A transparent solid cylindrical rod has a refractive index of $\frac{2}{\sqrt{3}}$ . It is surrounded by air. A light ray is incident at the midpoint of one end of the rod as shown in the figure. The incident angle $θ$ for which the light ray grazes along the wall of the rod is

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From Snell's law, μ air sin θ = μ cylider sin 90 ° – θ C ⇒1×sin θ = 2 3 × cos θ C But, sin ⁡ θ C = 1 μ c y l i n d e r = 3 2 Therefore cos ⁡ θ C = 1 2 ⇒sin θ = 2 3 . 1 2 = 1 3 ⇒ θ = sin – 1 ⁡ 1 3

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