# Curved mirror equations

#### Prerequisites

From a careful analysis of the easy-to-draw rays for the formation of a real image in front of a curved mirror, we can straightforwardly figure out the relationship among the distances and the sizes of the objects.

## Converging mirror equations

The converging (concave) mirror was discussed in the section on Curved mirrors. Suppose we have an object standing on the center line as shown below.

The spray of rays from a point on the object are shown and three of the rays (the easy-to-find ones) are followed through explicitly. Where they cross is the image point corresponding to the object point where the rays actually started. Now as we slide down the object, the purple ray angle will get less and the green ray will also slide down, so the image point will move up on the object and the image point of the base will be on the center line as shown. Now let's use the geometry to figure out the relationship of the object and image positions and sizes. Let's call

• distance the object is from the mirror $o$,
• the distance the image is from the mirror $i$,
• the height of the object $h$,
• the height of the image $h'$.

and let's begin by analyzing the purple ray.

The triangles consisting of [the center line, the object, and the upper purple ray] and the one consisting of [the center line, the image, and the lower purple ray] are similar. (These two: ) Therefore we have the equation:

$$\frac{h'}{h} = \frac{i}{o}$$

This says that the ratio of the sizes of the object and image are the same as the ratio of their distances from the mirror. If we now look at the green ray (the one through the center of the sphere that the mirror is a part of) we have the following diagram:

Now the triangles that we want consist of [the object, the center line, the upper part of the green line] and [the image, the center line, and the little bit of the green line below the center line up to the image]. (These two: ) If you look carefully, you will see that the similar triangles now tell us that the following ratios are equal:

$$\frac{h'}{h} = \frac{R-i}{o-R}$$

Equating the right sides of these two equations lets us eliminate the heights and get a relation for the distances alone.

$$\frac{i}{o} = \frac{R-i}{o-R}$$

Cross multiplying and rearranging yields

$$io - iR = oR - io$$

or

$$2io = iR + oR$$

This is an interesting result. We have three symbols and three terms but each term only contains two of the symbols. If we divide by the product $ioR$ and define $f$ (the focal length) as $R/2$, we get the result in its standard form:

$\frac{2}{R} = \frac{1}{i} + \frac{1}{o}$    or   $\frac{1}{f} = \frac{1}{i} + \frac{1}{o}$

So the reciprocal distances add to give the reciprocal focal length. (These reciprocal distances are given unit diopters = m-1. They are used in optometry.)

You will notice that if the object is inside the focal length the image distance becomes negative — and the height of the image also becomes negative. If you draw the ray diagram you will see that the situation is that the image becomes virtual, goes behind the mirror, and flips rightside up. This will all be made consistent in the section "Unified equations and sign Conventions" below, but it would be useful to play with a mirror simulation program to see how this works. Click on the image on the left below to try out the Physlet Physics sim. Another good sim is the Physics Classroom's Optics Benchshown on the right. Choose click the blue "Lens" button to change to a converging mirror.

## Diverging mirror equations

If the mirror is turned the other way, it spreads out rays that hit it rather than bringing them together, so it is called a diverging mirror. The ray pattern looks something like this.

The three easy-to-figure-out rays are still easy, even in this case. The green ray aimed towards the center of the sphere (the rightmost x) will bounce straight back. The purple ray aimed at the place the center line hits the sphere will bounce down at the same angle. And the red ray parallel to the center line will bounce back and up so that its extension through the sphere will go to the focal point (the second x from the right — a distance half-way between the center of the sphere and the mirror).

You can see that the rays that hit the mirror all diverge. They will never cross so we will never get a real image. But if we put a couple of eyes (and a brain) to collect those diverging rays, we see that extending them they would cross on the other side of the mirror. Since the brain has no way to tell that the rays it is receiving have been reflected, it assumes they travel straight and the brain creates an image — a virtual image — just as it did with the flat mirror.

We don't want to go through all that geometry and algebra again, but if you do you get the same result — except now f, i, and h' all come in with minus signs. This suggests that we can unify our three situations — concave mirror with real image, concave mirror with virtual image, and convex mirror with virtual image — with a single equation. Let's see what it is and how to use it.

## Unified equations and sign conventions

We've derived one set of the curved mirror equations — for a real image and a concave (converging) mirror — but there are two more sets; for a virtual image and a converging mirror and a virtual image with a diverging (convex) mirror. (A diverging mirror never makes a real image.) Furthermore, there are an equal set of equations for lenses. This is a pretty serious mess! But it turns out that if we choose a slightly different convention from the one we are used to, it all reduces to a pair of equations: one for the distances and one for the heights. How do we accomplish this seeming magic?

The trick is that instead of choosing a coordinate system to describe what's going on, we choose our signs by a standard orientation. Notice that in our treatment of the real image at the top of the page we described all the distances as positive — even though the object was above the center line and the image was below it. We took $h$ and $h'$ as both being positive. If we had chosen y coordinates one should have been positive and the other negative. Some books actually do this. We prefer to consider this the standard situation and make everything positive. Here are the orientations for the real image situation:

• object is to the left of the mirror → $o$ (object distance) is +
• object is above the center line  → $h$ (object height) is +
• image is to the left of the mirror → $i$ (image distance) is +
• image is below the center line → $h'$ (image height) is +
• mirror is concave   → $f$ (focal length) is +

With this, we have the pair of equations with everything positive and all the signs in the equations positive.

$$\frac{1}{f} = \frac{1}{i} + \frac{1}{o} \quad\quad \frac{h'}{h} = \frac{i}{o}$$

The sums of the reciprocals of the image and object distances equals the reciprocal of the focal length ($R/2$); and the ratio of the sizes of the image and object is the same as the ratio of their distances. These should be straightforward to remember.

Now for all the other situations (and even for lenses), the exact same equations also hold. The only trick is, when anything flips from the standard situation, it changes sign. (This is NOT a choice of coordinate system; it is a choice of standard. In some sense we are using different coordinate systems for different variables, but it may be better not to think about that!)

• object is to the right of the mirror →$o$ (object distance) is -
• object is below the center line  → $h$ (object height) is -
• image is to the right of the mirror → $i$ (image distance) is -
• image is above the center line → $h'$  (image height) is -
• mirror is convex (diverging)    → $f$ (focal length) is -

This method has a number of great advantages. There is only one pair of equations to learn and one standard situation to remember. And if we make a wrong assumption about whether an image is real or virtual, up or down, the equation will handle things for us. If the sign comes out differently from what we expected it means we just have to flip something!

But there is a bit of a "dangerous bend".  Students often treat symbols as if they were numbers instead of as containers that can hold values. This means that if we write "$-f$", for example, we do NOT mean that $f$ is negative. We mean the negative of whatever $f$ is. So if $f$ is negative, then $-f$ is positive. Get used to using your variables this way. Use the equations starting with symbols and all + signs. Then put in the values for the quantities with parentheses. So if we are using a convex mirror with a radius of 20 cm, we use the equation $1/f = 1/i + 1/o$, with all + signs, but we replace $f$ by (-10 cm). (Remember that $f$ is half the radius.)  Interesting physics happens when the sign of something changes in these equations, so treat them with care!

The conventions are displayed in the figure below. They make sense when you keep the standard situation — the real image from a converging mirror with the object in the top half.

What will be even better, is that the same equations will work for lenses, as long as the converging lens with a real image is taken as the "standard" situation!

Workout: Curved mirror equations

Joe Redish 4/11/12

Article 710