine our wave coming in here like this well this waves going to hit here every point it's the source of another wave so this point is going to start spreading out this point is going to start spreading out when we have a single slit we really have infinitely many sources of waves here and since some of them are blocked we could see an interference pattern over here on the wall because these can interact and interfere with each other what interference pattern are we going to see well on the wall over here we see a big ol bright spot right in the middle and if I were guessing I would have thought that was it big old bright spot because you shine a light through a small hole single hole you'll get a big bright spot there the weird thing is this jumps back up goes to a minimum zero point then it jumps back up and then it comes back up again and you get this these are going to be not very pronounced these aren't very pronounced you get a big bright spot in the middle these are relatively weak compared to other interference patterns that we've looked at and down here jumps up a little bit again over and over here so this is the pattern you see how can we get this how do we analyze it that's we're going to try to figure out to figure that out okay well this is a I said there's infinitely many sources here with when this wave gets to here that would take a long time to draw I'm going to draw eight so let's say there's one two eight sources let's just imagine there's eight here to make this a little bit easier to think about and the weird part is that this jumps back up here so let's look at this minimum right here let's look at this point where it goes to zero this destructive point so the wave from this topmost point this way from the topmost uppermost point has to travel a certain distance to get there I'm going to also look at the fifth one down this one that's basically halfway how about these two if these two interfere destructively the argument I'm going to make is if these two interfere destructively all the rest of them are going to have to interfere destructively why well we not play this game let's draw our right angle line here there we go and so we know that okay if these are going to interfere destructively this is the extra path length this extra path length this second wave this lower middle wave has to travel has to be what if I want destructive over here it's got to be a half wavelength three halves wavelength 5 halves wavelength that's how much it's got to be in order to be destructive so if this is the first point let's just say that's one half of a wavelength and what's the relationship between the angle that this is at on the wall compared to the center line well we already figured that out remember that relationship was d sine theta equals the path length difference between these that we derived the screen had to be very far away compared the width of the hole but that that relationship still applies what would D be in this case now we have to be pretty careful you have to be careful because this hole has a certain width we'll call that with W so if this hole has a certain width W how far apart are these these are not W apart these are W over two apart and so what's the relationship here for the path length difference between these two well if they're W over two apart I'd have d sine-theta the path length difference so D would be W over two times sine of the angle that this makes to this point on the wall and if their path length difference is lambda over two then that would be destructive so equals lambda over two and this is a little weird already because look I can cancel off the twos and what do I get I get the W the width of the entire width of the slit times sine of theta equals lambda this is giving me destructive remember before all the points that were integer wavelengths were giving me constructive this time it's giving me a destructive point over here and the reason is we played this game where W is the whole width these are only W over two apart that two cancels with that - okay but I really proved that this whole that they should all be destructive yet this is just for these two we've got infinitely many more in here how are we ever going to show that if these two cancel the rest of them cancel well we'll just pair them off look at this now imagine you come down one I go to this one I consider this wave that makes it over to here and the next wave down from this other one here so I move this one down a smidge and I move this one down a smidgen and I imagine these two waves traveling a certain distance to get over to this point what relationship holds between these two I can do the same thing these are also W over two apart so this here is also W over two so I get the same relationship I get W over two sign up is that going to be the same angle yeah it's the same angle same point on the wall this is really far away so these approximations are whole where these lines are supposed to be approximately parallel because this screen or this walls very far away compared to the width that equals well that's going to be the same thing I got W over two times sine of the same angle shoot that's got to equal the same thing that it did up here if the angle is the same my W over two is the same that's also going to equal half a wavelength that's also going to be destructive these two will also interfere destructively and I could keep playing this game I can pick this point here over to here and the next one down these two would have to be destructive I can pair them off and keep pairing them off I get destructive for all of them I can annihilate all of them by pairing them off and finding a partner that's destructive to that one and so this really is a destructive point this point over here all the light is gone completely annihilated gives you destructive so the short of it is that this relationship here this relationship that W the slit width times sine of theta the angle same angle we've always been defining it as equals integer wavelengths this time got to be careful though this time this gives you destructive points not the constructive points it was always constructive before this gives you destructive points now and you might be upset you might say hold on a minute we only proved this for this was just for N equals 1 or M equals one one way funny even prove this for anything besides M equals one well you can just as easily show that 3 lambda over 2 would also give destructive or 5 lambda over 2 that would give us all the odd integers here so M and here can be it can't be zero we'll talk about that in a minute it could be 1 2 3 4 5 and so on 1 we already showed 3 you get well if you made this three-halves wave that's also destructive that b3 five-halves wavelength the twos are always canceling so five halves light wavelength would work what about the even integers how do we get these well those come from the fact that I didn't have to pair these off with the top one and the middle one that's dividing this into W over two so pairing them off by lengths of W over two I can pair them off I can divide this by any even integer I can imagine pairing off instead of doing the topmost one in the middle one I can do the topmost one and skip one down here and so I can pair these off if I divide this into this distance right here that distance would be what that be W over four and so I can imagine pairing off okay if these two cancel if those two points cancel then the next one down so this one here and this one here would also cancel by the same reasoning and so I can play the same game now but W over four would be how I divided I can't divide it by anything I can't divide it by three or like two point five because I always want to pair these off in twos always twos thats it my whole plan that's my whole strategy here is to cancel these in twos and I can do that by dividing this by any even integer so W over four would work what would that give us okay W before the distance between these times sine theta equals this just say is the first one-half of a wavelength well if I solve this if I move the four over I get W sine theta equals to lambda so the twos also give us destructive interference I could divide by eight that would give us four once I move it over I can divide by any even integer any integer here is going to give us destructive points on the wall so this would be M equals one this would be M equals two and so on upwards so this relationship right here gives you all the destructive points how come n equals 0 is not a destructive point well M equals zero is right in the middle that's the most constructive points the brightest spot so M equals zero is not a destructive point but any other integer does give you a destructive point so this is the formula for the destructive points W is the entire width of the single slit theta is the angle the way we normally measure angle here you imagine a center line like that imagine a line up to your point on the wall this angle here would be theta and M is any integer that's not zero lambda is the wavelength of the actual light that you're sending in here now this just gives you the destructive points you might wonder hey I'm clever if the integers are giving us destructive points then the half integers should give us the constructive points if W sine theta equals you know lambda over 2 or 3 lambda over 2 is this going to give us constructive points and a not really so there are some complications here and if you're interested in why this does not give the constructive points I'm going to make another video watch that one because if you've been paying close attention you should be upset about something else too you should be upset about something earlier I've said might make it seem like we can prove this does not happen with the diffraction grading if you were paying close attention we proved quote-unquote that these do not occur and if you're upset by any of that or you want to know why the constructive formula does not exactly give you constructive points watch that video if you're happy with what we do know that this gives you the destructive points on the wall then you're good
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