Here, $f _0 = 20 \,m \,\, and \,\, f _e = 2 \, cm = 0.02 \, m$ In normal adjustment, Length of telescope tube, $L = f _0 + f _e = 20 +0.02 = 20.02m$ and magnification, $m = \dfrac{f _0}{f _e} = \dfrac{20}{0.02} = 1000$ The image formed is inverted with respect to the object.