Question

The angle of a prism is 60${}^{\circ }$. If the refractive index of the material of the prism is $\sqrt{2}$ the angle of minimum deviation is :

• A. 45${}^{\circ }$
• B. 60${}^{\circ }$
• C. 30${}^{\circ }$
• D. sin${}^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is C 30${}^{\circ }$

$\mu =\frac{sin\left(\frac{D+A}{2}\right)}{sin\left(A/2\right)}$

$\mu =2sin\left(\frac{D+60}{2}\right)$

$=2sin\left(\frac{D+60}{2}\right)$

$\frac{1}{\sqrt{2}}=sin\frac{D+{60}^{\circ }}{2}$

${45}^{\circ }=\frac{D+{60}^{\circ }}{2}$

$D=(90-60{)}^{\circ }$

$={30}^{\circ }$