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For constructive interference, I look at two of the same phase shift. They may arise form a thin interference layer of $\lambda/2$ thickness.

The result of the E component should be double the initial value. This can be seen from the wave equation $$E_1 = e_0 cos [ 2π ( x/λ -t/T)] . $$

If we add two of them with the same amplitude $e_0$, we get $$E_1 + E_2 = 2 e_0 cos [ 2π ( x/λ -t/T)].$$

The same is true for the B component $$B_1 + B_2 = 2 b_0 cos [ 2π ( x/λ -t/T)].$$

The intensity / energy I of the wave is proportional to the square of E, because B is proportional to E and I = E · B

So the intensity of the interfering waves should be 4x of a single wave. This seems faulty, because 2x of a source cannot give 4x of a result.

May you please find out where my mistake is?

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It is fine that some parts of a light wave have 4 times the intensity because other parts of the wave will undergo destructive interference and have less intensity. If you set up a detector that captures all of the emitted light, you will find that the total energy is equal to the total energy of the light sources. Interference only moves energy around.

It is impossible to have light only constructively interfere. Real light waves (even laser beams) are extended in spce and interference always has both constructive and destructive components, which conserves energy.

Here's a great demonstration: Where does the light go?

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This is what interference does. The sum of amplitudes of 1 can give intensity 4, 0 or anything in between.

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