In Young's double slit experiment, a beam of light which is a mixture of three wavelengths λ1,λ2 and λ2, is incident normally on the on the plane of the slits. If third maximum of λ1, fourth minimum of λ2 and second maximum of λ3 coincide on the screen, then λ1:λ2:λ3 is
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a
12 : 9 : 8
b
11 : 8 : 9
c
14 : 12 : 21
d
16 : 12 : 23
answer is C.
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Detailed Solution
3λ1Dd=3+12λ2Dd=2λ3Dd⇒λ2=6λ17 and λ3=3λ12∴ λ1:λ2:λ3=λ1:6λ17:3λ12=14:12:21