Derivation of the mirror equation | Geometric optics | Physics | Khan Academy

So imagine you've got an object sitting in front of this concave mirror. If you wanted to figure out where the image is formed, you can draw ray tracings. One ray you can draw is a parallel ray that goes through the focal point, but these rays are reversible. I don't have to draw that one.

Turns out, if you send rays back along the way they came, they'll just retrace the path they came along the other way. So these rays are reversible. I send a ray parallel; it gets sent through the focal point. But if I send a ray through the focal point, it will get sent parallel. In other words, I don't have to draw this ray. I can draw this ray right here, the one that goes from the tip of the object through the focal point; that's going to get sent parallel. So I'm going to draw this one just for fun, and then I'm going to draw another one.

You need to, in order to find where the image is. I'll draw this one here. So this white line is called the principal axis; it's drawn through the center of this curved mirror. I'm going to draw a ray that goes from the tip of the object to that center point because I know the law of reflection says that the angle in has to equal the angle out, and the angles are measured from the normal line.

It just so happens that this principal axis, that's usually drawn here, anyways, is serving as a perfectly good normal line for this center of the mirror since it's passing through that center of the mirror in a perpendicular way. I can just use that to my advantage. I know that the angle in's got to equal the angle out. I just have to make sure that this angle out is about looks like that, about equal to the angle that it came in at. So these two have to be the same angle, Theta.

Now I can find where my image is. The image of this object is going to be at the point where they cross. So the tip of this object gets mapped to this point right here, and we get an image that's upside down. It looks like that. So ray tracing is cool; it lets you find where the image is. But, I mean, I just kind of eyeballed this angle here. This might have been off by a little bit; it might have been off by a degree or two.

If I wanted to get exactly where the image is, I'd want an equation that I could just plug into. In other words, that I can plug into how far was the object from the mirror and how long is this focal length, and it would spit out exactly where the image is going to be. So that's what we're going to derive in this video. It's called the mirror equation, and it'll tell us how to relate the object distance, the image distance, and the focal length.

So let's do this. How do you derive it? If you look in a textbook, it looks complicated. It's not actually as bad as it looks. When I used to look at it the first time, I was like, “This is some sort of mathematical witchcraft. I don't want to deal with.” But it's not nearly as bad as it looks.

We're going to start with drawing triangles. So we'll notice that these two angles are the same. We're going to use this to our advantage. So we're going to make two triangles that have these as one of their angles. So for the first one, let's consider this one. Let's say one of the triangles will be from the base of this image to the center of the mirror, and then from the mirror to the tip of the image, and then from the tip of the image back down to the base of the image.

So imagine this pink triangle right here. It's a right triangle 'cause that angle right here is a right angle, and it's got Theta as one of its angles. But I could draw another triangle that has this angle Theta. I can go from the tip of the object to the center of the mirror, and then from the center of the mirror to the base of the object, and then from the base of the object to the tip of the object.

I get this blue triangle that's also a right triangle since this angle here is a right angle. So in other words, these triangles are similar. They both have an angle Theta; they both have a right angle. Or in other words, if you don't like similar triangles, just think about it this way: you could use a trig function. Pick your favorite trig function; I'm going to pick tangent.

So let's take tangent Theta. I know tangent Theta by definition is always the opposite over the adjacent. The opposite to this angle—we'll take this angle down here first—the opposite to that Theta is this side. What does that side mean? That's the height of the image, so I'm going to call that hi for image height.

That's how tall the image is. So I know that this is going to equal height of the image divided by the adjacent side to this angle here; that's this distance right here. We're going to give that a name; that's just how far the image is from the mirror. So we give this a name; we call this distance from the mirror to the image; we call that the image distance, and it's measured from the center of the mirror.

So not from like this end right here, this little tip part, but from wherever the center of the mirror is. It's measured from this point right here, and that's this adjacent side since this is how far that image is from the center of the mirror. So I'll call that di since it's the image distance.

But that was for this Theta down here. I know that tangent of theta—this is also a Theta up here—I can use the same relationship for this Theta. I know that tangent of this Theta also has to be opposite over adjacent, but the opposite of this Theta is this side right here. And what is that side? That's just the height of the object, so I'll call that ho—that's the object height.

So the opposite side for this Theta is the object height, and then you divide by what? You divide by the adjacent side; that's just going to be this distance from the mirror to the object. We'll give that a name, and if you guess object distance, then you guessed the right. This is going to be the object distance, and again we measure it from the center of the mirror, not some curved portion that sticks out over here, but from the center part of the mirror.

So that's the adjacent side to this Theta down here, so I'm going to write that as Do. This is an important relationship; this is actually given a name just by itself. This isn't the equation we're hunting for, but it's so important it gets its own name. Just part of this derivation gets its own name; it's called the magnification equation.

But it's usually not written this way; people usually write it as Hi and then they divide by Ho. So you get Hi/H0 equals... and then you multiply both sides by Di, and you get Di/Do. So a lot of times this is called the magnification equation; it gives you a way to find what the height of the image is.

So we were looking for a way to find how far the image is from the mirror, but this lets you figure out okay, you also need to know how big it's going to be. So if we solve this for the height of the image, we get height of the image is the height of the object times some factor, and that factor is just going to be the image distance divided by the object distance.

Here's the thing, though. Look, this image got flipped over. So if we define this image distance as positive, if it's on the same side as the object, we get a flipped over image—that'd be like a negative height. So since we want to represent flipped over images with a negative value, we actually write this equation down with a negative inside of here.

We say that the height of the image is equal to negative Ho Di/Do or you could put the negative up here too. So we stick a negative over here; that way we know that if you get a negative value for the image height, you know it's flipped over. So in other words, if the value I got for Hi was -3 cm, that means I'd get an image that had a height of 3 cm, but it'd be flipped over, and that's what the negative represents.

Okay, so that's kind of a side note. This is not what we're trying to derive; we're trying to derive a formula that would give us the image distance based on object distance and focal length. So we need to do another set of triangles. So what we'll do now is instead of considering these Thetas here, we'll consider these angles here.

Now I can't call them Theta 'cause I already called those Theta; so call these Fi. So these angles also have to be the same because anytime you have a line and then you cut another line through it, these angles here will always be the same. So we could do the same game; we could play the same game we played for Theta with these Fi.

We'll form two triangles; each triangle is going to have Fi as one of the angles. So for the first one, we'll do this object height as one side over here to the base of Fi, and then back up over to here. And then for the other one, we need to also have a triangle that has Fi in it.

So we'll do this from here to here, down to there, and then back up to Fi. So we've got two triangles: this triangle and this triangle, and they both have Fi, and they both have a right angle. So these are also similar triangles.

In other words, we'll play the same game; we'll say that tangent of this Fi is going to have to equal the opposite over the adjacent. The opposite to this Fi is this side, which is just Ho. So we can write Ho divided by the adjacent. Now is not Do, 'cause this side only goes to here; it doesn't go the whole way to the mirror. It only goes that far.

So that's the entire object distance minus this piece right over here. So if I subtracted this much from the object distance, I'd get the remaining amount, which is the adjacent part of this triangle. So this distance from the mirror to the focal point is given a name; it's called the focal length, and we represent it with an F.

So a little confusing 'cause F represents both a point, and it represents the length from the mirror to that point. So F is going to represent that length as well. So this adjacent side—we could write as the object distance minus the focal length since this remaining part right here is the adjacent side, which is object distance minus the focal length.

But we know that this Fi is also equal to this Fi, so I can do tangent of theta for this F. The opposite side would now be this side. What is that? What is this side of the triangle? That's just equal to the height of the image.

This side is the same as the image height, so I can say that this whole thing, tangent of Fi, has to equal opposite over adjacent. This time, the opposite of Fi is the image height, and we divide by this distance right here, which is just the focal length. So this adjacent side for this triangle is simply the focal length, so I'll just divide by F.

And so what I've got are two equations that I'm going to put together, and we will get the mirror equation out of this. There are different ways to proceed at this point; what I'm going to do is—I don't want Ho or Hi in either of these. So I'm just going to solve this one here for Ho over Hi.

I get Ho/Hi. So imagine dividing both sides by Hi and then multiplying both sides by Do minus F, and I'd get Ho/Hi equal to Do minus F, the focal length, divided by the focal length. But I can do the same thing up here; I can get Ho/Hi is just going to equal Do over Di since all you have to do is divide both sides by Hi and then multiply both sides by D.

But this left-hand side right here is the same as this left-hand side right here, so we know that Ho/Hi already equals Do minus F over F. And up here, we know that Ho/Hi equals Do/Di, so that means that Do minus F over F has to also equal Do over Di since both of these expressions equal Ho/Hi, so we can set them all equal.

CU, they're all equal to the same thing, Ho/Hi, and now we're going to solve this. We're just going to clean it up. The left-hand side I can write... I'll just stop using colors here. The left-hand side is going to be Do over F minus 1 since F over F just equals 1, and that's got to equal Do over Di.

Now we can divide both sides by Do. If I divide both sides by Do, I get 1/F since the Do cancels minus 1/Do equals—and then the Do will cancel with this Do on top and I just get 1/Di.

Now we made it, except it's usually written with this 1/Do added on the right. If we add 1/Do to both sides, we finally get the expression we wanted, which is 1/Do + 1/Di = 1/F. It's a pretty simple formula; it took a little bit of effort to get to this, but this is called the mirror equation, and it relates the focal length of the mirror to the image distance and the object distance.

In other words, if you know how far you put the object from the mirror and you know the focal length of the mirror, that lets you figure out exactly where the image distance is instead of just kind of eyeballing it and using a protractor. This will let you solve for where that image is exactly, and if you couple it with this magnification equation up here, you can also figure out the exact height of the image.

Now you might be feeling a little sketchy about this negative sign. I mean, I kind of threw it in quick over here. What's going on with this negative? In fact, how do we know if any of this stuff is negative or positive? Well, the convention that I use, that a lot of textbooks use now, is the one where focal lengths will be positive for concave mirrors.

Just like this mirror right here, this is a concave mirror. But if it was bent the other way, if the mirror looked like this, it'd be a convex mirror, and that would be a negative focal length. It kind of makes sense if the mirror was bent this way; the focal point is kind of back this way, which is kind of behind it.

So it makes sense that it's sort of negative. And the object distance over here, if it's on the same side as your eye, if you're using this correctly, your eye should be over on this side; you would see an object right here, but you'd also see an image of that object right here. If this object is on the same side as the mirror as your eye, which it should be, then this object distance is positive.

It's basically always positive unless you've got some double set of mirrors or something weird happening. If you've got a single mirror and there's no craziness going on, this object distance is just always defined to be positive using the conventions that we're using up here.

Again, there are other conventions you can use, but this is the one used in a lot of textbooks these days. The Di is a little trickier. The Di is also positive if that image is on the same side as your eye, like it is right here. So this image would be considered a positive image distance since it's on this side.

If the image got formed on this side of the mirror, so you formed an image right here, that would be a negative image distance. If this was 5 cm behind the mirror, we'd consider that a -5 cm image distance. If you use that convention with this mirror equation, you'll get a correct relationship between the object distance, image distance, and the focal length.

And if you use that same sign convention with this magnification equation, you'll also get the exact height of the image. And if that image height comes out negative, you'll know that it got flipped upside down; and if the image height comes out positive, you know that the image will stay right side up.

So recapping, using a bunch of similar triangles, we were able to derive a mirror equation that relates the object distance, image distance, and focal length. Along the way, we derived a magnification equation that relates the heights of the image and object to the distances of the image and object.

And you have to be very careful with signs; even though the object distance is basically always positive, the focal length can be positive or negative. The focal length will be positive for concave mirrors, and it'll be negative for convex mirrors. Image distances will be positive if they're on the same side of the mirror as your eye, if they're in front of the mirror, but they'll be negative if they're behind the mirror.

Again, this is not the only sign convention you could use, but it's as good as any other sign convention, and it's the one used in a lot of textbooks these days.