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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. 3736. Similarly for m = 2, the intensity is I = 2.9 mW/cm , also in
agreement with Fig. 3736. ...
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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.
 Fall '08
 SPRUNGER
 Physics

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