A plano-convex lens of refractive index $1.5$ and radius of curvature $30\text{cm}$ is silvered at the curved surface. At what distance from this lens should an object be placed in order to have a real image of the same size as the object?

A silvered lens behaves as a mirror with effective power $P_{eq}=2P_l+P_m$.Power of lens $P_l$: $\frac{1}{f_l}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{-30}\right)=\frac{0.5}{30}=\frac{1}{60}{\text{cm}}^{-1}$.Power of mirror $P_m$: $P_m=-\frac{1}{f_m}=-\frac{1}{-R/2}=\frac{2}{R}=\frac{2}{30}{\text{cm}}^{-1}$.$P_{eq}=2\left(\frac{1}{60}\right)+\frac{2}{30}=\frac{1}{30}+\frac{2}{30}=\frac{3}{30}=\frac{1}{10}{\text{cm}}^{-1}$.Effective focal length $F=-\frac{1}{P_{eq}}=-10\text{cm}$.The combination acts as a concave mirror of focal length $10\text{cm}$ (Radius of curvature $=20\text{cm}$).To get a real image of the same size, the object must be at the center of curvature: $u=2F=20\text{cm}$.