15. As shown in Figure 6, in the right triangle $\triangle ABC$, $\angle C=90^{\circ}$, $AC=4$, $BC=3$, where $\odot O_{1}, \odot O_{2}, \cdots, \odot O_{n}$ are $n(n \geqslant 2)$ equal circles, $\odot O_{1}$ is externally tangent to $\odot O_{2}$, $\odot O_{2}$ is externally tangent to $\odot O_{3}$, $\cdots \cdots$, $\odot O_{n-1}$ is externally tangent to $\odot O_{n}$, $\odot O_{1}, \odot O_{2}, \cdots, \odot O_{n}$ are all tangent to $AB$, and $\odot O_{1}$ is tangent to $AC$, $\odot O_{n}$ is tangent to $BC$. Find the radius $r$ of these equal circles (expressed in terms of $n$).
15. As shown in Figure 12, connect $A O_{1}$ and extend it to intersect $B C$ at point $D$. Connect $B O_{n}$ and extend it to intersect $A D$ and $A C$ at points $O$ and $E$, respectively. Draw perpendiculars from points $O_{1}$ and $O_{n}$ to $A B$, and let $F$ and $G$ be the feet of these perpendiculars.

Since $\odot O_{1}$ and $\odot O_{n}$ are equal circles tangent to $A B$, and $O_{1} F \perp A B, O_{n} G \perp A B$, we have $O_{1} F = O_{n} G = r$.

Also, since $\odot O_{1}$ is tangent to $A B$ and $A C$, we have $\angle O_{1} A F = \angle D A C$. Therefore, $\frac{A C}{A B} = \frac{C D}{D B}$.
Given $A C = 4, B C = 3$, then $A B = 5$. Hence, $C D = \frac{4}{3}$.
Since $\triangle O_{1} A F \sim \triangle D A C$, we have
$\frac{O_{1} F}{C D} = \frac{A F}{A C}$, which means $\frac{r}{\frac{4}{3}} = \frac{A F}{4}$. Thus, $A F = 3 r$.
Similarly, $B G = 2 r$.
Clearly, $F G = 2(n-1) r$.
Therefore, $3 r + 2(n-1) r + 2 r = 5$.
Hence, $r = \frac{5}{2 n + 3}$.