I just want to ask, is electric power equal to radiant flux? In calculating intensity, it is the total light power emitted (radiant flux) divided by the surface unit area.
closed as unclear what you're asking by Dmitry Grigoryev, Arsenal, Voltage Spike, Daniel Grillo, pipe Sep 9 '16 at 15:20
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It says that the power consume by the light source is equal to the light power emitted
That's probably not a true statement: at best it's intuitively misleading.
By the law of conservation of energy we know that all the electrical power consumed must go somewhere, but it doesn't need to go into light. Certainly not all of it goes into visible light: the LED also gets hot. There are other kinds of non-visible light that might be radiated: infrared light from the heat, and RF light from variations in the current in the device.
In practice the LED is probably in air, and is attached to wires. Some of the thermal energy is lost to the air by convection, and the wires by conduction. In practice this is where most of the energy that wasn't radiated as visible light will go.
Is electric power equal to radiant flux?
Both of these things are measured in watts, so it's possible for a device that emits light and nothing else. But since there's convective and conductive cooling taking energy out of the system by a mechanism that isn't light, we can deduce there will always be some degree of discrepancy between input electric power (current times voltage), and output radiant flux (power of photons leaving the LED).
No, it's not. Most of the energy you put into a diode goes into heat rather than light. Heat can remain as energy within the device or it can radiate as well. So any LED will have energy going into the frequency of light you want, radiated energy in the infrared range due to thermal radiation and then you still have conductive heat that will move and be released by touching other objects (air, metal contacts, pcb, etc.)
See here for a typical table for luminous efficiency of LEDs.