Step 1: The minimum deviation formula.
For a prism of angle $A$, the refractive index links to the minimum deviation $D$ by \[ \mu = \frac{\sin\!\left(\frac{A+D}{2}\right)}{\sin\!\left(\frac{A}{2}\right)}. \]
Step 2: Use the equilateral prism angle.
Both prisms are equilateral, so $A = 60^\circ$ and $\frac{A}{2} = 30^\circ$, where $\sin 30^\circ = 0.5$.
Step 3: First prism, $\mu = \sqrt{2}$.
\[ \sqrt{2} = \frac{\sin(30^\circ + D/2)}{0.5} \;\Rightarrow\; \sin(30^\circ + D/2) = \frac{\sqrt{2}}{2} = \sin 45^\circ. \] So $30^\circ + D/2 = 45^\circ$, giving $D = 30^\circ$.
Step 4: Second prism, $\mu = \sqrt{3}$.
\[ \sqrt{3} = \frac{\sin(30^\circ + D'/2)}{0.5} \;\Rightarrow\; \sin(30^\circ + D'/2) = \frac{\sqrt{3}}{2} = \sin 60^\circ. \] So $30^\circ + D'/2 = 60^\circ$, giving $D' = 60^\circ$.
Step 5: Compare the two deviations.
We found $D = 30^\circ$ and $D' = 60^\circ$, so $D' = 2D$.
Step 6: Conclusion.
The new minimum deviation is twice the first one. \[ \boxed{2D} \]