hence B β = BA²2 AX
But as above AX = BAtan ¹⁄₃ BFC
and substituting this value of AX we obtain:
B β = BA² 2 BAtan ¹⁄₃ BFC
hence we have, as in the text,
B β = ¹⁄₂ BA . tan ¹⁄₃ BFC
The second mode proposed by M. Fresnel, and that which I found most convenient in practice, consists in forming successive hypotheses as to the length of the side BC, and tracing the path of the incident ray FB, which being refracted at B, so as to make with the normal BK an angle = BKY = y′, and finally reflected in the direction BC, must make the angle YBZ = MBC. I shall describe it as follows: In the annexed figure ([fig. 75]) MBZ is a tangent to the reflecting surface at B, and KBF is the angle of incidence of the ray BF before its refraction at B. If KBF = x´, and the angle of incidence of FC = ECF = x, we have BFC (which is the inclination of those rays to each other, and must be equal to the difference of their angles of incidence to the same surface) = x - x´, whence knowing x, we easily find a value of x´ corresponding to the length of BC. Then for finding the angle of refraction KBY = y´ we have:
sin y´ = sin x´m
Fig. 75.
