Correct Answer - C
Magniifying power of an astronomical telescope is given by
`|m|(f_(o))/(f_(e))(1+(f_(e))/(D))`
where`f_(o)` is focal length of objective , `f_(e)` is focal length of eyepiece, `D` is least distance of distinct vision.
From this formula we observe that
Magnifying power `prop(1)/("focal length of eye-piece")`
Hence, to produce the largest magnification the focal length of eye- piece must be smallest.