Question

A Spherical surface of radius of curvature R separates air ( Refractive Index 1.0) from glass ( R.I 1.5). The center of curvature is in the glass. A point P is placed in Air is found to have a real image Q in the glass. The line PQ cuts the surface at Point O and PO=OQ. Then, the distance PO is equal to = ?

Answer 1

The magnitude of the distance PO will be 5 times the radius of curvature.

Explanation :

Given, Radius of curvature = R

Refractive index, μ₁ = 1

Refractive index, μ₂ = 1.5

since it is given,

PO = OQ

=> u = v

For spherical surfaces we know that,

(μ₂/v) + (μ₁/u) = (μ₂-μ₁)/R

=> 1.5/v + 1/u = (1.5-1)/R

since u = v

=> 1.5/u + 1/u = (1.5-1)/R

=> 2.5/u = 0.5/R

=> u = 2.5R/0.5 = 5R

=> PO = 5R

hence the magnitude of the distance PO will be 5 times the radius of curvature.