For a prism of refractive index $\sqrt{2}$ the angle of prism is equal to the angle of minimum devia

Question

For a prism of refractive index $\sqrt{2}$ the angle of prism is equal to the angle of minimum deviation. Then the angle of prism is\

Professor PC Thomas Classes · Answer

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