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(C) Let the refractive index of the slab be $\mu$. Applying Snell's law at the first interface (air to slab): $1 \cdot \sin 60^{\circ} = \mu \cdot \si

Vedclass · Accepted Answer

$1.32$

Vedclass · Answer

(C) Let the refractive index of the slab be $\mu$. Applying Snell's law at the first interface (air to slab): $1 \cdot \sin 60^{\circ} = \mu \cdot \sin r_1$,where $r_1$ is the angle of refraction. $\sin r_1 = \frac{\sin 60^{\circ}}{\mu} = \frac{\sqrt{3}}{2\mu}$. Inside the slab,the angle of incidence at the vertical face is $i_2 = 90^{\circ} - r_1$. For total internal reflection to occur,$i_2 \ge C$,where $C$ is the critical angle. $\sin i_2 \ge \sin C = \frac{1}{\mu}$. Since $\sin(90^{\circ} - r_1) = \cos r_1$,we have $\cos r_1 \ge \frac{1}{\mu}$. Using $\cos r_1 = \sqrt{1 - \sin^2 r_1} = \sqrt{1 - \frac{3}{4\mu^2}} = \frac{\sqrt{4\mu^2 - 3}}{2\mu}$. Substituting this into the inequality: $\frac{\sqrt{4\mu^2 - 3}}{2\mu} \ge \frac{1}{\mu}$. $\sqrt{4\mu^2 - 3} \ge 2 \Rightarrow 4\mu^2 - 3 \ge 4 \Rightarrow 4\mu^2 \ge 7 \Rightarrow \mu \ge \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \approx 1.32$. Thus,the minimum refractive index is $1.32$.

As shown in the figure,a light ray $P$ enters a slab at an angle of $60^{\circ}$ with the normal. Inside the slab,the light ray $Q$ undergoes total internal reflection at the vertical face. Find the minimum refractive index of the slab.

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