Problem — Image with a converging lens

Problem

A converging lens has focal length $f = 10\;\text{cm}$. An object is placed at $p = 15\;\text{cm}$ in front of the lens. Find the position and magnification of the image.

Solution

Setup. Thin-lens equation $\tfrac{1}{p}+\tfrac{1}{q}=\tfrac{1}{f}$.

Position.

$\frac{1}{q} = \frac{1}{f} - \frac{1}{p} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$

$\ev{q = 30\;\mathrm{cm}}$

Magnification.

$I = -\frac{q}{p} = -\frac{30}{15} = \ev{-2}$

Real image ($q>0$, on the other side of the lens), inverted ($I<0$) and magnified $2$ times.

Links

Topics: Ottica Concepts: Lenti e specchi