A transparent block A having refractive index $\mu=1.25$ is surrounded by another medium of refractive index $\mu=1.0$ as shown in figure. A light ray is incident on the flat face of the block with incident angle $\theta$ as shown in figure. What is the maximum value of $\theta$ for which light suffers total internal reflection at the top surface of the block?

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{r}+	heta_{\mathrm{C}}=90^{\circ} \ & \mu_1 \sin 	heta=\mu_2 \sin \mathrm{r} \ & \sin 	heta=\frac{\mu_2}{\mu_1} \sin \left(90-	heta_{\mathrm{C}}ight) \ & \sin 	heta=\frac{\mu_2}{\mu_1} \cos 	heta_{\mathrm{C}} \ & \sin 	heta_{\mathrm{C}}=\frac{\mu_1}{\mu_2}\end{aligned}$ $\begin{aligned} & \sin 	heta=\frac{\mu_2}{\mu_1} \sqrt{1-\frac{\mu_1^2}{\mu_2^2}} \ & \sin 	heta=\sqrt{\frac{\mu_2^2-\mu_1^2}{\mu_1^2}}=\sqrt{\frac{\frac{25}{16}-1}{1}} \ & \sin 	heta=\frac{3}{4} \ & 	heta=\sin ^{-1}\left(\frac{3}{4}ight)\end{aligned}$

A transparent block A having refractive index $\mu=1.25$ is surrounded by another medium of refractive index…

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