At the interface between two materials having refractive indices n_1 and n_2, the critical angle for reflection of an em wave is θ_1 C. The n_2…

At the interface between two materials having refractive indices $\displaystyle \mathrm{n}_1$ and $\displaystyle \mathrm{n}_2$, the critical angle for reflection of an em wave is $\displaystyle \theta_{1 \mathrm{C}}$. The $\displaystyle \mathrm{n}_2$ material is replaced by another material having refractive index $\displaystyle \mathrm{n}_3$ such that the critical angle at the interface between $\displaystyle \mathrm{n}_1$ and $\displaystyle \mathrm{n}_3$ materials is $\displaystyle \theta_{2 C}$. If $\displaystyle n_3>n_2>n_1 ; \frac{n_2}{n_3}=\frac{2}{5}$ and $\displaystyle \sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2}$, then $\displaystyle \theta_{1 C}$ is