Source-Bound Boundary Scans

Path Difference Boundary Three-Row Scan

A one-metre path change can reject a bright row and accept a dark row. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Bright row closes at one whole wavelength

Path B is 12 m while path A stays ten metres. The path difference is 2 m. The bright target is 2 m, so the row is accepted.

Δ=12 m−10 m=2 mT=(1+0)2 m=2 mA=1\Delta=12\ \text{m}\mathbin{-}10\ \text{m}=2\ \text{m}\quad T=\left(1+0\right)2\ \text{m}=2\ \text{m}\quad A=1
Path boundary rowThe path vectors, wavelength, mode, half-wave bit, and acceptance bit are checked.path Apath BpathA=10 mpathB=12 mpathDifference=2 mwavelength=2 morder=1 counthalfWaveBit=0 bitmode=brightacceptedBit=1 bit

The same path misses the bright target

Path B is 13 m while path A stays ten metres. The path difference is 3 m. The bright target is 2 m, so the row is rejected.

Δ=13 m−10 m=3 mT=(1+0)2 m=2 mA=0\Delta=13\ \text{m}\mathbin{-}10\ \text{m}=3\ \text{m}\quad T=\left(1+0\right)2\ \text{m}=2\ \text{m}\quad A=0
Path boundary rowThe path vectors, wavelength, mode, half-wave bit, and acceptance bit are checked.path Apath BpathA=10 mpathB=13 mpathDifference=3 mwavelength=2 morder=1 counthalfWaveBit=0 bitmode=brightacceptedBit=0 bit

The missed bright row is a dark row

Path B is 13 m while path A stays ten metres. The path difference is 3 m. The dark target is 3 m, so the row is accepted.

Δ=13 m−10 m=3 mT=(1+12)2 m=3 mA=1\Delta=13\ \text{m}\mathbin{-}10\ \text{m}=3\ \text{m}\quad T=\left(1+\frac{1}{2}\right)2\ \text{m}=3\ \text{m}\quad A=1
Path boundary rowThe path vectors, wavelength, mode, half-wave bit, and acceptance bit are checked.path Apath BpathA=10 mpathB=13 mpathDifference=3 mwavelength=2 morder=1 counthalfWaveBit=1 bitmode=darkacceptedBit=1 bit