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For a nonmagnetic sample bordering air, the normal-incidence Fresnel equations giveAt short infrared wavelengths the carrier response becomes small and $R$ approaches the background value $R_{\infty }$. Extrapolating the short-wavelength trend gives about $0.29$, henceThe leftmost point at $4\mu \mathrm{m}$ is about $0.27$ and has not quite reached that asymptote. As a consistency check, using the minimum below and $R(4\mu \mathrm{m})\simeq 0.27$ gives ${ϵ}_{\infty }\simeq 11$ without treating that point as the limiting value.

The minimum is near ${\lambda }_{\min }\simeq 13\mu \mathrm{m}$. For weak damping, zero optical reflectance means refractive-index matching, ${ϵ}_r=1$, rather than ${ϵ}_r=0$. Thususing $N=5.4\times 10^{24}{\mathrm{m}}^{-3}$. Given graphical uncertainty, about $0.08m_e$ is an appropriate estimate. This is a carrier effective mass from a reflectivity minimum, not an independent determination of the bare electron mass.

Just beyond the minimum, $\mathrm{Re}{ϵ}_r$ passes through zero near ${\lambda }_p={\lambda }_{\min }/\sqrt{1-1/{ϵ}_{\infty }}\simeq 13.6\mu \mathrm{m}$. At longer wavelengths it becomes negative and reflection rises sharply toward the conducting regime. Finite damping rounds the edge, prevents a strictly exact zero minimum, and keeps the plotted long-wavelength reflectivity below unity.

In a direct optical transition, a photon promotes an electron between the valence band and conduction band at essentially the same crystal wavevector: optical photon momentum is small on the scale of the Brillouin zone. An indirect optical transition also needs a phonon or another momentum source. Near an indirect gap, phonon emission requires photon energy approximately $E_g+\hslash \Omega$, while phonon absorption permits $E_g-\hslash \Omega$ with a thermally occupied phonon.

Strong direct interband absorption needs both many allowed transitions and a nonzero optical matrix element. The joint density of states counts pairs with $E_c(\mathbf{k})-E_v(\mathbf{k})=\hslash \omega$; nearly parallel bands and critical points of this energy difference can enhance it. Dipole selection rules and polarization weight that count, so a large joint density of states alone does not guarantee a strong absorption peak.

The supplied infrared spectrum lies below the interband threshold and instead probes free-carrier response. In the Drude model, $ϵ(\omega )={ϵ}_{\infty }-{\omega }_p^2/[\omega (\omega +i/\tau )]$, with ${\omega }_p^2=n{e}^2/({ϵ}_0{m}^{\ast })$. At long wavelengths the mobile electrons screen the field and the reflectivity is high. As wavelength decreases through the screened plasma edge, the real dielectric function becomes positive and propagating light can enter. The reflectivity reaches its minimum near index matching, $ϵ\simeq 1$, then rises toward the bound-electron background reflectance.

For background ${ϵ}_{\infty }=16$, its refractive index is four and the normal-incidence Fresnel equations give $R_{\infty }=[(4-1)/(4+1){]}^2=9/25=0.36$. The short-wavelength side of the figure is approaching a reflectance of this order, consistent with that background; its shortest plotted wavelength still has a carrier correction. Here “infinity” means high frequency relative to the free-carrier scale while remaining below the remote interband resonances.

The Drude-Lorentz reflectivity minimum is near $ϵ=1$, not the plasma-edge zero $ϵ=0$. Reading the deep minimum as ${\lambda }_{\min }\simeq 22\mu \mathrm{m}$ and neglecting damping there gives ${\omega }_p^2=({ϵ}_{\infty }-1)(2\pi c/{\lambda }_{\min }{)}^2$. The carrier effective mass from a reflectivity minimum is thereforeThe graph supports only a rough read-off: shifting the minimum by $0.5\mu \mathrm{m}$ changes this mass by about five percent. The deep narrow minimum supports the weak-loss assumption $\omega \tau \gg 1$ in this region, although finite scattering explains why the long-wavelength reflectivity is below unity. The estimate also assumes one dominant carrier species, a nonmagnetic bulk sample and a nearly constant background dielectric function. If the plasma-edge zero is used instead, its denominator is ${ϵ}_{\infty }$ and its wavelength is slightly longer; the two wavelengths must not be interchanged.