What is the width of the central maximum on a screen 2.0 m behind the slit?

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A 0.50 mm-wide slit is illuminated by light of wavelength 500 nm.

  • Fornoob Team’s answer

    The halfwidth is the position of the 1st diffraction minimum, solved as follows:

    Θ = arcsin(m*lambda/(d)) = 1.00000016666674E-3 rad, 5.72957890623833E-2 deg

    y (exact) = D*tan(Θ) = 2.00000100000075E-3 m

    width = 2y = 4.0000020000015E-3 m

  • Monique

    w= (2*lambda*L)/a

    2(500×10^-9m)(2m) / .0005m

    w= .004m = 4mm

  • Joseph

    delta y=4.9mm

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