A spherical surface of curvature R separates air (m=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  

 5R

$ \text{ Let u = -x , v = x } $

$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$

$ \frac{1.5}{x} - \frac{1}{-x} = \frac{1.5 -1}{R}$

$ \frac{2.5}{x} = \frac{0.5}{R}$

$ x = 5R$