The thin lens equation: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Medical Physics | AQA A-Level Physics

Key Definition

Thin lens equation: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, relating the focal length $f$, object distance $u$, and image distance $v$.

Sign convention (real-is-positive): real object distances and real image distances are positive; virtual image distances are negative.
For a converging lens, $f$ is positive. For a diverging lens, $f$ is negative.
MagnificationThe ratio of image height to object height, or equivalently the ratio of image distance to object distance. $m = \frac{v}{u}$. is given by $m = \frac{v}{u} = \frac{h_i}{h_o}$, where $h_i$ is image height and $h_o$ is object height.
Crucially, when $u = f$ the rays emerge parallel and no image is formed (image at infinity).
When $u < f$ for a converging lens, the image is virtual, upright, and magnified. This is how a magnifying glass works.

Worked Example

An object is placed 30 cm from a converging lens of focal length 10 cm. Calculate the image distance and the magnification.

Show Solution

Write the thin lens equation

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

Rearrange for 1/v

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30}$$

Calculate v

$$v = \frac{30}{2} = 15 \text{ cm}$$

Calculate magnification

$$m = \frac{v}{u} = \frac{15}{30} = 0.5$$

Answer

$v = 15$ cm (real, inverted, diminished). Magnification $= 0.5$.

Common Mistake HIGH

Students often: Forgetting the sign convention and plugging in a negative $f$ for a converging lens.
Instead: Converging lenses always have positive $f$. Only diverging lenses have negative $f$.