When a convex lens produces a real image of an object, the minimum distance between the object and image is equal to
[amp_mcq option1=”the focal length of the convex lens” option2=”twice the focal length of the convex lens” option3=”four times the focal length of the convex lens” option4=”one half of the focal length of the convex lens” correct=”option3″]
This question was previously asked in
UPSC CDS-2 – 2018
For a convex lens, a real image is formed when the object is placed outside the focal point (object distance |u| > f). The image formed is real and inverted. The lens formula is 1/v – 1/u = 1/f. Using distances as positive values, 1/v + 1/|u| = 1/f. Let D be the distance between the object and the image, D = |u| + v. To minimize D, we can express v in terms of |u| and f: 1/v = 1/f – 1/|u| = (|u|-f)/(f|u|), so v = f|u|/(|u|-f). Thus, D = |u| + f|u|/(|u|-f) = (|u|(|u|-f) + f|u|)/(|u|-f) = (|u|² – f|u| + f|u|)/(|u|-f) = |u|²/(|u|-f). Let x = |u|-f, so |u| = x+f. D = (x+f)²/x = (x² + 2xf + f²)/x = x + 2f + f²/x. For a real image, |u| must be greater than f, so x > 0. By AM-GM inequality, x + f²/x ≥ 2√(x * f²/x) = 2f. The minimum value occurs when x = f²/x, i.e., x² = f², so x = f (since x>0). This means |u|-f = f, so |u| = 2f. When |u|=2f, v = f(2f)/(2f-f) = 2f²/f = 2f. The minimum distance D = |u| + v = 2f + 2f = 4f. This happens when the object is placed at 2f, forming a real image at 2f.