# Michelson Interferometer ring contraction

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.

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blue_leaf77
Homework Helper
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.
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.