The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s double-slit experiment is

(2)For maxima $\mathrm{\Delta x}=\mathrm{dsin}\mathrm{\theta }=\mathrm{n\lambda }$
$\Rightarrow 2\mathrm{\lambda sin}\mathrm{\theta }=\mathrm{n\lambda }\Rightarrow \sin \mathrm{\theta }=\frac{\mathrm{n}}{2}$
since value of sin$\mathrm{\theta }$ can not be greater 1. 
$\therefore$n = 0, 1, 2 
Therefore only five maximas can be obtained on both side of the screen.