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The magnifying power of a telescope in normal adjustment is given by

A$m=\dfrac{f_e}{f_o}$
B$m=\dfrac{f_o}{f_e}$
C$m=1+\dfrac{D}{f_e}$
D$m=\dfrac{L}{f_o}\times\dfrac{D}{f_e}$
Answer & Solution
Correct answer: B. $m=\dfrac{f_o}{f_e}$
For a telescope, the magnifying power is the ratio of the angle subtended by the image at the eye to that subtended by the object, and in normal adjustment it is $m=f_o/f_e$. Options C and D are microscope relations.
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