Path Difference Decides Fringes

half-wave-path-difference-makes-a-dark-fringe

A dark fringe appears when the path difference is a half-integer number of wavelengths, delta = (m + 1/2) * lambda, so the waves arrive 180 degrees out of phase and cancel. The half-wave offset is what turns constructive addition into destructive cancellation. The clean zero assumes perfectly coherent monochromatic waves and ideal visibility; measured dark fringes are lifted by detector noise, stray light, and partial coherence.

A dark interference claim carries an explicit half-wave offset. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

The dark case measures a half step

The path difference is 3 m. With wavelength 2 m, that is halfway between one and two wavelengths.

Δ=3 mλ=2 m\Delta=3\ \text{m}\qquad \lambda=2\ \text{m}
Measured half-wave pathThe ledger records the half-wave offset explicitly.path Apath BpathA=10 mpathB=13 mpathDifference=3 mwavelength=2 morder=1 counthalfWaveBit=1 bitmode=darkacceptedBit=1 bit

Only the half-integer row is dark

The dark rule accepts the half-integer target and rejects the neighboring integer rows. That keeps bright and dark tests separate.

Δλtargetaccepted2 m2 m103 m2 m3214 m2 m20\begin{array}{c|c|c|c}\Delta&\lambda&\text{target}&\text{accepted}\\2\ \text{m}&2\ \text{m}&1&0\\3\ \text{m}&2\ \text{m}&\frac{3}{2}&1\\4\ \text{m}&2\ \text{m}&2&0\\\end{array}

A half-wave offset accepts a dark fringe

The dark claim is checked by a half-wave bit, so the path difference is not rounded into an integer count.

Δ=3 m=(1+12)λ;accepted=1\Delta=3\ \text{m}=\left(1+\frac{1}{2}\right)\lambda;\quad \text{accepted}=1
Dark half-wave ledgerThe half-wave bit is part of the checked path claim.path Apath BpathA=10 mpathB=13 mpathDifference=3 mwavelength=2 morder=1 counthalfWaveBit=1 bitmode=darkacceptedBit=1 bit