For an isosceles prism of angle A and refractive index $\mu$ , it is found that the angle of minimum deviation  ${\delta }_m=A$. Which of the following options is/are correct?

detailed solution

Correct option is A

δm=A & μ=sin⁡A+δ¯m2sin⁡A/2⇒μ=sin⁡Asin⁡A/2⇒μ=2sin⁡A/2cos⁡A/2sin⁡A/2=2cos⁡A/2⇒cos⁡A/2=(μ/2)⇒(A/2)=cos−1⁡(μ/2)⇒A=2cos−1⁡(μ/2) (D) is not correct. δm=i1+i2−A⇒i1+i2=2A and i1=i2=ASo option A is correct.r1=r2 and μ=sin⁡i1sin⁡r12cos⁡A/2=sin⁡Asin⁡r1⇒sin⁡r1=2sin⁡A/2cos⁡A/22cos⁡A/2⇒r1=A/2=i1/2 option (B) is correct Emergent ray tangential to the surface meansi2=90∘⇒r2=C (critical angle) r1=(A−C)μ=sin⁡i1sin⁡r1sin⁡i1=μsin⁡r1=2cos⁡A/2[sin⁡(A−C)]=2cos⁡A/2(sin⁡Acos⁡C−cos⁡Asin⁡C)=2cos⁡A/2sin⁡A1−sin2⁡C−cos⁡A1μ=2cos⁡A/2sin⁡A1−1μ2−cos⁡A1μ=2cos⁡A/2sin⁡A1−14cos2⁡A/2−cos⁡A12cos⁡A/2=2cos⁡A/2sin⁡A4cos2⁡A/2−12cos⁡A/2−cos⁡A12cos⁡A/2=sin⁡A4cos2⁡A/2−1−cos⁡A⇒i1=sin−1⁡sin−1⁡A4cos2⁡A/2−1−cos⁡AOption C is correct