0
$\begingroup$

In a Michelson Interferometer, a light beam is split into 2 different beams that travel different optical paths, through the "arms" of the interferometer. Then, they are reflected and, finally, recombined to form a single light beam again. But because they traveled optical paths of different lengths these beams will have a phase diference between them, which will result in an interference pattern. The straight-forward method makes the device easy to build and use.

However, it is said that one of the disadvantages of this interferometer is that it doesn't allow for accurate measures when the light source is white light, but I don't really understand why that's the case.

After a quick search on Wikipedia, they essentially say that the difference in the optical paths followed by the two light-beams needs to be inferior to the coherence length of the light source to produce significant interference contrast, so since white light has a low coherence length (of the order of $10^{-6}$m), the optical paths need to be practically the same for both light beams.

But in order to get interference, wasn't the point to get the light beams to follow different optical paths? This all just seems slightly contradictory to me.

$\endgroup$
5
  • 1
    $\begingroup$ " But in order to get interference, wasn't the point to get the light beams to follow different optical paths? " Interference will occur even if light beams follow same optical path. In fact, the experiment tries to ensure that the 2 beams follow the same optical path. Then , it tries to measure if the light beams along the 2 identical paths have different speed or not $\endgroup$ Apr 9 '21 at 10:29
  • $\begingroup$ But then where does the bit about how the difference in the optical paths needs to be inferior to the coherence length of the light source come into place? $\endgroup$
    – Rye
    Apr 9 '21 at 10:47
  • $\begingroup$ Where does it say that ? can you copy paste the full sentence you are referring to ? $\endgroup$ Apr 9 '21 at 10:51
  • $\begingroup$ On the Wikipedia link I mentioned: "Narrowband spectral light from a discharge or even white light can also be used, however to obtain significant interference contrast it is required that the differential pathlength is reduced below the coherence length of the light source. That can be only micrometers for white light, as discussed below." $\endgroup$
    – Rye
    Apr 9 '21 at 10:54
  • 1
    $\begingroup$ FTIR spectrometers use white light in a Michelson interferometer: en.m.wikipedia.org/wiki/Fourier-transform_infrared_spectroscopy $\endgroup$
    – Gilbert
    Apr 9 '21 at 17:59
2
$\begingroup$

Interference pattern has a visibility, meaning how better can the fringes distinguished. The visibility is related to the degree of first order coherence $g^{(1)}$, a measure of amplitude-amplitude correlation. It indicates the fact that exist a fixed relationship between the phases at different instant of time, or different position along the beam.

White light is broadband and larger the bandwidth, lesser the coherence time. The last represent the time in which there is a fixed phase relationship. Since interference depends upon phases, it's important to have a fixed phase relationship.

Coherence length is defined: $l_c = c\tau_c$, where c is the speed of light. When the fringe pattern is computed there is an oscillatory term that depends on the path lengths difference and a prefactor that is the modulus of the first order coherence. Taking for example chaotic light with collision broadening you have an exponential decay of the degree of first order coherence, like $\exp(-\frac{\tau}{\tau_c}) $. If you take the lengths path difference $|z_1 - z_2|$ much smaller than the coherence length $l_c$, this term can be high and so the visibility, since $\tau = \frac{|z_1 - z_2|}{c}$. Extremely small coherence time $\tau_c$ implies necessarily to take impractical extremely small paths difference.

$\endgroup$
1
$\begingroup$

The reason a white light source is more difficult to use than a monochromatic source is really very simple: visible fringes are easier to obtain when a monochromatic source is used. This is because monochromatic light has a longer coherence length. Fringes are only visible if the difference in path lengths is less than the coherence length. So, if the coherence length is just a few microns as in the case of white light, the two path lengths must be adjusted to be equal within a few microns - which is not always easy. An LED typically has a coherence length of 12 microns or so; and a diode laser can have a coherence length of meters. It's very easy to obtain fringes immediately after setting up a Michaelson interferometer with a laser source, but can be very tedious to obtain fringes with a white light source.

Yes, the difference in path length is indeed what is measured in an interferometer, but if fringes can only be seen when the path length is less than the coherence length, white light only allows a very small range of path length difference to be measured: a few microns. With a coherence length of meters, a laser can be used in an interferometer to measure path length changes that vary from a fraction of a micron to several meters, all with sub-micron precision.

$\endgroup$
1
$\begingroup$

The Michelson interferometer compares the lengths of two light paths.

One source produces two light beams which appear to come from two "coherent" (virtual) sources which traverse two paths and then superpose to produce an interference pattern consisting regions of different intensities - light and dark fringes.

Amongst other factors the separation of the fringes depends on the wavelength of the light.
The shape of the fringes depends on the orientation of the mirrors which if exactly at right angles to one another produces circular fringes and if at other angles to one another can produce straight line fringes.

White light contains a continuous range of wavelengths each of which produce an interference pattern with a different fringe separation.
It is only when the path lengths of the two beams which form the interference pattern are equal that constructive interference occurs at the same region for all wavelengths and that is often called the zero order fringe and is shown below in the middle of the diagram.

enter image description here

What you see beyond the zero fringe is the fringes pattern of different wavelengths overlapping one another.
So for the first bright fringe beyond the centre (first order fringe) it is blue (short wavelength) on the "inside" and violet" (longer wavelength) on the outside because the blue light has the smallest fringe separation and the violet light the largest fringe separation.

As one moves away from the centre you could have a bright fringe due to one wavelength occurring at a position where there is a dark fringe of another wavelength.
Thus moving out from the centre the fringe patterns due to each of the wavelengths which make up white light overlap so much that no further fringes are discernable.
With while light one might see a central white fringe and a few coloured fringes and then a uniform illumination.

To measure wavelength using the interferometer changes one of the path lengths by moving a mirror a measured distance and counting the fringes which cross a graticule in the field of view and each complete fringe movement corresponding to the mirror being moved half a wavelength of the light.
To get a reasonable estimate of wavelength one might measure the distance the mirror moves when $50$ fringes traversed the field of view.

The problem with trying to measure a wavelength of the light which makes up white light is that as soon as a mirror is moved only a little from a position when the light path lengths are equal the white light fringe disappears, coloured fringes replace it and then as the mirror moves further the fringes disappear.

One could use narrow width bandpass filters to measure the wavelengths of some of the components of white light.

One advantage of using a white light source is that it can be used to set up the interferometer with equal path length to within a fraction of a wavelength as only then will a zero order white fringe will be visible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.