What type of simulation do you have in mind? Is this a numerical or computer simulation, or a physical simulation where you replace the optical fibre by a different source?

If this is a numerical simulation, are you using a software package such as Zemax which includes optical building blocks which can be deployed directly, or will you construct the simulation from scratch using Matlab, Python, C, Fortran, or some other programming tool.

• What are you trying to simulate?
• What aspects of the fibre output are significant?
• How accurate does the simulation need to be?

What type of fibre is involved, and how is it excited?

• Single mode?
• Multimode graded index with under-filled excitation, such as by a VCSEL source as used for short-distance data links?
• Multimode graded index with all modes excited?
• Multimode step index

Is the illumination coherent or incoherent? Coherent near-monochromatic excitation of multimode fibres often results in an output intensity pattern with a strong speckle modulation. Is this significant for your application?

Illumination of multimode fibres by a diffuse extended source larger than the fibre core may excite "leaky" skew modes which exit the fibre at higher angles than are predicted by the fibre numerical aperture. This is more pronounced with short lengths of large core (>100 µm) step index fibres with high numerical aperture (>0.2).

For multimode step index fibre illuminated from a source with lower numerical aperture than the fibre, the output far-field radiation pattern can be very similar to that of the source.

In contrast, with a graded index fibre, both far field and near field intensity patterns will generally differ from what is launched into the fibre.

For single mode fibres, both near field and far field radiation patterns can often be approximated by a Gaussian field profile. This is simple to implement and in many circumstances will be good enough.

If higher accuracy is required, there are exact Bessel function solutions for the mode field of step index fibres. For single mode fibres, the far field radiation pattern can calculated via a Hankel transform of the near field amplitude.

Are the distances marked in your diagram representative? Is the intent to limit the numerical aperture captured from the fibre to 0.125?

If you use "a lens with the same NA as the fibre", what type of source would you use.

• Point source, uniform extended or apodised extended source?
• Angular distribution Isotropic, Lambertian, Gaussian, or some profile?

Where would you locate the source with respect to the lens?

I think you can use optisystem simulation tool for this application.

I agree with Alan Robinson.

Hi Alan Robinson ,

thanks for your detailed answer. I will try to answer here all of your questions.

1. It is a computer simulation. As you can see on the left of the diagram, there is a fiber tip. There I will need a source that reproduces the radiation coming out of a fiber tip. So far I don't have any experience with Zemax, but I do have with packages such as Comsol (with a GOP module), BeamProp and VPI.

2. This scenario is trying to simulate the light coming out of a fiber and entering in the eye of a technician (for example). So it is important to know how much of the input energy arrives at the detection plane.

3. Suppose it is the end of an SMF-28 fiber under coherent illumination at 500mW.

4. Initially it will be ok if we get an idea of the order of magnitude that arrives at the detection plane. So high accuracy is not a big deal here.

5. I yet don't have a clear idea about what source to use. Also I didn't get your question about the NA=0.125. But yes, that a reasonable value.

Thanks a lot for your help!

Hi Felipe Beltran-Mejia,

If you are unsure of the exact fibre type, and not too concerned about accuracy, a reasonable assumption is that the fibre is single mode, and emits an ideal Gaussian beam.

I would expect one or more of the simulation packages you mention to support free space propagation of Gaussian beams, but there is a lot you can do with pencil, paper and a pocket calculator. https://en.wikipedia.org/wiki/Gaussian_beam

For SMF 28e+ fibre, the nominal mode field diameter is 9.2 µm at 1310 nm and 10.4 µm at 1550 nm. https://www.corning.com/content/dam/corning/media/worldwide/coc/documents/Fiber/PI-1463-AEN.pdf

So at 1550 nm, the radiant exitance at the fibre exit facet falls to 1/e2 of the peak value at a radius, w0 = 5.2 µm.

At distances much greater than the Rayleigh range (0.055 mm), the angle at which the radiant intensity falls to 1/e2 of the peak is θ0 = λ/(π w0) = 0.095 radian. Note that the numerical aperture of 0.14 specified in the Corning data sheet is measured at 1% intensity, corresponding to a wider divergence and a lower intensity.

At a distance of 14 mm from the fibre tip, the 1/e2 radius of the irradiance is 1.33 mm, so most of the light from the fibre will pass through a 3.5 mm aperture.

(The half-angle subtended by a 3.5 mm circular aperture at a distance of 14 mm corresponds to a numerical aperture of 0.125 - which is why I mentioned that value).

More generally, for a Gaussian beam of radius w, the fraction of the total power transmitted through a circular aperture of radius R is [1 - exp(-2 R2/ w2)].

Peak irradiance is 2 P0 / (π w2) where P0 is the total power.

Specifically, if your technician holds the fibre 14 mm from his eye, 97% of the 500 mW power will fall inside a 3.5 mm circle aligned with the centre of the beam, so 484 mW. Peak irradiance of 18 W/cm2 is significantly higher than the 4 W/cm2 threshold for thermal corneal damage.

Note that at 1550 nm, most of the power will be absorbed in the cornea, and will not reach the retina of a human eye.

At 1310 nm, transmission is higher, but there is significant absorption in the lens of the eye.

At 850 nm, 980 nm or 1060 nm, transmission to the retina will be much higher.

The effective focal length of the human eye is approximately 17 mm, so light from a fibre only 14 mm from the subject will not be focused onto the retina. The retinal damage threshold will be reached much sooner with the fibre at a greater distance from the eye, such that the radiation converges to a smaller retinal spot. Is this one of the factors you intend to address in your model?

If you have not seen it already, this paper may be of interest:

D. Sliney et. al, "Adjustment of guidelines for exposure of the eye to optical radiation from ocular instruments: statement from a task group of the International Commission on Non-Ionizing Radiation Protection (ICNIRP)", Applied Optics, vol. 44, np 11, 2005.

Article Adjustment of guidelines for exposure of the eye to optical ...

Also: https://en.wikipedia.org/wiki/Laser_safety has some useful links.

Dear Alan Robinson , I was revisiting your answer and I'm not getting exactly the same results as you. Specifically where you wrote,

" At a distance of 14 mm from the fibre tip, the 1/e2 radius of the irradiance is 1.33 mm, so most of the light from the fibre will pass through a 3.5 mm aperture."

For me, at a distance of 14mm I get a w(z) = 1.88mm. Assuming the waist at the Rayleigh range is w(zR) = sqrt(2)*w0. On the other hand, I can get similar results to you if I assume w(zR) = w0, but I believe this is not correct.

Do you have any idea where could it be the error?

Thanks again for all your help.

Felipe Beltran-Mejia - You have not given enough information for me to understand where you get your 1.88 mm radius at 14 mm.

From the Corning data sheet, at wavelength, λ = 1.55·10-6 m:

1/e2 mode field radius of SMF-28e+ fibre is w0 = 5.2·10-6 m.

From https://en.wikipedia.org/wiki/Gaussian_beam#Evolving_beam_width

Rayleigh range zR = π w02 / λ = 0.055 mm

Radius at distance z; w(z) = w0 sqrt(1 + (z/zR)2)

At Rayleigh range: w(zR) = √2 w0 = 7.35·10-6 m

At 14 mm: w(14 mm) = 1.328 mm.

Are you assuming that the radius increases linearly from the Rayleigh distance?

There is an approximately linear increase for distances much greater than the Rayleigh distance, with divergence half-angle θ0 = λ / (π w0) = 0.095 radian

For shorter distances up to a few times the Rayleigh distance, you must use the square root formula (above) from the Wikipedia page.

Hope this helps.

Oh I'm sorry Alan, everything seems clear now. I was using my own expression for the cone, instead of using the hyperbolic relation as explained on the link you've sent.

Thanks again for your kind help!