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I have the following statement which I don't know how to explain:

Suppose I have 2 identical monochromatic waves (same intensity and phase) shooting into the same receiver. If each wave's intensity is I, based on energy conservation I would expect the 2 waves together will bring a total intensity of 2I. For example if each wave carries 100mW power, I'm expecting a 200mW total power on the receiver side.

However the summation using phasor gives a different result: if we consider that intensity is proportional to the square of wave's amplitude, the square root of 100 gives 10 for the amplitude of 1 wave (for simplicity I use 10 as amplitude which included the constants), adding the other wave of the same phase which gives 20 as amplitude of the new beam, then square it to get intensity which is 400mW instead of 200 in the above example.

If we keep going by summing 4 50mW waves with the same convention, we get (4 x sqrt(50))^2 = 800mW ... which the logic is obviously not correct.

My ultimate goal is to sum up power of beams with different intensities and phases. Phasor addition works great if I have amplitude and phase, but when I try to use intensity to get wave's amplitude I got the above dilemma. Could someone point out where my logic go wrong, and please explain the way to do power summation with intensity and phase known for each wave? Thanks!

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From Hecht optics textbook, intensity is actually the time averaged electric field amplitude squared

$I_1=E_1^2/2$

$I_2=E_2^2/2$

With total constructive interference, the resulting intensity is

$I_{max}=I_1+I_2+2\sqrt{I_1 I_2}$

With total destructive interference, the resulting intensity is

$I_{max}=I_1+I_2-2\sqrt{I_1 I_2}$

These equations can come from phasor addition, using the law of cosines.

Two beams will make a interference pattern, so the overall intensity is still just $I_1+I_2$

Uncertainty with number of photons and phase means that photons cannot have all the same phase, so you can't add two beams in an solely constructive manner.

https://en.wikipedia.org/wiki/Photon

With more beams you just add more phasors. Hecht optics has chapters on superposition of waves and interference.

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As explained here, you have to consider the fact, that your light waves' amplitude gets averaged over a certain amount of time, thus decreasing the intensity.

I quote from the website, just for the sake of having the information close to this question:

"Try this: Let $ψ=Acosθ_1$ and $ϕ=Bcosθ_2$.

Then: $I∝(ψ+ϕ)^2=(Acosθ_1+Bcosθ_2)^2$ $I∝A^2cos^2θ_1+B^2cos^2θ_2+2ABcosθ_1cosθ_2$.

The average value of $cos^2$ is 1/2; using a bit of trig, you should be able to convince yourself that the cross term averages to zero if the phases are random. Thus:

$I∝A_2/2+B_2/2$

$I_{total}=I_1+I_2$"

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  • $\begingroup$ The OP has stated that the phases are not random but the same, so that cross term average is not zero. $\endgroup$ – Farcher May 15 '17 at 14:45
  • $\begingroup$ Could you please provide more details about the case which the 2 phases are not random? thanks! $\endgroup$ – vvofdcx May 16 '17 at 3:45
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Your initial assumption that you can simply sum intensity (power) is incorrect.

The law of superposition of Maxwell's equations says that you can sum electric fields and magnetic fields -- or voltage and current in a transmission line or circuit context. So if you consider total voltage $V(t)=V_1(t)+V_2(t)$ and total current $I(t)=I_1(t)+I_2(t)$, then total power is $P(t)=VI=(V_1+V_2)(I_1+I_2)=V_1I_1+V_2I_2+V_1I_2+V_2I_1$. Notice those cross terms.

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    $\begingroup$ So no energy conservation for you? $\endgroup$ – lalala Feb 27 at 7:27

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