[66] The truth of the first of these equations (sin ξ = _m_ . sin γ) which merely expresses the ratio of the sines of the angles of incidence and refraction is obvious; but owing to the great number of small angles about C, a little consideration may be required to enable one to perceive the truth of the second. I therefore subjoin the steps by which I reached it. It is obvious (see fig. 72), that as ACH and BCF are equal, the line SC bisecting HCF must bisect ACB. But the production of AC clearly gives SCD opposite and equal to ACW and SCD is by construction = (α - ξ + γ) = (α + γ - ξ), and, therefore, ACB, which is twice ACW or SCD = (2 α + 2 γ - 2 ξ). Now, by construction OC is a normal to the refracting surface CB and its production C _g_ gives AC _g_ = γ. But γ = ACB - _g_ CB = (2 α + 2 γ - 2 ξ) - _g_ CB = (2 α + 2 γ - 2 ξ) - 90°, hence γ = {2 α + 2 γ - 2 ξ} - 90°, and γ - 2 γ = -γ = -2 ξ + (2 α - 90°) by transposition, and finally changing signs, we have as above: γ = 2 ξ - (2 α - 90°) = 2 ξ - θ. Eliminating γ between these two equations we obtain: sin ξ = _m_ . sin (2 ξ - θ) an expression, which, after various transformations of circular functions, assumes the form 1 ( 1 ) sin⁴ ξ - --- sin θ . sin³ ξ + (------ - 1) . sin² ξ + _m_ (4 _m_² ) 1 ------ sin θ . sin ξ + ¹⁄₄ sin² θ = 0[67] 2 _m_ [67] This expression is equivalent to that of M. Fresnel, but owing to a simplification in the fractional coefficients, it is not _literally_ the same. I was led to it by the following steps, starting from the original equation sin ξ = _m_ sin (2 ξ - θ) sin ξ = _m_ sin (2 ξ - θ) = _m_ {sin 2 ξ . cos θ - cos 2 ξ sin θ} = _m_ cos θ . sin 2 ξ - _m_ sin θ . cos 2 ξ = _m_ cos θ . 2 sin ξ . cos ξ - _m_ sin θ . {1 - 2 sin² ξ} = 2 _m_ cos θ . sin ξ . cos ξ - _m_ sin θ + 2 _m_ sin θ . sin² ξ. Therefore, _m_ sin θ + sin ξ - 2 _m_ sin θ sin² ξ = 2 _m_ cos θ . sin ξ . cos ξ. Then: _m_² sin² θ + 2 _m_ sin θ . sin ξ - 4 _m_² sin² θ sin² ξ + sin² ξ - 4 _m_ sin θ . sin³ ξ + 4 _m_² sin² θ . sin⁴ ξ = 4 _m_² cos² θ sin² ξ (1 - sin² ξ) = 4 _m_² . cos² θ . sin² ξ - 4 _m_² cos² θ sin⁴ ξ. Hence we have: _m_² sin² θ + 2 _m_ sin θ . sin ξ + (1 - 4 _m_²) . sin² ξ - 4 _m_ sin θ . sin³ ξ + 4 _m_² sin⁴ ξ = 0 Then dividing by 4 _m_² and arranging according to powers of ξ, we have as above: 1 ( 1 ) 1 sin⁴ ξ - --- sin θ . sin³ ξ + (------ - 1) . sin² ξ + ----- . sin θ . _m_ (4 _m_² ) 2 _m_ sin ξ + ¹⁄₄ sin² θ = 0