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How is energy conserved In Young's Double Slit Experiment ?
In destructive interference , energy is lost .
So what happens to that lost energy ?

Does it escape as heat ?

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  • $\begingroup$ Why do you think energy is lost? $\endgroup$
    – Andrew
    Mar 2, 2015 at 6:03
  • $\begingroup$ @Andrew Because intensity becomes zero at a dark fringe so I think light energy is lost . $\endgroup$
    – Klosew
    Mar 2, 2015 at 6:05
  • $\begingroup$ Ah I see. What is conserved is the total energy, which will be the integral of the energy density (~ intensity) over the whole screen. This integral will include regions of constructive and deconstructive interference. The total energy, computed by performing this integral, will equal the total energy emitted by the source. $\endgroup$
    – Andrew
    Mar 2, 2015 at 6:08
  • $\begingroup$ @Andrew But what happened exactly at the point of minima ? Energy got lost ? $\endgroup$
    – Klosew
    Mar 2, 2015 at 6:11
  • $\begingroup$ @Klosew: If there is dark fringe, there is also bright fringe. The energy lost due to destructive interference is taken up due to constructive interference. BTW, dark fringe is not really dark; photon can hit here also, but the probability is minuscule. $\endgroup$
    – user36790
    Mar 2, 2015 at 6:16

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I think your confusion is assuming that the energy of two waves add, when in reality there is an interference term.

The short, physics-y answer is that it is not that any energy has disappeared, rather the interference has caused some of the energy that 'would have been there' to show up in a different place. The energy got shifted but not destroyed.

This can be backed up with some math. Let's say we have an electromagenetic wave \begin{equation} \vec{E}_1(\vec{x},t)=\vec{\mathcal{E}}_1 \cos(\vec{k}\cdot\vec{x} - \omega t) \end{equation} where $\vec{\mathcal{E}}_1$ is a constant. Note $B=\frac{1}{c}\vec{k} \times \vec{E}$. I will suppose that I can neglect the magnetic field term in the energy below, to simplify things.

The energy is \begin{equation} U_1 = \frac{1}{2}\vec{E}^2 = \vec{E}_1^2 \cos^2(\vec{k}\cdot \vec{x} - \omega t) \end{equation}

Now let's add a second wave \begin{equation} \vec{E}_2(\vec{x},t)=\vec{\mathcal{E}}_2 \cos(\vec{k}\cdot\vec{x} - \omega t+\phi) \end{equation} For simplicity let's suppose that the wavenumber and frequency is the same, the only difference is the phase $\phi$.

Now, you might have thought that the energy is $U=U_1+U_2$. However this is not true! The energy is really \begin{equation} U=\frac{1}{2}\left(\mathcal{E}_1^2 \cos^2(\vec{k}\cdot\vec{x}-\omega t) + \mathcal{E}_2^2 \cos^2\left(\vec{k}\cdot\vec{x}-\omega t + \phi\right)+2\mathcal{E}_1\mathcal{E}_2 \cos(\vec{k}\cdot \vec{x}-\omega t)\cos(\vec{k}\cdot\vec{x}-\omega t + \phi)\right) \end{equation}

The last term above is the counterintuitive term that 'shifts the energy around.'

Finally, to be pedantic I should point out that above I am really talking about the energy density. The true, total energy is the integral of the energy density, and is not sensitive to these kinds of interference effects. In particular, (at least in electromagnetism), the total energy of the two waves is the same as the sum of the energy of each individual wave (the interference term will integrate to zero). Thus interference will shift the energy density around, but will leave the total energy unchanged.

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  • $\begingroup$ Perfect . Thanks . I have a similar question about interference pattern of electrons ( as in do electrons shift like over here energy density shifts ) . Should I open up a new question , edit this one or just try to get the answer in the comments ? $\endgroup$
    – Klosew
    Mar 2, 2015 at 7:05
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    $\begingroup$ @Klosew: the electron density shifts around in exactly the same way as the energy density. $\endgroup$ Mar 2, 2015 at 7:15
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    $\begingroup$ It is not so easy to prove energy conservation for interfering plane waves because they carry infinite energy. $\endgroup$
    – hyportnex
    Mar 2, 2015 at 12:31

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