Principal of reversibility is states that the position of object and image are interchangeable. So, By the versibility of u and v, as seen from the formula for lens .
`" " (1)/(f)=(1)/(v)-(1)/(u)`
It is clear that three are two position for which there shall be an image.
On the screen, let the first positions be when the lens is at O. Finding u and v substituting in lens, formula,
Given, `" " -u+v=D`
`rArr" " u=-(D-v)`
Placing it in the lens formula
`" " (1)/(D-v)+(1)/(v)=(1)/(v)`
On solging, we have
`rArr" " (v+D-v)/((D-v)v)=(1)/(f)`
`rArr" " v^(2)-Dv+Df=0`
` rArr" " v=(D)/(2)+-(sqrt(D^(2)-4Df))/(2)`
Hence, finding u
`" " u=-(D-v)=-((D)/(2)+-(sqrt(D^(2)-4Df))/(2))`
When , the object distance is `" " (D)/(2)+(sqrt(D^(2)-4Df))/(2)`
the image forms at `" " (D)/(2)-(sqrt(D^(2)-4Df))/(2)`
Similarly, when the object distance is
`" " (D)/(2)-(sqrt(D^(2)-4Df))/(2)`
The image forms at `" " (D)/(2)+(sqrt(D^(2)-4Df))/(2)`
The distance between the poles for these two object distance is
`" " (D)/(2)+(sqrt(D^(2)-4Df))/(2)-((D)/(2)-(sqrt(D^(2)-4Df))/(2))=sqrt(D^(2)-4Df)`
Let `" " d=sqrt(D^(2)-4Df)`
If `u=(D)/(2)+(d)/(2)`, then the image is at `v=(D)/(2)-(d)/(2)`
`:. " " ` The magnification `m_(1)=(D-d)/(D+d)`
Thus, `" " (m_(2))/(m_(1))=((D+d^(2))/(D-d))^(2)`
This is the required expression of magnification .