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Young's double slit Experiment

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


 Here, d=5λ,D=10d,y=d2

 Resultant Intensity at y=d2,Iy=?

 The path difference between two waves at y=d2 is


 Corresponding phase difference, ϕ=2πλΔx=π2 . 

 Now, maximum intensity in Young's double slit experiment, 

 Imax=I1+I2+2I1I2  I0=4II1=I2=II=I04 

intensity at y is  Iy=I1+I2+2I1I2cosπ2=2I=I02


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