13. Given the curve $x y^{2}=1(y>0)$, the slope $k$ of the tangent line at the point $\left(x_{0}, y_{0}\right)$ on the curve is $k=-\frac{1}{2 \sqrt{x_{0}^{3}}}$. A line parallel to the $y$-axis is drawn through the point $P_{1}(1,0)$, intersecting the curve at $Q_{1}$. A tangent line to the curve at $Q_{1}$ intersects the $x$-axis at point $P_{2}$. A line parallel to the $y$-axis is drawn through $P_{2}$, intersecting the curve at $Q_{2}$. This process is repeated (as shown in the figure), generating the sequences of points $P_{1}, P_{2}, \cdots$ and $Q_{1}, Q_{2}, \cdots$. Let $l_{n}=\left|P_{n} Q_{n}\right|$, find the value of $l_{2003}$.
13. Let $Q_{n}\left(x_{n}, y_{n}\right)$, then the slope $k$ of the line $Q_{n} P_{n+1}$ is $k=-\frac{1}{2 \sqrt{x_{n}^{3}}}$, so the equation of $Q_{n} P_{n+1}$ is $y-y_{n}=-\frac{1}{2 \sqrt{x_{n}^{3}}}\left(x-x_{n}\right)$. Setting $y=0$ gives $\quad x_{n+1}=x_{n}+2 y_{n} \sqrt{x_{n}^{3}}$. Since $\quad x_{n} y_{n}^{2}=1, x_{n+1} \cdot y_{n+1}^{2}=1$, we have $y_{n+1}=\frac{\sqrt{3}}{3} y_{n}$ and $y_{1}=1$, so $\left\{l_{n}\right\}$ is a geometric sequence with the first term 1 and common ratio $\frac{\sqrt{3}}{3}$. Therefore, $l_{2003}=\left(\frac{\sqrt{3}}{3}\right)^{2002}=\frac{1}{3^{1001}}$.