Consider two sources A and B as shown in the figure below . Let the two sources emit simple harmonic waves of same frequency but of different amplitudes , and both are in phase (same phase) . Let O be any point equidistant from A and B as shown in the figure . Calculate the intensity at points O , Y and X . ( X and Y are not equidistant from A & B)

Explanation

The distance between OA and OB are the same and hence , the waves starting from A and B reach O after covering equal distances (equal path lengths) . Thus , the path difference between two waves at O is zero .
`OA - OB = 0`
Since the waves are in the same phase , at the point O , the phase difference between two waves is also zero . thus ,the resultant intensity at the point O is maximum . Consider a point Y , such that the path difference between two waves is `lambda` . Then the phase difference at Y is
`Delta phi = (2pi)/(lambda) xx Delta r = (2pi)/(lambda) xx lambda = 2pi`
Therefore , at the point YH , the two waves from A and B are in phase , hence the intensity will be maximum .
Consider a point X , and let the path difference the between two waves be `(lambda)/(2)` . Then the phase difference at X is
`Delta phi = (2pi)/(lambda) (lambda)/(2) = pi`
Therefore , at the point X , the wave meet and are in out of phase . Hence , due to destructive interference , the intensity will be maximum .