The position of the point on the screen exactly opposite to one of the slits is at a distance y=2d from the central maximum.
The path difference Δx at this point is given by:
Δx=Dyd
Substituting y=2d and D=10d:
Δx=10d(2d)d=20dd2=20d
Given that the distance between the slits is d=5λ, we have:
Δx=205λ=4λ
The corresponding phase difference Δϕ is:
Δϕ=λ2πΔx=λ2π(4λ)=2π
The intensity at this point is given by:
I=I0cos2(2Δϕ)
Substituting Δϕ=2π:
I=I0cos2(4π)=I0(21)2=2I0
Answer: 2I0