# Does Beam Interference Depend on Longitudinal Distance Post-Beamsplitter?

• P3X-018
In summary, the interference term is z-dependent, but it does not depend on the longitudinal distance after the beamsplitter.
P3X-018
If you have 2 laser beams, A and B, at right angle with each other, that you send towards a beamsplitter (BS), such that you get 2 output beams from the BS, one being the sum of the reflected part of A and transmitted part of B, and the other the transmitted part of A and reflected part of B.

The total field at one output would be

$$\vert \vec{E}_1 + \vec{E}_2 \vert^2 = \vert E_1 \vert^2 + \vert E_2 \vert^2 + 2 Re\left( \vec{E}_1\cdot \vec{E}_2^{*} \right)$$

The question is now, does it make sense to say that the interference only happens at the BS, and hence the interference term $\vec{E}_1\cdot \vec{E}_2^{*}$ is independent of the longitudinal distance after the BS, so that it doesn't matter where a detector is placed to detect the total beam?
Or will it be dependent on the longitudinal distance?

If that is the case (that it does depend), would it then mean that if you place a lens after the BS, the interference term would be changed, because the beam waist size is changed due to the lens?
But how can a lens change (the area integral of) the interference term (power from the interference), if it only transforms everything one-to-one to a different scale?

Assuming the two beams have the same frequency and are coherent, and are perfectly aligned at the beamsplitter: then the interference will not change with longitudinal distance from the beamsplitter. The two beams will maintain a constant phase difference after the beamsplitter.

In practice, any misalignment of the beams, or slight difference in beam diameter, will cause some fringing effects.

The beams come from the same source, and the optical path difference is much less than the coherence length of the laser.
Though the 2 beams don't have exactly the same parameters (slightly different waist and waist location), and the two beams are different modes (one is TEM_00 the other TEM_01) so this would cause the beam interference to be depend on the distance from beamsplitter?

The interference term clearly contains z-dependence, it's just that whether this z is supposed to be interpreted as the position of the observation (detection) or the position where the beams meet (at beamsplitter). I'd definitely say the former, but some disagree and say that it's the later.

If the beams are not the same mode (TEM00 and TEM01), I'm not sure just how that would look.

How is the same source producing these different modes that can then be separated?

By sending part of the initial beam through a resonator and filtering out a desired mode.

P3X-018 said:
If you have 2 laser beams, A and B, at right angle with each other, that you send towards a beamsplitter (BS),
<snip>
The total field at one output would be

$$\vert \vec{E}_1 + \vec{E}_2 \vert^2 = \vert E_1 \vert^2 + \vert E_2 \vert^2 + 2 Re\left( \vec{E}_1\cdot \vec{E}_2^{*} \right)$$

The question is now, does it make sense to say that the interference only happens at the BS, and hence the interference term $\vec{E}_1\cdot \vec{E}_2^{*}$ is independent of the longitudinal distance after the BS, so that it doesn't matter where a detector is placed to detect the total beam?
Or will it be dependent on the longitudinal distance?

If that is the case (that it does depend), would it then mean that if you place a lens after the BS, the interference term would be changed, because the beam waist size is changed due to the lens?
But how can a lens change (the area integral of) the interference term (power from the interference), if it only transforms everything one-to-one to a different scale?

I think your question can most easily be answered by writing down explicit expressions for E1 and E2. Then, you can see that for a perfectly coherent pair of beams, the interference occurs everywhere. The issue is the coherence of each beam, and the mutual coherence between beams. The coherence length sets a longitudinal scale for interference- so a broadband output will exhibit interference after the beamsplitter for a shorter distance than a narrowband source.

The issue you raise with a lens doesn't apply to the coherence length (the bandwidth of light), but it does apply to the coherence area (transverse coherence length) which is also a coherence-driven effect. Here, the bandwidth is in the wavevector rather than the frequency, and so the relationship between entrance pupil diameter and coherence area is what matters. Except for high NA lenses- think immersion microscope objectives- then there are other issues due to how the polarization state of the light changes as the light encounters steep interfaces.

Try reading through Wolf's "Introduction to the theory of coherence and polarization of light", the specific example you are asking about is discussed in some detail.

I know that the coherence is one issue, but since the coherence length is so long due to the narrow linewidth of the laser, that the effects due to this can be ignored (coherence function being equated to 1), provided the optical path difference stays less than the coherence length.

That's also what I'm thinking, that the interference term depends on where it is measured (interference 'everywhere'), and not solely on where the beams intersect for the first time (at beamsplitter). Because the product E1E2*, clearly has z-dependence (z being longitudinal direction), through, for instance, the radius of curvature and beam width, which are functions on z. Hence the interference term would be affected by the placement of a lens, which changes the waist and it's position.

It's just that my professor made me confused when he claimed that it doesn't matter where (after the beamsplitter) the beams are detected, because the interference has already happened AT the beamsplitter. While I could see from the equations that the E1E2* term clearly has z-dependence and hence must depend on where you detect them! And also that it depends on the beam-parameters (curvature and width), so placing a lens would also affect the interference. So a lens can't be considered as something that 'just' rescales the interference term.

P3X-018 said:
By sending part of the initial beam through a resonator and filtering out a desired mode.

Okay, thanks.

## 1. What is beam interference detection?

Beam interference detection is a technique used to detect the presence of intersecting beams of light or other electromagnetic waves. It involves observing the patterns created by the interference of these beams to determine their direction, intensity, and other properties.

## 2. How does beam interference detection work?

Beam interference detection works by using a device called an interferometer, which splits a single beam of light into two and then recombines them. When the two beams intersect, they create a pattern of dark and light bands known as an interference pattern. This pattern can then be analyzed to gather information about the beams.

## 3. What are the applications of beam interference detection?

Beam interference detection has a wide range of applications in various fields such as optics, astronomy, and telecommunications. It is used to measure small changes in distance, detect tiny movements, and analyze the composition of materials. It is also used in the construction of precision instruments and devices.

## 4. What are the advantages of using beam interference detection?

One of the main advantages of beam interference detection is its high sensitivity and precision. It can detect very small changes in the properties of light beams, making it useful in various scientific and industrial applications. It is also a non-destructive and non-invasive technique, making it suitable for studying delicate samples or materials.

## 5. Are there any limitations to beam interference detection?

While beam interference detection has many advantages, it also has some limitations. It is highly dependent on the quality of the beams used and can be affected by external factors such as temperature and vibrations. Additionally, it requires sophisticated equipment and expertise to perform accurately, making it a relatively expensive technique.

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