Irodov Problem 5.23

The key insight is that minimum deviation depends on the refractive index of the medium the light travels through, not just the prism's material. When light enters water instead of air, the relative refractive index between the prism and surrounding medium changes, which shifts the angle at which minimum deviation occurs. Use the minimum deviation formula $n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ to find the prism's absolute refractive index from the air data, then recalculate $\delta_m$ using the effective refractive index in water.

The key insight is that minimum deviation depends on the refractive index of the medium the light travels through, not just the prism's material. When light enters water instead of air, the relative refractive index between the prism and surrounding medium changes, which shifts the angle at which minimum deviation occurs. Use the minimum deviation formula $n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ to find the prism's absolute refractive index from the air data, then recalculate $\delta_m$ using the effective refractive index in water.