One property of light sources that is usually stated, which is of particular importance when trying to create interference fringes, is the coherence length (or coherence time). The equation for the coherence length is given by $l_c = \frac{c}{n\Delta{f}}$ where $c$ is the speed of light, $n$ is the refractive index of the medium and $\Delta{f}$ is the bandwidth of the source.

An alternate description that I've seen (but unfortunately have lost a decent reference for) is that given a source emitting with wavelength range $\lambda \pm \Delta{\lambda}$, $l_c$ is the length that light with wavelength $\lambda$ and $\lambda + \Delta{\lambda}$ travels before they go from completely in-phase to completely out of phase. It leads from these definitions that an ideal, monochromatic source has an infinite coherence length and a broadband white light source has a very short coherence length.

This definition I am fairly happy with and can see what $l_c$ is trying to describe. The problem arises when we begin to talk about the limitations on when these non-idea light sources can produce interference.

Consider a Mach-Zinder interferometer (MZI):

a Mach-Zinder interferometer

it is always stated that in order to see an interference pattern through the MZI, (or for any other interference experiment) you must match the path-lengths of each arm of the interferometer to within the coherence length of the source.

My question is, quite simply, why? Considering the MZI and the definition of the coherence length above, I cannot seem to form an image in my mind as to why this should be so. I am able to follow the mathematics of the MZI (including the introduction of the degree of first-order temporal coherence: $g^{(1)}(\tau)$ - which is where this question first arose) but creating a physical image in my mind is proving very tricky.

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    $\begingroup$ Think of the two interfering beams as copies of one another, so that when they meet you are interfering $x(t)$ and $x(t+\tau)$, where $\tau$ is the time of flight mismatch. Does this help? $\endgroup$ Dec 28, 2014 at 11:44
  • $\begingroup$ Yes I see the image you're creating but I still can't see how there is a limit for $\tau$ such that beyond that limit there is no more interference of the beams. $\endgroup$
    – QuantumO
    Dec 28, 2014 at 12:26
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    $\begingroup$ The definition you quote ("they go from completely in-phase to completely out of phase") is incorrect. The key word that needs to be added is "random". When you compare two points separated by more than the coherence length, the phase relation changes rapidly and randomly. When you do an interference experiment using a slow light detector this change will be washed out, eliminating the interference effect. $\endgroup$
    – akrasia
    Dec 28, 2014 at 12:49

1 Answer 1


I hope that the image below may clarify the situation.

coherence length

Laving aside formulas, let me refer to the concept.

Coherence length $l_c$ of a wave-packet is the length of the wave-packet along which its wave-length is stable. The longer $l_c$, the better is for our interference experiments.

Let me explain. Please see the figure.

What we do in experiments as with the MZ interferometer, is to vary a bit the length of one of the arms and measure the intensity of the beam at one of the outputs of the upper beam-splitter, for instance at $O_{vert}$.

The coherence length of the packets A and B is the same, since we got the packets from the same "parent" packet at $BS_1$. Now, if $l_c$ is extremely short, and the arms $a$ and $b$ of the interferometer are slightly different, it is like the coherent region (constant $\lambda$) of the packet A, passed through $BS_2$ before he coherent region of B came to $BS_2$. So, no effect. But if $l_c$ is long, we will see that the intensity at $O_{vert}$ varies with the difference in path-length as cosine square.

Now, coherence time $\tau_c$ in void, is $\tau_c = l_c/c$.

If $\tau_c$ is long, we will see, for a fixed path-difference, a stable intensity at the examined output for a long time. As I said, varying the path-length difference the intensity for each path-length is stable, but comparing it for different path-lengths we get the cosine square dependence. Now, if $\tau_c$ is short, we will have difficulty in observing reliable results. Moreover, for the path-difference a bit longer, any effect will disappear, practically, we will see only the component of the packet that came later to the beam-splitter.

  • $\begingroup$ Good start, but I would add a discussion of partial coherence as mentioned in some of the comments. It isn't just a binary case of "coherent or not coherent" $\endgroup$ Dec 28, 2014 at 13:24
  • $\begingroup$ @CarlWitthoft: it's afternoon in my country, I have to leave the computer. I will return later. $\endgroup$
    – Sofia
    Dec 28, 2014 at 13:27
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    $\begingroup$ One missing piece (mentioned in other comments by @akrasia) is that of detection time. Every detector gathers light, or integrates, for some period of time before providing an output (could be a current, a charge, a digital readout ...). For many practical detectors the integration time is on the order of a second or a fraction of a second. If the coherence time is longer than the integration time, then the relative phases of the two beams is fixed, and interference emerges. If the coherence time is shorter than the integration time, the relative phases vary randomly: no interference. $\endgroup$
    – garyp
    Dec 28, 2014 at 14:36
  • $\begingroup$ @CarlWitthoft : when someone asks about a basic concept, let him/her first understand that concept well. After that, when the concept is understood, one can add details. If there are additional questions, the person will ask them. $\endgroup$
    – Sofia
    Dec 28, 2014 at 19:27
  • $\begingroup$ Sofia, I humbly disagree - your answer is misleading without a discussion of partial coherence. $\endgroup$ Dec 28, 2014 at 19:29

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