A rod of glass (mu = 1.5) with a square cross | vedclass

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(B) For the light to emerge from surface $B$, it must undergo total internal reflection at the inner curved surface. The condition for total internal

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$0.5$

Vedclass · Answer

(B) For the light to emerge from surface $B$, it must undergo total internal reflection at the inner curved surface. The condition for total internal reflection is that the angle of incidence $	heta$ must be greater than or equal to the critical angle $C$, where $\sin C = \frac{1}{\mu}$.From the geometry of the figure, consider the ray incident at the inner boundary. The angle of incidence $	heta$ is the angle between the ray and the normal to the surface. In the triangle formed by the center of the arc, the point of incidence, and the inner edge, we have $\sin 	heta = \frac{R}{R + d}$.For total internal reflection, we require $	heta \ge C$, which implies $\sin 	heta \ge \sin C = \frac{1}{\mu}$.Substituting the expression for $\sin 	heta$, we get $\frac{R}{R + d} \ge \frac{1}{\mu}$.Taking the reciprocal, we get $\frac{R + d}{R} \le \mu$, which simplifies to $1 + \frac{d}{R} \le \mu$.Therefore, $\frac{d}{R} \le \mu - 1$.Given $\mu = 1.5$, the maximum value is $\frac{d}{R} = 1.5 - 1 = 0.5$.

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