Note 58 — PART SECOND. NOTES ON THE ILLUMINATION OF LIGHTHOUSES, WITH SHORT NOTICES OF THEIR EARLY HISTORY. (continued 2)

The following steps will conduct us to this expression:

Fig. 60.

Let B b f E (fig. 60) represent the section of a zone by a plane passing through the axis of the lens AF, C the centre of curvature, F the radiant point, and FB′ b x, FE′m x′ the course of the extreme rays which are transmitted through the zone (and the latter of which passes from E′ to e through a portion of the zone or lens in contact with that under consideration). Then putting

AB = r₁; AB′ = r′₁; C b = ρ

AE = r₂; AE′ = r′₂; B b = t″; BE = r₁ - r₂ = l

ε = the first angle of refraction b B′ k

η = the second angle of refraction B′ b C

ε′ = the first angle of refraction e E k′

η′ = the second angle of refraction e m C

α = the angle of emergence b C q

α′ = the angle of emergence m C q

i′₁ = BFA; i′₂ = EFA; i₁ = B′FA; i₂ = E′FA

e₁ = B b B′; e₂ = E e E′.

Proceeding exactly as in the case of the central lens we shall have

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

b C = b m sin b m C sin b C m = l cos ε′ cos ¹⁄₂ (α - α′) (cos (η + ¹⁄₂ (α - α′)) sin (α - α′)

= l cos ε′ cos ¹⁄₂ (α - α′) cos {η + ¹⁄₂ (α - α′)} 2 sin ¹⁄₂ (α - α′) cos ¹⁄₂ (α - α′)

from which, putting b C = ρ

ρ = l cos ε′ 2 cos {η + ¹⁄₂ (α - α′)} sin ¹⁄₂ (α - α′)