Light
of wavelength 440 nm passes
through a double slit,
yielding the diffraction pattern of intensity
I versus deflection angle
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Solution: Click For PDF Version (a)
As the
boundary of the central envelope is at
The
wavelength of light used in obtaining the diffraction pattern is
Therefore, the width a of the slit will be
(b) From the intensity diagram, we note that the peak of the fourth fringe falls on the diffraction minimum of the central envelope and that are only 3 interference fringes other than the central fringe on each side of the central fringe. Let the two slits be separated by distance d. The 4th fringe is determined by the equation
We will next
calculate the intensity of the fringe that corresponds to
We have the condition that
For
calculating the intensity we have to find the value of
As the intensity at a fringe is given by
we find the
intensity of the fringe that corresponds to
where we have used that
The intensity
of the fringe, which corresponds to
For
Value of
Therefore,
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,
as shown in the figure.
We have to calculate
(a) the slit width and
(b) the slit separation.
We have to verify the displayed intensities of
the
and
interference fringes.
,
we note that
.
,
for
