My question is related to young's theorem and the double slit experiment.

The more light sources we add the sharper the peaks of the interference pattern become. However when we add more light sources, more energy is being radiated. So my question is if the peaks become sharper where does all this extra energy go?

enter image description here

(respectively for 2, 4, 8 and many equal brightness, and evenly spaced coherent, light sources)

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    $\begingroup$ Very loosely, the energy is proportional to the square of the amplitude of the electric field (modulo factors), and you'd have to integrate that over the entire distribution (read: interference pattern) to get the total received contribution. Sharp peaks does not mean less energy, but just sharp interference: the energy is more narrowly concentrated (i.e. in a smaller area) when the peaks are sharp. When the peaks spread out, you have large peaks and little peaks and there's energy (think of it as the area under the peaks...). $\endgroup$ Aug 21, 2016 at 22:05
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    $\begingroup$ Said another way, the plots in the question do not show absolute intensity. They show intensity normalized by peak height. If absolute intensity were plotted, then the peaks for the many sources would be much taller than for the few sources. $\endgroup$
    – John1024
    Aug 21, 2016 at 22:58
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    $\begingroup$ Nit picking: "For 2, 4, 8 and many equal brightness, and evenly spaced coherent, light sources." Not a big deal, of course, but it was bothering me. $\endgroup$ Aug 22, 2016 at 0:11
  • $\begingroup$ What are number of light sources and where you get the information about the peaks from? $\endgroup$ Aug 22, 2016 at 5:01
  • $\begingroup$ @HolgerFiedler light sources would be the number of slits or total number of edges. $\endgroup$ Aug 23, 2016 at 23:36

1 Answer 1


I will ignore the effects of finite slit widths which would modulate the intensity of the interference pattern to make the explantation easier and that would be in keeping with your diagrams. I will also try and avoid too much mathematics.

For the double slit (source) the amplitude of a peak is 2A where $A$ is the amplitude of the waves from a single slit.
Therefore the intensity at a peak of the double slit is $4I \propto (2A)^2$ as compared with the intensity from a single slit $I \propto A^2$.
The intensity of the fringes as a function of angular position $\theta$ are of the form $4A^2 \cos^2 \theta$ and this function when average out (think of root mean square voltage) has a value of $2A^2$ which is proportional to the intensity of the light from two slits.
So no energy is lost rather the energy flow is redirected to form the two slit interference pattern.

The first thing to note is that your diagrams are not to scale on the vertical (intensity) axis.
If the intensity of the peaks for the two slit pattern is $I_2$ then the intensity of the principal maxima for four slits is $\frac {4^2}{4}I_2 = 4I_2$, for eight slits $\frac {8^2}{4}I_2 = 32I_2$, for $N$ slits $\frac {N^2}{4}I_2$ and if $N=10^4$ the intensity of the principal maxima is $\frac {{10^4}^2}{4}I_2 = 2.5 \times 10^7I_2$.
So this perhaps gives you an indication of the energy flow being channelled into narrower and narrower corridors as the number of slits increases.
If one did the average from one principal maximum to the next one would find it to be $NI$ where $N$ is the number of slits and $I$ the intensity from one slit.

It might be worth you looking at another answer I wrote to a similar question? In that answer I tried to indicate using phasors how the pattern of principal and secondary maxima comes about.

With multiple slits you are adding waves which emanate from the slits which differ in phase because they travel different distance.

enter image description here

The path difference is $d \sin \theta$ and so the phase difference between the waves $\phi = \dfrac {2 \pi d \sin \theta}{\lambda}$

For your four slits the phasor diagrams at the maxima and minima look like this:

enter image description here

You will note that the principal maximum has an amplitude of $4A$ and hence an intensity of $16I$ where $I$ is the intensity from one slit.
The secondary maximum has an amplitude of just over $A$ and hence an intensity slightly greater than that of a single slit $I$.

To illustrate the reason for the principal maxima becoming narrower as the number of slits increases consider the following phasor diagram where the phase difference between adjacent slits is $20 ^\circ$.

enter image description here

If you have 2 slits then there are only two phasors with a resultant length of $1.97$ and hence the intensity of the light at this phase angle as compared with the intensity at a maximum is $\dfrac {1.97^2}{2^2} = 0.97$, ie the intensity has hardly dropped at all.

With four slits the length of the resultant phasor is $3.70$ and so the intensity, relative to a principal maximum has dropped $\dfrac {3.70^2}{4^2} = 0.86$ which is more than for two slits.

Doing the same for nine slits $\dfrac {5.76^2}{9^2} = 0.41$ and sixteen slits $\dfrac {1.97^2}{16^2} = 0.15$ shows that by 16 slits the intensity at a phase angle of $20^\circ$ is very small compared with the intensity of the principal maxima.
Finally for 18 slits the resultant is zero for this phase angle and so by $20^\circ$ the first minimum has been reached showing that the principal maxima are so much narrower and intense than for two slits.


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