8. The cross-sectional area of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ through the diagonal $B D_{1}$ is $S$, and $S_{1}$ and $S_{2}$ are the maximum and minimum values of $S$, respectively. Then $\frac{S_{1}}{S_{2}}$ is ( ).
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{2 \sqrt{3}}{3}$
(D) $\frac{2 \sqrt{6}}{3}$
8.C.
As shown in Figure 1, let the section through $B D_{1}$ be $\square B E D_{1} E_{1}$,
then $S=2 S_{\triangle B D_{1} E_{1}}=h$.
$B D_{1}$, where $h$ is the distance from $E_{1}$ to $B D_{1}$. It is easy to see that $h$ is minimized when it is the distance $h^{\prime}$ between the skew lines $B D_{1}$ and $B_{1} C_{1}$, at which point, $h=h^{\prime}=\frac{a}{\sqrt{2}}$.
($a$ is the edge length of the cube). Therefore,
$$
S_{2}=\frac{a}{\sqrt{2}} \cdot \sqrt{3} a=\frac{\sqrt{6}}{2} a^{2} .
$$

Also, $S_{1}=S_{\square B B_{1} D_{1} D}=\sqrt{2} a^{2}$, so, $\frac{S_{1}}{S_{2}}=\frac{2 \sqrt{3}}{3}$.