Answer:
Two beams (that we could think as sinusoidal waves) face destructive interference at a point P, if at that point one of the waves is in a peak, and the other wave is in a through, so when we add them, the waves will "cancel" each other (thus, we have destructive interference)
The case of constructive interference happens when at point P we have two peaks or two throughs, so the waves add up.
If both waves started at the same point and with the same phase and both traveled the same distance, they will always have constructive interference. But if the waves traveled different distances, the constructive interference will happen at points where the path difference of the waves is an integer multiple of the wavelength (remember that the distance between a peak and the next one is the wavelength).
If two beams of coherent light start at the same point and are in phase, then we will have the maximum of destructive interference at a point P if the path difference is exactly half of the wavelength (or (n/2) times the wavelength, with n an odd integer), this happens because the distance between a peak and the next thtough is exactly half of the wavelength.
Then we can conclude that the path difference between the two beams is of the form:
(n/2)*λ
where:
n = odd number = (2*k + 1) with k an integer.
λ = wavelength of the beams.
