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 gonna 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. But, 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 all, right. So we actually label these two angles differently. So we, we can call this one theta one because it has to do with material one. 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 two. All right. And we all have these two different indices of refraction because we have two different materials. This is N one and this is gonna be N two. Now, through experimentation, we've actually figured out the relationship between these indices and uh these indexes and these angles. And that relationship is this equation here, which is that N one times the sine of theta one is equal to N two times the sign of theta two, 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 this equation here is that instead of using theta I and theta R, that's what we used for reflection. We're actually just gonna use theta one and theta two for the two different materials that we're dealing with. All right, this theta one here is always going to correspond to your theta or your angle of incidence. In other words, it's always just the incoming lights and 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 data here is the angle of refraction. And that, that actually has to do with the outgoing lights once it's crossed the boundary into that new material. All right, that's how you always want to remember this. So basically, just as a note, you should always make sure to use N one and theta one for incident lights. All right, it'll just help keep track of your variables. 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 gonna have really two possibilities. You either have light that enters a material with a higher index of a fraction 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 gonna 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 gonna be theta one. We wanna find the angle of refraction. Now, the problem is, I don't know what that angle is. So I can't draw it here yet. But basically what happens is that this thing is gonna bend either this way or this way, it's gonna bend off in some direction. And I wanna find that. All right. Now, what other, what else do I know? I know the two materials that are involved. I know that air for N one is just uh equal to one. And I know that the N for water, which I'll call N two is equal to 1.33. So if I want to find the angle, I just have to use my new equation for Snell's Law which remember just says that N one times sine whoops and one times sine theta one equals N two, sine theta two. So this theta two here is my angle of a fraction. That's what I'm trying to solve for. But I have three out of four variables. I know the ends for both of them. And I also have the incident angle and that's usually how these problems are gonna go. You're gonna have three out of four variables and you'll just have to solve for the missing one. All right. So this N one here is just one, this sign of theta is just gonna be sign of 30 and this is gonna be 1.33 times the sign of the two. All right. So just to uh simplify this, this is gonna simplify down to 0.5 and we can also divide over the 1.33. So this is gonna be divided by 1.33 and this equals sign of theta two as one last step here which I just have to get rid of the sign to isolate that theta and I just take the inverse sign. So basically what happens is the two equals the inverse sign. And you could solve for this or you could just plug it in as a fraction 0.5 divided by 1.33. And you've got an angle of 22.1 degrees. So this is your angle of refraction here, notice how this number is actually smaller than the 30 degrees of the incident light. So that means that when you draw your line, this angle here has to actually be closer to the normal than it was over here. So basically, this line is gonna 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 this is 22.1 degrees. That's your angle of refraction. So basically what we found here is that when you have light that enters a material with a higher index of refraction, the light bends, does it bend toward or away from the normal? But what we just saw here is that it actually bends and gets closer to the normal. So light bends towards the normal. And what we end up with for our theta two is we end up with a number that is less than theta one. That's always how these things are gonna go so light bends towards the normal, uh for these kinds of problems. All right. So 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 a fraction. So let me ask you if this is basically just the opposite of this situation over here. Which way do you think light is going to bend? Is it gonna bend toward or away from the normal? Well, I guess that it's just gonna bend away from because it's the opposite of the left situation. And we're gonna end up with a situation where theta two is greater than theta one. But let's go ahead and calculate it just so we can make sure. So here we have a ray of light that's exiting glass into air, same angle. So same 30 degrees over here. This is gonna be my theta one. We know that the index of refraction for glass is 1.46 and for air, it's just N two that equals one. All right. So notice how, what happens is it doesn't matter which one is greater or smaller. The incoming light is always gonna labeled as N one. All right. So let's set up our Snell's law. We have the N one sin, theta one equals N two sine theta two. So we're looking for theta two, we just plug in all the rest of the numbers. So N one is equal to 1.46 times the sign of 30 then N two is just gonna equal one. So that just uh doesn't change anything. All right. So when you work this out, what you're gonna get here is 0.73. So this 0.73 equals sine of theta two. And all you have to do is just take the inverse sign. So this is gonna be that theta two is gonna be the inverse sign of 0.73. And which you'll end up with here is 44 sorry 46.9 degrees. So this is my new angle notice how this number is bigger than the 30 degrees. So that means when I go over to my diagram, I have to draw a an array that's gonna end up being farther away from the normal than this one. So here, what happens is that you end up with a large angle, this is 46.9 degrees. And what we can say here is that light bends away from the normal just as we expected. And we also got a number that is greater than our initial angle of incidence, which is exactly again what we expected. So that's really all there is to the uh the light refraction and Snell's law. Let's go ahead and take a look at some practice problems.