Two objects A and B are placed at 15cm and 25cm from the pole in front of a concave mirror having radius of curvature 40cm. The distance between images formed by the mirror is (in cm)

 
Using mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
For point A, u = -15 cm, f = -20cm
$\begin{array}{l}\frac{1}{v_A}+\frac{1}{-15}=\frac{1}{-20}\\ \frac{1}{v_A}=\frac{1}{15}-\frac{1}{20}=\frac{4-3}{60}=\frac{1}{60}\end{array}$
$v_A=+60cm$
For point B, u = -25cm, f = -20cm
$\begin{array}{l}\frac{1}{v_B}+\frac{1}{-25}=\frac{1}{-20}\\ \Rightarrow \frac{1}{v_B}=\frac{1}{25}-\frac{1}{20}=\frac{4-5}{100}=-\frac{1}{100}\end{array}$
$\Rightarrow v_B=-100cm$
Since, the sign of image formed by concave mirror is opposite i.e., both images lie opposite to pole. Hence, the distance between images formed by the mirror is $d=|v_A-v_B|=\left|60-(-100)\right|\Rightarrow d=160cm$