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 $\mathrm{n}_2$ 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 $n_3>n_2>n_1 ; \frac{n_2}{n_3}=\frac{2}{5}$ and $\sin \theta_{2 C}-\sin \theta_{1 C}=\frac{1}{2}$, then $\theta_{1 C}$ is :

<p>Let the wave be incident from the denser medium onto the rarer one in each case. Since $n_3>n_2>n_1$ the critical‐angle sines are</p> <p>– at the $n_2\kern1pt|\kern1ptn_1$ interface:</p> <p>$ \sin\theta_{1C}=\frac{n_1}{n_2}, $</p> <p>– at the $n_3\kern1pt|\kern1ptn_1$ interface:</p> <p>$ \sin\theta_{2C}=\frac{n_1}{n_3}. $</p> <p>We’re given</p> <p>$ \sin\theta_{1C}-\sin\theta_{2C}=\frac12 \quad\text{and}\quad \frac{n_2}{n_3}=\frac25. $</p> <p>Hence</p> <p>$ \frac{n_1}{n_2}-\frac{n_1}{n_3} =\;n_1\Bigl(\frac1{n_2}-\frac1{n_3}\Bigr) =\frac12, $</p> <p>and with $n_2=\tfrac25\,n_3$ this becomes</p> <p>$ n_1\Bigl(\frac1{n_2}-\frac1{n_3}\Bigr) =n_1\Bigl(\frac{5}{2n_3}-\frac1{n_3}\Bigr) =n_1\frac{3}{2n_3} =\frac12 \;\Longrightarrow\; n_3=3\,n_1,\quad n_2=\frac{2}{5}n_3=\frac{6}{5}n_1. $</p> <p>Therefore</p> <p>$ \sin\theta_{1C} =\frac{n_1}{n_2} =\frac{n_1}{\tfrac{6}{5}n_1} =\frac56, $</p> <p>so</p> <p>$ \theta_{1C}=\sin^{-1}\!\Bigl(\frac56\Bigr). $</p> <p>That corresponds to Option C.</p>