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In the picture above, black and blue lines are representing waves(mechanical) emerging from the slits(after diffraction). Let us consider that at point A, centre of a bright fringe is located. That means the path difference between the most upward black and blue line is an integral multiple of the wavelength of the wave. But the blue line before reaching the point A,it interacts with many blue lines. Aren't there any chance of cancellation of the wave(blue line)? Okay then for a while we consider that at point F both waves are in out of phase. Then how would this blue line reach to the point A? Shouldn't the particle at F remain static since the waves are cancelling out there?

Okay then we can again consider that somehow the wave propagates. However those interactions at point E and some other ones changes the intensity of blue line. Why do we then consider the intensity of resultant at A to be exacrly double of that of the black/blue line? Or is it just an approximation rather in reality it varies deoending on the position?

** All the waves that are represented by the black and blue lines are of same frequency and amplitude**

  • 1
    $\begingroup$ Most/all mediums we use in science (air, water, EM field, etc) are assumed lossless. Too many math teachers teach waves adding and then cancelling, in physics superposition is temporary, a brief moment or second in time, the waves actually continue forever until absorbed. All waves are eventually absorbed: water waves crash oh the beach, sound is absorbed by materials, EM waves (can be called photons)are produced by electrons in individual atoms/molecules and are only absorbed by atoms/molecules. MUST conserve energy! $\endgroup$ Commented Jun 3, 2021 at 21:32
  • $\begingroup$ But isn't it supposed to transform into some other forms of energy when superimpose in out of phase? Then how would the succeeding particles move in the wave form I mean oscillate? $\endgroup$
    – MSKB
    Commented Jun 3, 2021 at 22:00
  • $\begingroup$ see this question and my answer physics.stackexchange.com/questions/623648/… $\endgroup$ Commented Jun 4, 2021 at 0:32
  • $\begingroup$ also physics.stackexchange.com/questions/601335/… $\endgroup$ Commented Jun 4, 2021 at 1:39

2 Answers 2


The resultant displacement due to a wave is to do with the displacement of the 'vibrating medium', for example in a water wave the water goes to different heights as the wave passes.

Two waves, moving in different directions could pass a point in the water. If they interfere constructively the height of the resultant is double. If they interfere destructively the height could be zero (relative to the original water surface) - but that doesn't mean that the two waves have stopped. They would both carry on and be unaffected.

this link might be useful


  • $\begingroup$ But if we think waves like a string with some beads on it, and a transverse wave is produced then the oscillation of a bead will depend on the previous one. If the previous one doesn't oscillate then how will the next one oscillate? Considering the wave to be continuous $\endgroup$
    – MSKB
    Commented Jun 1, 2021 at 18:34
  • $\begingroup$ If the two waves meeting at F are continuous, then yes there could always be destructive interference there, e.g. a peak always meets a trough. Going from F towards E, there is a difference of how far from each slit we are moving. for example E might be 1.2cm further from the bottom slit (than F), but only 0.9cm further from the top slit. So the displacements of the two waves vary differently allowing a non zero resultant, tiny near F and growing as we approach E. For a line through F perpendicular to the red lines - there could always be destructive interference on that line. $\endgroup$ Commented Jun 1, 2021 at 18:52
  • $\begingroup$ Basically my question is that as we have noticed in the image there are innumerous interactions between waves. A lot of them might be destructive which means some pair of waves will eventually cancel. But since there are infinite amounts of these interactions, how are these waves able to proceed towards te screen? Even if some wavelets somehow makes a way to proceed, that number should have been less. And as a result it would turn out that the amoumt of bright fringes would be too low compared to that what is observed in real. $\endgroup$
    – MSKB
    Commented Jun 1, 2021 at 19:21
  • $\begingroup$ I know I am missing something or might have misunderstood the whole thing. Would be a great help if you could point out which points I am missing or needs some more understanding. $\endgroup$
    – MSKB
    Commented Jun 1, 2021 at 19:22
  • $\begingroup$ Sorry but I suspect there is some basic misunderstanding about waves. The way you have drawn the lines is confusing. Usually people daw concentric circles as well, centred on each slit, then the circles can represent e.g. peaks of the wave at an instant in time. The radial lines just show the direction of energy transfer. A link is being added to the answer to help you study waves. $\endgroup$ Commented Jun 1, 2021 at 19:25

Good question.

For water waves at points of out of phase (our point F) the energy doesn’t vanish, it gets distributed sideways and comes back to this point later. Water waves are not only transversal waves. The have also longitudinal components.

Not so for EM waves. Such waves consist of photons (for the QM theorists: how one make an EM wave without the emission of photons from excited subatomic particles?) Photons do not interact at low energies and any interference is a good mathematical tool to calculate fringes. In nature photons get deflected at edges in such a manner that some directions are preferred an one see an intensity pattern in the screen.

  • $\begingroup$ So basically we observe the propagation of longitudinal compoment of water wave during interference? This makes it intuitively correct than the transverse wave case. $\endgroup$
    – MSKB
    Commented Jun 2, 2021 at 6:03
  • $\begingroup$ While studying interference we get to see that at point A if both the waves meet at in phase, then there will be constructive interference. But I couldn't get that how does that blue line(wave) get to point A if it had already met another wave at out of phase at point F. The energy of the wave has been transformed into some other means how would the rest of the particle along FA get to oscillate then? $\endgroup$
    – MSKB
    Commented Jun 2, 2021 at 8:46

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