# Convex mirrors - image size

1. Apr 19, 2009

### IniquiTrance

1. The problem statement, all variables and given/known data

Prove that the virtual image in a convex mirror is always smaller than the real object.

2. Relevant equations

$$m = -\frac{d_{i}}{d_{O}}$$

3. The attempt at a solution

Not a hw problem. Something which is bothering me, and haven't been able to prove yet.

Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 19, 2009

### rl.bhat

When the object is at infinity, where the image is formed in convex mirror. Now write the relation between di, do and f with proper sign. Multiply by di to each term on both the side and find the relation for m. From the result, see whether you get your answer.

3. Apr 19, 2009

### IniquiTrance

Thanks for the response.

When the object is at infinity, the image is formed at the focus.

When I multiply both sides by di, I get $$-m + 1 = \frac{d_{i}}{f}$$

I still don't see why this proves $$|m|< 1$$ :(

4. Apr 19, 2009

### IniquiTrance

Ah ok I see it.

$$m = 1 - \frac{d_{i}}{f}$$

Since $$m = -\frac{d_{i}}{d_{O}}> 0$$ for convex mirrors, since the image is behind the mirror, while the object is in front, m is at max 1.

Thanks!