An object of size 5 cm is placed at a distance of 25 cm from the pole of a concave mirror of radius of curvature 30 cm. Calculate the distance and size of the image so formed. What will be the nature of the image?

We are given a concave mirror. 
Here, 
Object size, h = + 5.0 cm
Object distance, u = - 25 cm
Radius of curvature,  R = - 30 cm     [R is -ve for a concave mirror] 

Therefore, 
Focal length,  $\mathrm{f} = \frac{\mathrm{R}}{2} = -15 \mathrm{cm}$ 

Now, using the mirror formula, 

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

we have,
                  $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{-15} - \frac{1}{-25} \\ = \frac{-5+3}{75} \\ = \frac{-2}{75} \end{gathered}$$

i.e.,               $\mathrm{v} = -37.5 \mathrm{cm}.$

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

$\therefore$      Image size, $\mathrm{h}\text{'} = -\frac{v h}{\mathrm{u}}$ 

                                $$\begin{gathered} = -\frac{(-37.5 \mathrm{cm})\hspace{0.17em}(+5.0 \mathrm{cm})}{(-25 \mathrm{cm})} \\ = -7.5 \mathrm{cm}. \end{gathered}$$

As v is (-)ve, so a real, inverted image of height 7.5 cm is formed at a distance of 37.5 cm in front of the mirror.