It is only a part of this oval which serves for the refraction, namely, the part DK, ending at K, if AK is the tangent.  As to the, other part, Des Cartes has remarked that it could serve for reflexions, if there were some material of a mirror of such a nature that by its means the force of the rays (or, as we should say, the velocity of the light, which he could not say, since he held that the movement of light was instantaneous) could be augmented in the proportion of 3 to 2.  But we have shown that in our way of explaining reflexion, such a thing could not arise from the matter of the mirror, and it is entirely impossible.

[Illustration]

[Illustration]

From what has been demonstrated about this oval, it will be easy to find the figure which serves to collect to a point incident parallel rays.  For by supposing just the same construction, but the point A infinitely distant, giving parallel rays, our oval becomes a true Ellipse, the construction of which differs in no way from that of the oval, except that FC, which previously was an arc of a circle, is here a straight line, perpendicular to DB.  For the wave of light DN, being likewise represented by a straight line, it will be seen that all the points of this wave, travelling as far as the surface KD along lines parallel to DB, will advance subsequently towards the point B, and will arrive there at the same time.  As for the Ellipse which served for reflexion, it is evident that it will here become a parabola, since its focus A may be regarded as infinitely distant from the other, B, which is here the focus of the parabola, towards which all the reflexions of rays parallel to AB tend.  And the demonstration of these effects is just the same as the preceding.

But that this curved line CDE which serves for refraction is an Ellipse, and is such that its major diameter is to the distance between its foci as 3 to 2, which is the proportion of the refraction, can be easily found by the calculus of Algebra.  For DB, which is given, being called a; its undetermined perpendicular DT being called x; and TC y; FB will be a — y; CB will be sqrt(xx + aa -2ay + yy).  But the nature of the curve is such that 2/3 of TC together with CB is equal to DB, as was stated in the last construction:  then the equation will be between (2/3)y + sqrt(xx + aa - 2ay + yy) and a; which being reduced, gives (6/5)ay — yy equal to (9/5)xx; that is to say that having made DO equal to 6/5 of DB, the rectangle DFO is equal to 9/5 of the square on FC.  Whence it is seen that DC is an ellipse, of which the axis DO is to the parameter as 9 to 5; and therefore the square on DO is to the square of the distance between the foci as 9 to 9 — 5, that is to say 4; and finally the line DO will be to this distance as 3 to 2.

[Illustration]