Dioptrica nova, A treatise of dioptricks in two parts : wherein the various effects and appearances of spherick glasses, both convex and concave, single and combined, in telescopes and microscopes, together with their usefulness in many concerns of humane life, are explained / by William Molyneux of Dublin, Esq. ...

Demonstration.

In Tab. 11. Fig. 2. Let ed be a Plano-Convex Glass, whose absolute Focus we know is about a Diameter of the Convexity, Let sd be this Diameter. a is a Radiating Point, f the Centre of the Convexity, fe the Radius of the Con∣vexity produced directly to q. ae a Ray falling on the Glass, produced directly to y. Here we see the Angle of Inclina∣tion, or Incidence of the Ray a e is q e a.

Let it then be made— I say k is the respective Focus of the Ray a e.	1	a s : s d :: a d : d k
Let k l be made = ½ k d. I shall first Demonstrate, that by virtue of the first Refraction which the Ray suffers at its en∣trance on the Convex-Side of the Glass at e, 'tis directed as if it pro∣ceeded strait towards l.
For to the Consequents of the Analogy in the first step add their Halfs, and it shall be—	2	a s : s f :: a d : d l
And compounding the 2d—	3	a f : s f :: a l : d l
Here we see s f is equal to three Semidiameters of the Convexity, that is, to thrice f e.
The Angle of Refraction is y e l, if therefore we prove that y e l is of q e a the Angle of Incidence, it will be manifest, that by the first Refraction the Ray is directed to∣wards l.

In order to the Proof hereof, we lay down these Suppositions.
1. In the Triangle a e l we sup∣pose le and l d equal, because we suppose the Glass of the least Thickness imaginable, and the Segment of a large Sphere.
2. We suppose likewise, that the Angles of Incidence are all ve∣ry small, that so Sines and An∣gles may be proportional. Tho we could not express this truly in the Figure.
Wherefore, for the Demonstra∣tion of the forgoing Position, in the Triangle a e l it is—	4	a l:l e = l d :: s ∠ a e l:s ∠ a
But the Angle a e l is the Com∣plement of the Angle y e l to 180° Therefore the Sine of the Angle a e l is equal to the Sine of the Angle y e l.
Wherefore the 4th Step runs thus	5	a l:l d :: s. ∠ y e l: s.∠ a But in these small Angles, as the Sines are, so are the Angles.
Therefore the 5th stands thus—	6	a l: l d :: ∠ y e l : ∠ a
Then from 3 and 6 it follows—	7	∠ y e l: ∠ a :: a f:s f = 3 e f
Then in the 7 Triple the Ante∣cedent on this side, and subtriple the Consequent on the other side, and it will be—	8	3 ∠ y e l:∠ a :: a f:e f
Moreover, in the ▵aef it is—	9	a f:e f :: s ∠ a ef:s ∠ a
But the Sine of the Angle a e f is equal to the Sine of the Angle qea, being Complements to 180°.

Wherefore the 9 may stand thus	10	a f : e f :: s ∠ q e a: s ∠ a
But Sines and Angles in these small Angles being proportional, it follows from the 8th and 10th. Steps, That	11	3 ∠ y e l: ∠ a :: ∠ q e a: ∠ a.

Wherefore from the Analogy in the 11th, it is evident, that the Angle of Inclination or Incidence q e a is thrice the Angle of Refraction y e l; seeing three times the Angle y e l, and the Angle q e a, bear the same Proportion to the same Angle a. And this was the first thing to be proved; and consequently the Ray a e by its first Refraction at its Point of Incidence is di∣rected towards the Point l

It remains to be Demonstrated secondly, that the Ray, at its Eruption on the plain side of the Glass into Air, is refracted into e k.

For the Proof of this draw rem Parallel to the Axis. Now the Angle of Incidence from Glass to Air shall be m e l, the Angle of Refraction l e k, which we shall prove to be half the Angle of Inclination m e l; or we shall prove that the Angle m e k is thrice the Angle of Refraction l e k. Which being De∣monstrated, 'tis certain the Ray is refracted into e k.

For the Demonstration hereof we retain the Series of our former Steps, and in the Tri∣angle l e k we have it	12	l e = l d:l k :: s. •• e k l:s •• l e k
But s. •• e k l is equal to s. •• e k f being Complements to 180. Therefore the 12th. Analogy may stand thus	13	l d : l k :: s •• e k f:s •• l e k
And the Angle e k f is = •• m e k Wherefore	14	l xd:l k :: s •• m e k:s •• l e k

Now the Angles being as the Sines, and l k being by Con∣struction the third Part of l d, it follows from the 14th Step,

that the Angle l e k is the third Part of the Angle m e k. Where∣fore the Angle l e k is the Angle of Refraction agreeable to the Angle of Inclination m e l. Which was to be Demonstrated.