11. (This question is worth 20 points) Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with the right focus at $F(c, 0)$, 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 of the ellipse.
Let the coordinates of the right focus $\mathrm{F}$ of the ellipse be $(c, 0)$. Clearly, 1 is not a horizontal line, so let the equation of the line $\mathrm{l}$ be $x = k y + c$, and the coordinates of points $\mathrm{A}$ and $\mathrm{B}$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, respectively. By combining the equation of the line $\mathrm{l}$ with the equation of the ellipse and eliminating $x$, we get $\left(b^{2} k^{2} + a^{2}\right) y^{2} + 24 k b^{2} c y + b^{2} \left(c^{2} - a^{2}\right) = 0$.
$$
\begin{array}{l}
\text{By Vieta's formulas, }\left\{\begin{array}{l}
y_{1} + y_{2} = -\frac{24 k b^{2} c}{b^{2} k^{2} + a^{2}}, \\
y_{1} y_{2} = \frac{b^{2} \left(c^{2} - a^{2}\right)}{b^{2} k^{2} + a^{2}} = -\frac{b^{4}}{b^{2} k^{2} + a^{2}} .
\end{array}\right. \\
\overrightarrow{O A} \cdot \overrightarrow{O B} = x_{1} x_{2} + y_{1} y_{2} = \left(k y_{1} + c\right)\left(k y_{2} + c\right) + y_{1} y_{2} = \left(k^{2} + 1\right) y_{1} y_{2} + k c \left(y_{1} + y_{2}\right) + c^{2} \\
= \left(k^{2} + 1\right) \left(-\frac{b^{4}}{b^{2} k^{2} + a^{2}}\right) + k c \left(-\frac{24 k b^{2} c}{b^{2} k^{2} + a^{2}}\right) + c^{2} = \frac{-k^{2} b^{2} + a^{2} c^{2} - b^{4}}{b^{2} k^{2} + a^{2}} . \text{5 points}
\end{array}
$$

Since $O A \perp O B$ is equivalent to $\overrightarrow{O A} \cdot \overrightarrow{O B} = 0$, it follows from the above that there exists a line $\mathrm{l}$ satisfying the condition if and only if there exists a real number $k$ such that $\frac{-k^{2} b^{2} + a^{2} c^{2} - b^{4}}{b^{2} k^{2} + a^{2}} = 0$, which implies $k^{2} = \frac{a^{2} c^{2} - b^{4}}{b^{2} \left(1 + c^{2}\right)}$. Clearly, the existence of $k$ satisfying (1) is equivalent to $a^{2} c^{2} - b^{4} \geq 0$. (2) 15 points
Furthermore, since $b^{2} = a^{2} - c^{2}$, (2) is equivalent to $a^{2} c^{2} - \left(a^{2} - c^{2}\right)^{2} \geq 0$. Dividing both sides by $a^{4}$, we get $\frac{c^{2}}{a^{2}} - \left(1 - \frac{c^{2}}{a^{2}}\right)^{2} \geq 0$, which simplifies to $e^{2} - \left(1 - e^{2}\right)^{2} \geq 0$.
Since $e < 1$, solving this inequality yields: $e \in \left[\frac{\sqrt{5} - 1}{2}, 1\right)$. 20 points