Hey guys, in this video we're going to talk about an equation for mirrors that'll tell us pretty much everything that we need to know about the images produced by mirrors. Before we have to rely on ray diagrams, which are notoriously difficult to provide definite quantitative information such as the image location, whether it's real or virtual, whether it's upright or inverted, etcetera. Okay? Let's get to it. What we're going to focus on are spherical mirrors, and that's basically all that you ever see. Okay? What a spherical mirror is, it's a mirror that's cut from a sphere. Okay? It's not really cut from a sphere. These are shaped to exist as if they had been part of a sphere. Okay? So if you see the solid piece right here, the shaded-in piece, 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 a sphere. Okay? Now the focal length only depends upon this radius of curvature and it's just \( r \) divided by 2. K? Really easy to remember. The focal length of a mirror, half the radius of its curvature. K? 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 1object distance+1image distance=1r/2. Okay? In order to use the mirror equation properly, there are some sign conventions that we need to memorize. Okay? A concave mirror, which is a converging mirror. Right? It's a mirror that when collimated light comes into it, that collimated light eventually focuses onto a single point. That converging mirror, the concave 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 collimated light coming at it, that light is spread apart. That light is never focused. Okay? And by convention, we assign convex mirrors negative focal lengths. And they can only 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 going to use it in the mirror equation, and it too can only produce negative image distances. Okay? So let's talk about these image distances and what it means. If we have a positive image distance, this image is real and inverted. Okay? Those 2 are always paired. All real images are always inverted. Okay? And they always have a positive image distance. If the image distance is negative, this image is virtual, and it's upright. This always goes together. All virtual images are always upright, and they always have a negative image distance. Okay? So which of the 3 types of mirrors can produce a real image? The only one that can produce a real image is a concave mirror, and this should have seemed obvious. A real image is only formed by a convergence of light, and the only mirror that can converge light is the concave mirror. Any mirror that doesn't converge light 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 object's 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 resources you have. What that negative sign is there for is to tell you whether or not the image is upright or inverted. Okay? If you have a positive magnification, the image is upright. If you have a negative magnification, the image is inverted. How do you get a positive magnification? \(Si\) has to be negative, which means the image has to be virtual. How do you get a negative magnification? \(Si\) has to be positive, which means the image has to be real. Because the information is already given right here, you don't actually need the negative sign of this equation. It's just a convention. But what the equation tells you is the size of the image relative to the object. If the magnification is 2, the image is twice as tall. If the magnification is half, the image is half as tall. K? Sorry, the image. If the magnification is 2, the image is twice as tall. If the magnification is half, the image is half as tall. Okay? Let's do a quick example. A 1.4-meter-tall person stands 1 meter in front of a plain 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 1 divided by infinity is 0. What this means is if I move the image distance to the other side, we have that one over the image distance is the negative of one over the object distance, or that the image distance is the negative of the object distance so it's negative 1 meter, right? A person stands 1 meter in front of a mirror. So if the object distance is 1 meter, the image distance is negative 1 meter. What does that mean about the image? Is it real or is it virtual? It's clearly virtual. And we knew this already that a plain 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 \(Si\) over \(So\), which is negative 1 meter over 1 meter, which is positive 1. 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 1 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. Alright guys, thanks for watching.

# Mirror Equation - Online Tutor, Practice Problems & Exam Prep

### Mirror Equation

#### Video transcript

### Example 1

#### Video transcript

Hey, guys. Let's do an example. A 5 centimeter tall object is placed 10 centimeters in front of a convex mirror. If the radius of curvature of the mirror is 2 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. Alright?

In order to find that image distance, we need to use our mirror equation:

1 s + 1 s' = 1 fQuestion is now, what is the focal length? Well the focal length is going to be r over 2 in magnitude. But since this is a convex mirror, we know by convention the focal length has to be negative. So I'm going to 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 2 centimeters, so this is negative 2 over 2, which is negative 1 centimeter.

Now I can use the mirror equation. So let me rewrite it to isolate for the image distance:

1 s - 1 s' = 1 ( - 1 ) - 1 10And if you want, you can simplify this to use the least common denominator. This is negative 10 over 10 minus 1 over 10, which is negative 11 over 10. And then that makes finding the image distance as simple as just reciprocating the answer.

K? This is where a lot of students make mistakes. You are finding one over s'. You are not finding s'. This is not the final answer. The reciprocal of it is the final answer, and that is negative 0.91 centimeters. K? That's the image distance.

Now is this image real or virtual? Virtual? Well, it's negative, so it is virtual. Is it upright or inverted? Well, since it's virtual, it has to be upright. So this is virtual and upright. K? Those 2 have to go together and they always go with a negative image distance.

The only thing left is to find the height of the image which is given to us by the magnification:

m ; = -s' sSo the magnification and I'm just going to 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 the magnification. The image distance is 0.91. The object distance is 10. So this is going to be 0.09. That's the magnification. That means that the image height is 0.09 times the object height. And the object is 5 centimeters tall. So this is 0.45 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 a height of 0.45 centimeters. Alright guys, that wraps up this problem. Thanks for watching.

A 4 cm tall object is placed 15 cm in 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?

s_{i} = 7.5cm; Real; Inverted, 2cm

s_{i} = 15cm; Real; Inverted, 4cm

s_{i} = 0.13 cm; Real; Inverted, 0.0087cm

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

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?

concave; 20cm

convex; 20cm

concave; 10cm

convex; 10 cm

It is not possible

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