Thin Lens And Lens Maker Equations - Video Tutorials & Practice Problems
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1
concept
Thin Lens Equation
Video duration:
7m
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Hey guys in this video, we're gonna 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 gonna replace ray diagrams for lenses and allow us to find numeric answers to where the image is located, the height of the image, et cetera. All right, let's get to it for images produced by lenses. Really, all we're gonna consider are thin lenses. All right. What a thin lens is is it's a lens composed of two pieces of glass that are spherical. OK. And all that it means to be spherical is that it's part of some imaginary sphere with some radius of curvature. OK. 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. OK. So it's very, very thin Finland's is come in five basic types. OK. We have our by convex, we have our convex concave rights convex from one side, concave from the other. We have our plano convex, 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 from one side plane 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. OK? And lenses that are 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 lens thin in the middle, thick in the middle, sorry thin at the edges, that's a converging lens. So the biconcave 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 lens. The only 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 radii of curvature for the two pieces of glass that make it up. OK. For a lens with some focal length F, the image location is given by the thin lens equation. OK. And if you notice this is the exact equation as the mirror equation. OK. 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 too is identical. OK. 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. OK. 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. OK. 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. OK. We also have a magnification equation for thin lenses that is identical to that of mirrors. OK. 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. OK. Let's do a quick example. A biconcave 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 real or virtual? Is it upright or inverted? OK. Now, this biconcave 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, right? Biconvex, sorry biconcave means that the image is concave on sorry, the lens is concave on both sides. So this is thin in the middle and thick at the edges. So this is a diverging liens 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 lens equation. The object is placed seven centimeters in front of the lens. So the object distance is seven centimeters. So one over si plus one over. So equals one over F we're gonna isolate the one over SI which is one over F minus one over. So this is one over negative two, 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 only the solution for one over si it's not the solution for SI. So if I reciprocate the answer, we get negative 1.6 centimeters. OK. 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. All right guys, that wraps up our discussion on the Fin Linz Equation. Thanks for watching.
2
Problem
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
si = -8.5 cm; Real; Upright; 1.7 cm
B
si = -0.117 cm; Virtual; Upright; 1.7 cm
C
si = -8.5 cm; Real; Inverted; 3.4 cm
D
si = -8.5 cm; Virtual; Upright; 3.4 cm
3
concept
Lens Maker Equation
Video duration:
5m
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Hey guys in this video, we're gonna 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 of the two pieces of glass that make it up. All right, let's get to it. The focal length of the thin lens depends upon three things. OK. 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. OK. 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 a fraction of the glass itself, whatever that index is. OK. The lens maker equation is going to tell us what the focal length of this thin lens is going to be. And it is N minus one times one over R one where R one is the radius of the near glass minus one over R two or R two is the radius of the far glass. OK. 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. OK. 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. OK. If the center of curvature is behind the lens like this guy, the near glass, then the radius is positive. OK. So this radius is positive, this radius is negative. All right, let's do a quick example to illustrate this point. The following lens is formed by glass with a refractive index of 1.52. What is the focal length of the following lens? When an object is placed in front of the convex side? What if an object is placed in front of the concave side? OK. So first I'll apply the lens maker equation to find it if an object is placed here in front of the convex side. So that's gonna be N minus one, one over R one minus one over R two. The index of a fraction of the glass is 1.52 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 0.022. 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 were to happen if we were to place an object here? What would the focal length be? Well, we're gonna use the same lens maker equation one over F two N minus 11 over R one minus one over R two. OK. The N is the same 1.52 minus one. What about R one? Now, what's the near glass? The near glass is 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 lens, it's negative. So this is negative seven minus one over negative 10 for that 10 centimeter piece of glass, the center of curvature is also in front of the lens plugging this into a calculator. We get negative 0.022. Well, 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. OK. 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 of the lens maker equation. All right guys that wraps up this talk on the lens maker equation. Thanks for watching.
4
example
Example 1
Video duration:
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? OK. 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 to isolate one over si and this becomes one over F minus one over. So OK, by 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. OK. So the image distance is positive if one over F is larger than one over so or if so is larger than not a capital F, it's a lowercase F. If that is true, we have positive image distances. So we have real 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 then one over F then we're gonna have virtual images. OK. So if your object is placed outside of the focus, if the object of distance is greater than the focal length, you will always get real 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 only virtual images can be formed by divergent lenses exactly as we would expect regardless of what the actual numbers are. OK. That wraps up this problem. Thanks for watching guys.
5
example
Example 2
Video duration:
4m
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Hey guys, let's do an example. A biconcave lens has two different radii of curvature. If the radius of curvature of one piece of glass with a refractive index of 1.52 is four centimeters and the radius of curvature of the other piece is seven centimeters. What is the focal length of the lens? If an object is 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? OK. 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 is seven centimeters. Now, the lens maker equation tells us that one over F is N minus one times one over R one minus one over R two. OK. The index of refraction is 1.52 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 a 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 2.5 centimeters. OK. 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 either side. If we place it five centimeters from the lens, where will the image be formed? Now we need to use the thin lens equation that one over. So plus one over SS I equals one over F one over si is therefore gonna be one over F sorry minus one over. So which is one over negative 2.5 minus 1/5, which is negative 0.6. OK. So if I reciprocate this answer because once again, negative 0.6 is not the answer, it's the reciprocal then I get an image distance of negative 1.7 centimeters. OK. 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. OK? 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 si over. So 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 that's 0.34. 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 image is 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 is 0.34 which means that the image is roughly one third the height of the object or 0.34 centimeters. All right guys, that wraps up this problem. Thanks for watching.
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