p37_032 - 32. (a) The first minimum of the diffraction...

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Unformatted text preview: 32. (a) The first minimum of the diffraction pattern is at 5.00◦ , so a= λ 0.440 µm = 5.05 µm . = sin θ sin 5.00◦ (b) Since the fourth bright fringe is missing, d = 4a = 4(5.05 µm) = 20.2 µm. (c) For the m = 1 bright fringe, α= πa sin θ π (5.05 µm) sin 1.25◦ = = 0.787 rad . λ 0.440 µm Consequently, the intensity of the m = 1 fringe is 2 I = Im sin α α 2 = (7.0 mW/cm ) sin 0.787 rad 0.787 2 2 = 5.7 mW/cm , 2 which agrees with Fig. 37-36. Similarly for m = 2, the intensity is I = 2.9 mW/cm , also in agreement with Fig. 37-36. ...
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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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