Sol. [Insert diagram of concave mirror with object O at 40 cm and focal length f = 15 cm]

$\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u}$

$-\frac{1}{15} = \frac{1}{v_1} - \frac{1}{40}$

$\Rightarrow \frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40}$

$v_1 = -24 \text{ cm}$

When object is displaced by 20 cm towards mirror.

Now, $u_2 = -20$

$\frac{1}{f} = \frac{1}{v_2} + \frac{1}{u_2}$

$\frac{1}{-15} = \frac{1}{v_2} - \frac{1}{20}$

$\frac{1}{v_2} = \frac{1}{20} - \frac{1}{15}$

$v_2 = -60 \text{ cm}$

So, image shifts away from mirror by $= 60 - 24 = 36 \text{ cm}$.