A beam of light of a single wavelength is incident perpendicularly on a double-s

Question

A beam of light of a single wavelength is incident perpendicularly on a double-slit arrangement, as in Fig. 35-10. The slit widths are each 36 µm and the slit separation is 0.27 mm. How many complete bright fringes appear between the two first-order minima of the diffraction pattern?
figure:
http://img585.imageshack.us/img585/3167/fig3510.gif

First I need to find lambda to use the equation:
a*sin(theta) = lambda*m
m = 2 because the problem stated: "the TWO first-order minima of the diffraction pattern"
... right? please help. @.@;

Explanation / Answer

For a single slit diffraction, here is that the limits of the central peak are the first minima in the diffraction pattern due to either slit individually the angular locations of those minima are given by
                                  a sin = m
here let us rewrite this equation as
                                   a sin = m1 1
with the subscript 1 referring to the one slit diffraction for the first minima in the diffraction pattern we substitute m1 = 1
for a double slit interference, that the angular locations of the bright fringes of the double slit interference pattern are given by
                                  d sin = m
here let us rewrite this equation as
                                  d sin = m2 2 form m2 = 0, 1, 2, ........
with the subscript 2 referring to the double-slit interference

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we locate the first diffraction minimum with in the double slit fringe pattern by dividing the above two equations and
solving for m2
so that we get
                       m2 = d / a
                            = 0.27 mm / 36 µm

                            = 0.27*10-3 m / 36*10-6 m
= 7.5
taking the interference maximum m = 0 as well as five five interference maxima on each side of it we get the total fringes as
                       n = 7.5 + 1 + 7.5
                          = 16 bright fringes

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