Let then (in Fig. 7. Tab. 5.) BEβ be a double Convex Lens, C the Center of the Segment EB, and K the Center of the Segment Eβ, Bβ the thickness of the Lens, D a Point in the Axis of the Lens; and it is required to find the Point F, at which the Beams proceeding from the Point D, are collected therein, the Ratio of Refraction being as m to n. Let the distance of the Object DB = DA = d (the Point A being supposed the same with B, but taken at a distance therefrom, to prevent the coincidence of so many Lines) the Radius of the Segment towards the Object CB or CA = r, and the Radius of the Segment from the Object Kβ or K = ρ; and let Bβ the thickness of the Lens be = t, and then let the Sine of the Angle of Incidence DAG be to the Sine of the refracted Angle HAG or CAφ as m to n: And in very small Angles, the Angles themselves will be in the same proportion; whence it will follow that,
As d to r, so the Angle at C to the Angle at D, and d + r will be as the Angle of Incidence GAD; and again as m to n, so d + r to dn + rn / m, which will be as the Angle GAH = CAφ; This being taken from ACD which is as d, will leave m - nd - nr / m analogous to the Angle AφD; and the Sides being in this Case proportional to the Angles they subtend, it will follow, that as the Angle AφD is to the Angle ADφ, so is the Side AD or BD to Aφ or Bφ: That is, Bφ will be = mdr / m - nd - nr, which shews in what Point the Beams proceeding from D, would be collected by means of the first Refraction; but if nr cannot be subtracted from m - nd, it follows that the Beams after Refraction do still pass on diverging, and the Point φ is on the same side of the Lens beyond D. But if nr be equal to m - nd, then they proceed parallel to the Axis, and the Point φ is infinitely distant.
The Point φ being found as before, and Bφ - Bβ being given, which we will call δ, it follows by a Process like the former, that βF, or the focal Distance sought, is equal to δρn / m - δ + mρ = f. And in the room of δ substituting Bφ - Bβ = mdr / m - nd - nr - t, putting p for n / m - n, after due Reduction this following Equation will arise, mpdrρ - ndρt + nprρt / mdr + mdρ - mprρ - m - ndt + nrt = f. Which Theorem, however it may seem operose, is not so, considering the great Number of Data that enter the Question; and that one half of the Terms arise from our taking in the thickness of the Lens, which in most Cases can produce no great Effect; however it was necessary to consider it, to make our Rule perfect. If therefore the Lens consist of Glass, whose Refraction is as 3 to 2 'twill be 6drρ - 2dρt + 4rρt / 3dr + 3dρ - 6rρ - dt + 2rt = f. If of Water, whose Refraction is as 4 to 3, the Theorem will stand thus 12drρ - 3dρt + 9rρt / 4dr + 4dρ - 12rρ - dt + 3rt = f. If it could be made of Diamant, whose Refraction is as 5 to 2, it would be 10⁄3drρ - 2dρt + 4⁄3rρt / 5dr + 5dρ - 10⁄3rρ - 3dt + 2rt = f. And this is the universal Rule for the Foci of double Convex Glasses exposed to diverging Rays. But if the thickness of the Lens be rejected, as not sensible, the Rule will be much shorter, viz. pdrρ / dr + dρ - prt = f, or in Glass 2drρ / dr + dρ - 2rρ = f, all the Terms wherein t is found being omitted, as equal to nothing. In this Case, if d be so small, as that 2rρ exceed dr + dρ, then will it be - f, or the Focus will be Negative, which shews that the Beams after both Refractions still proceed diverging.
To bring this to the other Cases, as of converging Beams, or of Concave Glasses, the Rule is ever composed of the same Terms, only changing the Signs of + and -; for the distance of the Point of Concourse of converging Beams, from the Point B, or the first Surface of the Lens, I call a negative Distance or - d; and the Radius of a Concave Lens I call a negative Radius, or - r if it be the first Surface, and - ρ if it be the second Surface. Let then converging Beams fall on a double Convex of Glass, and the Theorem will stand thus - 2drρ / - dr - dρ - 2rt = + f, which shews that in this Case the Focus is always affirmative.
If the Lens were a Meniscus of Glass, exposed to diverging Beams, the Rule is - 2drρ / - dr + dρ + 2rρ = f, which is affirmative when 2rρ is less than dr - dρ otherwise negative: But in the Case of converging Beams falling on the same Meniscus, 'twill be + 2drρ / + dr - dρ + 2rp = f, and it will be + f, whilst dρ - dr is less than 2rρ; but if it be greater than 2rρ, it will always be found negative or - f. If the Lens be double Concave, the Focus of converging Beams is negative, where it was affirmative in the Case of diverging Beams on a double Convex, viz. - 2drρ / + dr + dρ - 2rρ = f, which is affirmative only when 2rρ exceeds dr + dρ: But diverging Beams passing a double Concave, have always a negative Focus, viz. - 2drρ / + dr + dρ + 2rρ = - f.
The Theorems for converging Beams, are principally of use to determine the Focus resulting from any sort of Lens placed in a Telescope, between the Focus of the Object-Glass and the Glass it self; the distance between the said Focus of the Object-Glass, and the interposed Lens being made = - d.
I here suppose my Reader acquainted with the Rules of Analytical Multiplication and Division, as that + multiplied by + makes the Product +, + by - makes -, and - by - makes +; so dividing + by + makes the Quote +, + by - makes -, and - by - makes +; which will be necessary to be understood in the preceding Examples.
In case the Beams are parallel, as coming from an infinite distance, (which is supposed in the Case of Telescopes) then will d be supposed Infinite, and in the Theorem pdρr / dr + dρ - prρ the Term prρ vanishes, as being finite, which is no part of the other infinite Terms; and dividing the Remainder by the infinite Part d, the Theorem will stand thus pρr / r + ρ = f, or in Glass, 2rρ / r + ρ = f.
In case the Lens were Plano-Convex exposed to diverging Beams, instead of pdρr / dr + dρ - prρ, r being infinite, it will be pdρ / d - pρ = f, or 2dρ / d - 2ρ if the Lens be Glass.
If the Lens be Double-Convex, and r be equal to ρ, as being formed of Segments of equal Spheres, then will pdρr / dr + dρ - prρ be reduced to pdr / 2d - pr f; and in case d be infinite, then it will yet be farther contracted to ½pr, and p being = n / m - n, the focal distance in Glass will be = r, in Water 1½r, but in Diamant ⅓r.