JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Geometrical Optics

The incident ray runs parallel to the base and meets the left face, which is inclined at $45^{\circ }$ to the base; the angle between this ray and the normal to that face is $45^{\circ }$, so the angle of incidence is $i=45^{\circ }$. Snell's law at the first face gives $\sin 45^{\circ }=\sqrt{2}\sin r_1$, hence $\sin r_1=\frac{1}{2}$ and $r_1=30^{\circ }$. At the second face the ray just grazes, so the internal angle there is the critical angle: $\sin r_2=\frac{1}{\sqrt{2}}$, giving $r_2=45^{\circ }$. The refracting angle of the prism is $A=r_1+r_2=30^{\circ }+45^{\circ }=75^{\circ }$. The interior angles of the triangular section sum to $180^{\circ }$: $45^{\circ }+75^{\circ }+\theta =180^{\circ }$, so $\theta =60^{\circ }$.