If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Physics library>Unit 15

Lesson 2: Mirrors

# 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?

• Shouldn't the "vertical" side of the triangle actually be vertical and go along the dotted line instead of ending at the curved mirror?
• Yes, you're right. Where the light ray hits the mirror is not directly below the principal axis. So, at , 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.
• At , they are two point where ray is intersecting , meaning image will form at two different place? Why did he chose last instead of first?
• 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)
• At how did you take Do - f/f as Do/f - 1?
• 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.
• 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.
• Can object distance be negative? And if so when does this happen?
• 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.
• In this video at you 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!