The maximum intensity in Young's double slit experiment is I0. Distance between the slits is d = 5λ, where  λ is the wavelength of monochromatic light used in the experiment. The intensity of light in front of one of the slits on a screen at a distance D = 10d is

Suppose P is a point in front of one slit at which intensity is I0. From figure, it is clear that y=d2. Path difference between the waves reaching at P isΔx=ydD=d2d10d=d20=5λ20=λ4Hence corresponding phase difference is given by         ϕ=2πλ×λ4=π2Resultant intensity at P is    I=Imaxcos2⁡ϕ2=I0cos2⁡π4=I02