Page:Optics.djvu/58

 of the different series will coincide; for if $$ \frac{\pi}{\iota}$$ be an even number, some one of the distances $$OO$$ or $$OO_{} \ (2 \iota), \ OO^{iv}$$ or $$OO_{iv} \ (4 \iota),$$ &c. will become equal to $$\pi$$, so that $$O^{(2n)}$$ and $$O_{(2n)}$$ meet at the opposite point of the circle from $$O$$; and if $$\frac{\pi}{\iota}$$ be an odd number, we must have $$(2 n+1) \iota = \pi,$$

that is, $$4 n \iota + 2 \iota = 2 \pi,$$

that is, $$2 n \iota + 2 \theta + 2n \iota + 2 \theta' = 2 \pi,$$

that is, $$OO^{(2n+1)} + OO_{(2n+1)} = 2 \pi.$$

It appears then upon the whole, that taking in the object $$O$$, the whole number of points visible will be $$ \frac{2 \pi}{\iota}$$.

42.Let us now consider the images produced by spherical reflectors. We must of course find the focus of reflected rays corresponding to each point in the object, and consider the figure which all such foci compose by their aggregation.

As a first instance, suppose there be presented to a spherical mirror, a portion of a sphere concentric with itself, (Fig. 38.). All the points of this are equidistant from the centre of the sphere, and if they be considered as foci of incident rays, the foci of reflected rays will all be at some other distance from the centre, on the same radii with them, so that the image will be a portion of a sphere like the object.

43.Let a plane object be placed in front of a spherical mirror, so as to be perpendicular to its axis; required the form of the image.

Fig. 39, represents a section of the object and mirror, (a concave one) through the axis. $$PQ$$ is a part of the section of the object having for its image $$pq, \ q$$ being the focus of reflected rays answering to $$Q,$$ and consequently lying on the same diameter with it.

In order to determine the form of the image $$pq,$$ we will consider it as a spiral curve referred to the centre $$E$$ as a pole, and $$EA$$ as an axis.