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 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

Let $PO=OQ=x$

Here $u=-x$ ; $v=+x$

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

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

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

$\Rightarrow$$x=5R$

So, $OP=OQ=5R$