An object $5cm$ is placed at a distance of $20cm$ in front of a convex mirror of radius of curvature $30cm$. Find the position of the image, its nature and size.

Formula:

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

Given,

$\frac{1}{v}+\frac{1}{(-20)}=\frac{1}{15}$

$\frac{1}{v}=\frac{7}{60}\Rightarrow v=\frac{60}{7}$

$v=8.57\approx 8.6cm$

The image is formed behind the mirror at a distance of $8.6cm$

$m=\frac{h_2}{h_1}=-\frac{v}{u}$

$\to \frac{h_2}{5}=-\frac{8.57}{20}$

$\therefore h_2=2.175cm$

Thus the image is $2.175cm$ high, virtual and erect.

Hence, this is the answer.