A screen is located at a distance of 40 cm from an illuminated...

Let the distance between the object and the screen be $D$, and the object distance from the lens be $x$.

The image distance from the lens is $D - x$.

Using the lens formula, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{1}{D-x} - \dfrac{1}{-x} = \dfrac{1}{f}$

$\dfrac{x + D - x}{x(D-x)} = \dfrac{1}{f}$

$x^2 - Dx + Df = 0$

For a real position of the lens to exist, the roots of this quadratic equation must be real, which requires the discriminant to be greater than or equal to zero:

$D^2 - 4Df \ge 0 \Rightarrow D \ge 4f$

Since no such position can be found, the roots must be imaginary, which means:

$D < 4f$

Given $D = 40 \text{ cm}$, we have:

$40 < 4f \Rightarrow f > 10 \text{ cm}$

Therefore, the focal length of the lens must be greater than $10 \text{ cm}$.