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1086_Physics ProblemsTechnical Physics

# 1086_Physics ProblemsTechnical Physics - Chapter 38(b 427 L...

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Chapter 38 427 (b) From Equation 38.4, I I max sin = L N M M O Q P P β β 2 2 2 b g where β π θ λ = 2 a sin . When θ = ° 15 0 . , β π = ° = 2 0 060 0 15 0 0 040 0 2 44 . sin . . . m m rad b g and I I max sin . . . = L N M O Q P = 1 22 1 22 0 593 2 rad rad a f . (c) sin θ λ = a so θ = ° 41 8 . : This is the minimum angle subtended by the two sources at the slit. Let α be the half angle between the sources, each a distance A = 0 100 . m from the center line and a distance L from the slit plane. Then, L = = ° F H G I K J = A cot . cot . . α 0 100 41 8 2 0 262 m m a f . FIG. P38.63(c) P38.64 I I max cos . cos . = ° ° = 1 2 45 0 45 0 1 8 2 2 e je j P38.65 d m sin θ λ = and, differentiating, d d md cos θ θ λ a f = or d m 1 2 sin θ θ λ d m d m 1 2 2 2 λ θ λ so θ λ λ d m 2 2 2 e j . *P38.66 (a) The angles of bright beams diffracted from the grating are given by d m a f sin θ λ = . The
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