HW10_Solution - Phys 322 HW10 Solution 1 x F x F d for fringe separation where x F 2d x 2d x 2 Number of fringes =(length(separation x x So x x 2d

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1 Phys 322 HW10 Solution 1. 2 F x for fringe separation where x d .   d x x d x F F 2 2 . Number of fringes = (length) / (separation) x x / So,       304 10 00 . 5 / 10 618 . 7 2 2 / 7 5 m m d x x F . 2.   2 0 N d ;   x n x n vacuum air ;        97 10 6 / 1 . 0 1 00029 . 1 2 2 7 0 m m N 3.   4 1 2 cos F t m d . Let 0 t , 0 m , (minimum thickness).     nm n d 15 . 96 30 . 1 4 10 5 4 7 0 . 4. 2 sin 2 0 R E , 2 sin 2 N R E chord length;     2 sin 2 sin 0 N E E ,         2 sin 2 sin 2 sin 2 sin 2 2 0 2 2 2 0 2 N I N E E
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This note was uploaded on 12/07/2011 for the course PHYS 322 taught by Professor Na during the Fall '11 term at Purdue University-West Lafayette.

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