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 $C$.Here,t

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $C$.Here,the ray travels parallel to the base. From the geometry of the right-angled prism,the angle of incidence $i$ at the hypotenuse is $(90^{\circ} - 	heta)$,where $	heta$ is the base angle.For total internal reflection,$i \geq C$,where $\sin C = \frac{1}{\mu}$.Thus,the condition is $i \geq \sin^{-1}\left(\frac{1}{\mu}ight)$.Substituting $i = 90^{\circ} - 	heta$,we get $90^{\circ} - 	heta \geq \sin^{-1}\left(\frac{1}{\mu}ight)$.Taking the sine of both sides: $\sin(90^{\circ} - 	heta) \geq \frac{1}{\mu}$.This simplifies to $\cos 	heta \geq \frac{1}{\mu}$.Therefore,the maximum value of the base angle $	heta$ is $\cos^{-1}\left(\frac{1}{\mu}ight)$.

$A$ 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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