At the interface between two materials having refractive indices \(n_1\) and \(n_2\), the critical angle for reflection of an em wave is \(\theta_{1 C}\). The \(n_2\) 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 \(\theta_{2 \mathrm{C}}\). If \(\mathrm{n}_3\gt\mathrm{n}_2\gt\mathrm{n}_1 ; \frac{\mathrm{n}_2}{\mathrm{n}_3}=\frac{2}{5}\) and \(\sin \theta_{2 \mathrm{C}}-\sin \theta_{1 \mathrm{C}}=\frac{1}{2}\), then \(\theta_{1 \mathrm{C}}\) is :

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(C) \(\sin ^{-1}\left(\frac{-5}{6}\right)\)

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\(\begin{aligned} & \sin \theta_{1 C}=\frac{n_1}{n_2} \\ & \sin \theta_{2 C}=\frac{n_1}{n_3} \\ & \sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2}\end{aligned}\)
\(\begin{aligned} & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{1}{2} \\ & \mathrm{n}_1 \frac{\mathrm{n}_2}{\mathrm{n}_3}-\mathrm{n}_1=\frac{\mathrm{n}_2}{2} \\ & \mathrm{n}_1\left(\frac{2}{5}-1\right)=\frac{\mathrm{n}_2}{2} \\ & \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{-5}{6} \\ & =\sin ^{-1}\left(-\frac{5}{6}\right)\end{aligned}\)

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