JEE Main 2026 Apr 8 Shift 2, Physics Q17: Refraction at Plane and Spherical surfaces

The interface is horizontal (along $\hat{i}$), so the normal is along $\hat{j}$. The sine of the angle each ray makes with the normal equals the magnitude of the $\hat{i}$-component of the unit ray vector.

For $\overrightarrow{AO}=2\hat{i}-3\hat{j}$: $|\overrightarrow{AO}|=\sqrt{4+9}=\sqrt{13}$, so $\sin \alpha =\frac{2}{\sqrt{13}}$.

For $\overrightarrow{OB}=C\hat{i}-4\hat{j}$: $|\overrightarrow{OB}|=\sqrt{C^2+16}$, so $\sin \beta =\frac{C}{\sqrt{C^2+16}}$.

Snell's law: ${\mu }_1\sin \alpha ={\mu }_2\sin \beta$.

$(1)\frac{2}{\sqrt{13}}=(1.5)\frac{C}{\sqrt{C^2+16}}$

Squaring: $\frac{4}{13}=\frac{2.25C^2}{C^2+16}$

$4(C^2+16)=13(2.25C^2)=29.25C^2$

$4C^2+64=29.25C^2⟹25.25C^2=64⟹C^2=2.535⟹C\approx 1.59\approx 1.6$.