10. Given that the area of the section through the body diagonal $B D_{1}$ of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is $S$, and the maximum and minimum values of $S$ are denoted as $S_{\text {max }}$ and $S_{\text {min }}$ respectively. Then $\frac{S_{\text {max }}}{S_{\text {min }}}=(\quad)$.
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{\sqrt{6}}{2}$
(C) $\frac{2 \sqrt{3}}{3}$
(D) $\frac{2 \sqrt{6}}{3}$
10. C.

Notice that the section is divided into two identical triangles by $B D_{1}$, that is, the ratio of heights is the ratio of areas with $B D_{1}$ as the base. The maximum height is $\frac{\sqrt{6}}{3}$, and the minimum height is $\frac{\sqrt{2}}{2}$. Therefore, $\frac{S_{\max }}{S_{\min }}=\frac{2 \sqrt{3}}{3}$.