In many experimental set-ups,the source and screen are fixed at a distance $D$ 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.

(N/A) Let the distance between the object and the screen be $D$. Let the distance of the lens from the object be $u$ and from the screen be $v$. Then $v + |u| = D$. Since $u$ is negative,$v - u = D$,or $v = D + u$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,we substitute $v = D + u$:
$\frac{1}{f} = \frac{1}{D+u} - \frac{1}{u} = \frac{u - (D+u)}{u(D+u)} = \frac{-D}{uD + u^2}$.
Rearranging gives the quadratic equation $u^2 + Du + fD = 0$.
The roots are $u = \frac{-D \pm \sqrt{D^2 - 4fD}}{2}$.
For real roots,$D^2 - 4fD \ge 0$,which means $D \ge 4f$. Thus,there are two positions for the lens.
Let the two positions be $u_1$ and $u_2$. The distance between them is $d = |u_1 - u_2| = \sqrt{D^2 - 4fD}$.
The magnification $m = \frac{v}{u}$. For the two positions,$m_1 = \frac{v_1}{u_1}$ and $m_2 = \frac{v_2}{u_2}$. Since $v_1 = |u_2|$ and $v_2 = |u_1|$,we have $m_1 = \frac{|u_2|}{u_1}$ and $m_2 = \frac{|u_1|}{u_2}$. The product $m_1 m_2 = 1$,so the ratio of image sizes is $m_1^2$ or $m_2^2$.