 Solution

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.