4. The base of a right quadrilateral prism is a rhombus with side length 2 and an angle of $30^{\circ}$, and the height is 1. A section is made through the base edge at an angle of $60^{\circ}$ to the base. The area of this section is $\qquad$ .
4. $\frac{4}{\sqrt{3}}$.

As shown in the figure, let the section be $A B E F$. Draw $F G \perp A D$, then $F G$ is the perpendicular line to the plane $A B C D$. Draw $F H \perp A B$, and connect $H G$, then $F G \perp A B$. According to the problem, $\angle F H G$ is the plane angle of the dihedral angle. In the right triangle $\triangle F G H$, $F G=1, \angle F H G=60^{\circ}, F H=\frac{F G}{\sin 60^{\circ}}=\frac{2}{\sqrt{3}}$, thus the area $S=A B \cdot F H=\frac{4}{\sqrt{3}}=\frac{4 \sqrt{3}}{3}$.