A
monochromatic beam of light is incident on a
“collimating”
hole of diameter
|
Solution: Click For PDF Version As the obstacles complement each other and when superimposed over each other will be identical with the circular hole of diameter .
We can thus imagine that the circular hole consists of superposition
of A and B,
as shown in the figure. Let the amplitude at P
due to the holes A
and B, when placed
one by one at the position of the slit, be
and
 ,
respectively. By the superposition principle the amplitude of the
light wave at P due
to the circular hole will be
If the point
P is the diffraction
minima of the collimating hole, the amplitude
As the intensity is proportional to the square modulus of the amplitude, the intensity at P due to the obstacle A will be equal to the intensity at P due to the complementary obstacle B. This property is known as the Babinet’s principle. |
.
Point P lies in the geometrical shadow region
on a distant screen, as
shown in the figure.
Two obstacles, shown in the figure below, are
placed in turn over the collimating hole. A
is an opaque circle with a hole in it and
B is
the “photographic
negative” of
A. Using
the superposition concepts, we
have to show that the intensity at P is
identical for the two diffracting objects A
and B (Babinet’s
principle). In this
connection it can be shown that the diffraction pattern of a wire is
that of a slit of equal
width.
.
at P will be zero. This can happen only if