[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 adjust FD up to the straight line VG in such a way that FD together with 3/2 of DG is equal to a given straight line, which is a quite easy plane problem:  and the point D will be one of those through which the curve BDK ought to pass.  And similarly, having drawn another ray LM, and found its refraction MO, the point N will be found in this line, and so on as many times as one desires.

To demonstrate the effect of the curve, let there be described about the centre L the circular arc AH, cutting LG at H; and about the centre F the arc BP; and in AB let AS be taken equal to 2/3 of HG; and SE equal to GD.  Then considering AH as a wave of light emanating from the point L, it is certain that during the time in which its piece H arrives at G the piece A will have advanced within the transparent body only along AS; for I suppose, as above, the proportion of the refraction to be as 3 to 2.  Now we know that the piece of wave which is incident on G, advances thence along the line GD, since GV is the refraction of the ray LG.  Then during the time that this piece of wave has taken from G to D, the other piece which was at S has reached E, since GD, SE are equal.  But while the latter will advance from E to B, the piece of wave which was at D will have spread into the air its partial wave, the semi-diameter of which, DC (supposing this wave to cut the line DF at C), will be 3/2 of EB, since the velocity of light outside the medium is to that inside as 3 to 2.  Now it is easy to show that this wave will touch the arc BP at this point C. For since, by construction, FD + 3/2 DG + GL are equal to FB + 3/2 BA + AL; on deducting the equals LH, LA, there will remain FD + 3/2 DG + GH equal to FB + 3/2 BA.  And, again, deducting from one side GH, and from the other side 3/2 of AS, which are equal, there will remain FD with 3/2 DG equal to FB with 3/2 of BS.  But 3/2 of DG are equal to 3/2 of ES; then FD is equal to FB with 3/2 of BE.  But DC was equal to 3/2 of EB; then deducting these equal lengths from one side and from the other, there will remain CF equal to FB.  And thus it appears that the wave, the semi-diameter of which is DC, touches the arc BP at the moment when the light coming from the point L has arrived at B along the line LB.  It can be demonstrated similarly that at this same moment the light that has come along any other ray, such as LM, MN, will have propagated the movement which is terminated at the arc BP.  Whence it follows, as has been often said, that the propagation of the wave AH, after it has passed through the thickness of the glass, will be the spherical wave BP, all the pieces of which ought to advance along straight lines, which are the rays of light, to the centre F. Which was to be proved.  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.