Eye piece of an astronomical telescope has focal length of 5 cm. If angular magnification in normal adjustment is 10, then distance between eye piece and objective should be
Answer & explanation
Correct answer: option 3
$\text{In normal adjustments } L = f_0 + f_e $
$ M = \frac{f_0}{f_e} = 10$
$ 10 = \frac{f_0}{5} \Rightarrow f_0 = 50cm$
$ L = f_0 + f_e = 50cm + 5 cm = 55cm$