(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 July 15, 2018, 6:57 pm ad1c9bdddf
Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here.
Path difference between two rays = d sin @. For destructive ...
Step by step solution is provided along with two diagrams.