JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 18)
A transparent block A having refractive index $\mu = 1.25$ is surrounded by another medium of refractive index $\mu = 1.0$ as shown in the figure. A light ray is incident on the flat face of the block with incident angle $\theta$ as shown in the 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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Explanation
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$$\begin{aligned} & \mathrm{r}+\theta_{\mathrm{C}}=90^{\circ} \\ & \mu_1 \sin \theta=\mu_2 \operatorname{sinr} \\ & \sin \theta=\frac{\mu_2}{\mu_1} \sin \left(90-\theta_{\mathrm{C}}\right) \\ & \sin \theta=\frac{\mu_2}{\mu_1} \cos \theta_{\mathrm{C}} \\ & \sin \theta_{\mathrm{C}}=\frac{\mu_1}{\mu_2} \end{aligned}$$
$$\begin{aligned} & \sin \theta=\frac{\mu_2}{\mu_1} \sqrt{1-\frac{\mu_1^2}{\mu_2^2}} \\ & \sin \theta=\sqrt{\frac{\mu_2^2-\mu_1^2}{\mu_1^2}}=\sqrt{\frac{\frac{25}{16}-1}{1}} \\ & \sin \theta=\frac{3}{4} \\ & \theta=\sin ^{-1}\left(\frac{3}{4}\right) \end{aligned}$$
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