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Keeping the average power constant, why some materials are more eager to be damaged by pulsed laser with respect of C.W. lasers, or viceversa?

When i talk about pulsed lasers i think for examples of duty cycles in the order of $10^5$.

For example optical elements (such as a vortex phase plate for donut-shaping the beam) have different tolleration regimes regarding the incident power not simple dependent on the average power but also on the peak power for pulsed beam.

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What do you mean by a duty cycle greater than 1? –  nibot Nov 11 '10 at 19:48
For example the laser I usually use has an average output power of approx 1W but it produced pulses of 100fs every 12.5 ns (80MHz). Tha means a peak power of some $10^5$ W... –  Steve Nov 12 '10 at 9:00
So it's not the duty cycle but the peak power. Duty cycle is "time on" over "total time". –  Cedric H. Nov 12 '10 at 11:50
100fs/12.5ns$\approx 10^{-5}$ in my book :P and in an approximation of square pulses "time_on" over "total time" is the inverse to "pulse_power" over "average power"... so ok that was my mistake in notation –  Steve Nov 12 '10 at 11:53

3 Answers 3

up vote 7 down vote accepted

Typically, a laser will damage an optical surface in one of two ways. The first is just what you would expect: the laser heats the material up until something bad happens. The second is also pretty simple, but less common because (AFAIK) it is really only a problem with very short pulses (on the order of femtoseconds). In this case a small but rapidly changing mechanical stress is produced by the beam, either as an effect of heating or the electrostrictive effect. The resulting shock is strong enough to damage the optical surface, usually by producing microscopic chips or causing optical coatings to de-laminate.

Now, the important issue here, which should answer your question, is that average power doesn't really mean much in the context of optical damage. It can be an issue in cases where heat dissipation over long time scales becomes a problem, but in cases where the laser is pulsed, damage mechanics are primarily determined by the peak power and the pulse width.

To make analogy: I have never been shot with a gun. If I get shot in the head tomorrow, my lifetime average bullet flux will be very low, just one bullet in 26 years. Yet I'll still be dead.

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Not being an exhausting list. Some phenomena that allows for this to happens is non-linear effects. A short pulse allow a higher intensity for a small period to time which enhance non-linear effects such as two-photon absorption and self-focusing. Two photon absorption will make a material absorb light which it wouldn't if light was low intensity. Self-focusing would make the light focus more which would decrease area which the heat can be diffused, or increase intensity for two or more photon absorption. The self-defocusing is possible too which would allow for decrease damage in the material if pulsed.

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the self focusing/defocusing was beyond my knowledge actually do you know any review i can read? –  Steve Nov 15 '10 at 13:55
Sorry, that's about the knowledge on the subject that I have. Z scan by Sheik-Bahae, is a very popular method for quantifying self-(de)focusing. Check some general non-linear optics books too. Authoritative names are, Yariv and Boyd. –  Bernardo Kyotoku Nov 15 '10 at 17:11

One thing that ought to matter is how much laser light gets reflected versus how much gets absorbed and transmitted.

I don't know much about this, but my naïve guess is that materials get damaged by lasers primarily because they are heated by the intensity of the light, and then they melt (or burn!). Therefore thermal conductivity or response properties of the material will come into play as well. For instance, if heat conducts well in a material, the local heating of the material by the laser will be able to diffuse to a larger region.

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if it was as simple as you say i guess that what comes into play is the amount of energy deposited in the materia by the laser and this energy is proportional to the average power of the beam regardless of it being CW or pulsed with a $10^5$ duty cicle.. i guess.. –  Steve Nov 11 '10 at 17:35
Right, I see now that I missed the distinction that you were trying to make. Do you have any references that discuss this phenomenon? –  j.c. Nov 11 '10 at 18:14
This depends on more than just the duty cycle; it also depends on the absolute pulse length. To see why, just assume a laser with a 1 hour/10000 hour cycle. If a single pulse deposits enough energy to damage the material, it doesn't matter how long it will take for the second pulse to arrive. –  MSalters Nov 16 '10 at 10:19

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