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Homework Help: Interference problem

  1. Feb 9, 2014 #1
    1. The problem statement, all variables and given/known data
    A system illustrated in the figure consists of two coherent point sources 1 and 2 located in a certain plane so that their dipole moments are oriented at right angles to that plane. The sources are separated by a distance d, the radiation wavelength is equal to ##\lambda##. Taking into account that the oscillations of source 2 lag in phase behind the oscillations of source 1 by ##\phi## (##\phi<\pi##), find the angles ##\theta## at which the radiation intensity is maximum.

    2. Relevant equations

    3. The attempt at a solution
    Let the radiations from the sources ##S_1## and ##S_2## interfere at a point P. As the radiation from ##S_1## is already ahead by the given phase, I have
    $$S_2P=S_1 P+d\cos\theta-\frac{\phi \lambda}{2\pi}$$
    For constructive interference,
    But solving the above equations doesn't give the right answer. :confused:

    Attached Files:

  2. jcsd
  3. Feb 9, 2014 #2


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    If you mean that S2P and S1P are distances of the point P from the sources, S2P - S1P≈dcosθ (when d<<S1P and d<<S2P). The phase difference between the rays arriving at P is (2π/λS2P-Φ -2π/λS1P), the effective path difference is S2P-S1P-Φλ/(2π) and that must be equal to integer times lambda.

  4. Feb 9, 2014 #3
    Hi ehild! :)

    Your method makes sense to me but solving it further doesn't seem to give the correct answer. Since ##S_2P-S_1P=d\cos\theta##, for constructive interference, I get:
    $$d\cos\theta-\frac{\phi \lambda}{2\pi}=n\lambda \Rightarrow \cos\theta=\left(n+\frac{\phi}{2\pi}\right)\frac{\lambda}{d}$$
    But the given answer is:
  5. Feb 9, 2014 #4


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    It appears to me that the problem is worded such that dipole 2 lags dipole 1 in time. So, the phase lag in wave 2 is a phase lag in time, not distance.
  6. Feb 9, 2014 #5


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    You are right, the phase of the light from source 2 is (2pi/λ)s2-(ωt-Φ) which results in a positive phase difference in distance.

  7. Feb 9, 2014 #6
    Sorry if this is a stupid question but what is the difference between the two? :confused:
  8. Feb 9, 2014 #7


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    Read my previous post. -Φ is associated with the time term in the phase of the wave. You get "+Φ" added to the space part.

    [tex]d\cos\theta+\frac{\phi \lambda}{2\pi}=n\lambda [/tex]

  9. Feb 9, 2014 #8
    I still don't seem to understand the difference between "phase lag in time" and "phase lag in distance". :(

    Working backwards from the answer, the phase difference must be
    But I don't get why we add ##\phi##. What is the difference in answer if it is mentioned that "phase lag is in distance"? :confused:
    Last edited: Feb 9, 2014
  10. Feb 9, 2014 #9


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    A travelling wave is of the form E=Eosin(ωt-kr). If the oscillation of S2 lags behind S1, the time lag appears in the ωt term : it becomes ω(t-Δt). The phase changes by -ωΔt=-Φ, but you can incorporate that Φ into the -kr term, with positive sign.

  11. Feb 9, 2014 #10
    Sorry if this is going to sound stupid but in this case, would it be correct to say that oscillation of S2 lags by ##\phi## in time but is ahead by ##\phi## in distance?

    Also, how do you conclude from the problem statement that the phase lag is in time not in distance? :confused:
  12. Feb 9, 2014 #11


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    No, I do not think so. Lag means something that happens later. There is phase difference because of time lag, and because of longer distance travelled.

    That was the problem statement:
    The sources are oscillating dipoles. Their motion is not in phase.
    The interference happens between the waves. Their phase difference is due to the time lag and the path difference.

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