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Mar 14, 2018 at 7:07 answer added Matt timeline score: 1
Oct 27, 2015 at 18:31 comment added Solomon Slow Are you describing an experiment that somebody actually performed? Perhaps they did not control the conditions well enough. For example, unless the experiment is performed in an absolutely anechoic chamber, then the contribution of each speaker is going to be much more complicated than just a pure, spherical wave.
May 14, 2015 at 10:02 history edited innisfree CC BY-SA 3.0
title - this seems to be more about sound than Young's experiment, so I changed it
May 14, 2015 at 8:13 answer added Guill timeline score: 2
May 11, 2015 at 18:51 comment added dmckee --- ex-moderator kitten 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 16:40 comment added Bill N 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.
S May 11, 2015 at 14:34 history suggested Gonenc
added homework tag
May 11, 2015 at 14:33 comment added Carl Witthoft 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 14:10 comment added Jon Custer The size of the sources relative to wavelength and distance to the 'screen' also play a role.
May 11, 2015 at 13:55 review Suggested edits
S May 11, 2015 at 14:34
May 11, 2015 at 13:55 comment added Gonenc 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 13:48 review First posts
May 11, 2015 at 13:55
May 11, 2015 at 13:47 history asked Choy CC BY-SA 3.0