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
$\frac{\mu_{2}}{ v }-\frac{\mu_{1}}{ u }=\frac{\mu_{2}-\mu_{1}}{ R }$
$\frac{ n }{ AI ^{1}}-\frac{1}{\infty}=\frac{n-1}{ R }$
$AI ^{\prime}=\frac{n R }{n-1}$
for 2 nd refraction
$u = BI ^{\prime}$
$= A I^{\prime}-2 R$
$=\frac{n R }{n-1}-2 R$
$=\frac{(n-2 n+2) R }{n-1}$
$u =\frac{(2-n)}{(n-1)} R$
apply $\frac{\mu_{2}}{ v }-\frac{\mu_{1}}{ u }=\frac{\mu_{2}-\mu_{1}}{ R }$
$\frac{1}{ R }-\frac{n(n-1)}{(2-n) R }=\frac{1-n}{- R }$
$-\frac{n(n-1)}{(2-n) R }=\frac{n-1}{ R }-\frac{1}{ R }$
$\frac{n(n-1)}{(n-2)}=(n-2)$
$n ( n -1)=( n -2)^{2}$
$n^{2}-n=n^{2}+4-4 n$
$3 n=4$
$n=\frac{4}{3}$