Why do the Fringes in Michelson Interferometer move inward/outward?

Question

We did this experiment using Michelson Interferometer of monochromatic source $\lambda$ to find the refractive index $n$ of dry air in a chamber of length $L$ by counting the number of fringes moved radially outwards as the pressure in the chamber decreases.

I am struggling to explain why do the fringes move outward when the pressure in the chamber decreases (We started from 290 torr and stopped at 120 torr).

This is my argument: As the pressure decreases, the refractive index is supposed to decrease (as $n \propto p$ for air $\approx$ ideal gas). This should reduce the path difference $2(n-1)L$ between the two interfering beams. The path difference between the incident and reflected beams is $2d\cos\theta$. $\theta$ is the angular separation of our fringe from the center. $2d$ is the mirror displacement but for our concern $2d = 2(n-1)L$ and thus we have the relation between $n$ and $\theta$ as $P.D. = 2L(n-1)\cos\theta$

For a bright fringe, $m\lambda = 2L(n-1)\cos\theta$. So for a given fringe, m is fixed and hence P.D is fixed. As $n$ decreases, $\cos\theta$ should increase in a way to keep P.D. constant.

But according to my argument, $\theta$ should decrease. That means I am wrong somewhere. But where?

Answer 1

if you look at the circular fringes and they seem to go outward, if you look well you can also see them going in ward, it is just the change from dark to brightwich changes , but you are right, the first impression is allways to see it going outward. Maybe you can repeat the experiment starting with low pressure, and you will also see the fringes moving outward.