Michelson inferometer

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chiddler

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Video. Towards the end.

N7XEz.png


The question: If the path of light is equidistant from center to M1 and center to M2, is there destructive or constructive interference?

I'm trying to figure out the path of light and what's going on with phase changes. So the following is my attempt:

Starting with M1. Source --> beam splitter, reflection upwards --> reflection downwards from M1 mirror --> refraction, phase change, as it goes through beam splitter again. So there is one phase change.

One phase change = 180 degree flip.

And M2. Source --> refraction at beam splitter, phase change --> compensator change plate, refraction, phase change --> M2 reflection --> compensator change plate, refraction phase change --> reflection off beam splitter.

So three phase changes = 180 degree flip.

So.

Are the pathways that I wrote accurate? I feel like i'm missing something. And i'm not sure if video actually answers said question. At least I missed it.
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I'm gonna say it's constructive but only b/c the path of light back and forth is the same length since it's "equidistant from center to M1 and center to M2".

Is that correct?
 
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SaintJude said:
I'm gonna say it's constructive but only b/c the path of light back and forth is the same length since it's "equidistant from center to M1 and center to M2".

Is that correct?
Click to expand...

I'm not sure what the answer is. But you didn't account for changing phases! If you are right, it's coincidental.
 
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I'd say destructive.

One beam (the one that goes up) hits the inside reflective surface of the glass and does not undergo a phase change because it is going from a more dense medium to a less dense one, it then hits the mirror, and does a 180 degree phase change and hits the screen at the end.

The beam that goes to the right, hits the mirror, does a 180 then comes back and hits the reflective surface and does another 180, then hits the screen.

If the mirrors are equidistant, there will be complete destructive interference. **** changes when you move the mirror though and I haven't quite sussed that out yet.

If you increase delta x by half a wavelength I think it is back in phase, which would make it entirely constructive
 
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Edit: Sorry, didn't see Class1P's post before I posted this.

I don't think this is something you can answer only with the information given. B/c now that I look at the paths of light, it's quite complex, passing through several lenses. And I've only seen questions that ask you to determine the results from simple paths dealing with 1 or 2 interfaces. Of course, it's likely I'm wrong since I don't know anything about the inferometer, so I'll let someone else answer.
 
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chiddler said:
I'm not sure what the answer is. But you didn't account for changing phases! If you are right, it's coincidental.
Click to expand...

I am somewhat rusty on when the phase change would happen but if I remember correctly, it would be only when you have a reflection from an interface between two medias. Refraction and reflection from normal mirrors do no lead to phase changes. There should be no phase changes in your diagram.

Disclaimer: I might be wrong but I don't think I am.
 
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milski said:
I am somewhat rusty on when the phase change would happen but if I remember correctly, it would be only when you have a reflection from an interface between two medias. Refraction and reflection from normal mirrors do no lead to phase changes. There should be no phase changes in your diagram.

Disclaimer: I might be wrong but I don't think I am.
Click to expand...

yes you're right about that. then i recant my supposed solution

BUT there is one instance of reflection as you describe within the beam splitter.
 
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what the...? If this shows up on my MCAT I'm walking out. :laugh: I'll look at this tomorrow when im not ridiculously tired
 
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chiddler said:
Video. Towards the end.

N7XEz.png


The question: If the path of light is equidistant from center to M1 and center to M2, is there destructive or constructive interference?

I'm trying to figure out the path of light and what's going on with phase changes. So the following is my attempt:

Starting with M1. Source --> beam splitter, reflection upwards --> reflection downwards from M1 mirror --> refraction, phase change, as it goes through beam splitter again. So there is one phase change.

One phase change = 180 degree flip.

And M2. Source --> refraction at beam splitter, phase change --> compensator change plate, refraction, phase change --> M2 reflection --> compensator change plate, refraction phase change --> reflection off beam splitter.

So three phase changes = 180 degree flip.

So.

Are the pathways that I wrote accurate? I feel like i'm missing something. And i'm not sure if video actually answers said question. At least I missed it.
Click to expand...

pathways are but locations of phase changes are not accurate

phase changes are occurring at reflection points
m1
source > glass refraction > no phase change reflection off silvered back > reflection off M1 = phase change > refraction through beam splitter

m2
source > glass refraction > compensator refraction > reflection off M2 = phase change > refraction through compensator reflection off beam splitter = phase change

m1 = 1 phase change = 180 degrees out of phase
m2 = 2 phase changes = 360 degrees out of phase

if the length of the paths are both the same it will result in destructive interference because the two paths are out of phase

if the length of Δx is moved 1/2λ the spot in the middle will go from destructive to constructive back to destructive interference

critical points (aka when you'll see complete constructive and full destructive interference)
1st: destructive = original

2nd: constructive = moved 1/4λ because its moving 1/4 to M1 and 1/4 back towards the splitter traveled an extra 1/2λ = 180 degrees
hence the 1/4λ Δx has now made path m1 360 degrees out of phase so it will now constructively interfere with the path from M2

3rd: destructive = moved 1/2λ so it has now traveled an extra λ making it 360 degrees out of phase on top of the original 180 so it is now 540 degrees out of phase which will lead to destructive interference.
 
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chiddler said:
yes you're right about that. then i recant my supposed solution

BUT there is one instance of reflection as you describe within the beam splitter.
Click to expand...

The reflections in the beam splitter are reflections from a silver mirror - there is no phase change there. If the semi-transparency is tripping you off, it is achieved having "holes" in the silver coating, not by making silver transparent.
 
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milski said:
The reflections in the beam splitter are reflections from a silver mirror - there is no phase change there. If the semi-transparency is tripping you off, it is achieved having "holes" in the silver coating, not by making silver transparent.
Click to expand...

i misinterpreted what you wrote earlier and did some reading.

OK. But I still came to the same conclusion that MrNeuro did. Do you not agree with that? I can see one phase change with the vertical light beam and two with the horizontal.
 
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chiddler said:
i misinterpreted what you wrote earlier and did some reading.

OK. But I still came to the same conclusion that MrNeuro did. Do you not agree with that? I can see one phase change with the vertical light beam and two with the horizontal.
Click to expand...

Hm, let's try to clear it up then. Why do you think refraction causes phase change? And why do you think M1/M2 mirrors cause phase change?
 
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So I think Milksi is right. I got some bad information from my old text books.

Most of the things I am reading only say a phase change happens when your incident light wave that is reflected goes from a medium with a lower index of refraction to one with a higher index of refraction. In this case a mirror would not count.
 
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milski said:
Hm, let's try to clear it up then. Why do you think refraction causes phase change? And why do you think M1/M2 mirrors cause phase change?
Click to expand...

i'm looking at every reflection in air/glass interface. these lead to phase changes, right? because air's n value is ~1 and glass is ~1.5.

no sorry for the misunderstanding. i'm not saying that refraction leads to phase changes as I initially did. only reflections from a low n to high n interface.
 
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Class1P said:
So I think Milksi is right. I got some bad information from my old text books.

Most of the things I am reading only say a phase change happens when your incident light wave that is reflected goes from a medium with a lower index of refraction to one with a higher index of refraction. In this case a mirror would not count.
Click to expand...

oh? why would the mirror not count?
 
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There is no reflection from the air/glass boundary. Well, there is, but it's very minor and you can ignore it here - most of the light goes through the glass without bouncing back. The only reflections that you are getting are from the silver mirrors on the splitter and M1/M2. I assume that M1/M2 are metal mirrors as well, since there would not be much point in letting most of the light go through them.

Reflection from metal mirrors does not have phase change - the phase change happens only when you have a partial reflection from an interface where the light can continue in a media in which its speed is lower.

To summarize:
- metal mirrors - no phase change
- refraction - no phase change
- reflection from air/glass transition - insignificant, so the phase change does not matter.
 
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chiddler said:
i'm looking at every reflection in air/glass interface. these lead to phase changes, right? because air's n value is ~1 and glass is ~1.5.

no sorry for the misunderstanding. i'm not saying that refraction leads to phase changes as I initially did. only reflections from a low n to high n interface.
Click to expand...

There are none of these in the experiment. Metals don't have a refraction index - the reflection in them does not create a phase change.
 
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oh. then the only variable, as the equation tells, is the deltaX value which can change the two light path distances and phase.

The only conditions what i wrote would be correct is if the beam splitter was not silvered and, rather, is capable of that partial reflection on its own plastic or glass self. Right?
 
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chiddler said:
oh. then the only variable, as the equation tells, is the deltaX value which can change the two light path distances and phase.

The only conditions what i wrote would be correct is if the beam splitter was not silvered and, rather, is capable of that partial reflection on its own plastic or glass self. Right?
Click to expand...

Yes, that's the whole trick - it is used to measure Δx. And you are correct about the plastic version of the splitter.
 
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hmmm so your saying that there are no phase changes

can you walk me through why there is only a little phase change in the air to glass region and as to why it can be ignored? and why is it that it isn't ignored in thin film calculations?

Answer to Milski's question.....highlight to see.... is this right?
and the compensator is there to decrease the velocity of the second path so it is moving @ the same pace as the first path?
 
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milski said:
No problem. 😉 Here is a bonus question for you - why is the compensator plate there? What does it compensate for?
Click to expand...

i looked it up before you asked me lol

mr neuro, kinda sorta, not exactly. velocity only decreases within the medium.
 
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The phase change happens due to reflection at the interface between air and glass. It is still 180 degrees but the amount of the reflected light is minimal.

Don't quote me on the numbers but regular glass reflects about 10%, high quality optics lower that to less than 1%.

You don't ignore it when talking about thin film interference because it's the subject of what you are talking about in that case. It's a bit like friction - for a lot of cases you can ignore it but when you are talking about static/kinetic friction obviously you cannot.

And yes on the compensator. There is a better way to phrase it but it's the same general idea.


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milski said:
The phase change happens due to reflection at the interface between air and glass. It is still 180 degrees but the amount of the reflected light is minimal.

Don't quote me on the numbers but regular glass reflects about 10%, high quality optics lower that to less than 1%.

You don't ignore it when talking about thin film interference because it's the subject of what you are talking about in that case. It's a bit like friction - for a lot of cases you can ignore it but when you are talking about static/kinetic friction obviously you cannot.

And yes on the compensator. There is a better way to phrase it but it's the same general idea.


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Click to expand...


when you say minimal are you referring to its intensity?

if so im assuming the intensity should still be significant as its reflecting back on to the final surface and interfering with the other path...?

help....i thought i had this down but now i CLEARLY do not

hmmm wiki says that phase changes do occur off mirrors and half silvered mirrors?

Given the above rules, mirrors, including half-silvered mirrors, have the following properties:

A ½ wavelength phase shift occurs upon reflection from the front of a mirror, since the medium behind the mirror (glass) has a higher refractive index than the medium the light is traveling in (air).
Click to expand...

i guess im having trouble seeing your side why would no phase change occur at the metal mirror?

some other things i think may aid in the discussion of this RIDICULOUS INTERFEROMETER
When a light ray is incident on a surface and the material on the other side of the surface has a higher index of refraction (i.e. a lower speed of light than the medium that the light is travelling in), then the reflected light ray is shifted in its phase by exactly one half a wavelength.

The index of refraction of a perfect mirror can be thought of as infinite. Thus light reflected by a mirror has its phase changed by one half a wavelength.

When a light ray is incident on a surface and the material on the other side of the surface has a lower index of refraction, the reflected light ray does not have its phase changed.

When a light ray goes from one medium into another, its direction changes due to refraction but no phase change occurs at the surfaces of the two mediums.

When a light ray travels through a medium, such as a glass plate, its phase will be shifted by an amount that depends on the index of refraction of the medium and the path length of the light ray through the medium.
Click to expand...

the last point i believe answers why you use a compensator right?
 
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I might be wrong after all. I see different information from different sources, including the phase change depending on the thickness of the film.

I was basing it on light not having any speed in the silver but that may work in a different way than I thought.
 
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"a beam of light reflected by a material with index of refraction greater than that of the material in which it is traveling changes phase by 1 cycle".

so is the implication that metals have an index of refraction? that doesn't sound possible.

oh hey! maybe since it's 50% silvered, this makes it translucent.
 
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Yes, I mean intensity. In thin layer interference the intensities of the two rays are comparable - both are reflections from an interface of the same media, which can give you full constructive/destructive interference. Here you have the reflected ray as only a few percent of the other - you may see some sort of interference but not full black/bright contrast that you would get from the main rays.

And for the compensator - yes, you want light to travel through the same thickness of glass.
 
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chiddler said:
"a beam of light reflected by a material with index of refraction greater than that of the material in which it is traveling changes phase by 1 cycle".

so is the implication that metals have an index of refraction? that doesn't sound possible.

oh hey! maybe since it's 50% silvered, this makes it translucent.
Click to expand...

You can consider n being infinity. Not sure how good idea it is. I need to read more on this.
 
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Item 17: If the frequency of the monochromatic radiation is increased (all else being the same), which of the following would result?

Why is I false?
I. The maximum intensity registered by the detector would be lower. Why is this false?
II. The minimum intensity registered by the detector would be lower. The minimum intensity of light in diffraction cannot be lowered. If total destructive interference occurs, the intensity is zero. Intensity cannot be negative, i.e. lower than zero.
III. More maxima and minima will be encountered for a given distance.

And also: Is there anything that WOULD affect the value of maximum and minimum intensities?
 
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SaintJude said:
Item 17: If the frequency of the monochromatic radiation is increased (all else being the same), which of the following would result?

Why is I false?
I. The maximum intensity registered by the detector would be lower. Why is this false?
II. The minimum intensity registered by the detector would be lower. The minimum intensity of light in diffraction cannot be lowered. If total destructive interference occurs, the intensity is zero. Intensity cannot be negative, i.e. lower than zero.
III. More maxima and minima will be encountered for a given distance.

And also: Is there anything that WOULD affect the value of maximum and minimum intensities?
Click to expand...

Why do you think intensity would be different?

A good comparison is to photoelectric effect. Changing intensity is completely different from changing frequency of light such that one causes electrons to be propelled and one does not.
 
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If the beamsplitter did not split the beam exactly in half:

constructive interference would have a lower maximum amplitude.
destructive interference would not be total
the arm with the stronger intensity would need to have its distance increased
the arm with the stronger intensity would need to have its distance decreased

Answer: B.
 
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you're right...intensity can be relevant because of interference.

in that case: changing frequency without changing phase of two light beams doesn't change their intensity.

er, did you need help with that question or did i interpret it correctly as a refutation?
 
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help 🙂 someone show me the light!!

But, uuum, yeah thanks for seeing my "point" that I apparently was making (?)...
 
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