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I am trying to work out the relation between a $\mu$step and the distance moved by the moveable mirror in the Michaelson Interferometer:


The mirror on the stage is moved by a stepper motor. Inside the stepper motor there are 200 steps for a complete revolution. Each step itself is made up of 256 µsteps. The drive shaft from the motor then drives a gear box which has a reduction ratio of 100:1, this in turn rotates a micrometer which requires two revolutions to move 1mm. The micrometer then pushes a lever arm that moves the mirror and has a reduction ratio of 6.25:1.


From the information above, I worked out that $$\text{1 stepper motor revolution}=200 \times 256\mu step=51200 \mu step$$ $$\text{1 gear box revolution}=100 \times \text{1 stepper motor revolution}=5120000 \mu step$$ $$\text{1 revolution of micrometer=1 gear box revolution}=5120000\mu step$$ $$\text{1mm moved by micrometer=2 revolutions of micrometer}=2 \times 5120000 \mu step=10.24\times 10^6 \mu step$$ But here's the part I cannot make sense of: "The micrometer then pushes a lever arm that moves the mirror and has a reduction ratio of 6.25:1."

Does this mean that every $6.25mm$ moved by the micrometer corresponds to $1mm$ of $mirror$ movement? In which case $$\text{1mm mirror moved}=6.25 \times \text{1mm micrometer moved}$$ $$=6.25 \times 10.24 \times 10^6 \mu step=64 \times 10^6 \mu step \tag{1}$$?

Or every $6.25 $ revolutions of the micrometer correspond to $1mm$ of mirror movement? $$\text{1 mm mirror moved}=6.25 \times 5120000=32 \times 10^6 \mu step \tag{2}$$ ? $$$$

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Yes, a reduction ratio of 6.25:1 means for 6.25mm of input the output will be 1mm.

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