I would like to compare the UV dose of a UV lamp 365 nm, 6 mW/cm2 at a distance of 0.5 cm with the UV dose of a UV lamp 365 nm, 250 mW/cm2 at a distance of 19 cm.
Actually the point is: does the bigger distance (19 cm vs 0.5 cm) compensate the higher energies (250 mW/cm2 vs 6 mW/cm2)? (of course this is intended for the same exposure time).
Best Regards,
Marcello
Before try to reply your questiopn I have to ask you:
1) Are you writing that you have a lamp that at a distance of 0,5 cm irradiate a surface with 6 mW/cm^2 and another lamp that at a distance of 19 cm irradiate with 250 mW/cm^2?
Have you measured such irradiance?
Because if it is so, it is evident that the dose will be different.
2) Or you want to say that you have two lamps, which specifications indicate at a specific distance (the same distance for both lamps, but not different distance of 0,5 cm and 19 cm) an output irradiance respectively of 6 mW/cm^2 and 250 mW/cm^2?
If so, you have to consider that the intensity of light is inversely proportional to the square of the distance. This means that as the distance from a light source increases, the intensity of light is equal to a value multiplied by 1/d^2
(https://en.wikipedia.org/wiki/Inverse-square_law).
In my opinion, to have the same irradiance, the lamps emitting 250 mW/cm^2 has to be at 10,4 cm to have the same dose.
I suggest you to verify the calculation, I am not sure to have not done errors (using the formula in https://en.wikipedia.org/wiki/Inverse-square_law), but first of all you have to verify if you are in the situation 1 or 2.
Best regards
Dear Dr. Grifoni,
Thank you for your attention to this topic.
In my case, I did the irradiation using 250 mW/cm2 at 19 cm (distance light source to irradiated surface).
The UV dose = Irradiance X time
In doing comparations in the literature data, I noticed that very often, in my context of UV thiol-ene chemistry, lamps of 6 mW/cm2 at 5 mm (0,5 cm) for 10 s o 60 min are used.
So: 1/(0.19^2) = 27.70 => my UV (dose is 1/27.70 X 250 mW/cm2)
1/(0.05^2) = 40000 => literature UV dose (my dose is 1/40000 X 6 mW/cm2)
My problem is exactly to compare the UV dose in these two cases.
Please, can you clarify me? 1 s of UV radiation at 365 nm at 250 mW/com at 19 cm is hiher or lower than 1 s UV irradiation at 365 nm at 6 mW/cm2 at 0.5 cm of distance?
Best Regards,
250 mW/cm2 and 6 mW/cm2 are obtained from the lamp specifications? If yes referred to the same distance? Generally when a lamp is described its intensity is defined in mW/cm2 (but not only), but the value is based on a specific distance. The important is to know if the values you mention are referred at the same distance. Is it so?
250 mW/cm2 is derived from the metrological report. I have several all the numbers at 3 or 4 distances.
6 mW/cm2 is at 5 mmm as per publication. Now you cast a doubt … the nominal fluence (6 mW/cm2) is only a complete parameter if there is the distance! You are right... I will check with the technical bulletin … thank you!
If the 250 e 6 mW/cm2 are the standard irradiance at the same distance, by my calculation, positioning the 250 mW/cm2 lamp at 19 cm you irradiate with a dose 38 times lower than the 6 mW/cm2 lamp at 0.5 cm.
In fact the ratio between the intensity of the lamps is 250/6= 41.6667.
You positioned the 250 W lamp ad a distance resulting about 38 times that of the 6 Watt/cm2 lamp. Doing so, you have considered the variation of the light almost linear with distance (38 against 41,667).
But the light intensity varies with the square of the distance, so to obtain the same irradiation you have to put the 250 W lamps at a distance such that the square (^2) of the ratio between the two distances is equal to the ratio between the two irradiances.
by my calculation the 250 W lamp distance has to be 3,2277486 cm. In fact (3,2277486/0,5)^2=41,667.
Best Regards
Dear Dr. Grifoni,
this is based on an answer on ResearchGate from Sohan Jheeta (popular answer number two),
https://www.researchgate.net/post/How_can_I_calculate_the_UV_dose_in_J_m22
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case one: nominal value 250/mW/cm2
250 mW/cm2 = 2.5 * 10^7 W/m2
at a lamp-to-surface distance of 0.19 m, I have (0.19)^2 = 3.61 * 10^-2
normalizing 1/3.61 * 10^-2 = 27.7
so, (2.5 * 10^7 W/m2)/27.7 = 9.02 * 10^5 W/m2
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case two: nominal value 6 /mW/cm2
6 mW/cm2 = 60 W/m2
at a lamp-to-surface distance of 0.005 m, I have (0.005)^2 = 2.5 * 10^-5
normalizing 1/2.5 * 10^-5 = 4 * 10^4
so, (60 W/m2)/4 * 10^4 = 1.5 * 10^-3 W/m2
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So, assumed that the nominal distance is a data at a standard surface-to-lamp distance (to be verified), I have ca. 9 order of magnitude of difference in the two cases.
Please, can you tell me where I am wrong?
Dear Marcello Iacono ;
Let me do two remarks.
a) Solving the '9 orders of magnitude' issue
As you have done, let us compare the ratio of the optical powers of the UV sources by comparing the UV fluxes at the same distance, such as for example at 1m.
The ‘low’ power source providing 6mW/cm² at 0,5cm will give
6mW/cm² *(0,005m)²/(1m)² = 1,5*10-4 mW/cm² = 1,5*10-3 W/m² at 1m distance when applying the 1/r² law. That’s the value you already have estimated above.
The ‘high’ power source providing 250mW/cm² at 19cm will give
250mW/cm² *(0,19m)²/(1m)² = 9,02mW/cm² ~ 90,2 W/m² at 1m distance when applying the 1/r² law. Here significant differences in the orders of magnitude show up compared to your estimation.
The ratio of ‘high’ to ‘low’ power will give: 90/(1,5*10-3) = 60000. This is only 4 to 5 orders of magnitude compared to your result. There are some troubles with the orders of magnitudes in your calculation.
But nevertheless a ratio of about 60000 still shows up yet.
b) validity of 1/r² law.
The 1/r² law *) is valid for the case that all lateral dimensions dx and dy of the spot of the light source are being much smaller that the distance r: dx
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