Refractive index of an equilateral prism is $\sqrt{2}$. What is the angle of minimum deviation\

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Professor PC Thomas Classes · Answer

$n=\frac{\sin \left(\frac{A+D}{2}ight)}{\sin \left(\frac{A}{2}ight)} \quad \begin{aligned} & A=60 \ & n=\sqrt{2}\end{aligned}$\ $\sqrt{2}=\frac{\sin \left(\frac{60+D}{2}ight)}{\sin 30}$\ $\sin \left(\frac{60+D}{2}ight)=\sqrt{2} 	imes \sin 30=\sqrt{2} 	imes \frac{1}{2}=\frac{1}{\sqrt{2}} \sin 45=\frac{1}{\sqrt{2}}$\ $	herefore \frac{60+\mathrm{D}}{2}=45$\ D =\