= Doppler-free absorption line model
{c}
{title2=$A_s(x)=B(x)-\frac{C_s}{1+4x^2/w_s^2}$}
Construct an original illustrative spectrum using ordinary <frequency> detuning $x=\nu_L-\nu_0$. Write the longitudinal <Doppler shift> in frequency units as $u=v/\lambda_0$, where $\lambda_0=c/\nu_0$. Choose a <normal distribution> with <standard deviation> $\sigma_D$ and a weak-probe <Lorentzian line shape> with half width at half maximum $\gamma_\nu$:
$$
g(u)=\frac{e^{-u^2/(2\sigma_D^2)}}{\sqrt{2\pi}\sigma_D},\qquad
L(y)=\frac{\gamma_\nu^2}{y^2+\gamma_\nu^2},\qquad
B(x)=\frac{\int_{-\infty}^{\infty}g(u)L(x-u)\,du}
{\int_{-\infty}^{\infty}g(u)L(-u)\,du}.
$$
Thus $B(0)=1$. Choose $\sigma_D=120\ \mathrm{MHz}$ and <natural linewidth> $\Gamma/(2\pi)=6\ \mathrm{MHz}$, so $\gamma_\nu=3\ \mathrm{MHz}$. The Gaussian-only <full width at half maximum> is $2\sqrt{2\ln2}\,\sigma_D\simeq282.58\ \mathrm{MHz}$; the averaged profile also includes the finite <Lorentzian line shape> width.
Add a deliberately phenomenological <Lamb dip>:
$$
A_s(x)=B(x)-\frac{C_s}{1+4x^2/w_s^2},\qquad
C_s=0.24\frac{s}{1+s},\qquad
w_s=\frac{\Gamma}{2\pi}\sqrt{1+s}.
$$
Here $A_s$ is the <absorption coefficient> normalized to the pump-off line-centre value. The factor $0.24$ is chosen, not inferred from populations or a fit. The dip's assigned <full width at half maximum> $w_s$ borrows the single-atom <power broadening> scale purely to illustrate contrast versus width. It is not the preceding overlap integral or a fit to historical spectra. There are no crossover features in this isolated-transition toy spectrum.
For $s=1$, the subtracted dip has depth $0.12$ and width $8.49\ \mathrm{MHz}$; for $s=9$, depth $0.216$ and width $18.97\ \mathrm{MHz}$. Its half-depth points are $x=\pm w_s/2$. Measure that width relative to the curved background $B(x)$, not a horizontal line through the total absorption minimum.
\Image[/saturation-spectroscopy-chosen-doppler-envelope-and-dip.png]
{title=Chosen Doppler envelope and phenomenological Lamb dips}
{description=Original numerical Gaussian velocity average with two explicitly chosen Lorentzian absorption dips. Every numerical setting is illustrative. The right panel magnifies line centre; these curves are not experimental data or fits.}
For a chosen cell <optical depth> $\tau_s(x)=0.4A_s(x)$, the line-centre pump-off transmission is $e^{-0.4}\simeq0.6703$. At $s=1$ it is $e^{-0.352}\simeq0.7033$. The transmitted probe therefore increases by about $4.92$ percent relative to pump off. Absorption contrast, transmission contrast and spectral width are three different quantities.
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