Consider the figure that I found from an answer here:
Michelson interferometer circular fringes
Now the path difference in the image in clearly $$\Delta d= 2d cos\theta$$ and I know a phase difference arises due to internal reflection by the beam splitter. Considering that all together, we have the condition for dark fringes to be: $$2d cos \theta=m\lambda$$ Here, I see two variables $m$ and $\theta$. My question is what is this $\theta$? It is simply the angle of incidence of the light on the mirrors, isn't it?
Moreover, how can I allow two variables to be present in my equation? For example, if I consider $m=3$, then it gives me some value of $\theta$ for a given $\lambda$ and $d$. What exactly is that value of $\theta$? If this is the angle of incidence for the light, then am I not considering the contributions of other light fallen at a different incidence angle to create the interference pattern?
Where do I lack in these concepts? Please note that the mirrors are exactly perpendicular to each other, creating circular fringes on the detector. Why are the fringes circular- is still a question to me.
