WAEC 2013 Physics · Question 41

A diverging lens of focal length 30cm produces an image 20cm from the lens. Determine the object distance

Options

A.10cm
B.12cm
C.50cm
D.60cm

Answer: D. 60cm

Explanation written with AI and checked against the answer key

Analyze the question: A diverging (concave) lens always forms a virtual, upright, diminished image on the same side as the object. We're given f = 30 cm and image distance v = 20 cm, and we need the object distance u. Concept: Use the lens formula: 1/f = 1/u + 1/v With the real-is-positive sign convention, a diverging lens has a negative focal length (f = −30 cm) and a virtual image has a negative image distance (v = −20 cm). Solution: 1/u = 1/f − 1/v 1/u = 1/(−30) − 1/(−20) 1/u = −1/30 + 1/20 1/u = (−2 + 3)/60 = 1/60 u = 60 cm Check: The object is 60 cm away and the virtual image forms at 20 cm, closer to the lens than the object. This is what a diverging lens does, so the result is consistent. Magnification = v/u = 20/60 = 1/3, so the image is diminished, as expected. Answer: The object distance is 60 cm, which is option D. Tip: For a diverging lens, the image is always virtual and closer to the lens than the object, so the object distance must be larger than the image distance. That eliminates A and B right away (10 and 12 are less than 20). Then remember to put in negative signs for both f and v, and the algebra gives you 60 cm.

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