# Example: Curved mirror equations

## Understanding the situation

Converging mirrors allow the creation of real images: images that look as if they are floating in mid-air. You can see this either with a ray diagram or a calculation as in curved mirror equations. Let's do an example to see how these two methods work together.

## Presenting a sample problem

As part of a Physics Magic Show, the demonstrator is showing the audience a trick in which a switch appears to light a bulb, but the actual bulb is somewhere else. The setup is shown in the figure below.

The actual lit bulb is hidden from the audience by a black cloth.

A. How far from the mirror should the demonstrator place the unlit bulb if it is to be at the same place as the real image created from the lit bulb? Show both by explicit calculation and a ray diagram.

B. In order for the image to work best, the size of the unlit bulb and the lit bulb's image should be the same size. The demonstrator is intending to use a bulb of the same size as the lit bulb. Is he making a mistake?

## Solving this problem

A. Let's first use the equation that relates the focal length with the object and image distances. Since the object is 5 cm beyond the focal point, the object is a distance $o = 15\;\mathrm{cm}$ from the mirror. What is the image distance, $i$?

$$\frac{1}{f} = \frac{1}{i} + \frac{1}{o}$$

Putting in numbers (all in cm) gives

$$\frac{1}{10} = \frac{1}{i} + \frac{1}{15}$$

so

$$\frac{1}{i} = \frac{1}{10} - \frac{1}{15} = \frac{15}{150} - \frac{10}{150} = \frac{5}{150} = \frac{1}{30}$$

so the image distance is

$$i = 30\;\mathrm{cm}$$.

Now let's draw the ray diagram. We'll consider a spot of the middle of the bright bulb.

We don't have the center point marked so the "easy ray" through the center is a bit tricky to draw. But we know that all the rays that are parallel coming in or going out go through the focal point. So if we take a ray parallel to the axis from the bulb, when it hits the mirror it goes out heading toward the focal point. This ray in  drawn in red. And the ray heading toward the focal point hits the mirror and goes out parallel to the axis. This ray is drawn in blue. The last easy ray, the one hitting the point at the axis goes off so angle of incidence equals angle of refraction. This ray is drawn in purple and confirms the image point. (Note that there is a bit of fuzziness in the meeting point due to the fact that in our derivation of the equation we ignored the fact that the curviness of the mirror reduces the distance. This is called spherical aberration.)

B. We note that the image point in our diagram is well above the midpoint of the unlit bulb. This suggests that the demonstrator might want to use a larger bulb for the unlit one. The equation for the size of the image and object will tell us for sure.

$$\frac{h'}{h} = \frac{i}{o} = \frac{30}{15} = 2$$

This tells us that the image will be twice as tall as the object, so the demonstrator should use a larger bulb for the unlit one.

Joe Redish 7/2/19

Article 729