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As we know that in lasers there are excited atoms. When energy is provided in the form of light, heat or electricity to these atoms, these excited atoms after sometime go to a lower state of energy and releases a photon and when this photon hits another electron which goes to a lower state of energy and releases two daughter photon and in every hit it amplifies the photon production by twice. So like when all the atoms have amplified photons and are in a ground state then why can't those photons provide energy to bring an atom to an excited state and repeat the whole process without the use of any extra electric current?

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    $\begingroup$ Um...(part of) the photons leave the laser after the spontaneous emission (that's where the beam comes from, after all"!), so how could that process repeat ad infinitum if energy is continuously leaving the system? I don't get your question. $\endgroup$
    – ACuriousMind
    Apr 21 '15 at 9:37
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Even if the laser had perfectly reflecting, i.e. lossless, mirrors at either end of the cavity, and both ends were sealed so no light could escape it would still require a continual power input. That's because excited atoms/molecules can decay by mechanisms that don't involve a photon e.g. collisional de-excitation. The lost energy goes into heating up the laser, so you'd need to supply energy to replace the energy lost as heat.

In the real world the mirrors aren't perfectly reflecting so you need to supply energy to make up for the energy lost by absorption at the mirrors (this also ends up heating up the laser).

And of course a laser isn't much use unless you make a hole in one for the beam to shine out of. That removes energy from the system, so you need continual power to replace the lost energy.

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    $\begingroup$ +1 I love it when people willingly indulged another before clearing things up a win-win situation. $\endgroup$
    – user6760
    Apr 21 '15 at 10:55
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    $\begingroup$ None of the mirrors in a laser resonator has a 'hole' in it. One of them is only partially reflecting and this is where the beam comes out. Good answer otherwise (+1). $\endgroup$ Apr 21 '15 at 17:08

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