I am reading some introduction on Fourier optics. In the text, the simplest example would be a single len system (focal length f) with an object f(x,y) sitting on the front focal plane of the len while the image is the corresponding spatial Fourier transformation [tex]F(\frac{x}{\lambda f}, \frac{y}{\lambda f})[/tex] where [tex]\lambda[/tex] is the wavelength.(adsbygoogle = window.adsbygoogle || []).push({});

This spatial transformation is quit confusing while being compared temporal case. In temporal case, Fourier transform of f(t) is given as [tex]F(\omega)[/tex] which is defined in frequency domain. Hence, we cannot directly observe that in physical space, we need to put [tex]F(\omega)[/tex] into some instrument to observe the frequency spectrum. But

for the spatial case, it seems that the Fourier transform is nothing but just a rearrangement or mapping and scaling from the source to the image, but anyway the image are still in the physical space because the image can be directly observed on the

rear focal plane with screen or film. I wonder if my understand is correct?

Another question on the complex function. In most cases, the Fourier transform will be complex-valued. And for len system, [tex]F(\frac{x}{\lambda f}, \frac{y}{\lambda f})[/tex] could also be complex, but I think physically the film will only record the intensity instead. But what about the phase? What's the role of the phase in this case? Or I ask in this way. If I keep the intensity for every point on the image unchanged but I shift the phase by a number or randomly change the phase, how difference the image will be? in what way?

Thanks.

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# How fourier optics works?

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