# total intensity measurement (of the whole visible light spectrum; from 400 - 800 nm) with a powermeter PM100?

I was using a power meter PM100, with a Si-photodiode sensor head, to measure the intensity of the white light source (250 W quartz halogen lamp). I have recorded the power, shown by the sensor, by setting the wavelength equal to 633 nm (i.e. the maximum intensity is supposed to be at this wavelength, in the spectrum); however, this value for the intensity (power divided by area of the sensor) is only true for the monochromatic light (supposed to use for laser light?) and want to estimate the intensity (of the white light source) from the measured value. I have the responsivity curve for the senor, I used, and the intensity at a wavelength. How can I estimate the total power/intensity of the light source?

Your help will be much appreciated.

## 1 Answer

You need to know the spectral distribution of the source, then fit that curve to the measured reading at the given wavelength. Then the total power output will be the integral under this curve, per solid angle. That is you also need to take into account the total spread of the light through a some surface in space. And then integrate over this surface as well.

I suggest this is done via some appropriate software, MATLAB would be my first choice, but you might have a different preference.

• Hello Nick, Thank you so much for your help. The spectral distribution of the source is like a Gaussian distribution (peak at about 633 nm; I am sorry I could not attach the spectra of the source here!). For now, it would be fine if I can estimate the total power of light (b/c the surface was pretty small, about 2.43 cm^2 rectangular area of glass slide, on which the light was illuminated for photocatlytic reaction). Could you, please, elaborate more about the calculation? (P.S. I have calculated the area under the responsivity curve in the wavelength region 400 nm - 800 nm). Thank you. Commented Sep 6, 2011 at 4:24
• You also need to know the response curve of the photodiode. (amps per watt of light-power, as a function of wavelength.) If this curve is very flat in your range, you can use the 633nm reading directly. Otherwise, use the 633nm value on the response curve to figure out how many amps are coming out of the detector, then that has to equal the integral of (spectrum * response curve) [integrated over wavelength] :-) Commented Sep 11, 2011 at 1:56