An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.

We are given a concave mirror.

Here,
Object size, h = + 4.0 cm
Object distance, u = -25.0 cm
Focal length, f = - 15.0 cm
Image distance, v = ?
Image size, h' = ? 

Now, using the mirror formula,  

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

$\therefore$                $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}} \\ = \frac{1}{-15}-\frac{1}{-25} \\ =-\frac{1}{15}+\frac{1}{25} \end{gathered}$$ 

                   $$\begin{gathered} \frac{1}{\mathrm{v}} = \frac{-5+3}{75} \\ = \frac{-2}{75} \\ \mathrm{v} = -37.5 \mathrm{cm} \end{gathered}$$

The screen should be placed at a distance of 37.5 cm on the object side of the mirror, to obtain a sharp image of the object. 

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

Image size,
                   $$\begin{gathered} \mathrm{h}\text{'} = -\frac{\mathrm{vh}}{\mathrm{u}} \\ = -\frac{(-37.5 \mathrm{cm}) (+4.0 \mathrm{cm})}{(-25 \mathrm{cm})} \\ = -6.0 \mathrm{cm}. \end{gathered}$$ 

The image formed is real, inverted (because h' is negative) and enlarged in size.