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Question

Two wavelengths λ1 and λ2 are used in bi-prism experiment. If λ1 is 430 nm, the value of the wavelength λ2 such that its fourth order bright fringe falls over the sixth order bright fringe of λ1 would be

A
645nm
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B
64.5nm
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C
6.45nm
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D
0.645nm
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Solution

The correct option is A 645nm
Fourth order bright fringe of λ2= sixth order bright fringe of λ1
4Dλ2d=6Dλ1d
4λ2=6λ1
λ2=64λ1
=64(430)
=645nm

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