A parallel beam of light is incident on a glass prism in the shape of a quarter cylinder of radius $R=0.05 \mathrm{m}$ and refractive index $n=1.5,$ placed on a horizontal table as shown in the figure. Beyond the cylinder, a patch of light is found whose the nearest distance $x$ from the cylinder is

• A

  $(3 \sqrt{3}-4) \times 10^{-2} \mathrm{m}$
• B

  $(2 \sqrt{3}-2) \times 10^{-2} \mathrm{m}$
• Correct answer
  C

  $(3 \sqrt{5}-5) \times 10^{-2} \mathrm{m}$
• D

  $(3 \sqrt{2}-3) \times 10^{-2} \mathrm{m}$

Solution

Given, radius $R=0.05 \mathrm{m}$ Refractive index $n=1 \cdot 5$ The critical angle, $\sin c=\frac{1}{n}=\frac{1}{1 \cdot 5}$ $$ =\frac{10}{15}=\frac{2}{3} $$ From above figure, $\cos c=\frac{R}{R+X}$ $\sqrt{1-\sin ^{2} c}=\frac{R}{R+X}$ $\sqrt{1-\frac{4}{9}}=\frac{R}{R+X}$ $\therefore$ $$ \begin{aligned} \frac{\sqrt{5}}{3} &=\frac{R}{R+X} \\ \sqrt{5}(R+X) &=3 R \\ \sqrt{5} R+\sqrt{5} X &=3 R \\ X &=(3 \sqrt{5}-5) \times 10^{-2} \mathrm{m} \end{aligned} $$

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