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Question
The intensity at the central maximum (O) in a Young’s double slit experimental set-up shown in the figure is IO. If the distance OP equals one-third of the fringe width of the pattern, show that the intensity at point P, would equal `(I_0)/4`.
Solution
Intensity at the central maxima = I0
Distance OP is one-third of the fringe width, ` x = 1/3 beta ` (given)
Fringe width is given by`(lambdaD)/d`
`therefore x= (lambda D)/(3d)`
Path difference , ` DeltaP = (xd)/D` and ,
Phase difference, `phi = (2 pi) / lambda ( Delta P)`
`= 2 pi /3`
Intensity at point P , I = ` I _0 cos^2 phi/2`
` = I_0 cos^2 ((2pi)/3) = I_0/4`
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