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Question

In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength λ. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.

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Solution

Given:
Separation between the two slits = d
Wavelength of the light = λ
Distance of the screen = D
(a) When the intensity is half the maximum:
Let Imax be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So, I=a2+a2+2a2cosϕ
Here, ϕ is the phase difference in the waves coming from the two slits.
So, I=4a2cos2ϕ2
IImax=124a2cos2ϕ24a2=12cos2ϕ2=12cosϕ2=12ϕ2=π4ϕ=π2Corrosponding path difference, x=λ4y=xDd=λD4d

(b) When the intensity is one-fourth of the maximum:

IImax=144a2cos2ϕ2=14cos2 ϕ2=14cosϕ2=12ϕ2=π3So, corrosponding path difference, x=λ3and position, y=xDd=λD3d .

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