Question:

For an isosceles prism of angle A and refractive index μ, it is found that the angle of minimum deviation δm = A. Which of the following option(s) is/are correct?

For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is i1 = sin⁻¹[sinA√(4cos²A/2)⁻¹ - cosA]

At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are related by r1 = (i1/2)

For the angle of incidence i1 = A, the ray inside the prism is parallel to the base of the prism

For this prism the refractive index μ and the angle of prism A are related as A = 1/2cos⁻¹(μ²/2)

Solution:

Option A
We know for minimum deviation, i1 = A + δm/2 (δm: angle of minimum deviation)
Given δm = A
Hence i1 = A
Now, for minimum deviation condition r1 = A/2
Hence r1 = i1/2
Correct
Option B
μ = sin(i1)/sin(r1)
μ = sin(A + δm/2)/Sin (A/2)
μ = sin A/sin A/2 = 2 cos A/2
Incorrect
Option C
Emergence = 90°
sin r2 = 1/μ
r2 = Sin⁻¹ (1/μ)
Also r1 = A - r2
sin i1 = μsinr1
sin i1 = μsin(A - r2)
sin i1 = μ(sinA cosr2 - cosA sinr2)
But μ sinr2 = 1
sin i1 = μ sinA cosr2 - cosA = sinA√(4cos²A/2)⁻¹ - cosA
Correct