\(\mathrm{u}=-60\, \mathrm{~cm}\)

\(\mathrm{f}=+30\, \mathrm{~cm}\)

\(\Rightarrow \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}+\mathrm{f}}=\frac{-60 \times 30}{-60+30}=60\, \mathrm{~cm}\)

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,

\(\mathrm{u}=-20\, \mathrm{~cm}\)

\(\mathrm{f}=+30\, \mathrm{~cm}\)

\(v=\frac{\mathrm{uf}}{\mathrm{u}+\mathrm{f}}=\frac{-20 \times 30}{-20+30}=-60\, \mathrm{~cm}\)

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