(a) A double slit experiment is performed with 589 nm light and slits-to-screen distance of 2.00 nm. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits.
(b) If the slit is replaced by a circular aperture with a diameter equal to the spacing of the double slit, find the width of the central maximum of the resulting pattern on the screen.© BrainMass Inc. brainmass.com October 24, 2018, 7:53 pm ad1c9bdddf
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Path difference between two rays = d sin @. For destructive ...
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