In young double slit experiment, The light incident on the slits contains two lights having wavelengths λ1 and λ2. If 4th maxima of λ1 coincides with third minima of λ2 and β1 and β2 be the fringe widths of λ1 and λ2 respectively, Then
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a
β1=β2
b
β1>β2
c
β1<β2
d
Informations are not sufficient to find the relation between β1 and β2
answer is C.
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Detailed Solution
4λ1Dd=2+12λ2Dd⇒λ1λ2=58Since β=λDd and λ1<λ2,we can conclude, β2>β1