An object $5.0 cm$ in length is placed at a distance of $20 cm$ in front of a convex mirror of radius of curvature $30 cm$. Find the position of the image.

Given: $u=-20\ cm$

$f=R/2=15\ cm$

Size of image, $h=5\ cm$

Let size of image be $h'$

From mirror formula, $\cfrac{1}{u}+\cfrac{1}{v}=\cfrac{1}{f}$

$\cfrac{1}{-20}+\cfrac{1}{v}=\cfrac{1}{15}$

$v=\cfrac{60}{7}=8.57\ cm$

The image is formed $8.57 cm$ behind the mirror. It is virtual and erect.

Magnification, $m=-v/u=8.57/20=0.428$

Also, $m=h'/h=h'/5$

From above, $h'/5=0.428$

$h'=2.14\ cm$