# Why does a double-slit experiment with sound waves never result in complete destructive interference?

Today, I find a question that if the speakers are set in parallel towards the wall. It has no complete destructive interference due to difference in distance from the point to each speakers, which leads difference in intensity, amplitude...(Answer is D)

However, for the young's double slit there is also difference between the point and the slits. So, there should be difference in intensity,amplitude, isn't it? However, we consider it as they are having complete destructive interference so they have no intensity, dark region, at some point.

Is it because the electromagnetic wave has high frequency, so we assume that there is no decrease in intensity, amplitude is negligible??

• It is because the person who has written the question didn't think of it that way! Of course the intensity goes to zero if we assume a perfect setup. May 11, 2015 at 13:55
• The size of the sources relative to wavelength and distance to the 'screen' also play a role. May 11, 2015 at 14:10
• Given that the first speaker's coil probably induces a phase shift in the current reaching the second speaker, I would bet that the pattern is not symmetric about the midpoint of line $XY$ . The person who wrote the question was rather sloppy. I'm guessing he wants the decaying maxima to look like the envelope of a single speaker, but why there's a nonzero floor is quite unclear. May 11, 2015 at 14:33
• If D is the answer they expect, the problem is poorly-worded. That's the down side of multiple choice tests. Maybe they intend for you to consider the reflections from the other walls will prevent a complete destructive interference. That's what will happen if you do the experiment in a real classroom. May 11, 2015 at 16:40
• Note that there is as assumption in the elementary treatment of the two-slit experiment that the distance $l$ between the screens is much larger than the distance $d$ between the slits. When those conditions are met, there is negligible intensity variation between the two sources across one fringe, so you don't notice the non-zeroness of the dark fringes. In the image shown above that condition $l \gg d$ is not present and you must do a more complete analysis. May 11, 2015 at 18:51