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# Snell's law example 1

## Video transcript

- [Voiceover] As promised, let's do a couple of simple Snell's Law examples. So let's say that I have two media, I guess, the plural of "mediums." So let's say I have air right here. Air, and then right here is the surface. That is the surface, let me do that in a more appropriate color, that is the surface of the water. So this right here is the surface of water. And I know that I have a light ray coming in with an incident angle of... So it has an incident angle. So, relative to the perpendicular, it has an incident angle of 35 degrees. What I want to know is what the angle of refraction will be. So it will refract a little bit, it will bend inwards a little bit since this outside is going to be in the air a little longer if you buy into my car travelling into the mud analogy. So it will then bend a little bit, and I want to figure out what this new angle will be. I want to figure out the angle of refraction. I'll call that theta two. What is this? So this is just straight up applying Snell's Law and I'm going to use the version using the refraction indices since we have a table here from the ck12.org flexbook on the refraction indices. You can go get it for free if you like. That just tells us that the refraction index for the first medium, so that is air, the refraction index for air times the sign of the incident angle, in this case it's 35 degrees, is going to be equal to the refraction index for water times the sine of this angle right over here, times the sine of theta two. And we know what the refraction index for air and for water is, and we just have to solve for theta two, so let's just do that. The fraction index for air is this number right over here: 1.00029, so it's going to be 1.00029 times the sine of 35 degrees is going to be equal to the refraction index for water which is 1.33. So it's 1.33 times sine of theta two. Now, we can divide both sides of the equation by 1.33. We divide both sides by 1.33. On this side, we're just left with sine of theta two, on the left hand side I get, I'll switch colors, 0.4314 is equal to sine of theta two. Now, to solve for theta, you just have to take the inverse sine of both sides of this. Theta two is equal to 25.6. Approximately equal to 25.6 degrees. So, Snell's Law goes with our little car driving into the mud analogy, it's going to be a narrow degree, it's going to come inwards a little bit closer to vertical. Theta two is equal to 25.6 degrees. And you could do the other way. Let's do another example. Let's say that we have, just to make things simpler, let's say that I have some surface over here, so this is some unknown material, and we're traveling in space, we're on the space shuttle, and so this is a vacuum. This is a vacuum right over here, or pretty darn close to a vacuum, and I have light coming in at some angle, I have light coming in at some angle, just like that, let me drop a vertical, so it's coming in at some angle. Actually, let me make it interesting, let me make the light go from the slower medium to the faster medium. Just because last time we went from the faster to the slower. So it's in a vacuum, so let's add some light traveling like this, and once again, if you view the, just to get whether it's going to bend inward or bend outward, the left side is going to get out first, so it's going to travel faster first, so it will bend inwards when it goes into the faster material. So this is some unknown material where light travels slower. And, let's say we're able to measure the angles. Let me drop a vertical right here. Let me drop a vertical right over here. And so, let's say that this right here, that right there is 30 degrees. Let's say that we're able to measure the angle of refraction, and the angle of refraction over here is, let's say that this is 40 degrees. So, given that we're able to measure the incident angle and the angle of refraction, can we figure out the refraction index for this material? Even better, can we figure out the speed of light in that material? So let's figure out the refraction index first. So, the refraction index for this questionable material times the sine of 30 degrees, is going to be equal to the refraction index for a vacuum. Well, that's just the ratio of the speed of light in the vacuum to the speed of light in the vacuum, so just going to be one. That's going to be one. This is the same thing as "n" for a vacuum. And I'll just write a one there times the sine of 40 degrees. Times the sine of 40 degrees. Or, if we wanted to solve for this unknown refraction index, we just divide both sides of the equation by sine of 30. So our unknown refraction index is going to be, this is just the sine of 40 degrees, sine of 40 degrees, over this: over the sine of 30 degrees. Over the sine of 30 degrees, so our unknown refraction index for our material is equal to 1.29. So, we were able to figure out the unknown refraction index, and we can actually use this to figure out the velocity of light in this material, because remember, this unknown refraction index is equal to the velocity of light in a vacuum, which is 300 million meters per second, divided by the velocity in this unknown material. So we know that 1.29 is equal to the velocity of light in a vacuum, so we could write 300 million meters per second divided by the unknown velocity in this material, I'll put up a question mark. And so, we can multiply both sides times our unknown velocity, I'm running out of space over here, I have other stuff written over here. So I could multiply both sides by this "V," and I'll get 1.29 times this V with a question mark is going to be equal to 300 million meters per second and then I can divide both sides by 1.29. The question mark is going to be this whole thing: 300 million divided by 1.29. Or, another way to think of it is, light travels 1.29 times faster in a vacuum than it does in this material right over here. Let's figure out it's velocity. Approximately 232 million meters per second. If we had to guess what this material is, let's see; I just made up these numbers, let's see if there's something that has a refraction index close to 1.29. So, that's pretty close to 1.29 here, so maybe this is some kind of interface with water in a vacuum where the water somehow isn't evaporating because of the lack of pressure. Or maybe it's some other material, let's keep it that way, something that wouldn't, maybe some type of solid material. But anyway, those are hopefully two fairly straightforward Snell's Law problems. In the next video, I'll do a slightly more involved one.