A spherical surface of curvature R separates air (µ = 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to

5 R

As $\frac{\mu 2}{v}-\frac{\mu 1}{u}=\frac{\mu-1}{R}$

or $\frac{1.5}{x}-\frac{1}{x}=\frac{1.5-1}{R} \Rightarrow x=\frac{2.5 R}{0.5}$ = 5 R