In this video, we’re talking about the diffraction of light waves. This picture we see here on screen is an example of that. But before we get into the diffraction of light waves specifically, let’s talk about diffraction in general. The reason we can do this is because all waves regardless of what type are capable of experiencing this phenomenon, called diffraction.
To get a sense for diffraction, imagine that we have a gigantic tank of water and we’re looking down on the water tank from above here. And let’s imagine that at one end of the tank, we have this big flipper arm that’s capable of moving down and up and down and up again. If this flipper arm starts to move up and down at this end of the tank, we know what will happen; it will start to generate waves. Since we’re looking down on these water waves from above, let’s say that we can identify the crest — the peak — of a wave with a line. So right here, this will be one wave crest and it’s moving from left to right. And as our flipper arm or wave generator continues to work, these crests will make their way down to the other end of the tank.
If we were to take a side-on view of these waves as they move along, they would look something like this. We know that if we continue to let this scenario unfold, these wave fronts will move on down towards the end of the tank and then ultimately reflect back. But what if before they get there, we put a barrier in their way? For example, what if we block off one-half of the waves path along this direction? If we do this, what will happen to these waves as they reach the barrier? Well, one idea — one hypothesis — is that the half of the waves that are blocked will be stopped in their tracks and the other half of the wave will continue on down towards the end of the tank.
But if we actually were to run this experiment, we would find that something else goes on. Rather than there being a line perpendicular to our barrier showing where the waves are on one side of the line and where there aren’t on the other, what we find instead is that the wave fronts that make it through this gap start to curl around the barrier. The wave fronts in a sense bend around the barrier so that for the area where we might assume that there are no waves, we actually do find some. It’s this bending around the barrier to the wave that’s called diffraction.
Thanks to diffraction, we see waves appearing in places where we wouldn’t otherwise expect them to be. This idea of diffraction may seem a bit strange. But actually, it’s behind a phenomenon that we’re all familiar with. Imagine that you and a friend are standing in a house, but you’re in different rooms in the house. Let’s say this is you in one room and this is your friend in another. These two rooms are connected by a doorway. And we know that even though you and your friend aren’t in line of sight with one another, you can’t see each other. By experience, we know that if you say something loud enough, your friend will still be able to hear you even though you can’t see one another. Since this happens all the time, we don’t think of it as unusual. But really it is kind of interesting.
In order for the sound waves of your voice to be able to reach your friend and be audible by them, those waves need to bend; they need to diffract through the opening of this doorway. If they didn’t, the only chance your friend will have of hearing you is if the sound waves reflected off the walls and then reached them. But this sound signal we know will be much weaker and much less likely to be heard. So the diffraction of sound waves, the bending of these waves around barriers, is the reason we’re able to hear one another in certain circumstances.
But then, this may raise a question: if sound waves are able to bend around barriers, why not light waves? In other words, if my friend out of sight and in a different room in the house can speak and I can hear them, then why can’t I see them as well, that is see their light reflected off them? The answer to this question comes down to a quality of the waves that are diffracting and the barrier or barriers that they’re diffracting around. We know that one way to represent a wave is the way we’ve done it here in the sketch using wave fronts. Each one of these wave fronts we could say indicates a peak or a crest on that wave form.
If we were to measure the distance between adjacent crests, that is wave fronts that are next to one another, then that measure distance is equal to the wavelength of this wave. In our sketch then, that wavelength would be equal to this distance or this distance or this distance. All of these distances are the measured distance between one wave crest and the next one that’s a wavelength. When it comes to diffraction, there’s a very special relationship between the wavelengths of the waves that are being diffracted and the distance between barriers — we could call it a gap — that the waves are diffracting through. Note that in this example, this gap is the open doorway. And the barrier is the wall that separates these two rooms of the house.
As we look at the distance of this gap and as we look at the distance of the wavelength of this wave that’s diffracting through the gap, we can see that they’re about equal with one another. And this is a plausible scenario. The wavelength of sound waves that we would speak — we could call that wavelength 𝜆 sub 𝑠 — is on the order of one meter long. That precise wavelength depends on what frequency we’re using with our voice. But roughly speaking, when we talk, the wavelength of the sound waves we produce is about one meter. One meter is also the approximate width of an average doorway in a house. And this brings us to a very useful principle of wave diffraction.
This principle is that for waves diffracting around the barrier, these waves are bent the most when the width of the gap that the waves travel through is equal to the wavelength. In other words, when the wavelength of a wave and the width of a gap that the wave travels through are the same, as the wave passes through that gap, it will bend more than it would bend for any other relationship between wavelength and gap width. This helps us understand why when we’re with our friend in different rooms of the house, we’re able to speak to one another. That is we can hear sound waves that are diffracted from the open doorway. But we’re not able to see one another.
In other words, light waves don’t bend as much when they travel around his barrier because the wavelength of a ray of light is thousands and thousands of times shorter than the wavelength of a sound wave. So while sound bends very easily around this one-meter wide doorway, light — visible light — that our eyes can see does not. Now at this point, we have the sense for what diffraction is. But it may still be a mystery just why diffraction occurs. That is what’s going on physically that makes this wave bend around the barrier? Why does that happen?
One of the best explanations for why it is that waves bend around barriers has to do with a certain way of thinking about his wave fronts. As we look at these wave fronts now, we may picture them or imagine them as a solid line moving along the crest of a wave. It turns out though there is another way to imagine these wave fronts as they move along. That method involves looking very closely at a wave front and saying that any point along a wave front — any of these dots we’re drawing in here — is actually its own source of a wave front. So for example, this top point we’ve marked out might create a wave front like this, whereas the next point would create a wave front like this, and so forth and so on for all the points we’ve identified. Note that we’ve only drawn in five points on his wave front with. But the theory says that every single point on the front can be considered its own wave source. But anyway, if we stick with these five we see that as these waves propagate out, what happens is the wave fronts from these individual points start to mix with one another. That’s called interference.
Now when we have a nice flat wave front like this one, the result of all this interference — all this adding and subtracting of wave amplitudes — boils down to another nice flat wave front. That is out of all the incredible complexity of all this mixing, we get a fairly simple result, a flat wave front. But that changes when our wave encounters a barrier. If we again put just a few dots on this wave front and consider the waves that come from these sources as we know all the points on this wave front would be, then when the fronts from these individual points mix and interfere, the result is not another flat wave front with the same width as this one. But instead, some of the wave fronts out at this end of things have nothing to cancel them out. And as a result, they continue to propagate out. And the direction of their propagation is around this barrier.
Basically, to explain why diffraction happens, we can use the language of wave interference. When two waves — this one here and this one here for example — interfere with one another and when those waves are in phase, that is the peaks of one wave line up with the peaks of the other and the troughs of one wave line up with the troughs of the other, then their sum — their combination — is a wave with an amplitude equal to the sum of the amplitudes of each of the individual waves that made it up. That’s constructive interference. And we know there’s an opposite type, called destructive interference. Destructive interference happens when waves mix but they’re completely out of phase. The peaks of one wave line up with the troughs of the other. When this happens, if the two original waves started out with the same amplitude, then the resultant wave has an amplitude of zero.
Now, getting back to all the waves created by these points along our wave fronts, these waves are all mixing, some constructively and some destructively. And the result of all that interference is the next wave front. Where that wave front exists, we see that there’s been constructive interference going on. And where the wave front doesn’t exist, say off to the sides of this plane wave, that’s where destructive interference has kept the wave fronts from expanding in that direction. And we see that destructive interference happens here, here, and here, but not over here. For this portion of the wave, something different is going on. Where before there was destructive interference, now there is more constructive going on. And as a result, the wave front is expanded. It bends we’ve been saying around this barrier. Let’s now get some practice with these ideas of wave diffraction through an example.
The diagram shows the wave fronts of two waves that have been diffracted through equally narrow gaps. Both waves have the same speed, wavelength, frequency, and initial displacement as each other.
Before we get to our questions, let’s take a look at this diagram. We see in this diagram the left gap as well as the right gap which are openings in his barrier for light waves to travel through. Now even though these two different light waves have different colours, we’re told that they have the same speed, wavelength, frequency, and initial displacement as one another. This means that for all intents and purposes, the light travelling to the left gap is identical to the light travelling to the right gap. After the light makes it through these gaps, it diffracts; it spreads out. And we see that it starts to mix, the light from the left gap with the light from the right gap.
In this diagram, a particular notation is used to represent wave fronts of this diffracted light. Each wave front is represented using a line. And the curvature of this line shows where the wave is travelling. Along with all this, we have these four points 𝐴, 𝐵, 𝐶, and 𝐷 marked out on the sketch. In this exercise, we’re going to answer questions about each one of these points starting out with point 𝐴. Regarding this point, we want to know how many wavelengths of this light is the left-hand gap from point 𝐴. How many wavelengths of this light is the right-hand gap from point 𝐴? Is the interference between the two light waves at point 𝐴 constructive or destructive?
As we start answering these questions, let’s clear some space on screen. Now in this first question, we’re asked how many wavelengths of this light is the left-hand gap from point 𝐴. An important point to realise here is that the light that goes through the left gap and the light that goes though the right gap is the same light. So again, even though these waves are coloured differently, they have the same wavelength, frequency, phase, and so on. So when the question asks about this light, it’s referring simply to the light in the diagram which is all the same light.
Knowing that, we want to know how many wavelengths of this light is the left-hand gap from the point marked out as 𝐴. Looking at our diagram, we see where this point is. It’s located on a wave front for the waves coming from the left gap and the waves coming from the right gap. And it’s at this point that we can remember an important fact about wave fronts. And that is that the distance between adjacent wave fronts that is two wave fronts that are next to each other is equal to a wavelength.
Going back to our diagram then, that means that point 𝐴 which is on the second wave front of the light that went through the left gap is a distance of one wavelength and then two wavelengths from that left-hand gap. That’s because we have two complete wave front cycles between that point and the gap. So that’s our answer to this first question. There are two wavelengths of light from the left-hand gap to point 𝐴 in the diagram.
Moving on to the next question, this one asks how many wavelengths of this light is the right-hand gap from point 𝐴. To figure this out, we’ll do something similar. We see that point 𝐴 is on the second wave front of the light that came through the right-hand gap. And that means for this light as well, there are one and then two complete wavelengths between that gap and point 𝐴. Once again then, our answer is two. That’s the number of wavelengths of this light the right-hand gap is from point 𝐴.
And then we have this last question about point 𝐴: is the interference that occurs there constructive or destructive? To answer this question, we can recall something helpful about wave interference. We can recall that when the crest of one wave overlaps with the crest of another wave, then that wave interference is constructive. On the other hand, when the crest of one wave overlaps with the trough of another wave, then that is destructive interference. So looking again at point 𝐴, we see that this point lies along the wave crests of both the light from the left-hand and the light from the right-hand gaps. So we have crest overlapping with crest. By our definition, that’s constructive interference.
Now that we’ve answered these questions about point 𝐴 in the diagram, let’s answer the same questions but this time about point 𝐵 rather than point 𝐴. So looking at point 𝐵 on our diagram, we first want to know how many wavelengths of the light the left-hand gap is from this point. To figure this out, we can again count wave fronts. Starting from the left-hand gap, we have one, two, three, four wave fronts, which means there are four wavelengths of this light from that gap to point 𝐵.
Next, we want to answer the same question, except now our start point is the right-hand gap. Once again, we’ll count the wave fronts starting from this gap. We count one, two, three, four wave fronts. And that tells us that there are four wavelengths of the light from the right-hand gap to point 𝐵.
Next, we want to know whether the interference at point 𝐵 is constructive or destructive. We see this point is at the overlap of two wave crests. Therefore, the interference at this point is constructive. That finishes point 𝐵. So now let’s answer these questions about point 𝐶.
Locating this point on the diagram, we see this it’s on a crest of a wave coming from the left gap. But it’s in between crests of the wave coming from the right gap. So answering this first question, how many wavelengths of light is the left-hand gap from this point, if we count wave fronts starting from that gap, we count one, two, three. But then moving on to the question about how many wavelengths this point is from the right-hand gap, if we count wave fronts once more, we get one, two, three, and then three and a half to point 𝐶.
The fact that point 𝐶 is three and a half wavelength from the right-hand gap tells us that for the wave coming from this gap, this point, point 𝐶, is at a low point, a trough of that wave. This means when we consider the last question is the interference at this point constructive or destructive, we see that point 𝐶 lies along a crest of the wave coming from the left-hand gap. But it lies along a trough of the wave coming from the right-hand gap. In other words, we have wave crest overlapping with wave trough. This is destructive interference.
And finally, we’ll do this all once more with point 𝐷. Starting at the left-hand gap, if we count wave fronts, we count one, two, three, four to point 𝐷. That’s the number of wavelengths of light from that gap to the point. And then from the right-hand gap, we count one, two, three wave fronts. Notice though that point 𝐷 lies along a wave crest of both waves. Therefore, we have crest overlapping with crest and the interference is constructive.
Let’s summarize now what we’ve learned about the diffraction of light waves. First off, we saw that diffraction is the bending of waves through an opening or around a barrier. We learned that all waves whether sound waves or light waves or water waves diffract. And we furthermore saw that diffraction, wave bending, is strongest when waves pass through an opening that is as wide as the wavelength is one. And finally, we saw that diffraction occurs due to wave interference or wave mixing with itself.