Max Intensity of 2 Waves: E^2(E_01 +- E_02)

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SUMMARY

The maximum intensity of two waves is calculated as (E_{01} + E_{02})^2 when the waves are in phase, and as (E_{01} - E_{02})^2 when they are out of phase by π radians. The intensity is derived from the superposition principle, where the intensity is proportional to the square of the amplitude. The relevant equations include U(x) = √(2/η)E(x) and η = √(μ₀/ε₀)/n, where n is the refractive index of the medium. This analysis confirms that the waves must share the same frequency for phase considerations to be valid.

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Homework Statement


Show that the maximum intensity of 2 waves is worth [tex](E_{01}+E_{02})^2[/tex] while for 2 waves out of phase by pi rad, it's worth [tex](E_{01}-E_{02})^2[/tex].


Homework Equations


Not sure about intensity. According to my notes it's worth [tex](U_1+U_2)(U_1+U_2)*[/tex] where * denotes the complex conjugate. And [tex]U(x)=\sqrt{\frac{2}{\eta}}E(x)[/tex].
Also [tex]\eta=\frac{\sqrt{\frac{\mu _0}{\varepsilon _0}}}{n}[/tex] where n is the refractive index of the medium. Is there some easier formula for the intensity that I could use?

The Attempt at a Solution


As they don't say anything if the 2 waves have the same frequency, I think that indeed they have the same frequency otherwise it's senseless to talk of a phase. Am I right?
 
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If we talk of superposition of waves a phasor treatment provides
Ar2 = A12 + A22 + 2A1A2cosT
where T is the phase difference. Now knowing that intensity is proportional to amplitude squared and using conditions for constructive and destructive interference you can easily show the required results.
 

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