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Unformatted text preview: 13. (a) The intensity for a singleslit diﬀraction pattern is given by
I = Im sin2 α
α2 where α is described in the text (see Eq. 376). To locate the extrema, we set the derivative of I
with respect to α equal to zero and solve for α. The derivative is
sin α
dI
= 2Im 3 (α cos α − sin α) .
dα
α
The derivative vanishes if α = 0 but sin α = 0. This yields α = mπ , where m is a nonzero
integer. These are the intensity minima: I = 0 for α = mπ . The derivative also vanishes for
α cos α − sin α = 0. This condition can be written tan α = α. These implicitly locate the maxima.
(b) The values of α that satisfy tan α = α can be found by trial and error on a pocket calculator or
computer. Each of them is slightly less than one of the values (m + 1 )π rad, so we start with these
2
values. The ﬁrst few are 0, 4.4934, 7.7252, 10.9041, 14.0662, and 17.2207. They can also be found
graphically. As in the diagram below, we plot y = tan α and y = α on the same graph. The
intersections of the line with the tan α curves are the solutions. The ﬁrst two solutions listed above
are shown on the diagram.
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.. y = tan α y = tan α y=α 0 π/2 π 3π/2 α (rad) (c) We write α = (m + 1 )π for the maxima. For the central maximum, α = 0 and m = − 1 . For the
2
2
next, α = 4.4934 and m = 0.930. For the next, α = 7.7252 and m = 1.959. ...
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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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