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The typical textbook explanation of the interference fringes in a double slit experiment goes like this. If the difference between the path lengths travelled by waves from each slit to a point on the screen is a whole number of wavelengths then the two waves interfere constructively at the point and we have a bright spot.

But is this right? Consider two waves which begin from the two slits in phase with the amplitudes of their electric fields equal to zero. They arrive at a point on the screen at which the difference of their path lengths is one wavelength. This also means that the difference between the time taken to reach the point is one period. So they arrive in phase at the point but since the amplitude of their electric fields is zero the point is a dark spot, not a bright spot.

What is the error in this argument?

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T/2 later you have amplitudes at max . The intensity of light is proportional to $A^2$ . when you "see" light you dont see the maxima and minima, you just get the mean energy. So as in the two slits you have at the point oscillation of the electric field with double amplitude.

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  • $\begingroup$ At a point on the screen the amplitude does not vary with time. $\endgroup$
    – phil342
    Commented Nov 5, 2022 at 1:36
  • $\begingroup$ You are right, I used the wrong expression, instead of maximal amplitude it should have said maximal E, but in your post you also spoke of amplitude zero at the slits? $\endgroup$
    – trula
    Commented Nov 5, 2022 at 15:16
  • $\begingroup$ I said the two waves have zero amplitude at the slits only for convenience of calculation. The point is that the phase difference between waves at the two slits and at the screens is constant in time. Even if the waves are in phase with each other but have zero amplitudes at the screen you have a dark spot. This contradicts the classical argument that a bright spot is where the path difference between the two waves is an integer number of wave lengths. $\endgroup$
    – phil342
    Commented Nov 7, 2022 at 1:29
  • $\begingroup$ I don't understand you. If there is no amplitude, there is no wave, so no light. So if no light is coming from the slits you have no light on the screen. If you have light thru the slits you have light on the screen, when the two waves are in phase. $\endgroup$
    – trula
    Commented Nov 7, 2022 at 18:19
  • $\begingroup$ The slit has magnitude, so while there are points at which the amplitude of the electric field is zero, there are other points at which the amplitude is not zero. In the classical analysis of diffraction the slit is assumed to have magnitude. In the analysis of interference the slit is assumed to have zero magnitude. The interference fringes occur on the diffraction fringes. What I am suggesting is that there is something wrong with the classical analysis. $\endgroup$
    – phil342
    Commented Nov 9, 2022 at 1:56

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