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Superposition of light waves: 90 degrees out of phase, amplitude of resultant?

  1. Mar 8, 2012 #1
    Two waves are perfectly superposed (traveling same direction), but are 90 degrees out of phase. Does this result in destructive or constructive interference, as the waves could be seen as either half anti-phase or half in-phase. For example, the first wave has an amplitude of 1. The second wave, 90 degrees out of phase with the first, has an amplitude of 2. Which of the following scenarios is true?

    A.) Total cancellation, as they are half anti-phase, so the wave with amplitude of 2 decreases the first wave's amplitude (of 1) by half it's magnitude, half of 2 is one, thus full cancellation.


    B.) Partial constructive interference, the resultant has an amplitude of 2, b.c the second wave adds half of its magnitude to the first wave, half of 2 is 1, thus the resultant has an amplitude of 2.

    Which is correct? Can you provide a proof using the equations governing wave superposition?

    My confusion stems from the fact that 90 degrees phase difference between 2 waves can be seen as either half in phase or as half out of phase.
  2. jcsd
  3. Mar 8, 2012 #2


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    hi franco1991! :smile:

    the important point is …
    … if you do use the equations, you'll get the right result, and then you can decide which terminology (if either) is approppriate! :wink:
  4. Mar 8, 2012 #3
    Okay, I used the formula (A3)^2 = (A1)^2 + (A2)^2 + 2(A1)(A2)COS(P) where P is the phase difference.

    Using 90 as P, 1 as A1 and 2 as A2 (as in my example), and I got 2.2 as the resultant's amp.

    I'm now thinking that the only way the resultant's Amplitude will ever be 0 is at 1/2 a wavelength/180 degrees phase difference, because the result is only 0 if 180 is used as P.

    I was considering the waves a being half anti-phase. Thus, I reasoned that to make two waves with a 90 degrees (or 1/2 wavelength) phase difference, you just need to make the second wave double the amplitude of the first (because they are halfway anti-phase) to fully negate each-other.

    This is wrong then, correct? There is no combination of waves with non-equal amplitudes and a phase difference that IS NOT 1/2 a wavelength/180 degrees that will cancel each-other?
  5. Mar 8, 2012 #4


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    That is correct. The only way to get complete cancellation with two waves is for them to have equal amplitudes and a 180° (π radians) phase difference. Otherwise, if the waves have the same frequency, you always get another wave with the same frequency, but an amplitude and phase which depends on the amplitudes and phases of the original waves.
  6. Mar 9, 2012 #5


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    hi franco1991! :smile:

    (just got up :zzz:)
    i think they meant, to use cosx + 2sinx = (cosAcosx + sinAsinx)secA (where tanA = 2) = cos(x- A)secA :wink:
  7. Mar 13, 2012 #6
    So did I use the wrong formula? And if so, how would I plug my numbers into the one you gave? What does x denote, and would I plug something into x and rearrange to solve for A, in your example?
  8. Mar 13, 2012 #7
    If two monochromatic plane waves are traveling in the same direction with the same frequency and have the same polarization but are out of phase by some phase θ, then the superposition is especially easy:

    Wave 1: E = a cos(kx - ωt)
    Wave 2: E = b cos(kx - ωt+θ)
    Total wave: E = a cos(kx - ωt) + b cos(kx - ωt+θ)

    Using some trigonometry identities, we get the total wave in the form:

    E = (a + b cos θ)cos(kx - ωt) - b sin θ sin(kx - ωt)

    If we want no total wave, then (a + b cos θ) = 0 and b sin θ = 0 because cos(kx - ωt) and sin(kx - ωt) are linearly independent functions. These two equations are only satisfied if a = b and θ = 180°.

    If the phase difference is 90°, the total wave reduces to:

    E = a cos(kx - ωt) - b sin(kx - ωt)

    The exact shape of this will depend on a and b, but you can't think of it as a simple cosine wave anymore. The peak amplitude occurs when kx - ωt = 135° and is:

    |E|peak = (a + b) √2/2
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