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 isΔx=dtanθ=d×yD=d×d210d=d20=5λ20=λ4 Corresponding phase difference, ϕ=2πλΔx=π2 .  Now, maximum intensity in Young's double slit experiment,  Imax=I1+I2+2I1I2  I0=4I∵I1=I2=I∴I=I04 intensity at y is  Iy=I1+I2+2I1I2cosπ2=2I=I02