tan i′₁ = BAAF = r₁φ; tan i′₂ = EAAF = r₂φ
sin e₁ = sin i′₁μ; sin e₂ = sin i′₂μ
r′₁ = r₁ - t″ sin e₁; r′₂ = r₂ - t″ sin e₂
tan i₁ = r′₁φ; tan i₂ = r′₂φ
sin ε = sin i₁μ; sin ε′ = sin i′₂μ
sin α = μ sin ε √μ² - 2 μ cos ε - 1; and sin α′ = μ sin ε′ √μ² - 2 μ cos ε′ + 1
Now, the angle b C m = α - α′ from which (since the triangle b m C is isosceles) b m C = 90° - ¹⁄₂ (α - α′); also, in the triangle b m e, the angle b m e = b m C - e m C = 90° - ¹⁄₂ (α - α′) - η and b e m = k′ e E′ = 90° - ε′
We have therefore in the triangle b m e
b m = b e sin b e m sin b m e = l cos ε′cos {η + ¹⁄₂ (α - α′)}
and in b m C