When a convex lens produces a real image of an object, the minimum distance between the object and image is equal to
the focal length of the convex lens
twice the focal length of the convex lens
four times the focal length of the convex lens
one half of the focal length of the convex lens
Answer is Right!
Answer is Wrong!
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.