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

For maxima Δ=dsinθ=nλ⇒2λsinθ=nλ⇒sinθ=n2since value of sin θ can not be greater 1 .∴ n=0,1,2Therefore only five maximas can be obtained on both side of the screen.