Having in this manner determined the angles of BCF, the obtuse angle BCA of the generating triangle of the zone is easily and directly deduced by the following expression, which results from the obvious relations existing among the known angles about C; and we have (see [fig. 73]),
BCA = 90° + γ = 90° + 2 ξ - θ.
We next proceed to consider the form of BA, the reflecting side of the zone, which is a point of the greatest consequence, as an error in the inclination of any part of its surface is doubled in the resulting direction of the reflected rays. The conditions of the question require, that every ray EW, after reflection at the surface AB, shall, like WI, be parallel to the first ray, which is reflected in the direction BC, and after a second refraction at C, emerges horizontally in CH. But, let us trace backwards the rays as they emerge in their horizontal directions IK, and it is obvious that if BA be made a straight line, then will every ray EW meet the first refracting side BC at the same angle, and there suffering the same refraction, they will go on parallel to each other, and never meet in the focus F. This convergence to F, which is a necessary condition of the problem, may, however, be produced by a curvature of AB, such that all the rays shall have a degree of convergence before falling on BC, sufficient to cause them to be finally refracted, so as to meet in F. On this account, they will occupy less space in passing through BC, than they did in passing through AC; and thus BC will be shorter than AC by some quantity which shall give to that part of AB which is at B the amount of downward inclination required for causing the ray BF finally to converge to F; and the line joining B and A must be a curve, every point of which has its tangent inclined so as to serve the same purpose.
Fig. 73.
To trace tangents to this curve, is therefore the next step in the process. The direction of the first tangent AZ depends upon very simple considerations; and all that is necessary to be done is to draw a line AU ([fig. 73]), parallel to BC (which is the parallel to the direction of the reflected rays), and forming an angle CAU, which is, of course, equal to the inclination of the extreme rays refracted by CB at C, with rays reflected from the arc which we have yet to trace. The line AX bisecting this angle, must therefore be a normal to the reflecting surface at A, and AB drawn perpendicular to AX, is consequently a tangent to the reflecting arc.
We must next find the direction of the second tangent Z b, which must be so inclined that the ray F b will, after refraction at b, be reflected into the direction, b C; but as the rigorous determination of this is difficult, I shall describe two approximations suggested to me by M. Leonor Fresnel. The first method is based upon assuming the inclination of the ray refracted at b to the ray refracted at C as equal to:
b FCm
(in which expression, m is the refractive index of the glass); a supposition which obviously differs very little from the truth, as small arcs may be assumed as nearly equal to their sines. Now, it will be recollected, that the rays refracted at C and b, must be reflected at A and b, in a direction parallel to C b, and therefore the inclination of the reflecting surfaces, or that which should be formed by the tangents ZA and Z b, being half that of the incident rays, is, according to the assumption, equal to b FC2 m, which may be expressed by ¹⁄₃ b FC, m being equal to 1·51. But as the inclination of the two radii AX and BX is equal to the inclination of the tangents of the reflecting surfaces to which they are normals, we obtain for the excess B β of the secant of the reflecting arc over its radius the following expression:
Bβ = ¹⁄₂ AB . tan ¹⁄₃ BFC.[68]