Question

A screen is placed a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and its is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
(a) must be less than 10 cm
(b) must be greater than 20 cm
(c) must not be greater than 20 cm
(d) must not be less than 10 cm.

Answer 1

(b) must be greater than 20 cm

Let the image be formed at a distance of x cm from the lens.
Therefore, the distance of the object from the lens, u​, will be = (40 − x) cm
From lens formula:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{f}=\frac{1}{x}-\frac{1}{40-x} \\ \Rightarrow \frac{1}{f}=\frac{40-x-x}{40x-x^2} \\ \Rightarrow f=\frac{40x-x^2}{40-2x} \\ \Rightarrow f(40-2x)=40x-x^2 \\ \Rightarrow x^2-2fx-40x+40f=0 \\ \Rightarrow x^2-(2f+40)x+40f=0 \end{gathered}$$

Therefore, we get x as:

$x=\frac{(2f+40)\pm \sqrt{(2f+40{)}^2-160f}}{2}$