[Solution by Dr. Abubacker] For a silvered lens $\frac1{f_{eq}}=\frac2{f_L}+\frac1{f_m}=\frac{4(\mu-1)}R+\frac2R=\frac{2(2\mu-1)}R$. The image falls on the object when the object is at the centre of curvature of the equivalent mirror, i.e. at $2f_{eq}=\frac R{2\mu-1}$. Answer (4).