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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.
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.
Then also as 3 to 2 will the sum of the one set be to the sum of the other; that is to say, TF to AS, and DE to AK, and BE to SK or DV, supposing V to be the intersection of the curve EK and the ray FO. But, making FB perpendicular to DE, the ratio of 3 to 2 is also that of BE to the semi-diameter of the spherical wave which emanated from the point F while the light outside the transparent body traversed the space BE. Then it appears that this wave will intersect the ray FM at the same point V where it is intersected at right angles by the curve EK, and consequently that the wave will touch this curve.
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.
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