The point D thus found will be one of those through which the curve ought to pass; and the proof will be the same as before.

And by this it will be proved that the waves which come from the point L, after having passed through the glass KAKB, will take the form of straight lines, as BC; which is the same thing as saying that the rays will become parallel.

Whence it follows reciprocally that parallel rays falling on the surface KDB will be reassembled at the point L.
[Illustration] Again, let there be given the surface AK, of any desired 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.