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B Question about frequency of light

  1. Dec 1, 2016 #1
    I flicker an LED on and off at a rate on the order of hundreds of kHz. The light, even though it is sent in pulses, still has its own frequency, correct? The frequency at which I turn the light on and off is not related to the frequency of the photons. Is this correct?
     
    Last edited: Dec 1, 2016
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  3. Dec 1, 2016 #2

    Orodruin

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    You cannot do that. The response time of a light bulb is much longer than that.

    For an incandescent bulb, you will get a spectrum based on the temprature of the filament.
     
  4. Dec 1, 2016 #3
    *For an LED, which responds to current fluctuations on the order of microseconds
     
  5. Dec 1, 2016 #4

    Drakkith

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    The LED itself generates light through the excitation and relaxation of electrons between electronic states. In other words, when you apply a voltage and send a current through an LED, you are exciting electrons from lower states of energy into higher states. When they fall to a lower energy state, they release that energy as a photon of light whose energy is the same as the difference in the energy states. So the rate at which you switch the LED on and off has little to do with how the light is generated.
     
  6. Dec 1, 2016 #5

    Nugatory

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    That is correct as long as the frequency of the light is large compared with the frequency with which you flash the light, which is the case here - the frequency of visible light is measured in the hundreds of terahertz and you are cycling the light at hundreds of kilohertz.

    You might try imagining the same experiment with sound waves. Sound waves typically have a frequency of hundreds of hz, so your experiment is analogous to switching a sound generator on and off every few months.

    (As an aside, there aren't any photons involved here. This is a purely classical thought experiment, and the light is best thought of as electromagnetic waves. Photons only come into the picture when there are significant quantum mechanical effects, and that's not the case here.).
     
    Last edited: Dec 1, 2016
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