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?

#### Solution 1

Phase difference = `(2pi)/lambda` × Path difference

ϕ_{1} =`(2pi)/lambda` × λ = 2π

where ϕ_{1} is the phase difference when the path difference is λ and the corresonding frequency is I_{1}=K

`phi_2=(2pi)/lambdaxxlambda/3=(2pi)/3`

where ϕ2 is the phase difference when the path difference is the `lambda/3` and the corresponding frequency is I_{2}_{ }

Using equation (iv), we get:

`I_1/I_2=(4a^2cos^2(phi_1/2))/(4a^2cos^2(phi_2/2))=(cos^2((2pi)/2))/cos^2(((2pi)/3)/2)=(cos^2(ph))/cos^2(pi/3)=1/(1/(2^2))=4`

`K/I_2=4`

`I_2=K/4`

#### Solution 2

Let *I*_{1} and *I*_{2} be the intensity of the two light waves. Their resultant intensities can be obtained as:

`I' = I_1 +` I_2 + 2sqrt(I_1I_2) cos phi`

Where,

`phi` = Phase difference between the two waves

For monochromatic light waves,

`I_1 = I_2`

`:. 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 = 2pi`

`:. I' = 2I_1 + 2I_1 = 4I_1`

Given

*I*’ = *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) = I_1`

Using equation (1), we can write:

`I_R = I_1 = K/4`

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