3. Given the four vertices of a rectangle $A(0,0)$, $B(2,0)$, $C(2,1)$, $D(0,1)$. A particle starts from the midpoint $P_{0}$ of $A B$ and moves in a direction making an angle $\theta$ with $A B$ to hit point $P_{1}$ on $B C$, and then reflects sequentially to points $P_{2}$ on $C D$, $P_{3}$ on $D A$, and $P_{4}$ on $A B$ (the angle of incidence equals the angle of reflection). Let the coordinates of $P_{4}$ be $\left(x_{4}, 0\right)$. If $1<x_{4}<2$, then the range of $\tan \theta$ is ( ).
(A) $\left(\frac{1}{3}, 1\right)$
(B) $\left(\frac{1}{3}, \frac{2}{3}\right)$
(C) $\left(\frac{2}{5}, \frac{1}{2}\right)$
(D) $\left(\frac{2}{5}, \frac{2}{3}\right)$
3. C.

As shown in Figure 1, let $P_{1}(2, a)$ $(a>0), \angle P_{1} P_{0} B=\theta$. Then
$$
\begin{array}{l}
\tan \theta=\tan \angle P_{1} P_{0} B \\
=\frac{P_{1} B}{P_{0} B}=a .
\end{array}
$$

From $\angle P_{1} P_{2} C$
$$
\begin{array}{l}
=\angle P_{3} P_{2} D \\
=\angle P_{3} P_{4} A=\theta,
\end{array}
$$

we get $\tan \angle P_{1} P_{2} C=\frac{C P_{1}}{C P_{2}}=\frac{1-a}{C P_{2}}=a$.
Thus, $C P_{2}=\frac{1}{a}-1$.
Furthermore, $\tan \angle P_{3} P_{2} D=\frac{P_{3} D}{P_{2} D}=\frac{P_{3} D}{2-C P_{2}}=a$.
Solving this, we get $P_{3} D=3 a-1$.
Similarly, we can solve for $A P_{4}=\frac{2-3 a}{a}$.
Given that $1<A P_{4}<2$, therefore, $\frac{2}{5}<a<\frac{1}{2}$.