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Unformatted text preview: 50. (a) We ﬁnd m from the last formula obtained in problem 49:
m= r2
1
(10 × 10−3 m)2
1
−=
−
−9 m)
Rλ 2
(5.0 m)(589 × 10
2 which (rounding down) yields m = 33. Since the ﬁrst bright fringe corresponds to m = 0, m = 33
corresponds to the thirtyfourth bright fringe.
(b) We now replace λ by λn = λ/nw . Thus,
mn = r2
1
nw r 2
1
(1.33)(10 × 10−3 m)2
1
−=
−=
− = 45 .
−9 m)
Rλn
2
Rλ
2
(5.0 m)(589 × 10
2 This corresponds to the fortysixth bright fringe (see remark at the end of our solution in part (a)). ...
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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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