As
shown in the figure,
|
Solution: Click For PDF Version Let the distance of the detector D, which is moved along the x-axis, from the source
be
 .
As the source
is at a distance d from
,
as shown in the figure, the distance from it to the detector D
will be
 .
As the two sources emit equal amount of radiations are coherent, the
phase difference between the waves reaching at D from
and
will be
The intensity maxima at the detector will occur at locations where
Therefore, three maxima will be at
We solve these equations algebraically, and note that the first, second, and the third maxima will occur when detector D is at locations given below.
Substituting the values
and
The next one
corresponds to
And the third
one corresponds to
Note for
phase difference
(b) As
the intensity varies as the inverse of square of the distance from
the source, at the first minimum the amplitude of the wave from the
source
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apart, they
emit equal amount of power in the form of
wavelength electromagnetic waves.
(a) We have to find the positions of the first
(that is, the
nearest), the second,
and the third maxima of the received signal,
as the detector D is moved out along Ox.
We have to answer whether the intensity at the
nearest minimum will be zero and give justification.
.
,
we find that the nearest maximum corresponds to
,
and
.
,
and
,
,
and
.
,
there is no interference as the detector will be on top of the
source