A ray of light is incident on the hypotenuse | vedclass

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(D) For total internal reflection to occur at the hypotenuse, the angle of incidence $i$ must be greater than or equal to the critical angle $C$.From

Vedclass · Accepted Answer

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

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(D) For total internal reflection to occur at the hypotenuse, the angle of incidence $i$ must be greater than or equal to the critical angle $C$.From the geometry of the prism, the light ray travels parallel to the base $BC$. The angle between the incident ray and the normal to the hypotenuse is the angle of incidence $i$.In the right-angled triangle, if the base angle is $	heta$, the angle of incidence $i$ at the hypotenuse is $90^{\circ} - 	heta$.For total internal reflection, $i \geq C$, where $\sin C = \frac{1}{\mu}$.Thus, $90^{\circ} - 	heta \geq C \Rightarrow 	heta \leq 90^{\circ} - C$.To find the maximum value of $	heta$, we set $	heta = 90^{\circ} - C$.Taking the cosine on both sides: $\cos 	heta = \cos(90^{\circ} - C) = \sin C$.Since $\sin C = \frac{1}{\mu}$, we have $\cos 	heta = \frac{1}{\mu}$.Therefore, $	heta = \cos ^{-1}\left(\frac{1}{\mu}ight)$.

$A$ 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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