Example 8 Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, O$ is the center of the base $A B C D$, and points $M, N$ are the midpoints of edges $C C_{1}, A_{1} D_{1}$, respectively. Then the volume of the tetrahedron $O-M N B_{1}$ is ( ).
(A) $\frac{1}{6}$
(B) $\frac{5}{48}$
(C) $\frac{1}{8}$
(D) $\frac{7}{48}$
Analysis: As shown in Figure 6, the four faces of the required pyramid are oblique sections of a cube, making it difficult to find the volume. However, using Basic Conclusion 14, the volume of the tetrahedron $O-M N B_{1}$ can be converted to a multiple of the volume of the triangular pyramid $C_{1}-M N B_{1}$. The key is to find the ratio of the distances from point $O$ and point $C_{1}$ to the plane $M N B_{1}$. As shown in Figure 7, connect $A_{1} C_{1}$ to intersect $B_{1} N$ at point $E$, and connect $E M$ and $O C_{1}$ to intersect at point $F$ (note that both lines are in the plane $A_{1} C$), then the problem is converted to finding the ratio $O F: C_{1} F$. Noting that in the rectangle $4 A_{1} C_{1} C$, $C C_{1}=1, A_{1} C_{1}=\sqrt{2}, A_{1} E$ : $E C_{1}=1: 2$, and points $M$ and $O$ are the midpoints of $C C_{1}$ and $A C$ respectively. By constructing a parallel geometric figure, it is easy to calculate $O F: C_{1} F=7: 4$ (using parallel geometric knowledge). Therefore, it is easy to get $V_{n-M N B_{1}}=$ $\frac{7}{4} V_{r_{1}-N B_{1}}=\frac{7}{48}$. Hence, the answer is (D).