Consider the situation shown in figure (17-E6). The two slits S1 and S2 placed symmetrically around the central line are illuminated by a monochromatic light of wavelength λ. The separation between the slits is d. The light transmitted by the slits falls on a screen ∑1 placed at a distance D from the slits. The slit S3 is at the central line and the slit S4 is at a distance z from S3. Another screen ∑2 is placed a further distance D away from ∑1. Find the ratio of the maximum to minimum intensity observed on ∑2 if z is equal to
(a) 
(b) 
(c) 
Let the intensity of slits 
and 
be 
. The amplitude through each slit be 
.
For 
, the phase difference is zero as it lies in the center of the screen. So here, the intensity(
) would be maximum.
For 
, the path difference is 
And the corresponding phase difference is
(a) For 
,
This is the condition of destructive interference, so the intensity at 
is zero. Hence, only one of the slits is illuminated and so the pattern on Σ₂ would be uniform with no maxima or minima. As maxima and minima are same, the ratio is one.
(b)For 
,
In this case, constructive interference is taking place at 
. So the intensity is maximum and is equal to 
that is 
.
In this case, slits 
and 
will form an interference pattern with alternate dark and bright fringes. Let the maximum intensity be 
while the minimum will be zero for a dark fringe.
Hence the ratio is 
.
(c)For 
,
The intensity at 
is thus 
.
And intensity at 
is 
Let the amplitude at 
be 
and at 
be 
.
Therefore,
And
So maximum amplitude at Σ₂ would be 
and minimum amplitude would be 
.
Hence, the ratio of intensities is
Rationalizing,
