= Solution
{c}
For a nonmagnetic sample bordering air, the normal-incidence <Fresnel equations> give
$$
R=\left|\frac{\sqrt{\epsilon_r}-1}{\sqrt{\epsilon_r}+1}\right|^2.
$$
At short infrared wavelengths the carrier response becomes small and $R$ approaches the background value $R_\infty$. Extrapolating the short-wavelength trend gives about $0.29$, hence
$$
\epsilon_\infty=\left(\frac{1+\sqrt{R_\infty}}{1-\sqrt{R_\infty}}\right)^2\simeq\boxed{11}.
$$
The 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 $R(4\,\mu\mathrm m)\simeq0.27$ gives $\epsilon_\infty\simeq11$ without treating that point as the limiting value.
The minimum is near $\lambda_{\min}\simeq13\,\mu\mathrm m$. For weak damping, zero <optical reflectance> means refractive-index matching, $\epsilon_r=1$, rather than $\epsilon_r=0$. Thus
$$
\omega_p^2=\left(1-\frac1{\epsilon_\infty}\right)\omega_{\min}^2,
\qquad
\boxed{m_e^*=\frac{Ne^2\lambda_{\min}^2}{4\pi^2\epsilon_0c^2(\epsilon_\infty-1)}\simeq0.082m_e,}
$$
using $N=5.4\times10^{24}\,\mathrm m^{-3}$. 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, $\operatorname{Re}\epsilon_r$ passes through zero near $\lambda_p=\lambda_{\min}/\sqrt{1-1/\epsilon_\infty}\simeq13.6\,\mu\mathrm m$. 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.
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