11. (16 points) As shown in Figure 2, given the parabola $y^{2}=2 p x$ $(p>0), A B$ is a chord passing through the focus $F$. If $A B$ makes an angle $\theta(0<\theta \leqslant \frac{\pi}{2})$ with the $x$-axis, find $\angle A O B$.
11. When $0<\theta<\frac{\pi}{2}$, the equation of $AB$ can be written as $y=\tan \theta\left(x-\frac{p}{2}\right)$, i.e.,
$$
x=\cot \theta \cdot y+\frac{p}{2} \text {. }
$$

This result also holds for $\theta=\frac{\pi}{2}$.
Substituting equation (1) into the parabola equation yields
$$
y^{2}-2 p \cot \theta \cdot y-p^{2}=0 \text {. }
$$

Then $y_{A}=\frac{p(\cos \theta+1)}{\sin \theta}, y_{B}=\frac{p(\cos \theta-1)}{\sin \theta}$.
Thus, $x_{A}=\cot \theta \cdot \frac{p(\cos \theta+1)}{\sin \theta}+\frac{p}{2}=\frac{p(1+\cos \theta)^{2}}{2 \sin ^{2} \theta}$,
$$
x_{B}=\frac{p(\cos \theta-1)^{2}}{2 \sin ^{2} \theta} \text {. }
$$

As shown in Figure 6, draw perpendiculars from points $A$ and $B$ to the $y$-axis, denoted as $AP$ and $BQ$ respectively. Then
$$
\begin{array}{l}
\tan \angle A O P=\frac{|A P|}{|O P|} \\
=\frac{x_{A}}{y_{A}}=\frac{1+\cos \theta}{2 \sin \theta}, \\
\tan \angle B O Q=\frac{x_{B}}{-y_{B}} \\
=\frac{1-\cos \theta}{2 \sin \theta} .
\end{array}
$$

Therefore, $\tan (\angle A O P+\angle B O Q)=\frac{4}{3 \sin \theta}$.
Hence, $\angle A O B=\pi-(\angle A O P+\angle B O Q)$
$$
=\pi-\arctan \frac{4}{3 \sin \theta} \text {. }
$$