A convex mirror has a radius of curvature of 30 cm. A 5 cm tall light bulb is placed at a distance of 20 cm from the mirror. Determine the nature of the image.

Object's distance u = -20 cm

The radius of curvature = 30 cm, therefore

Focal length, f = 15 cm

Substituting values in equation 1/f = 1/u + 1/v
$\frac{1}{\mathrm{f}} = \frac{1}{\mathrm{v}} + \frac{1}{\mathrm{u}}$    
$\frac{1}{15} = \frac{1}{\mathrm{v}} + \frac{1}{(-20)}$

$\frac{1}{\mathrm{v}} = \frac{1}{15} + \frac{1}{20}$

$\frac{1}{\mathrm{v}} = \frac{4+3}{60} = \frac{7}{60}$
$\mathrm{v} = 8.57 \mathrm{cm}$

Since the value is positive, it is on the right of the mirror, and the image formed by the convex mirror is always virtual.