Fourier optics - Transfer function of free space

Question

In all consulted literature, the transfer function of the free space is given as follows: $$\exp(-i k_z d) = \exp(-i2 \pi d \sqrt{1/\lambda^2 -\nu_x^2-\nu_y^2})$$

When referring to this source, they derive the transfer function from the following equation: $$H(\nu_x,\nu_y) = \frac{f_{in}(x,y)}{f_{out}(x,y)}$$

I'm wondering why they do it this way (note: I've already seen this in other sources). I thought the transfer function is defined in the frequency domain rather than in the time domain (frequency and space for the fourier optics respectively).

Answer 1

The source is specifically referring to the input and output being a plane wave. The key point on that slide is that complex exponentials in the form $\exp(i 2\pi\nu_x x)$ are eigenfunctions of linear, shift-invariant optical systems. Thus, a complex exponential signal goes into the system, and another complex exponential with the same frequency goes out. A plane wave has the same form as this complex exponential. As a plane wave propagates through a linear system, it remains a plane wave. It can only be amplified/attenuated and phase-delayed. New spatial frequencies (plane wave components) cannot appear in the output.