Find the smallest distance OP such that the intensity at P is 50% of the intensity at O.
Wavelength is 600nm; Intensity of both sources is same (Io)
Let the intensity of the two sources, !!S_1!! and !!S_2!! be !!I_o!!.
Therefore, the maximum intensity in the screen will be:
!!I_(max) = (sqrt(I_1) + sqrt(I_2))\^2!!
!!I_(max) = 4I_o!!
If intensity at point P is !!50%!! of !!I_(max)!!, it means
!!I_P = 2I_o!!
Intensity at any point P on the screen is given by:
!!I_P = I_1 + I_2 + 2sqrt(I_1 * I_2) cos (Δφ)!!
where, !!Δφ!! is the phase difference between the two rays, !!S_1P!! and !!S_2P!!.
!!2I_o = I_o + I_o + 2sqrt(I_o * I_o) cos(Δφ)!!
!!2I_o = 2I_o + 2I_o cos(Δφ)!!
!!1 = 1 + cos(Δφ)!!
!!cos(Δφ) = 0!!
!!Δφ = π/2!!
(since we have to find the smallest distance)
We know that,
!!("Phase difference") = (2π)/λ * ("Path difference")!!
!!Δφ = (2π)/λ * (Δx)!!
where, !!Δx!! is the path difference.
!!Δx = λ/4!!
We know that in a YDSE set up,
!!Δx = (y * d)/D!!
where, !!y!! is the distance on the screen from the point O.
Subsituting the values of !!d!! and !!D!! and solving for !!y!!, we get:
!!y = 1.5mm!!
Hence, the smallest distance OP comes out to be !!1.5mm!!