Hey guys, in this video, we're going to cover something called the thin lens equation. Just like for mirrors, we used ray diagrams to find qualitative information about the images, but when it came down to actual numbers, we relied on the mirror equation. The thin lens equation is going to replace ray diagrams for lenses and allow us to find numeric answers to where the image is located, the height of the image, etc. Alright, let's get to it.

For images produced by lenses, really all we're going to consider are thin lenses. What a thin lens is, is it's a lens composed of 2 pieces of glass that are spherical. And all that it means to be spherical is that it's part of some imaginary sphere with some radius of curvature. And to be thin, the radius of curvature has to be a lot larger than the thickness of the lens. For instance, maybe the radius of curvature is 10 centimeters for a particular piece of glass but when you put 2 of them together, the lens is actually only a couple millimeters thick. So, it's very very thin. Thin lenses come in 5 basic types. We have our biconvex. We have our convex concave. Right? Convex from one side, concave from the other. We have our planoconvex. It's flat or plain from one side and it's convex from the other. We have our biconcave lenses which are concave from both sides. And lastly, we have our plano concave lenses, concave lenses. Concave from one side, plain from the other.

Now the rule of thumb for deciding which of the lenses are converging and which of the lenses are diverging are the lenses that are the thickest in the middle are converging lenses. And lenses that are the thinnest in the middle are diverging lenses. So, you can look at this lens right here. It's thick in the middle, thin at the edges. That's a converging lens. This lens, thin in the middle, thick at the edges. That's a converging lens. For the biconcave, thin in the middle, thick at the edges. That is a diverging lens. For the planoconcave lens. That one could be either converging or diverging, it just depends on the radius or the radii of curvature for the 2 pieces of glass that make it up.

For a lens with some focal length f, the image location is given by the thin lens equation. And if you notice, this is the exact equation as the mirror equation. Because this is all based on geometry and it actually doesn't matter whether you have a spherical mirror or a spherical lens. That was why our rules for drawing ray diagrams were almost identical for lenses and for mirrors. The equation is identical. Now the sign conventions that are important are if the lens is converging, remember the rule of thumb for a converging lens is it's thick in the middle thin at the edges, the focal length is positive. If the lens is diverging, the focal length is negative. And just like always, if the image distance is positive, it is a real image and it is an inverted image. You guys should have this memorized, down packed by now. There have been so many videos that cover this exact concept. And if the image distance is negative, it's virtual and it's upright.

We also have a magnification equation for thin lenses that is identical to that of mirrors. So luckily, you don't have to memorize a new set of equations. The thin lens equation is identical to the mirror equation, and magnification for images produced by thin lenses is identical to the magnification for images produced by mirrors.

Let's do a quick example. A biconcave lens has a focal length of 2 centimeters. If an object is placed 7 centimeters in front of it, where is the image located? Is this image real or virtual? Is it upright or inverted? Now, this biconcave lens looks like this. Actually, I'm going to minimize myself and I'm going to draw off to the side where we don't really need it except to illustrate this point. Biconcave means that the image is concave on, the lens is concave on both sides. So this is thin in the middle and thick at the edges, so this is a diverging lens. If it has a focal length of 2 centimeters, since it's diverging, that focal length has to be negative. So, the focal length is going to object is placed 7 centimeters in front of the lens so the object is placed 7 centimeters in front of the lens, so the object distance is 7 centimeters. So 1/s_{I}+1/s_{o}=1/f. We're going to isolate 1/s_{I} which is 1/f-1/s_{o}. This is -0.64 but we still have to reciprocate our answer. Alright, this is only the solution for 1/s_{I}. It's not the solution for s_{I}. So if I reciprocate the answer, we get -1.6 centimeters. So is this image real or is it virtual? Well, the image distance is negative, so you should know automatically that this is a virtual image. And since it's virtual, automatically it's upright.

Alright guys, that wraps up our discussion on the thin lens equation. Thanks for watching.