Question

One end of a glass rod of refractive index $n=1.5$ is 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 figure). 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. None of these

Answer 1

The correct option is A $8R$

Use formula for refraction at spherical surface

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\frac{1.5}{v}-\frac{1}{u}=\frac{1.5-1}{+R}$

Also, $OP=2\left(OQ\right)orOQ=\frac{OP}{2}orv=\frac{-u}{2}$

$\therefore \frac{1.5}{\frac{-4}{2}}-\frac{1}{u}=\frac{0.5}{R}\frac{-3}{4}-\frac{1}{u}=\frac{1}{2R}\frac{-4}{u}=\frac{1}{2R}\therefore u=-8RorPO=8R$