4. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be two points on the ellipse $\frac{y^{2}}{a^{2}}+$ $\frac{x^{2}}{b^{2}}=1(a>b>0)$, $m=\left(\frac{x_{1}}{b}, \frac{y_{1}}{a}\right)$, $n$ $=\left(\frac{x_{2}}{b}, \frac{y_{2}}{a}\right)$, and $\boldsymbol{m} \cdot \boldsymbol{n}=0$. The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$, the length of the minor axis is $2$, and $O$ is the origin. Then the area of $\triangle A O B$ is - $\qquad$
4.1.

Given $e^{2}=\frac{c^{2}}{a^{2}}=\frac{3}{4}, b=1$, then $a^{2}=b^{2}+c^{2}=4$.
Also, $\boldsymbol{m} \cdot \boldsymbol{n}=0 \Rightarrow y_{1} y_{2}=-4 x_{1} x_{2}$.
Substituting the coordinates of points $A$ and $B$ into the ellipse equation, we get
$$
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{4}+x_{1}^{2}=1 \\
\frac{y_{2}^{2}}{4}+x_{2}^{2}=1 .
\end{array}\right.
$$
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2}|O A| \cdot|O B| \sin \angle A O B \\
=\frac{1}{2}|O A| \cdot|O B| \sqrt{1-\cos ^{2} \angle A O B} \\
=\frac{1}{2}|O A| \cdot|O B| \sqrt{1-\left(\frac{O A \cdot O B}{|O A| \cdot|O B|}\right)^{2}} \\
=\frac{1}{2} \sqrt{|O A|^{2} \cdot|O B|^{2}-(O A \cdot O B)^{2}} \\
=\frac{1}{2} \sqrt{\left(x_{1}^{2}+y_{1}^{2}\right)\left(x_{2}^{2}+y_{2}^{2}\right)-\left(x_{1} x_{2}+y_{1} y_{2}\right)^{2}} .
\end{array}
$$

Substituting equations (1), (2), and (3) into the above equation, we get
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{1}{2} \sqrt{\left(4-3 x_{1}^{2}\right)\left(4-3 x_{2}^{2}\right)-\left(-3 x_{1} x_{2}\right)^{2}} \\
=\sqrt{4-3\left(x_{1}^{2}+x_{2}^{2}\right)} .
\end{array}
$$

From (1) $)^{2}$, we easily get $y_{1}^{2} y_{2}^{2}=16 x_{1}^{2} x_{2}^{2}$, i.e.,
$$
16\left(1-x_{1}^{2}\right)\left(1-x_{2}^{2}\right)=16 x_{1}^{2} x_{2}^{2} \text {. }
$$

Thus, $x_{1}^{2}+x_{2}^{2}=1$.
Therefore, $S_{\triangle A O B}=1$.