LED WaveLenght Change

[LaserCutZ]

Hi all
is it possible to change wavelength of LED in similar manner as laser? ex if i have 450mn and i want to "double" it BBO xtal to 225nm? Way i see it only difference is that LED light not focused so reflected output wave will be all over the place instead of one spot but then if i apply pass through filter of 225nm then i should achieve my goal (well perhaps with big loses but some result? Any comments on that?
tnx!

[Phil Hobbs]

Post by LaserCutZ
Hi all
is it possible to change wavelength of LED in similar manner as laser? ex if i have 450mn and i want to "double" it BBO xtal to 225nm? Way i see it only difference is that LED light not focused so reflected output wave will be all over the place instead of one spot but then if i apply pass through filter of 225nm then i should achieve my goal (well perhaps with big loses but some result? Any comments on that?
tnx!

The first difficulty is that the nonlinear response of the crystal is,
well, nonlinear (quadratic, mostly). That means that to get any
significant amount of second harmonic output, you need a really high E
field. That means either short pulses of high energy or tight focusing.

You can't get either out of a LED, unfortunately.

There are other reasons as well, such as phase matching being impossible
with diffuse light (as from a LED). The electric polarization at the
second harmonic has half the wavelength of the pump light, which in
general is not equal to the wavelength of a propagating beam of the
second harmonic, due to crystal dispersion. If the two aren't equal,
you don't get a second harmonic beam. (The length of the interaction
region sets how nearly equal they have to be--if it's a centimetre, you
have to be within a wave number [30 GHz] or so.)

Thus to be a good doubler, a crystal needs two attributes: first, a
reasonably large nonlinear susceptibility (chi''), and enough
birefringence to compensate for its dispersion, so that you can find a
crystal orientation where the quadratic polarization phase matches with
a propagating wave at the second harmonic. (You can also play games
with the crystal, such as making it into a waveguide and putting a
periodic structure on it, e.g. a photorefractive grating.)

LED light is diffuse, i.e. it radiates into a whole hemisphere, so the k
vectors of the field components are all over the place and can't all
match to anything.


Cheers

Phil Hobbs