Thin Lens And Lens Maker Equations - Online Tutor, Practice Problems & Exam Prep

The thin lens equation, similar to the mirror equation, allows for calculating image location and height for thin lenses, which are composed of two spherical glass pieces. Types include biconvex, biconcave, and plano-convex lenses. Converging lenses are thickest in the middle, while diverging lenses are thinnest. The lens maker equation determines focal length based on the radii of curvature and the index of refraction. Key equations include $1/f=(n-1)(1/r₁)-(1/r₂))$ and magnification equations, which are identical to those for mirrors.

Thin Lens Equation

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.

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
• C. s_i = -8.5 cm; Real; Inverted; 3.4 cm
• D. s_i = -8.5 cm; Virtual; Upright; 3.4 cm

Lens Maker Equation

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 of the 2 pieces of glass that make it up. Alright? Let's get to it. The focal length of the thin lens depends upon 3 things. Okay? 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. Okay? 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 of the glass itself. Whatever that index is. Okay?

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 - 1 × 1 r 1 - 1 r 2 where r1 is the radius of the near glass and r2 is the radius of the far glass. Okay? Now remember that this equation is 1f; it's not the focal length so don't forget to reciprocate if If the center of curvature is in front of the lens, like this guy right here, the far, 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 lens like this guy, the near glass, then the radius is positive. Okay? So this radius is positive, this radius is negative. Alright? 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? Okay. So first, I'll apply the Lensmaker equation to find it if an object is placed here in front of the convex side. So that's going to be 1.52 - 1 × 1 10 - 1 7 . The index of refraction of the glass is 1.52 minus 1. 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 1 over positive 10 minus 1 over positive 7, right? That radius of curvature is also behind the lens. Plugging this into a calculator we're going to get negative 0.022 but don't forget we have to reciprocate our answer. So f1 is negative 45 centimeters. That's what the focal length is if you place an object in front of the convex side. Alright now for the second part I'm going to 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 going to use the same lens maker equation, 1f2, 1.52 - 1 × 1 7 - 1 10 . Okay? The n is the same, 1.52 minus 1. What about r1? Now, what's the near glass? The near glass is the 7 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 7 minus 1 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 going to 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. Okay? This is a fundamental result of the lens maker equation. Alright guys, that wraps up this talk on the Lens Maker Equation. Thanks for watching.

Example 1

Hey guys, let's do an example. What types of images can be formed by converging lenses? What about diverging lenses? Okay? 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 1 over s_i and this becomes 1 over f minus 1 over s_o. Okay? 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_o or if it's smaller. Okay? So the image distance is positive if one over f is larger than 1 over s_o or if s_o is larger than 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 1 over s_o or if s_o is less than f. Then we're going to have virtual images. Okay? So if your object is placed outside of the focus, if the object 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 going to 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 object distance is always going to 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 diverging lenses exactly as we would expect regardless of what the actual numbers are. Okay? That wraps up this problem. Thanks for watching, guys.

Example 2

Hey guys, let's do an example. A biconcave lens has 2 different radii of curvature. If the radius of curvature of one piece of glass with a refractive index of 1.52 is 4 centimeters and the radius of curvature of the other piece is 7 centimeters, what is the focal length of the lens? If an object is placed 5 centimeters from the lens, where will the image be formed? And is this image real or virtual? And finally, if the object is 1 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 going to get the same focal length. So I'm just going to choose an orientation so we can assign a near radius and a far radius. So I'll say the near radius is 4 centimeters and the far radius is 7 centimeters. Now the lens maker equation tells us that 1f = (n − 1) (1r1 − 1r2). Okay, the index of refraction is 1.52, the near radius is 4 centimeters, but the center of curvature appears in front of the lens so by convention, it's negative. The far radius is 7 centimeters and the center of curvature appears 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 2.5 centimeters. Okay. One answer done. Now if we place an object 5 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. Where will the image be formed? Now we need to use the thin lens equation. 1s0 + 1si = 1f. 1si = 1f − 1so which is 1−2.5 minus 15 which is negative 0.6. Okay? 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. Okay. Two answers down, 2 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 1 centimeter tall, what is the height of the image? So for that, we need to use the magnification, siso. Once again, not going to mess with the negative sign because we know that since this is a virtual image, it's going to be upright. That negative sign will just tell us whether it's upright or inverted and we don't need that information. This is 1.7 centimeters over the object distance which was 5 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 1 centimeter. So the height of our image is 0.34 centimeters. Alright? 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 1 third the height of the object or 0.34 centimeters. Alright guys, that wraps up this problem. Thanks for watching.