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On optical interference

  1. Jan 31, 2016 #1

    Charles Link

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    This post is related to a posting I did a couple of days ago, but I will go into more detail on this one, and it should be more easy to see the concept that is presented. A number of years ago, I was explaining interference concepts to a physics Ph.D. who had very little background in optics. In discussing the two-slit experiment, he was rather puzzled how a non-linearity had surfaced in that the sum of the individual intensity patterns was not the resultant intensity pattern that emerged. His argument, which really contained a good deal of validity, was that the wave equations we were using were Maxwell's and were completely linear, and the superposition of solutions should thereby be a solution. Thereby, how could any non-linear features ever emerge in a system, (such as intensity patterns that didn't superimpose), from a system that was defined by completely linear equations. My instincts at the time were that there must be some hidden non-linearity somewhere, but I was unable to pinpoint it. It was only quite a number of years later, upon solving a Fabry-Perot type interference problem, that the "hidden non-linearity" became apparent. The energy equation, which is basically intensity I=n*E^2, is also used in the computations along with the wave equation (that comes from Maxwell's equations), and thereby we have a governing equation in our system that is non-linear in the E-field. The equations that determine the E-fields (Maxwell's) are all linear in the E-fields and so the E-fields will always obey linear principles, but the energy equations are not of a linear form, so that we can expect to see cases where linear principles are not obeyed in regards to the energy. e.g. MTF (modulation transfer function) computations work in the incoherent cases, but do not work for coherent sources. Another example where we see linear behavior in the E-fields but not in the energy is when two plane waves are incident on a single interface from opposite directions. (such as in an optical interferometer). In this case, the fresnel coefficients (E-field coefficients for transmission and reflection) remain good numbers when both sources are present, but the energy reflection (e.g. R=rho^2) and transmission coefficients are no longer valid in this coherent case with the two incident plane waves interfering. There is no requirement that the system needs to be linear in the energy response, and in the case with two mutually coherent sources present, (one from each direction), the energy coefficients "R" and "T" can not be used to determine the outcome. Instead the individual E fields from each source need to be split using the fresnel coefficients, and the resulting E fields added (emerging from the junction in each direction), before computing the energy of each. The result turns out to depend on the relative phases of the incident plane waves, and energy is 100% conserved. It is actually possible in the case of a 50-50 energy split from the junction for each individual beam to have 100% of the energy emerge to the right with two sources present. By proper adjustment of the phase, 100% of the energy can alternatively be made to emerge to the left. i.e. the two sources are completely recombined by the beamsplitter. (Note for R=1/2 (50-50 split), the fresnel rho=+/-1/(sqrt(2))
     
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  3. Jan 31, 2016 #2

    Dale

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    That depends on how you define "governing equation". The equations of motion are linear.
     
  4. Jan 31, 2016 #3

    Charles Link

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    Dale, It's quite well known in optics that any measured intensity is proportional to the square of the E-field amplitude. What may puzzle you, I can explain how the pendulum and mass on a spring were connected to all of this from my previous post. The "physics" connection was weak or non-existent, but they were mathematically similar in that these were linear systems which had sinusoidal solutions, just like the electromagnetic sources. Once I solved the optics question here , it answered for me a similar puzzle, ( a mathematical one), that was present in the mechanical system.
     
  5. Jan 31, 2016 #4

    Drakkith

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    So you're saying that non-linearity appears in the form of the energy imparted by an EM wave to an object? Forgive me but I don't see the significance of this. Is there something I'm missing?
     
  6. Feb 1, 2016 #5

    blue_leaf77

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    Hi Charles, the energy equation is not exactly the same as intensity equation. It's true that the intensity is non-linear with respect to the field from each slit in the double slit experiment, but the energy, which is the integral of intensity over the entire screen is linear as you can see from Parseval's theorem. Suppose initially, only one slit is open, then Parseval's theorem applied to this slit reads as
    $$
    \iint_{\textrm{slit plane}} |E_1(x-a,y)|^2 dxdy = \iint_{\textrm{screen}} |\tilde{E}_1(x',y')|^2 dx'dy'
    $$
    Next, the previous slit is closed and the other one is opened, Parseval's theorem for this case is
    $$
    \iint_{\textrm{slit plane}} |E_2(x+a,y)|^2 dxdy = \iint_{\textrm{screen}} |\tilde{E}_2(x',y')|^2 dx'dy'
    $$
    If now both slit are open and they do not overlap such that ##E_{\textrm{both}}(x,y) = E_2(x+a,y)+ E_1(x-a,y)## , we have
    $$
    \begin{aligned}
    \iint_{\textrm{slit plane}} |E_{\textrm{both}}(x,y)|^2 dxdy &= \iint_{\textrm{slit plane}} \left( |E_2(x+a,y)|^2 + |E_1(x-a,y)|^2 \right)dxdy \\
    &= \iint_{\textrm{screen}} \left( |\tilde{E}_2(x',y')|^2 + |\tilde{E}_1(x',y')|^2 \right) dx'dy' \\
    &= 2\iint_{\textrm{screen}} |\tilde{E}_1(x',y')|^2 dx'dy'
    \end{aligned}
    $$
    where in the last line, I have made use the fact that ##|\tilde{E}_2(x',y')|^2 = |\tilde{E}_1(x',y')|^2## for their respective object is merely a translation of one another. Therefore, the energy equation is linear - the energy measured in the screen is simply the sum of energy from individual slit.
     
    Last edited: Feb 1, 2016
  7. Feb 1, 2016 #6

    Charles Link

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    If I followed your calcul
    By energy, I was referring to the energy intensity. The energy of the entire pattern, and the photon count, is conserved in an interference pattern. I am in complete agreement with the result of energy conservation in an interference pattern, but I think Parseval's theorem already assumes Fourier transforms (and MTF techniques) apply to the system's energy distribution across any plane. MTF methods will not give the intensity distribution of an interference pattern, because MTF methods use a simple energy intensity summation from the two sources. My post is pointing out that the creation of an interference pattern does involve a second-order (power of two exponent) process and the wave-particle duality contains both linear equations along with a second order equation.
     
  8. Feb 1, 2016 #7

    blue_leaf77

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    There is no such a thing as energy intensity, energy and intensity are two different quantities and I have never heard of these two terms combined together to form a composite word.
    Yes, Parseval's theorem is an implication of Fourier transform.
    I am not sure whether my understanding of MTF is similar to yours, but Fourier transform does give the intensity of a diffraction pattern (by taking its square modulus to be exact) in far-field limit.
    My posting in #5 addresses the repeated improper use of the word "energy": Some of which are
    Energy in the diffraction pattern is linear in terms of the individual energy of any opening in the object plane. What you should have used in place of "energy" is the "intensity".
     
  9. Feb 1, 2016 #8

    Charles Link

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    On further consideration, your use of the fourier transform and the E-field with Parseval's theorem may in fact be completely correct. MTF methods consider the energy distribution across a plane, in more formal terms what is referred to as the "irradiance" (Watts/cm^2) and take fourier transforms of the energy distribution and linearly convert the energy (by using fourier transforms and linear response) from one plane to another. ... Calling it "energy intensity" is perhaps poor grammar, but I have done many upon many calculations of both multi-source and Fabry-Perot type interference calculations. The mathematical detail and accuracy is my primary focus. In radiometrics, the term "intensity" is in watts/solid angle. The "irradiance" of radiometrics is often referred to as "intensity" in many optics texts. It can be tricky to pick the correct terms when addressing physics people with different specializations. I used the term "energy intensity" somewhat loosely and hoped it was understood what I was referring to.
     
  10. Feb 1, 2016 #9

    blue_leaf77

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    Perhaps "energy density" would be a better choice for use to cover a broader range of field specializations.
     
  11. Feb 1, 2016 #10

    Charles Link

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    There doesn't seem to be a perfect terminology. The term "energy density" I like to use when referring to the U=E^2/(8*pi) +B^2/(8*pi) (c.g.s. units). "Intensity" would have perhaps been a better choice than "energy intensity", but I'm hoping there are at least a few readers who made a good effort to follow the mathematical content. Oftentimes in an interference calculation, I routinely compute intensity"I"=n*E^2 where E is computed from summing terms containing Fresnel coefficients (for the E-fields) along with phase terms. (e.g. Fresnel coefficients r=(n1-n2)/((n1+n2) and t=2*n1/(n1+n2)). The precise name that is used for "I" does not alter the mathematics.
     
    Last edited: Feb 1, 2016
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