An object is kept in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate the two possible distances of the object from the mirror.

We are given a concave mirror.
So, focal length, f = - 20 cm
Also given the image formed is three times the size of the object. 

(a) When the image formed is real:

Given, 
Magnification,  $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}} = -3$ 

$\Rightarrow$ Image distance, $\mathrm{v} = + 34$ 

Now, using the mirror formula,
                  $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}$ 
we have, 

$\therefore$            $\frac{1}{\mathrm{u}}+\frac{1}{3\mathrm{u}} = -\frac{1}{20}$ 


$\Rightarrow$ $\mathrm{u} = \frac{-20 \times 4}{3} = \frac{-80}{3} = -26.67 \mathrm{cm}$ is the object distance measured from the left side of the mirror.

(b) When the image formed is virtual:

Magnification, $\mathrm{m} = \frac{\mathrm{h}\text{'}}{\mathrm{h}} = -\frac{\mathrm{v}}{\mathrm{u}} = +3$ 

$\Rightarrow$ Image distance, $\mathrm{v} = -3\mathrm{u}$ 

Now, using the mirror formula, 

                      $\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}} = \frac{1}{\mathrm{f}}$ 
we have,
                  $\frac{1}{\mathrm{u}}-\frac{1}{3\mathrm{u}} = -\frac{1}{20}$

$\Rightarrow$      $\mathrm{u} = \frac{-20 \times 2}{3} = \frac{-40}{3} = -13.33 \mathrm{cm}$

Therefore, the two possible distance of objects from the mirror are 26.67 cm and 13.33 cm from the left side.