# Finding intensity in YDSE set up

**Question**

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)

**Solution**

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!!.

Therefore,

!!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")!!

or,

!!Δφ = (2π)/λ * (Δx)!!

where, !!Δx!! is the path difference.

Therefore,

!!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!!