2
$\begingroup$

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) enter image description here

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??

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ 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. $\endgroup$
    – Gonenc
    Commented May 11, 2015 at 13:55
  • $\begingroup$ The size of the sources relative to wavelength and distance to the 'screen' also play a role. $\endgroup$
    – Jon Custer
    Commented May 11, 2015 at 14:10
  • 1
    $\begingroup$ 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. $\endgroup$
    – Carl Witthoft
    Commented May 11, 2015 at 14:33
  • 2
    $\begingroup$ 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. $\endgroup$
    – Bill N
    Commented May 11, 2015 at 16:40
  • $\begingroup$ 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. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented May 11, 2015 at 18:51

2 Answers 2

Reset to default
2
$\begingroup$

The reason D is the "correct" answer, is because the key word best, is used.
Although any answer may not be 100% correct, if at least one of the answers is better than the others, then that would make it the correct answer. A and B are eliminated because the amplitude should decrease as one moves away from the XY center line. C is eliminated because it shows no interference. D is the only one that shows both decrease in amplitude and signal interference.

Improve this answer
$\endgroup$
1
$\begingroup$

Could it be that the speakers are not point sources so that there is a slight spreading of each maxima so adjacent maxima overlap slightly?

Improve this answer
$\endgroup$
1
  • $\begingroup$ To add to my comment above: the fact that the wavelength of sound is much bigger than light means the effect is much more exaggerated for sound in the interference pattern $\endgroup$
    – Matt
    Commented Mar 14, 2018 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.