If the LWS fringing is caused by an interference effect due to two parallel surfaces in the optical path, then one can model the minima and maxima of the fringes as due to constructive and destructive interference in the transmission. For a wave of wavelength λ in a gap of size d, the conditions for constructive interference are met when the distance 2d is equal to an integer number n of wavelengths λ:

2d = nλ

Successive constructive interference peaks in the passband are related to consecutive integers n such that λ and n form pairs. Call one pair λ_{1}, n_{1}. The next, shorter sinusoid that fits exactly in the gap is λ_{2}, n_{2} where n_{2} = n_{1} + 1. We have:

2d = n_{1}λ_{1}

2d = (n_{1} + 1)λ_{2}

n_{1}λ_{1} = (n_{1} + 1)λ_{2}

\frac{n_{1}}{λ_{2}} = \frac{n_{1}}{λ_{1}} + \frac{1}{λ_{1}}

\frac{1}{λ_{2}} − \frac{1}{λ_{1}} = \frac{1}{n_{1}λ_{1}}

\frac{1}{λ_{2}} − \frac{1}{λ_{1}} = \frac{1}{2d}

If we want d in mm and λ in μm:

\frac{1}{λ_{2}} − \frac{1}{λ_{1}} = \frac{10^{−3}}{2d}

and if we want wavenumbers, σ in cm^{−1}

σ_{1} − σ_{2} = \frac{10}{2d}

d has been measured in the case of the LWS gap by taking the difference of successive peaks in the ripples. d ∼ 1.45 mm. The resulting waveform has peaks spaced by P = \frac{10}{2d}. The frequency of the waveform is ν = \frac{2π}{P}. This function is A = cos(2πdσ/5), where d = 1.45 and σ = \frac{10000}{λ} Note, technically, d should be replaced by the true gap width times the index of refraction of the gap material.