I did some math to figure out how much ITMX, ITMY and BS may have been moving (in a frequency band of 0.1-1 ish Hz) in their angles according to fluctuation of the DC light observed at the dark port when the Michelson was locked.

(Summary)

Wednesday night (August 27th)

• The dark port DC (ASAIR_B_LF) flucutated by 16% of the bright fringe when the Michelson was locked on a dark fringe.
• ITMX(Y) may have been moving by 11 urad or 22 urad in peak-to-peak.
• BS may have been moving by 5.7 urad or 11 urad in peak-to-peak.

Friday night (August 29th)

• The dark port DC (ASAIR_B_LF) flucutated by 1% of the bright fringe when the Michelson was locked on a dark fringe.
• ITMX(Y) may have been moving by 2.8 urad or 5.6 urad in peak-to-peak.
• BS may have been moving by 1.4 urad or 2.8 urad in peak-to-peak.

(Some math behind it)

Suppose the Michelson is locked on a dark fringe. If an ITM is misaligned by Ψ, this introduces a displacement and tilt in the reflected beam with respect to the one from the other ITM at the BS. The displacement is x = 2 L Ψ and the title is ϑ = 2 Ψ where L is the distance from the BS to the ITM. So we get a small amount of 01 or 10 mode at the dark port on top of the 00 modes. Since the effect on the resultant 00 mode in its power is proportial to 4th power of the displacement and tilt, we assume the 00 mode to vanish because of the locking loop. The only residual we obtain at the dark port is the 01 or 10 mode whose field can be written as

E₁₀ = 1/sqrt(2) ( x/w₀ + i * ϑ / ϑ₀),

where w₀ is the waist size and ϑ₀ is the divergence angle respectively. A factor of 1/sqrt(2) upfront comes from the BS reflection. If we plug the definition of x and ϑ into the equation, we get

E₁₀ = sqrt(2) ( L/w₀ + i / ϑ₀)  Ψ.

Squaring the above, one can get the dark port power as

P = 2 ( (L/w₀)²+ (1/ ϑ₀)²)  Ψ²

Note that P is already normalized by the input beam power or equivallently the bright fringe. The Rayleigh range of the beam around the BS is roughly 210 m (if my math is correct). This gives a waist size of 8.4 mm and divergence angle of 40 urad.  The ITM-BS distance L is about 5.34 m where I averaged out the Schnupp asymmetry. So the dark port power can be now explicitly written as

P = 1.24 x 10⁹  Ψ²

This is the equation I used for deriving the numbers listed at the very top.

For example, if one wants to explain a 16% DC light fluctuation observed at the dark port by an angle deviation in ITMX(Y), the misalignment should be Ψ = sqrt(0.16 / 1.24 x 10⁹ ) = 11.4 urad. In the case of the BS, the effect gets twice bigger due to the fact it affects both X and Y beams at the same time in a constructive manner.