Given, fe=2 cm, and fo=20 m=2000 cm For normal adjustment, Magnification power Length of telescope =fe−fo=2−2000=−1000=fo+fe=2000+2=2002 cm=20.02 m The image formed is inverted and magnified, and the aperture of objective is larger than eye piece of the telescope. (fo & fe are focal lengths of objective & eye piece respectively.)