Let's start out with the simple application of the lensmaker's equation. So I have a lens here that is now a relatively thick lens. The thickness of the lens is on the same order as the radius of curvature of the curved surface here. So I've got a thick lens. And I don't actually know what the index refraction of this material is. I can measure the radius curvature. And I think I can measure the focal length here. So let's go ahead and do a couple of quick measurements on this lens, and see if we can calculate the index of refraction of this material, and use that for some more experiments. So I start out with three rays. I have a ray box with three parallel rays. And I'm going to put the lens in here. I'm going to put it in so that the flat surface is facing the parallel rays. So the rays are not going to bend on that front surface. And I'm going to try to align this up so that the front surface is right at the 0 line, to make these measurements a little bit easier for me. You'll kind of notice if I misalign things, I go off the optic axis and my rays don't quite overlap all in the same spot. I'll line this up as well as I can, right there in the 0. We'll see where our rays overlap. I've got parallel rays coming in. And I'm going to have them coming to a focus right about out here. And so this focal point is about 7.5. It looks like 7.5 centimeters from the back surface, I guess, in this case. The back surface of the lens, the curved surface out from the lens, about 7.5 centimeters. I think the other measurement that I'm going to want to have, I'm going to go ahead and turn the light off for a minute, is I want to know what the radius of curvature of this lens is. Looks like it's about a circle. So let's go ahead and measure its diameter. Align this up right here in the 0. Actually, I'm going to align it up over here because it's going to be easier to measure. So I line this up on the 4. It looks like I have about 4 centimeters plus perhaps 4.3 centimeters. So 8.3 centimeters diameter. Now let's see if we can use that information to find the index refraction of the lens. So the problem we're trying to solve is how do I find the index of refraction of this lens. And we've measured the focal length at 7.5 centimeters. We've measured the diameter of the lens at 8.3 centimeters. We don't know what the index of refraction is. How do we approach this problem? First of all, let's think about what we know about principle planes. We know that the light is entering, in our example, from the left here. And it doesn't bend at the front surface. It's only going to bend at the back surface. Because the rays only bend at the back surface, the effect of focal length is going to be the same as the back focal distance that we measured. So this back focal distance, in this case, is going to be your effective focal length. And you don't have to worry about finding the principal planes inside the lens. So how do we use the effective focal length that we've measured to find the index of refraction? Well, you remember from the lensmaker's equation, we have two surfaces of our lens, like we do. It's the same as an equivalent lens system here. And we know that the power of the two lenses, the power of the whole lens, is equal to the power over the first surface plus the power of the second surface minus this d over n times the powers of the two surfaces. So if I put this into focal lengths, 1 over the focal length of my entire system. So that's the effective focal length we measured was related to the radius of curvature of the two surfaces, and the index of refraction of the material, as well as the thickness of the lens. Let's apply that to our problem. In our case, the radius of curvature of the front surface, this R1 is infinity. It's a flat surface, so we have a radius of curvature of infinity. And the radius of the back curvature, the R2, the back surface is pointing to the left so it's negative. The diameter divided by 2, that's our radius. And we know what our focal length is, so we can solve this equation. This last term goes to 0, because R1 is infinity. So 1 over infinity is 0. So now I don't have to worry about the thickness of my lens. And now I can plug in some numbers. And I can find that n is 1 minus R2 over f, if I do some algebra. And that gives me an index refraction of 1.55. So after I do this calculation, the very first thing I should say is, does that make sense? So I know the index refraction of air is 1, and water is 1.33. Glasses and clear plastics that I often use are generally on the order of 1.4 to 1.7. So 1.55 is a perfectly reasonable number, so this does make sense. We know if we send in three parallel rays into this lens in this orientation, where they go through the flat surface and don't bend. And then they bend at the second surface. The distance between the lens and the focal point is the effective focal length, as well as the back focal distance. But if I turn this lens around, like we normally do, now we've got bending at, see if i can get this aligned. There's some bending at the front surface, as well as bending at the back surface. And so the distance from the back of this lens to this focal point is much shorter, because the back focal distance is not the same as the effective focal length. Because the effective focal length is measured from the principle planes inside the lens. So one thing I want to do here is figure out where the principle plane is in this lens. And one way we can do that is I have this lined up pretty well, I'm going to draw a line through these rays going through the focus. And I kind of have to bump it, For the second line through the focus. I'm also going to draw the rays coming in here, these parallel rays that are entering the lens. What I want to do is, we want to see where these rays overlap, These focusing rays and the parallel rays that enter the lens. In the spot where those overlap, if you can see here, right back about here. This is going to be our principle plane of our lens. And so if we turn these rays back on, we'll try to line this up approximately the same place. I do not have my rays on my optical axis. We align these up so that the focus is in about the same spot. And we see that this principle plane is actually quite close to the front of the lens. And this distance now, what should this distance be? The distance between my principle plane and my back focal point. Well, this should be now my effective focal length. So this should be the same distance that we measured before. And we can see that the distance of this line line to that focus in centimeters, once again, is about 7.5 centimeters. So this effective focal length measured from the principle plane to the back focal point is the same as the focal length we measure as the back focal distance if we have the rays coming in this flat surface. And so the effective focal length is a property of the lens. But where we measure this back focal point depends on what orientation we use.