Now, since BB′ = b b′ nearly, AB′ = AB - b b′ or r′ = r - t″ tan e.
From this is obtained the angle of incidence i, and the first angle of refraction ε; for tan i = r′φ and sin ε = sin iμ.
| Next B′ b C = B b C - B b B′ or η = α - ε and sin α = μ sin η = μ sin (α - ε) |
| from which, sin α cos ε - cos α sin ε = sin α μ |
| whence sin α (cos ε - 1 μ) = cos α sin ε; and |
| sin² α (cos² ε - 2 cos εμ + 1μ²) = cos² α sin² ε = (1 - sin² α) sin² ε = sin² ε - sin² α sin² ε |
Then transposing we have
sin² α {(cos² ε + sin² ε) - 2 cos εμ + 1μ²} = sin² ε
and because (cos² ε + sin² ε) = 1 we have, by dividing,
sin² α = sin² ε {1 - 2 cos εμ + 1μ²} = μ² sin² ε μ² - 2 μ cos ε + 1
and sin α = μ sin ε √1 - 2 μ cos ε + μ²
Next, since b C = a′ b sin AC b = rsin α, putting C b = ρ″, and substituting we have ρ″ = rμ sin ε √μ² - 2 μ cos ε + 1
and, taking for the radius of curvature, the mean of ρ′ and ρ″ the values calculated for the central and marginal rays, we have finally ρ = ρ′ + ρ″ 2