Why are $\rm ZnSe$ beam-combiners typically "optimized" for a 45º angle of incidence? Everywhere I look it seems that $\rm ZnSe$ beam combiners (as the ones sometimes used in $\rm CO_2$ [10600nm IR] laser cutters) are "optimized" or designed to work at a 45º angle of incidence.
I am not sure if this is simply for general convenience's sake (as I imagine it might make working with the angles involved more convenient in a lab setting), or if there is really a technical reason behind this with how they apply the coatings.
My understanding is that for a given plate thickness, the beam displacement (offset from the original beam-path) will be smaller the smaller the angle of incidence:
For a 2mm plate and a refractive index of 2.3953 for $\rm ZnSe$, this gives 0.98mm for 45º, or 0.20mm for 10º.
Why, then, is minimizing said beam offset not seen as a desirable property in lieu of the "convenience" of working with 45º angles?
Is there a technical reason for why these beam-combiners work specifically well at 45º?
Is this even true?
(I do see it mentioned quite a bit on optical equipment manufacturer sites...)
Will a $\rm ZnSe$ combiner work sub-optimally or with extraneous issues if set at 20º, 30º or even 10º? Will it reflect/absorb more and transmit less?
Some sources:
partially_reflect_co2_optics.pdf

ZnSe-Beam-Combiner.html

 A: As @JonCuster says, Because it makes optical setup easier with nice $90$ degree beam direction changes.
Beyond that, they do have to be designed for some angle, and $90$ degrees is the favorite.
ZnSe has an index of refraction of $2.4$ at $10.6 \mu$m. This makes for a fairly high reflectance. With a powerful beam, this is dangerous. In a system with a lot of surfaces, there wouldn't be much beam left. An antireflection coating on the front surface is a requirement.
Likewise the back surface must have a coating to make it highly transmissive at $10.6 \mu$m and highly reflective at $633$ nm.
Dielectric coatings are a series of very closely spaced thin layers. Light is partially reflected and partially transmitted at each surface.
A simple antireflection coating might be a single layer with a thickness of $1/4$ wavelength of light in the medium. (The wavelength changes when the $n > 1$.) Light reflects off both the front and back surfaces of the coating. Light off the back has traveled $1/2$ wavelength farther than light off the front. It is therefore $180$ degrees out of phase. The reflected wave undergoes destructive interference.
Coatings get more complex when you have multiple wavelengths, but the idea is the same. You have to design the thicknesses to be some number of wavelengths.
Changing the angle to $45$ means you have to make the coating thinner to get the same travel distance.
