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
