Similarly these curved lines can be found in all the cases which can be proposed, as will be sufficiently shown by one or two examples which I will add.
Let there be given the surface of the glass AK, made by the revolution about the axis BA of the line AK, which may be straight or curved.

Let there be also given in the axis the point L and the thickness BA of the glass; and let it be required to find the other surface KDB, which receiving rays that are parallel to AB will direct them in such wise that after being again refracted at the given surface AK they will all be reassembled at the point L.
[Illustration] From the point L let there be drawn to some point of the given line AK the straight line LG, which, being considered as a ray of light, its refraction GD will then be found.

And this line being then prolonged at one side or the other will meet the straight line BL, as here at V.Let there then be erected on AB the perpendicular BC, which will represent a wave of light coming from the infinitely distant point F, since we have supposed the rays to be parallel.

Then all the parts of this wave BC must arrive at the same time at the point L; or rather all the parts of a wave emanating from the point L must arrive at the same time at the straight line BC.

And for that, it is necessary to find in the line VGD the point D such that having drawn DC parallel to AB, the sum of CD, plus 3/2 of DG, plus GL may be equal to 3/2 of AB, plus AL: or rather, on deducting from both sides GL, which is given, CD plus 3/2 of DG must be equal to a given length; which is a still easier problem than the preceding construction.