6. Given that the line $l$ passes through the left focus $F$ of the ellipse $C: \frac{x^{2}}{2}+y^{2}=1$ and intersects the ellipse $C$ at points $A$ and $B$, and $O$ is the origin. If $O A \perp O B$, then the distance from point $O$ to the line $A B$ is $\qquad$
6. $\frac{\sqrt{6}}{3}$.

It is known that $F(-1,0)$.
Let $l_{A B}: x=t y-1$.
From $\left\{\begin{array}{l}x=t y-1, \\ \frac{x^{2}}{2}+y^{2}=1\end{array}\right.$
$\Rightarrow\left(t^{2}+2\right) y^{2}-2 t y-1=0$.
Note that, the discriminant of equation (1) is greater than 0.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
Then $y_{1}+y_{2}=\frac{2 t}{t^{2}+2}, y_{1} y_{2}=\frac{-1}{t^{2}+2}$.
By $O A \perp O B$, we get
$$
\begin{array}{l}
x_{1} x_{2}+y_{1} y_{2}=\left(t y_{1}-1\right)\left(t y_{2}-1\right)+y_{1} y_{2} \\
=\left(t^{2}+1\right) y_{1} y_{2}-t\left(y_{1}+y_{2}\right)+1 \\
=\frac{-\left(t^{2}+1\right)}{t^{2}+2}-t \cdot \frac{2 t}{t^{2}+2}+1=0 \\
\Rightarrow-\left(t^{2}+1\right)-2 t^{2}+t^{2}+2=0 \\
\Rightarrow t^{2}=\frac{1}{2} .
\end{array}
$$

Therefore, the distance from point $O$ to line $A B$ is
$$
\frac{1}{\sqrt{1+t^{2}}}=\frac{\sqrt{6}}{3} \text {. }
$$