An object 0.5 cm high is placed 30 cm from a convex mirror whose focal length is 20 cm. Find the position, size and nature of the image.

We have,
Distance of object from the mirror, u = –30 cm;
Distance of image, v = ?
Focal length, f = + 20 cm 

We know, using the mirror formula,
                     $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}} =\frac{1}{\mathrm{f}}$

$\Rightarrow$                   $$\begin{gathered} \mathrm{v} =\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}} \\ =\frac{(-30) (+20)}{-30-20} \\ = + 12 \mathrm{cm} \end{gathered}$$ 

The image is virtual and erect as, it is formed at a distance of 12 cm behind the mirror. 

Now, size if the object, O = 0.5 cm 
sixe of the image = I
Magnification,  $\mathrm{m} = \frac{\mathrm{I}}{\mathrm{O}} = -\frac{\mathrm{v}}{\mathrm{u}}=-\frac{12}{-30}$
$\Rightarrow$                             $$\begin{gathered} \mathrm{I} = \frac{2}{5}\times \mathrm{O} \\ = \frac{2}{5}\times 0.5 \\ = +0.2 \mathrm{cm} \end{gathered}$$ 

Hence, the height of the image = + 0.2 cm.
The positive sign indicates that the image is erect is smaller in size