Question

Angle of minimum deviation is?

• A. ${45}^o$
• B. ${90}^o$
• C. ${30}^o$
• D. None

Answer 1

The correct option is C ${30}^o$

Angle of minimum deviation $\left({\delta }_m\right)$,

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$ where $\mu$ is refractive index and A is angle of prism.

$\sqrt{2}=\frac{\sin \left(\frac{60+{\delta }_m}{2}\right)}{\frac{1}{2}}$

$\frac{1}{\sqrt{2}}=\sin \left(\frac{60+{\delta }_m}{2}\right)$

$60+{\delta }_m={90}^{\circ }$

${\delta }_m={30}^{\circ }$