Page:Optics.djvu/102

 which is the focus of the opposite hyperboloid, and conversely, that rays diverging from a point may be refracted by such a lens so as to become parallel.

104.Sir I. Newton has given in the 14th section of the first volume of his Principia, two curious propositions relating to the present subject, which I will insert here, to save the trouble of referring to the book itself.

The first is:

To find the form of the surface of a medium, which will refract rays diverging from a point without it accurately to a point within itself.

Let $$A$$ (Fig. 101.) be the focus of the incident rays; $$B$$ that of the refracted; $$CD$$ the section of the surface; $$AD, \ DB$$ an incident and refracted ray, $$DE$$ an infinitely small portion of the curve; $$EF, \ EG$$ perpendiculars on $$AD, \ DB. $$

Now $$\frac{DF}{DE}= \cos EDF = \mathrm{sin~incid.}$$

$$\frac{DG}{DE}= \cos EDG = \mathrm{sin~refr.}$$

therefore if $$m$$ be as usual the refracting power, $$DF=m.DG;$$ but $$DF, \ DG$$ are the corresponding increments of $$AD, \ BD$$ so that if we call these $$u, \ v,$$ we have this differential equation to the curve,

$$du=-m.dv, \quad \mathrm{or}du+mdv=0$$

When $$B$$ is removed to an infinite distance, the equation becomes $$du=mdx$$ if $$x=AN,$$ $$DN$$ being perpendicular to $$AC;$$ now we shall see that this equation belongs to an hyperbola, or an ellipse, according as $$m$$ is greater or less than unity,

$$du=mdx$$ gives, by integration, $$u=mx+n,$$

that is, $$x^2+y^2=(mx+n)^2;$$

$$\therefore y^2=(m^2-1) x^2 +2 mx+n^2.$$

Now the equation to an hyperbola, when the abscissa is measured from the farther focus, is