## Abstract

No abstract available.

• View by:
• |
• |
• |

### Cited By

OSA participates in Crossref's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

### Figures (2)

Fig. 1

Die Brechkraftkomponenten ϕ1, ϕ2, ⋯, ϕk, die Höhenverhältnisse ω1, ω2, ⋯, ωk und die Dingschnittweite s1.

Fig. 2

Die Brechkraftkomponenten ϕ1, ϕ2, ⋯, ϕk, die Abstände e1′, e2′, ⋯, ek−1′ und die Dingschnittweite s1.

### Equations (29)

$ω k = h k h 1 = ∏ i = 2 k s i s i - 1 ′ ,$
$β ′ = n 1 n k ′ ∏ i = 1 k s i ′ s i$
$f = n 1 n k ′ ( s 1 ′ ∏ i = 2 k s i ′ s i ) s 1 = ∞ ,$
$ϕ = 1 f = n k ′ n 1 ( 1 s 1 ′ ∏ i = 2 k s i s i ′ ) s 1 = ∞ .$
$ϕ = ( ∑ i = 1 k ω i ϕ i - S 1 ) ( 1 - S 1 ∑ i = 2 k e ′ i - 1 ω i - 1 ω i ) + S 1 ω k$
$S 1 = - 1 / s 1$
$ϕ = ∑ i = 1 k ω i ϕ i .$
$t ϕ = k ,$
$ω 1 = 1 , ω 2 = 1 - ϕ 1 e 1 ′ , ω 3 = 1 - ϕ 1 ( e 1 ′ + e 2 ′ ) - ϕ 2 e 2 ′ + ϕ 1 ϕ 2 e 1 ′ e 2 ′ , ω 4 = 1 - ϕ 1 ( e 1 ′ + e 2 ′ + e 3 ′ ) - ϕ 2 ( e 2 ′ + e 3 ′ ) - ϕ 3 e 3 ′ + ϕ 1 ϕ 2 ( e 1 ′ e 2 ′ + e 1 ′ e 3 ′ ) + ϕ 1 ϕ 3 ( e 1 ′ e 3 ′ + e 2 ′ e 3 ′ ) + ϕ 2 ϕ 3 e 2 ′ e 3 ′ - ϕ 1 ϕ 2 ϕ 3 e 1 ′ e 2 ′ e 3 ′ , ⋮$
$ϕ = 1 / f$
$ϕ = ∑ i = 1 k ϕ i ,$
$ϕ = ∑ i = 1 k ω i ϕ i$
$M = | 0 0 0 0 0 0 0 · 0 0 0 1 ϕ k 0 0 0 0 0 0 0 · 0 0 1 d k - 1 1 0 0 0 0 0 0 0 · 0 1 ϕ k - 1 1 0 0 0 0 0 0 0 0 · 1 d k - 2 1 0 0 0 0 0 0 0 0 0 · ϕ k - 2 1 0 0 0 · · · · · · · · · · · · · 0 0 0 0 0 1 ϕ 3 · 0 0 0 0 0 0 0 0 0 1 d 2 1 · 0 0 0 0 0 0 0 0 1 ϕ 2 1 0 · 0 0 0 0 0 0 0 1 d 1 1 0 0 · 0 0 0 0 0 0 S 1 ϕ 1 1 0 0 0 · 0 0 0 0 0 1 d 0 1 0 0 0 0 · 0 0 0 0 0 ϕ 0 1 0 0 0 0 0 · 0 0 0 0 0 | .$
$S 1 = - n 1 / s 1 .$
$im Falle einer Brechfläche : ϕ i = ( n i ′ - n i ) / r i , im Falle einer Spiegelfläche : ϕ i = - 2 n i / r i , im Falle einzelner Linsen : ϕ i = 1 / f i , im Falle von Teilsystemen : ( ϕ i ) i j = 1 / ( f i ) i j .$
$M I = ∂ M / ∂ ϕ 0 , M II = - ∂ 2 M ∂ ϕ 0 ∂ d 0 = - ∂ M I ∂ d 0 , M III = - ∂ 2 M ∂ ϕ 0 ∂ ϕ k = - ∂ M I ∂ ϕ k , M IV = - ∂ 3 M ∂ ϕ 0 ∂ d 0 ∂ ϕ k = ∂ M II ∂ ϕ k = ∂ M III ∂ d 0 , M V = - ∂ 3 M ∂ ϕ 0 ∂ d k - 1 ∂ ϕ k = ∂ M III ∂ d k - 1 , M VI = ∂ 4 M ∂ ϕ 0 ∂ d 0 ∂ d k - 1 ∂ ϕ k = - ∂ M IV ∂ d k - 1 = - ∂ M V ∂ d 0 .$
$| M I 1 M II M III 0 - M IV M V 1 M VI | = 0.$
$t 2 k = ( 2 k 0 ) + ( 2 k - 1 1 ) + ( 2 k - 2 2 ) + ⋯ + ( k + 2 k - 2 ) + ( k + 1 k - 1 ) + ( k k ) , t 2 k - 1 = ( 2 k - 1 0 ) + ( 2 k - 2 1 ) + ( 2 k - 3 2 ) + ⋯ + ( k + 2 k - 3 ) + ( k + 1 k - 2 ) + ( k k - 1 ) , t 2 k - 2 = ( 2 k - 2 0 ) + ( 2 k - 3 1 ) + ( 2 k - 4 2 ) + ⋯ + ( k + 1 k - 3 ) + ( k k - 2 ) + ( k - 1 k - 1 ) , t 2 k - 3 = ( 2 k - 3 0 ) + ( 2 k - 4 1 ) + ( 2 k - 5 2 ) + ⋯ + ( k + 1 k - 4 ) + ( k k - 3 ) + ( k - 1 k - 2 ) .$
$( n m ) + ( n m - 1 ) = ( n + 1 m )$
$t 2 k = t 2 k - 1 + t 2 k - 2 , t 2 k - 1 = t 2 k - 2 + t 2 k - 3 ,$
$ω k = M IV , s k / n k = M IV / M VI , s k ′ / n k ′ = M IV / M II , ϕ = M II , z 1 = 1 - ∂ M II / ∂ ϕ 1 M II , z k ′ = ∂ M II / ∂ ϕ k - 1 M II = M IV - 1 M II ,$
$ω k = M III , s k / n k = - M III / M V , s k ′ / n k ′ = - M III / M I , β ′ = S 1 / M I .$
$ϕ i = M II - ( M II ) ϕ i = 0 ∂ M II / ∂ ϕ i = ϕ - ( M II ) ϕ i = 0 ∂ M II / ∂ ϕ i ,$
$d i = M II - ( M II ) d i = 0 ∂ M II / ∂ d i = ϕ - ( M II ) d i = 0 ∂ M II / ∂ d i ,$
$s 1 = 1 / β ′ - ∂ M II / ∂ ϕ 1 ϕ .$
$M I = | 0 0 1 ϕ 2 0 1 d 1 1 S 1 ϕ 1 1 0 1 1 0 0 | = - ( ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 - S 1 + S 1 d 1 ϕ 2 ) , M II = | 0 1 ϕ 2 1 d 1 1 ϕ 1 1 0 | = ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 , M III = | 0 1 d 1 S 1 ϕ 1 1 1 1 0 | = 1 - d 1 ϕ 1 + S 1 d 1 , M IV = | 1 d 1 ϕ 1 1 | = 1 - d 1 ϕ 1 , M V = | S 1 ϕ 1 1 1 | = S 1 - ϕ 1 , M VI = | ϕ 1 | = ϕ 1$
$ω 2 = 1 - d 1 ϕ 1 , s 2 n 2 = 1 - d 1 ϕ 1 ϕ 1 s 2 ′ n 2 ′ = 1 - d 1 ϕ 1 ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 , ϕ = ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 , z 1 = d 1 ϕ 2 ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 , z 2 ′ = - d 1 ϕ 2 ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 ,$
$ω 2 = 1 - d 1 ϕ 1 + S 1 d 1 , s 2 n 2 = 1 - d 1 ϕ 1 + S 1 d 1 ϕ 1 - S 1 , s 2 ′ n 2 ′ = 1 - d 1 ϕ 1 + S 1 d 1 ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 - S 1 + S 1 d 1 ϕ 2 , β ′ = S 1 - ( ϕ 1 + ϕ 2 - d 1 ϕ 1 ϕ 2 - S 1 + S 1 d 1 ϕ 2 ) .$
$ϕ 1 = ϕ - | 0 1 ϕ 2 1 d 1 1 0 1 0 | | 1 ϕ 2 d 1 1 | = ϕ - ϕ 2 1 - d 1 ϕ 2 , ϕ 2 = ϕ - | 0 1 0 1 d 1 1 ϕ 1 1 0 | | 1 d 1 ϕ 1 1 | = ϕ - ϕ 1 1 - d 1 ϕ 1 , d 1 = ϕ - | 0 1 ϕ 2 1 0 1 ϕ 0 1 0 | | 0 ϕ 2 ϕ 1 0 | = 1 ϕ 1 + 1 ϕ 2 - ϕ ϕ 1 ϕ 2 , s 1 = 1 β ′ - | 1 d 1 ϕ 2 1 | ϕ = 1 β ′ + d 1 ϕ 2 - 1 ϕ .$