This is a manifestation of "Frequency Aliasing" which shows up in lots of situations where you have data which is sampled at discrete intervals. For a more in-depth explanation of aliasing in general, you should look for references on the Shannon-Nyquist Sampling Theorem. In brief however, it is sufficient to explain that when data is sampled at some given resolution, called the sampling period $T_s$, then any component of the signal with a frequency above $\frac{1}{2T_s}$ (this is called the Nyquist frequency) will be "aliased", meaning they will appear to have a frequency of $\frac{1}{T_s}-f$ where $f$ is the frequency of the signal component in question. In this case the sampling period $T_s$ is the spacing of the pixels in the camera.
As you can see from the un-aliased central rings, the frequency of the rings increases radially. This is a spatial frequency, as opposed to temporal frequency which people are more used to, but the mathematics is the same. As the ring frequency increases, it crosses the Nyquist frequency of the camera, and you see the pattern repeated as the frequency components that compose it are aliased below the Nyquist frequency of the camera.
In fact, it is pretty straightforward to calculate where the secondary rings will show up (if you allow me to heavily gloss over some of the details). The spherical surface of the lens can be approximated as a parabolic surface $$y \propto r^2 $$ where $y$ is the height of the surface from the reference plate, and $r$ is the radial distance from the center of the primary rings. The frequency of the rings is related to the slope of the surface, which in this case is clearly linear in $r$, so without bothering to compute any constant factors, we can define $$f =Ar$$ where $f$ is the ring frequency and $A$ is a constant with units of $\frac{1}{\mbox{length}^2}$ or $\frac{\mbox{frequency}}{\mbox{length}}$.
Now, we could do some Fourier transforms and derive the exact form of the aliased ring image, but that is a bit more $\LaTeX$ than I feel like typing, and I suspect it is a bit beyond the scope of the question. Instead, I'll just calculate the location of the centers of the aliased ring patterns. So where are the centers of the ring patterns? They are the locations where the apparent ring frequency is zero. This is trivial for the central pattern, of course. Now where will the first set of aliased rings lie? We want to set the aliased frequency to zero, and solve for $r$; so
$$\begin{array}{lcr}
\frac{1}{T_s}-f = 0 &\mbox{and} &f=Ar\\
\frac{1}{T_s}-Ar = 0&\\
r=\frac{1}{AT_s}& \mbox{or} & r=\frac{2f_n}{A}
\end{array}$$
Computing the value of $A$ is a bit unnecessary for this explanation, so I'll simply say that it will depend on the wavelength of the light and the radius of curvature of the lens you are using. The resulting expression gives a simple description of where the rings will show up, and could serve as a guide for which lenses your microscope could measure without aliasing (i.e. when the aliasing won't appear until $r$ is larger than your field of view). Clearly, as the pixel spacing of your camera grows, aliasing becomes more of a problem, as it occurs closer and closer to the central rings; and as the radius of curvature of the lens increases (as the lens becomes less curved) aliasing becomes less problematic.
Also interesting is that the radial position of the rings in the diagonal direction is larger. This would seem counter-intuitive given the expression derived above, but keep in mind that the explanation I've given is vastly simplified. If you endeavor to derive a more general result using the Fourier transform, the diagonal rings will be correctly described.
The color effects are due to the Bayer filter on the camera. This filter is what allows the camera to capture color images, but it also has the effect of giving a slightly different sampling rate for each of the different colors. This means that the aliased pattern will have some color effects, but this is not "real", in the sense that it is only an effect of the camera itself. Bayer color effects can be seen in other situations as well, including situations that are more commonly described as Moiré effects than aliasing. For example:
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Moiré and aliasing effects are, of course, closely related, and are in some sense the same thing; but frequency aliasing is a very general concept that is applicable to any sort of sampled data, while the Moiré effect is generally the name given to the patters resulting from overlapping periodic patterns, rather than sampling specifically.
The reflected light you refer to will not give you interference fringes, because the path difference between the two routes (via top of lens and via another surface) is too big. If the path difference is greater than a few wavelengths (perhaps about 50* but see Farcher's comments below) – more if you are using laser light) the wavelength-spread of the light will mean that some wavelengths will interfere constructively and others destructively (and everything in between), so the reflected light that you refer to will just contribute to a brighter background, without giving its own fringes (that is rings).
On the other hand, light reflected from the top surface of the flat glass and from the bottom surface of the lens will produce fringes, because near the centre (the point of contact) the path difference is so small. For example, if the radius of curvature of the bottom surface of the lens is 1.0 m, then at 5 mm from the centre, $\Delta y = 125 \times 10^{-7} \text{m},\ $ which is probably about 20 wavelengths. Note that further out, even these fringes start to blur out, for the reason given in my first paragraph.
Best Answer
If the source was a laser you may well see fringes due to the interference of waves reflected off the top of the lens and the bottom of the glass plate however it is unlikely that observable fringes are seen if you use a conventional light source.
The important factor is the size of the optical path difference between the top of the lens and the bottom of the glass plate.
The path difference between the bottom of the lens and the top of the glass plate is of the order of a few (tens of) wavelengths of light whereas that from the top of the lens and the bottom of the glass plate is many tens (hundreds of) thousands wavelengths of light.
The coherence length of light from a conventional light source even it is called monochromatic (eg sodium) is much less than the optical path difference between the top of the lens and the bottom of the glass plate and so although the waves overlap they do not form a visible interference pattern.
Other contributory factors are that monochromatic light from a conventional light source is composed of a band of wavelengths and in the case of light from a Sodium source consists of two wavelengths of roughly equal intensity about $0.6 \,\rm nm$ apart.
All these preclude to observation of an interference pattern when there is a large optical path difference between the overlapping waves.