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Question: Refractive index of the solid ($\mu$) in terms of $\mu_0$ and i is-...

Refractive index of the solid (μ\mu) in terms of μ0\mu_0 and i is-

A

μ02+sin⁡2i\sqrt{\mu_0^2 + \sin^2 i}

B

μ0+sin⁡2i\mu_0 + \sin^2 i

C

μ02−2sin⁡2i\sqrt{\mu_0^2 - 2\sin^2 i}

D

μ02−sin⁡2i\sqrt{\mu_0^2 - \sin^2 i}

Answer

μ02−sin⁡2i\sqrt{\mu_0^2 - \sin^2 i}

Explanation

Solution

Applying Snell’s law at the interface between the solid and the prism, and considering the critical angle C and the emergent angle i, we derive the relationship:

μ=μ02−sin⁡2i\mu = \sqrt{{\mu_0}^2 - \sin^2 i}.