A light ray is incident normally on one face | vedclass

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(D) Let the angle of the prism at the base be $\alpha$. The light ray enters normally,so it passes undeviated and travels parallel to the base.At the

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$\cos^{-1}\left(\frac{1}{\mu}\right)$

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(D) Let the angle of the prism at the base be $\alpha$. The light ray enters normally,so it passes undeviated and travels parallel to the base.At the hypotenuse,the angle of incidence $i$ is the angle between the normal to the hypotenuse and the incident ray.From the geometry of the triangle,the angle between the hypotenuse and the base is $\alpha$. The angle between the normal to the hypotenuse and the base is $(90^\circ - \alpha)$.Since the ray is parallel to the base,the angle of incidence $i$ at the hypotenuse is $(90^\circ - \alpha)$.For total internal reflection,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin C = \frac{1}{\mu}$.Thus,$i \ge C \implies (90^\circ - \alpha) \ge C$.For the limiting case,$90^\circ - \alpha = C$.Taking the sine on both sides: $\sin(90^\circ - \alpha) = \sin C$.$\cos \alpha = \frac{1}{\mu}$.Therefore,$\alpha = \cos^{-1}\left(\frac{1}{\mu}\right)$.

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