1. Given data

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

We need the angle of minimum deviation $\delta_m$.

2. Formula for minimum deviation in a prism

For a prism at minimum deviation, $\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

Substitute $A=60^\circ$: $\sqrt{2} = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{\sin 30^\circ}$

Since $\sin 30^\circ = \frac{1}{2}$ we get $\sqrt{2} = \frac{\sin\left(\frac{60^\circ+\delta_m}{2}\right)}{1/2}$

So, $\sin\left(\frac{60^\circ+\delta_m}{2}\right)=\frac{\sqrt{2}}{2}$

3. Solve the trigonometric equation

We know $\frac{\sqrt{2}}{2} = \sin 45^\circ$

Hence, $\frac{60^\circ+\delta_m}{2}=45^\circ$

Therefore, $60^\circ+\delta_m=90^\circ$ $\delta_m=30^\circ$

4. Final answer

The angle of minimum deviation is $\boxed{30^\circ}$