An erect image 2.0 cm high is formed 12 cm from a lens, the object being 0.5 cm high. Find the focal length of the lens.

Since the size of image is greater than that of the object and a virtual image is being formed, the lens is convex in nature.

Now, $v=-12cm$, $h_i=2cm$ and $h_o=0.5cm$

Magnification, $m=\frac{h_i}{h_o}=\frac{v}{u}$

$\Rightarrow \frac{2}{0.5}=\frac{v}{u}$

$\Rightarrow u=\frac{v}{4}=-3cm$

Using the lens formula, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-12}-\frac{1}{-3}=\frac{1}{4}$

$\therefore f=4cm$