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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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# Single slit interference

What happens when there's only one hole? Created by David SantoPietro.

## Want to join the conversation?

- 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)
- that is such a cool question.

The answer is that you do see it. Its a little difficult to explain face to face even, so on here, may be a bit tough, but, here goes...

start with a single slit pattern. looks like a big hill in the middle with smaller hills on each side ok??

http://www.clemson.edu/ces/phoenix/labs/224/diffraction/fringes.jpg

The look at double slit. lots of equally sized hills (max) and valleys. (Min)OK so far??

http://images.fineartamerica.com/images-medium-large/1-double-slit-diffraction-pattern-omikron.jpg

So, where is the single slit pattern in the doubles slit pattern??

Well, if you look again at the double slit patter, the hills are not all he same size. They get smaller as you move away from the centre line. Now, how quickly they get smaller depends upon the size of the individual single slits that make up the double slit.

As the slits of the double slit arrangement get narrower, what happens to the double slit pattern? You will see that if the slits are very narrow, then the hills (maxima) in the double slit pattern get smaller but much slower.

One way to say it is this....

The interaciton of the light from the two slits produces the regular max-min pattern

The size of the individual slits determines the shape of the 'envelope' that controls the size of each of the maxima. Let me see if I can find a diagram to help...

OK; This is pretty good

http://physics.stackexchange.com/questions/109243/why-are-there-interference-patterns-inside-a-diffraction-envelope

hope that helps...if not, shout

But its a great question; shows good thinking(64 votes)

- 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)
- 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.(28 votes)

- What is the difference between interference and diffraction?(4 votes)
- Diffraction occurs when light bends around an obstacle. When light diffracts, it will create an interference pattern since the waves will no longer all be in phase from the coherent light source. Essentially the diffraction causes the interference to occur.(12 votes)

- are you sure it is the same angle?(9 votes)
- Yes, The angle measured in the video is correct.(1 vote)

- 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)
- Below is my understanding, and someone can correct me if I'm mistaken:

At5:27, 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)

- Why is w/2 chosen as the distance? Why not w/3?(7 votes)
- If you choose w/2, you can pair every point with another. If you choose w/3 you'll have a third of all the points left over(2 votes)

- 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)- 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

http://images.fineartamerica.com/images-medium-large/1-double-slit-diffraction-pattern-omikron.jpg

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

http://img.sparknotes.com/figures/C/c33e2bffc162212e1d9aa769ad3ae54f/envelope.gif

hope that is clear.... not an easy concept to grasp, so please keep asking f its not making sense yet...(9 votes)

- Why must the width between the two paths be w/2 for them to destructively interfere as demonstrated at6:55?(6 votes)
- if width of slit is made double , how does it affect size and intensity of central diffraction band ?(6 votes)
- 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)

## 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.