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I don't quite understand why no interference would occur when the separation between the two sources is less than the wavelength of the wave. I can't find demonstrations about such special case on the internet.

The following is the homework which raised my question:

Question + Solution

Suppose there are two speakers set up to demonstrate a two-source wave interference. The speakers are connected to the same signal generator to generate coherent sound waves at frequency $500Hz$ and wavelength $0.68m$. The distance between the two speakers is $0.45m$. The distance between the speakers and the observing line XY is $2m$.

Explain why neither a maximum or minimum is detected along XY.

The homework solution is:

Since the separation between the two speakers is $0.45m$ which is shorter than the wavelength, the path difference at any point on XY must be less than one wavelength. Thus no interference occurs.

But how and why??

Does this statement mean the observer would hear a constant amplitude (amplitude of a single speaker) of sound along the XY? This doesn't make sense to me.

Is there any graphical illustration I can refer to?

Also, shouldn't there always be an anti-nodal line with path difference of $0$ at the centre? Shouldn't there always be constructive interference along this line?

Please enlighten me! Thanks in advance!!

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    $\begingroup$ What is the orientation of line $XY$ relative to the line joining the two slits? $\endgroup$
    – Farcher
    Feb 20, 2020 at 18:48
  • $\begingroup$ @Farcher Aw my bad for not detailing XY, the original homework contained a diagram. XY is as usual, being perpendicular to the line joining the speakers. Also, it is a two sources set up not a double slits, not sure if this will make any difference? $\endgroup$
    – LemuelKL
    Feb 21, 2020 at 1:59
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    $\begingroup$ @LemuelL XY is as usual perpendicular to the line joining the speakers? Are you sure you don't mean parallel to the line joining the speakers? $\endgroup$
    – sammy gerbil
    Feb 21, 2020 at 4:36
  • $\begingroup$ @sammygerbil Oops sorry! English isn't my first language... I now attach the diagram from the original homework for clearer understanding. $\endgroup$
    – LemuelKL
    Feb 21, 2020 at 5:23

2 Answers 2

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I agree that the speakers are supposed to be in phase, and that the observing line is parallel to the line joining the speakers. The homework question should definitely have included these points.

Assuming these conditions, then the condition for antinodal lines (maxima) in the interference pattern is given by $\theta = \sin^{-1} (n \lambda/D)$, where $n = \pm 1, \pm 2 \dots$ labels the lines, and $D$ is the distance between the speakers. Similarly the condition for the nodal lines is given by $\theta = \sin^{-1} ( (n+1/2) \lambda / D)$, where $n = 0, \pm 1, \pm 2 \dots$.

The lowest order antinodal line is given by $\theta = \sin^{-1} (\lambda/D)$; clearly if $\lambda > D$ this equation cannot be satisfied and so no antinodal lines will occur. Similarly if $\lambda > 2 D$ no nodal lines are present.

For the parameters given, $\lambda / D = 1.551$, and so no antinodal lines will be visible. However two nodal lines, which are the lines which are easiest to see in interference patterns, should be visible at $\pm 49^\circ$.

I attach some inteference patterns for various values of $D$, for $\lambda = 1$. When $D > \lambda$ several nodal lines are clearly visible. At $\lambda = D$, only a pair of lines appear. For much smaller separations, no nodal lines are present - in fact the pattern looks very similar to that of a single point source. This can be understood qualitatively by saying that since the separation is smaller than $\lambda$, the two sources cannot be distinguished. Hopefully this gives you an idea of how the pattern evolves as the two sources are moved closer to each other.

enter image description here

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    $\begingroup$ I think there are some contradictions in the 3rd and 4th paragraphs of your answer. Shouldn't it be: For λ=0.68m, D=0.45m 1st nodal line: θ=sin−1(0.5λ/D) = ±49∘ and 1st antinodal line: θ=sin−1(λ/D) = None $\endgroup$
    – LemuelKL
    Feb 21, 2020 at 11:01
  • $\begingroup$ Yes, you're absolutely right. I'll change the text. $\endgroup$
    – Clara Diaz Sanchez
    Feb 21, 2020 at 11:16
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Several things seem to me to be suspect about this question and the answer provided.

(a) I think that the speakers are supposed to be in phase, not merely coherent (constant phase difference). You'll get interference either way, but to give definite answers to the specific questions, you need to know the phase difference; I think that it's intended to be zero, that is the sources are in phase.

(b) As Farcher points out, you ought to be told that the 'observing line' XY is parallel to the line joining the speakers, if that is indeed the case. Edit: Your new extract from the original question makes this clear.

(c) "shouldn't there always be an anti-nodal line with path difference=0 at the centre? Shouldn't there always be constructive interference along this line?" Yes and Yes, if the assumptions (a) is correct.

(d) And I don't see why you shouldn't get destructive interference at points along XY for which the difference between the paths to the speakers is 0.34 m. Such points will exist, though you'll never get constructive interference again, however far you go from the centre of XY, because you can never get a path difference greater than 0.45 m, and you need 0.68 m.

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