Wave Optics Part 2

For total internal reflection, the wave is incident from the denser medium, so $\sin\theta_{1c}=\frac{n_1}{n_2},\qquad \sin\theta_{2c}=\frac{n_1}{n_3}.$ Given $\frac{n_2}{n_3}=\frac25,$ we get $\sin\theta_{2c}=\frac{n_1}{n_3}=\frac{n_2}{n_3}\cdot\frac{n_1}{n_2}=\frac25\sin\theta_{1c}.$ Since $n_3>n_2>n_1$, we have $\sin\theta_{2c} < \sin\theta_{1c}$, so $\sin\theta_{2c}-\sin\theta_{1c}=-\frac35\sin\theta_{1c}<0.$ This cannot equal $\frac12>0$. Equivalently, the stated equation would require $\sin\theta_{1c}=-\frac56$, which is impossible for a critical angle. Hence no real $\theta_{1c}$ satisfies the given condition, so as written the correct choice is None of Above (seq 4). The proposed solution replaces the condition with a different equation and is therefore not valid for the question as given.