Question

A narrow parallel beam of light is incident on a transparent sphere of refractive index $n$ . If the beam finally gets focussed at a point situated at a distance $=2\times$ (radius of sphere) from the centre of the sphere, then find $n$ ?

• A. $\frac{4}{3}$
• B. $\frac{3}{2}$
• C. $-\frac{4}{6}$
• D. $\frac{-2}{3}$

Answer 1

Answer: (A)

From first refraction
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\frac{n}{A{I}^1}-\frac{1}{\infty }=\frac{n-1}{R}$
$A{I}^{\prime }=\frac{nR}{n-1}$
for 2 nd refraction
$u=B{I}^{\prime }$
$=A{I}^{\prime }-2R$
$=\frac{nR}{n-1}-2R$
$=\frac{(n-2n+2)R}{n-1}$
$u=\frac{(2-n)}{(n-1)}R$
apply $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$
$\frac{1}{R}-\frac{n(n-1)}{(2-n)R}=\frac{1-n}{-R}$
$-\frac{n(n-1)}{(2-n)R}=\frac{n-1}{R}-\frac{1}{R}$
$\frac{n(n-1)}{(n-2)}=(n-2)$
$n(n-1)=(n-2{)}^2$
$n^2-n=n^2+4-4n$
$3n=4$
$n=\frac{4}{3}$