JEE Main 2026 — Ray Optics Question with Solution

From the geometry of the prism, the angle of the prism is $A=180^{\circ }-(45^{\circ }+\theta )$.
The incident ray is parallel to the base, so the angle of incidence $i$ at the first surface is equal to the base angle $45^{\circ }$. Thus, $i=45^{\circ }$.
Applying Snell's law at the first surface: $1\cdot \sin (45^{\circ })=\mu \sin (r_1)$.
Given $\mu =\sqrt{2}$, we have $\frac{1}{\sqrt{2}}=\sqrt{2}\sin (r_1)⟹\sin (r_1)=\frac{1}{2}⟹r_1=30^{\circ }$.
At the second surface, the emergent ray grazes the surface, so the angle of emergence $e=90^{\circ }$.
Applying Snell's law at the second surface: $\mu \sin (r_2)=1\cdot \sin (90^{\circ })$.
$\sqrt{2}\sin (r_2)=1⟹\sin (r_2)=\frac{1}{\sqrt{2}}⟹r_2=45^{\circ }$.
The relation between prism angle and internal angles is $A=r_1+r_2$.
$A=30^{\circ }+45^{\circ }=75^{\circ }$.
Substituting $A$ in the first equation: $75^{\circ }=180^{\circ }-(45^{\circ }+\theta )$.
$75^{\circ }=135^{\circ }-\theta ⟹\theta =135^{\circ }-75^{\circ }=60^{\circ }$.