Show that an amplitude modulator gives a mirror symmetry in optics

Question

In optics, we can use amplitude modulators as optical components to alter the amplitude of a wave. I'm currently working with simulating some propagations through space, and realize that an amplitude modulator gives a mirror symmetry in some instances. I'll try to explain the situation below:

Suppose we have an incident plane $E$ wave that's moving through an amplitude modulator $T$. Both $E$ and $T$ can have spatial variation. I then want to show that the wave $E_f$ at a plane in the far field zone is a mirror image and symmetric around the origin. In order to try to show this, I wanted to use the Huygen Fresnel integral in the far field for a propagation distance $L$:

$$E_f(u,v) = \int \int E(x,y) T(x,y) e^{-jk (xu+yv)/L} dxdy$$

Now, I somehow want to relate $E_f(-u,-v)$ to the result above. So, by definition:

$$E_f(-u,-v) = \int \int E(x,y) T(x,y) e^{jk (xu+yv)/L} dxdy = \overline{\int \int E(x,y) T(x,y) e^{-jk (xu+yv)/L} dxdy}$$

$$ = \overline{E_f(u,v)}$$

Where I use that $E(x,y) T(x,y) \in \mathbb{R}$. However, this doesn't seem as the right answer, maybe since my assumption is wrong. Either way, I hope someone could give me some insights along the way to help me with this derivation, since I'm very mathematical of myself and would like to prove this rather than just accept that it's how it is.

This is an example:

Thanks.

Answer 1

Your calculation is correct! Why do you say that it doesn't seem right? If your confusion is in the assumption that $E$ and $T$ are real, here is the explanation. Before starting, remember that any complex variable can be decomposed as $re^{i\phi}$ where $r$ is the amplitude and $\phi$ the phase.

Firstly, you say that the incident wave is plane, which means its wavefronts are parallel to the input plane. Let us decompose the field in the input plane as $E(x,y)=E_0(x,y)e^{i\phi(x,y)}$. By definition, a wavefront is defined as a surface of constant phase, so for the phase to be constant in this plane we require that $\phi(x,y)=\phi_0$ is constant. Furthermore, as overall phase is arbitrary, we can simply take this to be zero and so $E(x,y)=E_0(x,y)$ is real.

Even more simply, $T(x,y)=T_0(x,y)$ must be real because it is an amplitude modulator and therefore cannot change the phase of the wave (again apart from some arbitrary constant). Thus we get your result that $E_f(-u,-v)=E_f(u,v)^*$. This will in general be complex, but if we only care about the amplitude in the far field then it shows that there should be symmetry about the origin (as shown in your picture, which I am guessing shows the intensity, proportional to the squared amplitude of the field).