Lens Magnification — Camera Lens (Worked Example)

Medium

Problem

A camera's converging lens has f = 5.0 cm. The subject is 50 cm away. Find the image distance and magnification.

Given

f = 5.0 cm
dₒ = 50 cm

Find

dᵢ and m.

Principle

A distant object forms a small inverted real image just beyond the focal point — the camera sensor sits there.

Formula

\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i};\quad m = -\frac{d_i}{d_o}

1/f = 1/dₒ + 1/dᵢ; m = −dᵢ/dₒ

Solution

$1/d_i = 1/5.0 - 1/50 = 0.20 - 0.02 = 0.18\ \mathrm{cm^{-1}}$
Invert: $d_i = 1/0.18 \approx 5.56\ \mathrm{cm}$
Magnify: $m = -5.56/50 \approx -0.111$

dᵢ ≈ 5.56 cm; m ≈ −0.11

Unit check

cm and dimensionless — consistent.

What this means physically

The sensor sits just 5.56 cm behind the lens (barely past f), capturing an image about one-ninth life size and upside down — camera firmware flips it for you. Focus the lens closer and dᵢ grows: that's what the focusing ring does.

Common mistake

Expecting magnification greater than 1 for a camera. Cameras reduce: |m| < 1 is the whole point, squeezing a person onto a centimeter-wide sensor.