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

Suppose P is a point in front of one slit at which intensity is to be calculated. From figure it is clear that  x=d2. Path difference between the waves reaching at P  Δ=xdD=d2d10d=d20=5λ20=λ4   Hence corresponding phase difference   φ=2πλ×λ4=π2Resultant intensity at PI=Imaxcos2φ2=I0cos2π4=I02