JEE Main 2026 — Ray Optics Question with Solution

The refractive index of a medium is inversely proportional to the speed of light in that medium, given by $n=\frac{c}{v}$.
For light travelling from medium $A$ to medium $B$, the critical angle ${\theta }_c$ is defined when light travels from a denser to a rarer medium.
Given speeds are $v_A=2.4\times 10^8$ m/s and $v_B=2.7\times 10^8$ m/s.
The refractive indices are $n_A=\frac{c}{v_A}$ and $n_B=\frac{c}{v_B}$.
The critical angle ${\theta }_c$ is given by $\sin {\theta }_c=\frac{n_B}{n_A}=\frac{v_A}{v_B}$.
Substituting the values: $\sin {\theta }_c=\frac{2.4\times 10^8}{2.7\times 10^8}=\frac{24}{27}=\frac{8}{9}$.
To find the value in terms of ${\tan }^{-1}$, we use the trigonometric identity for a right-angled triangle where perpendicular $p=8$ and hypotenuse $h=9$.
The base $b=\sqrt{h^2-p^2}=\sqrt{9^2-8^2}=\sqrt{81-64}=\sqrt{17}$.
Thus, $\tan {\theta }_c=\frac{p}{b}=\frac{8}{\sqrt{17}}$.
Therefore, ${\theta }_c={\tan }^{-1}\left(\frac{8}{\sqrt{17}}\right)$.