# Double Fraunhofer , Fourier transform to the image plane?

Consider the diagram below:

I know that from Franhofer diffraction we have: $$u_f(x_f)\propto \int u_o(x_o) e^{-k\frac{x_f}{f}x_o} dx_o$$ Assuming small angles.

I also know that $u_i(x_i)$ is related to the Fourier transform of $u_f(x_f)$ and therefore that: $$u_i(x_i) \propto \int u_f(x_f) e^{-\gamma x_i} dx_i$$ In the most general case what form does $\gamma$ take and why?

I would guess that in the small angle approximation we have: $$\gamma=k \frac{x_i}{d_{fi}}$$ But I think the above equation is more general and therefore their sould be a more general form of $\gamma$.

• I think the imaginary $i$ is missing in the integrals. Moreover, in the first integral, the exponential only represents the phase shift caused by the lens, not the propagation towards the focus. So I don't quite get what you're trying to ask. – Han-Kwang Nienhuys Jun 11 '16 at 18:10

First, I'll call $z$ the horizontal axis in your graph, and set the origin of this axis where we have the field distribution $u_o$. The plane at $z=0$ will have an axis (entering the screen) called $η$ and the remaining axis called $ξ$. Then, at $z=z_1$, there will the input of the lens and, in that position, a field distribution that we will call $u_l$. Behind the lens (whose focal length will be $f$), on its exit, we have a field $u'_l$. Then, at a distance $z_2$ from the exit of the lens, we have the final distribution of the field $u_i$ on a plane of coordinates $(u, v)$. To be consistent with your drawing, $z_1$ is the not specified distance between object and lens and $z_2$ is the sum of the focal length and the unknown distance $d_{fi}$.
From linearity of the wave propagation, we can have that the image field $u_i$ can be seen as the superposition integral between the object field $u_o$ and a certain impulse response $h$ of the whole system; note that this response will depend both on the position on the plane of the object and on the image's one. $$u_i (u, v) =\int\int_{-\infty}^{+\infty} h(u, v;ξ, η)\ u_o(ξ, η)\ dξ\ dη$$ Then the problem is on determining the expression of $h$. To find it, suppose that the object is point source (a δ function). Then, we'll have an incident spherical wave on the lens, diverging from the point $(ξ, η)$. In paraxial approximation, it can be written as: $$u_l = \frac{1}{jλz_1} e^{j\frac{k}{2z_1}[(x-ξ)^2 +(y-η)^2]}$$ Supposing that the lens has a pupil function $P(x, y)$ and that it introduces a phase shift depending on its thickness, can be demonstrated that after the lens we have a field: $$u'_l (x, y) = u_l (x, y)\ P(x, y)\ e^{-j\frac{k}{2f}(x^2+y^2)}$$ Then, we can use Fresnel diffraction equation to take into account the $z_2$ propagation and get: $$h(u, v;ξ, η) =\frac{1}{jλz_2} \int\int_{-\infty}^{+\infty} u'_l(x, y)\ e^{j\frac{k}{2z_2} [(u-x)^2 + (v-y)^2]}\ dx\ dy$$ Doing substitutions: $$h(u, v;ξ, η) =\frac{1}{λ^2 z_1 z_2} e^{j\frac{k}{2z_2}(u^2+v^2)} e^{j\frac{k}{2z_1}(ξ^2+η^2)} \int\int_{-\infty}^{+\infty} P(x, y)\ e^{j\frac{k}{2} (\frac{1}{z_1}+\frac{1}{z_2}-\frac{1}{f}) (x^2+y^2)} \ e^{-jk [(\frac{ξ}{z_1}+\frac{u}{z_2})x+(\frac{η}{z_1}+\frac{v}{z_2})y]}\ dx\ dy$$ This formula is quite ugly, but it can be simplified with some approximations and restrictions.