Page:Optics.djvu/101

 rays may be refracted so as to meet accurately in one point without any aberration.

This will easily be seen from simple Geometrical considerations.

Let $$AM$$ (Fig. 99.) be the axis major of an ellipse $$ARM; \ S, \ H$$ the foci, $$nRN$$ a normal, $$QR$$ parallel to $$AM$$.

Then since the angles $$HRN,$$ $$SRN$$ are equal,

$$\frac{\sin QRn}{\sin NRS} = \frac{\sin RNH}{\sin NRS} = \frac{\sin RNS +\sin RNH}{\sin NRS +\sin NRH}$$

$$=\frac{RS + RH}{NS+NH}$$

$$=\frac{AM}{SH}. $$

From this it appears that if a transparent spheroid have the ratio of its axis major to its eccentricity equal to its retracting power, rays entering it in a direction parallel to the axis major will be refracted accurately to the farther focus. Moreover if the surface of the ellipsoid be cut by a spherical surface, having that focus for its centre, a lens will be formed which will refract parallel rays accurately to the focus as there will be no refraction in their passage through the spherical surface.

For the same reason, rays diverging from the focus of this lens will be refracted so as to become parallel to the axis.

103.Again, let $$AM,$$ (Fig. 100.) be the axis major of a pair of hyperbolas, $$QR$$ a line parallel to it, $$S, \ H$$ the foci, $$RN$$ a normal,

$$\frac{\sin QRN}{\sin nRS} = \frac{\sin RNS}{\sin NRS} = \frac{\sin RNS -\sin RNH}{\sin NRS -\sin NRH}$$

$$=\frac{RS - RH}{SN-NH}$$

$$=\frac{AM}{SH} $$ as before.

Hence it appears that a plano-convex lens having its convex surface hyperboloidal, will refract parallel rays accurately to a point