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## Physics library

### Course: Physics library > Unit 14

Lesson 2: Interference of electromagnetic waves- Constructive and Destructive interference
- Young's double slit introduction
- Young's double slit equation
- Young's double slit problem solving
- Diffraction grating
- Single slit interference
- More on single slit interference
- Thin Film Interference part 1
- Thin Film Interference part 2

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# More on single slit interference

See how there's actually constructive (rather than just destructive) interference at some points on the screen. Created by David SantoPietro.

## Want to join the conversation?

- so is there an actual constructive spot in the very center where you have that large large peak?(9 votes)
- Correct. You can see this by looking at the formula.

Where theta=0 there will be no path difference. Consequently, there will be no way for the two wave fronts to get out of phase, and they therefore interfere constructively.(20 votes)

- What is the reason we get any additional constructive points besides the middle peak? If there are destructive interferences everywhere, why do we still see a pattern of peaks rather than just one peak in the middle?(15 votes)
- Because they are not completely destructive everywhere. Where they are completely destructive there we see a dark spot. And in other places we find mild bright places.(7 votes)

- At5:30why are the waves in different phases? Aren't they from the same source?

In the double slit experiment we assumed that they are from the same source since they originate from the same laser, so why not here as well?(10 votes)- Because they are travelling different distances.(3 votes)

- Oh! I'm super cooper CONFUSED!!

Is the bright spot in the middle made of constructive or destructive waves? If destructive waves cancel each other out as said in the first video,then how do they form a spot out there on the screen?

CAN SOMEONE PLEASE HELP ME?(8 votes)- Good question

Lets start by saying that ALL waves can be constructive and destructive; it depends on the way that they meet. Have a look at the topic again just to get really clear,

For the bright spot (or 'central maximum' this is where the waves will be meeting in a constructive way. (ie they are IN PHASE). If they were meeting in a totally destructve way, then there would be a dark spot. (ie a MINIMUM)

Hope that help your confusion

:-)(7 votes)

- if one peak meets one crest then they cancel out each other,is this violate the law that energy cannot be created nor destroyed since light is a form of energy?(3 votes)
- Nope ,

When a crest and a trough from two travelling wave meets , they just cancel out each other at that moment , after that specific moment there will be two waves again.

In a interference pattern , there are constructive and destructive interference points , making the energy(amplitude) at some points being " cancelled " while some being " doubled " , make the system's net energy not changing.(11 votes)

- what's the differrance between single slit interference and diffraction grating?

they both have independant waves in very short distance and i cant find the differance between placing many holes on the board to make many waves and considering wave as a group of independant waves(3 votes)- A diffraction grating has many, many slits. Why don't you watch the videos about the double slit and the diffraction grating.(2 votes)

- what happens if we take consecutive points instead of alternate ones?(3 votes)
- I wanna know that if the points are at the same wavefront. Don't they also have the same phases?(2 votes)
- yes, I think they would be in phase. if they belonged to the same wave front. interesting question

you could almost use that as a definition for a wave front...(3 votes)

- I am full on lost(3 votes)
- With the variable m, are we actually using it to multiply by its number and lambda or what is its purpose?(2 votes)
- It identifies which dark spot you are talking about when you calculate a location of a dark spot. There are many dark spots. The first one closest to the center corresponds to m=1. The next one is m=2, etc.(3 votes)

## Video transcript

- So I got to be honest with you. Single slit interference is confusing. In fact, this argument I gave earlier, first time I heard about this, I thought this was just
mathematical mumbo-jumbo. I was, like, "What are you talking about? "This makes no sense." It seems like you could
argue anything like this, But you can't. And I'm going to try to
show you that in this video. Specifically, I'm going
to try to show you that this same argument that we
made for the destructive points won't work for the constructive points. And I think that will help a little bit. In other words, this half
wavelength relationship won't give the constructive
points, not exactly. It'll give it approximately,
but it won't work exactly. Why? All right, well, let's look at this. So let's get rid of all of that. So let's say we were going to try to derive the formula for
the constructive points. The first thing I'd do is,
I'd say, well, all right, again, each point on here
defracts when it gets to the hole. I'll have infinitely many sources, but I can't draw infinitely many so let's just consider eight again. One, two, three, four, eight, I get my interference pattern on the wall, and this is the graphical representation. I mean, they're really,
like, smudgy lines over here, but the graphical representation
looks something like this. You'd have a big old
bright spot in the middle. These points on the end,
you keep getting them, but I can't keep drawing them over here. So it keeps on going. And I'd pick a constructive point. Right here, this is a bright spot. So this, if I had to give a guess for what point would be totally
constructive for all the waves, I'd say it's that point. So I'd take my topmost wave, I'd say it travels a certain
distance to this bring spot. I'd take my wave in the middle, travels a certain distance to get here. I'd imagine my line straight down here, to try to figure out the
path length difference. Now remember, this here is
the path length difference. What should that be in order
for these two purple waves to interfere constructively over here? It's got to be an integer wavelength. So one wavelength, two wavelengths ... since this is the first
one from the center, we'll just say that's one wavelength. What's the relationship? We know that. Remember, the relationship
for the path length difference and the theta, the angle that it's at, was just d sine theta. D is, well, the whole width
of the hole -- it's a w. So what width is this? This width between this source
of light and this source would be w over two. And the relationship I'd get? All right. D is w over two times sine
of theta, would equal ... For this first point, I'd
say it's equal to lambda. So let's just assume
these two are interfering constructively at this point. And that would give me: w times sine theta equals two lambda gives me a constructive point. Now, I'm already confused. What? W sine theta equals two lambda? Constructive? We already proved this
is a destructive point. Remember our relationship
for the destructive points that we derived was: w
sine theta equals m lambda as long as m is not zero, but
one, two, three, four, five. These are giving destructive
for any m equals one, two, it could even be negative if
you want to consider down here, any integer. Looks like we just proved
these are constructive. How are these constructive? Well, they're not, really. They kind of are, but watch what happens. Just, here we go. So if I follow this argument through, the thing that fails ... Our previous argument's fine. The argument that fails
is this current one with constructive, because yes, these two are constructive
there, but watch. This point here, now I ... just remember the game we played. We said if these two are constructive, then the rest of them
should all be constructive. Is that so? Well, let's go down
one, let's go down one. I imagine these two waves getting here. So far it's looking good. They're at the same angle. They're the same distance between them. I mean, this length here
is still w over two. So I'd still get w over
2, sine of the same angle, because it's the same point on the wall. So if w over two is the same,
sine of the theta's the same, then that's got to also be path
length difference of lambda, which means these two blue waves also interfere constructively. So this is looking pretty
good, which is kind of bad. I'll show you why. These two would also be constructive. Well, is this a constructive point? Is m equals two a constructive point or a destructive point in that end? It's a destructive point. This argument's failing,
and it fails because ... watch this ... Even though these two purple ones interfere constructively over
here -- here's a wave cycle. Even though the two purple
ones meet up constructively, let's say the top one was there, that means the one in the
middle, this one here, was also at the peak. So those two interfere constructively. How about the next two? Well, those two are going to interfere ... Now maybe those two are,
like, at this point. They're both constructive,
but they're not necessarily the same as the two purple ones. And how about the orange ones? Orange ones might be constructive, because they're both at the
same point in the phase, but they're not at the same
point as all the rest of them. You can have more. What about these down here? Oh, these might be down here. Those two together are also constructive, but you see the problem. Even though these two are constructive, this one's not constructive with this guy, and these all add up. In fact, for the most part, they cancel. That's why these are so little. You get these weak, you get
these really weak fringes on the sides of the single slit, because you're not going to get points where they all add up
really well necessarily. You get points where a lot
of them sort of cancel out. And it doesn't completely cancel those. Here's where I lied, for
the diffraction grating. Remember, for the diffraction grating ... Let me get rid of this. For the diffraction grating,
we had a single line and we made a ton of holes in it. And I said that diffraction
gratings are great because, if you come over to here, you make a ton of holes in here, instead of getting a
smudgy pattern on the wall, you get a big bright
spot right in the middle and then a well-defined,
well-defined, well-defined on each side of them, evenly
spaced, basically just zero and then extremely sharp. And then zero and then extremely sharp. And then zero and extremely sharp. And the whole argument I
made for defraction gradings was that the reason it's zero in between, the reason these are giving zero everywhere except these
constructive points was precisely because ... we come back over to here ... was precisely because of
this effect right here. This effect where they,
for the most part, cancel. Now I'm saying, "Eh, they don't actually
completely cancel necessarily." So these wiggles here are actually in a defraction grading pattern. They're just so small and unpronounced, compared to these, you
don't really notice them. What I'm saying is, if I wanted to draw
this more realistically, I would definitely have
this bright spot right here, but I'd have this in
between small variations, small points where it
becomes a little more, a little less, constructive
or destructive. What you have for a single slit is this: just one center bright spot. It's not going to be as well-defined because it's a defraction
grading, it's a single slit. But you still get these. You get these weird wiggles
that, for the most part, you ignore for a defraction
grading, but they're there. And for a single slit,
that's kind of all you got. So, can't really ignore it so much. Those are going to be there. It's because these
don't completely cancel. Our argument does not work. It works in the sense that two of these might be constructive. That means, you can pair
these off in constructive, but they won't all be at the
same point on their phase, which would give you a completely
constructive point there. So that's why we don't ... it's hard to find an exact formula. What's the formula for
the constructive points? Well, getting this formula
-- not quite as simple. You need to know a little
bit more physics to do that. And so, typically, in
introductory physics classes, you aren't asked to
find the exact locations of the most constructive points over here. Even these most constructive
points partially cancel. You do know how to find
the exact locations of the destructive points, though. And, if you wanted an approximate location of a constructive point, well, you can find the exact location of two neighboring destructive points, which, if you really wanted to, I mean, the constructive's in there, approximately in the middle, if you wanted to get a rough idea. I can still see some of
you being upset, though. You might say, "Wait. "Hold on a minute. "So we're saying this formula's good "for the destructive
points, but is this problem "we ran into for constructive points "also a problem for destructive points?" And it's not. I doesn't matter if
they're at different points in their phase for the destructive, because each pair cancels. In other words, when we
ran through this argument for the destructive points, look at -- if these two purple ones
cancel, then they cancel. I mean, if one was at the
peak, and then the other's at the trough or the valley,
those add up to zero. They're gone. Any effect they might have had on light hitting this point
on the screen is gone, completely negated. And so, what about the next two, These two blue ones? Well, those two, if these
two purple ones cancel, remember the argument went
that these two blue ones would have to cancel. So no matter where they're at, they're at some different
point on this cycle, let's say one's here and the other's ... well, that looked like the same. So let's say one's here and the other's at this corresponding 180
degree out-of-phase point. Well, they still cancel. That adds up to zero. And so it doesn't even matter that they're at different points in their phase. It doesn't matter. No matter where they're
at, one's 180 degrees out of phase with the other,
every contribution cancels out. You add up a bunch of zeros, you get zero. So destructive works fine. You don't run into the same
problem with constructive. It's a problem for the constructive points because these might add
up to some big number, and then the blue ones add
up to a different number, and the orange one adds
up to a different number, and then the red ones might
add up to a negative number, and you keep getting
these different numbers. You try to add them all
up, well, what do you get? That's why this formula's
not so easy to find. Adding up zeros, that's
easy -- just gives you zero. So, I hope I showed you that
this crazy mumbo-jumbo argument can't say anything whatsoever. And hopefully, that gives you
a little more justification, hopefully it makes you
believe a little more into this formula that we derived for
the destructive points. Those it does work for. And so, we can find
destructive points just fine. One more thing we can find is the width of this center bright fringe
here, the center bright spot. It's going to be wide. And since this goes to m equals one, now that first destructive
is over here, this is wide. This is, in fact, twice
as wide as all of these between these destructive points. And how wide is this? Well, you can find the angle to this first destructive
point up here, m equals one. You can find it to the
m equals negative one. You do a little trigonometry, you can actually get this length. That's another thing you can find exactly, is the width of this center bright spot, and the location is right in the center. But the location of these
constructive points up here, the exact location,
that's a little harder. You can find their width
again, because you can find the locations where they terminate. But finding where it
actually peaks in here, don't have an exact formula. We do have an exact
formula for the single slit destructive points. And that's typically what
you're going to have to find in these problems.