JEE Main 2026 — Ray Optics Question with Solution

From the given figure, the interface is horizontal and the normal is along the y-axis.

The incident ray is along the vector $\overrightarrow{AO}=2\hat{i}-3\hat{j}$.
The angle of incidence $\alpha$ is the angle between the incident ray and the normal.
$\sin \alpha =\frac{2}{\sqrt{2^2+(-3{)}^2}}=\frac{2}{\sqrt{13}}$

The refracted ray is along the vector $\overrightarrow{OB}=C\hat{i}-4\hat{j}$.
The angle of refraction $\beta$ is the angle between the refracted ray and the normal.
$\sin \beta =\frac{C}{\sqrt{C^2+(-4{)}^2}}=\frac{C}{\sqrt{C^2+16}}$

Applying Snell's law at the interface:
${\mu }_1\sin \alpha ={\mu }_2\sin \beta$
$1\times \frac{2}{\sqrt{13}}=1.5\times \frac{C}{\sqrt{C^2+16}}$
$\frac{2}{\sqrt{13}}=\frac{3}{2}\frac{C}{\sqrt{C^2+16}}$

Squaring both sides:
$\frac{4}{13}=\frac{9}{4}\left(\frac{C^2}{C^2+16}\right)$
$16(C^2+16)=117C^2$
$16C^2+256=117C^2$
$101C^2=256$
$C^2=\frac{256}{101}$
$C=\frac{16}{\sqrt{101}}\approx 1.592$

Approximating $\sqrt{101}\approx 10$, we get $C\approx 1.6$.

Answer: $1.6$