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