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Question
Draw a neat labelled diagram of Young’s Double Slit experiment. Show that `beta = (lambdaD)/d` , where the terms have their usual meanings (either for bright or dark fringe).
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Solution
`PM_1 = x - d/2 and PM_2 = x + d/2`
In `DeltaS_2M_2P`
`(S_2P)^2 = (S_2M_2)^2 + (PM_2)^2`
`therefore (S_2P)^2 = D^2 + (x + d/2)^2` ....(i)
In `DeltaS_1M_1P`
`(S_1P)^2 = (S_1M_1)^2 + (PM_1)^2`
`therefore (S_1P)^2 = D^2 + (x - d/2)^2` ....(ii)
`(S_2P)^2 - (S_1P)^2 = D^2 + (x + d/2)^2 - D^2 - (x - d/2)^2`
`(S_2P - S_1P)(S_2P + S_1P) = x^2 + xd + d^2/4 - x^2 + xd - d^2/4`
`therefore S_2P - S_1P = (2xd)/(S_2P + S_1P)`
In practice, the distances x and d are very small as compared to D. To a first approximation we can write
`S_1P = S_2P = D or S_2P + S_1P = 2D`
`therefore` The path difference between two waves is given by
`S_2P - S_1P = (2xd)/(2D) = (xd)/D` .....(iii)
Now the intensity at P will be maximum or minimum according to the path difference. The point P will be bright if the path difference is an even multiple of `lambda/2` .
i.e. `S_2P - S_1P = (xd)/D = 2n(lambda/2)` ....(iv)
where n = 0,1,2,3, ...........
Or `x = n(lambdaD)/d` ....(v)
The Point P will be dark if the path difference is an odd multiple of
i.e. `S_2P - S_1P = (xd)/D = (2m - 1)lambda/2` ...(vi)
where m = 1,2,3, ...............
Or `x = (2m - 1) (lambdaD)/(2d)` ....(vii)
Expression for the band (fringe) width
The distance between the center of two adjcent bright or dark bands is called band width or fringe width.
Let `X_n` and `X_n+1` denotes the distances of `n^(th)` and `(n+1)^(th)` bright band on the same side of central
bright band, then from equation (v)
`X_n = n(lambdaD)/d`
`X_(n+1) = (n+1)(lambdaD)/d`
`X_(n+1) - X_n = (n+1-n)(lambdaD)/d`
`therefore beta = (lambdaD)/d` ...(viii)
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