Ti:sapphire: laser propagation

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The discussion focuses on the beam propagation in Ti:sapphire lasers, addressing the roles of mirrors and birefringent plates in the laser cavity. It clarifies that the birefringent plate is not a filter but influences the cavity's properties, determining which frequencies achieve constructive interference. The pump beam does not circulate within the cavity, while the output pulses circulate among the mirrors. Additionally, the tuning plate is essential for phase matching due to the birefringence of Ti:sapphire, and it can sometimes be omitted by adjusting the crystal's orientation. Overall, understanding these components is crucial for optimizing laser performance.
frerk
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Hello :)

I have 2 questions about the beam propagation of an Ti:sapphire-laser:

Picture 1: The cw argon ion laser brings the beam to mirror (1). Then...? Then the beam (100% of it), go to the birefringent plate (4), gets then reflectet to the first mirror again (1) , which reflects it to the Ti:sapphire, then to mirror (2) , which brings the beam to the last mirror, that which is responsible for the outcoppling.?

But that's propably not right, because the birefringent plate is responsible for wavelength tuning. Because the resulting Laserbeam doesn`t "meet" this plate.

Picture 2: Ti:sapphire: femtosecond-laser. Here the same: what happens at mirror (2)? or is the birefringent plate useful for wavelength tuning of the pumping-laser, and not the resulting Laserbeam? And on this picture I miss the outcoppling beam...

Thank you for your help :)
 

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I'm not an experimentalist, but my understanding of this is as follows. The Brewster plate is not a filter, what it does is that it changes the properties of the entire cavity (made up of all the mirrors). Its setting will determine for which frequency there will be constructive interference (which frequency will "fit" inside the cavity). Therefore, it doesn't matter that it is not in the leg where the outcoupling occurs.
 
So I have to think about an standing wave which occurs between mirrors 1, 2, 3 and 4? (Because the entire cavity is made of all the mirrors). And the birefringent plate "choses" the waves, which can build up such a standing wave. (Results out of destructive and constructive interference).!?

Thank you :)

And on second picture the same. Just that mirror (5) is not 100% reflecting, so that a beam is coupled out there.?!
 
frerk said:
Hello :)

I have 2 questions about the beam propagation of an Ti:sapphire-laser:

Thank you for your help :)

Figure 1 is a bit unclear (no legend, for example). It seems that it corresponds to mode locked (pulsed) output. My understanding is that the pump beam propagates through the Ti:Saph crystal and then exits the cavity. Ti:Saph output pulses circulate through the x-cavity as: 1 -> 2 ->3 -> 2 ->1 -> 4 ->1 ->2 etc.. Mirror 3 is the output coupler, the pump beam does not circulate through the cavity. I am unclear about the function of the tuning plate, my guess is that it's required because Ti:Saph is birefringent, so a compensating plate is needed to control the phase matching of pulses as they circulate.

Figure 2 is very similar to figure 1, the addition of GVD correction (the prism pair) permits the generation of fs pulses. Again, the tuning plate may be required because of the intrinsic Ti:Saph birefringence. Apparently one can omit the tuning plate by rotating the Ti:Saph crystal:

https://www.researchgate.net/publication/26265365_Wavelength_tuning_of_Titanium_Sapphire_Laser_by_its_own_crystal_birefringence

Does this help?
 
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