To give an unmathematical catchy answer, let's look at Fraunhofer diffraction in double slit experiment.
Interference at the observation plane depends on slit parameter $d$. What is the frequency of slits? E.g. $1\,\text{mm}\frac{1}{d}$: number of slits per length. Concluding frequiency in the setup. The following argumentation links this frequency to the fourier transform. The **physical significance is in the real optics setup**. The setup is easier described, when transformed in fourier space.

The double slit link on high school level above gives all the math without integrals. After **visualizing the following concept**, you see that the integrals are just math to convert the diffraction pattern in fourier space via fourier transformation.

Using trigonometry first compute phase difference $\Delta\phi(\theta,d)$.
Go deeper in this concept using a sketch to visualize phase difference of $n\cdot\lambda$, $n\in\mathbf{N}$ as bright maxima in diffraction pattern. There is no magic in the next step. It's just a another point of view: Try to grasp $\frac{1}{d}$ as a parameter on its own: $\Delta\phi(\theta,\frac{1}{d})$.

Fourier space is a **synonym for frequency domain**. Acoustics examples are given in Eichenlaubs SE answer and ptomatoes optics explanation.

No literature: Calculate and understand yourself with the links above.

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.

## Best Answer

A fourier transform over spatial data gives a spectrum of spatial frequencies. Where a transform of temporal data gives the amplitude versus cycles per second, spatial frequency has units of cycles per meter (or whatever length unit).