From the reversibility of u and v, as seen from the formula for lens,

1/f=1/v-1/u

It is clear that there are two positions for which there shall be an image on the screen.

Let the first position be when the lens is at O.

Given –u + v = D

=>u = –(D – v)

Placing it in the lens formula

1/D-v+1/v=1/f

If the object distance is

then the image is at

The distance between the poles for these two object distances is