A biconvex thin lens of refractive index 1.5 has radii of curvature 20 cm and 30 cm. A plane mirror is placed perpendicular to the principal axis, 36…

Lens-maker's formula: $\dfrac1f=(1.5-1)\left(\dfrac1{20}+\dfrac1{30}\right)=\dfrac1{24}$, so $f=24\ \mathrm{cm}$. By the reversibility of light paths, the final image can coincide with the object only if the rays retrace themselves. That happens when the first image falls exactly on the plane mirror, since the rays then diverge from a point on the mirror, are reflected back along the same cone and return through the same points. (The only other self-retracing arrangement is the object at the focus, $x=f=24$ cm, where the rays leave the lens parallel and strike the mirror normally; it is excluded by the condition $x>f$.) So the image distance is $v=36\ \mathrm{cm}$. From $\dfrac1v-\dfrac1u=\dfrac1f$ with $u=-x$: $\dfrac1x=\dfrac1f-\dfrac1v=\dfrac1{24}-\dfrac1{36}=\dfrac1{72}$. Hence $x=72\ \mathrm{cm}$ (a real object, beyond $2f=48$ cm, giving a real image beyond $f$).