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Derive the expression for the fringewidth ofinterferencepatternin Young'sdouble-slitexperiment . |
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Answer» <P> Solution :Let A and B be twoslitsseparatedby a distance.d.. Let .`lambda`. be the wavelengthof light .Let .D.be the distance betweenthe screenand the double slit.Let .C. be a pointon BP such that `AP~~CP`.Path differencebetweenthe two wavesreaching.P. is GIVEN by BP-AP=BC=d From the D BFP , `BP^2 =BF^2 +FP^2`and from the `D AEP , AP^2 = AE^2 +EP^2` `BP^2 - AP^2=(D^2 + FP^2) -(D^2 - EP^2) ` i.e., `BP^2 -AP^2 =FP^2 -EP^2 ` `=(x+d/2)^2 - (x-d/2)^2 ` (BP-AP)(BP+AP)=`2(2x.d/2)` For a point .P. close to .O., BP > > AP=D (BP-AP)(2D)=2x.d But (BP-AP)=BC=d i.e.,`d=(xd)/D` or `x=(DELTAD)/d` For a constructiveinterference`d=nlambda` Distanceof `n^(th)`brightfringefrom the centralbrightfringe`x_n=n(lambdaD)/d` Distanceof `(n+1)^(th)`brightfringe from thecentralbrightfringe`x_(n+1)=(n+1)(lambdaD)/d` By definition,Fringewidthis the distancebetweentwo consecutivebrightor dark fringes . `x_(n+1)-x_(n)=b=lambda/d(n+1-n)` i.e,`beta=(lambdaD)/d` Fringewidth `beta prop D, beta prop lambda` and `beta prop 1/d`. We can show that the fringewidthbetweenany twodarkfringes is ALSO`beta=(lambdaD)/d` . |
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