## Abstract

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### Tables (3)

Table 1 Concentric Meniscus

Table 2 Form of Lens

Table A Exhibiting results of numerical calculations in various special cases, most of which are referred to in the text. (Distances all given in centimeters.)

### Equations (84)

$r 1 = A 1 C 1 , r 2 = A 2 C 2 , n c = d = A 1 A 2 ,$
$R 1 = 1 / r 1 and R 2 = 1 / r 2 .$
$F 1 = ( n − 1 ) R 1 , F 2 = − 2 n R 2 , F 3 = 2 n R 1 , F 4 = − ( n − 1 ) R 2 ,$
$A 1 F = Y 1 F , A 1 H = 1 − Y 1 F , A 2 F ′ = X 4 F , A 1 H ′ = X 4 − 1 F .$
$− 1 12$
$1 9$
$11 12$
$19 36$
$10 9$
$− 4 9$
$13 9$
$− 7 36$
$2 3$
$5 4$
$2 3$
$F = 10 , X 4 = 5 4 , Y 1 = 2 3 ,$
$A 1 F = − 1 15 , A 1 H = 1 30 , A 2 F ′ = 1 8 , A 2 H ′ = 1 40 ,$
$FH = FA 1 + A 1 H = H ′ A 2 + A 2 F ′ = 1 F = f ,$
$A 1 A 2 = 0.83 , A 2 C 1 = A 2 C 2 = A 2 H = A 2 H ′ = 2.50 , A 1 F = − 6.67 , A 2 F ′ = 12.50 ;$
$X 4 = 1 − c ( 3 F 1 + 2 F 2 + F 3 ) + c 2 ( 2 F 1 F 2 + F 2 F 3 + 2 F 3 F 1 ) − c 3 F 1 F 2 F 3 ; Y 1 = 1 − c ( 3 F 4 + 2 F 3 + F 2 ) + c 2 ( 2 F 4 F 3 + F 3 F 2 + 2 F 2 F 4 ) − c 3 F 4 F 3 F 2 ; F = F 1 + F 2 + F 3 + F 4 − c ( F 1 F 2 + 2 F 1 F 3 + F 2 F 3 + 3 F 1 F 4 + F 2 F 4 + F 3 F 4 ) + c 2 ( F 1 F 2 F 3 + 2 F 1 F 2 F 4 + 2 F 1 F 3 F 4 + F 2 F 3 F 4 ) − c 3 F 1 F 2 F 3 F 4 .$
$n = 1.5 ,$
$d = 1.5 c , F 1 = 0.5 R 1 , F 2 = − 3 R 2 , F 3 = 3 R 1 , F 4 = − 0.5 R 2 , ( n = 1.5 ) .$
$∊ = R 2 R 1 , ϕ = F R 1 , x = c ⋅ R 1 ,$
$ϕ ⋅ x = c ⋅ F , ϕ ⋅ R 2 = ∊ ⋅ F , ∊ x = c ⋅ R 2 .$
$X 4 = 1 + 3 ( 4 ∊ − 3 ) 2 x + 3 ( 1 − 4 ∊ ) x 2 + 9 ∊ 2 x 3 , Y 1 = 1 − 3 ( 4 − 3 ∊ ) 2 x + 3 ∊ ( ∊ − 4 ) x 2 − 9 ∊ 2 2 x 3 = X 4 − 3 ( 1 + ∊ ) x { 1 + 2 ( 1 − ∊ ) x + 3 ∊ x 2 } 2 ,$
$A ∊ 2 + B ∊ + C = 0 ,$
$A = 3 x ( 3 x − 2 ) ( x − 2 ) , B = 24 x 4 − 51 x + 14 , C = 2 ( 6 x − 7 + 2 ϕ ) .$
$Δ = B 2 − 4 A C ,$
$∊ = − B ± ( Δ ) 1 / 2 2 A .$
$F 1 = 0.25000 , F 2 = − 0.74820 , F 3 = 1.50000 , F 4 = − 0.12470.$
$A 1 A 2 = 0.30000 , A 1 C 1 = 2.00000 , A 2 C 2 = 4.00962 , A 1 F = − Y 1 / F = − 0.57095 , A 1 H = ( 1 − Y 1 ) / F = 0.42905 , A 2 F ′ = X 4 / F = 0.82167 , A 2 H ′ = ( X 4 − 1 ) / F = − 0.17833.$
$A 1 A 2 = 3.00 , A 1 C 1 = 20.00 , A 2 C 2 = 40.10 , A 1 H = 4.29 , A 1 F = − 5.71 , A 2 H ′ = − 1.78 , A 2 F ′ = 8.22 ;$
$ϕ x = − 1 30 , ϕ = 7 − 6 x 2 .$
$x = − 0.00944722 , ϕ = − r 1 = + 3.52837.$
$F = − 10 dptr : A 1 A 2 = 0.50 , A 1 C 1 = − 35.28 , A 1 C 2 = ∞ , A 1 H = 0.57 , A 1 F = 10.57 , A 2 H ′ = − 0.43 , A 2 F ′ = − 10 .43 ;$
$9 x 3 − 48 x 2 + 75 x − 28 + 4 ϕ = 0 .$
$135 x 4 − 720 x 3 + 1125 x 2 − 420 x + 2 = 0 .$
$ϕ = r 1 = − r 2 = 6.9099065.$
$X 4 = − Y 1 = 0.94970 , F = 1 .$
$9 x 3 + 24 x 2 − 42 x − 28 + 16 ϕ = 0.$
$135 x 4 + 360 x 3 − 360 x 2 − 420 x + 8 = 0 ,$
$X 4 = 0.971177 , Y 1 = 0.928672.$
$9 ∊ 2 x 3 + 24 ∊ ( 1 − ∊ ) x 2 + { 12 ( 1 + ∊ 2 ) − 51 ∊ } x − 14 ( 1 − ∊ ) + 4 ϕ = 0 .$
$12.15 x 4 + 75.6 x 3 − 33.3 x 2 − 147 x + 2 = 0.$
$2 ϕ = 7 − 6 x ,$
$X 4 = Y 1 = 1 , ( c = 0 ) ;$
$F = F 1 + F 2 + F 3 + F 4 = 7 ( R 1 − R 2 ) / 2 .$
$Δ ≡ ( p x 2 + q x + m ) 2 + μ x 4 + α x 3 + β x 2 + γ x + δ − 48 x ( x − 2 ) ( 3 x − 2 ) ϕ ,$
$μ = 144 − p 2 , α = − 2 ( p q + 396 ) β = 1353 − q 2 − 2 p m , γ = 2 ( q m + 378 ) , δ = 196 − m 2 .$
$ϕ = μ x 4 + a x 3 + β x 2 + γ x + δ 48 x ( x − 2 ) ( 3 x − 2 ) ,$
$Δ = ( p x 2 + q x + m ) 2$
$p = α = 0 , q = α = 0 , q = γ = 0 and m = δ = 0 ,$
$ϕ = α x 2 + β x + γ 48 ( x − 2 ) ( 3 x − 2 ) , p = 12 α = − 24 ( q + 33 ) ; and either { m = + 14 , β = 1017 − q 2 , γ = − 28 ( 27 + q ) , or { m = − 14 , β = 1689 − q 2 , γ = − 28 ( 27 − q ) ; Δ = ( 12 x 2 + q x + 14 ) 2 .$
$− 3 x 3 x − 2$
$− 2 ( 3 x − 7 ) 3 x − 2$
$− 3 x − 7 2 ( x − 2 ) ( 3 x − 2 )$
$− 10 x x − 2$
$3 x − 7 3 ( 3 x − 2 )$
$5 ( 5 x − 14 ) 2 ( x − 2 ) ( 3 x − 2 )$
$5 x − 14 3 ( x − 2 )$
$− 5 ( 6 x 2 − 4 x − 49 ) 6 ( x − 2 ) ( 3 x − 2 )$
$ϕ = B x 2 + γ x + δ 48 x ( x − 2 ) ( 3 x − 2 ) , p = ± 12 , q = ∓ 33 , β = 24 ( 11 ∓ m ) γ = 6 ( ± 11 m − 126 ) , δ = 196 − m 2 ,$
$ϕ = x ( μ x 2 + α x + β ) 48 ( x − 2 ) ( 3 x − 2 ) , q = ± 27 , m = ∓ 14 , α = − 18 ( 44 ± 3 p ) , β = 4 ( 15 ± 7 p ) , μ = 144 − p 2 ,$
$144 ϕ = u x + v ,$
$7 − 6 x 2$
$5 ( 7 − 3 x ) 12$
$3 ( 5 x − 14 ) 16$
$45 x − 154 48$
$ϕ = − 3 x 3 x − 2 ,$
$∊ = − 2 3 x − 2 .$
$ϕ = − 10 x x − 2 , ∊ = − 2 x − 2 ,$
$c = 2 R 1 ( 1 + 10 R 1 ) , R 2 = 1 + 10 R 1 10 , Y 1 = 50 R 1 2 − 15 R 1 − 12 50 R 1 2 X 4 = 10 ( 10 R 1 2 + 3 R 1 − 2 ) ( 10 R 2 1 + 1 ) .$
$A 1 C 1 = 5.95 , A 2 C 2 = 5.61 , A 1 H = 2.63 , A 1 F = − 7.37 , A 2 H ′ = 0.93 , A 2 F ′ = 10.93$
$ϕ = − 2 ( 3 x − 7 ) 3 x − 2 ,$
$c = 2 ( 7 R 1 + 1 ) 3 R 1 ( 2 R 1 + 1 ) , R 2 = − 3 ( 2 R 1 + 1 ) 5 , Y 1 = 6 R 1 + 1 R 1 , X 4 = − 2 R 1 2 − 143 R 1 − 22 3 ( 2 R 1 + 1 ) 2 ;$
$A 1 C 1 = − 69.29 , A 2 C 2 = − 23.43 , A 1 H = 0.71 , A 1 F = − 9.29 , A 2 H ′ = 270.60 , A 2 F ′ = 280.60 ;$
$c = H ′ 1 A 2 n$
$X 1 = 1 , F 11 = F 1 = F 3 , X 2 = X 1 − c ⋅ F 11 , F 12 = F 11 + X 2 ⋅ F 2 , X 3 = X 2 − c ⋅ F 12 , F 13 = F 12 + X 3 ⋅ F 3 ; F = F 13 , H 1 F = H 1 F ′ = − X 3 / F , H 1 H = H 1 H ′ = ( 1 − X 3 ) / F , H 1 K = ( 1 − X 3 ) / F .$
$F 1 = F 3 = ( n − 1 ) R 1 , F 2 = − 2 n R 2 , A 1 F = A 1 F ′ = − X 3 / F , A 1 H = ( 1 − X 3 ) / F , A 1 K = − ( 1 + X 3 ) / F .$
$F 1 = F 3 = R 1 / 2 , F 2 = − 3 R 2 , ( n = 1.5 ) .$
$∊ = R 2 R 1 , ϕ = F R 1 , x = c R 1 ,$
$ϕ x = c F , ∊ = 2 ( 2 − x − 2 ϕ ) 3 ( x − 2 ) 2 .$
$3 ∊ x 3 − 2 ( 6 ∊ − 1 ) x 2 + 4 ( 3 ∊ − 1 ) x + 4 / k = 0 .$
$A 1 A 2 = 1 cm , A 1 C 1 = − 50.725 cm , A 2 C 2 = − 25.363 cm , A 1 H = 0.662 cm , A 1 K = − 19.338 cm .$