Correct Answer - b.
Let the objec distance be x. The, the image distance is `D-x`.
From lens equation,
`(1)/(x)+(1)/(D-x)=(1)/(f)`
On algebraic rearrangement, we get
`x^(2)-Dx+Df=0`
On solving for x, we get
`x_(1)=(D-sqrt(D(D-4f)))/(2)`
`x_(2)=(D+sqrt(D(D-4f)))/(2)`
The distance between the two object position is
`d=x_(2)-x_(1)=sqrt(D(D-4f))`