The title says it all but I will add some details. I believe that interference takes place only in waves which are parallel to each other. See the picture to understand what I mean by parallel.

enter image description here

We see that in the image the interfering waves are parallel. If interference can happen in non-parallel waves then please give me an explanation but if it does not consider the double slit experiment. How are the waves parallel in the double slit experiment?

enter image description here

We see that in the double-slit experiment the waves are not parallel. Please answer accordingly as I have given both the situations.

  • $\begingroup$ You are on the right track in one sense: You will not see an interference pattern from two waves that have polarization perpendicular to each other. $\endgroup$ – The Photon Apr 7 '14 at 16:09

Put simply, yes waves interfere even if they are not directly aligned. In fact all waves interfere of a given type.

Interference is really just a re-statement of the superposition principle; that is, given 2 waves, the resulting pattern is simply the sum of the two waves at all positions in the space.

The first figure you provide shows how, in 1D, 2 waves that are in the same phase (which you are calling 'parallel') will interfere constructively; i.e. the sum of the two waves has a larger magnitude than either wave individually. The second shows how two waves in opposite phase will interfere destructively, i.e. the sum of the waves has a smaller magnitude than either.

The phase difference for the first is $0^o$, and the second is $180^o$. Consider all the other values, for example, $30^o$. We can see the general behaviour from the sum+difference trig identities. For $\sin$, we have:

$$\sin(u) + \sin(v) = 2 \, \sin\left(\frac{u+v}{2}\right) \; \cos\left(\frac{u-v}{2}\right) $$

But we have a constant phase difference, so

$$v = u+2\gamma \\ u-v = -2\gamma$$

which is independent of $u$, so for convenience create some constant $\eta$:

$$\eta = 2\cos\left(\frac{u-v}{2}\right) = 2\cos(\gamma) $$

Switching to $\theta$, this leaves us with:

$$\sin(\theta) + \sin(\theta+2\gamma) = \eta \sin\left(\frac{\theta+\theta+2\gamma}{2}\right) \\ = \eta \sin(\theta+\gamma)$$

What does that mean? It means that the sum of two waves with the same period but a phase difference of $2\gamma$ will be exactly the same as one wave shifted by $1\gamma$, scaled in amplitude by $\eta = 2\cos(\gamma)$. When $\cos(\gamma) = 0$, i.e. $\gamma = 90^o$, and the phase difference is $2\gamma = 180^o$, the scaling is zero and you get no wave out, perfect destructive interference.

In 2D, you draw a triangle formed by each source and a chosen point on the 2D plane. We now have a 1D wave along each line from a source to the point. The phase difference between the two waves depends on the difference between the lengths of the two wave paths. This gives you, yes, a phase difference, and the waves will constructively or destructively interfere. Looking at your diagram, the bold lines are say the peaks, so where they cross will be peaks. Where the thin lines cross will be troughs. Where thin and thick lines meet will be zero, so a straight line connecting those can be all zero, the destructive interference of the 1D example.

To the illustrations!

The path difference, i.e. the greater or lesser distance travelled along a line from one source compared to the other looks like this:

path difference between sources at x=+1 and x=-1.

The contours are at $0$ and $\pm1$. As you can see, the path difference is zero (i.e. the locus) in a straight line along the centre.

Our scaling value $\eta$ depended on $\cos(\gamma)$, and the phase difference there was $2\gamma$. Here, the phase difference is determined by the path difference; if the path lengths are the same, and the path difference is zero, the phase difference is zero too. Since $\gamma$ is a phase difference, we need the path difference as a proportion of the period of the wave. So if the period is 1/3, the phase difference $p$ for a given path difference $d$ is:

$$p = \frac{d}{1/3} \times 360^o = 3d \times 360^o$$

$\gamma$ is only half this, so $\gamma = 1.5d \times 360^o$. We can go back and plot, instead of the path difference $d$, the scaling factor $2\cos(540d^o)$ (I'm really plotting in radians, i.e. $2\cos(4.5\pi d)$):

scaling from phase diff plot

So we can see that the centre line, where the phase difference is zero, has constructive interferece - the scaling is 2x here. The black bands are where the path differences give destructive interference, as the sources cancel out.

Hopefully you can see, then, that interference, both constructive and destructive, happens at all angles - not just along the 'parallel' line between the two sources.

  • $\begingroup$ This is informative but doesn't answer the OP's question. $\endgroup$ – user42733 Apr 7 '14 at 16:10
  • $\begingroup$ @Phil H sorry but I do not understand the math as I am only 15. But instead could you provide me with a yes or no answer to the question whether non parallel waves can interfere with each other or not. In your words can interference occur when there is not a phase difference of 180 degree. $\endgroup$ – rahulgarg12342 Apr 7 '14 at 16:23
  • $\begingroup$ I'm saying that as you slide one wave next to the other, you increase the phase difference and go smoothly from making the wave bigger to making it smaller and eventually zero. Moving it further it becomes bigger again, in a cycle. $\endgroup$ – Phil H Apr 7 '14 at 18:48
  • $\begingroup$ I've added something hopefully clearer at the top, might get a chance to illustrate it a bit later. $\endgroup$ – Phil H Apr 7 '14 at 19:18
  • $\begingroup$ @ParthVader: I had gone down a bit of a rabbit hole, but hopefully the edited answer is a better fit. $\endgroup$ – Phil H Apr 8 '14 at 14:50

Yes, the waves do interfere when they are not parallel. In fact, there's no such thing as parallel waves: their direction is always fuzzy.

See this image of one wave going through an aperture with size comparable to wavelength:

enter image description here

You can see that it goes not only upward, but also expands to the left and right. Now let's add another source such that leftward-going part of right wave and rightward-going part of left one cross:

enter image description here

Interference is clearly visible only in the region where the waves cross, but not earlier (i.e. not lower in the image).

To make lack of parallelism even more apparent, I've tilted the sources to see that we still have interference:

enter image description here


The waves in the first picture are not necessarily parallel. Their amplitudes are simply projected onto a screen, which makes them appear to propagate only in the $x$ direction, while really they could be propagating in any direction at all (still assuming they are confined to a plane, so no propagation in the '$z$' direction.

Without loss of generality we can assume that only one of the waves moves at an angle with respect to the plane of projection. This process of projection will the cause then give rise to a distortion of the wave, which will cause the perceived wavelength to be multiplied by a factor, I believe it should be $\cos\theta$.

  • $\begingroup$ Sorry I did not get the second part of your answer. By parallel I mean in any plane and not just the x axis. The main part is only the parallel part. And please do answer the second part of my question. $\endgroup$ – rahulgarg12342 Apr 7 '14 at 15:45
  • 1
    $\begingroup$ The main message is: For the first picture to be true the waves don't actually have to be parallel. Therefore, there is no contradiction with the second picture. $\endgroup$ – Danu Apr 7 '14 at 16:36
  • $\begingroup$ So in short the answer is that waves can interfere if they are not parallel. Can you now please explain how do they interfere if not parallel. Thanks. $\endgroup$ – rahulgarg12342 Apr 7 '14 at 16:39
  • $\begingroup$ @rahulgarg12342 in exactly the way you see in the pictures. ALL of those can be viewed as non-parallel waves interfering $\endgroup$ – Danu Apr 8 '14 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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