WAEC 2011 Physics · Question 7

An object is placed 20.0cm in front of a converging lens of focal length 15.0cm. Calculate its image distance

Options

A.1.3cm
B.8.6cm
C.35.0cm
D.60.0cm

Answer: D. 60.0cm

Explanation written with AI and checked against the answer key

Analyze the question: An object is 20.0 cm from a converging (convex) lens with f = 15.0 cm. We need the image distance v. Concept: Use the lens formula (real-is-positive convention): 1/f = 1/u + 1/v Because u = 20 cm is greater than f = 15 cm but less than 2f = 30 cm, the object lies between F and 2F. So the image should be real, inverted, magnified, and formed beyond 2F (further than 30 cm). Solution: 1/v = 1/f − 1/u 1/v = 1/15 − 1/20 Common denominator is 60: 1/v = 4/60 − 3/60 = 1/60 v = 60 cm Check: v = 60 cm is beyond 2F (30 cm), which matches the prediction. Magnification = v/u = 60/20 = 3, so the image is 3 times larger than the object, which also fits an object between F and 2F. Answer: The image distance is 60.0 cm, which is option D. Tip: For a quick check, remember that the lens formula can be rewritten as v = uf/(u − f). Here: v = (20 × 15)/(20 − 15) = 300/5 = 60 cm. Also, if the object is between F and 2F of a converging lens, the image is always beyond 2F, so you can eliminate A, B, and C straight away (they are all less than 30 cm except C, which is 35 cm, but the calculation rules that out).

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