In this PDF on the derivation of intensity at any given point on the screen from single slit diffraction, on the first page, the amplitude of the wave produced by a small element is taken as $E_0/N$. I do not get the logic behind dividing by $N$ since why would the amplitude change on considering the wave produced by a single wavelet?

I thought that it had something to do with considering only a single wave out of many waves produced by the wavelet, but, while deriving the intensity in the double slit experiment, we do not divide the amplitude of the waves produced by the single wavelets at the two slits by anything.


1 Answer 1


Perhaps it is easier to start with $N$ wavelets each of amplitude $\dfrac {E_0}{N}$ and see what happens in the straight through direction.
The waves from these wavelets overlap at infinity (or at the focal point of a convex lens) to form a wave of amplitude $\dfrac{E_0}{N}\times N = E_0$.

So the $E_0$ refers to what happens at the centre of the diffraction pattern.
In your PDF where it refers to the phasor representation this is with $N$ phasors each of magnitude $\dfrac{E_0}{N}$ being along a straight line.

For an angle slightly different from the straight though one the addition will result is a slight smaller value of amplitude because the waves from each secondary source do not arrive exactly in phase with one another.

So this is a way of being able to compare the size of the addition of the secondary wavelets in various directions and having got the amplitude the intensity can be compared.

A more detailed analysis incorporates the idea that the amplitude in the forward direction is larger than in other directions (obliquity factor) as is described in this link so the simple analysis gives you a "flavour" of what the diffraction pattern will look like.

When it comes to the double slit the simplest analysis assumes a slit width which is comparable with the wavelength the diffraction envelopes from each slit are so wide that when they overlap the resulting two slit interference pattern is such that the maxima are assumed to be of the same intensity.

However for slit widths which are greater than the wavelength of light the overlapping diffraction patterns due to the two slits pattern are modulated by the two slit interference pattern.
So to do the full analysis you have to again cut up each of the slits into a number of secondary sources and evaluated the result of the wavelets from these secondary sources overlapping.

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  • $\begingroup$ Sir i have a question, the slit is divided into large number of point sources and the waves superposed due to all of them gives the resultant amplitude on any point on the screen, the amplitude E° of the wave due to point source decreases with distance r, why haven't you considered it( in your first paragraph of your answer) $\endgroup$
    – Shivansh J
    Commented Oct 29, 2019 at 4:36
  • $\begingroup$ @ShivanshJ You are correct to comment on text the the path lengths $r$ will differ but by how much? A few wavelength which are very much smaller than the path length. This means that the change in amplitude is very small and is neglected in this type of analysis. More rigorous treatments do not necessarily make such an approximation. $\endgroup$
    – Farcher
    Commented Oct 29, 2019 at 7:10

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