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problem36_25 - 36.25 a)Interference maxima Diffraction...

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36.25: a) Interference maxima: Diffraction minima: θ d i = sin and . λ sin n θ a d = If the th m interference maximum corresponds to the th n diffraction minimum then . d i θ θ = or n m a d = so mm. 0.280 mm) 840 . 0 ( 3 1 = = = d m n a b) The diffraction minima will squelch the interference maxima for all 3 = n m up to the highest seen order. For nm, 630 λ = the largest value of m will be when . 90 ° = θ . 1333 m 10 30 . 6 m 10 40 . 8 λ 7 4 max = × × = = - - d m . 444 m) 10 30 . 6 ( 3 m 10 40 . 8 λ 7 4 max = ×
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