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 \({n}_{2}\) 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 \({\theta }_{2C}\). If \({n}_{3}>{n}_{2}>{n}_{1};\frac{{n}_{2}}{{n}_{3}}=\frac{2}{5}\) and \(\sin {\theta }_{2C}-\sin {\theta }_{1C}=\frac{–1}{2}\), then \({\theta }_{1C}\) is:

[JEE Main 2025, 29 Jan (Shift 1)]

Question

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 \({n}_{2}\) 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 \({\theta }_{2C}\). If \({n}_{3}>{n}_{2}>{n}_{1};\frac{{n}_{2}}{{n}_{3}}=\frac{2}{5}\) and \(\sin {\theta }_{2C}-\sin {\theta }_{1C}=\frac{–1}{2}\), then \({\theta }_{1C}\) is:

[JEE Main 2025, 29 Jan (Shift 1)]

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