4. As shown in Figure 1, in the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$, $E$ and $F$ are the midpoints of $A B$ and $A A_{1}$, respectively. The sine value of the plane angle formed by the dihedral angle between plane $C E B_{1}$ and plane $D_{1} F B_{1}$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) 1
4.C.

As shown in Figure 4, extend $CE$, $D_{1}F$, and $DA$. In the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$, since $E$ and $F$ are the midpoints of $AB$ and $AA_{1}$ respectively, it follows that $CE$, $D_{1}F$, and $DA$ intersect at a single point $G$. Connect $B_{1}G$. Given that the edge length of the cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ is 1, we have:
$$
B_{1}C=\sqrt{2}, B_{1}G=\sqrt{3}, CG=\sqrt{5} \text{, }
$$

which satisfies $B_{1}C^{2}+B_{1}G^{2}=CG^{2}$. Therefore, $B_{1}C \perp B_{1}G$.
Similarly, $B_{1}D_{1} \perp B_{1}G$.
Thus, $\angle CB_{1}D_{1}$ is the plane angle of the dihedral angle $C-B_{1}G-D_{1}$ formed by the planes $CEB_{1}$ and $D_{1}FB_{1}$.

Connect $CD_{1}$. In $\triangle B_{1}CD_{1}$, since $B_{1}C=B_{1}D_{1}=CD_{1}=\sqrt{2}$, it follows that $\angle CB_{1}D_{1}=60^{\circ}$.

Therefore, the sine value of the plane angle of the dihedral angle formed by the planes $CEB_{1}$ and $D_{1}FB_{1}$ is $\frac{\sqrt{3}}{2}$.