d form,
generated by revolution about the axis AB, and let the thickness of
the glass at the middle be AB. Also let the point L be given in the
axis behind the glass; and let it be supposed that the rays which fall
on the surface AK tend to this point, and that it is required to find
the surface BD, which on their emergence from the glass turns them as
if they came from the point F in front of the glass.

Having taken any point G in the line AK, and drawing the straight line
IGL, its part GI will represent one of the incident rays, the
refraction of which, GV, will then be found: and it is in this line
that we must find the point D, one of those through which the curve DG
ought to pass. Let us suppose that it has been found: and about L as
centre let there be described GT, the arc of a circle cutting the
straight line AB at T, in case the distance LG is greater than LA; for
otherwise the arc AH must be described about the same centre, cutting
the straight line LG at H. This arc GT (or AH, in the other case) will
represent an incident wave of light, the rays of which tend towards
L. Similarly, about the centre F let there be described the circular
arc DQ, which will represent a wave emanating from the point F.

Then the wave TG, after having passed through the glass, must form the
wave QD; and for this I observe that the time taken by the light along
GD in the glass must be equal to that taken along the three, TA, AB,
and BQ, of which AB alone is within the glass. Or rather, having taken
AS equal to 2/3 of AT, I observe that 3/2 of GD ought to be equal to
3/2 of SB, plus BQ; and, deducting both of them from FD or FQ, that FD
less 3/2 of GD ought to be equal to FB less 3/2 of SB. And this last
difference is a given length: and all that is required is to draw the
straight line FD from the given point F to meet VG so that it may be
thus. Which is a problem quite similar to that which served for the
first of these constructions, where FD plus 3/2 of GD had to be equal
to a given length.

In the demonstration it is to be observed that, since the arc BC falls
within the glass, there must be conceived an arc RX, concentric with
it and on the other side of QD. Then after it shall have been shown
that the piece G of the wave GT arrives at D at the same time that the
piece T arrives at Q, which is easily deduced from the construction,
it will be evident as a consequence that the partial wave generated at
the point D will touch