Q 12-09-150JEE MainJEE Main 2022 (28 Jul, Shift 1)Easy
In normal adjustment, for a refracting telescope, the distance between objective and eye piece is $30$ cm. The focal length of the objective, when the angular magnification of the telescope is $2$, will be:
Answer: (A) $20$ cm
In normal adjustment $f_o + f_e = 30$ cm and $m = \dfrac{f_o}{f_e} = 2$.
So $f_o = 2f_e$, giving $3f_e = 30$, $f_e = 10$ cm and $f_o = 20$ cm.
Solution by Sreeraj P, M.Sc Physics