JAMB 2018 Physics · Question 80

A lens of focal length 15cm forms an erect image which is three times the size of the object. The distance between the object and the image is ___.

Options

A.10cm
B.20cm
C.30cm
D.40cm

Answer: B. 20cm

Explanation written with AI and checked against the answer key

Analyze the question An erect image that is three times the object's size means the lens is a converging (convex) lens used as a magnifying glass. The image is virtual and on the same side as the object. We need the distance between object and image. Concept Lens formula (real-is-positive convention): 1/f = 1/u + 1/v, and magnification m = v/u For a virtual image, v is negative, so the magnification is m = -v/u in magnitude terms. Here the image is erect and magnified, so |v| = 3u. Solution Using the convention where the virtual image distance is negative: v = -3u 1/f = 1/u + 1/v 1/15 = 1/u - 1/(3u) 1/15 = (3 - 1)/(3u) = 2/(3u) So 3u = 30, giving u = 10 cm. Then the image distance is |v| = 3 × 10 = 30 cm (virtual, on the same side as the object). Check with the magnification formula: m = f/(f - u) = 15/(15 - 10) = 3 ✓ Distance between object and image Both are on the same side of the lens, so the separation is: 30 - 10 = 20 cm Answer: B (20 cm) Tip: For an erect magnified image from a convex lens, the image is virtual and on the same side as the object, so subtract the distances (v - u). For a real image, they are on opposite sides, so you would add them. Also, a quick shortcut for virtual images: u = f(m - 1)/m = 15 × 2/3 = 10 cm.

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