Convex Mirror — Car Side Mirror Image (Worked Example)

Medium

Problem

A car's convex side mirror has focal length f = −0.40 m. Another car is 6.0 m behind the mirror. (a) Where does its image form? (b) What is the magnification?

Given

f = −0.40 m (convex → negative)
dₒ = 6.0 m

Find

Image distance dᵢ and magnification m.

Principle

The mirror equation works like the lens equation. Convex mirrors have negative focal lengths and always form virtual, upright, diminished images.

Formula

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, \qquad m = -\frac{d_i}{d_o}

1/f = 1/dₒ + 1/dᵢ; m = −dᵢ/dₒ

Solution

Rearrange: $1/d_i = 1/f - 1/d_o = 1/(-0.40) - 1/6.0$
Compute: $-2.5 - 0.1667 = -2.6667\ \mathrm{m^{-1}}$
Invert: $d_i = 1/(-2.6667) = -0.375\ \mathrm{m}$
Magnification: $m = -(-0.375)/6.0 = +0.0625$

dᵢ = −0.375 m (0.375 m behind the mirror, virtual); m = +0.0625

Unit check

1/m terms throughout; magnification unitless. Consistent.

What this means physically

The image sits 37.5 cm behind the mirror surface — virtual, upright, and 1/16 actual size. That shrinking is why mirrors warn 'objects are closer than they appear': the small image tricks your brain about distance.

Common mistake

Using f = +0.40 m. Convex mirrors diverge light, so f is negative by sign convention — the sign is what makes the image virtual.