Abstract

It is shown that all the structure present in the interferogram of an “ideal” rotational spectrum consisting of many equally-spaced, identical lines is contained in a sequence of “signatures” at path differences of 0, 1/(2B), 2/(2B),…, where B is the reciprocal of inertia. For a symmetric line shape the signatures are all symmetric and homologous, the central one being upright and all successive ones inverted; but asymmetry in the line shape introduces increasing asymmetry in the successive signatures. This agrees with experience. Further, by measuring two vertical distances on each signature one may determine as many harmonics of the line shape as there are signatures. As an example, a typical run of the large interferometric modulator at The Johns Hopkins University is so analyzed.

© 1960 Optical Society of America

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References

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  1. John Strong and G. A. Vanasse, J. Opt. Soc. Am. 49, 844, (1959).
    [Crossref]
  2. G. A. Vanasse, J. Opt. Soc. Am.50, to be published.
  3. E. V. Lowenstein, J. Opt. Soc. Am. 50, 1163 (1960).
    [Crossref]
  4. Gerhard Herzberg, Molecular Spectra and Molecular Structure, (D. Van Nostrand Company, Inc., Princeton, New Jersey, 1950), Vol. I.

1960 (1)

1959 (1)

Herzberg, Gerhard

Gerhard Herzberg, Molecular Spectra and Molecular Structure, (D. Van Nostrand Company, Inc., Princeton, New Jersey, 1950), Vol. I.

Lowenstein, E. V.

Strong, John

Vanasse, G. A.

John Strong and G. A. Vanasse, J. Opt. Soc. Am. 49, 844, (1959).
[Crossref]

G. A. Vanasse, J. Opt. Soc. Am.50, to be published.

J. Opt. Soc. Am. (2)

Other (2)

Gerhard Herzberg, Molecular Spectra and Molecular Structure, (D. Van Nostrand Company, Inc., Princeton, New Jersey, 1950), Vol. I.

G. A. Vanasse, J. Opt. Soc. Am.50, to be published.

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Figures (5)

Fig. 1
Fig. 1

A typical interferogram and the HCN absorption spectrum giving rise to it. The first “signature” in the interferogram is its most prominent feature; it is large and upright. The next signature is over five times smaller and inverted. Two more signatures can be made out at equal intervals; both are slightly lopsided. The signatures owe their existence to the fact that the spectrum is approximately periodic.

Fig. 2
Fig. 2

Dependence of the signature on the phase-angle of the line-shape harmonic producing it. The longest leg has been given a dash, the second longest a double dash. The principal peak M lies between these two legs and the subsidiary peaks m1 and m2 lie at their ends. At ϕ=45°, the two vertical distances that must be measured have been shown. The ratio R=Mm2/Mm1 determines the phase angle up to its quadrant. The latter may be found from the altitude of M and the orientation of m1, with respect to M.

Fig. 3
Fig. 3

The phase angle appropriate to the first quadrant as a function of the quantity R defined in Fig. 2. The curve is very slightly convex toward the axis.

Fig. 4
Fig. 4

The “maximum sweep” Mm1 (indicated in Fig. 2) as a function of the phase angle. This is the scale factor by which the corresponding quantity read from the interferogram must be reduced to obtain the amplitude of the line-shape harmonic.

Fig. 5
Fig. 5

HCN line shape calculated from the signatures in Fig. 1. The line shape is seen to be skewed to the right and its center is displaced to the right by B/40.

Equations (17)

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f ( t ) = 0 2 N B cos ν t g ( ν ) d ν .
f ( t ) = 0 2 B g ( ν ) J = 0 N - 1 cos ( ν + 2 J B ) t d ν ,
f ( t ) = sin N B t sin B t 0 2 B g ( ν ) cos [ ( N - 1 ) B + ν ] t d ν .
t = μ π / B + τ / N B ,             f = N F ,
F ( τ ) = sin τ N sin τ / N 0 2 B g ( ν ) × cos [ μ ν π / B + ( 1 - 1 / N + ν / N B ) τ ] d ν .
F ( τ ) = sin τ τ 0 2 B g ( ν ) cos ( μ ν π / B + τ ) d ν ,
F ( τ ) = sin τ τ ( α μ cos τ - β μ sin τ ) ,
and             α μ = 0 2 B g ( ν ) cos μ ν π / B d ν β μ = 0 2 B g ( ν ) sin μ ν π / B d ν .
g ( ν ) = μ = 0 ( α μ cos μ ν π / B + β μ sin μ ν π / B ) ,
α μ = A μ cos ϕ μ             and             β μ = A μ sin ϕ μ .
F ( τ ) = A μ sin τ τ cos ( τ + ϕ μ )
g ( ν ) = μ = 0 A μ cos ( μ ν π / B - ϕ μ ) .
sin τ τ cos ( τ + ϕ )
R = M m 2 / M m 1
tan ϕ = sin 2 τ - 2 τ cos 2 τ cos 2 τ - 1 + 2 τ sin 2 τ .
g ( ν ) = μ = 0 A μ cos ( μ ν π / B + ϕ μ ) ,
g ( ν ) = 29.3 - 11.3 cos ( 180 ν B ) ° - 7.87 cos ( 360 ν B - 25 ) ° - 3.71 cos ( 540 ν B - 5 ) ° - 1.23 cos ( 720 ν B ) ° .