JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Geometrical Optics

At minimum deviation the refracted ray inside the prism runs parallel to the base, so the two internal refraction angles are equal and each is $A/2$, giving $r=A/2$. At the coated exit face the ray meets the interface between the glass of index $n$ and the coating of index $n/2$; total internal reflection just sets in when the angle of incidence there equals the critical angle, $\sin {\theta }_c=\frac{n/2}{n}=\frac{1}{2}$, so ${\theta }_c=30^{\circ }$. The stated condition $r={\theta }_c$ then gives $A/2=30^{\circ }$. The prism angle is $60^{\circ }$.