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 × (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
$\frac{{\mu }_2}{\text{v}}-\frac{{\mu }_1}{\text{u}}=\frac{{\mu }_2-{\mu }_1}{\text{R}}$
$\frac{\text{n}}{{\text{AI}}^1}-\frac{1}{\infty }=\frac{n-1}{\text{R}}$
${\text{AI}}^{\text{’}}=\frac{n\text{R}}{\text{n}-1}$
for 2nd refraction
$\text{u}={\text{BI}}^{{}^{\prime }}$
$=\text{A}{\text{I}}^{{}^{\prime }}-2\text{R}$
$=\frac{n\text{R}}{\text{n}-1}-2\text{R}$
$=\frac{(n-2n+2)\text{R}}{\text{n}-1}$
$\text{u}=\frac{(2-n)}{(\text{n}-1)}\text{R}$
$\text{apply}\frac{{\mu }_2}{\text{v}}-\frac{{\mu }_1}{\text{u}}=\frac{{\mu }_2-{\mu }_1}{\text{R}}$
$\frac{1}{\text{R}}-\frac{n(n-1)}{(2-\text{n})\text{R}}=\frac{1-n}{-\text{R}}$
$-\frac{n(n-1)}{(2-\text{n})\text{R}}=\frac{n-1}{\text{R}}-\frac{1}{\text{R}}$
$\frac{n(n-1)}{(\text{n}-2)}=\left(n-2\right)$
$\text{n}\left(\text{n}-1\right)={\left(\text{n}-2\right)}^2$
${\text{n}}^2-\text{n}={\text{n}}^2+4-4\text{n}$
$3\text{n}=4$
$n=\frac{4}{3}$