For an isosceles prism of angle A and refractive index μ it is found that, the angle of minimum…

For an isosceles prism of angle A and refractive index μ it is found that, the angle of minimum deviation is  δm=A. Which of the following options is/are correct?
  1. 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⁡[sin⁡A4cos⁡A22-1-cos⁡A]
  2. For the angle of incidence i1=A, the ray inside the prism is parallel to the base of the prism.
  3. At minimum deviation, the incident angle  i1 and the refracting angle r1 at the first refracting surface are related by r1=i12
  4. For this prism, the refractive index μ and the angle of prism A are related as A=12cos-1⁡μ2

Solution

i=e (for minimum deviation)

r1+r2=A, r1=r2

i δm=2i-A=A (given)

⇒ i=A

⇒ r1=A2=i2

ii μ=sin⁡Asin⁡A2=2cos⁡A2⇒ A=2cos-1⁡μ2

iii μ sin⁡r2=1

sin⁡r2=1μ

r1+r2=A

r1=A-r2

=A-sin-1⁡1μ

sin⁡i=μsin⁡r1

i=sin-1⁡μsin⁡A-sin-1⁡1μ

i1=sin-1⁡μ2-1sin⁡A-cos⁡A=sin-1⁡μsin⁡A-θC

Here,  μ=2cos⁡A2

iv Condition of min. deviation i=e and r1=r2=A2

Asked in: JEE Advanced 2017 (Paper 1)

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