Constructive interference

AP Physics 2 EK 14.6.A.4.i: when the displacements of the superposed wave pulses or waves are in the same direction, the interaction is called constructive interference. The signs agree, so the sum is bigger than either part.

The path-difference condition. For two coherent sources emitting in phase, constructive interference arrives wherever the two path lengths differ by a whole number of wavelengths. That is the printed AP Physics 2 equation $\Delta D = m\lambda$, with $m$ an integer. Zero path difference is $m = 0$, the central maximum.

Check the geometry before using it. The sheet prints $\Delta D = m\lambda$ once, but the CED attaches it to two opposite outcomes. For a double slit, $d\,y_{\text{max}}/L \approx m\lambda$ locates the bright fringes (EK 14.8.A.1.v). For a single slit, $a\,y_{\text{min}}/L \approx m\lambda$ locates the dark ones (EK 14.7.A.4.iv). Same algebra, opposite answer.

Amplitudes do not have to match. Unlike destructive interference, where total cancellation needs equal amplitudes, constructive interference simply adds whatever two amplitudes turn up.

If the two sources start half a cycle out of phase, the whole condition inverts and a whole-wavelength path difference gives cancellation instead. Check the phase of the sources, not just the distances.