#18 bench companion · the arrow picture
Each cell of a table holds an arrow. Detector 0 collects the arrows that end there, and its share is the length² of their sum. Three stages below, same numbers as the bench (toy turn, all dials at 45° unless you move them).
stage 1 · no marker
Both arms end at detector 0. The upper arm brings a fixed arrow (blue). The lower arm's arrow (orange) is turned by the delay φ; the join's two i's flip it by 180° on top, so at φ = 0 the two point opposite and cancel: dark. At φ = 180° they line up and add: bright. The black arrow is the sum. Drag φ and the dot walks along the brown curve, sin²(φ/2). The fringe is the angle between two arrows.
stage 2 · M present, before the late choice
The meeting copied the arm into M. So at detector 0 the upper arrow sits in the cell with M = 0 and the lower arrow in the cell with M = 1. Arrows in different cells never add. Each is worth 0.25 whatever φ is, total 0.5: the flat black line. Nothing cancelled. The angle between the arrows is still exactly what it was in stage 1; the table just never puts them together.
stage 3 · late turn on M, then sort by D2
The late turn mixes M's two cells, so each D2 answer gets a piece of both arrows and they can add again. The group D2 = 0 gets cos θ of the upper arrow and i·sin θ of the lower one. That i is a quarter turn (+90°) of the crossed-over arrow, and it is the whole ±90° shift. Try it: τ = −90° twists M's lower cell back before the turn and D2 = 0 lands on brown. θ = −45° makes the i a −i, so the groups just swap shifts. θ = 0° mixes nothing and both groups go flat. (Group arrows are drawn ×√2, since each group holds half the photons, so length² reads as the group's share.)