An object is 5 m to the left of a flat screen. A converging lens for which the focal length is 0.8m is placed between object and screen. 

(a) Show that for two positions of lens form images on the screen and determine how far these positions are from the object? 

(b) How do the two images differ from each other?

a) Using the lens formula, $\frac{1}{\mathrm{V}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
we have, $\frac{1}{5-\mathrm{u}}+\frac{1}{\mathrm{u}}=\frac{1}{0.8}\text{ or }\frac{1}{5-\mathrm{u}}+\frac{1}{\mathrm{u}}=1.25$
$\therefore \mathrm{u}+5-\mathrm{u}=1.25\mathrm{u}(5-\mathrm{u})$
or $1.25{\mathrm{u}}^2-6.25\mathrm{u}+5=0\Rightarrow \mathrm{u}=4\mathrm{m}$ and 1m
Both the values are real, which means there exist two positions of lens that form images of object on the screen.
b) $\begin{array}{l}\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}\\ \therefore {\mathrm{m}}_1=\frac{(5-4)}{(-4)}=-0.25\text{ and }{\mathrm{m}}_2=\frac{(5-1)}{(-1)}=-4.00\end{array}$
Hence, both the images are real and inverted, the first has magnification –0.25 and the second –4.00.