Mirror equation example problems | Geometric optics | Physics | Khan Academy

Mere equation problems can be intimidating when you first deal with them, and that's not because the mere equation is all that difficult. It's kind of easy; it's just a few fractions added together. The place where it gets tricky is deciding whether these should be positive or negative.

So there's a bunch of positive and negative sign decisions that you have to make, and if you make even one of those incorrectly, you can get the wrong answer. So let's do a few mirror equation problems and you can see how the signs work.

Now, I have to warn you, everyone has their own sign convention. There's a lot of different sign conventions when you deal with optics. The one I'm going to use is the one that I feel like most textbooks are using these days, and it's the one where anything on the front side of the mirror—so by front side, your eye should be over here somewhere. So let's say your eye's over here, right? You're looking at some object, maybe it's an arrow or a crown right here, a blue crayon you're holding in front of this mirror right here.

So this is the mirror right here, and the convention I'm using is that anything on this side of the mirror is going to be counted as positive. So if it's a focal point on that side, positive; if it's an object distance on this side, it's positive; and if the image distance comes out positive, you'll know that it's also on this front side of the mirror.

Anything that comes out negative—so if we get a negative image distance after we do our calculation—we'll know that that thing is behind the mirror. That kind of makes sense: negative like behind, positive like in front. So this side over here is in front of the mirror, positive, and back here would be behind the mirror, and that would be negative.

So I've got some numbers in here already; let's just solve this one. So what do we do? We're going to use this mirror equation. We're going to say that one over the focal length—and already we have to decide on a positive or negative sign.

So this mirror, the way it's shaped right here, based on how we're looking at it, is a concave mirror. And with the sign conventions we just discussed and the signs I'm using in this formula, concave mirrors always have a positive focal length. So in other words, since this focal point is four centimeters from the center of the mirror, I'm gonna have to plug in the focal length as positive four centimeters.

Notice I'm not converting; that's okay. If you leave everything in centimeters, you'll just get an answer in centimeters, so it's okay. You don't have to convert as long as everything's in the same units. And we'll set that equal to one over the object distance; just one other warning: sometimes instead of (d_o), you'll see this as (s_o) for object distance, or you might see (d_i) as (s_i) for image distance. It's the same thing, it's just a different letter.

So the object distance again is on this side, in front of the mirror, so it's going to be positive twelve centimeters, since it's located twelve centimeters from the center of the mirror. So it's also going to be positive twelve centimeters. And now we're going to add to that one over the image distance.

The image distance we don't know; I haven't drawn the image on here. It's going to be a surprise; we don't know what this is going to be, but we can solve for it. So we can solve for image distance. I'll subtract one over twelve from both sides, which will give us one over four centimeters minus one over twelve centimeters, and that's going to have to equal one over the image distance.

So one over four, you could rewrite as three twelfths. So three twelfths minus one twelfth is just going to be two twelfths, and that's going to equal one over the image distance. But two twelfths is just one sixth, so one over the image distance is just going to be one sixth. But that's what one over the image distance is. Sometimes people forget to flip this at the end. We don't want one over the image distance; we want the image distance.

So finally, you take one over each side, and we solve, and we get that the image distance is going to be six centimeters. And it came out positive; that's important. This came out to be positive six centimeters. So where's our image going to be?

Since this image distance came out to be positive, our image is going to be in front of the mirror. So it's going to be over here; it's going to be six centimeters from the mirror, somewhere around here. So at this point right here is where our image is going to be, but we don't know how big it's going to be or whether it's right side up or upside down.

To figure that out, we have to use a different equation, and that equation is called the magnification equation. It says that the magnification is equal to the height of the image divided by the height of the object, and that's equal to negative the image distance divided by the object distance. So it turns out this ratio of negative image distance over object distance is always equal to the ratio of the height of the image over the height of the object.

So what's the height of our image going to be? Let's just solve for it. If we solve for the height of our image, we get that the height of the image is going to be—multiply both sides by (h_0)—we'll have the negative signs already here. So negative height of the object times this ratio of the image distance over the object distance.

And now we can just plug in numbers: the height of the object says it's three centimeters tall right here, so it's three centimeters. So negative three centimeters times the ratio of the image distance was six, the object distance was twelve. And so if you solve this, you'll get one half of negative three, which is negative one point five centimeters.

The negative sign means that this image got inverted, so it got flipped over. It's going to be upside down compared to what it was before, and the one point five is how tall it's going to be. So what we end up with is an image six centimeters from the mirror, and it's going to have a height half as tall, so one point five centimeters, and it's going to be upside down because of this negative sign.

So it's going to be one point five centimeters tall and upside down. That's what you're gonna see if you look into this mirror. It's like a fun house mirror; it's a weird curved mirror. You'd see an upside-down image right here. It might look like you could reach out and grab it, but it's gonna be an optical illusion. There's going to be no object there; it's just going to be the image of this object here.

So that's an example using a concave mirror. What would change? What if we did this? What if we took our object, so say we don't put it here anymore, we move it inside here? So instead of being at twelve centimeters, we moved it to like three centimeters. What would we do differently? Everything would be the same; we just would plug in instead of positive twelve down here, we'd plug in positive three.

So the mirror equation works exactly the same. You still plug in whatever that object distance is, you solve for your image distance. We're of course going to get a different image distance, but whatever you get, that would tell you where the image is. And then you would take that, plug it into the magnification equation if you wanted to decide how big the image is going to be and whether it's going to be right side up or upside down.

So these numbers are going to be different, but you would use this equation the exact same way. Now, what would happen if instead of using a concave mirror, we used a convex mirror? Let's say we use the mirror shaped like this. So imagine our eye again is over here, looking at this object inside of the mirror, and we're going to see an image of the object.

We're going to see the object right here, but we're also going to see the image of the object. This mirror, this time instead of concave, this is a convex mirror. So its focal length is behind the mirror. So what do we do now to figure out where the image is? We again use the mirror equation. We're going to use the same equation.

We're going to have one over the focal length, and again, I immediately have to make a decision on the sign. With the convention that I'm using, this focal length is behind the mirror, so this focal length for a convex mirror is going to be negative. So this would be negative four centimeters, and that's going to equal one over the object distance. Well, again, the object is in front of the mirror, twelve centimeters, so this object distance is going to be positive twelve centimeters.

And then we add to that the image distance, which we don't know; this is what we're going to solve for. So this time, if we solve, we're going to have one over negative four, and again, we have to subtract one over twelve, and that's all going to have to be equal to one over the image distance.

So now, on the left-hand side, we have negative one fourth, but that's the same as negative three twelfths. So negative three twelfths minus one twelfth is the same as negative four twelfths, and that's gotta equal one over the image distance. But negative four twelfths is the same as negative one third, so one over the image distance has to be equal to negative a third.

And if we flip that over, we get that the image distance finally is going to be negative three centimeters. So in other words, this here is negative one third. So when you flip that over, you get that the image distance is negative three centimeters. Where is this image going to be?

Well, since it came out negative, that means it's behind the mirror because that's the sign convention we're using. So it's going to be three centimeters behind the mirror, so it's going to be like over here, about at this point, right around here somewhere, three centimeters behind this mirror.

And again, if we want to figure out how tall it's going to be, whether it's right side up or upside down, we're going to have to use the magnification equation. That magnification equation looked like this: it said that the height of the image over the height of the object had to be negative image distance over object distance.

So let's just see what this right-hand side is going to be. If we just set this equal to—it’s going to be negative of negative three as the image distance. So I have to plug in the negative three, and you keep that negative in there. This negative out here comes along always, but now we have another negative inside of this image distance of negative three, and the object distance again was twelve centimeters.

So what are we left with? We get negative of negative three, which is positive three over twelve, which is positive one fourth, and the centimeters cancel. So what this ratio tells you, which is the magnification, is that the image is not going to be inverted. This positive means it's going to be right side up, and the one-fourth ratio means it's going to be one-fourth as large.

The image, I should say, is going to be one-fourth as large as the object is going to be, and the reason is this ratio is equal to height of the image over the height of the object. So in other words, if I can multiply—if I want to multiply both sides, I can say that height of the image is equal to positive one-fourth times the height of the object.

Well, the height of our object was three centimeters in this case, but whatever your height of the object is, you multiply it by this ratio of negative (d_i) over (d_o), and you get what the height of the image is going to be. So we're going to get one-fourth of three, so we get positive three-fourths of a centimeter. It's gonna be—it’s going to be tiny. This is gonna be a little, little image that's gonna be right side up.

So it's gonna be right here, but it's gonna be teeny; it's only gonna be like this three-fourths of a centimeter tall. That's what our image is going to look like. So recapping, you can use the mirror equation to figure out where the images are going to be located. The sign convention we're using is that objects, images, and focal lengths in front of the mirror are going to be positive. Anything behind the mirror is going to be negative, and you could use the magnification equation to figure out how tall the image is going to be relative to the object by taking negative the image distance over the object distance.