Coherence pattern of HeNe laser in interferometer

  • A
  • Thread starter Marisa5
  • Start date
  • #1
54
41

Main Question or Discussion Point

Hello, i hope i posted this in the correct forum!

When measuring coherence of a HeNe laser using a Michelson interferometer, there are key points along the moving interferometer arm where coherence is highest. The distance between each of these peaks corresponds with the length of the laser cavity. Based on your knowledge of gas laser fundamentals, why is this the case?

I can't find anything online to answer this question and was hoping someone here might know. Thank you for helping, it means a lot!
 

Answers and Replies

  • #2
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
What is your criterion for determining that coherence is highest? What is the geometric layout of the interferometer? Are you using a diffuse plate at the source and getting a ring pattern, or do you have it in the plane wave=Twyman-Green configuration? ## \\ ## Additions: Just a guess, if it is a ring pattern is that the cavity is producing a center spectral line plus perhaps two (prominent) side lobes that have wavelengths that depend on the length of the cavity. For the interference that occurs in a Michelson, and the fringe contrast in the rings for the case of the sodium doublet, (two wavelengths that are closely spaced), see https://www.physicsforums.com/threads/fringe-visibility-of-sodium-doublet-in-michelson-interferometer.933638/#post-5902650 and https://www.physicsforums.com/threads/interferometer-2-spectral-lines.944019/#post-5973375
 
Last edited:
  • #3
54
41
What is your criterion for determining that coherence is highest? What is the geometric layout of the interferometer? Are you using a diffuse plate at the source and getting a ring pattern, or do you have it in the plane wave=Twyman-Green configuration?
The criteria is that the difference between the maximum amplitude and minimum amplitude of the interference signal is largest (it fluctuates as the cavity expands when it heats up), giving it the highest visibility. This value for visibility is combined with the known intensity of each arm of the interferometer to provide a coherence value. The layout is the classic setup. I do not believe there is a diffuse plate at the source but I think a ring pattern is present at the detector.
 
  • #4
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
The criteria is that the difference between the maximum amplitude and minimum amplitude of the interference signal is largest (it fluctuates as the cavity expands when it heats up), giving it the highest visibility. This value for visibility is combined with the known intensity of each arm of the interferometer to provide a coherence value. The layout is the classic setup. I do not believe there is a diffuse plate at the source but I think a ring pattern is present at the detector.
With a ring pattern, this usually means it would have a diffuser plate. In any case, please see the additions I made to post 2.
 
  • #5
54
41
With a ring pattern, this usually means it would have a diffuser plate. In any case, please see the additions I made to post 2.
Thank you so much. If I come up with more I'll post back in here.
 
  • #6
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
Thank you so much. If I come up with more I'll post back in here.
One additional note: I once ran a very high resolution spectrum of a HeNe laser with a diffraction grating spectrometer, and as I recall, there were a couple of=(several) lobes in the spectrum. Let me google this and see if the lobes are indeed part of a typical spectrum. Yes=see https://www.google.com/imgres?imgurl=https://www.repairfaq.org/sam/henemd8.gif&imgrefurl=https://www.repairfaq.org/sam/laserhen.htm&h=470&w=870&tbnid=Zq6ZfZhuuY3fMM:&q=HeNe+laser+modes&tbnh=113&tbnw=211&usg=__H5q0qKQ2uFVriV4d2F2V6b_Eu4g=&vet=10ahUKEwjl1s_auqncAhUDnKwKHR-vACIQ9QEIOzAA..i&docid=jVeBLht_7G_6aM&sa=X&ved=0ahUKEwjl1s_auqncAhUDnKwKHR-vACIQ9QEIOzAA I think this, along with the equations in the sodium doublet experiments would explain your results.
 
  • #7
54
41
One additional note: I once ran a very high resolution spectrum of a HeNe laser with a diffraction grating spectrometer, and as I recall, there were a couple of=(several) lobes in the spectrum. Let me google this and see if the lobes are indeed part of a typical spectrum. Yes=see https://www.google.com/imgres?imgurl=https://www.repairfaq.org/sam/henemd8.gif&imgrefurl=https://www.repairfaq.org/sam/laserhen.htm&h=470&w=870&tbnid=Zq6ZfZhuuY3fMM:&q=HeNe+laser+modes&tbnh=113&tbnw=211&usg=__H5q0qKQ2uFVriV4d2F2V6b_Eu4g=&vet=10ahUKEwjl1s_auqncAhUDnKwKHR-vACIQ9QEIOzAA..i&docid=jVeBLht_7G_6aM&sa=X&ved=0ahUKEwjl1s_auqncAhUDnKwKHR-vACIQ9QEIOzAA I think this, along with the equations in the sodium doublet experiments would explain your results.
What exactly am I looking at here? Sorry, I'm new to this and need my hand held.
 
  • #8
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
Look carefully through the first sodium doublet experiment with the Michelson=the first "link" of post 2. Post 7 in particular. Each separate wavelength makes a very well defined ring pattern at the receiver with 100% visibility. When two slightly different wavelengths are input in the interferometer at the same time, there will be places, (path differences), where the fringe (ring) patterns line up, and places where they don't. In addition, each of the individual wavelengths could have a finite spectral width to them, making for less than 100% visibility, as in the case of this sodium doublet experiment. This sodium doublet was a somewhat lengthy exercise=you might find the second link of the two links in post 2 to be easier to follow.
 
  • #9
54
41
Look carefully through the first sodium doublet experiment with the Michelson=the first "link" of post 2. Post 7 in particular. Each separate wavelength makes a very well defined ring pattern at the receiver with 100% visibility. When two slightly different wavelengths are input in the interferometer at the same time, there will be places, (path differences), where the fringe (ring) patterns line up, and places where they don't. In addition, each of the individual wavelengths could have a finite spectral width to them, making for less than 100% visibility, as in the case of this sodium doublet experiment. This sodium doublet was a somewhat lengthy exercise=you might find the second link of the two links in post 2 to be easier to follow.
Great. Again, I appreciate your help.
 
  • #10
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
Great. Again, I appreciate your help.
To simplify matters, you could assume your HeNe source contained just two closely spaced wavelengths, with the spectral spacing being a result of the cavity length. You will find with the Michelson ring patterns that for some path distances, the ring maxima will occur at the same ## \theta##, and in other positions, the bright spots in the interference rings are the dark spots of the other wavelength.## \\ ## Edit: Each pattern is of the form ## I(\theta)=I_o \, \cos^2(\frac{\pi D \cos(\theta)}{\lambda}+\phi_o) ## with ## \lambda ## being different for each. (See the first link=post 7). When ## D ## is selected properly, the ring patterns for both ## \lambda ## will coincide, and fringe visibility will be maximized. ## \\ ## See post 4 of the second link where this calculation is performed. The first link gets too complicated.
 
Last edited:
  • #11
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
@Marisa5 Please see post 10, especially the "Edit" part. Hopefully this helps answer your question. ## \\ ## And one additional part: Using the equation in post 4 of the second "link", ##\frac{\pi D \cos(\theta) \, \Delta \lambda}{\lambda_o^2}=m \pi ##, I believe it is also the case that ## \frac{\lambda_o^2}{(2)L}=\Delta \lambda ## for the HeNe cavity of length ## L ##. Let's sketch a proof of this:
Let ## m_o \lambda_1=2L ## and ## (m_o-1) \lambda_2=2L ##. This gives ## m_o \Delta \lambda=\lambda_2 ##. But ## m_o=\frac{2L}{\lambda_1} ##, so that ## 2L \, \Delta \lambda=\lambda_1 \lambda_2 ##. ## \\ ## This gives the result that ## D=2mL ##. Looks like your observation that the maxima in visibility are spaced, according to the cavity length.## \\ ## (Note: The distance ## D ## in the links is path difference distance=Michelson mirror distance x2. This final factor of 2 is a minor detail=it looks like we're on the right track, edit: And in fact, we got it to work). ## \\ ## See also: https://en.wikipedia.org/wiki/Longitudinal_mode I originally omitted the factor of 2 in the proof I sketched. I have now corrected it.
 
Last edited:
  • #12
Andy Resnick
Science Advisor
Education Advisor
Insights Author
7,387
1,838
Hello, i hope i posted this in the correct forum!

When measuring coherence of a HeNe laser using a Michelson interferometer, there are key points along the moving interferometer arm where coherence is highest. The distance between each of these peaks corresponds with the length of the laser cavity. Based on your knowledge of gas laser fundamentals, why is this the case?

I can't find anything online to answer this question and was hoping someone here might know. Thank you for helping, it means a lot!
Just wanted to mention that in the way you phrase the question, 'coherence' would refer to the fringe contrast ('fringe visibility')- comparing neighboring bright and dark regions. The fringe visibility will be highest when the path length difference is small, then smoothly decrease and reach the 1/e point (IIRC) at the coherence length, which is set by the source. Beyond this length, the fringe visibility is very low.

HeNe lasers can have very long coherence lengths (cm to meters); meaning the fringes are quite visible even for fairly large path length differences.

https://en.wikipedia.org/wiki/Coherence_length
 
  • #13
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
Just wanted to mention that in the way you phrase the question, 'coherence' would refer to the fringe contrast ('fringe visibility')- comparing neighboring bright and dark regions. The fringe visibility will be highest when the path length difference is small, then smoothly decrease and reach the 1/e point (IIRC) at the coherence length, which is set by the source. Beyond this length, the fringe visibility is very low.

HeNe lasers can have very long coherence lengths (cm to meters); meaning the fringes are quite visible even for fairly large path length differences.

https://en.wikipedia.org/wiki/Coherence_length
It looks like the OP may have stumbled across another feature to the "coherence" results. From what the OP claims to have observed, it does look like it has a reasonable explanation in that the fringe patterns of two or more very closely spaced wavelengths are responsible for the results.
 
  • #14
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
On p.19 of this article, http://web.physics.ucsb.edu/~phys128/experiments/laser/LaserFall06.pdf , they seem to suggest the coherence length that results from two wavelengths is where the fringe patterns start to cancel each other. They don't go into a more in-depth description of what the OP has found, where the fringe patterns will subsequently line up again, and then cancel again, in intervals. ## \\ ## Note to the OP: In this paper, they use a plane wave description of the interferometer, in which case you do not get rings, but instead, the intensity varies between the two receivers, with each receiver getting a uniform intensity. The plane-wave version of the Michelson interferometer is also referred to as the Twyman-Green configuration. The mathematics is a little simpler for the plane wave case, (than for the case of a ring pattern), but the results are similar, in that the coherence should show a periodicity that depends upon the length of the laser cavity, at least for the case of two separate wavelengths that are closely spaced.
 
Last edited:
  • #15
54
41
@Charles Link

The laser I work with has a cavity length of ~14 cm, which is consistent with the distance between coherence maxima along the interferometer arm. I have recently learned that the gain profile at this cavity length will have no more than 2 modes sweeping through at any given time. The wavelength separation between these modes is approximately 1 pm (very tiny!), so am I understanding correctly from your article (LaserFall06.pdf) that coherence length is contingent on two separate wavelengths in the interferometer?

As the laser cavity expands due to increasing thermal pressure from the gas, modes of increasing wavelength sweep under the gain profile. Each mode is orthogonally polarized to the mode before it which I've observed using a linear polarizer directly in front of the laser. The best description I've found so far is this:

"The tendency for alternate modes to run in crossed polarizations is a bit more complex and has to do with the fact that most simple gas laser transitions actually have multiple upper and lower levels which are slightly split by small Zeeman splitting effects. Each transition is thus a superposition of several slightly shifted transitions between upper and lower Zeeman levels, with these individual transitions having different polarization selection rules (Section 3.3, pp. 135-142, including a very simple example in Fig. 3.7). All the modes basically share or compete for gain from all the transitions. The analytical description of laser action then becomes a bit complex - each axial mode is trying to extract the most gain from all the subtransitions, while doing its best to suppress all the other modes - but the bottom line is that each mode usually comes out best, or suffers the least competition with adjacent modes, if adjacent modes are orthogonally polarized."
https://www.repairfaq.org/sam/laserhen.htm#henpolm

Sorry if this post is slightly disjointed, I just found this info useful and relevant to my question about coherence length. Thank you.
 
  • #16
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
@Marisa5 Interesting, but this is starting to go beyond my level of expertise. I'm presuming, (but not completely sure), that the magnetic field that causes the Zeeman splittings in the HeNe laser is, in fact, the earth's magnetic field. To begin to really get a good answer to your question, it might take several laser experts, and they may not completely agree on the answer without additional experimentation. ## \\ ## For the Michelson interferometer, I think perhaps the most common wavelengths that are studied are the sodium doublet at ## \lambda=##589.0 and 589.6 nm. ## \Delta \lambda=6 ## pm for the sodium doublet, so I think the Michelson interferometer may easily resolve and show fringe patterns that line up and don't line up if the interference consists of two wavelengths with ## \Delta \lambda \approx 1 ## pm. (I believe when the spectal lines are closer together, the distance ## d ## may be greater between successive maxima points where the contast is maximized). ##\\ ## The homework problems that appeared on Physics Forums regarding the Michelson interferometer and the sodium doublet were actually slightly more advanced than what I had in an Optics course as an advanced undergraduate student, where we did the Michelson interferometer sodium doublet experiment. With additional optics work that I did as a graduate student as well as afterwards, I was able to work out some of the quantitative details that the students needed for their homework questions.
 
Last edited:
  • #17
54
41
I appreciate your time on this, perhaps I'll have to work through those problems thoroughly to develop a better understanding of the interference.
 
  • #18
Charles Link
Homework Helper
Insights Author
Gold Member
4,529
1,929
I appreciate your time on this, perhaps I'll have to work through those problems thoroughly to develop a better understanding of the interference.
@Marisa5 I do recommend you work thoroughly through the derivations that I did for the students. I'd be happy to try to answer any questions you might have regarding any of the formulas that I derived. I do think the HeNe laser that you have just might consist primarily of two closely spaced wavelengths, making the sodium doublet to be one with very similar results.## \\ ## [This item is one that I don't know for sure whether the laser experts would be in agreement on=there could be multiple wavelengths equally spaced with spacing correlated to the cavity dimensions=the problem of the resulting interference patterns could get slightly more complex.]
 
Last edited:

Related Threads on Coherence pattern of HeNe laser in interferometer

Replies
6
Views
10K
  • Last Post
Replies
8
Views
5K
Replies
5
Views
4K
  • Last Post
Replies
7
Views
7K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
574
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
Top