JEE Main 2026 Apr 5 Shift 1, Physics Q15: Prism

Refractive index: $\mu =\frac{c}{v}=\frac{3\times 10^8}{2.12\times 10^8}\approx \sqrt{2}$.

For an equilateral prism, $A=60^{\circ }$. The minimum-deviation relation $\mu =\frac{\sin \left(\frac{A+{\delta }_{\min }}{2}\right)}{\sin (A/2)}$ gives

$\sqrt{2}=\frac{\sin \left(\frac{60^{\circ }+{\delta }_{\min }}{2}\right)}{\sin 30^{\circ }}$, so $\sin \left(\frac{60^{\circ }+{\delta }_{\min }}{2}\right)=\frac{1}{\sqrt{2}}$.

Therefore $\frac{60^{\circ }+{\delta }_{\min }}{2}=45^{\circ }\Rightarrow {\delta }_{\min }=30^{\circ }$.