The maxima that is farthest from the slits is infinitely up or infinitely down the screen and corresponds to \(\theta=\frac{\pi}{2}\).
So, with the given condition, we have, \(n =\frac{ d \sin \left(90^{\circ}\right)}{\lambda}=\frac{ d }{\lambda}=2\)
Thus, we have two maxima on the screen on either side of the central maxima. Thus, the maximum number of possible maxima observed are \(2+1+2=5\).