| When C = d, MP = | DC |
| 2C |
As the summit plane op rises above the base mn by the successive convergency of the optic axes to different points in the line onp, it may be asked how it happens that the conical frustum still appears a solid, and the plane op where it is, when the optic axes are converged to points in the line mmn, so as to see the base distinctly? The reason of this is that the rays emanate from op exactly in the same manner, and form exactly the same image of it, on the two retinas as if it were the summit cd, [Fig. 19], of the real solid when seen with both eyes. The only effect of the advance of the point of convergence from n to m is to throw the image of n a little to the right side of the optic axis of the left eye, and a little to the left of the optic axis of the right eye. The summit plane op will therefore retain its place, and will be seen slightly doubled and indistinct till the point of convergence again returns to it.
It has been already stated that the two dissimilar pictures may be united by converging the optical axes to a point beyond them. In order to do this, the distance ss′ of the pictures, [Fig. 21], must be greatly less than the distance of the eyes l, r, in order that the optic axes, in passing through similar points of the two plane pictures, may meet at a moderate distance beyond them. In order to explain how the relief is produced in this case, let ab, cd, ab, cd, [Fig. 22], be the dissimilar pictures of the frustum of a cone whose summit is cd, as seen by the right eye, and cd as seen by the left eye. From l and r, as before, draw lines through all the leading points of the pictures, and we shall have the points a, a united at m, the points b, b at n, the points c, c at o, and the points d, d at p, the points s, s at m, and the points s′, s′ at n, forming the cone mnop, with its base mn towards the observer, and its summit op more remote. If the cone had been formed of lines drawn from the outline of the summit to the outline of the base, it would now appear hollow, the inside of it being seen in place of the outside as before. If the pictures ab, ab are made to change places the combined picture would be in relief, while in the case shewn in [Fig. 21] it would have been hollow. Hence the right-eye view of any solid must be placed on the left hand, and the left-eye view of it on the right hand, when we wish to obtain it in relief by converging the optic axes to a point between the pictures and the eye, and vice versa when we wish to obtain it in relief by converging the optic axes to a point beyond the pictures. In every case when we wish the combined pictures to represent a hollow, or the converse of relief, their places must be exchanged.
Fig. 22.
In order to find the height mn, or rather the depth of the cone in [Fig. 22], let d, d, c, c, represent the same quantities as before, and we shall have
| MP = | Dd |
| C - d |
| NP = | Dd′ | , | and |
| C - d′ | |||
| OP = | Dd′ | - | Dd |
| C - d′ | C - d | ||
When d, c, d, d′ have the same values as before, we shall have MN = 18·7 feet!
When c = d, mp will be infinite.