2. In a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, $E$ is the midpoint of $A B$, and $F$ is the midpoint of $C C_{1}$. Then the distance from $D$ to the plane passing through points $D_{1}, E, F$ is $\qquad$.
Consider the plane $D_{1} G E H F$, then $G E / / D_{1} F$
$$
\Rightarrow \triangle G A E \sim \triangle F C_{1} D_{1} \Rightarrow G A=\frac{1}{2} A E=\frac{1}{4} \text {, }
$$

Also, $D_{1} G / / F H \Rightarrow \triangle G A_{1} D_{1} \sim \triangle F C H$
$$
\Rightarrow \frac{C H}{F C}=\frac{A_{1} D_{1}}{G A_{1}}=\frac{4}{3} \Rightarrow C H=\frac{2}{3}, B H=\frac{1}{3} \text {. }
$$

Then $S_{\triangle D E H}=1-\frac{1}{4}-\frac{1}{3}-\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3}=\frac{1}{3}$,
Thus $E H \cdot D M=\frac{2}{3} \Rightarrow D M=\frac{\frac{2}{3}}{\sqrt{\frac{1}{4}+\frac{1}{9}}}=\frac{4}{\sqrt{13}}$,
Therefore $D D_{1} \cdot D M=D N \cdot D_{1} M \Rightarrow D N=\frac{\frac{4}{\sqrt{13}}}{\sqrt{1+\frac{16}{13}}}=\frac{4}{\sqrt{29}}$.
So the distance from $D$ to the plane passing through points $D_{1}, E, F$ is $\frac{4 \sqrt{29}}{29}$.