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.
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
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