Question

For a glass prism $(\mu =\sqrt{3})$ the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

• A. 45
• B. 90
• C. 60
• D. 30

Answer 1

Answer: (C)

Using, $\mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\frac{A}{2}}$

According to question, ${\delta }_m=A$

$\therefore \sqrt{3}=\frac{sin\frac{A+A}{2}}{sin\frac{A}{2}}$

(Given $\mu =\sqrt{3})$

$\sqrt{3}=\frac{sinA}{sin\frac{A}{2}}=\frac{2sin\frac{A}{2}cos\frac{A}{2}}{sin\frac{A}{2}}orcos\frac{A}{2}=\frac{\sqrt{3}}{2}=cos{30}^{\circ }$

$\Rightarrow \frac{A}{2}={30}^{\circ }\Rightarrow A={60}^{\circ }$