A ray of light is incident on the hypotenuse | vedclass

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(D) Let $\alpha$ be the maximum value of the base angle for which light is totally reflected from the hypotenuse.From the geometry of the right-angled

Vedclass · Accepted Answer

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

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(D) Let $\alpha$ be the maximum value of the base angle for which light is totally reflected from the hypotenuse.From the geometry of the right-angled prism,the angle of incidence $i$ at the hypotenuse is $(90^\circ - \alpha)$.For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin C = \frac{1}{\mu}$.Thus,for the limiting case (maximum angle $\alpha$): $i = C$$90^\circ - \alpha = C$Taking the sine of both sides:$\sin(90^\circ - \alpha) = \sin C$$\cos \alpha = \frac{1}{\mu}$Therefore,the maximum value of the base angle is:$\alpha = \cos^{-1}\left(\frac{1}{\mu}\right)$

$A$ ray of light is incident on the hypotenuse of a right-angled prism after travelling parallel to the base inside the prism. If $\mu$ is the refractive index of the material of the prism,the maximum value of the base angle for which light is totally reflected from the hypotenuse is

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