The same optical path can accept or reject a bright label depending on the checked phase-flip bit. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Hold the optical path fixed

Both comparison films use the same optical path. That isolates the phase-flip bit as the change being tested.

path=3 m\text{path}=3\ \text{m}
Accepted bright filmThis row has the phase flip required for the bright verdict.thicknessrefractiveIndex=3/2thickness=1 mopticalPath=3 mwavelength=2 mphaseFlipBit=1 bitmode=brightorder=1 countacceptedBit=1 bit

Changing only the phase bit changes the verdict

The first two rows have the same path. The accepted bit changes because the phase-flip bit changes, not because the path changed.

ntpathflipaccepted321 m3 m11321 m3 m00322 m6 m10\begin{array}{c|c|c|c|c}n&t&\text{path}&\text{flip}&\text{accepted}\\\frac{3}{2}&1\ \text{m}&3\ \text{m}&1&1\\\frac{3}{2}&1\ \text{m}&3\ \text{m}&0&0\\\frac{3}{2}&2\ \text{m}&6\ \text{m}&1&0\\\end{array}

The phase-flip bit changes the bright verdict

Both films have the same optical path, but only the one with the checked phase-flip bit accepts the bright label.

path=3 m;flip=11;no flip0\text{path}=3\ \text{m};\quad \text{flip}=1\Rightarrow1;\quad \text{no flip}\Rightarrow0
Thin-film phase-flip contrastSame optical path, different phase-flip bit, different verdict.thicknessthicknessrefractiveIndex=3/2thickness=1 mopticalPath=3 mwavelength=2 mphaseFlipBit=1 bitmode=brightorder=1 countacceptedBit=1 bitrefractiveIndex=3/2thickness=1 mopticalPath=3 mwavelength=2 mphaseFlipBit=0 bitmode=brightorder=1 countacceptedBit=0 bit