A ray of light travelling in a transparent me | vedclass

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(C) For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$ $(i > C)$.Given $i = 45^o$,we requir

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$\mu = 1.5$

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(C) For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$ $(i > C)$.Given $i = 45^o$,we require $45^o > C$,which implies $\sin 45^o > \sin C$.We know that $\sin C = \frac{1}{\mu}$,so $\sin 45^o > \frac{1}{\mu}$.Substituting $\sin 45^o = \frac{1}{\sqrt{2}}$,we get $\frac{1}{\sqrt{2}} > \frac{1}{\mu}$,which simplifies to $\mu > \sqrt{2}$.Since $\sqrt{2} \approx 1.414$,we need $\mu > 1.414$.Among the given options,only $\mu = 1.5$ satisfies this condition.

$A$ ray of light travelling in a transparent medium of refractive index $\mu$ falls on a surface separating the medium from air at an angle of incidence of $45^o$. For which of the following values of $\mu$ can the ray undergo total internal reflection?

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