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For unresolved, uniformly bright blackbodies at a common distance $d$, the monochromatic observed radiative fluxes are $F_{s,\lambda }=\pi B_{\lambda }(T_s)R_s^2/d^2$ and $F_{p,\lambda }=\pi B_{\lambda }(T_p)R_p^2/d^2$. Define the planet-star contrast $r_{\lambda }=F_{p,\lambda }/F_{s,\lambda }$. The Planck law givesOutside the exoplanet secondary eclipse, the system flux is $F_s+F_p$; during complete occultation it is $F_s$. Consequently the exact thermal eclipse depth normalized to the out-of-eclipse light isOnly in the faint-planet limit is $d_{\lambda }\simeq r_{\lambda }$. Reflected starlight, partial occultation and spatial temperature variations are omitted under the stated blackbody model.

For a cooler hot Jupiter with $T_p<T_s$, the planet-star contrast is exponentially small at short wavelengths and increases towards a constant plateau. This is the monotonicity of blackbody planet-star contrast, rather than the peaked shape of the planet's absolute blackbody flux. An illustrative example uses $T_p=1500\mathrm{K}$, $T_s=5800\mathrm{K}$ and $R_p/R_s=0.1$:

At wavelengths long enough for the Rayleigh-Jeans law to apply to both bodies, $B_{\lambda }\simeq 2c{k}_{B}T/{\lambda }^4$. ThereforeThe common ${\lambda }^{-4}$ dependence cancels. In the illustrative case, the contrast plateau is about $2.6\times 10^{-3}$, or $2600$ parts per million.

At 10 parsecs, $0.01$ arcseconds corresponds to $a=0.10$ astronomical units. Take a solar-radius star with $T_{\star }=5772\mathrm{K}$, zero Bond albedo, full day-night heat redistribution, negligible internal heating, and a hydrogen-helium atmosphere with mean particle mass $2.3m_H$. The planetary equilibrium temperature isFor Jupiter mass and radius, $g\simeq 24.8\mathrm{m}{\mathrm{s}}^{-2}$. Its atmospheric scale height isAssume a strong band spans $N_H=5$ atmospheric scale heights and saturates in the annulus model for transmission spectroscopy. The atmospheric spectral-feature amplitude isFor a five-standard-deviation detection, the uncertainty of the measured differential contrast must satisfyThis estimate scales linearly with the assumed feature height; one atmospheric scale height would require about $7.5\mathrm{ppm}$. The question gives no opacity or abundance from which to fix $N_H$, so an atmospheric detection threshold is necessarily assumption-dependent.

For thermal emission at $10\mu \mathrm{m}$, assume the planet and star emit as blackbodies. The thermal eclipse depth from the Planck law isThus the uncertainty required for a five-standard-deviation exoplanet secondary eclipse detection isThese are uncertainties of the final transit or eclipse contrasts, including the uncertainty of their reference levels. The distance affects photon counts and observing time but cancels from the flux ratios. An opaque exactly isothermal atmosphere emits a featureless blackbody spectrum: an eclipse detects its thermal light, while identifying atmospheric composition requires spectral features and a nonisothermal structure or other diagnostics.

Assume that both objects emit as blackbodies, take $T_{\ast }=5778\mathrm{K}$ and $R_p/R_{\ast }\simeq R_J/R_{\odot }=0.100$, and use $T_p=1250\mathrm{K}$ because the line-free window sees the $1$-bar continuum photosphere. Across a narrow bin, the thermal eclipse depth isAt $\lambda =17\mu \mathrm{m}$ this givesIntegrating the Planck law over the full $1\mu \mathrm{m}$ bin changes this narrow-bin estimate only slightly.

For $R_p=R_{\oplus }$, $R_{\ast }=0.1R_{\odot }$, $T_p=600\mathrm{K}$, and $T_{\ast }=3500\mathrm{K}$, the same thermal eclipse depth at $17\mu \mathrm{m}$ isThus a $100\mathrm{ppm}$ uncertainty givesThis assumes one independent eclipse measurement with that precision, negligible reflected light in the infrared, no atmosphere, and blackbody emission from the bare surface.

Assume full heat redistribution, soFor $T_{\ast }=5778\mathrm{K}$, $T_{\mathrm{eq}}=600\mathrm{K}$, and Bond albedo $A_B=0.5$,Taking $R_p=R_{\mathrm{Nep}}=0.0354R_{\ast }$ givesThe reported $50\mathrm{ppm}$ visible eclipse would therefore implywhich is impossible. At least one assumption, the measurement, or the stated system parameters must fail; even $A_g=1$ gives only about $1.2\mathrm{ppm}$.

At $20\mu \mathrm{m}$, reflected light is negligible. The thermal eclipse depth isWith $T_p=600\mathrm{K}$ this is

Let ${\tau }_1$ and ${\tau }_2$ be the top-down optical depths at $P_1$ and $P_2$, and let $\mu$ be the outward direction cosine. The formal solution of the radiative transfer equation givesFor a semi-infinite lower layer in local thermodynamic equilibrium, $I_{\nu ,0}=B_{\nu }(T_3)$.

DefineThe emergent planetary surface flux isHence the band-centre planet-star ratio isIf $T_1=T_2=T_3=T_p$, the weights telescope to $1/2$, so $F_{p,\nu }=\pi B_{\nu }(T_p)$ and the expression reduces to the blackbody thermal eclipse depth, about $72\mathrm{ppm}$ for $T_p=600\mathrm{K}$.

For blackbody star and planet observed at one common wavelength sufficiently long for the Rayleigh-Jeans law to hold for both,This follows by substituting $B_{\lambda }=2c{k}_{B}T/{\lambda }^4$ in the thermal eclipse depth. A radiative flux ratio alone does not determine the radius without a planetary brightness temperature and a stellar temperature. Spectral peaks interpreted with Wien's displacement law must refer to $F_{\lambda }$, not $F_{\nu }$ or $\lambda F_{\lambda }$.