Question

A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The center of curvature is in 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 is equal to

Answer 1

Let us assume

OP = PQ = a (say)

According to sign conventions,

Object distance (u) = -x

Image distance (v) = +x

We know the formula

$\frac{{\mu }_{\text{2}}}{\text{v}}\text{-}\frac{{\mu }_1}{\text{u}}\text{=}\frac{{\mu }_{\text{2}}\text{-}{\mu }_{\text{1}}}{\text{R}}$

Substituting the values in the above formula,

$\frac{\text{1}\text{.5}}{\text{x}}\text{-}\frac{1}{\text{- x}}\text{=}\frac{\text{1}\text{.5 - 1}}{\text{R}}$

$\frac{\text{1}\text{.5}}{\text{x}}\text{+}\frac{1}{\text{x}}\text{=}\frac{\text{0}\text{.5}}{\text{R}}$

$\frac{\text{2}\text{.5}}{\text{x}}\text{=}\frac{\text{0}\text{.5}}{\text{R}}$

$\text{x = 5R}$

The distance PO is equal to 5R.