Answer
$\Delta \phi=0$
Work Step by Step
We know that the phase change due to the path difference is :
$\Delta \phi_1=(-\frac{100}{400})\times(2\pi)=-\frac{\pi}{2}$
As the initial phase difference is $\frac{\pi}{2}$, hence
$\Delta \phi=\Delta \phi_1+\Delta \phi_2$
We plug in the known values to obtain:
$\Delta \phi=\frac{\pi}{2}+({-\frac{\pi}{2}})=0$
