In may 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

Let object be placed at a diatnce x from the lens then image distance becomes `(D-x)`. We can use lens formula as follows :
`(1)/(v)-(1)/(u)=(1)/(f)`
`rArr" "(1)/(D-x)-(1)/(-x)=(1)/(+f)`
`rArr" "(1)/(D-x)+(1)/(x)=(1)/(f)`
`rArr" "(x+D-x)/(x(D-x))=(1)/(f)`
`rArr" "fD=xD-x^(2)`
`rArr" "x^(2)-Dx+fD=0`
`x=(Dpmsqrt(D^(2)-4fD))/(2)`
Both the roots of x are valid when `D gt 4f`. Hence there are two positions of lens for which image can be formed on screen.
`x_(1)=(D+sqrt(D^(2)-4fD))/(2)`
`x_(2)=(D-sqrt(D^(2)-4fD))/(2)`
Separation between these two locations of lens can be written as :
`d=x_(1)-x_(2)`
`rArr" "d=sqrt(D^(2)-4fD)`

`"When "u=(D)/(2)+(d)/(2)`
`v=(D)/(2)-(d)/(2)`
`"Magnification , m"=(v)/(u)=(D-d)/(D+d)`
`"When "u=(D)/(2)-(d)/(2)`
`"Magnification, m."=(v)/(u)=(D+d)/(D-d)`
`therefore" "(m.)/(m)=((D+d)^(2))/((D-d)^(2))=((D+d)/(D-d))^(2)`