What is seen on the screen when the flame is very close (at about 5cm) to the lens?

Given, Distance between the object and lens u=2m=200cm Focal length of the lens f=10cm We know that, 1/v - 1/u = 1/f ⇒ 1/v - 1/200 = 1/10 1/v = 1/10 + 1/200 ⇒ v = 200/19 = 10.52cm Now, Distance between object and screen t(d) = |u| + |v| = 200 + 10.52 = 210.52cm Now, u = d - v We know that, 1/v - 1/u = 1/f 1/v = 1/f + 1/d - v (≈u = d - v) 1/v = (d - v + f)/f(d - v) v² - dv + df = 0 So, after solving, v = d ± √(d² - 4df)/2 We know that m = v/u So it is clear that, when flame moves towards the lens, the image size is gradually increasing.