Video: Total Internal Reflection

In this lesson, we will learn how to define total internal reflection (TIR) by describing rays produced by light rays incident at or past the critical angle for TIR.

13:49

Video Transcript

In this video, weโ€™re talking about the concept in optics known as total internal reflection. As weโ€™ll see, total internal reflection is an idea that relates refracted light, reflected light, and incident light at an interface. Letโ€™s jump right in.

As we get started, say that we have an interface between two different materials. The first material, the one on the bottom side of the interface, has an index of refraction we can call ๐‘› sub i. And as a side note, we can recall that index of refraction for a material is equal to the ratio of the speed of light in vacuum, ๐‘, to the speed that light takes on in that particular material. The index of refraction of a material then is a measure of just how much light slows down when it enters that material.

So on the underside of our interface, we have a material with index of refraction weโ€™re calling ๐‘› sub i. And weโ€™ll say that, above that interface, we have a different index, ๐‘› sub f. And letโ€™s say that, in addition to this setup, so far we also have a ray of light which is incident on this interface. Now we ask a question. What will happen to this ray when it reaches this point?

Well, really, there are two things. And to help us see what they are more clearly, we can sketch in this dashed line, which is called the normal line, thatโ€™s perpendicular to this interface. When this ray of light reaches this boundary between these two materials, some of the light will be reflected off this interface. That reflected light ray will look a bit like this.

And if we call the angle of our incident light ray measured from the normal line to that ray ๐œƒ sub i, the angle of incidence. And say we want to relate that angle of incidence to what we can call the angle of reflection. That angle is measured in a similar way starting from the normal line. And we can refer to it as ๐œƒ sub r.

At this point, we can recall whatโ€™s called the law of reflection, which tells us, using the symbology weโ€™ve adopted, that ๐œƒ sub i, the angle of incidence, is equal to ๐œƒ sub r, the reflected angle. In other words, these two angles weโ€™ve highlighted here are equal. So thatโ€™s one thing that will happen to this light beam. Part of it will be reflected back from the interface. But in most cases, not all of it will be reflected. Some will be refracted, that is, transmitted into this second optical material, identified with the index of refraction ๐‘› sub f.

In order to show that happening, weโ€™ll need to know something about ๐‘› sub f compared to ๐‘› sub i. In particular, weโ€™ll need to know which one of these two values is greater. Letโ€™s say that, in this case, ๐‘› sub f is less than ๐‘› sub i. In other words, our ray is going from an area of higher index of refraction to lower. When this happens, the refracted portion of this ray bends away from the normal line drawn perpendicular to our interface. We can also label this angle from the normal line to the refracted ray. And letโ€™s say that we call it ๐œƒ sub f.

The way we figure out ๐œƒ sub f given ๐œƒ sub i as well as the index of refraction, ๐‘› sub i, and index of refraction, ๐‘› sub f, is through a law called Snellโ€™s law. This law tells us that the initial index of refraction, ๐‘› sub i, multiplied by the sin of the angle of incidence, ๐œƒ sub i, is equal to ๐‘› sub f times the sin of ๐œƒ sub f, the refraction angle. In other words, given an angle of incidence as well as an initial and final index of refraction, we can solve for ๐œƒ sub r, the angle of reflection, and ๐œƒ sub f, the angle of refraction.

Given this setup, we may start to wonder what happens if we increase this angle of incidence. In other words, say that our new angle of incidence, ๐œƒ sub i, looked like this. If we change ๐œƒ sub i, we can see that, based on the law of reflection as well as Snellโ€™s law, that ๐œƒ sub r and ๐œƒ sub f will change too. And based on those two laws, hereโ€™s what our new reflected and refracted rays might look like.

Now letโ€™s imagine that we keep going in this direction. We keep extending or increasing this angle of incidence of our ray. Each time we do this and then recalculate ๐œƒ sub r and ๐œƒ sub f, we start to notice something interesting. As our incident angle, ๐œƒ sub i, gets bigger and bigger, our refracted angle, ๐œƒ sub f, is even larger than ๐œƒ sub i. And in fact, ๐œƒ sub i wouldnโ€™t have to get too much bigger than what weโ€™ve shown in order for ๐œƒ sub f, the refracted angle, to go all the way to 90 degrees. And indeed, there is such an incident angle, ๐œƒ sub i, for which that happens, through which the refracted ray is at 90 degrees to the normal line.

Now this is a really very interesting point weโ€™ve reached here because it means that, for this angle of incidence, no light, whether reflected or refracted, is making it into our second medium identified by ๐‘› sub f. And notice something else going on too. 90 degrees, which is currently the angle of refraction of our refracted ray, is the biggest that angle could ever possibly be. The reason is that if that angle ever exceeded 90 degrees, then that means the ray will never actually enter the second material. It would be almost reflected back into the first. In that case, itโ€™s no longer a refracted or bent ray.

In the situation weโ€™re saying now where the angle of refraction is 90 degrees, none of the light is making it into the second material. Under this condition, the angle of incidence of our incident ray is given a special name. Itโ€™s called the critical angle and typically symbolized ๐œƒ sub c. This is the angle of incidence for which the angle of refraction is 90 degrees.

Now like we said, thatโ€™s the most the angle of refraction can ever possibly be. If we were to increase our angle of incidence even beyond the critical angle, then in that case we would no longer have any refracted light. It will all be reflected. At this point, weโ€™ve reached a condition which is called total internal reflection. This is the point at which thereโ€™s no refracted light because that refracted light ray has reached the 90-degree point and then been extinguished.

By now, 100 percent of the light that we send into this interface at the incident angle is reflected back at the reflected angle. This is the condition known as total internal reflection. Generally speaking, there are two necessary conditions for total internal reflection to occur. The first one is a condition we specified earlier on. We said that the index of refraction of the material the light ray is trying to get into is less than the index of refraction of the material itโ€™s already in.

As an example of this, our initial material could be water and the final material could be air. The index of refraction of water is greater than that of air. So thatโ€™s one necessary condition for total internal reflection to occur.

The other necessary condition is that our angle of incidence, ๐œƒ sub i, needs to be greater than what we called the critical angle, ๐œƒ sub c. We saw that, at ๐œƒ sub c, at the critical angle, we still had a refracted ray, granted the ray was being refracted at 90 degrees. But it still existed. Itโ€™s only beyond that critical angle where we have no refraction at all and we have 100 percent reflection, total internal reflection.

Now if we bring our angle of incidence, ๐œƒ sub i, back down to ๐œƒ sub c, the critical angle, we can get a view into how to calculate this angle. All right, so here we are back at our critical angle of incidence, ๐œƒ sub c. In addition to our reflected ray, we have a refracted ray where that angle of refraction is 90 degrees. If we look over to the bottom right of our screen, we see this equation called Snellโ€™s law. We can use the information in this equation as well the information in our sketch to give us an expression for this critical angle, ๐œƒ sub c.

Hereโ€™s how this will work. In Snellโ€™s law, we have the angle of incidence normally called ๐œƒ sub i. Well, we know that, in this special case, that angle of incidence is the critical angle, ๐œƒ sub c. So we make that substitution into the equation. Additionally, on the right-hand side, we know something about ๐œƒ sub f, the angle of refraction. This angle, according to our diagram, is 90 degrees. And so we can substitute that value in for ๐œƒ sub f. Now the sin of 90 degrees is a value we may have memorized. Itโ€™s equal to one. And this means we can simplify the right-hand side of this whole expression to read ๐‘› sub f, the final index of refraction.

Now our goal here is to get an expression for ๐œƒ sub c, the critical angle. To help us do that, letโ€™s divide both sides by ๐‘› sub i. This has the effect of canceling that term on the left-hand side. And then as a final step, we can take the inverse or arc sin of both sides of this equation. When we do that, the entire left-hand side of this expression simplifies to just ๐œƒ sub c, the critical angle. And by this point, weโ€™ve arrived at an expression for the critical angle. Itโ€™s equal to the inverse sin of ๐‘› sub f divided by ๐‘› sub i. So the critical angle we see just depends on the ratio of the indices of refraction of the two materials weโ€™re working with. Thatโ€™s all that matters in determining this angle. At this point, letโ€™s get some practice with these ideas through an example.

Which of the following definitions best describes the critical angle for total internal reflection? A) The critical angle is the angle at which all incident light at a boundary is reflected. B) The critical angle is the angle of incidence minus the angle of refraction. C) The critical angle is the angle of incidence plus the angle of refraction. D) The critical angle is the angle at which the refracted ray travels along the boundary from which the incident ray is reflected. And then, last but not least, option E) the critical angle is the angle of refraction minus the angle of incidence.

Now in these answer options, weโ€™ve seen a lot of descriptions of angles: angles of refraction, angles of incidence, angles of reflection. To get some clarity on what these answer options mean, letโ€™s sketch out a scenario that involves the critical angle. Say we have an interface here between two materials. The index of refraction of the material on top weโ€™ll call ๐‘› sub f, and that on bottom weโ€™ll call ๐‘› sub i. And letโ€™s say further that we have a ray of light coming up through this first material and reaching the interface between the two.

Now we donโ€™t need to specify exactly what ๐‘› sub f and ๐‘› sub i are. But it is important that ๐‘› sub i be greater than ๐‘› sub f. The index of refraction of the material the ray is trying to get into must be less than the index of refraction that the material is already in. This is a necessary condition for total internal reflection. In addition to all this, letโ€™s say that our incoming ray has an angle of incidence that is the critical angle โ€” weโ€™ll call it ๐œƒ sub c โ€” that has a very specific meaning in terms of what the refracted ray looks like.

But before we get to that ray, letโ€™s look at the reflected ray, the one that bounces off this interface between the two materials. For this reflected ray, if we call the angle of reflection ๐œƒ sub r, then by the law of reflection, we know that that reflection angle, ๐œƒ sub r, is equal to the angle of incidence, which in this case is the critical angle. So thatโ€™s the reflected ray in this scenario.

But we also know thereโ€™s a refracted ray. And because our angle of incidence is the critical angle, we know that the angle of refraction โ€” we can call it ๐œƒ sub f โ€” is equal to 90 degrees. This is what it means for our incoming ray to be at the critical angle. So we now have these three angles marked out: the incident angle, the reflected angle, and the refracted angle. Weโ€™re now ready to evaluate these five answer options to see which one best describes the critical angle.

We already have answer option E) on screen. But letโ€™s recall answer options A), B), C), and D). Answer option A) said that ๐œƒ sub c, the critical angle, is the angle at which all incident light is reflected. But looking at our diagram, we see this isnโ€™t so. At the critical angle, we still have a refracted ray. Itโ€™s at 90 degrees, but it still exists. So answer option A) isnโ€™t our choice.

Option B) says that the critical angle is equal to the angle of incidence minus the angle of refraction. But remember, in the case of the critical angle, that angle ๐œƒ sub c is equal to ๐œƒ sub i. And the angle ๐œƒ sub f, the angle of refraction, is 90 degrees. So this equation is essentially saying that one quantity, ๐œƒ sub c, is equal to another quantity, which weโ€™ve said is identical to ๐œƒ sub c, minus 90 degrees. But mathematically, that canโ€™t be the case. So B) isnโ€™t our choice either.

And for a similar reason, we wonโ€™t choose option C). This option says that ๐œƒ sub c, the critical angle, is equal to the angle of incidence plus the angle of refraction. But again, at ๐œƒ sub c, the critical angle, the angle of incidence is the critical angle. So this option is saying that a particular value is equal to itself, in this case, plus 90 degrees. So option C) isnโ€™t our definition for critical angle either.

Option D) is interesting. It says that the critical angle is the angle at which the refracted ray travels along the boundary from which the incident ray is reflected. Now as we look at our sketch, we see that indeed this is happening. Our refracted ray drawn here is moving along the boundary at which our incident ray is reflected. So option D) looks like a great choice.

Letโ€™s just consider option E) to make sure weโ€™re covering all the bases. This option says that the critical angle is the angle of refraction minus the angle of incidence. But for the same reason that options B) and C) were incorrect, option E) isnโ€™t correct either. At the critical angle, the angle of incidence is equal to the critical angle. So if we subtract this angle from 90 degrees, the angle of refraction, we wonโ€™t get the original angle back. This confirms that option D) is our best answer.

The critical angle is the angle at which the refracted ray travels along the boundary from which the incident ray is reflected.

Letโ€™s summarize now what weโ€™ve learned so far in this lesson about total internal reflection. Weโ€™ve seen that total internal reflection, sometimes abbreviated TIR, is when all the light incident on a boundary is reflected from it. We saw that total internal reflection happens when the angle of incidence of a ray exceeds, is greater than, whatโ€™s called the critical angle, ๐œƒ sub c. At the critical angle, shown here in this sketch as ๐œƒ sub c, the angle of refraction, shown here as ๐œƒ sub f, is equal to 90 degrees.

And finally, we saw that if ๐‘› sub f is the index of refraction of the material the incident ray is trying to get into and ๐‘› sub i is the index of the material itโ€™s already in, then the critical angle, ๐œƒ sub c, is given mathematically as the inverse sin of ๐‘› sub f over ๐‘› sub i.

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