= Solution
Assume full heat redistribution, so
$$
T_{\rm eq}=T_*\sqrt{\frac{R_*}{2a_{\rm orb}}}
(1-A_B)^{1/4}.
$$
For $T_*=5778\,\mathrm K$, $T_{\rm eq}=600\,\mathrm K$, and <Bond albedo> $A_B=0.5$,
$$
\frac{a_{\rm orb}}{R_*}
=\frac12\left(\frac{T_*}{T_{\rm eq}}\right)^2
\sqrt{1-A_B}\simeq32.8.
$$
Taking $R_p=R_{\rm Nep}=0.0354R_*$ gives
$$
\left(\frac{R_p}{a_{\rm orb}}\right)^2
\simeq1.17\times10^{-6}.
$$
The reported $50\,\mathrm{ppm}$ visible eclipse would therefore imply
$$
\boxed{A_g\simeq\frac{50\times10^{-6}}{1.17\times10^{-6}}simeq43},
$$
which 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> is
$$
\left(\frac{R_p}{R_*}\right)^2
\frac{e^{hc/(\lambda k_BT_*)}-1}
{e^{hc/(\lambda k_BT_p)}-1}.
$$
With $T_p=600\,\mathrm K$ this is
$$
\boxed{F_p/F_*\simeq7.2\times10^{-5}\simeq72\,\mathrm{ppm}}.
$$
Solved by gpt-5.6-sol high.
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