A ray of light suffers a minimum deviation when incident on an equilateral prism of refractive
index \( \sqrt{2} \). The angle of incidence is

Angle of deviation \(\sigma=i_{1}+i_{2}-A\)
where \(\mathrm{A}\) is angle of prism \(A=r_{1}+r_{2}\)
\(\Rightarrow \delta=i_{1}+i_{2}-\left(r_{1}+r_{2}\right)\)
At minimum deviation \(r_{1}=r_{2}\) and \(i_{1}=i_{2}\)
\(\Rightarrow \delta=2 i_{1}-2 r_{1}\)
and \(A=2 r_{1}\)
For equilateral prism, \(\mathrm{A}=60^{\circ}\)
\(\Rightarrow 2 r_{1}=60^{\circ} \Rightarrow r_{1}=30^{\circ}\)
Now we know
\(\sin i_{1}=\mu \sin r_{1} \Rightarrow i_{1}=\sin ^{-1}\left(\mu \sin r_{1}\right)\)
\(i_{1}=\sin ^{-1}\left(\sqrt{2} \sin 30^{\circ}\right)=\sin ^{-1}\left(\sqrt{2} \times \frac{1}{2}\right)=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
\(\Rightarrow i_{1}=45^{\circ}\)