A Keplerian telescope is adjusted in its normal setting for parallel rays. Mounting of the objective has diameter $$D$$ and the diameter of the image of that mounting formed by the eye piece of the telescope is $$d$$. Magnifying power of the telescope is

A

$$\frac{D}{d}-1$$

B

$$\frac{D}{d}+1$$

C

$$\frac{D+d}{D-d}$$

D

$$\frac{D}{d}$$