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
$\frac{\mu _{2}}{\text{v}} - \frac{\mu _{1}}{\text{u}} = \frac{\mu _{2} - \mu _{1}}{\text{R}}$
$\frac{\text{n}}{\text{AI}^{1}} - \frac{1}{\infty } = \frac{n - 1}{\text{R}}$
$\text{AI}^{\text{'}} = \frac{n \text{R}}{\textit{n} - 1}$
for 2nd refraction
$\text{u} = \text{BI}^{'}$
$= \text{A} \text{I}^{'} - 2 \text{R}$
$= \frac{n \text{R}}{\textit{n} - 1} - 2 \text{R}$
$= \frac{\left(\right. n - 2 n + 2 \left.\right) \text{R}}{\textit{n} - 1}$
$\text{u} = \frac{\left(\right. 2 - n \left.\right)}{\left(\right. \textit{n} - 1 \left.\right)} \text{R}$
$\text{apply} \frac{\mu _{2}}{\text{v}} - \frac{\mu _{1}}{\text{u}} = \frac{\mu _{2} - \mu _{1}}{\text{R}}$
$\frac{1}{\text{R}} - \frac{n \left(\right. n - 1 \left.\right)}{\left(\right. 2 - \textit{n} \left.\right) \text{R}} = \frac{1 - n}{- \text{R}}$
$- \frac{n \left(\right. n - 1 \left.\right)}{\left(\right. 2 - \textit{n} \left.\right) \text{R}} = \frac{n - 1}{\text{R}} - \frac{1}{\text{R}}$
$\frac{n \left(\right. n - 1 \left.\right)}{\left(\right. \textit{n} - 2 \left.\right)} = \left(n - 2\right)$
$\text{n} \left(\text{n} - 1\right) = \left(\text{n} - 2\right)^{2}$
$\text{n}^{2} - \text{n} = \text{n}^{2} + 4 - 4 \text{n}$
$3 \text{n} = 4$
$n = \frac{4}{3}$