For the central bright fringe, the path difference is 0.
S2O
- S1O = 0.
For the first bright fringe, S2Q
- S1Q = l
Notice that in the triangle S1S2Z, 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.
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 width.
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