For a prism of prism angle θ = 60 ° , the refractive indices of the left half and the right half are, respectively, n 1 and n 2 , n 2 ≥ n 1 as shown in the figure. The angle of incidence i is chosen such that the incident light rays will have minimum deviation if n 1 = n 2 = n = 1 . 5 . For the case of unequal refractive indices, n 1 = n and n 2 = n + ∆ n (where ∆ n < < n ), the an

Step-by-step solution

Given, n 1 = n 2 = n = 3 2 In case of minimum deviation, r m = A 2 = 60 ° 2 = 30 ° and i = e Applying expression for minimum deviation, n = sin δ min + A 2 sin A 2 ⇒ 3 2 = sin ( i ) sin 60 ° 2 ⇒ sin i = 3 4 ,    cos i = 7 4 If n 1 = n = 3 2 and n 2 = n + ∆ n : Angle of emergence, e = i + Δ e Applying Snell's law, we get, 1 × sin i = n sin 30 ° Then, n + Δ n sin 30 ° = 1 sin i + Δ e Solving both equations we get, 1 2 Δ n = sin i +