11. In the Cartesian coordinate system $x O y$, for the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the right focus is $F(c, 0)$. If there exists a line $l$ passing through point $F$ intersecting the ellipse at points $A, B$, such that $O A \perp O B$. Find the range of the eccentricity $e=\frac{c}{a}$ of the ellipse.
According to the problem, the slope of line $l$ is not 0. Let $l: x=k y+c$,
$A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, combining $\left\{\begin{array}{l}b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}, \\ x=k y+c\end{array}\right.$
$$
\Rightarrow\left(b^{2} k^{2}+a^{2}\right) y^{2}+2 b^{2} k c y-b^{4}=0,
$$

then $y_{1}+y_{2}=-\frac{2 b^{2} k c}{b^{2} k^{2}+a^{2}}, y_{1} y_{2}=-\frac{b^{4}}{b^{2} k^{2}+a^{2}}$.
Thus, $O A \perp O B \Rightarrow x_{1} x_{2}+y_{1} y_{2}=\left(k y_{1}+c\right)\left(k y_{2}+c\right)+y_{1} y_{2}=0$
$\Rightarrow\left(k^{2}+1\right) y_{1} y_{2}+k c\left(y_{1}+y_{2}\right)+c^{2}=0 \Rightarrow-\left(k^{2}+1\right) b^{4}-2 b^{2} k^{2} c^{2}+c^{2}\left(b^{2} k^{2}+a^{2}\right)=0$
$\Rightarrow a^{2} c^{2}-b^{4}=b^{2} k^{2}\left(b^{2}+c^{2}\right)=a^{2} b^{2} k^{2} \Rightarrow k^{2}=\frac{a^{2} c^{2}-b^{4}}{a^{2} b^{2}} \geqslant 0$,
thus $a^{2} c^{2}-b^{4}=a^{2} c^{2}-\left(a^{2}-c^{2}\right)^{2} \geqslant 0 \Rightarrow a^{4}-3 a^{2} c^{2}+c^{4} \leqslant 0 \Rightarrow e^{4}-3 e^{2}+1 \leqslant 0$ $\Rightarrow \frac{3-\sqrt{5}}{2} \leqslant e^{2} \leqslant \frac{3+\sqrt{5}}{2} \Rightarrow \frac{3-\sqrt{5}}{2} \leqslant e^{2}<1 \Rightarrow e \in\left[\frac{\sqrt{5}-1}{2}, 1\right)$.
Therefore, the range of the eccentricity of the ellipse is $\left[\frac{\sqrt{5}-1}{2}, 1\right)$.