A point object placed in front of a convex lens (of focal length 0.2m) at a distance of 0.3m. The lens is cut into halves each part is displaced by 0.0005m symmetrically as shown. Distance between the two images that are formed is

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{0.2}-\frac{1}{0.3}=\frac{1}{0.6}$

$\Rightarrow \mathrm{v}=60\mathrm{cm}$

$\mathrm{m}=\frac{60}{-30}=-2$.

The point image is at a distance of 0.0005 m from the principle axis of the lens, considering this as an extended object.

Thus, two inverted images are formed at a distance of 2x0.0005=0.001m on either sides of the principle axis.

So the total distance between the images will be $2\left(0.001\right)+0.001=0.003\mathrm{m}$