p36_063 - 63 For the fth maximum y5 = D sin 5 = D(5/d and...

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63. For the fifth maximum y 5 = D sin θ 5 = D (5 λ/d ), and for the seventh minimum y 0 7 = D sin θ 0 7 = D [(6 + 1 / 2) λ/d ]. Thus, y = y 0 7 y 5 = D · (6 + 1 / 2) λ d ¸ D µ 5 λ d = 3 λD 2 d = 3(546
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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