Using Mirror formula

$$ \begin{aligned} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\\\ & \frac{1}{v}=\frac{1}{f}-\frac{1}{f} \\\\ & v=\frac{u f}{u-f} \end{aligned} $$

For object $\mathrm{A}, \mathrm{u}_{\mathrm{i}}=-15 \mathrm{~cm}, \mathrm{f}=-20 \mathrm{~cm}, \mathrm{~v}_1=$ ?

$$ \begin{aligned} & \mathrm{v}_1=\frac{\mathrm{u}_1 \mathrm{f}}{\mathrm{u}_1-\mathrm{f}}=\frac{(-15)(-20)}{(-15)-(20)}=\frac{+300}{5} \\\\ & \mathrm{v}_1=+60 \mathrm{~cm}(+ ve, virtual) \end{aligned} $$

For object $\mathrm{B}, \mathrm{u}_2=-25 \mathrm{~cm}, \mathrm{f}=-20 \mathrm{~cm} \mathrm{v}_2=$ ?

$$ \begin{aligned} & \mathrm{v}_2=\frac{\mathrm{u}_2 \mathrm{f}}{\mathrm{u}_2-\mathrm{f}}=\frac{(-25)(-20)}{(-25)-(-20)}=\frac{500}{-5} \\\\ & \mathrm{v}_2=-100 \mathrm{~cm}(- ve, real) \end{aligned} $$

Hence, the distance between images formed by the mirror is,

d = 60 – (–100) = 160 cm