If now one considers the point I as at the bottom of the crystal, and that it is viewed by the rays ICR, Icr, refracted equally at the points Cc, which should be equally distant from D, and that these rays meet the two eyes at Rr; it is certain that the point I will appear raised to S where the straight lines RC, rc, meet; which point S is in DP, perpendicular to Qq. And if upon DP there is drawn the perpendicular IP, which will lie at the bottom of the crystal, the length SP will be the apparent elevation of the point I above the bottom.
Let there be described on Qq a semicircle cutting the ray CR at B, from which BV is drawn perpendicular to Qq; and let the proportion of the refraction for this section be, as before, that of the line N to the semi-diameter CQ.
Then as N is to CQ so is VC to CD, as appears by the method of finding the refraction which we have shown above, Article 31; but as VC is to CD, so is VB to DS. Then as N is to CQ, so is VB to DS. Let ML be perpendicular to CL. And because I suppose the eyes Rr to be distant about a foot or so from the crystal, and consequently the angle RSr very small, VB may be considered as equal to the semi-diameter CQ, and DP as equal to CL; then as N is to CQ so is CQ to DS. But N is valued at 156,962 parts, of which CM contains 100,000 and CQ 105,032. Then DS will have 70,283. But CL is 99,324, being the sine of the complement of the angle MCL which is 6 degrees 40 minutes; CM being supposed as radius. Then DP, considered as equal to CL, will be to DS as 99,324 to 70,283. And so the elevation of the point I by the refraction of this section is known.
41. Now let there be represented the other section through EF in the figure before the preceding one; and let CMg be the semi-ellipse, considered in Articles 27 and 28, which is made by cutting a spheroidal wave having centre C. Let the point I, taken in this ellipse, be imagined again at the bottom of the Crystal; and let it be viewed by the refracted rays ICR, Icr, which go to the two eyes; CR and cr being equally inclined to the surface of the crystal Gg. This being so, if one draws ID parallel to CM, which I suppose to be the refraction of the perpendicular ray incident at the point C, the distances DC, Dc, will be equal, as is easy to see by that which has been demonstrated in Article 28. Now it is certain that the point I should appear at S where the straight lines RC, rc, meet when prolonged; and that this point will fall in the line DP perpendicular to Gg. If one draws IP perpendicular to this DP, it will be the distance PS which will mark the apparent elevation of the point I. Let there be described on Gg a semicircle cutting CR at B, from which let BV be drawn perpendicular to Gg; and let N to GC be the proportion of the refraction in this section, as in Article 28. Since then CI is the refraction of the radius BC, and DI is parallel to CM, VC must be to CD as N to GC, according to what has been demonstrated in Article 31. But as VC is to CD so is BV to DS. Let ML be drawn perpendicular to CL. And because I consider, again, the eyes to be distant above the crystal, BV is deemed equal to the semi-diameter CG; and hence DS will be a third proportional to the lines N and CG: also DP will be deemed equal to CL. Now CG consisting of 98,778 parts, of which CM contains 100,000, N is taken as 156,962. Then DS will be 62,163. But CL is also determined, and contains 99,324 parts, as has been said in Articles 34 and 40. Then the ratio of PD to DS will be as 99,324 to 62,163. And thus one knows the elevation of the point at the bottom I by the refraction of this section; and it appears that this elevation is greater than that by the refraction of the preceding section, since the ratio of PD to DS was there as 99,324 to 70,283.
But by the regular refraction of the crystal, of which we have above said that the proportion is 5 to 3, the elevation of the point I, or P, from the bottom, will be 2/5 of the height DP; as appears by this figure, where the point P being viewed by the rays PCR, Pcr, refracted equally at the surface Cc, this point must needs appear to be at S, in the perpendicular PD where the lines RC, rc, meet when prolonged: and one knows that the line PC is to CS as 5 to 3, since they are to one another as the sine of the angle CSP or DSC is to the sine of the angle SPC. And because the ratio of PD to DS is deemed the same as that of PC to CS, the two eyes Rr being supposed very far above the crystal, the elevation PS will thus be 2/5 of PD.
42. If one takes a straight line AB for the thickness of the crystal, its point B being at the bottom, and if one divides it at the points C, D, E, according to the proportions of the elevations found, making AE 3/5 of AB, AB to AC as 99,324 to 70,283, and AB to AD as 99,324 to 62,163, these points will divide AB as in this figure. And it will be found that this agrees perfectly with experiment; that is to say by placing the eyes above in the plane which cuts the crystal according to the shorter diameter of the rhombus, the regular refraction will lift up the letters to E; and one will see the bottom, and the letters over which it is placed, lifted up to D by the irregular refraction. But by placing the eyes above in the plane which cuts the crystal according to the longer diameter of the rhombus, the regular refraction will lift the letters to E as before; but the irregular refraction will make them, at the same time, appear lifted up only to C; and in such a way that the interval CE will be quadruple the interval ED, which one previously saw.