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# Total internal reflection

## Video transcript

- [Voiceover] We know from the last few videos, if we have light exiting a slow medium, so let's say so I have a light ray that's exiting a slow medium right over there, and let me draw, this is its angle, its incident angle right over there. And the way to visualize this, the way to reason this out if you can, we're obviously not talking about the true mechanics of light, is to imagine if a car was coming from a slow medium to a fast medium. If it was going from the mud to the road. If the car was moving in the direction of this ray, the left tires are going to get out of the mud before the right tires, and they're going to be able to travel faster. And so this is going to move the direction of the car to the right, so the car is going to travel in a direction like that, where this angle right over here, right over there, is the angle of refraction. And if this is a slower medium than that, if this is a fast medium over here, we get theta two is going to be greater than theta one. Now what I want to ask, or what I want to figure out in this video, is, is there some angle depending on the two substances that the light travels in, where if this angle is big enough, because we know that this angle's always larger than this angle, that the refraction angle is always bigger than the incident angle, moving from a slow to a fast medium, is there some angle, if I were to approach it right over here, if I were to approach at this angle, let's just call this theta three. Is there some angle, theta three, where that is large enough that the refracted angle is going to be 90 degrees? If that light is actually never going to escape into the fast medium? And if I had an incident angle larger than that, so if I had an incident angle larger than theta three, so let's add an incident angle like that, so whatever that is. That the light won't actually even travel along the surface, it definitely won't escape, it won't even travel along the surface, but it'll actually reflect back. So you actually have something called total internal reflection. And to figure that out, what we need to do is figure out at what angle, what angle theta three, do we have a refraction angle of 90 degrees? At what angle, theta three, do we have a refraction angle at 90 degrees? And then that angle, that incident angle, is going to be called our "critical angle", because if you have an angle anything larger than that, then you're actually not going to have refraction, you're actually not going to escape the slow medium, you're just going to reflect at the boundary back into the slow medium. So let's try to figure that out, and I'll do it with an actual example. So let's say I have water. Let's say this is water. It has an index of refraction of 1.33. And let's say I have air up here, and air is pretty darn close to a vacuum, and you know we saw this index of refraction is 1.00029 or whatever, but let's just, for the sake of simplicity, say its index of refraction is 1.00. And what I want to do for light that's coming out of the water, for light that's coming out of the water, let me find another color. For light that's coming out of the color, I want to find some critical angle. Some critical angle here, I'll call it theta critical, where the angle of refraction is 90 degrees. Where it will actually never escape. So this right over here, this is 90 degrees. And so if I have any angle less, any incident angle less than this critical angle I will escape, at the critical angle I just kind of travel at the surface, and anything larger than that critical angle I will actually reflect. I'll actually have total internal reflection. So let's think about what this theta, this critical angle, could be. So we'll just break out Snell's Law again. We have the index of refraction of the water, 1.33, times the sign of our critical angle, times the sign of theta critical, is going to be equal to the index of refraction of the air, which is just one. Times the sine of this refraction angle. So times the sine of 90 degrees. Now, what is the sine of 90 degrees? And to figure that out, you have to think about the unit circle, you can't just do the "sohcahtoa" because this is actually why the unit circle definition is useful. But if you think of the unit circle, if you go 90 degrees, we are now here on the unit circle and the sine is the Y coordinate. So this is a unit circle, so that is a one, that's a one right over here. So the Y coordinate, when you're right over here, is one. So this right over here, is going to be one. So if we want to figure this out, we can just divide both sides by 1.33, so we get the sine of our critical angle is going to be equal to well, this is just going to be one over 1.33. And if you wanted to generalize it, this is going to be the index of refraction, this right here is the index of refraction of the faster medium. That right there, we could call that index of refraction of the faster medium. And this right here is the index of refraction of the slower medium. I'll call that the slower medium. So that is N S. Because you're always going to find the sine of 90 degrees, this is always going to simplify to one when you're finding that critical angle. So just to keep solving it, getting the inverse sine of both sides, and we get our critical angle. Our critical angle is going to be the inverse sine of one over 1.33. And so this right here, so our critical angle is 48.8 degrees, which tells us, so this here is 48.8 degrees, which tells us if we have light leaving water at an incident angle of more than 48.8 degrees, at more than 48.8 degrees, it actually won't even be able to refract, it won't be able to escape in the air, it's actually just going to reflect at that boundary. If you have angles less than 48.8 degrees, it will refract. So if you have an angle right over there, it will be able to escape and refract a little bit. And then right at 48.8, and I've rounded a little bit, but right at that critical angle, you're just going to travel, you're going to have a refraction angle at 90 degrees or really just travel at the surface of the water. And this is actually how fiber optic cables work. Fiber optic cables are just, you can just view them as kind of glass pipes and the light is traveling the incident angles are so large here, that the light will just keep reflecting within the fiber optics. So this is the light ray, they travel at larger than the critical angle, so instead of escaping into the surrounding air or whatever, it'll just keep reflecting within the glass tube, allowing that light information to actually travel. So anyway, hopefully you found that reasonably interesting.