For the central bright fringe, the two long sides of the triangle are the same length; the path difference is 0. S2O - S1O = 0.
For the first bright fringe, one side of the triangle is a one wavelength longer; S2Q - S1Q = l
Notice that in the triangle S1S2Z, that S2Z is l
We know that S1S2 is s. Therefore sin q = l/s
For the triangle OPQ, tan f = w/D. Although in this diagram, it is clear that q ¹ f , in the real thing, we can assume that q = f , as the real set up is very much longer. Also, we know that for small angles sin q = tan q
Therefore:
l = w
s D
=> l = ws
D
To produce easily measurable fringes, D must be large i.e. in metres while s is small (<1 mm). The larger D is the less bright the fringes. We can increase the width of the fringes by increasing the wavelength, or by decreasing the slit spacing, s.
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