This value is the intensity at (15cm) and the intensity change with the distance

Example: VL-215.L with emitted wavelength of 365 nm (it means each photon has energy of E = 3.4 eV  or 5.44E-19 Joule.

The intensity is IE:

IE = ET / T*A = E*N/T*A = 2300E-6 W/cm2 = 2.3E-3 W/cm2   (at 15 cm)

where ET is the total energy , namely ET = E*N assuming that N is the total number of photons impinging per unit area A over time T.

In terms of number of photons (particle flux) you may calculated that

IN = IE/E = N/T*A = 4.22E+15 photons/sec/cm2        (at 15 cm)

Unit uW/m2 means it is not intensity but monochromatic irradiance. Irradiance obeys the inverse distnace square law. Intensity describes the distribution of radiation within very small solid angles. Do you want to calculate intensity or irradiance?

As Urszula mentions, you're seemingly looking for irradiance rather than intensity, though the convention on radiometric units if often sloppy between fields! To calculate irradiance, this depends on your source; if it is just a 'perfect' bulb, then you can in theory use a point source. But if the source has a reasonable extent, then this is not an adequate assumption; typically, to use a point source approximation you must be at least 5x as far from your source as the greatest extent of your source for the simplification to hold. From your question, I'm guessing you're using UV lamps, which have a physical extent of at least a few centimetres, so I wouldn't use a point source model.

Instead, if accuracy matters, I'd use a line source model - I'll outline it here if it helps; I derived a model for this as part of my PhD in the context of UV phototherapy, as the lamps used in treatment are typically ~ 1.75 metres long and the patient is right on top of them, so point source was a no go. Equally, cylinder models aren't great as the orientation UV radiation is received at matters. Anyway, you can take a similar approach here; essentially you're extending a point source model to an array of point sources.

If your detector is directly focused on a lamp of length L and a length h,  then the irradiance E a distance d away is given by;

E = (SR/d) [ (L-h)/(sqrt(d^2 + (L-h)^2))  + h/sqrt(d^2 + h^2)  ]

where SR is a constant that is easily experimentally determined. If the detector and source are not directly inclined, and there is some surface normal angle, then this matters in irradiance calculations and the above equation becomes slightly more complicated; as Researchgate doesn't allow me to use nice math formatting I won't put it here but I will link you to some papers where we derive and use it; the model works within 1% for UVR irradiance at any orientation and if you're using UV lamps I'd suggest you give it a read - I'm more than happy to answer any questions you might have on it, hope it helps!

Article Dose modeling in ultraviolet phototherapy

Would the Wattage of a bulb (whether UV or normal lights) be the maximum radiant intensity the bulb has? Say a bulb is 15 W, would that be the maximum radiant power the bulb has? I am asking based on your description of irradiant intensity.@ R. Grimes

Radiometric units can be tricky; radiant intensity is the power per solid angle. The formula I give deals with irradiance, the power per unit area rather than radiant intensity. The SR quantity in the answer I gave above is in fact the linear power density of the source, with units of W/m. So if your bulb was 10cm long with a 15 W power, you'd expect an SR of 15W/0.1 = 150 W/m 

I suggest to measure the irradiance which reach the sample surface.

Article About the possibility of calibrating optical detectors by so...

This value is the intensity at (15cm) and the intensity change with the distance

Neither the Inverse Square Law nor the Keitz equation will give you a precise answer. You must model the lamp body using a radiation view factor for a cylinder radiating to a point. See one of my articles on UV lamp modeling or my UVGI Handbook (Springer), or Modest's book https://www.sciencedirect.com/science/book/9780123869449.

Here's another answer... Since intensity is the number of photons hitting per unit area, you can shine your UV light on a material with desirable band gap and do a photoelectric effect experiment to measure the current. With the current reading, you should be able to get the number of excited electrons, which should correspond to the number of photons that hit the given surface area of your material. Density of states calculation will shed some more light on the photoexcitation process, as direct correspondence of 1 photon-1 electron excitation need review at a deeper level.

Hi, my question is: if we know that the intensity at 15 cm is 350 µW/cm2, how we can calculate the intensity of the same bulb at a distance of 3 cm?

I posted the extended question here: https://www.researchgate.net/post/How_to_calculate_UVA_irradiation_power

In general inverse square distnace law is the solution for simplified calculations. From your data you can calculate the radiant intensity for the specific direction and then recalculate for the necessary distance. However this law has some limitations - it works properly for large distance (when it can be assumed that source is a point). The shorter the distance the bigger the discrepancies between the data calculated according to this law and measured. In your case to find out the irradiance at 3 cm distance more data are needed (first of all the size of the source, the radiance pattern).

Thank you Urszula Blaszczak , Sadanand Pandey

Hello!

I have this UV lamp to use in my water treatment experiment

https://www.pondteam.com/produkt/uv-c-9-watt-pond-friend/

The experiment is actually jar teests, each can is containing 1liter water that gous through the UV light apparatus.

Can somebody help me with estimating the intensity?

The dose = intensity * time (for the huge water treatment plant it is usually 400 J/m^2 and I will make the formula to be with that value at the end), so

400 J/m^2 = intensity * time

How m^2 are apllied / derived?

Wavelength is 254 nm? It is in the case of  UV tube that emits UV light at 254 nm, (mercury line) having strong germicidal activity. But do I have mercury line here?

Otherwise it has 9 Watt

230Volt / 50hz

here is also a link to the product:

https://www.pondteam.com/wp-content/uploads/2017/11/40305_I_PondFriend-UV-9000_31102015-1.pdf

How to calculate irradiation?

Thanks in advance!

Ultraviolet (UV) Measurement - EIT LLC

www.eit.com › sites › default › files › instruments › UV...

PDFUltraviolet radiation is present when a UV source is on. ... Simpler devices that measure UV emissions over ... In the real world the terms intensity and irradiance

Urszula Blaszczak tengo una pregunta como puedo hallar la intesidad y la dosis de radiación apartir de los siguientes datos: Lamparas de 15W , con una longitud de onda de 254nm, con una distancia de radiación de 30 cm?. Muchas gracias

Joahnna, is 15W the electrical power of the lamp or optical? For example manufacturer of HNS L 36 W 2G11 says it consumes 36 W of electrical power, while optical power around 254nm is 10,8W and this is the value that should be taken into account in terms of dissinfenction.

It is also important to know the dimensions of the lamp as well as the radiation pattern (radiant intensity distribution). If the distance r of the irradiated surface is 5 time bigger than the maximum dimension D of the lamp, you can use the inverse square distance law to estimate the irradaince at the point. For accurate calculations or smaller relations r/D simmulations are necessary. We are using Trace Pro but there are other possibilities.

When you have the irradiance distribution at the surface and the size of the surface it is easy to estimate the dose by adapting the time of operation of the lamp.

Please provide more details about lamp power

I have a UV- 254 nm source what will be the intensity

David Robert Grimes could you explain difference between LED power and LED radiant in this example? https://formlabs.com/wash-cure/tech-specs/

What I really want to know is whether or not this is suitable for curing cell-laden hydrogels. it's typically used for post-curing of 3D printed parts, so perhaps too powerful. cheers

Urszula Blaszczak , you have mentioned that - "If the distance r of the irradiated surface is 5 time bigger than the maximum dimension D of the lamp, you can use the inverse square distance law to estimate the irradaince at the point. For accurate calculations or smaller relations r/D simmulations are necessary. "

Can you please provide some source to understand this concept?

For the 5D distance discrepancy in irradiance (or illuminance) casued by difference in the shape of the source in relation to the point source is below 1%.

On my desk at the university I have a book in Polish with explanation - probably it won't be helpfull. But I suppose any book dealing with radiation could help as inverse square distance law is a general law.

A related question: what is the relationship between wavelength and energy consumption of a UV lamp?