Let the object be placed x m in front of lens the distance of image from the lens is(3 – x) m. Applying lens formula, $\frac{1}{v} - \frac{1}{u}$ = $\frac{1}{f}$ $\frac{1}{3-x} - \frac{1}{-x}$ or $\frac{1}{3-x} + \frac{1}{x}$ = $\frac{1}{f}$ or $\frac{x+3-x}{x(3-x)}$ = $\frac{1}{f}$ or 3f = 3x - $x^2$ so, $x^2$ - 3x + 3f = 0 Now x = $\frac{+3 \pm \sqrt{9 - 4 \times (3f)}}{2}$ or x = $\frac{+3 \pm \sqrt{9 - 12f}}{2}$ Condition for image to be obtained on the screen, i.e., real image. 9 – 12f ≥ 0 or 9 ≥ 12f or f ≤ 0.75 m. So, maximum focal length is 0.75 m.