In Young's Double Slit Experiment, the condition for the $n^{\text{th}}$ minimum (dark fringe) is that the path difference between the waves from the two slits is given by $\Delta x = (2n - 1)\frac{\lambda}{2}$, where $n = 1, 2, 3, \dots$ For the fifth minimum, we substitute $n = 5$: $\Delta x = (2(5) - 1)\frac{\lambda}{2} = (10 - 1)\frac{\lambda}{2} = \frac{9\lambda}{2}$. Thus, the path difference for the fifth minimum is $\frac{9\lambda}{2}$.