Why is the white light of the interferogram produced by using Michelson Interferometer necessarily symmetric? This is really hard to think. white http://upload.wikimedia.org/wikipedia/en/d/dc/Colored_and_monochrome_fringes.png
If we assume the wave-fronts that initially enter the system to produce this image are symmetric, than the result is intuitively symmetric as all the components used to produce it are too. And anyone who looks at the images you have shown, will clearly see a symmetry about the Y-axis. Assuming they do not look for minuscule errors due to wave-front discontinuity.
Let's make sure we have all the right ideas by verifying the definition of "symmetry". "Symmetry across a Y-axis" is a reflectional symmetry. Meaning that both sides, if split down the center and one side flipped, are equal to each-other.
I suppose the other question is why this symmetry is visible for such a system? It was made using a "Michelson Interferometer", if we look at such a device it is clearly also reflectively symmetric (Neat). Now, keep in mind that all wave fronts curve. Given the reflective path, the wave-fronts on the edge would start to fall further behind the center wave-front, giving it a different position. The shorter path will have edge wave-fronts in different positions than the longer path. This will create the interference we see. Both paths are almost equal, we are talking about a difference in distance on the order of fractions of wavelengths.
Hope that helps.
Due to the nature of any light source, the vibrating particles making the light can be out of sink with each-other. This is not true for a laser, once the laser is started the selected waveform for that laser will reinforce the motion of the particles, making them move in synchronization with each-other. If the particles producing the light are moving together the discontinuities will diminish.