A screen is placed at a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens

let the distance between the source and the lens be x , then the distance between the lens and the screen is 40-x 

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \frac{1}{(40-x)}+\frac{1}{x}=\frac{1}{f} \\ {x}^2-40x+40f = 0 \\ x = \frac{40\pm \sqrt{1600-160f}}{2} \end{gathered}$$

for x to take real values (1600-160f)>0

f<10 cm 

for the image to be formed on the screen the focal length should be less than 10 cm , but since no such positions were found hence the lens used has focal length greater than 10 cm