Measuring laser output with Lux meter

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SUMMARY

The discussion centers on measuring laser output using a lux meter, specifically for a green laser at 532nm. The user seeks to convert lux readings to milliwatts (mW) with an acceptable accuracy of 10-20%. Key calculations involve using the luminous efficacy curve, which for 532nm is approximately 0.86, and determining the appropriate detector size to achieve the desired lux reading. The recommendation is to use an optical power meter for more accurate measurements, as lux meters are primarily designed for photometric measurements.

PREREQUISITES
  • Understanding of photometric and radiometric units, specifically lux and watts/mW.
  • Familiarity with the concept of luminous efficacy and its application in light measurement.
  • Basic knowledge of laser wavelengths, particularly the characteristics of a 532nm green laser.
  • Ability to perform unit conversions between lux and lumens based on detector area.
NEXT STEPS
  • Research the use of optical power meters for accurate laser output measurement.
  • Study the "Introduction to Radiometry" by William Wolfe for in-depth understanding of light measurement.
  • Learn about the luminous efficacy curve and its implications for different wavelengths.
  • Explore the differences between photometric and radiometric measurements in practical applications.
USEFUL FOR

Individuals interested in laser measurement, including hobbyists, optical engineers, and researchers in photometry and radiometry. This discussion is particularly beneficial for those seeking to understand the limitations of lux meters in measuring laser output.

jodyflorian
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Hiya,

Question 1: (Main question)
I bought a lux meter while understanding very little about how light is measured. I'm hoping to measure the output of a laser in mW. I assume it's possible to do this at least to an accuracy of about 10-20% which is all I need? If so, please could someone help me out with the equation?


The lux meter came with a frequency response graph, so for example if I measure a green laser (532nm) I know that I need to reduce the reading to 95%... But I'm not sure if this is a graph for the sensitivity of the human eye for various frequencies, or if it's the sensor's sensitivity for frequencies. If it's the latter, then I would need a function for the human eye's sensitivity to frequencies to calculate the power of a laser as well!

Thanks!

PS: I've shone a supposed 20mw laser onto the lux meter (green, 532nm) and it reads around 4000 lux - so this the ballpark figure for double checking the equation :)
 
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1 Lux = 1 Lumen/m^2 and is a photometric unit. Watts (or mW) is a radiometric unit. So, not only do you need to convert a radiometric unit to a photometric unit (using the "luminous efficacy" curve), but also you need to convert a measurement of incidance to flux.

I highly recommend you get some reading material to familiarize yourself with the subject: NIST has some online material, I think, but for me the gold-standard reference is "Introduction to Radiometry" By William Wolfe.

Unfortunately, you should have obtained an optical power meter, not a luminometer for this measurement (unless there is a specific reason why you need to measure things photometrically)

In any case, let's run through your numbers: 4000 lux and 20 mW of 532 nm laser light. The luminous efficacy curve gives a value near 1 (0.86) at that wavelength, so the effective "eyeball watts" is 20*0.86 = 17 lm. In order to convert lm to lux, you need to specify the size of the beam or size of the detector: in order to get 4000 lux from 17 lm, I calculate a detector that is 0.004 m^2 or 6.5 cm on a side: large, but reasonable.

Does that help?
 
Thanks Andy!

I think I follow, or at least I think I will once I have a good look over it lol...

So do you think I could get some fairly reasonable accuracy by using the lux meter?

If it helps to know, the lux meter has a translucent circular aperture with diameter 40mm, and the laser beam is approx 2-3mm diameter.

Your explanation has already helped me follow the maths of it all but if you can, I'd be very grateful if you could clarify firstly what you meant by '6.5 cm on a side', (i.e. if the 40mm aperture is relevant?) and if you think I need to buy an optical power meter or if the lux meter should do the job?

Thanks!
 

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