An astronomical telescope is to be designed with a magnifying power...

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

Given $m = 50$, we have $\dfrac{f_o}{f_e} = 50 \Rightarrow f_o = 50 f_e$.

Given $L = 102 \text{ cm}$, substituting $f_o$ gives $50 f_e + f_e = 102 \text{ cm}$.

$51 f_e = 102 \text{ cm} \Rightarrow f_e = 2 \text{ cm} = 0.02 \text{ m}$.

The focal length of the objective is $f_o = 50 \times 2 \text{ cm} = 100 \text{ cm} = 1 \text{ m}$.

The power of the objective is $P_o = \dfrac{1}{f_o} = \dfrac{1}{1} = 1 \text{ D}$.

The power of the eyepiece is $P_e = \dfrac{1}{f_e} = \dfrac{1}{0.02} = 50 \text{ D}$.