In a modified Young's double slit experiment, the two coherent...

The total phase difference $\Delta \phi$ at any point on the screen is the sum of the initial phase difference between the sources and the phase difference due to the path difference.

$\Delta \phi = \Delta \phi_{\text{initial}} + \dfrac{2\pi}{\lambda} \Delta x$

At the geometric center of the screen, the path difference from the two slits is zero ($\Delta x = 0$).

Given that the initial phase difference $\Delta \phi_{\text{initial}} = \pi$, the total phase difference at the center is:

$\Delta \phi = \pi + 0 = \pi$

A phase difference of an odd multiple of $\pi$ results in destructive interference. Therefore, a completely dark fringe (minimum) will be formed at the geometric center instead of the usual central bright fringe.

Answer: A completely dark fringe (minimum)