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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**

MSKB
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2 Answers2

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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

https://courses.lumenlearning.com/boundless-physics/chapter/waves/

John Hunter
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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.

HolgerFiedler
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