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What is Fluoresence?


Pylon

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I noticed that most definitions/explanations of the word are relatively consistent, however reading a definition is not always the most practical way to predict/understand/explain certain examples of what will happen when shining a UV light onto a certain material. I couldn't find a post dedicated to explaining Fluorescence in simplistic terms on this forum so I thought I would create one. Maybe this post could be moved over to the tutorial section later.

 

When trying to write this out, I found myself using the keywords below, but being unable to differentiate/separate the definitions between them because they were too abstract for me to use with certainty when trying to visualize what they are referring to spatially/physically. They all seem to sort of blend together. Anyways:

 

What is the difference between energy, energy level, photons, waves, wavelengths, particles, matter/material, radiation, electromagnetic-radiation, heat, thermal radiation, infrared radiation, light, and space? I've heard some people say "everything is energy" or "energy is fundamental", so if that is true, is material a type of energy? How is "energy" different than waves or matter?

 

What is the difference between absorption, reflection, refraction, emission, and re-emission?

 

Regardless of all of those puzzle pieces not yet put into place to create a comprehensive whole, here is my definition/explanation of how fluorescence makes sense to me currently. These explanations are mainly used to get the point across to help people get results in photography and who want to actually use it in that way, and may not line up with a complete, universal, all-encompassing, "theory of everything":

 

 

 

If we take a look at the electromagnetic spectrum, we can see that visible light is only a small portion of it.

 

Our eyes can only see/detect/sense visible light, however there exists a much wider range of light, including ultraviolet light and infrared light which are invisible to the naked eye and cannot be detected very well -- if at all -- directly.

 

Ultraviolet light has a shorter wavelengths compared to visible light, while infrared light has a longer wavelengths compared to visible light.

 

The wavelength of any given category of light is physically very small, so the common unit of measurement used is nanometers, or nm for short, especially when measuring UV-VIS-IR radiation.

 

Fluorescence refers to the process that happens when you shine light of a particular wavelength onto a material object and then the material object absorbs part of that energy inward but then reflects/emits(?) part of it back outward, usually at a longer wavelength than what was originally shined onto it. All material contains different properties that make the material absorb/reflect the light in different ways (some material reflects/emits "brighter", which I see to be one variable to consider when working with UV lights, some with a different wavelength/color, which is another variable). Can the word "brightness" be replaced with "higher energy"?

 

For instance, If I shine a UV light onto a wall that has a certain type of white paint on it, the white paint would absorb almost all of the UV light and reflect almost nothing back outward towards me in the visible spectrum that I could see using the naked eye. If this were to happen, is it because the material is not dense enough to reflect any of the absorbed energy back out in the visible spectrum (400-700nm). (?)

 

If I take the exact same UV light and shine it onto an a white piece of paper, however, I will see a very bright blue spot on that piece of paper reflecting back at me.

 

If I take the exact same UV light and shine it onto a different type of material that also appears white under the same lighting conditions as in the previous two examples (lets say sunlight), depending on the properties of that material, it could fluoresce a different color when exposed exclusively to UV light, so long as it is a longer wavelength (ex. the material might appear more purple or more yellow compared to the other types of material previously mentioned), with the brightness also varying from the previous two examples. Some pink pigment/powder, for instance, fluoresces orange when exposed to UV light. Certain types of plastic that appear black under exclusively visible light, appear orange when under exclusively UV light.

 

The only difference between these three examples is the different type of material. When seeing with your naked eyes, all material will look different in exclusively visible light than it does when exposed exclusively to UV light. The material can appear to be a different color than it normally does under visible light or appear more or less bright than it does in visible light. Yes, some bright neon material that looks a certain color in visible light fluoresces very brightly and usually with the same color in UV, and that is what most people think of when they think of the word fluorescence. However, given enough exposure time, ISO sensitivity, and aperture size, you can photograph any material and view it properly exposed - this means you can remix the entire world around you with different colors! Paper and white cotton, for instance, fluoresces very brightly, so a shorter exposure would be used. Dirt and leafs, however, usually do not glow as bright, so you need either a brighter light or a longer exposure time to capture a proper exposure of that material.

 

Remember that you cannot see true UV light directly (it is invisible), you can only see the byproduct of UV (ie. florescence, the process where a shorter wavelength of light (like UV) gets re-emitted as a longer wavelength of light after some of its energy is lost after hitting an object) after it has been shined onto an object. If we exclude the idea of ultraviolet light and just think in terms of regular visible light, a green leaf appears green because it is absorbing all non-green visible light waves and therefore can only reflect back out green-only light waves towards your eye.

 

You also don't need to use UV light in order to observe fluorescence to happen, although because UV lies just outside visible light, it is the most commonly used light source to demonstrate fluorescent effects. You can have visible-induced infrared fluorescence, or blue-induced green fluorescence, for example (the latter is actually what a lot of divers use when photographing marine life underwater). Both are possible because you are simply taking a shorter waveform of light and converting it to a longer waveform of light after it hits an object.

 

 

 

...

 

With that written (containing any descriptions/wording that someone may want to correct/fix/improve).... Here are some questions:

 

In this video, at time mark 2:50, he shows how two different types of material in the leaf absorb light, and therefore anything that is not absorbed is reflected. Well, if that is true, is he taking into consideration how UV light causes visible emittence/reflectance via fluorescence? If there can be Green-Induced Red Fluorescence, for example, how can one tell what wavelengths an object is actually reflecting originate from? Is the object absorbing all wavelengths except red (and therefore reflecting red from sunlight or red light), or is the object being fluoresced by a shorter wavelength?

 

At 1:22, I don't understand when they show an illustration of the electrons getting excited to a higher energetic orbit level around the nucleus (that is what I am assuming that illustration is showing), then loose energy and then return to their ground state. I could picture that happening if the UV light were to turn on, shine on the electrons, and then turn off, but what I don't understand is how all the electrons get excited and then loose energy while the light is shining on them constantly. If the light were to shine on them constantly, it seems like the electrons would reach at and stay at the excited state until the light was turned off. Do they go back and forth and oscillate between getting excited and returning to their ground-state while the light is on?

 

I believe I heard somewhere that objects that are black (aka objects that absorb more light/energy/waves(?)) are more "dense" in their physical structure, which is why more energy gets absorbed/lost into the material than reflected. Is this true?

 

When the sun shines down on us, what is the ratio between UV waves, visible waves, and IR waves? Is the visible light more "energetically" "powerful" than UV?

 

If we are working with additive color theory, or with light/photons, then when I go out in the sunlight and look at an object that appears to be white, that means that the object is absorbing very little and reflecting out a lot, so much so that it is reflecting all colors, which add up to white. That is the same reason why wearing black clothing or riding in a black car when it is sunny warms up, it is because the light waves/photons/energy is getting absorbed into the material, and because energy cannot be created or destroyed but only converted, it gets converted into heat aka a longer electromagnetic wavelength aka infrared (is that true, or is there a difference between heat and infrared radiation?). White, on the other hand, is reflecting off all of the of the visible light, therefore the object doesn't get hot.

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Fluorescence occurs when an atom or molecule in an excited state decays into a lower-energy ground state, emitting a photon in the process. In the everyday world, most such events involve the electrons around the atoms. Something boosts them to a higher state (excitation) and some time later they decay back (emission.) Excitation may occur by electricity, thermal effects, charged particles, or most often the absorption of a photon. If excitation is by a photon, the emitted photon will be of a lower energy (by an amount called the Stokes shift.) The lost energy is dissipated as heat. This is why ultraviolet light can excite visible emission.

 

Atomic nuclei have excited states, too, but their transitions are so energetic that they usually involve hard X-rays or gamma rays.

 

Reflection occurs when a photon bounces off a target without being absorbed. The common things that can happen to a photon at a target are transmission, reflection, refraction, or absorption.

 

To find all the wavelengths which can possibly excite an emission, one must obtain an excitation spectrum (most fluorimeters can do this.)

 

White reflects all visible light, but that is not all light. An object can be heated by absorbing other wavelengths. Black jeans and blue jeans look similarly dark, but black jeans are much hotter because they absorb IR as well as visible.

 

Most black objects absorb light via electron excitation--it has nothing to do with density as such.

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+1, love that description, very efficient. Still wondering about how the electron goes from an excited state then decays to the ground state over time when the light is constantly shining on the atoms, however.
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This is a HUGE post, and I don't have time to go into all the possible details. Maybe others can answer what I leave out.

 

The way a material reflects light in one part of the spectrum is NOT correlated with what it might do somewhere else. Thus, something black to the eyes (non-reflective in visible) might be white in near infrared (reflective in near infrared) or vice versa.

 

The second thing is that emission and reflection are not the same. Reflection happens when there is a sudden change in refractive index as light goes from one material to another (an "interface"). Some percentage of the incoming light is reflected when the light hits the interface, and this depends on the square of the difference of refractive indices of the two materials. For example, air has refractive index ~1, and glass has refractive index ~1.5 in visible light, so light going from air to glass will reflect ((1.5-1)/(1.5+1))^2 = 0.04 = 4% at that interface. It gets more complicated when you have multiple interfaces as in a filter or lens, and you also have to account for angle and polarization.

 

Emission is when light is released by a material (no incoming light is required), and there are many possible processes that can lead to it including thermal, fluorescence, etc.

 

The third thing is that, again, fluorescence is a form of emitted light. The material absorbs some UV (or whatever), electrons are boosted to an excited state of higher energy, then (after some lifetime in that excited state) they fall down to another state that isn't the same as the one they started with, and the remaining energy can either end up as heat or can be turned into additional photons.

 

I could picture that happening if the UV light were to turn on, shine on the electrons, and then turn off, but what I don't understand is how all the electrons get excited and then loose energy while the light is shining on them constantly.

Okay, the misconception here is that light is a continuous wave. In reality light comes in discrete packets (photons) and when you have a light source on continuously, it's just making multiple discrete photons.

 

When the sun shines down on us, what is the ratio between UV waves, visible waves, and IR waves? Is the visible light more "energetically" "powerful" than UV?

The sun is a big hot object and the light it makes is (to first approximation) black body radiation for an object at 5,505C. Black body radiation is when an object glows because it's hot. ("Hot" is a relative term: everything solid or liquid above absolute zero radiates black body radiation.) Now, the solar spectrum is not perfectly black body because the gases that make up the sun absorb certain spectral lines, so these are subtracted out of the black body spectrum. Then the light gets to the atmosphere of Earth, and our atmosphere scatters and absorbs still more of the light. So the exact proportion depends on your altitude and location somewhat. That said, about 42-43% is visible, 3-5% UV, and 52-55% infrared according to wiki. (I am not going to agonize about the accuracy of wiki given the large spread in values one would expect. Plain old clouds will knock these figures willy-nilly.)

 

Oh, and UV has more energy than visible light. Typical values are 3.5 eV ("electron volts") for a UV photon, and 1.6 eV for red photons.

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The bright blue appearance of the sheet of white paper is due to the optical brighteners included in most papers nowadays to make them look whiter. So the blue colour is due to fluorescence of the OB, similar effect in fabrics washed in most washing powders - very hard to avoid OB's today.

 

Dave

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I couldn't find a post dedicated to explaining Fluorescence in simplistic terms on this forum so I thought I would create one.

 

We do have this small snippet in the Fluorescence Sticky:

 

The type of fluorescence most often encountered in UV photography is the visible light emission, or luminescence, from an object caused by its absorption of high energy ultraviolet light (shorter wavelengths). The absorbed UV photons excite the object's electrons to a higher quantum state. The electrons immediately relax to a lower quantum state and emit photons of lower energy in the form of visible light (longer wavelengths). This visible fluorescence may be successfully photographed.

 

I don't think I know how to make it any simpler, but welcome suggestions.

 

Later:

I suppose I should add that photon absorption can only occur when the photon's energy matches the energy gap between the high & low quantum states of the atom (or molecule) doing the absorbing. In other words, not every object visibly fluoresces under UV light.

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Later:

I suppose I should add that photon absorption can only occur when the photon's energy matches the energy gap between the high & low quantum states of the atom (or molecule) doing the absorbing. In other words, not every object visibly fluoresces under UV light.

Also, even when the gap between the states matches the photon energy, when it decays it may not emit a photon; it can also become just vibrational or rotational (thermal) energy. Whether a photon is emitted in addition to quantized vibrations/rotation depends on conservation of not just energy but also momentum. For example, in diodes it is possible to have excited electrons that decay in a totally thermal way (like in a typical silicon diode) or by light emission also (an LED). Conservation of momentum is what determines which will occur.

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Yep.

Good ole physics.

Love the stuff. :D

 

Sometimes I wish I had studied physics instead o' mathematics. But water under the bridge.

 

Turbulent water of course. Requiring lots of computation apps, yes?? :lol: :lol: :lol:

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Sit down and get some popcorn, this may take a while:

 

Think of it this way. Light is a traveling wave in the electromagnetic fabric of space, analogous to a sound wave (pressure difference) in air or a height wave (potential energy difference) in water or a rope. For brevity I will refer only to the electric portion of the EM wave since the magnetic contribution is trivial in the UV/Vis/NIR portion.

 

The amplitude (brightness) is a measurement of the degree of the charge separation between the positive and negative poles of the EM wave. This is proportional to the number of identical photons in the wave. It takes energy to separate the charges, compress the medium harder or raise the water/rope.

 

The frequency (color) is a measurement of how quickly the separation happens. It takes more energy in the case of light to flip the electric charge faster. This is again analogous to the greater amount of energy needed to make air vibrate faster for higher frequencies or to make a rope jump faster.

 

Atoms and molecules (particles) have at least one nucleus containing a postively charged proton and a cloud of negatively charged electrons. Overall the charges tend to balance out but there is still a separation of charge between the nucleus and the cloud.

As the light wave encounters the particle the electron cloud interacts with the electric field of the light wave. At this point there is a probability any given photon in the wave can transfer its energy to the electron cloud which will then oscillate at the same frequency as the photon. This probability is dependent on the relative alignment of the electric fields between the particle and the wave (polarization), and the amount of energy needed for the particle to "get excited". This last point needs some clarification:

 

Molecules have vibrational and rotational energy states not present in atoms. This literally means the molecule wiggles and spins around much as you may be reading this. Vibration and rotation are separate ways to store energy, another is to have an electron change the shape of its cloud (usually get biger). Energy can be stored in any or all of these modes. The vibrational and rotational modes are what we are talking about as “thermal” energy since these are the modes that generally correspond to the heat capacity of a medium. All these modes are quantized which means the amount of energy needed to make the molecule spin/wiggle/inflate is discreet, not continuous; however, these “degrees of freedom” can combine into “vibronic” and rovibraitional” transitions e.g. the energy can be split into just enough to make the particle spin AND/OR vibrate faster AND/OR inflate electron jump to a higher energy orbital, typically further from the nucleus (inflate the cloud) OR cause the outermost electron to flip its quantum “spin”. The latter is classically “forbidden” so the probability of this happening is low. This comes up later.

 

So as the light wave encounters the particle the electric field of the light wave “couples” with the electric field of the particle. ALL or NONE of the photon’s energy must be transferred. In the case of the former the energy of the photon has been captured (absorbed) and the particle is in a higher energy state.

 

IF this energy cannot be divided evenly and completely to the various energy storage modes of the particle then the particle just snaps back to its original state. This action creates a local disturbance in the local EM field of space inducing a wave identical to the original photon (emission). There is a slight time delay between the act of absorption and emission proportional to the mismatch between energy needed and that provided by the photon. The smaller the mismatch the longer the delay. In this time the particle may change its orientation slightly and the photon is emitted in a slightly different direction. This is a type of scattering. Most particles (e.g. water vapor) absorb higher energy light (UV) more readily than lower energy light (IR). Blue light is more likely to be thusly scattered than red and what makes the sky blue and sunsets red. It is also the source of the "index of refraction" and why blue light travels a bit slower through glass than red.

 

IF the energy (frequency) of a photon of the wave is the same to the frequency of the amount of energy needed to induce it to spin AND/OR vibrate faster AND/OR inflate then the energy of the photon is transferred to the particle and the photon is gone. In this case the particle has captured the energy of the photon and is in a higher energy state.

 

Now what?

 

The particle is now spinning and/or wiggling faster and/or has swelled up or has a flipped out electron. Please keep in mind the deflated and inflated particle have individual spinning and wiggling modes.

 

In a vacuum the particle can stay this way for a while but in a non-vacuum it will be constantly slamming into its neighboring particles. Energy is then transferred via inelastic collisions to the neighbors (heat) until the particle’s rotation and vibration are minimized, however the particle will still be swollen or have a flipped out electron.

 

Eventually the electron will spontaneously deflate (e.g. return to its former lower enegy state - picosecond to microsecond time scale). This again causes a shift in the local EM field and induces a photon but some of the energy can be put into making the "deflated" particle wiggle and vibrate faster. The rest becomes a new, lower energy photon than the one originally absorbed. THIS is fluorescence. This is why fluorescence is red shifted compared to the color of the original light and why fluorescence can come out with a very different direction, polarization and time delay compared to the original light.

 

I mentioned a flipped electron and how this was classically forbidden with a low probability. The same is true of the reverse process so it can be quite a while (microsecond to hours) before the electron manages to flip back over. This again releases energy as a photon in a manner analogous to fluorescence but because the mechanism is different it has a different name – phosphorescence. Both these processes fall under the descriptor luminescence.

 

Hope this helps :rolleyes:

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Yeah, dark energy is totally unrelated to all of this. Nobody knows what dark energy actually is, but it's got nothing to do (so far as I'm aware) with most optical processes we are concerned with here.

 

It might help to know that there are several different physical descriptions of reality, each successively more general, starting with classical physics.

 

In chronological order (which is also the order of increasing complexity...):

Classical physics- newton's equation of motion (F=ma) and Maxwell's equations (light is waves, photons don't exist)

 

"Old" quantum theory- light comes in photons (discrete lumps of energy), which are just assumed to exist, and electrons go in circles around the nucleus like little planets. This is the Bohr model of the atom.

 

Quantum theory- Replaces the arbitrary rules from the Bohr model (e.g. why do the electrons go in circles and not fall into nucleus?) with solutions of the Schroedinger equation. Schroedinger replaces F=ma. Maxwell's laws are left mostly unchanged except now the waves are assumed to have discrete energies instead of being able to have any arbitrary energy.

 

Subtopic: Dirac equation- is basically same as Schroedinger equation but modified to agree with special relativity. Replaces Schroedinger (but is much harder to work with, so people use Schroedinger whenever they can ignore relativistic effects). Maxwell's equations are still used with it.

 

Quantum Electrodynamics (QED)- replaces quantum theory/Dirac equation AND maxwell's equations.. I don't know this theory, as I've only gotten to Dirac in my reading. I'm told that it finally gets rid of the need to treat Maxwell's equations separately from Schroedinger/Dirac by replacing all of them with a new set of concepts. This is what Feynman and lots of others won a Nobel for.

 

Standard Model- replaces QED (or makes it a special case) by uniting it with similar theories of the weak force and strong force (which I've ignored up to this point as we've been talking optics). This is where we are now, with many possible candidates to replace it. Standard Model does not explain dark matter or dark energy. Also does not include general relativity.

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To explain fluorescence, you need to go to the quantum theory (or at least the "old" quantum theory) level.

 

Speaking generally, what people do is use the simplest of these models that is sufficient to explain the phenomenon they're interested in. Sometimes you see "quasi-classical" models that have some quantum aspects and some classical ones. For example, when modeling molecules, quantum chemists typically assume that the nuclei obey F=ma according to classical physics, but the electrons obey a modified form of the Schroedinger equation called the Kohn-Sham equation. The reason is that the nuclei are 1800 times heavier than the electrons, so the quantum behavior of the nuclei doesn't matter much in comparison to the more swift-moving electrons. The electrons provide the forces on the nuclei, and the nuclei are then updated with F=ma to their new locations. Then the forces are recalculated with Kohn-Sham, and the whole process is repeated.

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Photon bashes atom light emitted.

:D

 

That is pretty much what we are going for here for photographers.

Those with scientific backgrounds can go as deep as they are comfortable with.

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Yeah, dark energy is totally unrelated to all of this. Nobody knows what dark energy actually is, but it's got nothing to do (so far as I'm aware) with most optical processes we are concerned with here.

 

Its my understanding"dark" energy and/or matter is just the modern version of Here be Dragons!

 

https://en.wikipedia.org/wiki/Here_be_dragons

 

To explain fluorescence, you need to go to the quantum theory (or at least the "old" quantum theory) level.

 

Speaking generally, what people do is use the simplest of these models that is sufficient to explain the phenomenon they're interested in.

 

Its the only way to maintain some shred of sanity. Sometimes a sphere IS the best way to model a horse.

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I spent 15 years working with various forms of luminescence and it didn't take long to learn that there are terminology issues (especially when it comes to defining phosphorescence).

 

I suggest Andreas simple "sticky" definition for fluorescence is fine for photography purposes.

 

If you really want to know more you might try getting several excellent technical books on the subject from your local library.

Solid State Luminescence, Luminescence of Solid State Materials, Luminescent Materials etc

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I spent 15 years working with various forms of luminescence and it didn't take long to learn that there are terminology issues (especially when it comes to defining phosphorescence).

 

 

As have I. Longer in fact both academically and in industry. I can't remember any other accepted definition for phosphorescence than one involving an intersystem crossing (electron spin flip). Looking at Wikipedia however it seems that chemiluminescence is incorrectly called phosphorescence (presumably because the timescale is similar).

 

https://en.wikipedia...Phosphorescence

 

There is also Triboluminecence which is caused by a physical disturbance of a material.

 

https://en.wikipedia...iboluminescence

 

I suspect this is the source of light from radium based paint - alpha particles from decaying radium impacting a zinc sulfide phosphor crystal inducing an intersystem transfer which itself decays to emit light.

 

There's also Raman scattering which is similar to fluorescence.

 

I suggest Andreas simple "sticky" definition for fluorescence is fine for photography purposes.

 

 

That would depend on the timescale. Fluorescence happens much faster than phosphorescence. Knowing the specific mechanism involved would be important in setting up the shot properly e.g. time a shot for phosphorescence when the excitation flash has decayed away and the phosphor intensity is maximized.

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We just use a UV-LED torch and leave it on to photograph either fluorescence or phosphorescence in the dark.

Then no worry about decaying excitation wavelengths.

 

I have a chunk of calcite which fluoresces a nice pinkish-orange under my Nichia 365 UV-LED torch. Easy to photograph in the dark. (The calcite didn't fluoresce much under the 385 UV-LED.)

 

This particular calcite also phosphoresces a pretty blue under my cheezy little "UV sanitizing wand" from the drugstore which presumably is in the UV shortwave range** somewhere (254 nm?) else the calcite would not phosphoresce. The phosphorescent glow (steady under the wand) of the calcite is a little more difficult to photograph because the emitted light is rather dim, but it can be done. However I haven't been able to make a vid of it to show the fading phosphorescence when the wand is turned off.

 

**This sanitizing wand thinger is meant to be used with the UV light shining through a slot along the bottom. If the wand is turned or raised up, there is a mechanism which turns the UV light off so it does not shine in your eyes.

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Those things freak me out. I went to have my nails done once in one of those crazy nail places and was all like where are your goggles if this thing is really UV light !!! :blink: :P :P :lol:

The general public seems to have absolutely no awareness of what UV light can do.

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