Video: Critical Angle for Total Internal Reflection

In this video, we will learn how to relate the paths of refracted and internally reflected light rays to the refractive index of media they travel in.

13:05

Video Transcript

In this video, our topic is the critical angle for total internal reflection. As we’ll see, this angle shows us how light can approach a boundary but not cross that boundary. This leads to an important optical effect called total internal reflection.

As we get started, let’s recall a few things about a ray of light being incident on a boundary. One of the first things we can say is that what makes an optical boundary what it is is the fact that the index of refraction on either side of the boundary is different. So here we have these two indices, where 𝑛 one and 𝑛 two are not equal to each other.

Another thing we can recall is that our ray of light approaches the boundary at some angle. And we measure this angle relative to a line drawn perpendicular to the surface where the ray of light reaches it. It’s common to call this angle, at which a ray of light is incident on a boundary, the angle of incidence and represent it using πœƒ sub 𝑖.

Coming in at this angle, if our ray wasn’t bent or refracted at all as it crosses this boundary, then its path would look like this dashed line. But we know this is not the path the ray of light follows because 𝑛 one is different from 𝑛 two. We can say that this boundary line separates optically different materials. Because of that, if and when our ray of light crosses over into the material with refractive index 𝑛 two, it will refract. It will bend off of this dashed line that we’ve drawn in.

It’s worth pointing out that, in general, some of the light in our incoming ray will be reflected off this interface. But for now, we’re focusing only on the part that goes into the new material and gets refracted. There are two different ways that the refracted ray can bend. One way is for the ray to bend closer to this normal line here. This kind of bending takes place if the index of refraction of the material our light started in is less than the index of refraction of the material it moves into. So if 𝑛 one is less than 𝑛 two, we say that our ray bends toward the normal.

And then the second way our ray of light could bend is away from this normal line. This happens when 𝑛 one is greater than 𝑛 two. In other words, the index of refraction of the material the ray is leaving is greater than the refractive index of the material it’s entering.

Now, let’s say we indeed do you have a situation where 𝑛 one is greater than 𝑛 two. So our ray of light really does bend like this. And now we can identify another angle, this one, and recall that this is called the angle of refraction and often symbolized πœƒ sub π‘Ÿ. And as a last bit of background, let’s recall that there’s a mathematical relationship that connects these variables: πœƒ sub 𝑖, πœƒ sub π‘Ÿ, 𝑛 one, and 𝑛 two. That relationship is called Snell’s law. It tells us that the index of refraction of the material our ray begins in multiplied by the sine of the angle of incidence is equal to the index of refraction of the material the ray travels into times the sine of the angle of refraction.

Now, keeping in mind that the refractive index of the material our ray starts in is greater than the index of refraction of the material it moves into, let’s say that we start to change the angle of incidence of our incoming beam. Specifically, let’s say we start to increase it. So whereas before we had an angle of incidence like this, now let’s say that this is our greater angle of incidence.

Doing this will change the way the ray of light is refracted as it crosses the boundary. In particular, increasing angle of incidence also increases angle of refraction. So our new ray may be refracted to look like this.

Now that we have these new angles of incidence and refraction, let’s say we increase them even more. Now, this is our new angle of incidence, and this is our new angle of refraction. And we can see that we’re getting pretty close to a limit with our refracted angle. To reach that limit, say that we once again increase our ray’s angle of incidence and that, at this incidence angle, our refracted ray moves like this. That is, it moves along the boundary between the materials of our interface.

When this happens, when the angle of refraction of a beam is 90 degrees, then we’ve reached the limit of our incoming ray actually leaving the first material and entering the second material. At this angle of incidence, it doesn’t really do either. Rather, it moves along the boundary between the two. In recognition of reaching this limit, we give this incidence angle a special name. We call it the critical angle, and we often represent it using πœƒ sub 𝑐.

Now that we know this angle exists, let’s see what happens when we insert it for πœƒ sub 𝑖 in our Snell’s law equation. When we do this, on the left-hand side, we have 𝑛 one times the sin of πœƒ sub 𝑐. And on the right-hand side, we have 𝑛 two times the sine of the angle of refraction, corresponding to the critical angle of incidence. From our diagram, we can see that that angle is 90 degrees.

When a ray of light comes into an interface at the critical angle, the angle of refraction is always 90 degrees. Knowing this, we can recall that the sin of 90 degrees is equal to one. And that means that the right-hand side of our equation simplifies to 𝑛 two. If we then divide both sides of the equation by 𝑛 one, then that factor cancels out on the left. And lastly, if we apply the inverse sine or arc sine function to both sides of the equation, then the application of the arc sine in the left nullifies the application of the sine function. And so we are left with this equation here.

If we know the indices of refraction of both materials on either side of an interface, then we can use that information to solve for the critical angle of incidence. An important thing to keep in mind in all this is that we’ve required that 𝑛 one, the index of refraction on this side of our interface, is greater than 𝑛 two, the index of refraction on this side. If that wasn’t so, if say 𝑛 one was equal to 𝑛 two or 𝑛 one was less than 𝑛 two, then there would be no critical angle πœƒ sub 𝑐. It simply wouldn’t exist. That is, if, for example, 𝑛 two were less than 𝑛 one, then there’s no angle of incidence πœƒ sub 𝑖 we could use that would lead to a refracted ray being bent in an angle of 90 degrees.

So having a critical angle at all is only possible if our ray is moving from a higher to a lower index of refraction. This can be tricky to remember. But thankfully, the form of our equation for the critical angle helps us. If we forgot that 𝑛 one must be greater than 𝑛 two for πœƒ sub 𝑐 to exist, and say we plugged in an 𝑛 two that was greater than 𝑛 one. If we try to evaluate the inverse sine of a number larger than one on our calculator or computer, then our device would either return an error message or an imaginary number. Either result would indicate that no critical angle exists for the inputs as we’ve entered them.

Now, one reason this angle πœƒ sub 𝑐 is so helpful is that it tells us the smallest angle of incidence required for a ray of light not to leave the material that it starts out in. At any angle of incidence greater than πœƒ sub 𝑐, like this angle of incidence here, our incoming ray will follow a path where, instead of some of the ray being refracted into the material, like we’ve been seeing, all of it is reflected off its surface. When this happens, we say that a ray is totally internally reflected. That is, it hasn’t left the material it started out in.

This phenomenon of total internal reflection comes in very useful when we want to transmit optical signals over long distances. When we do this, it’s common to use what’s called a fiber optic cable. This cable is set up so that it has a relatively high index of refraction compared to its surroundings. Which means that when light enters the cable and then it encounters an interface, instead of being refracted out of the cable, it’s reflected back into it. And then the next time it reaches an interface, the same thing happens. The ray of light is reflected back internally. This takes place over and over until the ray reaches the other end of the optical cable.

A technology like this requires knowledge of the critical angle of light as it reaches each one of these boundaries within the optical fiber. So long as the angle of incidence is always greater than or equal to the critical angle, the light will remain within the fiber, even as the fiber bends.

Knowing all this about the critical angle, let’s get some practice now through an example.

Which of the following formulas correctly shows the relation between the critical angle for total internal reflection πœƒ sub 𝑐 for a light ray, the refractive index 𝑛 sub 𝑖 of the substance the light is propagating in, and the refractive index 𝑛 sub π‘Ÿ of the substance when the light is reflected from its surface? (a) sin of πœƒ sub 𝑐 is equal to 𝑛 sub π‘Ÿ divided by 𝑛 sub 𝑖. (b) sin of πœƒ sub 𝑐 is equal to 𝑛 sub 𝑖 divided by 𝑛 sub π‘Ÿ. (c) sin of πœƒ sub 𝑐 is equal to the sin of 𝑛 sub π‘Ÿ over 𝑛 sub 𝑖. (d) πœƒ sub 𝑐 is equal to 𝑛 sub π‘Ÿ over 𝑛 sub 𝑖. (e) πœƒ sub 𝑐 is equal to sin of 𝑛 sub π‘Ÿ over 𝑛 sub 𝑖.

Okay, in all these answer options, we see possible relations between these three variables. πœƒ sub 𝑐, the critical angle of incidence; 𝑛 sub 𝑖, the index of refraction of the substance a ray of light starts in; and 𝑛 sub π‘Ÿ, the refractive index of the substance that the ray of light reflects off of.

We can begin by making a sketch of what’s taking place here physically. We’re told that we have a surface. And let’s say that this is our surface. And it’s there because on one side of it, say above it, the index of refraction of that material is 𝑛 sub 𝑖. While below this boundary, the index of refraction of that material is called 𝑛 sub π‘Ÿ.

Our problem statement tells us that 𝑛 sub 𝑖 is the refractive index of the substance that a ray of light is initially propagating in. So let’s draw that ray of light like this. And then here’s what we know. If the angle of incidence of this incoming ray of light is πœƒ sub 𝑐, the critical angle, then in that case when this ray reaches the surface, the boundary between these two materials, it will travel along that surface. In other words, it will neither refract into this material nor reflect back into this one. Rather, it moves right along the boundary between the two. If we consider the boundary that a ray of light moves along and the line normal to that boundary, then we can see that this ray is at 90 degrees to that normal line.

Now, even though this ray of light is in some sense not refracting, in this limiting case, we can nonetheless say that the angle of refraction is equal to 90 degrees. Indeed, this is the condition for the angle of incidence to be the critical angle πœƒ sub 𝑐. So that’s what’s taking place in this situation. We have these two refractive indices and a light ray approaching the boundary between these materials at the critical angle.

Knowing all this though, we still need to choose from among these five candidates for the correct mathematical relationship between these three variables. Here’s how we can get started doing that. We can recall an optical law called Snell’s law. This law says that if we have a ray of light here incident on a boundary between two optically different materials. Then the angle of incidence πœƒ sub 𝑖 of that ray and the angle of refraction πœƒ sub π‘Ÿ, along with the indices of refraction of the two materials, are related according to this equation. 𝑛 sub 𝑖 times the sine of the angle of incidence is equal to 𝑛 sub π‘Ÿ times the sine of the angle of refraction.

Now, this law is generally true, but it’s also helpful to us in the specific case when the angle of incidence is equal to the critical angle πœƒ sub 𝑐. When that happens, when πœƒ sub 𝑖 is equal to πœƒ sub 𝑐, then we can see from our diagram here that πœƒ sub π‘Ÿ is equal to 90 degrees. So we have these two substitutions we can make into the variables in Snell’s law. If we write a version of Snell’s law using these values, then looking at the right-hand side, we see we can do a bit of simplification. The sin of 90 degrees is equal to one, which means we can write the equation this way.

And next, if we divide both sides of this equation by 𝑛 sub 𝑖, the refractive index of the substance the light is propagating in, then 𝑛 sub 𝑖 cancels on the left-hand side. And that leaves us with this expression. The sine of the critical angle πœƒ sub 𝑐 is equal to 𝑛 sub π‘Ÿ divided by 𝑛 sub 𝑖. We can now take this expression and compare it against our five answer options. And we see right away that it’s a match for option (a). So by using Snell’s law and then applying the specific conditions implied by a critical angle of incidence, we found that the sin of πœƒ sub 𝑐 is equal to 𝑛 sub π‘Ÿ divided by 𝑛 sub 𝑖.

Let’s summarize now what we’ve learned about critical angle for total internal reflection. In this lesson, we saw that when a ray of light incident on a boundary is refracted at 90 degrees, then its angle of incidence is called the critical angle. We learned further that, for a critical angle to exist, 𝑛 sub 𝑖, the index of refraction of the material the light ray begins in, must be greater than 𝑛 sub π‘Ÿ, the index of the material on the other side of the boundary. When this condition is met, we saw that the sine of the critical angle, represented πœƒ sub 𝑐, is equal to the ratio 𝑛 sub π‘Ÿ divided by 𝑛 sub 𝑖. And finally, we saw that light reaching a boundary at an angle equal to or greater than the critical angle is totally internally reflected.

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