Question

A ray of light incident on one of the face of a prism as shown. After refraction through this surface ray is incident on base of the prism. The refractive index of the prism is $\sqrt{3}$. The net deviation of the ray due to prism is-

• A. ${60}^o$ clockwise
• B. ${166}^o$ clockwise
• C. ${104}^o$ clockwise
• D. ${104}^o$ anticlockwise

Answer 1

The correct option is B ${166}^o$ clockwise

$\text{We know}$

$1\times \sin {60}^{\circ }=\sqrt{3}\times \sin r_1$

$\Rightarrow \sin r_1=\frac{1}{2}\Rightarrow r_1={30}^{\circ }$

$\begin{array}{c}\text{also, we know, in such}\\ \text{prism,}\\ r_1+r_2=A\\ {30}^{\circ }+r_2={67}^{\circ }\\ ⟹r_2={37}^{\circ }\\ {\text{for TIR at point 2 let critical angle is i}}_c\\ \text{we know ,}\\ \sin i_c=\frac{1}{n}\\ \sin i_c=\frac{1}{\sqrt{3}}\\ \text{and,}\sin r_2=\frac{3}{5}\\ \sin r_2>sin{i}_c\\ ⟹r_2>i_c\\ ⟹\text{TIR will take place for point 3 and vertices C}\\ {67}^{\circ }=r_2+r_3⟹r_3={67}^{\circ }-{37}^{\circ }\\ r_3={30}^{\circ }\\ \sqrt{3}\times \sin r_3=1\times \sin e\\ \sin e=\frac{\sqrt{3}}{2}⟹e={60}^{\circ }\\ \text{deviation at point 1}={30}^{\circ }\text{clockwise}\\ \text{deviation at point 2}={106}^{\circ }\text{clockwise}\\ \text{deviation at point 3}={30}^{\circ }\text{clockwise}\\ \text{net deviation}={30}^{\circ }+{106}^{\circ }+{30}^{\circ }={166}^{\circ }\text{clockwise}\end{array}$