In many experimental set-ups the source and screen are fixed at a distance say `D` and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

Explanation

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 .