JEE Main 2025 Apr 2 Shift 2, Physics Q23: Prism

The prism formula at minimum deviation reads

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$.

With $A=60^{\circ }$ and $\mu =\sqrt{2}$:

$\sqrt{2}=\frac{\sin \left(30^{\circ }+\frac{{\delta }_m}{2}\right)}{\sin 30^{\circ }}\Rightarrow \sin \left(30^{\circ }+\frac{{\delta }_m}{2}\right)=\frac{\sqrt{2}}{2}=\sin 45^{\circ }$.

So $30^{\circ }+\frac{{\delta }_m}{2}=45^{\circ }\Rightarrow {\delta }_m=30^{\circ }$.

At minimum deviation $i=e$, and ${\delta }_m=2i-A$:

$i=\frac{A+{\delta }_m}{2}=\frac{60^{\circ }+30^{\circ }}{2}=45^{\circ }$.