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 $(TIR)$ at the inner curved surface.Let $i$ be the angle of inc

Vedclass · Accepted Answer

$0.5$

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(B) For the light to emerge from surface $B$,it must undergo Total Internal Reflection $(TIR)$ at the inner curved surface.Let $i$ be the angle of incidence at the inner curved surface.For $TIR$,the angle of incidence $i$ must be greater than or equal to the critical angle $	heta_c$.$\sin i \geq \sin 	heta_c$From the geometry of the figure,the light ray at the inner surface makes an angle $i$ with the normal. The normal to the inner surface at the point of incidence passes through the center of curvature. Considering the triangle formed by the center of curvature,the point of incidence,and the inner edge of surface $A$,we have $\sin i = \frac{R}{R+d}$.The critical angle $	heta_c$ is given by $\sin 	heta_c = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3}$.For $TIR$ to occur,$\sin i \geq \sin 	heta_c$,so $\frac{R}{R+d} \geq \frac{2}{3}$.$3R \geq 2R + 2d$$R \geq 2d$$\frac{d}{R} \leq 0.5$.Thus,the maximum value of $\frac{d}{R}$ is $0.5$.

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