hw7_solution

# hw7_solution - Physics 3C Webassign Homework 7 Solutions:...

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Physics 3C Webassign Homework 7 Solutions: The small angle approximation does not work here. θ is not approximately sin θ , etc. We can use the formulas for a) At the m=2 maximum, tan θ = (400m/1000m) => θ = 21.8 degrees So λ = d sin θ b /m = 308 sin(21.8)/2 = 57.2 m b) Minimums happen when λ (m + ½ ) = d sin θ d And m=2 so sin θ d = (5/2) λ /d = 0.46 => θ d = 27.66 degrees So y = (1000m) tan(27.66) = 124m

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Location A = central maximum. B = first minimum. Difference in vertical distance is Minimums in intensity at d sin θ d = d (20/150) = λ (m + ½ ), where point B corresponds to m=0. Maximums in intensity at d sin θ b = d(20/150) = λ m, where point A corresponds to m=0 where d is the distance between the doors Now y_A = 0 so the total distance between doors is just d: d = (150m)(1.9m)(1/2)/20m = 7.12 m
Constructive interference 2nt = (m + ½ ) λ So reflected maxima are at λ = 2nt/(m+1/2) = 2(1.48)(278nm)/(m + ½ ) For m=0 => 1646 nm (infrared) For m =1 => 549 [visible] ** Only one we can see. For m =2 => 329nm [ultraviolet]

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## This note was uploaded on 09/08/2008 for the course PHYS 3B taught by Professor Wu during the Spring '08 term at UC Irvine.

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hw7_solution - Physics 3C Webassign Homework 7 Solutions:...

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