Example 2 Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with its right focus at $F(c, 0)$, there exists a line $l$ passing through point $F$ that intersects the ellipse at points $A$ and $B$, such that $O A \perp O B$. Find the range of the eccentricity of the ellipse.
(2015, National High School Mathematics League (B Volume))
On the one hand, if $O A \perp O B$, by Proposition 1, the distance from point $O$ to line $l$ is $\frac{a b}{\sqrt{a^{2}+b^{2}}}$.

On the other hand, if the distance from point $O$ to line $l$ is $\frac{a b}{\sqrt{a^{2}+b^{2}}}$, draw a perpendicular from $O$ to $O A$, intersecting the ellipse at points $B_{1}$ and $B_{2}$, then
$$
\begin{array}{l}
\frac{1}{|O A|^{2}}+\frac{1}{\left|O B_{1}\right|^{2}}=\frac{1}{|O A|^{2}}+\frac{1}{\left|O B_{2}\right|^{2}} \\
=\frac{1}{a^{2}}+\frac{1}{b^{2}} .
\end{array}
$$

Thus, the distance from point $O$ to lines $A B_{1}$ and $A B_{2}$ is also $\frac{a b}{\sqrt{a^{2}+b^{2}}}$.

Since there are at most two lines passing through point $A$ with a fixed distance $\frac{a b}{\sqrt{a^{2}+b^{2}}}$ from $O$, point $B$ must coincide with one of $B_{1}$ or $B_{2}$. Therefore, $O A \perp O B$.
Hence, $O A \perp O B$
$\Leftrightarrow$ line $l$ is tangent to the circle $x^{2}+y^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Thus, there exists a line $l$ passing through point $F$ that intersects the ellipse at $A$ and $B$, such that
$$
\begin{array}{l}
O A \perp O B \\
\Leftrightarrow|O F|=c \geqslant \frac{a b}{\sqrt{a^{2}+b^{2}}} \\
\Leftrightarrow c^{2}\left(2 a^{2}-c^{2}\right) \geqslant a^{2}\left(a^{2}-c^{2}\right) \\
\Rightarrow \frac{3-\sqrt{5}}{2} \leqslant e^{2} \leqslant \frac{3+\sqrt{5}}{2} .
\end{array}
$$

Since $e<1$, we have $e \in\left[\frac{\sqrt{5}-1}{2}, 1\right)$, which is the desired result.