Question

A ray of light passes through a prism of refractive index `sqrt2` as shown in the figure. Find:

1. The angle of incidence (∠r₂) at face AC.
2. The angle of minimum deviation for this prism.

Answer 1

(i) Since at point N, the angle of refraction is 90°, then ∠r₂ is the critical angle for the glass-air pair of media.

sin ∠r₂ = `1/μ = 1/sqrt2`

∴ ∠r₂ = `sin^-1(1/sqrt2)` = 45°

(ii) μ = `sin(("A" + δ_"m")/2)/(sin  "A"/2)`

Or, `sqrt2 = sin ((60^circ + δ_"m")/2)/(sin  60^circ/2)`

Or, `sqrt2 = sin (30^circ + δ_"m"/2)/(1/2)`

Or, 0.7 = `sin(30^circ + δ_"m"/2)`

Or, sin⁻¹ 0.7 = `30^circ + δ_"m"/2`

Or, 44.4° = `30^circ + δ_"m"/2`

∴ δ_m = 28.8°