Abstract

Rugate filters reflectance spectra were calculated for various index profiles with half-Gaussian modulation; graded-index as well as step-index profiles were used. The results show excellent sidelobe suppression around the stopbands, relatively high optical density, and good flatness in the reflectance band.

© 1993 Optical Society of America

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References

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  1. W. H. Southwell, “Spectral response calculations of rugate filters using coupled-wave theory,” J. Opt. Soc. Am. A 5, 1558–1564 (1988).
    [CrossRef]
  2. W. H. Southwell, “Coating design using very thin high- and low-index layers,” Appl. Opt. 24, 457–460 (1985).
    [CrossRef] [PubMed]
  3. P. G. Veryl, J. A. Dobrowolski, W. J. Wild, R. L. Burton, “Synthesis of high rejection filters with the Fourier transform method,” Appl. Opt. 28, 2864–2875 (1989).
    [CrossRef]
  4. M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980).
  5. A. Thelen, “Design of optical minus filters,” J. Opt. Soc. Am. 61, 365–369 (1971).
    [CrossRef]
  6. E. P. Donovan, D. Van Vechten, A. D. F. Kahn, C. A. Carosella, G. K. Hubler, “Near infrared rugate filter fabrication by ion beam assisted deposition of Si(1−x)Nx films,” Appl. Opt. 28, 2940–2944 (1989).
    [CrossRef] [PubMed]
  7. W. J. Gunning, R. L. Hall, F. J. Woodberry, W. H. Southwell, N. S. Gluck, “Codeposition of continuous composition rugate filters,” Appl. Opt. 28, 2945–2948 (1989).
    [CrossRef] [PubMed]
  8. W. H. Southwell, R. L. Hall, “Rugate filter sidelobe suppression using quintic and rugated quintic matching layers,” Appl. Opt. 28, 2949–2950 (1989).
    [CrossRef] [PubMed]
  9. W. H. Southwell, “Using apodization functions to reduce sidelobes in rugate filters,” Appl. Opt. 28, 5091–5094 (1989).
    [CrossRef] [PubMed]
  10. B. G. Bovard, “Rugate filter design: the modified Fourier transform technique,” Appl. Opt. 29, 24–30 (1990).
    [CrossRef] [PubMed]

1990 (1)

1989 (5)

1988 (1)

1985 (1)

1971 (1)

Born, M.

M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980).

Bovard, B. G.

Burton, R. L.

Carosella, C. A.

Dobrowolski, J. A.

Donovan, E. P.

Gluck, N. S.

Gunning, W. J.

Hall, R. L.

Hubler, G. K.

Kahn, A. D. F.

Southwell, W. H.

Thelen, A.

Van Vechten, D.

Veryl, P. G.

Wild, W. J.

Wolf, E.

M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980).

Woodberry, F. J.

Appl. Opt. (7)

J. Opt. Soc. Am. (1)

J. Opt. Soc. Am. A (1)

Other (1)

M. Born, E. Wolf, Principles of Optics, 6th ed. (Pergamon, New York, 1980).

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Figures (8)

Fig. 1
Fig. 1

(a) 50-cycle rugate filter with Gaussian apodization and Gaussian matching on each side. (b) Reflectance of the corresponding rugate, filter.

Fig. 2
Fig. 2

(a) 50-cycle rugate filter with Gaussian apodization and Gaussian matching on each side. (b) Reflectance of the corresponding rugate filter.

Fig. 3
Fig. 3

(a) 50-cycle rugate filter with Gaussian apodization and no matching regions. The substrate index and the medium index are the same as the average profile index, which is equal to 1.52. (b) Reflectance of the corresponding rugate filter.

Fig. 4
Fig. 4

(a) 50-cycle rugate filter with Gaussian half-apodization and no matching regions. The substrate index is the same as the lower index in the profile (n = 1.52), and the medium is air. (b) Reflectance of the corresponding rugate filter.

Fig. 5
Fig. 5

(a) 50-cycle rugate filter with Gaussian half-apodization and a LiF substrate (n = 1.38). (b) Reflectance of the corresponding rugate filter.

Fig. 6
Fig. 6

(a) Stack equivalence of the rugate filter shown in Fig. 4. (b) Reflectance of the corresponding stack.

Fig. 7
Fig. 7

(a) Stack equivalence filter of 50 layers with Gaussian half-apodization. (b) Reflectance of the corresponding stack.

Fig. 8
Fig. 8

(a) Same index profile as that in Fig. 6(a) but with np = 0.5 and ns = 1.38. (b) Reflectance of the corresponding stack.

Equations (5)

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n ( x ) = n a + 1 2 n p sin ( 2 π x p ) ,
BW = Δ λ λ = n p 2 n a ,
OD = log 10 ( 1 1 R ) = ( 1 . 36 × BW × N ) log 10 ( 4 / n s ) ,
n ( x ) = n L + 1 2 n p [ 1 + sin ( 2 π x p ) ] exp ( α x 2 )
n ( x ) = n a + 1 2 n p [ 1 + ( 1 ) i ] exp ( α x i 2 ) ,

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