Correct Answer - D
Here, `f_(o) = 20m` and `f_(e) = 2 cm = 0.02 m`
in normal adjustement,
Length of telescope tube, `L = f_(o) + f_(e) = 20 + 0.02 = 20.02 m`
and magnification, `m = (f_(o))/(f_(e)) = (20)/(0.20) = 1000`
The image formed is inverted with respect to the object.