WAEC Physics 2016 Paper

Concept:For an astronomical telescope in normal adjustment, the magnifying power is the ratio of the objective focal length to the eyepiece focal length.Explanation:Given focal length of objective, $f_o = 3.5\ \text{m} = 350\ \text{cm}$.Given focal length of eyepiece, $f_e = 5\ \text{cm}$.Formula for magnifying power in normal adjustment:$$M = \frac{f_o}{f_e}$$Substitute the values in the same unit:$$M = \frac{350}{5} = 70$$So, the magnifying power is $70$.Answer:$M = 70.0$Correct option: A. $70.0$