I have solar energy given by watts/m^2 for a different time of the day. I need to measure the lumen of sun and convert this data to lux. Is there any conversion factor from Watts per square meter to lux?
MATLAB: How to measure the lumen of sun from its energy given by w / m ^ 2
energyenergy transformationlumenlux
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vipul, there is no power in an image, by the strict definition, but there is energy. You may have some other layman's definition. If so, let's hear it. Let's look at the units. I'll use SI units here. Power is in watts, or joules per second. A light source has power and it emits it, and it lands on your scene which then reflects it onto your camera sensor. So what hits a given pixel? It's called irradiance and its units are power per unit area or watts/m^2 or joules/(seconds*m^2). By the way, you can look up any of these radiometric definitions in Wikipedia or here
So what does a gray level really mean? The gray level is a measure of energy? Why energy? Exposure is the irradiance integrated over time and has units of joules per square meter (the "per second" units got integrated away). It's the measure of the energy of all the photons that hit the area of sensor. When you look at exposure for one single pixel, then you sum up (integrate) the exposure over the area of that one pixel, so the meters squared go away and you are left with joules. So gray level is proportional to joules, or energy. It's also proportional to power since it's power integrated over the area of a pixel and over the time of exposure, but both of those are constants for a given snapshot. So the pixel value is proportional to power by a conversion factor (that has units), but it really has units of energy.
So when you see SNR in the Wikipedia article, the gray level goes in place of the P (Power) in the equation, not in place of the A (amplitude). And when you see formulas with 10Log or 20Log, you use the 10Log formula, not the 20Log formula.
I hope that explains it better.
Welcome to the very confusing field of optical units. An image does not have units of intensity. See the table of the bottom of this page: https://en.wikipedia.org/wiki/Candela I'm surprised you couldn't find anything because there is tons of information out there, unfortunately it will make your head spin unless you have a Ph.D. in optics (sometimes even if you do, speaking from personal experience). The units of an image are like joules, or in photometric units lux*m^2*seconds, which is lumens*seconds, which is candela*steradian*second (to get it into all base SI units). Anyway, think of it as a measure of energy (or luminous energy). Let's use regular radiometric units instead of photometric (luminous) units (which are restricted to the human visual range and a lot more complicated). So you have optical power hitting your sensor. Like 10 watts over an area of 1 cm by 1 cm. Now the CCD well integrates those photons. Each pixel might be 5 microns by 5 microns. So now we have watts per area multiplied by the area. That's how many watts are integrated by that pixel. But the pixel only integrates for a certain number of seconds, and watts is joules per second, so you have watts*seconds = (joules/second)*second = joules. That's why I say it's like joules or energy.
OK, that's more than you wanted to know, so I won't even bother to go into the "intensity" of a light source which is even more complicated. Even the "experts" don't agree. For example the American Institute of Physics says that the "intensity" of a light source is W/steradian, yet intensity is an SI Base quantity (like meter, kg, second, ampere, kelvin, and mole) and it's units are candela. My late optics professor, Jim Palmer of the College of Optics at the University of Arizona, got so worked up on the sloppy usage that he wrote a paper on it: "Getting intense about intensity", Metrologia, 1993, vol 30, pp. 371-372. I attach a partial screenshot of his paper, for educational purposes.
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