A $4.0$-cm-high object is placed at a distance of $60$ cm from a concave lens of focal length $20$ cm. Find the size of the image.

We have $f=-20cm$ and $u=-60cm$

For a lens,

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

or, $\frac{1}{v}=\frac{1}{u}+\frac{1}{f}$ $=\frac{1}{-60}+\frac{1}{-20}=-\frac{1}{15}$

The magnification is: $m=\frac{h_e}{h_o}=\frac{v}{u}$ $=\frac{-15}{-60}=\frac{1}{4}$

or, $h_e=\frac{h_o}{4}=\frac{4.0}{4}=1.0cm$

So the image is $1.0cm$ high. The positive sign shown that is erect.