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प्रश्न
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`?
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उत्तर १
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`
उत्तर २
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:
IR = I1 = `K/4`
Hence, the intensity of light at a point where the path difference is `pi/3` is `K/4` units.
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