The maximum intensity in Young's double slit experiment is I0. Distance between the slits is d = 5 λ, where λ 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 infront 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=λ4Hence corresponding phase difference φ=2πλ×λ4=π2Resultant intensity at PI=Imaxcos2φ2=I0cos2π4=I02