WAEC Physics 2017 Paper

Concept:For a converging lens, an erect and magnified image is virtual and is formed on the same side of the lens as the object.Magnification is $m = \frac{v}{u} = 3$.Explanation:Let $u$ be the object distance and $v$ be the image distance from the lens.Since the image is three times the size of the object, $v = 3u$.For a virtual image formed by a convex lens, using distances as magnitudes:$\frac{1}{f} = \frac{1}{u} - \frac{1}{v}$.Substitute $f = 10\, \text{cm}$ and $v = 3u$:$\frac{1}{10} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}$.So, $u = \frac{20}{3} = 6.67\, \text{cm}$.Then $v = 3 \times 6.67 = 20.0\, \text{cm}$.Distance between the image and the object is $v - u = 20.0 - 6.67 = 13.33\, \text{cm}$.Answer:C. 13.3 cm