Question

In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is `λ/3`.

In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is `λ /3`?

Answer 1

Phase difference = `(2pi)/lambda xx "Path difference"`

ϕ₁ = `(2pi)/lambda xx lambda` = 2π 

Where ϕ₁ is the phase difference when the path difference is λ and the corresponding frequency is I₁ = K 

ϕ₂ = `(2pi)/lambdaxxlambda/3=(2pi)/3`

Where ϕ₂ is the phase difference when the path difference is the `lambda/3` and the corresponding frequency is I₂. 
Using equation, we get:

`I_1/I_2 = (4a^2cos^2(phi_1/2))/(4a^2cos^2(phi_2/2))`

`K/I_2 = (cos^2((2pi)/2))/cos^2(((2pi)/3)/2)`

`K/I_2 = (cos^2(pi))/cos^2(pi/3)`

`K/I_2 = 1/(1/(2^2))`

`K/I_2=4`

I₂ = `K/4`

Answer 2

Let I₁ and I₂ be the intensity of the two light waves. Their resultant intensities can be obtained as:

I' = `I_1 + I_2 + 2sqrt(I_1 I_2) cos phi`

Where,

`phi` = Phase difference between the two waves

For monochromatic light waves,

I₁ = I₂

∴ I' = `I_1 + I_1 + 2sqrt(I_1I_1) cos phi`

= `2I_1 + 2I_1 cos phi`

Phase difference = `(2pi)/lambda xx "Path diffrence"`

Since path difference = λ,

Phase difference, `phi` = 2π

∴ I' = `2I_1 + 2I_1 = 4I_1`

Given

4I₁ = K

∴ `I_1 = "K"/4` .....(1)

When path difference = `pi/3`

Phase difference, `phi = (2pi)/3`

Hence, resultant intensity, `I_R^' = I_1 + I_1 + 2sqrt(I_1I_1) cos  (2pi)/3`

= `2I_1 + 2I_1(-1/2)`

= I₁

Using equation (1), we can write:

I_R = I₁ = `K/4`

Hence, the intensity of light at a point where the path difference is `pi/3` is `K/4` units.