If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Single slit interference

What happens when there's only one hole? Explore the concept of Single Slit Interference in light waves. Uncover how waves spread out at a hole due to Huygen's Principle, leading to diffraction. Understand how each point on a wavefront acts as a source of another wave, creating an interference pattern, and examine the math behind destructive points and the role of slit width in wave interference. Created by David SantoPietro.

Want to join the conversation?

  • blobby green style avatar for user rictoes
    what happens when you combine the concepts of single slit and double slit together? why cant you see the single slit pattern in double slit interference?
    (46 votes)
    Default Khan Academy avatar avatar for user
  • mr pink red style avatar for user Nart Barileva
    How come pairing the top point and 5th point from the top produces a distance of w/2 ? There are 4 inter-point intervals between those two points but only 3 inter-point intervals among the rest of the points. So it didn't divide the slit into half.
    (22 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user davidsantopietro
      Sorry for any confusion. Yeah, using just eight sources was a way to get the idea across. When the number of sources N is allowed to be infinite (or very, very large), the distance between the first source and N/2 + 1 source will approach w/2. I wanted to keep the number of sources even so we could pair each source in groups of two with none left over. Although, I suppose when the energy of the wave is distributed over infinitely many sources, having one infinitesimal source left over doesn't really matter anyways.
      (30 votes)
  • leafers tree style avatar for user Deya Chatterjee
    What is the difference between interference and diffraction?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leafers sapling style avatar for user Dominic Nguyen
    are you sure it is the same angle?
    (9 votes)
    Default Khan Academy avatar avatar for user
  • leaf yellow style avatar for user Guilherme Bortolotto
    By using the distance w/2 we find the θ of the first destructive interference. But, when we use a different distance (w/5, for example), a different value of θ can be found, which isn't necessarily multiple of the first one! Why does that happen?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Rebecca
      Below is my understanding, and someone can correct me if I'm mistaken:

      At , David says that IF the top and middle beams of light interfere destructively, then all of the other beams of light will interfere destructively (as every beam of light will have a "partner" that destructively interferes and thus cancels it out).

      In other words, in cases where the beams of light at the top and midpoint are destructive, then we will see this diffraction pattern. This will occur when the difference between the lengths of the beams of light are (lambda)/2.

      For other cases, the diffraction pattern will be less noticeable, or non-existent. Consider how we do not observe diffraction patterns when there are larger holes in, say, window curtains. These diffraction patterns occur under the assumption that the slit is small relative to the wavelength of the light - i.e. w<(lambda).

      Again, this is my understanding, as I was also confused by this and couldn't find a clear explanation on the internet.

      Additional note: for w>(lambda), see the next video, titled "More on single slit interference".
      (7 votes)
  • male robot johnny style avatar for user Moortal
    Why is w/2 chosen as the distance? Why not w/3?
    (7 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user rictoes
    Thanks Teacher Mackenzie! For some reasons I cant press the comment button, so I'll just post my question again here.

    Is it correct to say that double slit pattern is the sum of 2 single slit patterns combined together?
    If that is correct, then two slits that are very close will form a single slit pattern, and two slits that are very far apart will form two single slit patterns.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Teacher Mackenzie (UK)
      no; sorry to say that is not the case. The double slit pattern is not a combination of two singles slit patterns.
      But you can see BOTH patterns at the same time.
      For the double slit pattern, The waves from each slit 'combine' to form the regular maxima and minima of the double slit pattern. Here

      But you can see here that each of the slits has its own 'influence' on the pattern of maxima and minima.

      Here is a rough explanation:
      Imagine if we have two identical sound sources A and B. and combined, they produced a new sound C.
      The new sound C is due to the interaction of A and B.
      But then we find that A has its own independent effect on C
      And B also has its own independent effect on C
      So there are two effects going on here... the COMBINED effect of A and B
      and the independent effects due to A and the effect due to B.

      So thats what is happening here.

      See the green picture above....The combination of the two coherent sources produces the regular max/ min pattern of a double slit diffraction pattern. But look, it is gradually fading towards the edges...that is the effect of the single slit pattern.
      So the individual slits 'superimpose' their individual effects ONTO the double slit pattern. We say that the individual slits have imposed a single slit 'envelope' onto the double slit pattern

      looks like this

      hope that is clear.... not an easy concept to grasp, so please keep asking f its not making sense yet...
      (9 votes)
  • blobby green style avatar for user Andrew Wong
    Why must the width between the two paths be w/2 for them to destructively interfere as demonstrated at ?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Mukul Soni
    if width of slit is made double , how does it affect size and intensity of central diffraction band ?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user nilaykumar1
    Ok so, initially the equation is dsin(t) = m L/2 (t being theta, L being lambda). Unlike the double slit experiment, the d (the distance between two sources of wavefronts) can be varied. So theoretically I can pick any random point on the final screen, measure it's angle, t, from the point source being analyzed, and then pick a different point source that is a correct distance, d, from the original point source to destructively interfere with it at the final point on the screen. Basically, wouldn't it be true that for every point on the final screen, there will be two point sources that destructively interfere? And then using the logic of this video, couldn't i justify pairing up all the other points with their respective points of destructive interference? Obviously this isn't true, because then there would be no interference pattern at all, but what am i missing here? Why isn't this true?

    The decision to pick d = w/2 just seems so arbitrary to me. Why would the half-way point guarantee destructive interference?
    (4 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] 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, why do the waves spread out at a hole? Why does diffraction 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 answer to why they spread out at a hole is something called Huygen's Principle. And I can't say it. This is a Dutch physicist, scientist, who figured this out. Huygen's Principle. And I apologize right now to all the Dutch people out there, I'm butchering this name. Huygen's 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. Remember 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 can say, "Yeah, that wave front moves from here to there." That's what it does. Or, he realized, with a wave, you can 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 kind of mess is this 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 wave front right back. This is crazy, but true. If you let every point on this wave be another wave source, it will just add up to another wave front here. You will just get the same thing back. And this is the key to understanding why diffraction happens. It's because 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 and 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 re-join up with their buddies. You just get this one here spreading out. 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 of that, if we imagine our wave coming in here like this. Well, this wave's going to hit here. Every point's the source of another wave. So this point's going to start spreading out. this point's 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've thought that was it. Big ol' bright spot, because you're shining a light through a small hole. Single hole, you would get a big bright spot there. The weird thing is, this jumps back up goes to a minimum. A zero point. And then 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, it 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 what we're going to try to figure out. 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 big 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 top most point, this wave from the top most, upper most 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 half way. 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 know how to play this game. Let's draw out 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 of 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 wave length, three halves wave length, five halves wavelength. That's much it has be in order to be destructive. So if this is the first point, let's just say that's one half of a wave length. 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 sin theta equals the path length difference between these. That we derived. This screen had to be very far away compared to the width of the hole. But that relationship still applies. What would d be in this case? Now we have to be pretty careful. We have to be careful because this hole has a certain width. We'll call that width 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 distance between these two? Well if they're w over two apart, I have d sign theta as the path length difference, so d would be w over two. Times sin 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 two's. And what do I get? I get that w, the width of the entire width of the slit, times sin of theta equals lambda. This is giving me destructive. Remember before, all of 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 hole width. These are only w over two apart. That two cancels with that two. Okay, but I didn't really prove that this hole, 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. Okay, so I move this one down a smidgen, I move this one down a smidgen. 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'd get the same relationship. I'd get w over two. Sin of, 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 line are supposed to be approximately parallel because the screen or the wall's very far away in comparison to the width. That equals... well, that's going to be the same thing. I've got a w over two, times sin of the same angle. Shoot, that's got to equal the same thing that it did up here. If the angle's 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 can 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 could 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, this slit width, times sin 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 one. "Or m equals one. "One wavelength. "You didn't prove this for anything besides n equals one. Well, you can just as easily show that three lambda over two would also give destructive. Or five lambda over two. That would give us all the odd integers here. So m, m here can be... it can't be zero. We'll talk about that in a minute. It could be one, two, three, four, five, and so on. One we already showed. Three you get, well, if you made this three halves wavelength, that's also destructive. That'd be three. Five halves wavelength, the two's are always cancelling. So five halves 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 top most one and the middle one. I can do the top most 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'd 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 divide it. I can't divide it by anything. I can't divide it by three. Like 2.5, because I always want to pair these off in two's. Always two's, that's my whole plan. That's my whole strategy here, to cancel these in two's. 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 over four with the distance between these, times sin theta, equals, let's just say it's the first one, half of a wavelength. Well if I solve this, if I move the four over, I get w sin theta equals two lambda. So the two's also give us destructive interference. I can 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 a destructive point 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 m equals zero is not a destructive point? Well, m equals zero is right in the middle. That's the most constructive point. That's 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 is 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 sin theta equals, you know, lambda over two, or three lambda over two, is this going to give us constructive points? And eh, not really. So, there's 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, that 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.