WAEC Physics 2011 Paper

Concept:Use the thin lens formula to relate object distance $u$, image distance $v$, and focal length $f$ of a converging lens.Explanation:For a converging lens, the lens formula is:$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$Given: $u = 20.0\ \text{cm}$ and $f = 15.0\ \text{cm}$.Substitute the known values:$$\frac{1}{15.0} = \frac{1}{20.0} + \frac{1}{v}$$Rearrange to isolate $\frac{1}{v}$:$$\frac{1}{v} = \frac{1}{15.0} - \frac{1}{20.0}$$Take the LCM of $15.0$ and $20.0$, which is $60.0$:$$\frac{1}{v} = \frac{4 - 3}{60.0} = \frac{1}{60.0}$$Thus, $v = 60.0\ \text{cm}$.The positive sign indicates the image is formed on the opposite side of the lens.Answer:The image distance is $60.0\ \text{cm}$.Correct option: D. $60.0\ \text{cm}$.