Imagine we have a power source produce constants power in Watt. If I placed blue LED, then red LED, which color gives more intensity and wht color gives the most?
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1$\begingroup$ What do you mean by "gives the most?" $\endgroup$– Gradient137Commented Nov 18, 2018 at 8:07
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2$\begingroup$ How do you define intensity? $\endgroup$– my2ctsCommented Nov 18, 2018 at 9:36
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$\begingroup$ Yep. You need to specify what exactly you mean by "intensity" here. My hunch is what you really mean is how it looks to the eye, because physical intensity is just power by area, and the powers are the same in all cases by stipulation so there is no difference in the physical intensity of light received at any given distance between the different diodes. But you will need to actually say so we can know what you have in mind. $\endgroup$– The_SympathizerCommented Nov 18, 2018 at 12:54
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2$\begingroup$ You need to consider luminosity of the eye response, green is much brighter to the eye. Blue and red photons are not converted as efficiently by the eye. $\endgroup$– PhysicsDaveCommented Nov 18, 2018 at 13:18
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4 Answers
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Your human eye and brain are a calibrated system. Our eye is very sensitive to green, 1 watt of green will be about 100 times brighter than a watt of blue. In other words if you had a 1 watt flashlight at night in the woods a blue one would be useless. You can google or wiki luminosity function.
answered Nov 18, 2018 at 13:15
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4$\begingroup$ 100 is too much. I'd say about 16 times. Namely, the ratio of luminous efficacies of 555 nm light (peak sensitivity) to 461 nm (dominant wavelength of sRGB blue) is 16. You'd have to go down to 428 nm to get the ratio of 100, and this wavelength is closer to violet than to blue. $\endgroup$– RuslanCommented Feb 22, 2020 at 18:27
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Red light is produced by lower-energy photons than blue light is; if you use the same amount of energy for both red and blue sources, then the red light source will be brighter since less energy is used per photon. This means more photons can be produced. This is why your battery-drained headlamp emits such a low intensity white light but can emit a normally-bright red light once you switch modes.
The color that would give the most intensity is the largest wavelength of light possible: a wavelength the size of the universe.
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$\begingroup$ Again, how are brightness and intensity defined here? These statements can be meaningful only if that is clarified. $\endgroup$– my2ctsCommented Nov 18, 2018 at 9:39
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$\begingroup$ You need to consider luminosity of the eye response, green is much brighter to the eye. Blue and red photons are not converted as efficiently by the eye. $\endgroup$ Commented Nov 18, 2018 at 13:19
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$\begingroup$ The efficiency differences of the eye is negligible compared to the 2 to 1 energy difference between blue and red light. $\endgroup$ Commented Nov 19, 2018 at 11:27
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Light intensity should be same. However, the no. of photons will be more in case of red LED as the energy of single photon is less.
Here, the assumption is that efficiency of both led is same. which is fairly true in most cases.
but suppose two lights of same intensity is being shined..the no. of photons of lower frequency will be more than that of higher frequency photons.
so photon flux will be more for red than the blue.
let us check it from some experimental back up.A laser beam and a microwave beam can carry the same amount of energy. In this case the laser beam contains a smaller number of photons, but each photon in the laser beam has a higher energy than the photons in the microwave beam.for energy transfer the amplitude of the wave matters and square of the amplitude determines the intensity.
https://www.quora.com/Does-blue-light-have-more-intensity-as-compared-to-red-light
Removed first part of quote as it was creating some confusion. Though the statement that photon flux and energy flux are proportional was correct and did not mean equality, it is removed for the sake of clarity.
answered Nov 18, 2018 at 8:08
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$\begingroup$ This is incorrect. Energy flux and photon number flux are different. $\endgroup$– my2ctsCommented Nov 18, 2018 at 9:37
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1$\begingroup$ I think, you should read the answer again. I never equated energy flux and photon flux. $\endgroup$ Commented Nov 18, 2018 at 10:39
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1$\begingroup$ The intensity of light is commonly defined in terms of power emitted in a unit solid angle. Tha means it is related to energy emitted per second and not the number of photons. Thus I say that intensity will be same for both led. $\endgroup$ Commented Nov 18, 2018 at 10:48
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1$\begingroup$ In your quote it says "light intensity or energy flux is proportional to photon flux". This equates the two quantities. Perhaps this is unintended? $\endgroup$– my2ctsCommented Nov 18, 2018 at 10:49
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$\begingroup$ You should also consider luminosity. $\endgroup$ Commented Nov 18, 2018 at 13:16
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With the same power output any color gives the same intensity, if intensity is defined as power per surface or angle unit.
