For an isosceles prism of angle A and refractive index μ it is found that, the angle of minimum deviation is δ m = A . Which of the following options is/are correct?

Step-by-step solution

i = e (for minimum deviation) r 1 + r 2 = A , r 1 = r 2 i δ m = 2 i - A = A (given) ⇒ i = A ⇒ r 1 = A 2 = i 2 i i μ = sin ⁡ A sin ⁡ A 2 = 2 cos ⁡ A 2 ⇒ A = 2 cos - 1 ⁡ μ 2 i i i μ sin ⁡ r 2 = 1 sin ⁡ r 2 = 1 μ r 1 + r 2 = A r 1 = A - r 2 = A - sin - 1 ⁡ 1 μ sin ⁡ i = μ sin ⁡ r 1 i = sin - 1 ⁡ μ sin ⁡ A - sin - 1 ⁡ 1 μ i 1 = sin - 1 ⁡ μ 2 - 1 sin ⁡ A - cos ⁡ A = sin - 1 ⁡ μ sin ⁡ A - θ C H e r e , μ = 2 cos ⁡ A 2 i v Condition of min. deviation i = e a n d r 1 = r 2 = A 2