3. As shown in the figure, let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ have left and right foci $F_{1}, F_{2}$, respectively. A line through $F_{2}$ intersects the ellipse at points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. If the area of the incircle of $\triangle A F_{1} B$ is $\pi$, and $\left|y_{1}-y_{2}\right|=4$, then the eccentricity of the ellipse is $\qquad$
The perimeter of $\triangle A F_{1} B$ is $4a$, and the inradius $r=1$.

Thus, $S_{\triangle A F_{1} B}=2a \cdot r=c|y_{1}-y_{2}| \Rightarrow 2a=4c \Rightarrow e=\frac{c}{a}=\frac{1}{2}$.