A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 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`

Options

A. 5R
B. 3R
C. 2R
D. 1.5R

Answer

A

Explanation

`mu_(2)/v-mu_(1)/u=(mu_(2)-mu_(1))/R,1.5/(PO) + 1/(PO)=(1.5-1)/R`