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A glass prism has a right-triangular cross section $\mathrm{ABC}$, with $\angle \mathrm{A}=90^{\circ}$. A ray of light parallel to the hypotenuse BC and incident on the side AB emerges grazing the side AC. Another ray, again parallel to the hypotenuse BC, incident on the side AC suffers total internal reflection at the side $\mathrm{AB}$. Which one of the following must be true about the refractive index $\mu$ of the material of the prism?
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Verified Answer
The correct answer is:
$\sqrt{\frac{3}{2}} < \mu < \sqrt{2}$


$\mathrm{r}+\theta_{\mathrm{C}}=90^{\circ}......\mathrm{(1)}$
$1 \times \sin \left(90^{\circ}-\alpha\right)=\mu \sin \mathrm{r}$
$\cos \alpha=\mu \sin \mathrm{r}......\mathrm{(2)}$
$90^{\circ}-\mathrm{e}>\theta_{\mathrm{C}}......\mathrm{(3)}$
$\mu \sin \mathrm{e}=1 \times \sin \alpha......\mathrm{(4)}$
$(3) \&(4)$
$90^{\circ}-\theta_{C}>\mathrm{e}$
$\cos \theta_{\mathrm{C}}>\sin \mathrm{e}$
$\cos \theta_{\mathrm{C}}>\frac{\sin \alpha}{\mu}$
$1-\sin ^{2} \theta_{\mathrm{C}}>\frac{1}{\mu^{2}}\left[1-\mu^{2} \sin ^{2} \mathrm{r}\right]$
$1-\frac{1}{\mu^{2}}>\frac{1}{\mu^{2}}\left[1-\mu^{2} \sin ^{2}\left(90^{\circ}-\theta_{\mathrm{C}}\right)\right]$
$1-\frac{1}{\mu^{2}}>\frac{1}{\mu^{2}}-\cos ^{2} \theta_{\mathrm{C}}$
$1-\frac{2}{\mu^{2}}>-\left[1-\frac{1}{\mu^{2}}\right]$
$2>\frac{3}{\mu^{2}}$
$\mu>\sqrt{\frac{3}{2}}$
$(1) \&(2)$
$\cos \alpha=\mu \sin \left(90^{\circ}-\theta_{C}\right)$
$\cos \alpha=\mu \cos \theta_{C}$
$\cos \alpha < 1$
$\mu \cos \theta_{C} < 1$
$\sqrt{1-\frac{1}{\mu^{2}}} < \frac{1}{\mu}$
$1-\frac{1}{\mu^{2}} < \frac{1}{\mu^{2}}$
$1 < \frac{2}{\mu^{2}}$
$\mu < \sqrt{2}$
$\therefore \sqrt{\frac{3}{2}} < \mu < \sqrt{2}$
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