Michelson Interferometer ring contraction

[epsilon]

I have been trying for hours to understand what is physically causing the interferometric rings to contract when the separation of the mirrors is reduced.

From the equation: $m\lambda = 2Lcos\theta$, where $m$ is the number of fringes, if we consider just one fringe at a fixed wavelength, $m\lambda$ is constant and hence $2Lcos\theta$ is also constant.

Hence reducing $L$ causes $cos\theta$ to increase, which is analogous to reducing $\theta$. [Is this where I'm going wrong?]

Question 1: When reducing d in the image above, does it matter if we are moving $L_1$ towards $L_2$ or vice versa? (Is it directionally dependent?)

Question 2: The image suggests that $\theta$ is only linked to $L_1$. If I move $L_1$ towards $L_2$, the adjacent side of the right-angled triangle is getting shorter, and hence $\theta$ must be increasing. But this goes against what happens when it is considered mathematically.

How can we tell PHYSICALLY whether the rings are contracting or expanding? Thank you.

 

[blue_leaf77]

Hence reducing $L$ causes $cos\theta$ to increase, which is analogous to reducing $\theta$. [Is this where I'm going wrong?]

That's correct, and should confirm that the fringes contract when the mirror separation is reduced.

Question 1: When reducing d in the image above, does it matter if we are moving L1L_1 towards L2L_2 or vice versa? (Is it directionally dependent?)

It doesn't matter.

Question 2: The image suggests that θ\theta is only linked to L1L_1. If I move L1L_1 towards L2L_2, the adjacent side of the right-angled triangle is getting shorter, and hence θ\theta must be increasing. But this goes against what happens when it is considered mathematically.

I don't get your point.

How can we tell PHYSICALLY whether the rings are contracting or expanding? Thank you.

I think you can place a movable marker on a particular fringe, when you decrease $d$, this marker should go radially inward. That's when you would say the fringes are contracting.
Btw, notice that the zeroth order is located at infinity. This means, if you decrease $d$, all fringes will actually contract such that in the end there is only bright area on the screen when $d=0$, i.e. when the mirrors coincide.

 

[epsilon]

That's correct, and should confirm that the fringes contract when the mirror separation is reduced.

It doesn't matter.

I don't get your point.

I think you can place a movable marker on a particular fringe, when you decrease $d$, this marker should go radially inward. That's when you would say the fringes are contracting.
Btw, notice that the zeroth order is located at infinity. This means, if you decrease $d$, all fringes will actually contract such that in the end there is only bright area on the screen when $d=0$, i.e. when the mirrors coincide.

Thank you, that was quite helpful (still slight uncertainty but better understanding now than before!)