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

Answer

A

Explanation

Here, `mu_(1) = 1, mu_(2) = 1.5`, Let `PO = OQ = x`
As light is travelling from Rarer to Denser medium,
`:. (mu_(1))/(-u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`:. (1)/(-(-x)) + (1.5)/(x) = (1.5 - 1)/(R )`
`(1)/(x) + (1.5)/(x) = (0.5)/(R )`
or `(2.5)/(x) = (0.5)/(R )` or `x = 5 R`