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8 January, 21:17

In a Young's two-slit experiment it is found that an nth-order maximum for a wavelength of 680.0 nm coincides with the (n+1) th maximum of light of wavelength 510.0nm. Determine n.

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  1. 9 January, 00:33
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    n = 3

    Explanation:

    We know that the nth-order maximum for the Young's two-slit experiment is given by the following expression:

    dsin (θ) = nλ

    If two maxima concides we must have the relation:

    nλ1 = mλ2 - - - eq1

    In this particular experiment

    m = n+1

    λ1 = 680 nm

    λ2 = 510 nm

    Replacing m in eq1 we have

    nλ1 = (n+1) λ2

    Solving for n we have:

    n = λ2 / (λ1-λ2)

    Therefore:

    n = 510 / (680-510) = 3
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