KinexiaLEARN PHYSICS

Thin Lens Equation — Camera Focus (Worked Example)

Medium
Problem

A camera's 50 mm converging lens (f = +0.050 m) photographs a subject 2.0 m away. (a) How far behind the lens does the image form? (b) What is the magnification?

Ray diagram through a convex camera lens with object 2.0 m away and image forming 51.3 mm behindFFsubject 2.0 mimage 51.3 mm
Rays through the 50 mm lens: the image forms just behind the focal point.

Given

  • f = +0.050 m

  • dₒ = 2.0 m

  • converging lens

Find

Image distance dᵢ and magnification m.

Principle

The thin lens equation relates focal length, object distance and image distance; magnification is the ratio of image to object distance.

Formula

Solution

  1. Rearrange: 1/di=1/f−1/do=1/0.050−1/2.01/d_i = 1/f - 1/d_o = 1/0.050 - 1/2.0

  2. Compute: 20−0.5=19.5 m−120 - 0.5 = 19.5\ \mathrm{m^{-1}}

  3. Invert: di=1/19.5=0.0513 m=51.3 mmd_i = 1/19.5 = 0.0513\ \mathrm{m} = 51.3\ \mathrm{mm}

  4. Magnification: m=−0.0513/2.0=−0.0256m = -0.0513/2.0 = -0.0256

dᵢ ≈ 51.3 mm behind the lens; m ≈ −0.0256

Unit check

1/m − 1/m = 1/m, inverted to m. Magnification is a unitless ratio. Consistent.

What this means physically

The image forms just 1.3 mm behind the focal point — nearly at f, as expected for a distant object. The negative, tiny magnification means the sensor sees an inverted image about 1/39 life-size.

Common mistake

Mixing millimeters and meters. Convert f to meters (0.050 m) before combining with dₒ = 2.0 m.