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33. Geometric Optics

# Thin Lens And Lens Maker Equations

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concept

## Thin Lens Equation 7m
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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 etcetera. All right, let's get to it for images produced by lenses. Really, all we're going to consider are thin lenses. Alright, what a thin lens is is. It's a lens composed of two pieces of glass that are spherical. Okay, and all that it means to be spherical is that it's part of some imaginary sphere with some radius of curvature. Okay, 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 two of them together, the lens is actually only a couple millimeters thick, so it's very, very thin. Finland's has come in five basic types. Okay, we have our by convex. We have our con vex con cave rights convex from one side, con que from the other. We have our Plano convex. It's flat or plane from one side and its convex from the other. We have our by con cave lenses, which air con came from both sides. And lastly, we have our Plano con cave lenses con que from one side plane from the other. Now the rule of thumb for deciding which of the lenses are converging, in which the lenses air diverging are the lenses that are the thickest in the middle are converging lenses okay and lenses. The thinnest in the middle are diverging lenses. All right, 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 limbs thin in the middle, thick in the middle. Sorry, thin at the edges. That's a converging lens for the bike con cave thin in the middle, thick at the edges. That is a diverging lens for the Plano concave. Thin in the middle, thick at the edges, that is a diverging wins the Onley one up for grabs is the convex concave lens that one could be either converging or diverging. It just depends on the radius or the radius of curvature for the two pieces of glass that make it up. Okay for a lens with some focal length F, the image location is given by the thin lens equation. Okay, And if you notice this is the exact equation as the mirror equation, okay, because this is all based on geometry, and that 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 two is identical. Okay, now the sign conventions that are important are if the lenses 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 lenses diverging the focal length is negative. Okay? And just like always, if the image distances positive, it is a really image, and it is an inverted image. You guys should have this memorized down packed by now. Okay? There have been so many videos that covered this exact concept. And if the image distances negative, it's virtual and its upright. Okay, We also have a magnification equation for thin lenses that is identical to that of mirrors. Okay, so luckily, you don't have to memorize a new set of equations. The thin Linds 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. Okay, let's do a quick example. Abe icon Cave lens has a focal length of two centimeters. If an object is placed seven centimeters in front of it, where is the image located? Is this image riel or virtual? Is it upright or inverted? Okay, now this by con cave lens Looks like this. Actually, I'm gonna minimize myself, and I'm gonna draw off to the side where we don't really need it except to illustrate this point by convex. Sorry. By Con cave means that the images con cave on. Sorry, the lenses con cave on both sides. So this is thin in the middle and thick at the edges. So this is a diverging Linz, right? If it has a focal length of two centimeters since It's diverging that focal length has to be negative. So the focal length is gonna be negative. Two centimeters. Besides that, now we can use the thin Linds equation. The objects placed seven centimeters in front of the lens. So the object distances seven centimeters. So one over S I plus one over s O equals one over f. We're gonna isolate one over s I. Which is one over f minus one over s. Oh, this is one over. Negative too, right? That's the sign for sorry. Yeah, that's the sign for the focal length. Minus 1/7. Right. This whole thing is gonna be negative. 0.64 but we still have to reciprocate our answer. All right, this is on Lee. The solution for one over s I It's not the solution for s I. So if I were Sipa Kate the answer, we get negative centimeters. Okay, So is this image really or is it virtual? Well, that image distances 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 Linds equation. Thanks for watching
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Problem

A biconvex lens has a focal length of 12 cm. If an object is placed 5 cm from the lens, where is the image formed? Is it real or virtual? Is it upright or inverted? What’s the height of the image if the object is 2 cm tall?

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concept

## Lens Maker Equation 5m
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example

## What Types of Images Can be Formed by Lenses? 3m
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Hey, guys, let's do an example. What types of images can be formed by converging lenses? What about diverging lenses? So let's start with converging lenses for converging lenses. We know by convention the focal length is positive. So let's take a look at the thin lens equation and see what types of images we can produce. I'm going toe isolate one over s I. And this becomes one over F minus one over eso. Okay, Bye. Convention. That object distance is always positive for a converging lens that focal length is always positive. So this number could be either positive or negative, depending on if one over F is larger than one over s or if it's smaller. So the image distance is positive. If one over F is larger than one over eso or if eso is larger than not a capital F, it's a lower case F. If that is true, we have a positive image distances. So we have really images. But the image distance can also be negative. The image distance can be negative if one over f is less than one over s not, or if s not is less than one over f then we're gonna have virtual images. Okay, so if your object is placed outside of the focus, if the object distances greater than the focal length, you will always get riel images. But if your object is placed inside of the focus at a distance less than the focal length, you will always get virtual images. Now, what about diverging lenses before even doing any math? What do you guys think are gonna be the images produced by diverging lenses? They should always be virtual, because you can never have converging light from a diverging lens. By convention, the focal length of a diverging lens is negative. So applying the same equation and the same process that image Sorry that object distance is always gonna be positive. But now the focal length is negative, so you have a negative number minus a positive number. That's always going to be negative. So the image distance is always negative, right? That means on Lee. Virtual images can be formed by diverging lenses exactly as we would expect, regardless of what the actual numbers are. Okay, that wraps up this problem. Thanks for watching guys
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example

## Image Formation by a Biconcave Lens 4m
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