JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2025
- Medium
- MCQ
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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Answer & Solution
Correct Answer
(C) \(\sin ^{-1}(3 / 4)\)
Step-by-step Solution
Detailed explanation

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