The distance between the objective and eyepiece of an astronomical telescope in normal adjustment is $27\text{ cm}$ and its magnifying power is $8$. What is the focal length of the eyepiece?
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You can isolate $f_e$ directly from any telescope system data using the short form: $f_e = \frac{L}{m + 1}$. Here, $f_e = \frac{27}{8 + 1} = \frac{27}{9} = 3\text{ cm}$. This completely removes the intermediate substitution step!