Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

Table of contents

33. Geometric Optics

1

concept

7m

Play a video:

Was this helpful?

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

2

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?

A

s_{i} = -8.5 cm; Real; Upright; 1.7 cm

B

s_{i} = -0.117 cm; Virtual; Upright; 1.7 cm cm

C

s_{i} = -8.5 cm; Real; Inverted; 3.4 cm

D

s_{i} = -8.5 cm; Virtual; Upright; 3.4 cm

3

concept

5m

Play a video:

Was this helpful?

Hey, guys, in this video, we're going to talk about something called the lens maker equation, which is the equation that will tell us the focal length of a thin lens based on the shape off the two pieces of glass that make it up. All right, let's get to it. The focal length of the Thin Lin's depends upon three things. It depends on the radius of curvature of the near glass. What I mean by near is if I have an object over here, this face is the near glass, so it depends upon that radius of curvature. It depends upon the radius of curvature of the far glass, the glass on the opposite side. So that radius of curvature. And it depends on the index of refraction off the glass itself, whatever that index is. Okay, the lens maker equation is going to tell us what the focal length off this thin lens is going to be. And it is in minus one times one over R one. Where are one? Is the radius of the near glass minus one over R two. Where are two is the radius of the far glass. Now, remember that this equation is one over the focal length. It's not the focal length. So don't forget to reciprocate your answer. There is an important sign convention that we need to know in order to apply this equation. If the center of curvature is in front of the lens like this guy right here the near sorry. The far glass has a center of curvature on the front side of the lens. Then the radius is negative. Okay, If the center of curvature is behind the Linz like this guy, the near glass, then the radius is positive. Okay, so this radius is positive. This radius is negative. All right, let's do a quick example to illustrate this point. The following lenses for my glass with a refractive index of 1. What is the focal length of the following lens within objects placed in front of the convex side? What if in object, is placed in front of the con cave side. So first I'll apply the lens maker equation to find it. If in object is placed here in front of the convex side, so it's gonna be in minus one one over R one minus one over R two. The index of refraction of the glasses 15 to minus one. What is the radius of the near glass? That's 10 centimeters. Is it positive or negative? It's positive because the center of curvature is behind the lens. So this is one over positive. 10 minus one over positive seven. Right. That radius of curvature is also behind the lens. Plugging this into a calculator. We're gonna get negative 0.22 But don't forget, we have to reciprocate our answer. So F one is negative. 45 centimeters. That's what the focal length is if you place an object in front of the convex side, all right. Now, for the second part, I'm gonna minimize myself so that I don't get in the way. And what would happen if we were to place an object here? What would the focal length B. Well, we're gonna use the same lens maker Equation one over F two in minus one one, over R one minus one over R two. Okay, the end is the same. 15 to minus one. What about our one now? What's the near glass? The near glasses. The seven centimeters is the center of curvature in front of the lens or behind the lens. Now it's in front of the lens, and if it's in front of the Linz, its negative. So this is negative seven minus one over negative for that 10 centimeter piece of glass. The center of curvature is also in front, off the lens, plugging this into a calculator. We get negative. 0022 We'll look at that. We got the same answer regardless of which side of the lens we put our object on. And this is actually a fundamental result of the lens maker equation that no matter what side of the lens you put the object on, you're gonna have the same focal length. Okay, When we were drawing Ray diagrams for lenses, we assumed that the focus was at the same distance on either side of the lens. OK, this is a fundamental result off the lens maker equation. All right, guys, that wraps up this talk on the lens maker equation. Thanks for watching

4

example

3m

Play a video:

Was this helpful?

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

5

example

4m

Play a video:

Was this helpful?

Hey, guys, let's do an example. Ah, by Con Cave lens has two different radio of curvature. If the radius of curvature of one piece of glass with the refractive index of 1.52 is four centimeters and the radius of curvature of the other pieces seven centimeters, what is a focal length of the lens? If in objects placed five centimeters from the lens, where will the image be formed? And is this image real or virtual? And finally, if the option is one centimeter tall, what's the height of the image? Okay, so let's start all the way at the beginning. What's the focal length of this lens Now? The lens maker equation tells us as we know that it doesn't matter the orientation of this lens. We're gonna get the same focal length, so I'm just gonna choose an orientation so we can assign a near radius in a far radius. So I'll say the near radius is four centimeters and the far radius of seven centimeters. Now the lens maker equation tells us that one over f is in minus one times one over R one minus one over R two. Okay, the index of refraction is 152 The near radius is four centimeters, but the center of curvature appears in front of the lens. So by convention, it's negative. The far radius is seven centimeters, and the center of curvature appears in Sorry behind the lens. So by convention, that's positive. Plugging this into a calculator, we get negative 0.393 But that isn't our answer. We have to reciprocate this because this is one over F. So the focal length is going to be negative. 25 centimeters. Okay, One answer done. Now, if we place an object five centimeters from the lens once again, it doesn't matter the orientation of the lens because it's the same focal length on but either side. If we place it five centimeters from the lens, where will the image reformed? Now we need to use the thin lens equation. That one over s O plus one over s. I equals one over F one over s. I is therefore gonna be won over f Sorry. Minus one over S O, which is one over negative to five minus 1/5, which is negative. 06 Okay, so if I were sip a Kate. This answer, because once again, negative six is not the answer. It's the reciprocal. Then I get an image distance of negative. 1.7 centimeters. Okay, two answers down. Two more to go. Is this a real image or a virtual image? You guys should know this instantly by now. This is a virtual image. Okay? Why? Because the image distance is negative. And finally, we want to know if the object is one centimeter tall. What is the height of the image? So for that, we need to use the magnification s I over eso once again not gonna mess with the negative sign because we know that since this is a virtual image, it's gonna be upright. That negative sign will just tell us whether it's upright or inverted. And we don't need that information. So this is 1.7 centimeters over the object distance, which was five centimeters and thats So the height of the image is the magnification 0.34 times the height of the object, which is one centimeter. So the height of our images 0.34 centimeters. All right, so we know our focal length of this lens. Negative 2.5 centimeters. Image distance negative. 1.7 centimeters Which means it has to be a virtual image. The question wasn't asked, but this image is therefore upright. The magnification 0.34 which means that the images roughly one third the height of the object or 0. centimeters. Alright, guys, that wraps up this problem. Thanks for watching.

© 1996–2023 Pearson All rights reserved.