curve KDE was of such a nature that
having drawn from some point which had been assumed, such as K, the
straight lines KA, KB, the excess by which AK surpasses AD should be
to the excess of DB over KB, as 3 to 2. For it can similarly be
demonstrated, by taking any other point in the curve, such as G, that
the excess of AG over AD, namely VG, is to the excess of BD over DG,
namely DP, in this same ratio of 3 to 2. And following this principle
Mr. Des Cartes constructed these curves in his _Geometric_; and he
easily recognized that in the case of parallel rays, these curves
became Hyperbolas and Ellipses.

Let us now return to our method and let us see how it leads without
difficulty to the finding of the curves which one side of the glass
requires when the other side is of a given figure; a figure not only
plane or spherical, or made by one of the conic sections (which is the
restriction with which Des Cartes proposed this problem, leaving the
solution to those who should come after him) but generally any figure
whatever: that is to say, one made by the revolution of any given
curved line to which one must merely know how to draw straight lines
as tangents.

Let the given figure be that made by the revolution of some curve such
as AK about the axis AV, and that this side of the glass receives rays
coming from the point L. Furthermore, let the thickness AB of the
middle of the glass be given, and the point F at which one desires the
rays to be all perfectly reunited, whatever be the first refraction
occurring at the surface AK.

I say that for this the sole requirement is that the outline BDK which
constitutes the other surface shall be such that the path of the
light from the point L to the surface AK, and from thence to the
surface BDK, and from thence to the point F, shall be traversed
everywhere in equal times, and in each case in a time equal to that
which the light employs, to pass along the straight line LF of which
the part AB is within the glass.

[Illustration]

Let LG be a ray falling on the arc AK. Its refraction GV will be given
by means of the tangent which will be drawn at the point G. Now in GV
the point D must be found such that FD together with 3/2 of DG and the
straight line GL, may be equal to FB together with 3/2 of BA and the
straight line AL; which, as is clear, make up a given length. Or
rather, by deducting from each the length of LG, which is also given,
it will merely be needful to adjus