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Lasing experiment and the measurement of intensity
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[QUOTE="collinsmark, post: 4506866, member: 114325"] The key is to better model the detector. Your equation of [I]P = I[/I][SUP]2[/SUP][I]R[/I] is correct, but don't forget that in a typical photo-detector, the resistance [I]R[/I] of the photo-detector is not constant, and changes with the light's intensity. The fact that [I]R[/I] is not constant is typical of most photo-detectors. Many photo-detectors work at a constant voltage (roughly speaking) rather than have a constant resistance. Let's step back a moment and consider an ideal photo-detector. For example, let's consider an ideal solar cell that converts light power to electrical power with a constant efficiency. Also, let's assume that the potential difference (aka. voltage) between the terminals of this ideal solar cell is constant. (In practical solar cells this constant voltage [Edit: I should probably reword that as constant "[I]emf[/I]"] idea is a fairly good approximation of their behavior.) Now what is the relationship between the light intensity and the current, given that? Everything I've said in this post depends upon the characteristics of the detector though. If you do some investigation and find out that the detector's voltage doesn't change much over varying light intensity*, then I think you've found your answer. *(At least in the range of intensities that you were measuring in the experiment.) [/QUOTE]
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Lasing experiment and the measurement of intensity
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