In the case of two
slits, the application of Rayleigh's criterion that the maximum of
one peak should lie on or below the first minimum of the other
gives a dip in intensity between the peaks to 8/
of the
maximum. This follows because the separate intensity patterns vary as
(sin(
)/
)2 where
= (kd /2sin(
)) , and if
we set one peak at the point where
=
for the other the
dip in intensity corresponds to
=
/2 for each peak,
giving a ratio of dip to peak intensity of
2(1/(
/2))2 = 8/
.
If we keep to this criterion, and note that in this case the maximum
intensity is the central intensity of one peak plus the intensity of the
other peak at the same point, for peaks to be resolved we require two
values of
,

apart, which satisfy
But we know that at the peaks
and
are each equal to a multiple
of 2
, so we have

=
,
giving, for small

(expanding sins to first order, obtaining and
solving
a quadratic in
F
),

.
Now, from Equation L29.1,

= -
2dcos(
)
and
itself is a large multiple, p , of
, so making the reasonable
approximation of neglecting the phase change
compared with 2
p ,
,
which gives a resolving power
= 
,
since
p
2d.
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