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29 July, 22:46

Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00~/mu/text{m}3.00 μm.

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  1. 29 July, 23:48
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    The condition for a minimum of the intensity of light falling on double slits is given by:

    d · sinα = (n + 1/2) λ

    Therefore, we can solve for λ:

    λ = (d · sinα) / (n + 1/2)

    = (3 · sin30) / (3 + 1/2)

    = 0.429 μm

    Therefore, the light has a wavelength of 429nm.
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