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Question

In an interference experiment, 3rd bright fringe is obtained on the screen with a light of 700 nm. What should be the wavelength of light source in order to obtain 5th bright fringe at the same point -

A
420 nm
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B
500 nm
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C
750 nm
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D
630 nm
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Solution

The correct option is A 420 nmGiven, λ1=700 nm As we known that y=ΔxDd For the bright fringe, Δx=nλ Here, Δx1=Δx2 n1λ1=n2λ2 3Dλ1d=5Dλ2d 3λ1=5λ2 ∴λ2=3λ15=3×7005=420 nm <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.

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