Question

A spherical convex surface of radius of curvature $R$ separates air (${\mu }_a=1$) from glass(${}_a{\mu }_g=1.5$) .The centre of curvature is in the glass is $O$. 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. 1.5 R

Answer 1

The correct option is A 5R

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

where $u$ and $v$ is the object distance and image distance, respectively.

Given : ${\mu }_1=1$ and ${\mu }_2=1.5$

Let the distance of object from point O be $x$.

$⟹$ $u=-x$ and $v=x$

$\therefore$ $\frac{1.5}{x}-\frac{{\mu }_1}{(-x)}=\frac{1.5-1}{R}$

$\frac{3}{2x}+\frac{1}{x}=\frac{1}{2R}$

$\frac{3+2}{2x}=\frac{1}{2R}$

$⟹x=5R$