In the diagram shown,a ray of light is incide | vedclass

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(B) For total internal reflection $(T.I.R.)$ at the interface of $\mu_{1}$ and $\mu_{2}$:$i > C_{1} \Rightarrow \sin i > \sin C_{1}$ $..........(1)$Us

Vedclass · Accepted Answer

$\mu _1^2 - \mu _2^2 > \mu _3^2$

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(B) For total internal reflection $(T.I.R.)$ at the interface of $\mu_{1}$ and $\mu_{2}$:$i > C_{1} \Rightarrow \sin i > \sin C_{1}$ $..........(1)$Using Snell's law at the interface of $\mu_{1}$ and $\mu_{2}$ for the critical angle $C_{1}$:$\mu_{1} \sin C_{1} = \mu_{2} \sin 90^{\circ}$$\Rightarrow \sin C_{1} = \frac{\mu_{2}}{\mu_{1}}$ $..........(2)$From $(1)$ and $(2)$:$\sin i > \frac{\mu_{2}}{\mu_{1}}$ $..........(3)$For $T.I.R.$ at the interface of $\mu_{1}$ and $\mu_{3}$,the angle of incidence is $(90^{\circ} - i)$:$(90^{\circ} - i) > C_{2} \Rightarrow \sin(90^{\circ} - i) > \sin C_{2}$$\Rightarrow \cos i > \sin C_{2}$Using Snell's law at the interface of $\mu_{1}$ and $\mu_{3}$ for the critical angle $C_{2}$:$\mu_{1} \sin C_{2} = \mu_{3} \sin 90^{\circ}$$\Rightarrow \sin C_{2} = \frac{\mu_{3}}{\mu_{1}}$ $..........(4)$Substituting $(4)$ into the inequality $\cos i > \sin C_{2}$:$\cos i > \frac{\mu_{3}}{\mu_{1}}$Squaring both sides:$\cos^{2} i > \frac{\mu_{3}^{2}}{\mu_{1}^{2}}$$1 - \sin^{2} i > \frac{\mu_{3}^{2}}{\mu_{1}^{2}}$Since $\sin i > \frac{\mu_{2}}{\mu_{1}}$,then $\sin^{2} i > \frac{\mu_{2}^{2}}{\mu_{1}^{2}}$,which implies $-\sin^{2} i Therefore,$1 - \sin^{2} i > 1 - \frac{\mu_{2}^{2}}{\mu_{1}^{2}}$.Combining these,we get $1 - \frac{\mu_{2}^{2}}{\mu_{1}^{2}} > \frac{\mu_{3}^{2}}{\mu_{1}^{2}}$$\mu_{1}^{2} - \mu_{2}^{2} > \mu_{3}^{2}$.

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