An astronomical telescope is set in normal adjustment for viewing a...

For an astronomical telescope in normal adjustment, the magnifying power is given by $m = \dfrac{f_o}{f_e}$ and the length of the telescope tube is $L = f_o + f_e$.

Given $m = 14$ and $L = 105\text{ cm}$.

$\dfrac{f_o}{f_e} = 14 \Rightarrow f_o = 14 f_e$

Substituting this into the length equation:

$14 f_e + f_e = 105$

$15 f_e = 105 \Rightarrow f_e = 7\text{ cm}$

$f_o = 14 \times 7 = 98\text{ cm}$

The focal lengths of the objective and the eyepiece are $98\text{ cm}$ and $7\text{ cm}$ respectively.

Answer: $98\text{ cm}$ and $7\text{ cm}$