Given data

• Prism is equilateral, so prism angle: $A = 60^\circ$
• Refractive index: $\mu = \sqrt{2}$

Condition for minimum deviation For a prism at minimum deviation, $i=e$ and $r_1=r_2=\frac{A}{2}$

Hence, $r = \frac{60^\circ}{2} = 30^\circ$

Apply Snell's law at first face At air-to-prism interface, $\mu = \frac{\sin i}{\sin r}$

So, $\sqrt{2} = \frac{\sin i}{\sin 30^\circ}$

Since, $\sin 30^\circ = \frac{1}{2}$

Therefore, $\sin i = \sqrt{2}\cdot \frac{1}{2} = \frac{\sqrt{2}}{2}$

Thus, $i = 45^\circ$

Find angle of emergence Under minimum deviation, $e=i$

Therefore, $e = 45^\circ$

Final answer The angle of emergence is: $\boxed{45}$