How does light look like when it is 180° out of phase? When two lights are 180° out of phase, would it look like this?
In the photo below, the left side is flipped and it is a mirror image.

 A: When people talk about phases of a light field, they usually refer to the electric field $E$. To understand why this is of interest, consider  two light sources in 1D. Their (skalar) electric fields at the position $z_0$ are given by $E_1(z_0)=|E_1| e^{i\phi_1(z_0)}$ and $E_2(z_0)=|E_2| e^{i\phi_2(z_0)}$. The relative phase $\phi_2(z_0) - \phi_1(z_0)$ of these fields determines, whether they are constructively or destructively interfering. Since photographic plates and ccd chips do not measure the electric field, but measures the intensity, $I\propto |E_{total}|^2 = |E_1(z_0) + E_2(z_0)|^2$, the relative phase is important.
What you are doing is fundamentally different. You took the intensity of an image $I(x,y)$ for all negative $x$ values and just reused it for all positive $x$. Not only is it rather meaningless to ask  about a relative phase of two intensities in general, but you do not have two intensities at the same point in the image.
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I'm not sure whether or not you know how two electric fields, which are 180deg out of phase look like. Here are two simulations. The light sources are located at $x=\pm 12$ and $y=0$. In the left image the electric field of the sources are out-of-phase by 180deg, and in the right image they are in-phase

In both I plot the intensity.
