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:

We need to find the focal length of the objective lens $$f_o$$ of a refracting telescope in normal adjustment.

Since the telescope is in normal adjustment, the final image is formed at infinity, so its length is given by $$L = f_o + f_e$$ and its angular magnification by $$m = \frac{f_o}{f_e}$$.

We are given that $$L = f_o + f_e = 30$$ cm and $$m = \frac{f_o}{f_e} = 2$$, which implies $$f_o = 2f_e$$.

Substituting $$f_o = 2f_e$$ into the expression for the length yields $$2f_e + f_e = 30$$, so that $$3f_e = 30$$. This gives $$f_e = 10\text{ cm}$$ and hence $$f_o = 2 \times 10 = 20\text{ cm}$$.

Therefore, the correct answer is Option A: 20 cm.