Question

A prism is filled with liquid of refractive index of $\sqrt{2}$ If angle of prism is $60{}^{\circ }$ , find angle of minimum deviation

• A. $75{}^{\circ }$
• B. $60{}^{\circ }$
• C. $45{}^{\circ }$
• D. $30{}^{\circ }$

Answer 1

Answer: (D)

Applying the formula for refractive index $(\mu )$ of a prism having angle of prism A $\mu =\frac{\sin \frac{A+\delta m}{2}}{\sin \frac{A}{2}}$ ?(1) Putting the given values of $\mu =\sqrt{2}$ and angle of prism $\text{A = 6}{\text{0}}^{\text{o}}$ in the Eq. (1), we get $\sqrt{2}=\frac{\sin \frac{{60}^o+\delta m}{2}}{\sin \frac{{60}^o}{2}}=\frac{\sin \frac{{60}^o+\delta m}{2}}{\sin {30}^o}$ or $\sin \left(\frac{{60}^o+\delta m}{2}\right)=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}$ or $\sin \frac{{60}^o+\delta m}{2}=\sin {45}^o$ or ${60}^o+\delta m={90}^o$ or ${\delta }_m={90}^o-{60}^o={30}^o$