Welcome back, everyone. We've already talked a lot about light reflection. And in this video, we're going to cover a new concept called light refraction. I'm going to show you exactly what refraction is, and we'll go over the one equation that you need to solve problems, which is called Snell's law. We'll actually go ahead and solve a couple of examples together. So let's get started.

Now when we talked about light reflection, we said that when light comes in and hits a flat shiny surface, it bounces off at the exact same angle that it came in at. In other words, this angle was equal to this angle. But that's actually not the full story because we've also seen from our video on the index of refraction that light can transmit into another material and it will change speed. That's what refraction is about. Whenever light enters a new material at an angle, it changes speed. We've already seen that before, but it also has to change direction. That's what you should think about when you hear the word refraction. Light hits the boundary between two materials and it has to sort of bend at a different angle than the one that it came in at.

Alright. So we actually label these two angles differently. So we can call this one theta one, because it has to do with material 1, and that means that this is also theta one. And that means that this new angle with respect to the normal is equal to theta 2. Alright? And we also have these two different indices of refraction because we have two different materials. This is n1, and this is going to be n2. Now through experimentation, we've actually figured out the relationship between these indices and these angles. And that relationship is this equation here, which is that n1×sintheta1=n2×sintheta2. Your textbooks will refer to this as Snell's law of refraction, named after Snell, the scientist who came up with it.

Basically, what you should know out of this equation from here is that instead of using theta I and theta r, that's what we used for reflection, we're actually just going to use theta 1 and theta 2 for the two different materials that we're dealing with. Alright? This theta 1 here is always going to correspond to your angle of incidence. In other words, it's always just the incoming light, then that should make sense. So in other words, the left side of your equation should always be for the incoming light. The right side of the equation in this theta here is the angle of refraction. And that actually has to do with the outgoing light once it's crossed the boundary into that new material. Alright? That's how you always want to remember this. So basically, just as a note, you should always make sure to use n1 and theta 1 for incident lights. Alright? It'll just help keep track of your variables, and you won't get the wrong answer.

Now that's really it for the equation. Let me go ahead and show you how to use it using these examples. Now in your problems, you're going to have really two possibilities. You either have light that enters a material with a higher index of refraction or a lower one. Those are really only the two possibilities here. So let's take a look at the first one. In this first problem, we have that a ray of light is going to enter water at a 30 degree incident angle. So this is my 30 degrees, and because this is the incident light that's incoming, that's going to be theta 1. We want to find the angle of refraction.

n 1 × sin 30 ° = 1.33 × sin theta 2This simplifies down to 0.5, and dividing by 1.33 gives the equation sintheta2=0.51.33. Taking the inverse sine of this fraction gives us theta2=22.1°. This is your angle of refraction. Notice how this number is actually smaller than the 30 degrees of the incident light. So basically, this line is going to get a little bit steeper because remember angles are measured relative to the normal, and this normal here, this angle has to be smaller than the one over here.

So, what we found here is that when you have light that enters a material with a higher index of refraction, the light bends towards the normal, and what we end up with for our theta 2 is a number that is less than theta 1. That's always how these things are going to go.

Now let's go ahead and take a look at the second type of problem where we have light entering material with a lower index of refraction. So here we have a ray of light that's exiting glass into air, same angle, so same 30 degrees over here. This is going to be my theta 1. We know that the index of refraction for glass is 1.46, and for air, it's just n2=1. So, setting up our Snell's law equation:

1.46 × sin 30 ° = sin theta 2When you work this out, you get 0.73 equals sin of theta 2, and taking the inverse sin, theta 2 will be 46.9°. In this case, light bends away from the normal and we also get a number that is greater than our initial angle of incidence, which is exactly what we expected. So that's really all there is to light refraction and Snell's law. Let's go ahead and take a look at some practice problems.