## Abstract

No abstract available.

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1. We contemplate, of course, only beams to be collimated along the axis, not obliquely. Accordingly the (symmetrical) source is throughout assumed to be centered upon the axis of the optical system.
2. The case x< f to which corresponds a virtual image is without interest, the deviation Ω increasing steadily and rapidly when the disc moves from the focal plane towards the lens.
3. Ω retains, of course, its value if all linear dimensions are changed in the same ratio.

#### Other (3)

We contemplate, of course, only beams to be collimated along the axis, not obliquely. Accordingly the (symmetrical) source is throughout assumed to be centered upon the axis of the optical system.

The case x< f to which corresponds a virtual image is without interest, the deviation Ω increasing steadily and rapidly when the disc moves from the focal plane towards the lens.

Ω retains, of course, its value if all linear dimensions are changed in the same ratio.

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F. 1
F. 2
F. 3

### Equations (43)

$( x ′ − f ) ( x − f ) = f 2 , or 1 x ′ + 1 x = 1 f$
$y ′ y = f x − f .$
$y ′ x ′ = y x .$
$tan ω ′ 0 = y ′ x ′ = y x = tan ω 0$
$ω ′ 0 = y / f .$
$Ω = 2 π π r 2 ∫ 0 r y f ydy = 2 3 r f .$
$ω ′ ≑ tan ω ′ = y ′ x ′ − a = s a ,$
$ω ′ = y ′ + s x ′ = ω 0 + s x ′ ,$
$ω ′ 1 ¯ = ω 0 + 1 2 R x ′ .$
$ω ′ = y ′ − s x ′ = ω 0 − s x ′ , for s ≦ y ′$
$ω ′ = s x ′ − ω 0 , for s ≧ y ′ .$
$ω ′ 2 ¯ = 1 R ∫ 0 y ′ ( ω 0 − s x ′ ) d s + 1 R ∫ y ′ R ( s x ′ − ω 0 ) d s ,$
$ω ′ 2 ¯ = ω 0 ( y ′ R − 1 ) + 1 2 R x ′ .$
$ω ′ ¯ = 1 2 ( ω 0 y ′ R + R x ′ ) , for y ′ ≦ R .$
$s ≦ R ≦ y ′ , ω ′ = ω 0 − s x ′ ,$
$ω ′ 2 ¯ = ω 0 − 1 2 R x ′ ,$
$ω ′ 1 ¯ = ω 0 + 1 2 R x ′ ,$
$ω ′ ¯ = ω 0 = y x = y ′ x ′ , for y ′ ≧ R .$
$ω ′ ¯ = R 2 x ′ ( 1 + y ′ 2 R 2 ) , for y ′ ≦ R .$
$Ω = 2 π π r ′ 2 R 2 x ′ ∫ 0 r ′ ( 1 + y ′ 2 R 2 ) y ′ d y ′$
$Ω = 2 r ′ 2 [ R 2 x ′ ∫ 0 R ( 1 + y ′ 2 R 2 ) y ′ d y ′ + 1 x ′ ∫ R r ′ y ′ 2 d y ′ ] .$
$Ω = R 2 x ′ ( 1 + r ′ 2 2 R 2 ) , r ′ ≦ R ,$
$Ω = R 3 3 x ′ r ′ 2 ( 1 4 + 2 r ′ 3 R 3 ) , r ′ ≧ R ,$
$x ′ = f x x − f , r ′ = r f x − f ,$
$Ω = R ( x − f ) 2 f x [ 1 + 1 2 r 2 R 2 ( f x − f ) 2 ] , x ≧ f ( 1 + r R ) ,$
$Ω = 1 3 R 3 ( x − f ) 3 f 3 r 2 x [ 1 4 + 2 r 3 R 3 ( f x − f ) 3 ] , f ≦ x ≦ ( 1 + r R ) .$
$Ω ∞ = 2 3 r ′ x ′ = 2 3 r f$
$Ω R = 3 4 R x ′ = 3 4 r x ,$
$Ω R = 3 4 r f ( 1 + r / R ) .$
$Ω R : Ω ∞ = 9 8 : ( 1 + r R ) .$
$Ω R : Ω ∞ = 45 : 56 .$
$f = 100 , r = 5 , R = 12.5 ,$
$Ω ∞ = 1 / 30 = 1 ° 55 ′ , correspnding to x = f , Ω R = 1 ° 32 ′ correspnding to x = 1.4 f .$
$Ω ′ ( x ) = R 2 x 2 { 1 − 1 2 r 2 f R 2 · 2 x − f ( x − f ) 2 }$
$x f = α [ 1 ± 1 − 1 / α ] ,$
$3 x ( x − f ) 2 − ( x − f ) 3 = 8 ( r f R ) 3$
$( x f ) 3 − 3 2 ( x f ) 2 + β = 0 , β = 1 2 ( 1 − 8 r 3 R 3 ) ,$
$u 3 + 3 p u + 2 q = 0 ,$
$u 1 , u 2 , u 3 = 2 − p ( cos θ 3 , cos θ + 2 π 3 , cos θ + 4 π 3 ) ,$
$x f = 1 2 + 0.8700 = 1.3700 ,$
$Ω ″ ( x ) = R 3 2 r 2 f 3 x − f x ,$
$Ω min = 0.02673 = 1 ° 31 ′ .9 ( x / f = 1.37 ) .$
$Ω R = 0.02679 = 1 ° 3 2 ′ . 1 for x / f = 1.40 ,$