UV and irradiance


by hopeariipus
Tags: irradiance
hopeariipus
hopeariipus is offline
#1
Aug13-09, 03:52 PM
P: 1
Dear forum members,

I would like to ask some help regarding irradiance calculation. I am a high school student doing a project in biology in which I expose drosophila m. (fruit flies) to UVA radiation and I would like to mention in my paper the actual dose that was used.

I know that... UV dose (J/m^2) = W/m^2 x s

W/m^2 refers to irradiance, this is the formula that I found for calculating it:

E= dΦ / d Ad

1. Ad is the area which is exposed to the radiation.
What does the d in front of Ad stand for?

dΦ = E0 cosθ d Ad

So, the formula above is the radiation intensity (W)...

2. E0 is the irradiance of light incident upon material, how do I calculate that?

From here I found the information: http://metrology.hut.fi/courses/S-10...ADIOMETRIA.pdf (on the 5th page)

Lamp:

Original Philips UVA
Type HP3147/A
220V ~ 50 Hz 75W


Thank you for any advice and help!
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Lok
Lok is offline
#2
Aug13-09, 08:56 PM
P: 406
W/m2 means energy over a certain surface ( which in the case of a fruit fly is less than 1.5 mm2 )

Energy is computed by power and distance ( from source to target) with the illusive reflectivity taking it's tole.
Andy Resnick
Andy Resnick is offline
#3
Aug15-09, 10:20 AM
Sci Advisor
P: 5,468
Quote Quote by hopeariipus View Post
<snip>

I know that... UV dose (J/m^2) = W/m^2 x s

W/m^2 refers to irradiance, this is the formula that I found for calculating it:

E= dΦ / d Ad

1. Ad is the area which is exposed to the radiation.
What does the d in front of Ad stand for?

dΦ = E0 cosθ d Ad
The first formula is the definition of irradiance (also emittance and excitance), where 'Φ' is the flux and Ad is the projected area- loosely speaking, the amount of area that is perpendicular to the flux. That's where the 'cosθ' term comes from in the second formula- a flat surface perpendicular to the flux has the same projected area, but a surface parallel to the flux has zero projected area. The extra 'd's indicate the expression is in terms of differential areas (which is important if the surface is curved, for example).

Quote Quote by hopeariipus View Post
<snip>
Lamp:

Original Philips UVA
Type HP3147/A
220V ~ 50 Hz 75W

Thank you for any advice and help!
This is not sufficient information to calculate the irradiance. It's not enough to know the power of the bulb, one must know the *emitted* power in the waveband of interest and the geometry of the lighting situation. Phillips (or Osram, another manufacturer) can tell you how much light (in your waveband of interest) the bulb gives off, but you need to know what the projected area of your sample is.

Another wrinkle- if your source is a fluorescent bulb (long and thin, not like a short arc or an incandescent), and if there are optical elements that refract or reflect some of the light toward your sample, you need to first calculate how much flux is passing through your sample plane. The rough calculation is fairly simple- what fraction of the total output is directed at your sample is usually sufficient- and if the illumination is fairly uniform, that simplifies the calculation even more.

Hope this helps...

chiquinho
chiquinho is offline
#4
Dec14-10, 05:46 PM
P: 1

UV and irradiance


So, if I want to calculate the UV-B irradiance, I need to consider the sum the irradiance from 290-315 nm?


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