JAMB Physics 2016 Paper

Concept:For a convex lens producing a virtual image, the image is erect and magnified, but the image distance is taken as negative in the lens formula.Explanation:Object distance, $u = 10\text{ cm}$.Magnification, $M = 3$, so $|v| = 3u = 3 \times 10 = 30\text{ cm}$.Since the image is virtual and lies on the same side as the object, $v = -30\text{ cm}$ under the real-is-positive sign convention.Using the lens formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$.$\frac{1}{f} = \frac{1}{10} + \frac{1}{-30} = \frac{1}{10} - \frac{1}{30}$.$\frac{1}{f} = \frac{3 - 1}{30} = \frac{2}{30}$.Therefore, $f = \frac{30}{2} = 15\text{ cm}$.Answer:D. $15\text{ cm}$