Explanation

(b) The resultant intensity at any point P is
$I = 4I_{0}\cos^{2}\left( \frac{\phi}{2} \right)$
$\therefore\ \ I_{0} = 4I_{0}\cos^{2}{\phi/2}$
Or $\cos{\frac{\phi}{2} = \frac{1}{2}}$
$\therefore\ \frac{\phi}{2} = \frac{\pi}{3}\ or\ \phi = \frac{2\pi}{3}$
If $\Delta x$ is the corresponding value of path difference at P, then
$\phi = \frac{2\pi}{\lambda}(\Delta x)$
$\frac{2\pi}{3} = \frac{2\pi}{\lambda}\Delta x$.
As $\Delta x = \frac{xd}{D}$
$\therefore\ \frac{1}{3} = \frac{1}{\lambda}\frac{xd}{D}$
Or $x = \frac{\lambda}{3d/D} = \frac{6 \times 10^{- 7}}{3 \times 10^{- 4}} = 2 \times 10^{- 3}m$
$x = 2$mm
This is the difference of point P from central maximum.