(4) The focus of the parabola $y^{2}=4 x$ is $F$, points $A$ and $B$ are on the parabola, and $\angle A F B=$ $\frac{2}{3} \pi$. The midpoint of chord $A B$ is $M$, and its projection on the directrix $l$ is $M^{\prime}$. Then the maximum value of $\frac{\left|M M^{\prime}\right|}{|A B|}$ is ( ).
(A) $\frac{4 \sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{2 \sqrt{3}}{3}$
(D) $\sqrt{3}$
4 Let the projections of $A$ and $B$ on the directrix $l$ be $A'$ and $B'$, respectively, then
$$
\left|M M^{\prime}\right|=\frac{\left|A A^{\prime}\right|+\left|B B^{\prime}\right|}{2} .
$$

Since
$$
\left|A A^{\prime}\right|=|A F|,\left|B B^{\prime}\right|=|B F|,
$$

we have
$$
\left|M M^{\prime}\right|=\frac{|A F|+|B F|}{2} .
$$

Given that $\angle A F B=\frac{2}{3} \pi$, we have
$$
\begin{aligned}
|A B|^{2} & =|F A|^{2}+|F B|^{2}-2|F A| \cdot|F B| \cos \angle A F B \\
& =|F A|^{2}+|F B|^{2}+|F A| \cdot|F B|,
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
|A B|^{2}+\left|M M^{\prime}\right|^{2} & =|A B|^{2}+\left(\frac{|A F|+|B F|}{2}\right)^{2} \\
& \geqslant|A B|^{2}+|F A| \cdot|F B| \\
& =(|F A|+|F B|)^{2} \\
& =4\left|M M^{\prime}\right|^{2},
\end{aligned}
$$

Thus, $|A B|^{2} \geqslant 3\left|M M^{\prime}\right|^{2}$, which implies $\frac{\left|M M^{\prime}\right|}{|A B|} \leqslant \frac{\sqrt{3}}{3}$. Therefore, the correct choice is B.