= Solution
For a non-scattering atmosphere above a <blackbody> surface, the difference between the two curves is the transmitted surface term $B_\lambda(T_s)\mathcal T_{\lambda,s}$. The atmospheric-only curve is an integral of local thermal emission weighted by transmission to space. Molecular absorption blocks some wavelengths and permits narrow windows at others. In a transparent window, adding the hot surface substantially increases the emerging <radiance>; in an opaque interval, the two curves nearly coincide because the surface contribution is extinguished.
The original graph shows the clearest surface-dominated window around $0.99$–$1.04\,\mu\mathrm m$, with the blue curve near $5\,\mu\mathrm W\,\mathrm{cm^{-2}\,sr^{-1}\,\mu m^{-1}}$ at its maximum and the atmospheric-only contribution small. Weaker possibilities are around $0.85$–$0.92\,\mu\mathrm m$ and near $0.8\,\mu\mathrm m$, with less available signal. The $1.17$–$1.20\,\mu\mathrm m$ peak contains a measurable surface contribution but a larger atmospheric background. The near-equality around $1.27$–$1.29\,\mu\mathrm m$ makes that peak a poor surface thermometer despite its appreciable total radiance. These are approximate readings, not precisely tabulated band edges.
At $T_s\simeq766\,\mathrm K$, the <Wien displacement law> puts the thermal wavelength peak near $3.8\,\mu\mathrm m$. A $1\,\mu\mathrm m$ window is still in the <Wien approximation> regime, but its radiance is exponentially larger than for an Earth-like $300\,\mathrm K$ or colder Martian surface. For example, at $1\,\mu\mathrm m$ the <Planck function> ratio $B_\lambda(766)/B_\lambda(300)$ is of order $10^{12}$. A modest hot-spot temperature increase likewise gives a relatively large fractional radiance change. Inverting this signal for surface temperature requires an atmospheric transmission and emission model, not just identifying a peak.
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