problem35_25

Problem35_25 - R(0.900 m(5.50 10 7 m = = 3.81 10 3 m 4 d 1.30 10 m So the distance to the first minimum is one half this 1.91 mm 35.25:a)To the

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35.25: a) To the first maximum: m. 10 81 . 3 m 10 1.30 m) 10 (5.50 m) 900 . 0 ( λ 3 4 7 1 - - - × = × × = = d R y So the distance to the first minimum is one half this, 1.91 mm. b) The first maximum and minimum are where the waves have phase differences of zero and pi, respectively. Halfway between these points, the phase difference between the
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This note was uploaded on 01/31/2009 for the course PHYS 221 taught by Professor Hu during the Spring '08 term at Texas A&M.

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