32
| NO″EN=96=1.5 | NEO″=tan−11.5=56°19′12; |
| ∴ OEO″=65°46′. | |
The distances EO′, EO″, &c. are of course the secants of these angles to the radius EN.
41. Suppose now that the mirrors instead of being parallel are inclined to each other, as HI, IK. (Fig. 37.)
In this case, the images of an object O will no longer be in the same straight line, but it will easily be seen, that they will be all at the same distance from the intersection of the mirrors; for if IO, IO′, for example, be joined, the two right-angled triangles IHO, IHO′ are exactly equal in all respects.
There are of course here as before, two series of images,
O′, O″, O‴, &c.O′, O″, O‴, &c.
their number will not, however, be unlimited, as we shall see.
In order to determine their places, we must find the values of the angles OIO′, OIO″, &c. or of the arcs OO′, OO″, which measure them in the circle round I, containing all the images.
| Let HIO or HO | =θ, the radius begin taken as the unit, |
| OIK or OK | =θ′, |
| HIK or HK | =ι. |
| Then OO′ | = | 2HO=2θ, |
| OO″ | = | 2KO′−OO′=KO′+KO=2KH=2ι, |
| OO‴ | = | 2HO″−OO″=HO″+HO=OO″+2HO |
| 2HO =2ι+2θ, | ||
| OOiv. | = | 2KO‴−OO‴=KO‴+KO=OO‴+2KO |
| 2HO =2ι+2θ+2θ′=4ι, | ||
| &c.&c. | ||
