For an isosceles prism of angle A and refract | vedclass

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(C) At minimum deviation,$\delta_m = 2i - A = A$,which implies $i = A$. Since $i_1 = i_2 = i$ and $r_1 = r_2 = r$,we have $2r = A$,so $r = A/2$.Thus,$

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$A, C, D$

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(C) At minimum deviation,$\delta_m = 2i - A = A$,which implies $i = A$. Since $i_1 = i_2 = i$ and $r_1 = r_2 = r$,we have $2r = A$,so $r = A/2$.Thus,$r_1 = i_1/2$,making option [$A$] correct.Using Snell's law,$\sin i = \mu \sin r \implies \sin A = \mu \sin(A/2) \implies 2 \sin(A/2) \cos(A/2) = \mu \sin(A/2) \implies \mu = 2 \cos(A/2)$.Rearranging gives $\cos(A/2) = \mu/2$,so $A = 2 \cos^{-1}(\mu/2)$. Option [$B$] is incorrect.For the emergent ray to be tangential,$i_2 = 90^\circ$,so $r_2 = 	heta_c = \sin^{-1}(1/\mu)$. Then $r_1 = A - r_2 = A - \sin^{-1}(1/\mu)$.Using $\sin i_1 = \mu \sin r_1 = \mu \sin(A - \sin^{-1}(1/\mu)) = \mu [\sin A \cos(\sin^{-1}(1/\mu)) - \cos A \sin(\sin^{-1}(1/\mu))] = \mu [\sin A \sqrt{1 - 1/\mu^2} - \cos A (1/\mu)] = \sin A \sqrt{\mu^2 - 1} - \cos A$.Substituting $\mu = 2 \cos(A/2)$,$\mu^2 - 1 = 4 \cos^2(A/2) - 1$. Thus,$i_1 = \sin^{-1}[\sin A \sqrt{4 \cos^2(A/2) - 1} - \cos A]$. Option [$C$] is correct.At minimum deviation,the ray inside the prism is parallel to the base,which occurs when $i_1 = i_2 = A$. Option [$D$] is correct.

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