On Jul 7, 11:29 pm, Farsang <youzpal...@netscape.net> wrote:
Hi Michael:
This is primarily for image analysis, but would like also to do
Physical Optics to simulate better the real situation.
This is the application:
We'd like to get spot size of dia = 2 Cm - 5 Cm (absolute Max) at a
minimum distance of 100 feet (ideally 250 feet) from the fiber.
Fiber output power is 200 W and wavelength = 980 nm.
I was thinking to get the required spot size using Geometric Image
Analysis of ZEMAX and then do the physical optics
for closer simulation.
Because of the optical invariance it'll be difficult to achieve the
above spot size using catalog optics.
Do you have any idea how we could get this spot size, appreciate your suggestions and comments?
Ideally we'd like to have
Spot Size = 2 Cm - 3 Cm at 250 feetHi Farsang,
Power in the Spot = 200 W
If you are doing imaging analysis, you can model the output of a
fiber as follows: Go to the Field Data box and click on "Object
Height". Enter enough field points to represent your fiber, perhaps
by a point in the center of the fiber, at the .7 zone, and at the
fiber's edges. Set your wavelength in the wavelength box. In the
Lens Data Editor, set the distance from your fiber face to the next
lens or point of interest. Set the fiber's numerical aperture by
setting <General> <Aperture> <Aperture Type: Object Space NA>,
<Aperture Value: 0.22>. Make the fiber's output telecentric by
clicking on "Telecentric Object Space".
If you are looking for the footprint of the light through all of
the lens surfaces (and the image plane, if there is one), you'll have
to go to <Analysis> <Geometric Image Analysis> <Settings>, set the
Field Size: to 0.500 (if your units are set to millimeters), set the
File: to CIRCLE.IMA, set the Image Size: to the size of your area of
interest at a particular surface, then set Surface: to your surface of interest.
If you are coupling the fiber's light into another fiber, set the
NA in the <Analysis> <Geometric Image Analysis> <Settings> box to the numerical aperture of the fiber being coupled. That will let you see
the coupling efficiency.
If you are doing physical optics propagation, you'll have to wait
for Michael's reply.
But from your last post, I think I see what you are doing.
To model the optical system, if you need a 3 cm spot at 250 feet,
from a 0.5 mm diameter source, your optical system's focal length
needs to be about (250/30)* 0.5 (sorry about the mixed units) = 4.16
feet. At a numerical aperture of 0.22, your system's aperture needs
to be about 23 inches in diameter. It might be hard to come up with a
23" diameter f/2.27 diffraction-limited lens. It's not impossible,
but it would be expensive. Even the military might want to see a
better solution.
The problem arises from the size of your source. You'll get a lot
further if you use the output of the diode directly (whose source is effectively the size of a bacterium), without passing it through a
fiber, if that is possible.
Wade Kelman
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