WAEC 2015 Physics · Question 9

An object is placed 15cm from a diverging lens of focal length 12cm. The image of the object formed by the lens in

Options

A.real and 6.67cm from the lens
B.virtual and 6.67cm from the lens
C.real and 60.00cm from the lens
D.virtual and 60.00cm from the lens

Answer: B. virtual and 6.67cm from the lens

Explanation written with AI and checked against the answer key

Analyze the question: An object is 15 cm from a diverging (concave) lens with focal length 12 cm. We need the nature and position of the image. Concept: Use the lens formula: 1/f = 1/u + 1/v With the "real is positive" convention, a diverging lens has a negative focal length, so f = -12 cm. The object distance is u = +15 cm. Solution: 1/v = 1/f - 1/u 1/v = -1/12 - 1/15 The LCM of 12 and 15 is 60: 1/v = -5/60 - 4/60 = -9/60 v = -60/9 = -6.67 cm The negative sign means the image is virtual, and it forms on the same side of the lens as the object. It is 6.67 cm from the lens. Check: A diverging lens always forms a virtual, upright, diminished image. The image distance of 6.67 cm is less than the object distance of 15 cm, so the image is smaller, which agrees with this. Answer: The image is virtual and 6.67 cm from the lens, which is option B. Tip: A diverging lens never gives a real image. You can eliminate options A and C straight away. Then just compute v. For a diverging lens, the image distance is always smaller than both the object distance and |f|, so 60 cm (option D) is impossible.

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