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, its nature and size.

We are given a convex mirror.

Here, we have 

Object size, h = + 5 cm 
Object distance, u = -20 cm
Radius of curvature, R = + 3.0 cm [R is +ve for a convex mirror]
$\therefore$ Focal length ,  $\mathrm{f} = \frac{\mathrm{R}}{2} = +15 \mathrm{cm}$ 

From mirror formula,

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

we have, 

                      $$\begin{gathered} \frac{1}{v}= \frac{1}{+15}-\frac{1}{-20} \\ = \frac{4+3}{60} \\ = \frac{7}{60} \end{gathered}$$ 

Image distance, $\mathrm{v} = \frac{60}{7}\simeq 8.6 \mathrm{cm}.$ 

Magnification, $\mathrm{m} = -\frac{\mathrm{v}}{\mathrm{u}}= \frac{\mathrm{h}\text{'}}{\mathrm{h}}$ 
Therefore, 

 $$\begin{gathered} \mathrm{Image} \mathrm{size}, \mathrm{h}\text{'} = -\frac{\mathrm{vh}}{\mathrm{u}} \\ = -\frac{8.6 \times 5}{-20} \\ = 2.15 \simeq 2.2 \mathrm{cm}. \end{gathered}$$ 

A virtual and erect image of height 2.2 cm is formed behind the mirror (because v is positive) at a distance of 8.6 cm from the mirror.