problem36_11

problem36_11 - x(3.00 m(6.33 10 7 m 5.43 10 3 m So the...

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36.11: a) m. 10 5.43 m 10 3.50 m) 10 (6.33 m) (3.00 λ 3 4 7 1 - - - × = × × = = a x y So the width of the brightest fringe is twice this distance to the first minimum, 0.0109 m. b) The next dark fringe is at m 0.0109 m 10 3.50 m) 10 6.33 ( m) 2(3.00
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