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UV MineralsAlternative Optical Strategy for observing mineral fluorescence

8th Feb 2016 19:13 UTCHoward Heitner

The relatively expensive Hoya filter on a short wave light removes the bright blue and green light from the mercury vapor light. This allows the observation of less intense light from fluorescing minerals The same thing can be accomplished, at least for red fluorescing minerals, by putting an inexpensive red plastic filter over the observers eyes and using an unfiltered mercury lamp. I gave a brief talk about this at the Rochester Symposium, using eucryptite as an example. As I recall the audience was greatly amused.

9th Feb 2016 06:04 UTCJoel Dyer

Hi Howard,

Good testing and observations in your work: I've successfully carried out similar experiments myself and I applaud your approch. Personally, I think that similar cheap, DIY solutions to many tricky microscopy / photography problems can often be searched for and found. Working on many new adaptations myself at the moment, but experimentation takes a lot of time, patience and repetition and some failures at the various stages.

Below is a crude, very initial test of mine for using adapted DIY Rheinberg illumination to bring out crystal faces and details in inclusions in some quartz: similar kind of idea like your, and the material cost me perhaps 1€.

As a whole, a lot of amateurs in different fields of work simply cannot start to invest hundreds, thousands or more, on top notch equipment and creation of super-engineering lab conditions, which one often sees in pictures regarding stack photography, fluid inclusion studies etc. Surpisingly cheap materials can be adapted to a vast range of implementation, if tested properly.

Not sure what you meant by your audience being very amused: I hope in a positive, constructive way, and not just laughing at cheap, simple, yet effective methodology. I wish you happy photographing & adaptation in the future too!

Cheers,

18th Jun 2016 07:41 UTCFranklin Roberts

Ok, this has been a thought experiment of mine for a while now. I don't see why the following process wouldn't allow one to photograph fluorescence with little or no contamination from a filtered or unfiltered UV light source, or any other excitation wavelength for that matter, even if the excitation source is a visible wavelength.

Simply polarize the excitation source and place a cross polarized filter over the camera lens. This would serve to null out any of the light from the excitation source, but would have no effect on light produced by fluorescence of the specimen being photographed, since that light would not be polarized. That sounds simple enough, but Inexpensive plastic polarizing filters might not pass UV light, especially shortwave UV. The filters may themselves exhibit fluorescence, fogging the photograph.. There is however a very simple way to polarize light of any wavelength. This process, based on an effect discovered by Scottish Physicist Sir David Brewster, predates the invention of the plastic polarizing filter by about two centuries, and produces a beam of perfectly polarized light from any non polarized light source, in this case, an ultraviolet lamp.

Brewster observed that when a beam of light encounters a flat glass plate at an angle, the plate acts as a beamsplitter. Two separate beams of light are produced, one that passes through the plate and a second that is reflected from the surface of the plate. By varying the angle of incidence, that is, the angle at which the light beam strikes the plate, Brewster noticed that something very interesting occurred. Quite suddenly, at a specific angle of incidence, the reflected beam of light became perfectly linearly polarized. He confirmed this by placing a second glass plate in the reflected beam at the same angle of incidence as the first plate and then slowly rotating the second plate around the axis of the incoming beam. As the angle of second glass plate approached 90 degrees with respect to the first plate, the incoming beam of light was seen to go dark when observed through the second plate. What had happened was that the polarized beam was completely reflected from the second plate set at 90 degrees to the first. Since all the light was reflected, none was allowed to pass through the plate.

That magic angle at which the light reflected from the glass plate suddenly becomes polarized is known today as Brewster's angle (go figure). It's exact value is slightly different, depending on the material from which the reflective plate is made. The actual value of Brewster's angle is a function of the index of refraction of the plate material and is in fact the arctangent of the index of refraction of the plate. What the heck does all this theory mean? It means that if you bounce your UV light source off of a clean glass or fused quartz plate set at Brewster's angle before letting it strike your mineral specimen, you should be able to keep the light from the UV source out of your photograph by placing a linear polarizing filter over the camera lens and adjusting it to null out the incoming polarized UV light. The light of fluorescence from the specimen will not be affected.

For fused quartz (RI 1.46) Brewster's angle is 35.8 degrees. A glass plate (RI 1.5) gives a Brewster's angle of 56 degrees. You can even skip the light from a water surface, if you skip it at 53 degrees. The wavelength of the excitation light will have some effect on the exact value of Brewster's angle, so you might have to experiment a bit, but this method should work. The given values for the refractive index are specified at 589 nm, the wavelength of the yellow doublet lines of sodium, so RI will vary with wavelength. Keep that in mind if you try this.

Frank

18th Jun 2016 10:23 UTCAlfredo Petrov Manager

Thanks for the fascinating suggestion, Frank!

You might need to run the UV light through a narrow slit in an opaque sheet before it reaches the sloping glass plate, otherwise you'll have UV light hitting the glass at various different angles because of the area of the UV light not being a point source.

18th Jun 2016 12:13 UTCOwen Melfyn Lewis

Fun thinking, Frank :)-D

But is it proved that fluorescence excited by a linearly polarised UV source is always rendered unpolarised in the process?

It seems too that even given the above, the method might work only in the examination of isotropic monocrystalline or glass specimens. Polycrystalline/microcrystalline specimens destroy the linear polarisation of emr passing through them. Anisotropic and transparent single crystals divide incident illuminating radiation into two cross-polarised rays as it passes through the specimen.Reflection/re-radiation off other than a highly polished surface must always be randomly polarised?

18th Jun 2016 16:49 UTCRob Woodside 🌟 Manager

Cute! How much intensity would be lost by splitting the beam at the Brewster angle? By columnating as Alfredo suggests?

19th Jun 2016 04:46 UTCDoug Daniels

Franklin - interesting concept. However, doing the math on the two specific examples you give : glass, with an R.I. of 1.5 - the angle is 56.3 degrees (or, 56, rounded). With fused quartz, with R.I. of 1.46., the angle is 55.6 degrees (again, 56, rounded). Not sure where the 35.8 degrees comes from..... Am I missing something?

19th Jun 2016 17:04 UTCGerhard Niklasch Expert

The fluorescent response may be polarized for reasons of its own, quite independent of any polarization of the exciting UV radiation:

* It may come from a transition between electronic states / orbitals which are oriented in a particular way relative to the host crystal (and in general this is not the same transition that's responsible for absorbing the exciting photons),

* once emitted, when the medium is optically active and pleochroic, the light would immediately be decomposed into ordinary and extraordinary rays which may then suffer selective attenuation along their journey through the crystal, just like light shining through it from behind,

* refraction as they pass from the crystal into air would add polarization when the rays aren't traversing the face at right angles.

The third factor can play a role even with isometric minerals like fluorite where the first two cannot occur.

While I'm not aware of any systematic study of these effects, I've been having a lot of fun with an UV lamp, a variety of samples, and a hand-held analyzer (or a rotatable analyzer mounted on a microscope). For example, the yellow fluorescence from apatite (activated by traces of Mn and REEs) tends to have its E vector in the c plane, at right angles to the c axis, and is almost completely extinguished when the analyzer is turned parallel to this axis. Likewise with many zircons. The blue light from benitoite tends to be brighter when viewed with the analyzer parallel to the c axis, though in this case there's rarely a good prism face to view it through, and the third of the above factors might dominate for light emerging through the pyramid faces. I've found the effect to be weak but consistently present in scheelite, yet absent, to the limits of what I can discern, in powellite.

Enjoy,

Gerhard

20th Jun 2016 01:04 UTCOwen Melfyn Lewis

Thanks Gerhard. A man who understands.

2nd Sep 2016 02:20 UTCFranklin Roberts

Sorry Doug, I cheated and quoted from a chart that I found online. Should have just done the math myself.

Frank

2nd Sep 2016 03:44 UTCFranklin Roberts

Owen,

It really shouldn't matter much if the polarization of the small amount of UV striking the specimen is shifted, if it is illuminating a fluorescent substance, it is going to be heavily absorbed anyway. The purpose of this little thought experiment is to null out as much scattered ambient UV energy as possible before it can get to the camera lens. Once the ultraviolet light gets into the camera optics, it is possible for the optical glass, coatings, cements, etc to fluoresce and expose the film/sensor with the resultant visible light, fogging the image. In any case, polarizing the UV source while photographing the subject through a cross polarized UV filter is bound to reduce, or even eliminate, unwanted UV from screwing up the color rendition of the resulting photograph.

As I look at the spectrum of a low pressure UV mercury lamp, another idea occurs to me. The majority of the visible light that passes through the dark filter glass of a mineral lamp is constrained to only two very narrow lines. They are 404.7 nm (violet) and 435.8 nm (blue). There is also energy emitted in the green, but transmission by the filter glass at that portion of the visible spectrum is essentially zero, so those lines are already eliminated from the lamp output..

By stacking three filters in series over the lens aperture, it should be very easy to eliminate all the light capable of affecting the color rendition of the photographs. Those filters are:

#1: UV "haze" or "daylight" filter Essentially a low pass filter that blocks UV light while passing visible light.

#2 Notch filter, centered on 435.8 nm This is a narrow bandwidth filter that eliminates the strong blue line of low pressure mercury lamps.

#3 Notch filter, centered on 404.7 nm This is a narrow bandwidth filter that eliminates the weak violet line of low pressure mercury lamps.

The notch filters can be found at Anchor Optics. The low pass UV haze filter can be purchased at any camera store.

This filter stack should work on any UV lamp that consists of a clear "germicidal" low pressure mercury lamp with a black glass UV pass/visible block filter. This WILL NOT work with fluorescent "black light" bulbs with an integral purple or woods glass envelope. Those lamps employ an internal phosphor coating to downconvert UVB and UVC light into 365 nm UVA or "longwave" UV. The visible light output of the phosphor coating is very broadband and cannot be eliminated with notch filters.

Frank

24th Sep 2016 08:49 UTCAlexander Ringel

I watch UV minerals through a piece of intense yellow uranium glass. It removes any blue and allows so even to see really weak fluorescence under my 405 nm laser.

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