Question

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 is equal to :

• A. 5R
• B. 3R
• C. 2R
• D. 1.5R

Answer 1

The correct option is A 5R

Given, $PO=OQ$

$\Rightarrow$ $u=v$

We know

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{{\mu }_2}{v}-\frac{{\mu }_1}{-u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{{\mu }_2+{\mu }_1}{V}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{1.5+1}{v}=\frac{1.5-1}{R}$

$\Rightarrow \frac{2.5}{v}=\frac{0.5}{R}$

$\Rightarrow v=\frac{2.5R}{0.5}$

$v=5R$

$OP=OQ=u=v=5R$