JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 23)
At the interface between two materials having refractive indices $n_1$ and $n_2$, the critical angle for reflection of an em wave is $\theta_{1C}$. The $\mathrm{n}_2$ material is replaced by another material having refractive index $n_3$ such that the critical angle at the interface between $n_1$ and $n_3$ materials is $\theta_{2 C}$. If $n_3>n_2>n_1 ; \frac{n_2}{n_3}=\frac{2}{5}$ and $\sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2}$, then $\theta_{1 C}$ is :
$ \sin^{-1}\left( \frac{1}{6n_1} \right) $
$ \sin^{-1}\left( \frac{1}{3n_1} \right) $
$ \sin^{-1}\left( \frac{5}{6n_1} \right) $
$ \sin^{-1}\left( \frac{2}{3n_1} \right) $
Explanation
$$\begin{aligned} & \sin \theta_{1 C}=\frac{n_1}{n_2} \\ & \sin \theta_{2 C}=\frac{n_1}{n_3} \\ & \sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2} \end{aligned}$$
$$\begin{aligned} & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \\ & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\mathrm{n}_1=\frac{\mathrm{n}_2}{2} \\ & \mathrm{n}_1\left(\frac{2}{5}-1\right)=\frac{\mathrm{n}_2}{2} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{-5}{6} \\ & =\sin ^{-1}\left(-\frac{5}{6}\right) \end{aligned}$$
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