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problem36_19

problem36_19 - 5 4 = = m m within the first diffraction...

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36.19: a) , 3 If = a d then there are five fringes: . 2 , 1 , 0 ± ± = m b) The 6 = m interference fringe coincides with the second diffraction minimum, so there are two fringes
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Unformatted text preview: ) 5 , 4 ( + = + = m m within the first diffraction maximum on one side of the central maximum....
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