If I take the Fourier transform of the autocorrelation of a signal in time, I will get the power spectral density.
It so happens that the autocorrelation function is a Fourier transform pair of the power spectral density. This is not to say that the only way to calculate the power spectral density is from the autocorrelation function.
As I stated at https://physics.stackexchange.com/a/309544/59023, the power spectral density, $S_{k}$, is proportional to the square of the magnitude of the Fourier transform of a signal, i.e., $S_{k} \propto \lvert X_{k} \rvert^{2}$.
Although what is the physical meaning of the power spectrums of $E(t)$ and $I(t)$, respectively, and their differences?
First, let me use the generic symbol $X_{k}$ to represent the Fourier transform of the time domain signal $x_{n}$.
The words power spectrum are somewhat ambiguous here. In principle, one can compute a power spectrum (i.e., respective value vs. frequency) from each component of $\mathbf{E}$ or its magnitude. One can also compute the amplitude spectra, $A_{k} \propto \lvert X_{k} \rvert$, of the signal.
In the following, I will assume you are asking about $S_{k}$ and not $A_{k}$ for each of these.
The power spectrum of $\mathbf{E}$, whether of components ($E_{j}$) or vector magnitude ($\lvert \mathbf{E} \rvert$), describe the power of the field as functions of frequency with units (if properly normalized) of (V m-1)2 Hz-1. This is useful when trying to determine whether there exists, e.g., a wave at a given frequency which would show up as peak above the backgroun in $S_{k}$. If the oscillations exist only along the x-component of $\mathbf{E}$ (i.e., a longitudinal, electrostatic oscillation), then the spectrum of both $\lvert \mathbf{E} \rvert$ and $E_{x}$ would show a frequency peak but not $E_{y}$ or $E_{z}$.
The intensity, as you have written it, is just the field energy density multiplied by a constant. Thus, the power spectrum of $I$ would be qualitatively similar to that of $\lvert \mathbf{E} \rvert$.
Although why is this true? Wouldn't this power spectrum only indicate the frequency of the signal itself and not the photons comprising it?
I am not sure about what you were told and whether you are correctly conveying that information. A discrete Fourier transform or DFT (i.e., what you use in practice on real signals through algorithms like the FFT) is not the same as a continuous Fourier transform (CFT). In a DFT, the frequency bin width is defined as:
$$
\Delta f = \frac{ f_{s} }{ 2 \ N } \tag{1}
$$
where $f_{s}$ is the sample rate of the signal [e.g., vectors per second] and $N$ is the number of individual points used in the DFT.
In a CFT, the minimum $\Delta f$ is mathematically zero (i.e., infinitesimally small) but quantum shows us that energy/momentum are quantized and thus have discrete values. Therefore, there are physical limits on the lower bound of $\Delta f$. In this case, a variant of the uncertainty principle is applicable, called the time-energy uncertainty principle, which is roughly given as:
$$
\Delta E \ \Delta t \geq \frac{\hbar}{2} \tag{2}
$$
where $\hbar$ is the Planck constant and $\Delta Q$ is the minimum resolution of quantity $Q$.
Thus, the transition has a known energy change but we cannot know this better than that given by Equation 2. For photons, we can directly convert energy to frequency with some constants, i.e., $E = h \ \nu$, thus we have the limitation on the frequency resolution of the emitted photons.
There's a bit of a misconception here. A prism causes dispersion, which is the decomposition of a broad spectrum of light into its spectral components via the components' deviation angle from their original trajectory - but this is not in any way related to Fourier transform, rather it's because of Snell's law of refraction, and the fact that refraction changes with the frequency of the light (color).
You can talk about the Fourier transformation of light, but in a different context: spatial frequencies. Much like every sound signal is composed of temporal frequencies, every optical image is composed of spatial frequencies, and one can analyze the Fourier transform of an image to learn about the spatial composition.
One of the most useful cases of Fourier transform in optics is taking the Fourier transform of an optical system's impulse response, which is the image of a perfect point source of light, a.k.a the point spread function (which analogous to linear system's impulse response in signal processing). The real part of the normalized Fourier transform of the point spread function is called the modulation transfer function and is one of the most common metrics to evaluate the quality of an optical system.
Best Answer
The issue is that a sine wave on a finite interval is not the same function as a pure sine wave, and so the Fourier transforms will be different.
Note that in the rest of this answer, I'm going to assume by "Fourier transform" you mean "Fourier transform over the entire real line," and not "a Fourier series over a finite region of the real line." But I will mention just for completeness that if you have a sine wave with period $T$ and you do a Fourier series over a finite interval of length $kT$ for some positive integer $k$, then in fact you would find that the only non-zero Fourier coefficient is the one corresponding to the sine wave of period $T$.
Anyway back to your question. The Fourier transform $\tilde{f}(\omega)$ of $f(t)$ is
\begin{equation} \tilde{f}(\omega) = \int_{-\infty}^\infty {\rm d} t e^{-i \omega t}f(t) \end{equation} If we take $f(x) = e^{i \Omega t}$ for $t_1 \leq t \leq t_2$, and $0$ otherwise, then \begin{equation} \tilde{f}(\omega) = \int_{t_1}^{t_2} {\rm d} t e^{i (\Omega-\omega) t} = \frac{e^{i (\Omega-\omega) t_2} - e^{i(\Omega-\omega) t_1}}{i (\Omega-\omega) } \end{equation} There are a few useful special cases to know.
One is when $\Omega=0$ -- then $f(t)$ describes one pulse of a square wave. Taking $t_2=-t_1=T/2$ for simplicity, the Fourier transform is \begin{equation} \tilde{f}(\omega) = \frac{2 \sin (\omega T/2)}{\omega} \end{equation} This characteristic behavior $|\tilde{f}(\omega)|\sim 1/\omega$ is common to "sharp edges" in the time domain signal, which excite Fourier modes of arbitrarily large frequencies. The slow falloff $\sim 1/\omega$ can cause many issues in dealing with sharp edges in time domain signals, when transforming into the frequency domain, in practical applications.
Another interesting limit is $\omega \rightarrow \Omega$. In fact the Fourier transform is equal to $T$, and diverges in the limit of infinite times! This is precisely the frequency of the truncated sine wave. We can understand the $T\rightarrow \infty$ limit by proceeding carefully. For simplicity let's assume $t_2=T/2>0$ and $t_1=-t_2=-T/2$, and refer to $f_T(\omega)$ as the truncated sine wave, with the duration of the window being $T$. Then \begin{eqnarray} \tilde{f}_T(\omega) &=& \int_{-T/2}^{T/2} {\rm d} t e^{ -i (\omega-\Omega) t } \\ &=& \int_{-T/2}^0 {\rm d} t e^{ - i (\omega - \Omega ) t } + \int_{0}^{T/2} {\rm d} t e^{- i (\omega - \Omega )t} \\ &=& \frac{1 - \exp\left( \frac{-i T}{2}\left(\omega - \Omega \right) \right)}{i(\omega - \Omega )} + \frac{\exp\left( \frac{i T}{2}\left(\omega - \Omega \right) \right) - 1 }{i(\omega - \Omega )} \\ &=& \frac{2 \sin \bigl((\omega - \Omega) T/2\bigr)}{\omega - \Omega} \end{eqnarray} Now we can take the limit $T\rightarrow \infty$, using the delta function representation \begin{equation} \lim_{\epsilon\rightarrow 0} \frac{\sin (x/\epsilon)}{\pi x} = \delta(x) \end{equation} where $\delta(x)$ is a Dirac delta function.
This yields \begin{equation} \lim_{T \rightarrow \infty} \tilde{f}_T(\omega) = 2\pi \delta(\omega-\Omega) \end{equation} In the limit, the function $\sin T (\omega-\Omega)/(\omega-\Omega)$ becomes $\sim \delta(\omega-\Omega)$. This corresponds to your intuition that the Fourier transform should be dominated by the frequency of the truncated part of the sine wave.
Note: Thanks to @nanoman, who pointed out an error in an earlier version of this post.