Destructive interference

AP Physics 2 EK 14.6.A.4.ii: when the displacements of the superposed wave pulses or waves are in opposite directions, the interaction is called destructive interference. The signs disagree, so the sum shrinks.

The condition. For two coherent in-phase sources, destructive interference lands where the path difference is a half-integer number of wavelengths: $\Delta D = (m + \frac{1}{2})\lambda$, so $\lambda/2$, $3\lambda/2$, $5\lambda/2$ and onward. That extra half wavelength is what puts a crest on top of a trough.

Partial is the normal case. Total cancellation needs the two amplitudes equal as well as opposite. A 5 cm crest meeting a 3 cm trough leaves 2 cm, and that is still destructive interference. An answer that reasons "the interference is destructive, so the displacement is zero" is asserting something it has not checked.

Nothing is destroyed. By superposition the waves pass through each other and re-emerge unchanged (EK 14.6.A.2), and the energy missing from a dark fringe turns up in the bright ones. A dark fringe is a redistribution, not a loss.

One more sign trap. The printed $\Delta D = m\lambda$ is the bright-fringe condition for a double slit but the dark-fringe condition for a single slit (EK 14.7.A.4.iv), so read which opening you are dealing with before deciding what an integer $m$ means.