4. In $\triangle A B C$, $C A=2, C B=6, \angle A C B=$ $60^{\circ}$. If point $O$ is on the bisector of $\angle A C B$, satisfying $\overrightarrow{O C}=m \overrightarrow{O A}+n \overrightarrow{O B}(m, n \in \mathbf{R})$,

and $-\frac{1}{4} \leqslant n \leqslant-\frac{1}{20}$,
then the range of $|\overrightarrow{O C}|$ is . $\qquad$
4. $\left[\frac{\sqrt{3}}{4}, \frac{3 \sqrt{3}}{4}\right]$.

As shown in Figure 4, with $C$ as the origin and the line $CB$ as the $x$-axis, establish a Cartesian coordinate system.
Then $B(6,0), A(1, \sqrt{3}), l_{c o}: y=\frac{\sqrt{3}}{3} x$.
Let $O\left(x, \frac{\sqrt{3}}{3} x\right)(x>0)$.
$$
\begin{array}{l}
\text { By } \overrightarrow{O C}=m \overrightarrow{O A}+n \overrightarrow{O B} \\
\Rightarrow\left(-x,-\frac{\sqrt{3}}{3} x\right) \\
=m\left(1-x, \sqrt{3}-\frac{\sqrt{3}}{3} x\right)+n\left(6-x,-\frac{\sqrt{3}}{3} x\right) \\
\Rightarrow\left\{\begin{array}{l}
-x=m(1-x)+n(6-x), \\
-\frac{\sqrt{3}}{3} x=m\left(\sqrt{3}-\frac{\sqrt{3}}{3} x\right)+n\left(-\frac{\sqrt{3}}{3} x\right)
\end{array}\right. \\
\Rightarrow n=\frac{x}{4 x-9} . \\
\text { By }-\frac{1}{4} \leqslant n \leqslant-\frac{1}{20} \\
\Rightarrow-\frac{1}{4} \leqslant \frac{x}{4 x-9} \leqslant-\frac{1}{20} \\
\Rightarrow \frac{3}{8} \leqslant x \leqslant \frac{9}{8} .
\end{array}
$$

Therefore, $|\overrightarrow{O C}|=\sqrt{1+\frac{1}{3}} x=\frac{2 \sqrt{3}}{3} x \in\left[\frac{\sqrt{3}}{4}, \frac{3 \sqrt{3}}{4}\right]$.