Abstract

By replacing the silver layers in the conventional interference filter with seven dielectric layers having alternately high and low indices, it has been possible to obtain a filter having an 80-percent peak transmission and 65A half-width for a first-order transmission band centered at 5200A. The properties of the filter are determined from the theory of the parallel-plate interferometer.

© 1952 Optical Society of America

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References

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  1. P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983(A) (1947).
  2. H. D. Polster, J. Opt. Soc. Am. 39, 1038–43 (1949).
    [CrossRef]
  3. A. E. Gee and H. D. Polster, J. Opt. Soc. Am. 39, 1044–47 (1949).
    [CrossRef]
  4. G. N. Ranachandran, Proc. Indian Acad. Sci. 16, 336–48 (1942).
  5. M. Banning, J. Opt. Soc. Am. 37, 792–97 (1947).
    [CrossRef] [PubMed]
  6. L. N. Hadley and D. M. Dennison, J. Opt. Soc. Am. 37, 451–65 (1947).
    [CrossRef]

1949 (2)

1947 (3)

1942 (1)

G. N. Ranachandran, Proc. Indian Acad. Sci. 16, 336–48 (1942).

Banning, M.

Dennison, D. M.

Gee, A. E.

Hadley, L. N.

Leurgans, P.

P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983(A) (1947).

Polster, H. D.

Ranachandran, G. N.

G. N. Ranachandran, Proc. Indian Acad. Sci. 16, 336–48 (1942).

Turner, A. F.

P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983(A) (1947).

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Figures (3)

Fig. 1
Fig. 1

Reflectivity from seven dielectric layers all of 1 4 λ optical thickness at λ5500. (Four ZnS, index 2.4, three cryolite, index 1.36.)

Fig. 2
Fig. 2

Variation with wavelength of phase shift on reflection from (A) seven-layer filter of Fig. 1, and (B) an opaque silver layer (translated to coincide with A at δ = 90). 1 4 λ 0 is the optical thickness of each of the dielectric layers.

Fig. 3
Fig. 3

Transmission of a 15-layer all-dielectric filter deposited to give a first-order pass band. Broken curve is for a second-order conventional interference filter of comparable half-width.

Equations (5)

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T = [ ( 1 + A T ) 2 + 4 R T 2 sin 2 ( δ + φ ) ] - 1 ,
R + T + A = 1
δ = 2 π μ d cos ϑ / λ .
T max = ( 1 + A T ) - 2 ,             T min = T 2 / ( 1 + R ) 2 ,
d λ h ω = - λ d δ δ = - λ n π - φ 1 - R 2 R 1 2 ,             for             δ + φ = n π