Video Transcript
In this video, we’re talking about the refraction of light. A great way to see an example of refraction is to try out what we have on screen here. Take a glass, fill it up with water, and then put a pencil or pen in the glass. When we look at the boundary between the water and the glass and the air just above it, we see that light from this pencil or pen is being bent. And that’s what it means for light to be refracted. And that’s a great place for us to start with the definition of this term, refraction. It’s the bending of light when that light crosses from one material into another.
Now when we talk about light, one way to think of light is as a wave. In other words, light is oscillating electromagnetic energy. At some points along the wave, the wave displacement is high, such as at the peaks or the troughs of the wave. And at some places, it’s low, right when the wave crosses over its midpoint. Another way to represent a light wave like this is to consider what it might look like if we were looking down on the wave, say from this perspective.
Looking at it this way, we could represent our wave using a series of what are called wavefronts. Notice that each one of these wavefronts lines up with the peak on the wave viewed from the side. So we have this light wave. And we’re looking down on it from an aerial view. Or we could call it a bird’s eye view. And this wave is moving from left to right.
Now if nothing was in the way of this wave, then it would just keep on moving in this direction forever. But let’s say that, rather than that happening, we put an object in the path of this wave of light. And let’s say further that we don’t know exactly what this material is that we’ve put in front of the wave. But we do know that it’s more dense than the medium in which our wave is currently traveling. So the question is, what will happen to the wavefronts of our wave when they reach this material?
Well, if we were to draw in the very next wavefront on our wave as it moves from left to right and assume that the material wasn’t there. That is, the wave was still traveling in its regular medium. That wavefront would look like this. And in fact, part of what we’ve drawn here is correct. The part of the wavefront that isn’t yet inside the material does indeed look like that.
But for the part that is now inside this new material, it doesn’t behave in the same way. We said that this material, whatever it is, is more dense than the material the wave was originally traveling in. Practically, that means that the wave will slow down as it enters this material, which means that this part of the wavefront will actually lag behind the other part that isn’t yet in the material.
We could almost think of it this way. Imagine you’re driving a car along the road. And then at one point on one side of the road, there was a muddy patch. Well, as our car reached that part of the road, the tires on the right-hand side would still be getting good traction with the road. But the tires on the left-hand side would be encountering this mud and slowing down. It’s kind of like that with this wavefront, which is partially in and partially out of this new medium.
So like we said, the part of the wavefront that’s inside this material is moving more slowly than the part that’s outside. So it would look something like this. Notice that this wavefront now has a bend to it. And if we kept following this wave as it completely moves inside this new material, we would start to notice that the wave is now going off in a new direction compared to the direction it originally traveled in. In fact, there are two observations we can make as we watch these wavefronts travel through this new medium.
The first thing we notice is that when light enters a new material, it changes direction. And then, secondly, we can say that when light enters a new material, its wavelength changes. Take a look once more at the wavefronts of this wave as it moves along. We know that these wavefronts represent the distance between consecutive peaks in our light wave. In other words, the distance from one wavefront to another is the wavelength of the wave. This distance from one wavefront to another outside a material is clearly greater than the distance inside the material. This means that as our wave enters this material, its wavelength is getting compressed, shortened.
What we’re seeing then is this point number two. That when light is entering a new material, its wavelength isn’t the same. Now regarding this first observation we made, there’s one caveat that’s worth mentioning. We’ve said that when light enters a new material, it changes direction. Technically though, there is an exceptional case where this doesn’t happen, where the light doesn’t change direction even as it enters a new material. If the surface of our new material was oriented at exactly 90 degrees to the incoming light wave, then in that condition only this wave wouldn’t bend. It would keep on moving left to right in a straight line. But for any and every other angle of incidence besides 90 degrees, the light wave would change direction as it enters this new medium.
Now there’s a basic physical fact that’s behind both of these observations we’ve made. And that is that as we saw, when light enters a medium, its speed changes. In this case, we saw wave speed decrease as the wave entered this medium. And that’s why our wavefront that’s half in and half out of the medium looks the way it does. Part of it, the bottom part, is moving faster than the other part, the top half.
Now this fact that light speed changes as it moves from one medium to another is important enough to introduce to us a new equation and a new term. Considering our light wave once again, let’s say that while the wave is outside the medium, it’s moving at the speed of light, 𝑐. And let’s say further that once the wave enters this medium, it’s moving at a different speed that we’ll call 𝑣.
Now if we take the ratio of those two speeds, 𝑐 to 𝑣, the speed of the wave before and after it enters the medium, then this fraction is equal to what is called the index of refraction of the material. The reason this term has this name, index of refraction, is because it’s a measurement of just how much this material will make light bend when light enters it. The more a material slows down light compared to its speed outside the material, the more the light wave will bend as it enters that medium. As we see here, index of refraction is typically represented using a lowercase 𝑛.
Now let’s say that this particular material we’re putting in the path of this wavefront has an index of refraction such that it slows light down so that it’s moving at one-half the speed of light in vacuum. In that case, the index of refraction of the material would equal 𝑐 divided by 0.5𝑐. Or simplifying this fraction, it’s equal to two. In other words, an index of refraction being equal to two physically means that light slows down to half its speed in vacuum by entering that material.
It’s worth pointing out that any material, whether water or glass or diamond, has its own particular index of refraction. And in fact, the medium that our wave was traveling in before it reached this particular material has its own index of refraction. We can call that index 𝑛 sub 𝑖, the initial index of refraction. And then we’ll call our index of our material 𝑛 sub 𝑓.
Just as a side note, notice that if our wave is traveling at the speed of light in vacuum in a particular material, then that means that material must be vacuum. And the index of refraction of a vacuum is equal to the speed of light in vacuum, 𝑐, divided by the speed of light in that material, which is also 𝑐. So the refractive index, which is another name for the index of refraction of a vacuum, is one.
Every other medium we encounter, whether air or another gas or a solid, has an index which is higher at least slightly than this value of exactly one. We’ll come back to that point a bit later. But for now, let’s look at how these relative indices of refraction, 𝑛 sub 𝑖 and 𝑛 sub 𝑓, affects how much our light wave bends when it crosses this interface into the new material.
To understand this more clearly, let’s imagine another scenario where we have a ray of light running into an interface, the surface of a new material. Now in this case, unlike before, we’re not representing this ray of light using wavefronts. But instead, to keep our diagram simple, we’re just drawing it as a single ray.
Physically though, the same thing is going on now as before. There are a series of advancing wavefronts that are moving along in the direction of this wave. And they’re encountering this new material. Whenever we have a ray of light moving from one medium into another, one of the most helpful ways to orient ourselves to this scenario is to sketch in what’s called a normal line perpendicular to the plane of that surface. And in this instance, that word “normal” doesn’t mean ordinary or regular. But it literally means at 90 degrees to the surface.
Now let’s say that, above this surface, in what we could call the original medium, there’s an index of refraction we’ll call 𝑛 sub 𝑖. And then below the surface, inside this new material, there’s an index we’ll call 𝑛 sub 𝑓. If 𝑛 sub 𝑖 was equal to 𝑛 sub 𝑓, then here’s what would happen. Our ray of light would enter this new material. But it would keep going in a straight line without being bent or refracted at all. That’s because, from the perspective of this ray of light, it’s entering an optically similar material. But let’s say that’s not the case. Let’s say that 𝑛 sub 𝑖 and 𝑛 sub 𝑓 are actually different.
Now this brings up two possibilities for just how they’re different. It could be that 𝑛 sub 𝑖 is less than 𝑛 sub 𝑓. In other words, our ray is entering a region of higher index of refraction. If that’s the case, then our ray will bend. And the direction it bends will be toward the normal line that we’ve drawn here inside our new medium. So compared to the way the ray would’ve gone if it hadn’t bent at all, when 𝑛 sub 𝑓 is greater than 𝑛 sub 𝑖, it bends inward toward the normal line. That’s one possibility for how 𝑛 sub 𝑖 and 𝑛 sub 𝑓 relate.
But the other possibility is that 𝑛 sub 𝑖 is greater than 𝑛 sub 𝑓. And in this case, the ray bends in the opposite direction, away from that normal line. In this case then, our ray might look like this. Where we see that, compared to the direction it would’ve gone if it hadn’t refracted at all, it’s bent away from the normal line. Now this is refraction. It’s the bending of light as it passes from one material into another.
And in order to quantify this bending, it’s helpful to label two angles on this sketch that we made. The first angle we’ll label is called the angle of incidence. We’ll represent it as 𝜃 sub 𝑖. That’s from the normal line above our interface to the incoming ray. And then the second angle we’ll label is called the angle of refraction. And that goes from the normal line inside our new material to the refracted ray. One way to represent the angle of refraction using symbols is to call it 𝜃 sub 𝑟.
Just as a quick side note, we know that when we have a ray like this incident on a surface, it’s possible not only for that ray to be refracted into the surface. But it can also be that some of the ray is reflected from it. If that were to happen, then we would have what’s called an angle of reflection. And since that word also starts with R like refraction, we might be tempted to label this angle as 𝜃 sub 𝑟.
The point is, in those scenarios where we have reflection as well as refraction, we just need to be careful about our notation to keep the angle indicators straight. We wouldn’t want to have two quantities referred to by the same name, 𝜃 sub 𝑟. But in this case, life is simpler because we don’t have any reflection. So our 𝜃 sub 𝑟 indeed refers simply to the angle of refraction of this ray.
Now these four variables — 𝑛 sub 𝑖, 𝑛 sub 𝑓, 𝜃 sub 𝑖, and 𝜃 sub 𝑟 — are all connected to one another through a law known as Snell’s law. Here’s what this law says. Snell’s law holds that if we take the original index of refraction, 𝑛 sub 𝑖, and multiply it by the sine of the angle of incidence of our ray, 𝜃 sub 𝑖. Then that product is equal to the final index of refraction, 𝑛 sub 𝑓, multiplied by the sine of the angle of refraction, 𝜃 sub 𝑟.
One reason this law is so great is because if we know three of the four variables — 𝑛 sub 𝑖, 𝜃 sub 𝑖, 𝑛 sub 𝑓, and 𝜃 sub 𝑟 — then we can solve for the fourth one using this law. Let’s consider how this works through an example.
Let’s give ourselves some specific values for some of these variables. Let’s say that the initial index of refraction that our incoming ray is in is 1.3. That’s roughly the index of refraction of water. And then we’ll say our final index of refraction is one, which is approximately the index of refraction of air. And let’s say further that the angle of incidence, 𝜃 sub 𝑖, of our ray as it reaches this water–air interface is 30 degrees.
Knowing all this, we want to solve for the angle of refraction of our refracted or bent ray. And we can do this using Snell’s law. Looking at this law, we know that, in our case, 𝑛 sub 𝑖 is equal to 1.3, the index of refraction of water. That 𝜃 sub 𝑖, the angle of incidence, is 30 degrees. And that 𝑛 sub 𝑓, the final index of refraction, is equal to one, about the index of refraction of air. So we can substitute these values in to Snell’s law.
And we see that 1.3 times the sin of 30 degrees is equal to one times the sin of 𝜃 sub 𝑟, where 𝜃 sub 𝑟, the angle or refraction, is what we want to solve for. We can simplify this equation a bit because, on the right-hand side, one times any quantity is equal simply to that quantity. So we can erase the one.
And as a next step, what we’ll want to do is isolate 𝜃 sub 𝑟 by itself on this side of the equation. To do this, we’ll apply an operation to both sides of the equation that inverts this sine operation we have going on. The best way to invert the sine operation is to apply the arcsine or inverse sine to that value. When we do this on the right-hand side, the inverse sin of the sin of 𝜃 sub 𝑟 simply equals 𝜃 sub 𝑟.
So now to solve for 𝜃 sub 𝑟, we only need evaluate the left-hand side of this expression, the arcsin of 1.3 times the sin of 30 degrees. When we enter this expression on our calculator, to two significant figures, we find a result of 41 degrees. So that answers our question of what the angle of refraction of this ray would be given this angle of incidence and our initial and final indices of refraction.
Now let’s think back to the two observations we made earlier in the video. The first thing we observed was that when light enters a new medium, its direction changes. Snell’s law addresses that change in direction through this equation. But we also noticed that the wavelength of light entering a new material changes too. To see that a bit more clearly, we’ve sketched in here the wavefronts on both the incident part of the wave as well as the refracted part.
Recall that the distance between adjacent wavefronts represents the wavelength of the wave. And that this distance clearly increases once the wave has crossed over into our new medium, 𝑛 sub 𝑓. If we were to call the original wavelength of our wave 𝜆 sub 𝑖 and the final wavelength 𝜆 sub 𝑓, then it’s the case that 𝜆 sub 𝑓 divided by 𝜆 sub 𝑖 is equal to 𝑛 sub 𝑖 divided by 𝑛 sub 𝑓. In other words, the ratio of the indices of refraction tells us how much the wavelength of light changes as it crosses an interface. One thing to be careful about is the subscripts to these values. Notice that, on the left-hand side, we have a final value to an initial value, whereas on the right, we have initial to final.
Let’s take a moment now to summarize what we’ve learned about the refraction of light. The first thing we learned is that light refracts, that is, bends, whenever it enters a new material. We learned further that refraction is due to the change in wave speed that happens whenever a wave crosses an interface. That change in wave speed causes a change in wave direction as well as in wavelength. That change in wave direction is summarized by an equation known as Snell’s law. Where it relates the initial and final index of refraction with the angle of incidence and the angle of refraction. And that the change in wavelength is described by this equation, which says that the ratio of the final to the initial wavelength is equal to the ratio of the initial to the final index of refraction.
And lastly, we saw that this term, the index of refraction, represented using a lowercase 𝑛, is equal to the ratio of the speed of light and vacuum to the speed light has when it enters that particular material.
