 Solution

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.