Question

When light rays are incident on a prism at an angle of ${45}^o$, the minimum deviation is obtained. If the refractive index of the material of the prism is $\sqrt{2}$, then the angle of prism will be

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is D ${90}^o$

$\begin{array}{c}Let\angle idenoteincidenceangleand\angle rdenoterefractedangle.\\ Given:\\ \angle i=2\times \angle r\\ Nowbysnel{l}^{\prime }sLawapplingatinterfaceofism:\\ \frac{\sin i}{\sin r}=\frac{\sin \left(2r\right)}{\sin r}=\frac{2\sin r\times \cos r}{\sin r}=2\cos r=\sqrt{2}\\ wherewehaveusedtrignometricidentity;\\ \sin 2r=2\sin x\times \cos x\\ Thereforewehave;\\ \cos r=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\\ \Rightarrow r={45}^0\\ nowwehave3relationfromaminimumdeviation;\\ A=r_1+r_2\\ A=2r\\ A={90}^0\end{array}$