A larger transverse bucket keeps visibility as coherent divided by initial. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A larger bucket keeps the same fraction rule

Initial transverse budget is 15. The scan changes coherent count and recomputes the dephased complement.

FT=coherentinitialF_T={\text{coherent}\over\text{initial}}
Transverse scale rowThe visible fraction is coherent bucket divided by the initial bucket.initial 15coherent 6dephased 9fraction 2/5

Three coherent buckets give three exact fractions

The dephased bucket closes the same initial total in every row.

coherentdephasedFTvisible312153692561234512\begin{array}{c|c|c|c}\text{coherent}&\text{dephased}&F_T&\text{visible}\\3&12&\frac{1}{5}&3\\6&9&\frac{2}{5}&6\\12&3&\frac{4}{5}&12\\\end{array}

The middle row keeps two fifths visible

Coherent 6 out of initial 15 gives visible fraction 2/5.

6/15=256/15=\frac{2}{5}
Transverse scale rowThe visible fraction is coherent bucket divided by the initial bucket.initial 15coherent 6dephased 9fraction 2/5