2. A cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1 is cut by a plane passing through the body diagonal $A C_{1}$. Then the minimum value of the area of this section is $\qquad$ .
2. $\frac{\sqrt{6}}{2}$.

As shown in Figure 1, it is easy to see that the section $A F C_{1} G$ is a parallelogram. Draw $F E \perp A C_{1}$, with the foot of the perpendicular at $E$. Then the area of the section $S=A C_{1} \cdot E F$.

Since $A C_{1}=\sqrt{3}$ is a constant, the area $S$ is minimized when $F E$ is minimized. And when $F E$ is the common perpendicular of the skew lines $A C_{1}$ and $B B_{1}$, its length is the smallest, so $F$ is the midpoint of $B B_{1}$.
$$
\begin{array}{l}
\text { Therefore, } E F=\frac{\sqrt{2}}{2}, \\
S_{\min }=A C_{1} \cdot E F=\sqrt{3} \times \frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{2} .
\end{array}
$$