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. `1.5 R`

Answer

A

Explanation

Let us say `PO = OQ = X`
`mu_(1) = 1.0`

Applying `(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)`
Subsituting the values with sign
`(1.5)/(+X) - (1.0)/(-X) = (1.5 - 1.0)/(+R)`
(Distances are measured from O and are taken as positive in the direction of ray of light )
`therefore (2.5)/(X) = (0.5)/(R) " " therefore X = 5R`