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=nλÞ 2λsinθ=nλ Þ sinθ=n2  since value of sin q  can not be greater 1.\ n = 0, 1, 2Therefore only five maximas can be obtained on both side of the screen.