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I know that a higher color temperature is associated with cool light and a lower color temperature is associated with warm light (green and red)

However I'm trying to understand the correlation between color temperature and wavelength (or frequency) of light.

Most bulbs are rated with a color temperature in kelvin but contain no information regarding the wavelength of the light.

I need to produce a vast amount of light in the wavelength of 380 nm.

Is it possible to convert kelvin (color temp) into wavelength? If not how can I know the wavelength of a given bulb or light source? I suppose that light may be made up of multiple wavelengths not just one - if that is the case then how can I know which wavelength dominates or how much of each wavelength is emitted by a given source?

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    $\begingroup$ Possibly related: physics.stackexchange.com/q/297533/26969 $\endgroup$
    – Floris
    Sep 13, 2017 at 16:01
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    $\begingroup$ Have a look at Edgar Bonet's answer here. physics.stackexchange.com/questions/331899/… $\endgroup$
    – Farcher
    Sep 13, 2017 at 16:10
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    $\begingroup$ FYI: The phrase "color temperature" comes from photography. Professional photographers sometimes use photo flood lamps in their studio. If the bulb's envelope is not colored blue (as some of them are) then the "color temperature" listed on the package of an incandescent photo flood bulb is the actual temperature of the tungsten filament (expressed in Kelvins) when the lamp operates from the recommended voltage. $\endgroup$ Sep 13, 2017 at 17:14
  • $\begingroup$ If you need "vast amounts of UV light" you might want to look at indoor tanning solutions. UVA is typically in the range of 320 nm to 400 nm. At least these are lamps that are designed to produce "reasonable amounts of" UV in the approximate range you are interested in. What exactly do you need 380 nm for? $\endgroup$
    – Floris
    Sep 13, 2017 at 20:30
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    $\begingroup$ No, Donald… look at Answers like Solomon Slow's. How could whether Color Temperature and Wavelength were the same affect which wavelength dominated or how much of each wavelength was emitted? More usefully, why are you Asking? What could producing any amount of light at any wavelength have to do with colour temperature? Why would you buy a bulb or light source without knowing its wavelength or temperature? Sorry but you don't "suppose light may be made up of multiple wavelengths…" You either know that for a fact, or you go back to re-inventing at least 400 years of wheels. $\endgroup$ Dec 13, 2020 at 19:15

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Color temperature describes how a particular light source produces light (in the visible range) that "looks like" light from a black body radiator at a particular temperature.

Because it is only intended to help with understanding the apparent color, it will not (in general) help you with figuring out the amount of light in the UV, unless you are actually talking about an incandescent emitter, and the enclosure of the emitter (light bulb) has excellent transmission properties across the range of wavelengths of interest.

In general the spectrum of a black body radiator is given by the Planck Equation. This earlier answer may be helpful to give you some background and plots.

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Colour temperature is related to black-body radiation, in theory the distribution of wavelengths of a light source of a certain colour temperature should resemble that of a black body of that temperature.

So colour temperature is not exactly related to a wavelength, but to a distribution of wavelengths. You can calculate the wavelength of the highest intensity using Wien's displacement law.

The application to fluorescent lamps is more or less a marketing trick (describing the colour the lamp is perceived to have), as the emitted spectrum contains discrete peaks from the mercury and materials within the used phosphor, together with more or less broad emission spectra by the phosphor, stimulated by the mercury's emissions, very unlike to the theoretical black body radiation.

If you need only one wavelength (or light in the range around a single wavelength), you could use an LED, or lasers, or maybe blacklight tubes.

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I'm trying to understand the correlation between color temperature and wavelength (or frequency) of light.

This was answered in my answer to your previous question:

A lamp doesn't just produce one wavelength. It produces a wide range of wavelengths with different amounts of power in each one (characterized by a spectrum). For an incandescent lamp, the spectrum will span the entire human visible range, and well into the infrared (and a bit into the ultraviolet). This will more-or-less approximate the emission of a black-body radiator with a certain temperature. The color temperature tells us what temperature black-body this lamp most closely approximates.

Color temperature can't be used to describe a single wavelength source. It is best used to describe a source with a broad output spectrum approximating a black-body spectrum. As the other answers here have said, it is also used in marketing lamps that have discretely lined spectra (such as fluorescent and LED lamps), to say what temperature blackbody the lamp looks most like when viewed by the human eye.

I need to produce a vast amount of light in the wavelength of 380 nm.

Since this is ultraviolet, most lamps will be designed not to produce this wavelength.

You should probably be looking for a black light, which is a lamp specifically designed to produce UV. The Wiki article includes a table of different fluorescent phosphors used in blacklights, and their peak emission wavelengths. It looks like the closest to 380 nm is the $\rm SrB_{4}O_7, Eu$ type, with 370 nm center wavelength and 20 nm spectral width. Probably you want to look for a blacklight using this phosphor. From what's in Wiki, this type should be available from Osram, but might be hard to find in North America.

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  • $\begingroup$ Thanks. I took a look at Osram. I suppose a product like amazon.com/dp/B01DBZIL7Y would also do the trick? They allow you to choose from 365 nm to 425 nm in increments of 5-10nm. $\endgroup$
    – Donlad Lee
    Sep 13, 2017 at 18:53
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    $\begingroup$ That does look like a way to generate substantial UV light. Whether it meets your unstated requirements, I have no idea. $\endgroup$
    – The Photon
    Sep 13, 2017 at 18:57

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