Let #A=I# and #B=4I#
When two waves have a phase difference of #(2n+1)pi, ninZZ#, the peak of one wave is directly above the trough of another. Therefore destructive interference occurs. So, the magnitude of the intensity is #abs(A-B)=abs(I-4I)=abs(-3I)=3I#
However, if the two waves have a phase difference of #2npi, ninZZ#, then the peak of one wave lines up with the peak of another. And so, constructive interference occurs and the intensity becomes #A+B=I+4I=5I#
Matt Comments
Intensity is proportional to amplitude square (#IpropA^2#) so if wave of #I# has amplitude #A# then the wave of #4I# would have amplitude #2A#
When #2pi# out of phase, you have constructive interference (so amplitude #2A+A=3A# and intensity #9A^2"# or #9I#) and when #pi# out of phase destructive interference (so amplitude #2A-A=A# so intensity #I#)
