mW/nm?

Is it a typo? Should it read 'mW/♦m', perhaps?

Hi,

Can you please tell me how to calculate the energy output from an 8W UV lamp. Assuming I use the 8W lamp for 1hour with the intensity of 6mW/cm2, how much is the total energy output in joules? thank you

CJ

The power output of an 8-W lamp is 8-W, but not all in the UV. We would need more information than you have given to calculate the total power output in the UV.

However, the total output is of little interest unless you are either using the lamp in an integrating sphere or scanning the lamp (with suitable mirrors) past a very large receiver; so I suspect that this is not what you want to know either.

The questions we can answer are:

The total specified energy output of the lamp (heat plus UV plus other optical radiation) if left on for one hour is about 27800 Joules, and the corresponding energy density (presumably in the UV) is 21.6 Joules/cm² per hour. This latter is of course measured at the same location as the specification you transferred.

Total specified energy output of the lamp (heat plus UV plus other optical radiation) if left on for one hour is abt 27800 J, & energy density is 21.6 J/cm2 per hr. .....Can you tell me how to calculate the energy output (27800 J) & energy density is (21.6J/cm2)....

Thanks

CJ

I cannot see what use the former figure (27800 Joules) will be to you, as it is actually the energy input; all this is saying is that conservation of energy means the total energy output has to be the same as the input. It is not informative as to the UV output.

Similarly, the energy density 21.6/J/cm² apples only when the lamp is mounted the same way as for the manufacturer's tests, and is measured in the same position as during these tests.

If you are interested in the total UV output (which may be of interest in some applications) you need to know the measurement distance (this is typically but not universally 1-metre) and the angular distribution of the output intensity.

Hi

I am only interested in the total energy output of the lamp not about the UV output....So can you pls tell me how to calculate the energy output in Joules if I switch on the lamp (8W) for 100minutes with the UV intensity of 6mW/cm2. Thanks.

CJ

Sorry if I'm a blockhead...

If you don't want to know the UV portion of energy output but the "total energy output", in long term the power is 8 W and the energy is the product of power times the "switched on - time". It's just a matter of energy conservation.

8 W = 8 J/s and in 100 minutes (6000 seconds) → 48000 J

Of course in the 100 minutes may be a small delay between input and output energies: there is a small energy quantity which is used to heat the filament and the lamp from ambient to service temperature, which is emitted when the lamp is switched off.

Maybe there is some misunderstanding on terminology...

Kind regards

I suspect that is a typo as I have seen UV kill reqts. listed as mW/((mm)2

chaterpilar is correct; but because you give no detail of the context his answer may not address your needs.

Is this shown against a graph or table against wavelength? That would tell you what "colour" the uv is. If that is not the reason, please provide more context.

Remember that UV can be dangerous if misused and the types of damage depend on the wavelength of the radiation; so if you are not 100% confident you know what you are doing or the implications (as well as the immediate meaning) of the specifications, keep asking questions until you do (and/or until you have established that your adviser understands enough to keep you both safe).

It wouldn't hurt to read this:

http://en.wikipedia.org/wiki/UV_index

Light intensity is measured by power/unit surface.

i.e μW/cm².

If the units given by your supplier are mW/nm. Where "nano" is 10‾9. the square unit is missing. therefore wrong unit.

But if it reads nm², than the conversion is very simple:

cm² =0,01m² nm²=0,000000001m²

mW=0,001w µW= 0,000001w

Why don't you call them ask for explanation's?

Wangito.

A personal observation, One is an idiot until he asks.

mw/nm is not "wrong unit" - it just measures something different from power/unit_area. See chaterpilar's reply which covered this as well as your interpretation of "intensity". Because the value under this definition of intensity will change as light travels through a lossless optical system, intensity is equally commonly taken to mean radiance, which is (for example) mw/cm²/steradian.

Hi all,

I agree with Chaterpilar. The posted units refers to different parameters. UV light (if we can call "light" some electromagnetic waves with energies beyond the visible part of the spectrum) goes from just the visible upper frequency (lower wavelength) of visible spectrum, about 400 nm (nanometers) up to about 100 nm, where X rays spectrum begins.

From the point of view of safety, wavelengths about 365 nm is O. K. Skin or eye damage use to start at wavelengths of 320 nm. That's the reason for use 365 nm wavelength on fluorescent inspections (dye penetrants or magnetic particles)

Regarding the "light" intensity, is measured in μW/cm² and from the same field (Non destructive examination) the intensity required on the part surface is in the order of 800 ÷ 1000 μW/cm².

So, there's no formula to convert mW/nm to μW/cm².

You need:

• Define the wavelength you require.
• Define the intensity of UV radiation required, on what part and to what distance from the UV source.
• If you know the power and intensity of the source, remember the inverse square law: The intensity over any surface diminish with the square of the distance to the source.

Cheers.

Hi,

I was wondering if you could tell me how to convert mW/cm2 to nm? Thank you

Collin

It is not possible.

[Unless you mean how to read the output from the surface of a lamp from a graph of output versus wavelength (or how to calculate the same result for a black body at a specified temperature, etc.)]

thanks.

Can you pls tell me how to read the output from the surface of a lamp from a graph of output versus wavelength?

and

Can you pls tell how to calculate the same result for a black body at a specified temperature?

Thanks

Collin

In order to give a more usable answer, I'd need to see the data you are working with - and know what information you are trying to derive from it.

In the absence of that:
I'd assuming you are looking at a graph with the horizontal axis in nm, (though they sometimes use Angstoms or microns just for the **** of it as far as I'm concerned). The vertical axis would read mW/cm²/nm or something similar.

That will give you the output intensity within each nm region (these could be regarded as very narrow bands of colour). If you want the total intensity (in mw), you need to integrate across the spectrum - one relatively convenient is to take 50nm chunks, add the powers and multiply by 50.

You may need to check where this is measured; typical options are:
a) at the surface of the lamp
b) or at a surface 1-m away from the lamp

If you want the total visible power output (not usually very useful, which is why this is usually given in units that are averages of peoples' perceptions, and therefore incomprehensible):
In case a) you will need to multiply by the area of the lamp to get the total output in each nm of the spectrum;
In case B you will need to know the angular distribution of the output.

Hi,

How to convert mW/cm2 into nanometers?

I have a UV source, which reads the mW/cm2 and UV meter I have reads the same.

For my negative lithography purpose I need precise exposure of UV light.

e.g. 6.5 mW/cm2 = How many nm?

Thank you.

Ashish

pistonpowers@gmail.com

Hi,

I need a little help with energy generated from a UV lamp. Can you pls tell me how to calculate the energy output (consumed) by the a 4W lamp switch on for 100minutes with the intensity of 2mW/cm2 and a mAmp reading of 78mA. Thank you.

regards

CJ

You asked this before. Various people tried to help, and, given that the question lacked clarity, I couldn't improve on what they offered.

If you can clarify what you need to know, we will do our best - again.

The answer to the question you asked is as follows:
Over a period of an hour total energy output* and total energy input of a lamp will be remarkably similar. So, if it really is a 4W lamp and you leave it on for 100 minutes, the energy consumed will be ≈ 4*60*100 Joules. (24 kJ)
*(Total will include heat and other outputs that are not usually considered useful)

I doubt that is what you meant, but the remainder of the information you provide is in any case totally inadequate for any other answer. If you clarify what you really need to calculate, then we can tell you what additional information is required.

Thank you for trying to help ....that's exactly wht i want. As you said 4W x 60s x 100 = 24000J (for all the energy input and output including heat). Even if the intensity of the UV (UVA, BLB bulb) is varied from 2mW/cm2 to 6 mW/cm2, the energy consumed will still be 24000J? When the intensity is measured (using a radiometer) at 2mW/cm2, the mAmpmeter (from the UV lamp controller) reads 78mA and when it's 6mW/cm2, the mAmmeter reads 215mA. Is the input/output of the energy ( including heat, light etc) is still the same 24,000J? thank you so much for help me.

regards

CJ

Right, so now I understand that you are asking what happens when you vary the power you want from the lamp. But critical information is still lacking.

Do you want the power dissipated in the lamp, or the power in your system including the controller? Even if the former, we can't simply say how the input and output power are related - lamp efficiency generally changes with input power, and we don't know how it's being driven or the relationship for this particular type of lamp..

So anything more than that is largely guesswork, but here goes: the current and output power move in remarkably similar ways, which suggests (only suggests) that the lamp is being pulsed. So we might assume that the lamp Voltage is approximately constant while the current is flowing - so the power might have changed by approximately the same factor as the current: so power would be 4Wx78/215, or 1.5 Watt. But I wouldn't use this figure without knowing a great deal more about your lamp and control system - or what use you want to make of the data.

Hi,

Can you give me the method to calculate power in the system including the controller, when i vary the intensity from 2mw/cm2 to 6mw/cm2? thanks

regards

CJ

You don't seem even to be trying to follow the line of thought. So I can't think that you will be able to make use of the information, even when you have it.

However, assuming the data sheet doesn't give you the information you need, the only possible way is to borrow a power meter and measure the current and Voltage inputs to the controller, and their relative phases (assuming that this is a box that is powered off AC). If it's run from a DC power supply, all you need do is measure the input Voltage and current .

Dear Physicist,

I would like to convert 33mW/cm2 to W/cm2. Do you know how? thanks

Cheers

Hason

Yes, thank you.

Dear all,

Can anyone please tell me how to convert 33mW/cm2 to W/cm2. Thank you

Cheers

Hason

p/s: no jokers please

Apologies, I thought I was responding to your humour.

Nevertheless, as the rules of CR4 forbid this sort of assistance with schoolwork, I cannot do more than than provide this link.

Hi,

I was also wondering if you could help me out with another problem. Does anyone knows how to calculate the intensity of ultrasonic irradiation from a ultrasound machine (direct immersion horn type). Frequency at 20kHz, max watt at 750W. If the amplitude is at 30% with pulse 3 seconds (3s on and 3s off), energy consumed = 74, 733J; wht would the intensity of the ultrasound waves be (in mW/cm2)? thanks

regards

CJ