Physics - Geometrical Optics Question with Solution | TestHub

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 ${\theta }_{2\mathrm{C}}$. If ${\mathrm{n}}_3>{\mathrm{n}}_2>{\mathrm{n}}_1;\frac{{\mathrm{n}}_2}{{\mathrm{n}}_3}=\frac{2}{5}$ and $\sin {\theta }_{2\mathrm{C}}-\sin {\theta }_{1\mathrm{C}}=\frac{1}{2}$, then ${\theta }_{1\mathrm{C}}$ is