## Abstract

The conditions for transformation with uniform intensity from an annular to a cylindrical beam, as needed for mode control in high-power lasers, are analyzed. The equations for the shapes of the mirrors in two different arrangements are derived and evaluated for feasibility of fabrication. The shape of the inner mirrors as well as the major part of the outer mirrors should present no problem of fabrication. In either system, however, that edge of the outer mirror which reflects to the axis requires a sharp curvature, the machining of which might appear formidable. Nevertheless, recent work with diamond machining has accomplished even sharper curvatures than here required.

© 1978 Optical Society of America

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### Figures (7)

Fig. 1

Linear waxicon used for beam transformation from annular region (with intensity IS) to central cylindrical region resulting in steeply rising intensity (I) toward the axis.

Fig. 2

Schematic diagram of two-part mirror system.

Fig. 3

Two-part mirror system for transformation from annular beam to central cylindrical beam with uniform intensity.

Fig. 4

Schematic diagram of nonlinear waxicon system.

Fig. 5

Nonlinear waxicon mirror system for transformation from annular beam to central cylindrical beam with uniform intensity.

Fig. 6

Radius of curvature of outer mirror of both systems vs distance from the edge that reflects to the axis.

Fig. 7

Change κ needed in axial/radial feed ratio vs distance from the edge when machining the outer mirrors.

### Equations (55)

$a − y ′ + y + [ ( a − y ′ + y ) 2 + ( r ′ − r ) 2 ] 1 / 2 = c .$
$( d y ) / ( d r ) = tan θ = ( d y ′ ) / ( d r ′ ) ,$
$tan 2 θ = ( r ′ − r ) / ( a + y − y ′ ) .$
$r ′ 2 − r ′ 0 2 = k 2 r 2 ,$
$( r ′ − r ) ( 1 + cos 2 θ ) sin 2 θ = ( r ′ − r ) 2 cos 2 θ 2 sin θ cos θ = ( r ′ − r ) tan θ = c .$
$k r ′ − ( r ′ 2 − r ′ 0 2 ) 1 / 2 = k c tan θ .$
$d y ′ d r ′ = 1 k c [ k r ′ − ( r ′ 2 − r ′ 0 2 ) 1 / 2 ] ,$
$y ′ = ∫ 0 y ′ d y ′ = 1 k c ∫ r 0 ′ r ′ [ k r ′ − ( r ′ 2 − r ′ 0 2 ) 1 / 2 ] d r ′$
$y ′ = r ′ 2 − r ′ 0 2 2 c − 1 2 k c [ r ′ · ( r ′ 2 − r ′ 0 2 ) 1 / 2 − r ′ 0 2 ln r ′ + ( r ′ 2 − r ′ 0 2 ) 1 / 2 r 0 ′ ] ,$
$( k 2 r 2 + r ′ 0 2 ) 1 / 2 − r = c [ ( d y ) / ( d r ) ] ,$
$y = ∫ 0 y d y = 1 c ∫ 0 r [ ( k 2 r 2 − r ′ 0 2 ) 1 / 2 − r ] d r ,$
$y = − r 2 2 c + k 2 c × { r [ r 2 + ( r 0 ′ k ) 2 ] 1 / 2 + ( r 0 ′ k ) 2 ln r + [ r 2 + ( r 0 ′ k ) 2 ] 1 / 2 r 0 ′ / k } .$
$y ′ = 5 1 + √ 2 ( ( r 10 ) 2 − 1 − 9 ( 96 ) 1 / 2 { r ′ 10 [ ( r ′ 10 ) 2 − 1 ] 1 / 2 } − ln { r ′ 10 − [ ( r ′ 10 ) 2 − 1 ] 1 / 2 } ) .$
$y = 5 1 + √ 2 { r ( 96 ) 1 / 2 90 ( 96 r 2 8100 + 1 ) 1 / 2 + 9 ( 96 ) 1 / 2 ln [ r ( 96 ) 1 / 2 90 + ( 96 r 2 8100 + 1 ) 1 / 2 ] − ( r 10 ) 2 } .$
$l − y − y ′ = 0 ,$
$l = [ ( y − y ′ ) 2 + ( r ′ − r ) 2 ] 1 / 2 .$
$l = ( r ′ − r ) / ( sin 2 θ ) ,$
$l = ( y ′ − y ) / ( cos 2 θ ) ,$
$( − d y ) / ( d r ) = tan θ ,$
$( d y ′ ) / ( d r ′ ) = tan ( 90 ° − θ ) = cot θ .$
$r ′ 2 − r 0 2 = ( r 0 2 − r 2 ) k 2 ,$
$r = r ′ − 2 y ′ [ ( d r ′ ) / ( d y ′ ) ] .$
$r = { [ r 0 2 ( 1 + k 2 ) − r ′ 2 ] 1 / 2 / k }$
$∫ 0 y ′ d y ′ y ′ = ∫ r 0 ′ r ′ 2 k d r ′ k r ′ − [ r 0 2 ( 1 + k 2 ) − r ′ 2 ] 1 / 2 .$
$∫ d y ′ y ′ = ∫ 2 k d r ′ k r ′ − ( a 2 − r ′ 2 ) 1 / 2$
$a = r 0 ( 1 + k 2 ) 1 / 2 ,$
$∫ 2 k d r ′ k r ′ − ( a 2 − r ′ 2 ) 1 / 2 = 1 1 + k 2 [ 2 k arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ − k ( 1 − k 2 ) arctan k + 2 k 2 ln k r ′ − ( a 2 − r ′ 2 ) 1 / 2 a ( 1 + k 2 ) 1 / 2 ] + const .,$
$y ′ = C [ k r ′ − ( a 2 − r ′ 2 ) 1 / 2 ] 2 k 2 / ( 1 + k 2 ) × exp 2 k 1 + k 2 arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ .$
$y = y ′ 1 − cos 2 θ 1 + cos 2 θ = y ′ tan 2 θ = y ′ ( d r ′ d y ′ ) 2 .$
$y = y ′ [ k r ′ − ( a 2 − r ′ 2 ) 1 / 2 2 k y ′ ] 2 , y = [ k r ′ − ( a 2 − r ′ 2 ) 1 / 2 4 C k 2 ] 2 / ( 1 + k 2 ) × exp [ − 2 k 1 + k 2 arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ ] .$
$r ′ = ( a 2 − k 2 r 2 ) 1 / 2 ,$
$y = [ ( a 2 − k 2 r 2 ) 1 / 2 − r ] 2 / ( 1 + k 2 ) 4 C k 2 k 2 / ( 1 + k 2 ) × exp − 2 k 1 + k 2 arctan k r ( a 2 − k 2 r 2 ) 1 / 2 ] .$
$y 0 = a 2 / ( 1 + k 2 ) 4 C k 2 k 2 / ( 1 + k 2 ) = y a ′ = C ( k a ) 2 k 2 / ( 1 + k 2 ) ,$
$C = a ( 1 − k 2 ) / ( 1 + k 2 ) 2 k 2 k 2 / ( 1 + k 2 ) ,$
$y = [ ( a 2 − k 2 r 2 ) 1 / 2 − r ] 2 / ( 1 + k 2 ) / 2 a ( 1 − k 2 ) / ( 1 + k 2 ) × exp [ 2 k 1 + k 2 arctan k r ( a 2 − k 2 r 2 ) 1 / 2 ] ,$
$y ′ = a ( 1 − k 2 ) / ( 1 + k 2 ) 2 k 2 k 2 / ( 1 + k 2 ) [ k r ′ − ( a 2 − r ′ 2 ) 1 / 2 ] 2 k 2 / ( 1 + k 2 ) × exp [ 2 k / ( 1 + k 2 ) arctan ( a 2 − r ′ ) 1 / 2 r ′ ] .$
$ρ = { 1 + [ ( d y ) / ( d r ) 2 } 3 / 2 ( d 2 y ) / ( d r 2 ) .$
$d y ′ d r ′ = k r ′ − ( r ′ 2 − r ′ 0 2 ) 1 / 2 k c$
$d 2 y ′ d r ′ 2 = 1 c − r ′ k c ( r ′ 2 − r ′ 0 2 ) 1 / 2 ,$
$ρ ′ = { 1 + [ k r ′ − ( r ′ 2 − r ′ 0 2 ) 1 / 2 k c ] 2 } 3 / 2 1 c − r ′ k c ( r ′ 2 − r ′ 0 2 ) 1 / 2 .$
$ρ = { c 2 + [ ( k 2 r 2 + r ′ 0 2 ) 1 / 2 − r ] 2 } 3 / 2 c 2 [ k 2 r ( k 2 r 2 + r ′ 0 2 ) 1 / 2 − 1 ] .$
$d y ′ d r ′ = 2 y ′ r ′ − ( a 2 − r ′ 2 ) 1 / 2 .$
$y ′ = 1 2 [ r ′ − ( a 2 − r ′ 2 ) 1 / 2 ] exp arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ ,$
$d y ′ d r ′ = exp arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ .$
$d 2 y ′ d r ′ 2 = exp arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ ( a 2 − r ′ 2 ) 1 / 2 ,$
$ρ ′ = [ 1 + exp 2 arctan ( a 2 − r ′ 2 ) r ′ ] 3 / 2 exp arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ · ( a 2 − r ′ 2 ) 1 / 2 .$
$ρ ′ ≅ 4 ( a Δ ) 1 / 2 [ 1 + ( Δ 2 a ) 1 / 2 − Δ a ] .$
$d y d r = − tan θ = − d r ′ d y ′ = − 1 / exp arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ ,$
$d 2 y d r 2 = 1 / [ ( a 2 − r 2 ) 1 / 2 exp arctan r ( a 2 − r 2 ) 1 / 2 ] .$
$ρ = [ 1 + exp − 2 arctan r ( a 2 − r 2 ) 1 / 2 ] 3 / 2 ( a 2 − r 2 ) 1 / 2 × exp arctan r ( a 2 − r 2 ) 1 / 2 .$
$ρ 10 = 10 [ 1 + exp ( − π / 2 ) ] 3 / 2 · exp ( π / 4 ) = 29.12 cm .$
$FR = d y ′ d r ′ = k r ′ − ( r ′ 2 − r ′ 0 2 ) 1 / 2 k c = k r ′ − ( r ′ 2 − 100 ) 1 / 2 10 k ( 1 + 2 1 / 2 ) ,$
$κ = Δ 10 − 9 ( 20 Δ + Δ 2 ) 1 / 2 10 ( 96 ) 1 / 2 .$
$F R = d y ′ d r ′ = exp arctan ( a 2 − r ′ 2 ) 1 / 2 r ′ = exp arctan ( 28 Δ − Δ 2 ) 1 / 2 14 − Δ ,$
$κ = exp arctan ( 28 Δ − Δ 2 ) 1 / 2 14 − Δ − 1.$