problem35_28

problem35_28 - 35.28:The distance between maxima is l (6.56...

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35.28: The distance between maxima is cm. 0369 . 0 m) 10 2(8.00 cm) (9.00 m) 10 56 . 6 ( 2 λ 5 7 = × × = = - - h l x So the number of fringes per centimeter is
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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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