# Reduction Ratio Problem

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? $$\text{1 mm mirror moved}=6.25 \times 5120000=32 \times 10^6 \mu step \tag{2}$$ ?