An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

We are given a concave mirror. 

Here, 
Object size, h = + 7.0 cm
Object distance, u = - 27 cm
Focal length, f = - 18 cm 
Image distance, v = ?

Image size, h' = ? 

Now, using the mirror formula, 

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

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

$\Rightarrow$                      $$\begin{gathered} = \frac{1}{-18}-\frac{1}{-27} \\ = \frac{-3+2}{54} \\ = \frac{-1}{54} \end{gathered}$$

i.e.,                  $\mathrm{v} = -54 \mathrm{cm}$ 

The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image. 

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

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

                           $$\begin{gathered} =-\frac{(-54)\times (+7)}{(-27)} \\ = -14 \mathrm{cm} \end{gathered}$$ 

The image is real, inverted and enlarged in size.