MCQ
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 $\theta_{2 C}$. If $n_{3}>n_{2}>n_{1}$; $\frac{\mathrm{n}_{2}}{\mathrm{n}_{3}}=\frac{2}{5}$ and $\sin \theta_{2 \mathrm{C}}-\sin \theta_{1 \mathrm{C}}=\frac{1}{2}$, then $\theta_{1 \mathrm{C}}$ is
- A$\sin ^{-1}\left(\frac{1}{6 n_{1}}\right)$
- B$\sin ^{-1}\left(\frac{2}{3 n_{1}}\right)$
- C$\sin ^{-1}\left(\frac{5}{6 n_{1}}\right)$
- ✓$\sin ^{-1}\left(\frac{1}{3 \mathrm{n}_{1}}\right)$
