What is the concept of total cancellation in a Michelson interferometer?

[JimOlsen]

SETUP:
Michelson interferometer with laser source.
d1, d2 are the beam to mirror path leg distances.

Does anyone know how many times the light reaches the point total cancellation during a micrometer adjustment that passes from: d1 < d2 to d1 = d2 then d1 > d2 ?

I know that some believe total cancellation is not possible - but some claim they have achieved it. (lets ignore this point for sake of argument)

 

[Bobbywhy]

JimOlsen, Welcome to Physics Forums!

Can you please supply some reference(s) for your test setup/results? Thank you.

 

[JimOlsen]

I'm asking a question, so I don't have the results. The setup is the standard Michelson interferometer which I recommend using a cube beam splitter with. I looked through the internet but cannot find any data or videos of passing through the d1=d2 equal point.
If you know of any videos or data of this event please let me know.

I believe the mount will require a clock for hands off adjusting.
I have a setup but don't have a micrometer mirror mount slide. Guess I will have to make one somehow.

 

[tommw]

If I understand your setup correctly, cancellation is observed when d1 = d2, right?
Then you will also observe cancellation for every path-length-difference that equals a multiple of your laser wavelength λ:
d1 = n*λ+d2 (with n = ...-2,-1,1,2,...)
so depending on your laser you'll have none to three cancellation within one micrometer path-length-difference.

So far to the theory.. Now, whether you can observe that phenomenon in your setup or not depends on the accuracy of your micrometer-screw, the bandwidth of your laser and the overall stability of your setup. You might want to think about using a piezo-actuator instead of a micrometer screw to get more precise control over the path-length-difference.

 

[JimOlsen]

If I understand your setup correctly, cancellation is observed when d1 = d2, right?

This is what I am trying to figure out.
It seems to me that at d1 = d2 the waves will be in constructive interference so a very bright spot would appear instead of total black.

I believe the phase reverses at every mirror and there are two mirrors for each leg on a MM interferometer. So there should be two wave crests at d1 = d2.

Im just confused about how cancellation can occur at d1 = d2 - at this point the phase of each beam is the same and the waves add not cancel.

 

[tommw]

As I still don't know your setup, I have no idea how many phase shifts you will introduce at what points. But you're right, assuming that we introduce the same number of phase flips (three in the most simple case) in every path, then d1=d2 leads to constructive interference. In this case we will have cancellation at
d1 = (n+1/2)λ+d2

 

[JimOlsen]

But you're right, assuming that we introduce the same number of phase flips (three in the most simple case) in every path, then d1=d2 leads to constructive interference. In this case we will have cancellation at
d1 = (n+1/2)λ+d2 Then this is the point of total blackout.

This I agree with. n=1 in this case, right?
That is 1/2 λ from d1 = d2 and a trough meets a crest resulting in total cancellation.
What confuses me is that if this is the case then there should be two positions of total blackout the other occurring when d1 = (n-1/2)λ+d2. (notice the sign change) Its just comming from the other direction.

There should be two positions of blackout one on either side of d1 = d2 when a trough meets a crest. Still confused and does not make sense to me how there is only 1 blackout posiltion.

In all the videos and simulations I’ve seen there is only 1 blackout and they all reference it at d1 = d2. I don’t see how this is possible. Why is there not two blackouts?

 

[tommw]

Yes, there are many "black-outs" spaced a whole multiple of the wavelength from each other - in both directions + and minus. (or in other words, n can of course be negative).
The reason why you only see one maximum/minimum in some data or video is probably that the experimentalist did not change path-length over one whole period. Anyway in a real experiment you will never exactly reach d1=d2±λ/2 because you won't be able to position your mechanics accurate enough. And even if you did, you wouldn't know because there is no means to distinguish n=0 from n=-9 or n=7...
The question whether you have constructive oder destructive interference at d1=d2 really depends on the setup.. I can't further comment on this if I don't have further information about your or anyone else's setup.

 

[JimOlsen]

Were getting mixed up

Im using the term blackout to describe only complete and total cancellation where all the light is totally canceled none reaches the screen.

I originally was only referring to the point where total cancellation occurs and no light at all reaches the screen. The adjustment of one leg not beyond 1 wavelength d1+ λ to d1-λ
directly through the equal point or d1=d2.

After there are more than 1 fringe on the screen youve gone too far.

Anyway I think youve already answered it