WAEC Physics 2012 Paper

Concept:A diverging (concave) lens always forms a virtual, erect and diminished image on the same side of the lens as the object.When the object is between the focal point and the optical centre, the image is formed nearer to the lens than the object.Explanation:The object is placed at $X$, which lies between the focal point $F$ and the optical centre $C$.For a diverging lens, the focal length $f$ is taken as negative in the lens formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.The image distance $v$ is negative, so the image is formed on the same side as the object.The magnitude of the image distance is $|v| = \frac{|f|\,u}{|f| + u}$.Since $|f| + u > u$, this gives $|v| < u$.Therefore, the image is closer to the lens than the object $X$.So the image lies between $X$ and the optical centre $C$.Answer:B. Between $X$ and $C$.