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 $n$?

• Correct answer
  A

  $\frac{4}{3}$
• B

  $\frac{3}{2}$
• C

  $-\frac{4}{6}$
• D

  $\frac{-2}{3}$

Solution

From first refraction

$\frac{{\mathrm{\mu }}_2}{\text{v}}-\frac{{\mathrm{\mu }}_1}{\text{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\text{R}}$

$\frac{\text{n}}{{\text{AI}}^1}-\frac{1}{\infty }=\frac{n-1}{\text{R}}$

${\text{AI}}^{\text{'}}=\frac{n\text{R}}{\mathit{\text{n}}-1}$

for 2nd refraction

$\text{u}={\text{BI}}^{\prime }$

$=\text{A}{\text{I}}^{\prime }-2\text{R}$

$=\frac{n\text{R}}{\mathit{\text{n}}-1}-2\text{R}$

$=\frac{(n-2n+2)\text{R}}{\mathit{\text{n}}-1}$

$\text{u}=\frac{(2-n)}{(\mathit{\text{n}}-1)}\text{R}$

$\text{apply}\frac{{\mathrm{\mu }}_2}{\text{v}}-\frac{{\mathrm{\mu }}_1}{\text{u}}=\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{\text{R}}$

$\frac{1}{\text{R}}-\frac{n(n-1)}{(2-\mathit{\text{n}})\text{R}}=\frac{1-n}{-\text{R}}$

$-\frac{n(n-1)}{(2-\mathit{\text{n}})\text{R}}=\frac{n-1}{ \text{R}}-\frac{1}{\text{R}}$

$\frac{n(n-1)}{(\mathit{\text{n}}-2)}=\left(n-2\right)$

$\text{n}\left(\text{n}-1\right)={\left(\text{n}-2\right)}^2$

${\text{n}}^2-\text{n}={\text{n}}^2+4-4\text{n}$

$3\text{n}=4$

$n=\frac{4}{3}$

Hence, the correct answer is option: 'A'.