1

concept

## Mirror Equation

9m

Play a video:

Was this helpful?

Hey, guys, in this video, we're gonna talk about an equation for mirrors that will tell us pretty much everything that we need to know about the images produced by mirrors. Before we had to rely on ray diagrams, which are notoriously difficult to draw, then are really only good for qualitative information. With the mirror equation will be able to get definite quantitative information, such as the image, location, whether it's real or virtual, whether it's upright, inverted, etcetera. Okay, let's get to it. What? We're gonna focus on our spherical mirrors, and that's basically all that you ever see. Okay, what a spherical mirrors is is it's a mirror that's cut from a sphere. Okay, It's not really cut from a sphere thes air shaped to exist as if they had been part of a sphere. Okay, so you see the solid piece right here, the shaded in peace that's our actual mirror. What I've gone ahead and done is drawn with dashed lines. The sphere that this mirror appears to be part of okay, and that sphere has a radius R. Okay, now the mirror, because it's a piece of this imaginary sphere that doesn't actually exist it's still defined by that radius. And we call that for a curve. We call it the radius of curvature. That would be the radius of the sphere that that piece belonged to if it was part of this here. Okay, now the focal length Onley depends upon this radius of curvature, and it's just are divided by two. Okay, really easy to remember the focal length of a mirror half the radius of its curvature. Okay, Once you know the focal length for any spherical mirror, we have the mirror equation to determine where the image is located. Okay, we would say that one over the object distance how far the object is from the mirror, plus one over the image distance. How far the images from the mirror equals one over the focal length of that mirror. Right, which is just as we saw are over to In order to use the mirror equation properly, there are some sign conventions that we need to memorize. Okay. A con cave mirror, which is a converging mirror, right? It's a mirror that when column mated, light comes into it. That column aided like eventually focuses onto a single point that converging mirror. The con cave mirror has a positive focal length. Okay, always a positive focal length by convention, and it can produce images that have a positive image distance or a negative image distance. And we'll talk about what that means in a second. Convex mirrors, on the other hand, are diverging mirrors. When you have initially column aided light coming at it, that light is spread apart. That light is never focused, and by convention we assign convex mirrors negative focal lengths, and they can Onley produce negative image distances. Okay, for plain mirrors, which neither converge nor diverge light, we say by convention that the focal length is infinity. That's how we're gonna use it in the mirror equation. And it, too, can Onley produce negative image distances. So let's talk about these image distances and what it means. If we have a positive image. Distance this images riel and inverted. Okay, those two are always paired. All riel images are always inverted, okay, and they always have a positive image distance. If the image distances negative, this image is virtual and its upright. This always goes together. All virtual images air always upright, and they always have a negative image. Distance. Okay, so which of the three types of mirrors can produce a real image? The only one that could produce a real image is a con cave mirror, and this should have seemed obvious. A real images on Lee formed by a convergence of light and the Onley mirror that converge light is the con cave mirror. Any mirror that doesn't converge like cannot produce a real image. So the other two types of mirrors are stuck producing virtual images. Okay, lastly, straight as a result of the mirror equation, it's really easy to find the magnification of an image, which is how tall it is relative to the objects height, and this is given by this equation right here in this orange box. Now, the negative sign is a convention that may or may not be included in your textbooks and your lectures, and any other resource is you have what that negative sign is, therefore, is to tell you whether or not the images upright or inverted. Okay, if you have a positive magnification, the images upright. If you have a negative magnification, the images inverted. How do you get a positive magnification s I has to be negative. Which means the image has to be virtual. How do you get a negative magnification s? I has to be positive, Which means the image has to be inverted. Okay, Sorry. The image has to be riel because the information is already given right here. You don't actually need the negative sign of this equation is just a convention. But what the equation tells you is the size of the image relative to the object. If the magnification is to the objects twice as tall. But the magnification is half the objects, half is tall. Okay, Sorry. The image, if the magnification to the images twice as tall. If the magnification is half the images, half is tall. Okay, let's do a quick example. Ah, 1.4 m tall person stands 1 m in front of a plane mirror. Where is the person's image located? And how tall is it? Okay, so we have our mirror equation, which is always going to tell us where our image is located. Now, the thing here is that the focal length is actually infinity, right? And one divided it by infinity is zero. What this means is if I move the image distance to the other side, we have that one over. The image distances the negative off one over the object distance, or that the image distance is the negative of the object distance. So it's negative. 1 m, right? A person stands 1 m in front of a mirror. So if the object distances 1 m, the image instances negative 1 m. What does that mean about the images it riel? Or is it virtual? It's clearly virtual. And we knew this already that a plane mirror can only produce virtual images. Now, how tall is it? Well, to find the height, we need to use the magnification equation and I'll use the sign just to be proper about this. This is negative. S I over S O, which is negative. 1 m over 1 m, which is positive one. The positive means that this image is upright. This is why the sign is pretty much irrelevant in the magnification equation. We already knew that because this image distance was negative. This image was virtual and therefore it had to be upright. So we didn't gain any new information from the magnification or the sign of the magnification. But a magnification of one means that the height of the image has to equal the height of the object, which, as we're told, is 1.4 meters. Okay, this wraps up our discussion on the mirror equation. All right, guys, Thanks for watching.

2

example

## Object in Front of Convex Mirror

4m

Play a video:

Was this helpful?

Hey, guys, let's do an example of five centimeter tall. Object is placed 10 centimeters in front of a convex mirror. If the radius of curvature of the mirror is two centimeters, where is the image located? Is the image real or virtual? Is the image upright or inverted? And what is the height of the image? Okay, now the first three questions are all given to us by the image distance. The answer to those questions is given to us by the image distance right, based on the image distance, we know where the image is located. Obviously, we know whether it's real or virtual based on the sign, and we know whether it's upright or inverted, based on whether it's real or virtual. So. Really, those first three questions are answered by the same piece of information. All right, in order to find that image distance, we need to use our mirror equation. That one over the object distance plus one over the image distance is one over the focal length. Question is, now, what is the focal length? Well, the focal length is gonna be our over to in magnitude, but since this is a convex mirror we know by convention, the focal length has to be negative. So I'm gonna put a little negative sign in here so that I don't forget. And I don't mess up the problem because I used the wrong sign. The radius of curvature is two centimeters, so it's negative to over to which is negative. One centimeter. Now, I can use the mirror equation, so let me rewrite it to solve or sorry. Toe isolate for the image distance. This is one of us minus one of us. Not. This is gonna be one over. Negative one minus 1/10. Okay. And if you want, you can simplify this to use the least common denominator. This is negative. 10/10 minus 1/10 which is negative. 11/10. And then that makes finding the image distance as simple as just reciprocating the answer. Okay. This is where a lot of students make mistakes. You're finding one over s. I You are not finding, s I. This is not the final answer. The reciprocal of it is the final answer. And that is negative. 091 centimeters. Kate. That's the image distance now. Is this image real or virtual Well, it's negative. So it is virtual. Is it upright or inverted? Well, since its virtual, it has to be upright. So this is virtual and upright. Okay, those two have to go together and they always go with the negative image distance. The only thing left is defined the height of the image which is given to us by the magnification. So the magnification and I'm just gonna drop the sign because the sign is, frankly, a waste of time. We already know that it's upright, so we don't care about the sign of magnification. The image distances 0.91 the object distances 10. So this is gonna be 009 That's the magnification. That means that the image heights is 0.9 times the object type, and the object is five centimeters tall. So this is 0. centimeters. So if we want to sum up all the information about this image, it is located 0.91 centimeters from the mirror, technically behind the mirror, it is a virtual image which is upright and has ah, height of 0.45 centimeters. Alright, guys, that wraps up this problem. Thanks for watching

3

Problem

A 4 cm tall object is placed in 15 cm front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

A

s

_{i}= 7.5cm; Real; Inverted, 2cmB

s

_{i}= 15cm; Real; Inverted, 4cmC

s

_{i}= 0.13 cm; Real; Inverted, 0.0087cmD

s

_{i}= -7.5cm; Real; Inverted, -2cm4

Problem

You want to produce a mirror that can produce an upright image that would be twice as tall as the object when placed 5 cm in front of it. What shape should this mirror be? What radius of curvature should the mirror have?

A

concave; 20cm

B

convex; 20cm

C

concave; 10cm

D

convex; 10 cm

E

It is not possible