Babinet’s principle. A monochromatic bean of parallel light is incident on a “collimating” hole of diameter $\mathit{x}\mathbf{\gg }\mathit{\lambda }$ . Point P lies in the geometrical shadow region on a distant screen (Fig. 36-39a). Two diffracting objects, shown in Fig.36-39b, are placed in turn over the collimating hole. Object A is an opaque circle with a hole in it, and B is the “photographic negative” of A . Using superposition concepts, show that the intensity at P is identical for the two diffracting objects A and B .

Consider that a monochromatic bean of parallel light is incident on a “collimating” hole of diameter $\mathit{x}\mathbf{\gg }\mathit{\lambda }$.

Point P lies in the geometrical shadow region on distant screen. Over the collimating hole two different diffracting objects are placed. Object A is an opaque circle with a hole and B is the photographic negative of A .

Consider the Huygen’s principle of diffraction, when object A is in the place where, the huygen’s wavelets passes through the hole reaches the point P . Consider that it produces the electric field $E_A$.

When object B is placed at the point , where the light that was blocked by object A gets to the point P . The Electric field at P is $E_B$.

Thus, the resultant electric field at the condition when neither A nor B is present is,

$\overrightarrow{E_A}+\overrightarrow{E_B}=0$

The resultant is zero, since the point P is in geometric shadow.

Hence,

$\overrightarrow{E_A}=-\overrightarrow{E_B}$

And the intensity is proportional to the square of the electric field. So

$I\propto E^2$

Here, the intensity at point P is same when A and B are present. Therefore,

$I\propto E^2$

Hence, it has been proved that the intensity at P is identical for two diffracting objects A and B .