# Two slit diffraction and energy conservation

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1. Dec 13, 2015

### LmdL

Hi all,
I have a small misunderstanding about the energy conservation in diffraction from 2 slits.
First, I understand the energy conservation of interference from 2 slits.
If intensity from each slit is I, then I have intensity of 2I after slits plane.
Interference is given by:

So at bright fringes I get cos^2=1, so intensity is 4I. And in dark fringes I get cos^2=0, so intensity is 0.
Since interference just distributes the intensity over screen, the cos^2 pattern with 4I maximum and 0 minimum, on average, results in a 2I intensity, as just after the slits plane.
Now I add the diffraction, i.e. multiply by sinc^2. This leads to a lower intensity pattern compared to the case of pure interference, so averaged intensity will be lower then 2I. Where the energy is lost?

2. Dec 14, 2015

### Simon Bridge

You are comparing averages between two different situations... no special reason the averages have to be the same.
Instead, compare the incoming energy with that leaving the slits.

3. Dec 14, 2015

### blue_leaf77

The fields in the slit plane and in the observation plane is connected through Fourier transform in the case of far-field diffraction, therefore you don't need to worry about energy conservation because it is guaranteed by Parseval's theorem
$$\int |E(x)|^2 dx = \int |\tilde{E}(x')|^2 dx'$$
where $\tilde{E}(x') = FT[E(x)]$, $x$ is coordinate in the slit plane, and $x'$ the coordinate in the observation plane.

4. Dec 14, 2015

### LmdL

Still didn't get it. I understand that this must be true (energy is conserved) and even understand why it's true (intuitive) for the interference case. But when the diffraction is introduced, I cannot get it in intuitive way. The only explanation I can think of is that I don't multiply by sinc^2 in the diffraction case, but rather by Asinc^2, where A is some normalization constant which ensures energy conservation. So the overall picture will look like that:

And not like that: