Explanation of formation of Newton's rings I was asked to do a lab report in my University about Newton's rings experiment that we made in lab. I understand that the aim of the experiment is to measure the wavelength of a light after the formation of Newton's rings and I understand the mathematical derivation of formulae used for that (relation between radius and wavelength due to constructive or destructive interferences).
I am confused a little about the behaviour of the light inside the setup. When light enters the plano-convex lens what are the cases that lead the light to reflect and refract in a specific way that when interference occures bright fringes are formed due to a constructive one or dark fringes are formed due to a destructive one?
I hope somebody can help me to understand how the rings are formed exactly so that can help me understand what I am working on.
 A: The Wikipedia article gives a pretty good explanation of the principle of Newton's Rings, but maybe the lens is confusing you. What's important is just the air gap between the lens and the flat plate, and the fact that with a lens the thickness of the air gap increases with distance away from the contact point of the curved surface of the lens and the flat surface of the plate.  
If you simply stack an almost flat plate on top of a perfectly flat plate, you will see Newton's Rings, usually distorted into something resembling wood grain in a plank.  The rings amount to a topographic map of the air gap between the plates.  
What causes Newton's Rings is the interference of light. When you look at the stack (whether it's a lens or a plate on top), for each point in the stack your eye receives two rays: one that has passed through the top piece of glass and reflected off its bottom surface, and one that has passed through that bottom surface, traversed the air gap between the glass pieces, and reflected off the top surface of the bottom piece of glass. 
Both rays reach your eye, but the one that reflects off the bottom piece of glass has to travel farther, so its phase is shifted by an amount corresponding to the thickness of the air gap.  That means that the two rays arrive with a phase difference.
What you see depends on that phase difference.  If it corresponds to an odd integral multiple of half a wavelength of the light, you don't see any light because the two rays cancel out.  If it corresponds to an even integral multiple of half a wavelength of light, you see a bright point (at that point in the stack) because the two rays reinforce each other. 
If the top and bottom pieces of glass were perfectly flat the air gap would be uniform; so with collimated illumination and viewing it from a long distance away, you would see a uniform brightness across the assembly.  But in the case of the top piece being a lens with the curved surface down against a flat plate, you see a "map" of the air gap.  By counting the fringes out from the contact point, you can calculate the curvature of the lens surface.  If the fringes aren't smoothly curved, you know that the lens surface curvature is distorted.
A: From what you've written, I deduce that you know that the path difference between light reflected back up from a point P on the lower surface of the lens and reflected back up from the point Q immediately below it on the plane surface is $\frac{r^2}{R}.$ Here, $R$ is the radius of curvature of the lower face of the lens and $r$ is the 'ring radius', that is the distance from the point of contact between lens and surface to point Q. This path difference can be expressed as $\frac{r^2}{R\lambda}$ wavelengths, $\lambda$ being the wavelength of the light. So the path difference between the light reflected from P and Q gives rise to a phase difference of $2\pi\frac{r^2}{R\lambda}.$
There is another phase difference, independent of any path difference. Light travelling through air and reflected from glass (as at Q in our case) experiences a phase reversal, that is a phase change of $\pi$. There is no phase change for light travelling through glass and reflected from the glass-air interface (at P in our case). 
So the light reflected from P and Q will differ in phase by $2\pi\frac{r^2}{R\lambda}-\pi.$
Light reflected at P and Q will superpose and interfere constructively if the phase difference is $2\pi n,$ with $n = 0, 1, 2…,$ that is a whole number of cycles. The condition for a bright ring is therefore
$$2\pi\frac{r^2}{R\lambda}-\pi=2\pi n\ \ \ \ \ \text{in other words}\ \ \ \ \ \frac{r^2}{R\lambda}=n+\frac{1}{2}.$$
