Q152 / 170 · Manipal (MET) 2017 · Physics
The angle of minimum deviation for a prism of angle $A$ is $180-2 A$. The refractive index is
- A$\sin \frac{A}{2}$
- B$\cos \frac{A}{2}$
- C$\tan \frac{A}{2}$
- Correct answerD$\cot \frac{A}{2}$
Solution
Refractive index is given by
$n=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$
Given, $\delta_m=180^{\circ}-2 \mathrm{~A}$
$\begin{aligned}
\therefore & n=\frac{\sin \left(\frac{A+180^{\circ}-2 A}{2}\right)}{\sin \frac{A}{2}} \\
& n=\frac{\sin \left(\frac{180^{\circ}}{2}-\frac{A}{2}\right)}{\sin \frac{A}{2}} \\
\Rightarrow & n=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\cot \frac{A}{2} .
\end{aligned}$
Hence, the correct answer is option: 'D'.