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I've recently been learning about Fourier optics, specifically, that a thin lens can produce the Fourier transform of an object on a screen located in the focal plane. With this in mind, does the lens in a human eye produce a Fourier transform on the retina? Any help appreciated |
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No we don't see Fourier transforms--- we see classical (geometrical) optics, which is light propagating along geometric paths in the limit of small wavelengths. This limit makes it so that the light we get from a source is refocused into a point at a location corresponding to the source, there is no Fourier transform involved. The phenomenon you are talking about is a combination of the diffraction law together with the focusing law. To say that the lens produces the Fourier transform is a misleading way to say it--- all the lens does is focus the diffraction pattern in different directions onto different points on the photographic plate. The diffraction is what is doing the Fourier transform. If you place a diffracting object at one focus of the lens, the lens will project the diffraction pattern produced by a diffracting object at the focal distance on the other side of the lens onto a screen in such a way that different outgoing angles are each focused onto a different point. Since the diffraction intensity is equal to the Fourier transform of the transmission function, this will produce an image which performs a Fourier transform of the object at the focal point. To see an example of how diffraction produces Fourier transforms, see this answer: How does the Fraunhofer irradiance distribution look for a double slit aperture of different lengths? |
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Check Wikipedia on the subject. It says the image to be transformed has to be 1 focal length in front of the lens (not at infinity or at least further than a focal length). It says the image has to be in a transparent film, and be lighted from behind by plane waves, as from a point source at a distance. |
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The answer is no. A positive thin lens does have the property that the complex field amplitude at distance $f$ after the lens is the Fourier transform of the complex field amplitude at distance $f$ before the lens, where $f$ is the focal length of the lens. This is called a $2f$ system. However, it's wrong to say that that's an "image", because those distances don't match the condition for image-forming: $$ 1/a + 1/b = 1/f $$ Here, $a$ and $b$ are both $f$, and so obviously $1/2f \neq 1/f$. |
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As mentioned in the question, a thin lens will produce in its focal plane the Fourier transform of the optical field in its pupil, possibly multiplied by a quadratic phase term. However, to understand how this relates to imaging in the wave optics picture, we need to take a step back, and look at the situation more generally. Under the paraxial approximation, the propagation of an optical field can be modeled with the Fresnel diffraction integral: \begin{equation} U^\prime(x,y) = \frac{e^{i k z}}{i \lambda z} \mathrm{exp}\left[ \frac{i \pi (x^2 + y^2)}{\lambda z}\right] \ldots\\ \times \iint_{-\infty}^{\infty} U(\xi, \eta) \mathrm{exp}\left[ \frac{i \pi (\xi^2 + \eta^2)}{\lambda z}\right] \mathrm{exp}\left[ \frac{-i 2 \pi (x \xi + y \eta)}{\lambda z}\right] d\xi d\eta \end{equation} where $U(\xi,\eta)$ is an optical field, $U^\prime(x,y)$ is the field after propagation by a distance $z$, and $\lambda$ and $k$ are the wavelength and wave number, respectively. In the case of a thin lens, a transparency in contact with the lens, and a propagation distance equal to the focal length $f$, we can represent the input field as $$ U(\xi, \eta) = t_A(\xi, \eta) \mathrm{exp}\left[ \frac{-i \pi (\xi^2 + \eta^2)}{\lambda f}\right] $$ where $t_A$ is the amplitude transmission of the transparency, and the quadratic phase term is the wavefront curvature introduced by a thin lens of focal length $f$. If you plug this into the diffraction integral above, you see that, when $z = f$, the integral reduces to a Fourier transform and we have \begin{aligned} U^\prime(x,y) &= \frac{e^{i k z}}{i \lambda z} \mathrm{exp}\left[ \frac{i \pi (x^2 + y^2)}{\lambda z}\right] \iint_{-\infty}^{\infty} t_A(\xi, \eta) \mathrm{exp}\left[ \frac{-i 2 \pi (x \xi + y \eta)}{\lambda z}\right] d\xi d\eta \\ {} &= \frac{e^{i k z}}{i \lambda z} \mathrm{exp}\left[ \frac{i \pi (x^2 + y^2)}{\lambda z}\right] \mathcal{F}[t_A](x,y) \end{aligned} where $\mathcal{F}$ is the Fourier transform. I'm not explicitly stating it, but you can assume that the Fourier transforms I write are always appropriately scaled. In this case, if the FT is defined to take functions of $(\xi, \eta)$ and return functions of spatial frequency $(\alpha, \beta)$, you should assume the implied scaling $(\alpha, \beta) \rightarrow (\frac{x}{\lambda z},\frac{y}{\lambda z})$. Now, I won't derive it here because the integrals are huge, but if you use the first equation I wrote to propagate some object field by a distance $f$, then apply the wavefront modification by a thin lens of focal length $f$, and propagate another distance $f$, you will see that the quadratic phase terms all cancel each other, and the resulting field is exactly the Fourier transform of the object field, without even the quadratic phase term you get if the object is directly against the lens. If you have trouble with this, keep in mind the Fourier transform identity for the double FT of a function; this makes the derivation simple. More generally, this derivation can be applied to an arbitrary series of optical elements and propagation distances. With sufficient effort, it can be shown that, for a paraxial optical system described by an ABCD matrix, an optical field is propagated through the system by: \begin{equation} U^\prime(x,y) = \frac{e^{i k L_0}}{i \lambda B} \mathrm{exp}\left[ \frac{i \pi D(x^2 + y^2)}{\lambda B}\right] \ldots\\ \times \iint_{-\infty}^{\infty} U(\xi, \eta) \mathrm{exp}\left[ \frac{i \pi A (\xi^2 + \eta^2)}{\lambda B}\right] \mathrm{exp}\left[ \frac{-i 2 \pi (x \xi + y \eta)}{\lambda B}\right] d\xi d\eta \end{equation} where $L_0$ is the effective optical path length through the optical axis of the system. This, of course, is still valid only for a coherent optical system. One way of thinking about this in the context of an imaging system (like an eye or a camera) is that it only applies to the field due to a single point in the scene being imaged. The ultimate image can be obtained by coherently propagating the field from each object point, taking the magnitude squared of the resulting field to get its intensity, and then adding the intensities from each object point. Thus, I suppose one could claim that we see a superposition of Fourier transforms from each object point, rather than directly seeing a Fourier transform. Indeed, the image on your retina doesn't look like the picture you get if you take some everyday scene and Fourier transform it on your computer. Nonetheless, lenses do perform Fourier transforms on optical fields. When considering an imaging system however, you must consider where the field that is being transformed is, relative to the lens. In general, this field is not the field at the object you are looking at; it is the field some distance in front of your pupil, and in a real-world situation, it is not simply one coherent field from one source point, but an incoherent superposition of fields from every point in your field of view. As a practical matter, this means that incoherent imaging is rarely simulated with the ABCD integral above. This sort of computation is useful for coherent imaging systems (a telescope is a good example, if you're only talking about stars and not extended objects), but in the incoherent case it is much simpler to simulate imaging purely by applying the MTF/OTF as a convolution or linear filter. Even in this case, however, the computation is still based on a Fourier transform. |
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