(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_ FC ------, 2 _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: B β = ¹⁄₂ AB . tan ¹⁄₃ BFC.[68] The value of B _b_ gives, of course, the direction of the second tangent Z _b_ (which must be equal in length to AZ), whence we easily deduce the chord of the reflecting side A _b_. [68] The following steps will shew the mode of obtaining this expression: Suppose (fig. 74, on opposite page) F _n_ to be a ray incident on the surface BC very near _b_ or B (which, although exaggerated in the figure for more easy reference, are close together), and let this ray F _n_ be refracted in the direction _n_ O, and draw _n n′_ parallel to CA, the ray which is refracted at C, then will _n′_ _n_ O = _m_ . _b_ FC = ²⁄₃ _b_ FC. But the tangent AZ should make with the tangent _b_ Z an angle equal ¹⁄₃ _b_ FC, or _one-half_ the inclination of the rays refracted at _b_ and C, which are afterwards by the agency of those tangents, to be reflected in the directions parallel to _b_ C and to each other. Hence we have AX _b_ (which is the inclination of the normals to those tangents), _b_ FC _b_ FC or AX _b_ = BZ _b_ = ------ = ------ nearly. 2 _m_ 3 [Illustration: Fig. 74.] But putting AXB (fig. 73, p. 277) for AX _b_, and BFC for _b_ FC, a supposition which may be safely made when the differences are so small, and founding upon the analogy AX∶ AB ∷ R ∶ tan AXB, we have BA = AX . tan AXB = AX . tan ¹⁄₃ BFC. Then AB² = B β (B β + 2 AX) = B β² + 2 B β . AX and neglecting B β², which is very small, we have: BA² = B β . 2 AX nearly, BA² hence B β = ---- 2 AX BA But as above AX = ------------- tan (¹⁄₃ BFC) and substituting this value of AX we obtain: BA² B β = ----------------- ( BA ) (2 -------------) ( tan (¹⁄₃ BFC)) hence we have, as in the text, B β = ¹⁄₂ BA . tan(¹⁄₃ BFC)