Page:Optics.djvu/57



The number of images is not unbounded in this case, as in that of two parallel mirrors, for when any one of the images, as $OO_{′}$, or $=$ gets between the lines $2KO=2θ′$, $OO_{″}$ produced, no further reflexion can take place, as no rays proceeding from such a point could fall on the face of either of the mirrors.

In order to express this condition algebraically, we must observe, that of the first series of images, $=$, $2HO_{′}−OO_{′}=HO_{′}+HO=2HK=2ι$, $OO_{‴}$, &c. lie on one side of the mirrors, and $=$, $2KO_{″}−OO_{″}=KO_{″}+KO=OO_{″}+2KO$, $=2ι+2θ′$, &c. on the other, and that if $O^{(n)}$, for instance, be the last, the distance $O_{(n)}$, or $HI$, that is, $KI$, must be the first that is greater than $O′$ or $O‴$,

If $O^{v.}$ be the last image, we must have $O″$;

the same expression as before, $O^{iv.}$ being the number of images in the one case, and $O^{vi.}$ in the other.

In like manner we should find, that the number of images in the other series is the least whole number greater than $O^{(2n+1)}$.

If $KO^{(2n+1)}$ be a measure of $2nι+2θ+θ′$, since $2nι+ι+θ′$ and $Kk$ are proper fractions, the number of images in each series must be $π$, and therefore the whole number of images $(2n+1)ι+θ>π$. In this case, however, two images