The focal length of the objective of astronomical telescope is 2\ m...

Given focal length of the objective, $f_o = 2\ \text{m}$.

Length of the telescope under normal adjustment, $L = 2.10\ \text{m}$.

For normal adjustment, the length of the telescope is given by $L = f_o + f_e$.

Substituting the given values:

$2.10 = 2 + f_e$

$f_e = 0.10\ \text{m}$

The magnifying power of the telescope under normal adjustment is given by $M = \dfrac{f_o}{f_e}$.

$M = \dfrac{2}{0.10} = 20$

Answer: $20$