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

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

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

Question 2: The image suggests that [itex]\theta[/itex] is only linked to [itex]L_1[/itex]. If I move [itex]L_1[/itex] towards [itex]L_2[/itex], the adjacent side of the right-angled triangle is getting shorter, and hence [itex]\theta[/itex] 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 [itex]L[/itex] causes [itex]cos\theta[/itex] to increase, which is analogous to reducing [itex]\theta[/itex]. [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!)