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. `2 R`
D. `1.5 R`

Answer

A

Explanation

`(mu_(2))/(v)-(mu_(2))/(u)=(mu_(2)-mu_(1))/(R)implies (1.5)/(+OQ)-(1)/((-OP))-(1.5-1)/(+R)`
On putting `OQ=OP,OP=5R`