Fiber Core Imaging

Hi:

Can a fiber core be imaged down by a factor of 10, the idea is to make a large core ( 100 um) appear as close as to a point source.

I thought I could make reverse beam expander (1/10 reduction) using two aspheres and image the fiber core (located at focal point of the first lens). I couldn't find comerical lenses that could do this, because of the fiber NA (~0.2) the first lens dia. is large (depending on the selected focal length) and the second lens dia. because of its shorter focal length is small not being able to collect light coming out of the first lens.

Is there a way to do this?

thanks

Reply to
Farsang
Loading thread data ...

Not without losing light.

If you want to collect ALL the light from a 100 um, 0.2 NA fiber and stuff it into a 10 micron spot, you are effectively saying that you want the light to be diverging much more rapidly than it is when it exits the fiber. Since your fiber NA is already 0.2, achieving 10x more divergence (10x increase in NA for small NA) is not possible.

If you can start with a single-mode fiber instead, you will have a ~10 micron spot right away. But if your source is multimode to begin with, you won't get much of it into single-mode fiber (brightness theorem).

Frank

Reply to
Lineshape

Hi Frank:

1-We want to collimate the output of the 100 um fiber into a 30 mm dia beam, but this is not possible because of the size of the core ( it is an extended source), the core size and the effective focal length of the collimating lens set the min limit of the output beam divergance. So I thought by imaging the core size to a smaller size and using that as a source would help collimation.

2-Your argument about NA comes from " h1 x NA1 = h2 x NA2 ", the optical invariant, where h1 and h2 are object and image sizes, respectively. Isn't this equation valid only for paraxial case (i.e. NAs defined by sines of half angles) and exact equation is h1 x tan (half angle1) = h1 x tan(half angle2) in which case you could have larger output "half angle2" and achive 10 X more diverganc

Reply to
Farsang

1 - To turn an "extended source" into a "point source" to improve collimation is the same as spatially filtering it so it isn't extended (multimode) anymore. You have to strip modes and therefore light to improve collimation at a given diameter. 2 - Yes, it was why I said 10x for small NA. Phil answered it more precisely.

Frank

Reply to
Lineshape

It is not the tangent, but the sine. That is, n h sin(u) is conserved.

Reply to
Dave Schaack

The conservation of radiance doesn't depend on the Abbe sine condition, though. The sine condition is only approximately satisfied by any lens, and it's easy to build lenses that don't satisfy it even approximately--one of the aberrations shown in lens design programs is OSC, for "offense against the sine condition".

Radiance conservation can be derived from Maxwell's equations, or from diffraction theory, but the easy way to derive it is from the second law of thermodynamics, via a thought experiment. I posted the following a few years ago in this very boutique.

Cheers,

Phil Hobbs

---------------------------- Date: Wed, 16 Jun 2004 17:04:48 -0400

Andrew Resnick wrote: >

Just a thought experiment. Consider two reservoirs, well insulated except for a Sufficiently Small Heat Engine running off the temperature difference from Rhot to Rcold. Drill a hole in each reservoir's insulation, and place Magic Optical System in between (also suitably insulated). Add well-silvered baffles so that only light that will make it through M.O.S. gets out of the reservoir insulation.

Call the area solid-angle products looking into each end of the M.O.S. AOhot and AOcold. Then the net radiative energy transfer will be:

Qdot(hot->cold)= (sigma/pi)*(Thot**4*AOhot - Tcold**4*AOcold),

where sigma is the Stefan-Boltzmann constant, and I've assumed Lambertian radiation (which should be right, since it's cavity radiation we're considering).

If AOhot < AOcold, the hot reservoir will spontaneously heat up until Qdot=0, which will happen when

Thot/Tcold = (AOcold/AOhot)**0.25,

and the SSHE will run forever.

There may be a couple of Is to dot and Ts to cross, but that's the general flavour.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

The conservation of radiance doesn't depend on the Abbe sine condition, though. The sine condition is only approximately satisfied by any lens, and it's easy to build lenses that don't satisfy it even approximately--one of the aberrations shown in lens design programs is OSC, for "offense against the sine condition".

Radiance conservation can be derived from Maxwell's equations, or from diffraction theory, but the easy way to derive it is from the second law of thermodynamics, via a thought experiment. I posted the following a few years ago in this very boutique.

Cheers,

Phil Hobbs

---------------------------- Date: Wed, 16 Jun 2004 17:04:48 -0400

Andrew Resnick wrote: >

Just a thought experiment. Consider two reservoirs, well insulated except for a Sufficiently Small Heat Engine running off the temperature difference from Rhot to Rcold. Drill a hole in each reservoir's insulation, and place Magic Optical System in between (also suitably insulated). Add well-silvered baffles so that only light that will make it through M.O.S. gets out of the reservoir insulation.

Call the area solid-angle products looking into each end of the M.O.S. AOhot and AOcold. Then the net radiative energy transfer will be:

Qdot(hot->cold)= (sigma/pi)*(Thot**4*AOhot - Tcold**4*AOcold),

where sigma is the Stefan-Boltzmann constant, and I've assumed Lambertian radiation (which should be right, since it's cavity radiation we're considering).

If AOhot < AOcold, the hot reservoir will spontaneously heat up until Qdot=0, which will happen when

Thot/Tcold = (AOcold/AOhot)**0.25,

and the SSHE will run forever.

There may be a couple of Is to dot and Ts to cross, but that's the general flavour.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Maybe changing fibers is a(n expensive) way to go. If you need something better than your standard single-mode fiber, you can get some very low NA fibers with superb power handling that are "Large Mode Area" (LMA) fibers. Even though the 10 um core in this fiber may not seem large

formatting link
and is smaller than the 100 um core you're using now, the power handling is possibly better and the numerical aperture lower. The combination of spot size and NA may provide your beam with *much* better characteristics. It's not cheap but if you're willing to spend $1k on optics, maybe you can spend $1k on the fiber instead to get a better beam.

If you're dealing with low powers and can keep your launch spot size small, go with standard single mode fiber. If you want more power, look at these products.

Reply to
John_H

Right, but the presence of the sine in the Lagrange Invariant is not directly an expression of the Abbe sine condition. It is simply an expression of the relationship between the angular radius of a (cylindrically symmetric) beam and the projected solid angle of that beam. The presence of pi in the denominator of your radiometric expressions shows that you are using (correctly) the projected solid angle, which in angular terms is pi sin^2 (u).

The Lagrange Invariant states the best that one can do in transferring radiation from one plane to another. It is an additional fact that if the optical system is not corrected for coma (that is, does not meet the Abbe sine condition) that one will not do that well.

Reply to
Dave Schaack

But what you quoted _was_ the Abbe sine condition. The radiance conservation law depends on the differentials.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Hi John:

Thanks for the information.

We are using one of LIMO's High Power Diode Laser and don't think we could replace its output fiber. The problem becomes that of coupling LIMO fiber into LMA-10. If we want use an imaging optics to image down 100 um LIMO fiber core to 10 um core of LMA-10, the output NA of the imaging optics will be larger than LMA-10 NA = 0.1 and we'll loose power.

I'll check with LIMO see if they could use LMA-10 for output fiber

Reply to
Farsang

So what is the correct answer to Farsang's conjecture?

Reply to
Dave Schaack

The second law argument I posted earlier works for all optical systems, whether imaging or non-imaging. Its simplicity is partly due to the fact that it supplies an upper limit, rather than an exact value for the spectral radiance of the image. It's a conservation law, so properly speaking it ought to have an equals sign in it, but for optical purposes we can consider light misdirected e.g. by aberration or scattering to be lost, and so we lose the equality and the rule becomes an upper limit.

I haven't seen an expression for image radiance for general imaging systems, probably because it's complicated to take account of aberrations and defocus, as well as to the imaging geometry--telecentricity and isoplanatism and all the rest of the jargon-infested swamp where lens designers work. (Which I say with admiration--I've never designed a real lens in my life.)

One can use the thought experiment to give a differential version of the radiance conservation law. By arranging the baffles sufficiently cleverly, one can make heat flow spontaneously from cold to hot unless

n**2*dA_I*dOmega_I = n**2*dA_O*dOmega_O,

where Omega is the projected solid angle.

Since one can in principle image any spot on the object into any spot in the image, from whatever angle one likes, this rule must hold everywhere, and in all states of focus.

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

Cabling-Design.com Forums website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.