A right‑angle prism has an apex angle of 45°. When green light of wavelength 520 nm is incident, the minimum angle of deviation is 35°. Find the refractive index of the prism for green light.

Step-by-step solution

1. Use the relation for minimum deviation in a prism: $n=\frac{sin\frac{A+D_{min}}{2}}{sin\frac{A}{2}}$.
2. Insert $A=45^{c}irc$ and $D_{min}=35^{c}irc$.
3. Compute the numerator: $sin\frac{45+35}{2}=sin40^{c}ircapprox0.6428$.
4. Compute the denominator: $sin\frac{45}{2}=sin22.5^{c}ircapprox0.3827$.
5. Divide to get $n=\frac{0.6428}{0.3827}\approx 1.68$.
6. So the refractive index of the prism for green light is about 1.68.

Concept tested: Refractive index from minimum deviation

Published 4 October 2026 · How this solution was prepared and checked