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 _x´_ sin _y´_ = -------- _m_ [Illustration: Fig. 75.] Now, if FB be refracted, so as to make with the reflecting side an angle equal to ZBY, it must (if the position of B be rightly chosen), be reflected so as to follow BC, thus making MBC = YBZ, and calling each of these angles = μ, we have the right angle NBZ made up of μ + _y´_ + NBK. But NBK clearly equals μ, because it is the inclination of the normals to BC and BZ, and hence _y´_ + 2 μ = 90°. This, therefore, forms a crucial test for the length of BC. I may only remark, that we already know the numerical value of _y_; and that of μ is easily found, for μ = CBA + ABM = CBA + BAM = CBA + (MAC - BAC) = CBA + ¹⁄₂ (180° - υ) - BAC. Thus knowing μ and _y´_, we have only to see whether (_y´_ - 2 μ) - 90° = 0 We have now only to find the length of the radius AX or _b_ X (see fig. 73, p. 277), which will describe the reflecting surface or arc AZ _b_, and to determine the position of its centre X. We already know the values of _y′_ and _y_, the angles of refraction of C and _b_, and their difference _y_ - _y′_ gives us the inclination of the rays which are to be reflected (into directions parallel to C _b_) at _b_ and at A. This quantity is, of course, double the inclination of tangents to the reflecting surface AZ and _b_ Z, and of their normals AX and _b_ X. Again, we have the chord line sin (AC _b_) A _b_ = AC . ------------; sin (_b_ CA) and, as above, AX _b_ = ¹⁄₂ (_y_ - _y′_) = φ sin (¹⁄₂ (180° - φ)) And AX = _b_ X = ρ = A _b_ . ------------------- = sin φ ¹⁄₂ A _b_ . cosec(¹⁄₂ φ). And, lastly, for the co-ordinates to X, the centre of curvature for the reflecting arc, we have OX = ρ . sin OAX and OA = ρ . cos OAX.[69]