## Abstract

A refracting optical system is designed for the transformation of an annular laser beam to a uniform circular beam. The optical surfaces of the proposed system are easy to fabricate as they involve moderate values of radii of curvature. The change in the radii of curvature is found to be gradual throughout the surface.

© 1987 Optical Society of America

### References

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1. J. W. Ogland, “Mirror System for Uniform Beam Transformation in High-Power Annular Lasers,” Appl. Opt. 17, 2917 (1978).
[CrossRef] [PubMed]
3. P. Gillet, Calculus and Analytical Geometry (Heath, Lexington, MA, 1981).

#### Gillet, P.

P. Gillet, Calculus and Analytical Geometry (Heath, Lexington, MA, 1981).

#### Other (2)

P. Gillet, Calculus and Analytical Geometry (Heath, Lexington, MA, 1981).

### Cited By

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

### Figures (5)

Fig. 1

Schemetic diagram of half of the axially symmetric beam-transforming refracting system.

Fig. 2

Maximum surface slope vs a, D; b, c′; and c, ratio of the maximum and minimum slope.

Fig. 3

Surface slope (dy/dr) against radial distance when R′ = 13 cm, R = 10 cm, k = 2, n = 1.5172, and D = 16.59 cm.

Fig. 4

Final layout of half of the beam transforming system.

Fig. 5

### Equations (18)

$r i 2 - R 2 = k 2 r o 2 .$
$y i + n [ ( r i - r o ) 2 + ( D - y i + y o ) 2 ] 1 / 2 + d - y o = c ,$
$tan ( θ i i - θ r i ) = ( r i - r o ) / ( D - y i + y o ) = tan ( θ i o - θ r o ) ,$
$( r i - r o ) [ n - cos ( θ i i - θ r i ) ] / [ sin ( θ i i - θ r i ) ] = c ′ ,$
$d y i / d r i = tan θ i i = tan θ i o = d y o / d r o .$
$D = R / tan ( θ i i , R - θ r i , R ) ,$
$θ r i = sin - 1 [ ( sin θ i i ) / n ] .$
$r i = [ k 2 R 2 / ( k 2 - 1 ) ] 1 / 2 .$
$r i = [ 8 ( r i - r o ) ± [ 12 ( r i - r o ) 2 - 4800 ] 1 / 2 ] / 6 ,$
$sin ( θ i i - θ r i ) = [ sin θ i i ( n 2 - sin θ i i ) 1 / 2 - sin θ i i cos θ i i ] / n ,$
$cos ( θ i i - θ r i ) = [ cos θ i i ( n 2 - sin θ i i ) 1 / 2 - sin θ i i ] / n ,$
$d y i / d r i = d y o / d r o = n / { { c ′ / ( r i - r o ) ] 2 - n 2 + 1 } 1 / 2 .$
$max slope min slope = ( { 1 - n 2 + ( c ′ 2 k 2 ) / [ R 2 ( k 2 - 1 ) ] } / [ 1 - n 2 + ( c ′ 2 / R 2 ) ] ) 1 / 2 .$
$d y i / d r i = { 22.63165 - 2.02953 r i , 10 ≤ r i ≤ 10.3 , + 0.03114641 r i 3 + 0.008482869 r i 4 76.92947 - 7024512 r i - 0.7524784 r i 2 - 0.0004514059 r i 5 , 10.3 ≤ r i ≤ 13.$
$y i = { 22.63165 r i - 1.0147965 r i 2 - 124.833 , 10 ≤ r i ≤ 10.3 + 0.007866025 r i 4 + 0.00169657 r i 5 76.92947 r i - 3.512256 r i 2 - 0.250826 r i 3 - 0.000075234317 r i 6 - 339.5445 , 10.3 ≤ r i ≤ 13 ,$
$d y o / d r o = 2.615915 - 1.19055 r o + 0.657615 r o 2 - 0.2185367 r o 3 + 0.0405096 r o 4 - 0.003002334 r o 5 .$
$y o = 2.615915 r o - 0.59527 r o 2 + 0.219205 r o 3 - 0.054634175 r o 4 + 0.00810192 r o 5 - 0.000500389 r o 6 ,$
$ρ = [ 1 + ( d y / d r ) 2 ] 3 / 2 / d 2 y / d r 2 .$