What is the refractive index of the material of a prism if the angle of minimum deviation for the prism of angle $\frac{\pi}{3}$ is $\frac{\pi}{6}$?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\sqrt{2}$
$n=\frac{\sin\left(\frac{A+Δ_{min}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
where,
A, Angle of Prism = $\frac{\pi}{3}$
$Δ_{min}$, Angle of Minimum Deviation = $\frac{\pi}{6}$ and $\frac{\pi}{3}$
$n=\frac{\sin\left(\frac{A+\frac{\pi}{6}+\frac{\pi}{3}}{2}\right)}{\sin\left(\frac{\frac{\pi}{3}}{2}\right)}$
$=\frac{\sin(\frac{\pi}{4})}{\sin(\frac{\pi}{6})}=\frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}}=\sqrt{2}$