Question

A prism of refractive index $\sqrt{2}$ has refracting angle ${60}^{\circ }$ answer the question
Angle of maximum deviation is

• A. ${45}^{\circ }$
• B. ${\sin }^{-1}\left(\sqrt{2}\sin {15}^{\circ }\right)$
• C. ${30}^{\circ }+{\sin }^{-1}\left(\sqrt{2}\sin {15}^{\circ }\right)$
• D. none

Answer 1

The correct option is C ${30}^{\circ }+{\sin }^{-1}\left(\sqrt{2}\sin {15}^{\circ }\right)$

$\begin{array}{c}\text{For maximum deviation,}\\ \angle e={90}^{\circ }\text{angle of emergence}\end{array}$

$\begin{array}{c}\text{at point}2\\ \begin{array}{cc}{\mu }_1\sin {\theta }_1={\mu }_2\sin {\theta }_2& \\ \sqrt{2}\sin r_2=1\times 1\Rightarrow r_2& ={\sin }^{-1}\frac{1}{\sqrt{2}}\\ r_2& ={45}^{\circ }\end{array}\end{array}$

$\begin{array}{c}\text{now,}\\ \begin{array}{c}A=r_1+r_2\Rightarrow r_1=A-r_2={60}^{\circ }-{45}^{\circ }\\ ={15}^{\circ }\end{array}\\ \text{at point}1\\ \begin{array}{c}{\mu }_1\text{sin}{\theta }_1={\mu }_2\sin {\theta }_2\\ 1\times \sin i=\sqrt{2}\sin r_1\\ \sin i=\sqrt{2}\sin {15}^{\circ }\\ i={\sin }^{-1}\left(\sqrt{2}\sin {15}^{\circ }\right)\end{array}\end{array}$

$\begin{array}{cc}i& ={\sin }^{-1}\left(\sqrt{2}\sin {15}^{\circ }\right)\\ \delta =& \text{angle of deviation}=-A+i+e\\ & ={30}^{\circ }+{\sin }^{-1}\left(\sqrt{2}\sin 15\right)\end{array}$