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The detector manufacturer's technical note defines noise-equivalent power relative to unit signal-to-noise ratio and a 1 Hz detection bandwidth, and explains how integrated noise changes with bandwidth.

Detector noise and graphical reading. Noise-equivalent power specifies the incident optical power producing signal-to-noise ratio one, with the electrical bandwidth and averaging convention stated. A spectral NEP has units $\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1/2}$; a band-integrated NEP has units W. Specific detectivity normalizes the latter by detector area and noise bandwidth:Its usual unit, the Jones, is $\mathrm{cm}\mathrm{H}{\mathrm{z}}^{1/2}{\mathrm{W}}^{-1}$. Removing the expected square-root area and bandwidth dependence helps compare different detector sizes and readout bandwidths, provided noise conditions and spectral response are comparable. This does not normalize away every background, temperature or implementation difference.

Reading the native embedded voltage plot gives approximate means $V_C\simeq 0.331\mathrm{V}$, $V_S\simeq 0.406\mathrm{V}$ and $V_H\simeq 0.493\mathrm{V}$. The one-second root mean square scatter is about $0.006\mathrm{V}$; the plotted peak-to-peak excursions are larger and must not be substituted for an root mean square noise. Independent graphical digitization of the blue trace gives means near $0.331,0.406,0.493$ V and scatters about $0.006$–$0.007$ V, consistent with this estimate. These are readings from a drawing, not exact original data.

The supplied Planck approximation gives $B_C\simeq 0.08590$ and $B_H\simeq 0.12836$ in $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}(\mathrm{c}{\mathrm{m}}^{-1}{)}^{-1}$. Thus the noise-equivalent radiance of one one-second sample isHere the calibrating means are used to determine the response slope; their own uncertainty is included explicitly in the final scene estimate below.

Optical throughput and detectivity. A paraxial $f/F$ cone has half-angle approximately $1/(2F)$ and projected solid angle approximately $\pi /(4F^2)$. With $F=2$ and $A_d=10^{-6}{\mathrm{m}}^2$,Using the exact cone angle $\tan \theta =1/(2F)$ gives projected etendue $A_d\pi {\sin }^2\theta =1.85\times 10^{-7}{\mathrm{m}}^2\mathrm{sr}$, within the intended approximation. The radiance noise corresponds to detector power noiseThe wavenumber bandwidth is $1\mathrm{c}{\mathrm{m}}^{-1}$ and must match the radiance's per-wavenumber unit; it is not an electrical noise bandwidth. For the common order-of-magnitude assignment $\Delta f\sim 1\mathrm{Hz}$ to a one-second sample,Area in the last expression is $1\mathrm{m}{\mathrm{m}}^2=0.01\mathrm{c}{\mathrm{m}}^2$. For a specified one-second boxcar with white noise, the one-sided equivalent noise bandwidth is instead $1/(2t)=0.5\mathrm{Hz}$. That convention gives spectral NEP $\simeq 2.5\times 10^{-10}\mathrm{W}\mathrm{H}{\mathrm{z}}^{-1/2}$ and $D_{\ast }\simeq 3.9\times 10^8$ Jones. The source does not specify the electrical filter, so the two conventions should not be mixed or assigned spurious precision.

Two-point retrieval and its uncertainty. Assume linear, stable response for the three views. The radiometer two-point calibration isWith target radiances treated as known, differentiating givesNotice $b<0$ for an intermediate scene: raising the hot voltage while keeping the scene voltage fixed reduces the inferred scene radiance. Numerically $a\simeq 0.2621$, $b\simeq -0.1214$ in radiance units per volt.

The scene interpolation fraction is about $0.463$, giving $B_S\simeq 0.1056$ andFor independent white noise in each view, each hundred-sample mean has standard error ${\sigma }_{\bar{V}}=0.006/10=0.0006\mathrm{V}$. The radiance error isUsing the given temperature sensitivity,A scatter estimate closer to $0.007$ V changes this to roughly $0.14$ K. These are random standard errors under the averaging assumptions. Target-temperature/emissivity uncertainty, response drift and atmospheric or surface-emissivity corrections are separate: the retrieved quantity is a brightness temperature, not automatically the true sea-surface temperature.

Use one spectral resolution element, $\Delta \nu =0.5{\mathrm{cm}}^{-1}$, and assume comparable angular resolution horizontally and vertically, so the pixel solid angle is $\Omega \simeq {\theta }^2=6.67\times 10^{-6}\mathrm{sr}$. The question specifies only the vertical angle; without an assumption about horizontal field of view the diameter is not unique. Interpret throughput $\eta =0.5$ as dimensionless transmission. With primary collecting area $A_p=\pi d^2/4$, the power in that element iswhen $A_p$ is in square metres. There is no extra factor of $\pi$ multiplying the supplied spectral radiance.

For an integration time $t=1$ s, the order-of-magnitude effective noise bandwidth is $\Delta f\simeq 1/t=1$ Hz. The root-mean-square noise-equivalent power is then $P_n=\mathrm{NEP}\sqrt{\Delta f}=5\times 10^{-11}$ W.^[ref] A signal-to-noise ratio of 10 requires $P_s=5\times 10^{-10}$ W, and thereforeAn ideal rectangular average of a one-sided white noise spectral density has $\Delta f=1/(2t)$, giving $d\simeq 5.9$ cm instead. A circular field with diameter $\theta$ changes the diameter by another order-unity factor. These convention choices all require an aperture substantially larger than the $9.5$ mm diffraction estimate, so signal collection is the stronger constraint. The estimate treats the quoted NEP as applicable to the reconstructed channel; a detailed Fourier transform spectroscopy design would also specify scan weighting, background and how interferogram noise propagates into that channel.