mirror waves?

[ndrew2505]

what exactly does the wave rating mean for mirros? 1/4 wave, 1/10 wave etc...

ive wondered for a while but havent found out yet

 

[millirad]

I found this quote:
"A "quarter-wave" mirror is a mirror whose figure is an accurate parabola within one quarter wavelength of a specific color of visible light (green, I believe). In other words, the hills and valleys on the mirror's surface are smaller than about 100 nanometers from the ideal parabolic figure."

So the smaller the fraction the more accuracy the surface of the mirror has.
( I think )

 

[dr-ebert]

The same goes for planar and spherical mirrors as well, of course. 1/4 wave is pretty rough.

Generally, it's not important if a microscopic area of a mirror deviates farther from the optimal shape. For that reason, you don't specify the maximum deviation but average the error over the entire surface which results in a "root-mean-square" error value. A good optical mirror has an RMS value of 1/20 lambda (wavelength) or better.

 

[HIMNL9]

By the way, ndrew, are you referring to common layered mirrors, or about dielectric multilayered mirrors ?

Cause for dielectric multilayered mirrors, the wavelenght is also referred to the distance of the different layers (like, as example, in "resonant" mirrors for laser cavities, where the distance from the different layers are matched for cause a constructive interference at the desired wavelenght)

For some other types of mirrors, instead, the wavelenght is referred, how can say it in English ..... about the "cleaning / planarity" precision of the reflecting surface ..... as example, take solid copper mirrors for cutting CNC machines, that are discs of pure copper with a face plane and polished as a mirror ..... cause basically any surface have a certain granularity, at microscopic levels, more "fine" is the polishing and more "uniform" is the surface, and more percentual of the light is reflected from the surface (and at 2 or more Kilowatt, also a 0,something percent of adsorbed power can cause a lot of problems )

Ok, my english is bad, but, just for make a basic example (a bit extreme example), imagine one of those "mirrors" that are used in photography (the silver corrugated ones, for illuminate the scene), compared with a common mirror ..... the corrugated one can reflect the light, but disperse a lot and also loose part of it, where instead the common mirror can reflect more, in more precise way, and lost less ..... for "wavelenght grade polishing" is the same principle, only translated in microscopic fields

 

[Laser_Ben]

As millirad stated, it is a measure of the surface flatness of the optic. The wavelength used is usually the green line of a mercury lamp.

 

[Benm]

For laser related optics, its also common to use the 633 nm HeNe wavelength as reference.

 

[Laser_Ben]

Yes, usually it will state with the optic what the measure wavelength is.

 

[dr-ebert]

Contrary to what HIMNL9 wrote, surface roughness and reflectivity are completely unrelated. Reflection can be direct (mirror) or diffuse (white wall). The corrugated ("orange peel", OP) reflectors, commonly used in LED flashlights to create a more even beam, don't lose light, they just distribute it differently.

 

[Arenared]

Nice work

 

[HIMNL9]

Contrary to what HIMNL9 wrote, surface roughness and reflectivity are completely unrelated. Reflection can be direct (mirror) or diffuse (white wall). The corrugated ("orange peel", OP) reflectors, commonly used in LED flashlights to create a more even beam, don't lose light, they just distribute it differently.

I just suggest you to do a simple experiment, if you want to know about it.

Take two pieces of mirrors of the same material and size, one "first surface optic grade" , and one "orange peel" type.

Put them in the same place, repaired from air, at full sunlight, same illumination angle and rate, with 2 thermometers connected to them.

If your statement is true, the 2 thermometers must indicate the same temperature, also after an hour or more, cause if the two mirrors reflects in the same way and don't "loose" any light, adsorbing it, then there's no significant increase of temperature on the corrugated one, compared with the plain one (cause the only way that a mirror can "loose" light, is adsorbing the related energy and convert it in heat).

But i bet that you find the opposite, if you do the experiment