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