Q 12-09-039JEE MainJEE Main 2026 (8 Apr, Shift 2)Hard
Light ray incident along a vector $\overrightarrow{AO}$ ($\overrightarrow{AO} = 2\hat{i} - 3\hat{j}$) emerges out along vector $\overrightarrow{OB}$ ($\overrightarrow{OB} = C\hat{i} - 4\hat{j}$) as shown in the figure below. The value of $C$ is ______.
Answer: (A) $1.6$
The boundary is along the $x$-axis, so the normal is along $y$. The angle with the normal has $\sin = \dfrac{|x\text{-component}|}{\text{magnitude}}$:
$\sin\alpha = \dfrac{2}{\sqrt{13}}$, $\quad \sin\beta = \dfrac{C}{\sqrt{C^2 + 16}}$.
Snell's law, $1 \times \sin\alpha = 1.5\sin\beta$:
$$\frac{4}{13} = \frac{2.25C^2}{C^2 + 16} \;\Rightarrow\; 4C^2 + 64 = 29.25C^2 \;\Rightarrow\; C^2 = \frac{64}{25.25} \approx 2.53$$
$C \approx 1.6$.
Solution by Sreeraj P, M.Sc Physics