image by comsol model galleryIntensity mentioned in superposition does not require the formula power over area but the intensity is inversely proportional to r square. A simpler understanding like Intensity is lower because the distance is further is often sufficient for this part of the topic.
In fact to simplify the above fact, questions require you to take the Intensity as I (the same value) eventhough one source is nearer than the other source. If this is not the case , the calculation I would require power divide by area of sphere of radius r.
Straight forward example:
Constructive interference at P : Each source providing intensity I at P(eventhough P may not be same distance from the two sources) .Take amplitude of I to be A, constructive interference will result in waveform having 2A. Now using ratio, the resultant intensity will be 4I
Destructive interference at Q: Take amplitude to be A from each source, but cancel out due to destructive interference , resultant intensity 0.
Not so straight forward case:
If two coherent waves of intensities I and 2I meet in phase at a point, what is the resultant intensity at that point. This require perserverance mathematically!
If I has amplitude A, the 2I has an amplitude of 1.414A. Use ratio to deduce.
Now the resultant amplitude A' =2.414 A , for constructive interference.
Use ratio again , I has amplitude A , now A' has intensity I'.
The answer I' =5.83I
Another not so straight forward case:
If the wave from one source S1 is I having amplitude A and the other source S2 has amplitude 1.5A.
What is the resulant intensity at P and Q, where there is constructive interference and destructive interference respectively?
At P: Amplitude is 2.5A, intensity I'= 6.25I
At Q : Amplitude is 0.5A, intensity I"= o.25 I.
To sum up, intensity is dependent on r and amplitude.These two relationships are to be treated independently.
