= Solution
{c}
Let the two returning monochromatic fields at the chosen detector port be
$$
E_1=a\cos(\omega t),\qquad E_2=a\cos(\omega t+\varphi),\qquad
\varphi=2\pi\nu\Delta d.
$$
Here $\nu$ is <wavenumber>, rather than temporal <frequency>. An optical-cycle average uses $\langle\cos^2\omega t\rangle=1/2$ and $\langle\cos\omega t\cos(\omega t+\varphi)\rangle=\cos\varphi/2$. In units absorbing the detector's field-to-power conversion,
$$
\boxed{P=\langle(E_1+E_2)^2\rangle=a^2[1+\cos(2\pi\nu\Delta d)].}
$$
The choice of output port can reverse the cosine sign. For mutually incoherent components of the <radiation spectrum>, add powers rather than instantaneous fields, giving
$$
P(\Delta d)=\int_0^\infty Q(\nu)[1+\cos(2\pi\nu\Delta d)]\,d\nu.
$$
Subtracting the constant term leaves the <Fourier cosine transform> of the detected spectral weighting $Q(\nu)$, including throughput and detector response.
The printed prefactor requires a normalization convention. In the derivation, $a$ is the amplitude of each returning field at the detector, not an entrance amplitude with input power $a^2$. For a lossless balanced <beam splitter> and literal input power $P_{\rm in}$, <conservation of energy> instead gives $P_{\rm out}=P_{\rm in}[1+\cos\varphi]/2$, so the source formula identifies $a^2=P_{\rm in}/2$. Taking both $P_{\rm in}=a^2$ and the printed formula literally would give $P_{\rm out}=2P_{\rm in}$ at zero <optical path difference>, a concrete normalization contradiction.
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