Position of the image formed by lens.
$\begin{array}{l}\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{1}{\mathrm{v}}+\frac{1}{60}=\frac{1}{30}\\ \frac{1}{\mathrm{v}}=\frac{1}{30}-\frac{1}{60}=\frac{1}{60}\end{array}$
v = 60 cm
So, position of the image is (60 - 40) = 20 cm behind the plane mirror, it acts as a virtual object. So final image is a real image 20 cm from the plane mirror. This real image will act as an object for the lens and final image is
$\begin{array}{l}\frac{1}{\mathrm{v}}+\frac{1}{20}=\frac{1}{30}\\ \frac{1}{\mathrm{v}}=\frac{-1}{60}\end{array}$
v = -60cm from lens
i.e., 20 cm behind the plane mirror.