A ray of light incident parallel to the base | vedclass

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Question

(A) For total internal reflection $(TIR)$ to occur at the faces,the angle of incidence at each face must be greater than the critical angle $	heta_c$

Vedclass · Accepted Answer

$\mu > \sqrt{5}$

Vedclass · Answer

(A) For total internal reflection $(TIR)$ to occur at the faces,the angle of incidence at each face must be greater than the critical angle $	heta_c$.From the geometry of the prism,the angles of incidence at the two faces are $(45^{\circ} + r)$ and $(45^{\circ} - r)$,where $r$ is the angle of refraction at the first surface.For $TIR$ to occur at both faces:$45^{\circ} + r > 	heta_c$ and $45^{\circ} - r > 	heta_c$.The more restrictive condition is $45^{\circ} - r > 	heta_c$.Since $\sin r = \frac{\sin 45^{\circ}}{\mu} = \frac{1}{\mu\sqrt{2}}$,we have $r = \arcsin\left(\frac{1}{\mu\sqrt{2}}ight)$.Substituting this into $45^{\circ} - r > 	heta_c$,we get $45^{\circ} - 	heta_c > r$.Taking the sine of both sides: $\sin(45^{\circ} - 	heta_c) > \sin r$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and knowing $\sin 	heta_c = \frac{1}{\mu}$ and $\cos 	heta_c = \frac{\sqrt{\mu^2-1}}{\mu}$:$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{\mu^2-1}}{\mu} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\mu} > \frac{1}{\mu\sqrt{2}}$.Multiplying by $\mu\sqrt{2}$,we get $\sqrt{\mu^2-1} - 1 > 1$,which simplifies to $\sqrt{\mu^2-1} > 2$.Squaring both sides: $\mu^2 - 1 > 4$,so $\mu^2 > 5$,or $\mu > \sqrt{5}$.

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