cere oculis distantia sidera nostris,
AEtheraque ingenio supposuere suo.

In the first place, it is necessary that the beginner should rightly
know what is the nature of the instrument he is to use. And this is the
more necessary because, while it is perfectly easy to obtain such
knowledge without any profound acquaintance with the science of optics,
yet in many popular works on this subject the really important points
are omitted, and even in scientific works such points are too often left
to be gathered from a formula. When the observer has learnt what it is
that his instrument is actually to do for him, he will know how to
estimate its performance, and how to vary the application of its
powers--whether illuminating or magnifying--according to the nature of
the object to be observed.

Let us consider what it is that limits the range of _natural_ vision
applied to distant objects. What causes an object to become invisible as
its distance increases? Two things are necessary that an object should
be visible. It must be _large_ enough to be appreciated by the eye, and
it must _send light_ enough. Thus increase of distance may render an
object invisible, either through diminution of its apparent size, or
through diminution in the quantity of light it sends to the eye, or
through both these causes combined. A telescope, therefore, or (as its
name implies) an instrument to render distant objects visible, must be
both a magnifying and an illuminating instrument.

[Illustration: _Fig. 1._]

Let EF, fig. 1, be an object, not near to AB as in the figure, but so
far off that the bounding lines from A and B would meet at the point
corresponding to the point P. Then if a large convex glass AB (called an
_object-glass_) be interposed between the object and the eye, all those
rays which, proceeding from P, fall on AB, will be caused to converge
nearly to a point _p_. The same is true for every point of the object
EMF, and thus a small image, _emf_, will be formed. This image will not
lie exactly on a flat surface, but will be curved about the point midway
between A and B as a centre. Now if the lens AB is removed, and an eye
is placed at _m_ to view the distant object EMF, those rays only from
each point of the object which fall on the pupil of the eye (whose
diameter is about equal to _mp_ suppose) will serve to render the object
visible. On the other hand, every point of the image _emf_ has received
the whole of the light