JAMB Physics 2018 Paper

Concept:An erect image formed by a lens is virtual, so the image lies on the same side of the lens as the object.Explanation:Magnification is $M = \frac{\text{image distance}}{\text{object distance}} = 3$.Let the object distance be $u$. Then the magnitude of the image distance is $3u$.Since the image is virtual, the signed image distance in the lens formula is $v = -3u$.Use the lens formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.Substitute $f = 15\, \text{cm}$ and $v = -3u$: $\frac{1}{15} = \frac{1}{u} - \frac{1}{3u}$Thus $\frac{1}{15} = \frac{2}{3u}$, so $u = 10\, \text{cm}$.Hence the image is $3u = 30\, \text{cm}$ from the lens, on the same side as the object.The distance between the object and the image is therefore $30\, \text{cm} - 10\, \text{cm} = 20\, \text{cm}$.Answer:$20\, \text{cm}$ (Option B).