Video Transcript
In this video, our topic is two-slit interference. We’re going to see how light, when it’s passed through two equal openings, can interfere with itself. We’ll learn how to describe this interference mathematically and also why it is that the pattern this interfering light creates looks like bright spots of light that alternate with dark spots. All throughout our discussion, we’ll be working with this same basic setup. On the left, we have a barrier with two identical holes in it. These are our two slits or our two openings. And then over here on the right, we have a screen. Any light that passes through the two slits is projected onto this.
Speaking of light, in our setup, we’ll have coherent light waves, waves that have the same wavelength and maintain the same phase relationship, moving from left to right and being incident on this barrier and on the two slits. Now, if we had to guess what kind of pattern this light would create on our screen, we might think that it would be two bright spots on the screen that are in line with the openings. It turns out though that the pattern we do see is very different.
What actually appears when we pass coherent light through two parallel slits is a pattern of alternating bright spots like these here and dark spots, the spaces in between the bright spots. This pattern can seem very confusing until we remember that light is a wave. This fact has two important implications here.
First, because light is a wave, rather than passing straight through these openings like this, instead, it will spread out from each opening like this. This is called diffraction. And in the overlap between these two regions, this highlighted region here, those waves of light will interfere with one another. And they will do this according to what’s called the principle of superposition.
This principle tells us that when waves of any type overlap, then their displacements add together. So say, for example, that we had two waves like this and that they were overlapping. The principle of superposition tells us that when these waves overlap, that is, when they interfere, the resulting wave would look like this pink one. The amplitude of this resulting wave reflects the fact that the peaks of our two original waves lined up, as did the troughs or the low points of those waves.
We say that waves like this are in phase with one another. And therefore, this interference that’s going on is constructive interference. But what if our situation was different? What if the two waves that were interfering with one another had a peak of one wave that lined up with the trough of the other and, likewise, the trough of the original wave lined up with the peak of the other?
The principle of superposition still applies as always. But now the resultant wave would look like this. You would have an amplitude of zero. This is known as destructive interference. And it happens when two overlapping waves are one-half of a wave cycle out of phase.
Now, it’s exactly these kinds of interference, constructive and destructive, that are going on in this region. And that’s what creates this pattern of alternating bright and dark spots. The bright spots correspond to places on the screen where constructive interference takes place, whereas the dark spots show signs of destructive interference.
Let’s see how this might look for specific waves coming out of our two openings. Here, we’ve zoomed in a bit on our barrier with the two slits so that these two openings are now farther apart. And we’re going to look at a spot on the screen, this spot here, right in the screen’s center. Now, based on the way that waves of light spread out once they pass through these two holes, we know that light from both openings will be incident at this location on the screen. When that happens, those two waves might look like this. And notice that when these two waves arrive at the screen, they’re at the same point in their wave cycles. This means that the waves are in phase, and therefore they’ll interfere constructively. That tells us that this location on the screen will have a bright spot. The light is especially strong there because of this constructive overlap.
But now let’s pick another location on the screen, say this one up here. Once again, because of the way that light spreads out once it passes through the two openings, we know that light from both of our slits will reach this spot here. When it does, it can look like this. And if we look closely at these two waves, we can see that when they reach the screen, they’re one-half of a wave cycle out of phase. That is, the wave from our top opening is coming off of a minimum here and moving up towards its maximum, whereas the wave from the bottom opening is coming off of a maximum up here and moving toward a minimum, a trough.
When these waves meet then, they will interfere destructively, which means that there will be a dark spot at this point on the screen. So then, here on our screen, we have a bright spot, while here the screen is dark. We might wonder about the reason for this difference. It’s true that in the case of our dark spot, we have destructive interference going on, while the bright spot corresponds to constructive. But the real reason behind those two different types of interference has to do with what we call the path length difference of our two rays of light.
At this point, it’s important to note that even though we’ve drawn these waveforms to represent the light moving along from the two openings in our barrier, it’s not true that the light actually follows this wavy path. Instead, these waves represent an oscillating electromagnetic field. If we were to actually locate the light somewhere between the two openings it passes through and the screen where it lands, we would only ever find the light along straight lines like these. It’s these paths — and we can call their distances 𝑑 one and 𝑑 two, respectively — that the light actually moves along as it goes from the slits to the screen.
And so when we talk about the path length difference between these rays of light, we’re really talking about the magnitude of the difference between these distances. We could call that difference Δ𝑑. And that is the path length difference of our two rays of light. To figure out what this value actually is, let’s look at the rays of light as they leave the two openings.
Say that we draw a dot at the center of our top opening like this. And then from that dot, we draw a line down until it intersects 𝑑 two at a 90-degree angle. Now, let’s do this. Let’s say that the straight-line distance between the centers of our two openings is a distance 𝑑. And let’s also say that the angle between this dashed line we’ve drawn and our barrier is called 𝜃.
If we look carefully, we can see that we now have a right triangle, where the hypotenuse of that triangle is the distance 𝑑. Along with this, we can say that the lower side of the triangle is equal to Δ𝑑, the path length difference between these two rays. But then, looking at this right triangle, we can see that Δ𝑑 is equal to the hypotenuse 𝑑 multiplied by the sine of this angle we’ve defined, called 𝜃.
Now, in calling 𝑑 sin 𝜃 the path length difference of our two rays, we are making a bit of an assumption. We’re assuming that the two rays, characterized by their distances 𝑑 one and 𝑑 two, are parallel. Looking at our sketch, we can tell right away that that’s not technically true. We can see that these rays actually start out a distance 𝑑 apart. But they end up at the same point on the screen.
It turns out though that this assumption is not as bad as it may seem. That’s because in a typical two-slit interference scenario, the screen is actually much, much farther away from our two slits than it’s shown here. If that was the case, if our screen was much farther away, then our two rays still wouldn’t quite be parallel. But they’ll be much closer to it than we’re showing here.
Knowing that, let’s go with this assumption that our two rays really are parallel. If that is true, then that means this angle that we’ve called 𝜃 shows up elsewhere in the sketch. If we draw horizontal dashed lines out from the center of each of our two openings, then when our two light rays are parallel, the angles between those horizontal lines and the two rays are both the same. And they’re both 𝜃. This is a geometric thing to see, based on the fact that if our two rays really are parallel, then that means that this angle here is 90 degrees and this angle here is too.
Okay, so so far, we’ve seen that the path length difference between our two rays, assuming they’re parallel, is 𝑑 times the sin of 𝜃, where 𝑑 is the distance between our slit openings and 𝜃 is this angle that we’ve defined here. We can see that although 𝑑 will remain the same regardless of the direction the rays of light travel, the angle 𝜃 does depend on that direction. And therefore, the path length difference changes with the ray direction. It’s those changes that sometimes give rise to bright spots on the screen, while sometimes causing destructive interference, leading to dark spots.
Now, let’s consider the wave conditions required for a bright or a dark spot to be generated. Remember, we said earlier that if two waves are in phase with one another, then they’ll interfere constructively. So let’s say we have two waves and both of them have the same wavelength; we’ll call it 𝜆. Now, for these two waves we’re thinking of to be in phase, their phase difference could either be zero times 𝜆. In other words, they’re not shifted at all relative to one another. Or it could be one times 𝜆. Since the waves move in cycles, this is equivalent to not being separated by anything. Or the two waves could have a phase difference of two 𝜆 or three 𝜆 and so on. Really, any integer multiple of the wavelength 𝜆 would tell us that our two overlapping waves are in phase. And when they are in phase, that means they’ll interfere constructively. And that means a bright spot will appear at that point on the screen.
So here’s what we’re finding out. In order for a bright spot to appear on the screen, in other words, in order for constructive interference between our two waves to occur, the path length difference between them, 𝑑 times the sin of 𝜃, needs to be equal to some integer — we’ll call that integer 𝑛 — times the wavelength of the waves.
Now, our two waves, it’s worth pointing out, have the same exact wavelength because they had the same light source. This means that whatever wavelength our light waves have here before they reach the two slits, they’ll maintain, even as they pass through the openings and onto the screen. So then this relationship here tells us that we’ll get a bright spot on our screen whenever 𝑛 is equal to, say, zero or plus or minus one or plus or minus two and so on, any integer value. And just as a side note, when 𝑛 does equal zero, that means the path length difference between our two rays is zero. And that’s what we saw generating this bright spot here at the very center of our screen.
Now that we know the condition for bright spots, what about for dark spots? In other words, what path length difference gives rise to destructive interference between our rays? To see what that is, let’s look back up here at our two destructively interfering rays. We saw that the reason this interference is destructive is that the two waves are one-half wave cycle out of phase. We could represent that this way. We could say that one-half 𝜆 is the path length difference of the waves.
But then other conditions will lead to this same result too. For example, if the waves are three-halves of a wavelength out of phase or five-halves or seven-halves, and so on, for destructive interference, the path length difference of the rays needs to be some integer plus one-half all times the wavelength. We can see in this equation that if 𝑛 is zero, then it reduces to one-half 𝜆. So then the two waves are once again half a cycle out of phase. Or if 𝑛 is one, then we get three-halves in parentheses, once more leading to a half wave cycle difference.
Basically, to get this relationship for destructive interference, we copied the one for constructive interference and then added in a shift so that now the two waves are one-half wave cycle out of phase. That’s what makes the difference between constructive interference, leading to bright spots, and destructive interference, leading to dark ones.
Now, these two relationships we’ve developed do involve the assumption we mentioned earlier that our two rays of light are parallel. Strictly speaking, they’re not. But they’re so close to parallel, typically, that these equations are still quite useful for predicting when bright and dark spots will appear. So far, we’ve thought of the location of those spots based only on this angle 𝜃. But what if, instead, we described the position of, say, the bright spots on our screen in terms of a distance on the screen, starting at our center point?
To see how this works, let’s imagine that we zoom out a bit from our current view. Okay, so here we have our barrier, again with our two openings. And this time, they look to be much closer together than they did before. And also as before, we have our screen off to the right. Now, let’s say that our screen is separated from our barrier by a distance capital 𝐿 and also say that we marked the locations of a series of bright spots on our screen. In other words, these are the locations where constructive interference occurs.
Based on what we’ve seen so far, it’s possible to develop a relationship telling us the distances above the center point on the screen or below, as the case may be, that these different bright spots, these maxima, occur. To see how to do this, let’s consider this particular bright spot here. Even though we can see that this bright spot is one, two, three bright spots above the center point, we’re going to give it a general label. We’re simply going to call this the 𝑛th bright spot, which means that we can call the vertical distance from the center point of our screen to this spot 𝑦 sub 𝑛.
The question we want to answer is, what is this height 𝑦 sub 𝑛? Well, notice that from this perspective, our two slits, the openings in our barrier, are very close together. That’s fairly standard actually for a typical two-slit interference setup. The gap between these slits, that distance 𝑑, is often on the order of the wavelength of the incoming light 𝜆. For visible light, that wavelength is very small, and so 𝑑 is as well.
All this to say, at this scale, we can approximate the light coming from our two slits as coming from this point right here, at the center of those slits. So this is the path the light would follow to reach this spot on our screen. What we’ve got here then as we look at this shape between our barrier and our screen is a right triangle. And to define it a little bit further, let’s give a name to this angle of the triangle. Let’s call it 𝜃 sub 𝑛 to signify that it’s the angle corresponding to the 𝑛th bright spot.
Now, if we recall our trigonometry, we can say that the tangent of this angle, 𝜃 sub 𝑛, is equal to the opposite side of the triangle — that’s 𝑦 sub 𝑛 — divided by the adjacent side — that’s capital 𝐿. So we have this expression for the tangent of 𝜃 sub 𝑛. And now let’s do this. Let’s give a name to the hypotenuse of this right triangle. Let’s just call it capital 𝐻.
Thinking of another trigonometric function, the sine, we can say that the sin of 𝜃 sub 𝑛 is equal to the opposite side, 𝑦 sub 𝑛, divided by the hypotenuse, capital 𝐻. So notice that the only difference between the right side of these two expressions is what’s in the denominator of our fractions, in the one case capital 𝐿 and in the other case capital 𝐻.
Now, let’s think about this. What if this angle, 𝜃 sub 𝑛, was very small? We can see that the smaller this angle gets, the closer capital 𝐻 and 𝐿 will be in size. Indeed, if the angle went to zero, then 𝐻 and 𝐿 would be the same length.
We’re going to now make what’s called a small-angle approximation. We’re going to say that because this distance 𝐿 is typically very large compared to the other distances involved in this setup that this angle 𝜃 sub 𝑛 is quite small, which means that the hypotenuse of our triangle is approximately equal to the length 𝐿. And that means that the tan of 𝜃 sub 𝑛 is approximately equal to the sin of 𝜃 sub 𝑛.
Therefore, we can take this expression here and replace tan of 𝜃 sub 𝑛 with the sine of that angle. And now that we have this relationship here, which we derived for bright spots on our screen, we can find a similar sin 𝜃 term in our equation for constructive interference here. And moreover, this angle 𝜃 corresponds to some integer that we’ve already called 𝑛. So really, we could call our angle 𝜃 sub 𝑛 since it corresponds to this 𝑛th multiple of 𝜆.
So let’s do this. Let’s take this equation and we’ll write it up here. But when we do, we’re going to replace the sin of 𝜃 sub 𝑛 with 𝑦 sub 𝑛 over 𝐿. And the reason we can do that is because of this equation we’ve just developed here. We see that sin 𝜃 sub 𝑛 is equal to 𝑦 sub 𝑛 over 𝐿. So here’s what that gives us. 𝑑 times 𝑦 sub 𝑛 over 𝐿 is equal to 𝑛 times 𝜆. And if we multiply both sides of the equation by 𝐿 over 𝑑, then those two factors cancel on the left side. And we find that 𝑦 sub 𝑛 is equal to 𝑛𝜆𝐿 over 𝑑. This is the equation for the distance from the midpoint of the screen at which the 𝑛th bright spot, the 𝑛th maximum, occurs.
Knowing all this, let’s summarize what we’ve learned about two-slit interference. We saw in this lesson that when coherent light is passed through parallel slits, the light interferes with itself, creating a pattern of bright and dark spots. We saw further that constructive interference, leading to the creation of bright spots, occurs when the path length difference between the two rays of light, 𝑑 times the sin of 𝜃, is equal to an integer multiple of the wavelength of those rays.
On the other hand, destructive interference, creating dark spots on the screen, occurs when the path length difference between the two rays is equal to one-half of a wave cycle. And finally, the distance between the screen’s center and what we call the 𝑛th bright spot on the screen is equal to 𝑛 times the wavelength of the incoming light multiplied by the distance between the slits and the screen all divided by the distance 𝑑 between the slits. This is a summary of two-slit interference.
