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# Derivation of the mirror equation

In this video David derives the mirror equation and magnification equation. Created by David SantoPietro.

## Want to join the conversation?

- 7:29Shouldn't the "vertical" side of the triangle actually be vertical and go along the dotted line instead of ending at the curved mirror?(23 votes)
- Yes, you're right. Where the light ray hits the mirror is not directly below the principal axis. So, at9:05, the denominator is not really f but f minus a tiny bit. If the ray had hit the mirror much farther down the mirror it would be even easier to notice the difference. Therefore, the mirror equation only works well for light rays close to the principal axis. Rays that hit far away from the principal axis will not obey this mirror equation very well.

I also know that spherical mirrors experience spherical aberration, and parabolic mirrors experience coma, for non-parallel rays. What this means is that they are not "perfect" reflectors in that the image won't be perfect. But if the mirror is small in comparison to the radius of curvature (that is, you're only using the part of the sphere or parabola close to the principal axis, where it hasn't curved too much yet), then the difference in distance you've uncovered is small and won't affect the answer, or the appearance of the image very much.(19 votes)

- At1:58, they are two point where ray is intersecting , meaning image will form at two different place? Why did he chose last instead of first?(2 votes)
- One of the points in which the two rays intersect is during entry (before the rays are reflected by the mirror). In other words, we are looking for the point at which the reflected rays - reflected by the mirror - intersect, not where the two rays intersect before they are reflected. Thus, the second point of intersection, formed by the rays reflected from the mirror is the point the image will form.(1 vote)

- At10:20how did you take Do - f/f as Do/f - 1?(2 votes)
- I guess you meant (Do - f) / f. It's just a simple math. D0/f - f/f which is D0/f - 1.(1 vote)

- how the negative sign comes in mirror magnification formula is still not clear?(1 vote)
- There's no such thing as negative distance or negative height. So h_o and h_i are both positive. That's why you don't get a negative sign when you're trying to derive the equations. However, we don't want to just use a scalar distance, we want to use a 1D vector for the distance (1D vectors, in this case, just mean numbers where signs specify the direction). In other words, we want to be able to determine not just the height of the image but also the direction (up or down, + or -). So then, we're not actually looking for h_o and h_i, but vector versions of h_o and h_i which I will call H_o and H_i.

Using trigonometry, we can find out that h_i/h_o = d_i/d_o. Now if we pick a sign convention for our vectors (+ means on the left of the mirror and - mean on the right of the mirror), we can define:

H_o = + h_o

H_i = - h_i

D_o = + d_i

D_i = + d_o

Now if we want to plug those vectors into the equation, we will have to do this:

h_i/h_o = d_i/d_o

-h_i/h_o = -d_i/d_o // multiply both sides by -1

H_i/H_o = - D_i/D_o

So then, when you plug in the vector equivalents of the heights and distances, you have to include the negative sign on the right side of the equation since you're plugging in -1 times the height of the image.

The reason trigonometry didn't give us the negative sign was because it was working with scalar values for height and distance instead of vectors. If we tried plugging in a negative number for h_i without putting a negative sign on the right side of the equation, you'd find that the two sides are not equal because the equation wasn't fixed to work for vectors.(3 votes)

- Does a convex parabolic mirror always form a virtual image?(1 vote)
- Yes, a convex parabolic mirror will always form a virtual image as it is a diverging mirror, which means that it will diverge rays of light if they are parallel to the principal axis which will result in the rays of light never "really" meeting and will hence always form a virtual image which is diminished but erect.(2 votes)

- Can object distance be negative? And if so when does this happen?(0 votes)
- Whenever you are dealing with position and you want to keep track of where things are you establish a frame of reference. As part of this frame of reference you establish what direction is positive. Any distance in that direction is a positive distance and any distance in the opposite direction is negative. There is nothing strange or mysterious about a negative distance.(4 votes)

- In this video at12:17you use the convention that concave = +f and convex = -f. However, in the Thin lens equation video, you say that concave = -f and convex = +f... So which one is it?(1 vote)
- While dealing with reflection of of light by spherical mirrors we follow a set of sign conventions called the "NEW CARTESIAN SIGN CONVENTION". in this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis of the coordinate system. The conventions are as follows:-

i) The object is always placed to the left of the mirror. This implies that the light form the object falls on the mirror form the left -hand side.

ii) All distances measured to the "LEFT" of the origin (along the + x-axis) are taken as negative. All distances measured to the "RIGHT" of the origin (along the + x-axis) are taken as positive.

Hope that helps!(2 votes)

- At12:17you said that for convex mirror f is negative.So if I were given a convex mirror, its focal length and the image distance, do I need to take both of them as negative to calculate object distance?(1 vote)
- 9:05can't be right. The side has to be smaller than the focal length.(1 vote)
- Which distance are you referring to? None of the sides of either triangle have to be smaller than the focal length.(1 vote)

- How does the concave mirror reflect upon the convex? I don't understand(1 vote)

## Video transcript

- [Voiceover] 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. And so 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 gonna get sent parallel. So I'm gonna draw this one just for fun and then I'm gonna draw another one. You need two 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 gonna 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
form the normal line and 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
gotta 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. And now I can find where my image is. The image of this object is gonna 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 and it looks like that. So ray tracing is cool; it lets
you find where the image is. But I mean I just kinda
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 could 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 gonna be. So that's what we're gonna
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 the textbook,
it looks complicated. It's not actually as bad as it looks. When I used to look at 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 gonna start with drawing triangles. So we'll notice that these
two angles are the same. We're gonna use this to our advantage. So we're gonna make two triangles that have these as one of their angles. So 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 and
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 gonna 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 gonna call that
"hi" for image height. That's how tall the image is. So I know that this is gonna
equal height of the image divided by the adjacent
side to this angle here is this distance right here and we're gonna 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 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. And 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 gonna be this distance from the mirror to the object. We'll give that a name, and if you guessed object
distance, then you guessed right. This is gonna 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 gonna write that as "do". And so 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 divided by ho equals, and then you multiply both sides by di, you get di over 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 lookin' 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 is it gonna be? So if we solve this height of the image, we get the height of the image is the height of the object
times some factor, and that factor's just gonna be the image distance divided
by the object distance. Here's the thing, though. Look at. 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 over 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
negative three centimeters, that means I'd get an image that had a height of three centimeters, 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. 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 we already called those theta, so I'll call these phi. So these angles also have to be the same because any time 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 phis. We'll form two triangles, each triangle's gonna have
phi 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 phi and then back up over to here. And then for the other one we need to also have a
triangle that has phi in it, so we'll do this from here to here, down to there, and then back up to phi. So we've got two triangles. This triangle and this triangle, and they both have phi 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 phi is gonna have to equal the
opposite over the adjacent. The opposite to this phi 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 gonna 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 phi
is also equal to this phi. So I can do tangent of theta for this phi, 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 phi, has to equal opposite over adjacent. This time the opposite of
phi 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 gonna put together and we will get the mirror
equation out of this. There's different ways
to proceed at this point. What I'm gonna do is I don't want ho or hi in either of these, so I'm just gonna solve this
one here for ho over hi. And I get ho over hi, so imagine dividing both sides by hi, and then multiplying
both sides by do minus f, and I'd get ho over hi equals do minus f, the focal length, divided by the focal length. But I could do the same thing up here. I can get ho over hi. It's just gonna equal do over di. So that's all you have to do
is divide both sides by hi and then multiply both sides by do. But this left-hand side right here is the same as this
left-hand side right here. So we know that ho over hi already equals do minus f over f, and up here we know that ho
over hi equals do over di, so that means that do minus f over f has to also equal do over di since both of these
expressions equal ho over hi. So we can set them all equal 'cause they're all
equal to the same thing: ho over hi. And now we're gonna solve this. We're just gonna clean it up. The left-hand side I can write, I'll just stop using colors here, the left-hand side is gonna
be do over f minus one, since f over f just equals one. And that's gotta equal do over di. And now we can divide both sides by do. If I divide both sides
by do I get one over f, since the do cancels, minus one over do equals, and then the do will
cancel with this do on top, and I just get one over di. And now we made it except
it's usually written with this one over do added on the right. If we add one over do to both sides, we finally get the expression we wanted, which is one over the object distance, plus one over the image distance, equals one over the focal length. 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 now 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 kinda eyeballing
it and using a protractor. This'll 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 kinda 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. And it kinda makes sense if
the mirror was bent this way, the focal point is kinda back this way which is kinda 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. And 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 got a single mirror,
there's no craziness goin' on, this object distance is just
always defined to be positive using the convention
that we're using up here. And 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, say you formed an image right here, that would be a negative image distance. If this was five centimeters
behind the mirror, we'd consider that a negative five centimeter image distance. And 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, and 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, focal length can be positive or negative. 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. And 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.