Question

A small object struck on the surface of a glass sphere $\mu =1.5$ is viewed from diametrically opposite position. Calculate transverse magnification and position of image.

Answer 1

Let refractive index of glass $n_1=1.5$

Refractive index of air $n_2=1$

Radius of curvature $=R$

$u=-2R$

Using formula

$\begin{array}{c}\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\\ \frac{1}{v}=\frac{1.5}{R}=\frac{1-1.5}{-R}\\ \frac{1}{v}=\frac{0.5}{R}-\frac{1.5}{R}\\ \frac{1}{v}=\frac{0.5}{R}\left(1-\frac{3}{2}\right)\\ \frac{1}{v}=\frac{-1}{4R}\\ v=-4R\end{array}$

Transverse magnification $=\frac{-v}{u}$

$=\frac{-4R}{-2R}=2$