Physics Question from JEE Main 2025 · 29 January, Shift 1

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: