JEE Main 2025 — Wave Optics Question with Solution

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

– at the $n_2|n_1$ interface:

$\sin {\theta }_{1C}=\frac{n_1}{n_2},$

– at the $n_3|n_1$ interface:

$\sin {\theta }_{2C}=\frac{n_1}{n_3}.$

We’re given

$\sin {\theta }_{1C}-\sin {\theta }_{2C}=\frac{1}{2}\text{and}\frac{n_2}{n_3}=\frac{2}{5}.$

Hence

$\frac{n_1}{n_2}-\frac{n_1}{n_3}=n_1(\frac{1}{n_2}-\frac{1}{n_3})=\frac{1}{2},$

and with $n_2=\frac{2}{5}{n}_3$ this becomes

$n_1(\frac{1}{n_2}-\frac{1}{n_3})=n_1(\frac{5}{2n_3}-\frac{1}{n_3})=n_1\frac{3}{2n_3}=\frac{1}{2}⟹n_3=3n_1,n_2=\frac{2}{5}{n}_3=\frac{6}{5}{n}_1.$

Therefore

$\sin {\theta }_{1C}=\frac{n_1}{n_2}=\frac{n_1}{\frac{6}{5}{n}_1}=\frac{5}{6},$

so

${\theta }_{1C}={\sin }^{-1}\left(\frac{5}{6}\right).$

That corresponds to Option C.