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
ϕ1 =`(2pi)/lambda` × λ = 2π
where ϕ1 is the phase difference when the path difference is λ and the corresonding frequency is I1=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 I2
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 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,
`"I"_1 = "I"_2`
∴ I' = `"I"_1 + "I"_1 + 2sqrt("I"_1"I"_1) cos phi`
= `2"I"_1 + 2"I"_1 cos phi`
Phase difference = `(2pi)/lambda xx "Path diffrence"`
Since path difference = λ,
Phase difference, `phi` = 2π
∴ I' = `2"I"_1 + 2"I"_1 = 4"I"_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"_1"I"_1) cos (2pi)/3`
= `2"I"_1 + 2"I"_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.
