Question

One end of a glass rod of refractive index $n=1.5$ is a spherical surface of radius of curvature $R$. The centre of the spherical surface lies inside the glass. A point object placed in air on the axis of the rod at the point $P$ has its real image inside glass at the point $Q$ (see fig.). A line joining the points $P$ and $Q$ cuts the surface at $O$ such that $OP=2OQ$. The distance $PO$ is.

• A. $8R$
• B. $7R$
• C. $2R$
• D. Non of these

Answer 1

The correct option is A $8R$

For spherical surfaces, the following relation holds good

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

where,

${\mu }_2=$Refractive index of glass

${\mu }_1=$Refractive index of air

$v=$OP

$u=$ OQ

Given that $OP=2OQ$ (in magnitude, and opposite in sign)

$\frac{1.5}{OQ}-\frac{1}{OP}=\frac{1.5-1}{R}$

$\frac{1.5}{OQ}-\frac{1}{-2OQ}=\frac{1.5-1}{R}$

$4R=OQ$

$OP=2OQ=2\left(4R\right)$

$\Rightarrow OP=8R$