In a double-slit experiment, it is found that blue light of wavelength 460 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location?

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  1. For constructive interference
    d sinQ=ml=2x460nm=920nm
    For destructive interference of the other light, we have
    d sinQ=(m’+1/2)l
    When the two angle are equal, then
    l=1.84x103 nm for m’=0
    l=613 nm for m’=1
    l=368 nm for m’=2

    The only wavelength here that is visible is 613 nm

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