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Question

Obtain the expression for fringe width in the case of interference of light waves.

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Solution

Light rays coming from slits S1 and S2 interfere at point P.
From figure, S1P=(yd2)2+D2
Or S1P=y2+d24yd+D2
Or S1P=Dy2D2+d24DydD2+1
As D>>d and y<<D, we neglect d2/4D and y2/D2.
We get
Or S1P=D1ydD2
Or S1P=D[1yd2D2]
Or S1P=Dyd2D
Similarly we get S2P=D+yd2D
Path difference between two light rays Δx=S2PS1P
Or Δx=(D+yd2D)(Dyd2D)=ydD
For constructive interference, Δx=nλ
ydD=nλ
y=nλDd
Fringe width β=ynyn1
β=nλDd(n1)λDd
β=λDd

658749_623669_ans_c575e738fe3940f69134ad8c5db0e28b.png

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