Abstract

In Fabry–Perot interferometry a large fraction of light can be reflected by the etalon; thus, it is lost for spectroscopic purposes. The present article shows the possibility of recovering this light, collecting it by means of reflectors and sending it back to the etalon under a different angle. Some devices are examined and some experimental proofs of the device are reported. The gain achieved is of the order of five.

© 1967 Optical Society of America

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References

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  1. P. Jacquinot, Rept. Progr. Phys. 23, 268 (1960).
    [Crossref]
  2. J. M. Burch in Quantum Electronics III, N. Bloembergen, P. Grivet, Eds. (Columbia University Press, New York, 1965), Vol. 2, p. 1187.
  3. W. K. Kahn, Appl. Opt. 5, 407 (1966).
    [Crossref] [PubMed]

1966 (1)

1960 (1)

P. Jacquinot, Rept. Progr. Phys. 23, 268 (1960).
[Crossref]

Burch, J. M.

J. M. Burch in Quantum Electronics III, N. Bloembergen, P. Grivet, Eds. (Columbia University Press, New York, 1965), Vol. 2, p. 1187.

Jacquinot, P.

P. Jacquinot, Rept. Progr. Phys. 23, 268 (1960).
[Crossref]

Kahn, W. K.

Appl. Opt. (1)

Rept. Progr. Phys. (1)

P. Jacquinot, Rept. Progr. Phys. 23, 268 (1960).
[Crossref]

Other (1)

J. M. Burch in Quantum Electronics III, N. Bloembergen, P. Grivet, Eds. (Columbia University Press, New York, 1965), Vol. 2, p. 1187.

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Figures (8)

Fig. 1
Fig. 1

Ratio I(r)/I(i) of reflected to incident light as a function of the incidence angle θ.

Fig. 2
Fig. 2

Axicon reflector setup.

Fig. 3
Fig. 3

The nonconfocal reflector pair setup.

Fig. 4
Fig. 4

The resonator equivalent infinite array of lenses.

Fig. 5
Fig. 5

The cardinal points of the optical system.

Fig. 6
Fig. 6

The sink of light in the case K = 2, r = 1. f0 = focal length of lens L. b0 = f0 + [f02/2(a0f0)].

Fig. 7
Fig. 7

The sink of light in the case K = 3, r = 1. f0 = focal length of lens L. b0 = (4a0f0f02)/4(a0f0).

Fig. 8
Fig. 8

(a) The spectrum of the used light, (b) The spectrum of the Fabry–Perot output without mirror, (c) The effect of the resonator for different values of b. - - - Gain. — Output with mirror. ⋯ Difference with and without the mirror.

Equations (13)

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F = [ ( π / 2 ) ( 4 R ) ½ / ( 1 - R ) ] = π R ½ / ( 1 - R ) ,
τ = [ I ( t ) / I ( i ) ] max = ( 1 + F sin 2 δ ) - 1 ,
C = [ I ( r ) / I ( i ) ] max / [ I ( t ) / I ( i ) ] min .
lim R 1 [ I ( t ) ( θ ) / I ( i ) ( θ ) ] = F sin 2 δ / ( 1 + F sin 2 δ ) = 1 ,
Δ θ = θ 0 [ F ½ - ( F - 1 ) ½ ] ,
F ½ / [ F ½ - ( F - 1 ) ½ ] 2 F .
φ × H = const . = K ,
α < 2.
Z 0 Z 0 = - f 0 2 ,
c = 2 ( a 0 - f 0 ) .
F 1 F 1 = 2 b 0 - 2 f 0 - 2 f 0 2 / c ;
α = 2 [ ( 2 a 0 b 0 / f 0 2 ) - 2 ( b 0 / f 0 ) - 2 ( a 0 / f 0 ) + 1 ] .
α = 2 cos θ .

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