Physics - Geometrical Optics Question with Solution | TestHub

first, for image formation from lens
$u=-60 \mathrm{cm}$

$f=+30 \mathrm{cm}$

$\Rightarrow v=\frac{uf}{u+f}=\frac{-60\times 30}{-60+30}=60 \mathrm{cm}$

this real image formed by lens acts as virtual object for mirror

Real image from plane mirror is formed $20 \mathrm{cm}$ in front of mirror, hence at $20 \mathrm{cm}$ distance from lens. Now, for second refraction from lens, $u=-20 \mathrm{cm}$

$f=+30 \mathrm{cm}$

$v=\frac{uf}{u+f}=\frac{-20\times 30}{-20+30}=-60 \mathrm{cm}$

So, final virtual image is $60 \mathrm{cm}$ from lens, or $20 \mathrm{cm}$ behind mirror