Focusing light into an optical fiber cable

Question

I'm trying to focus light from a flashing light lamp throught two lens into a optical fiber cable which is around $0.8 \ \mathrm{cm}$ in diameter.

• flashing light : a bulb and then covered by a glass of $7.5 \ \mathrm{cm \times 7\ cm}$
• Fresnel Lens $310 \times 310 \ \mathrm{mm}$ , Thickness: $2 \ \mathrm{mm}$ , Focal length: $220 \ \mathrm{mm}$

I tried to use a piece of white paper to focus the image, however, my teacher said that the focal point might have more intense light. But focal point is not somewhere I can see. Is his theory right about focal point have more intense light than the image point?

I wonder if I could use more convex lens to focus the spot of light into a smaller diameter?

Answer 1

Ok, first of all, if you are dealing with fiber optics things are little bit more difficult than just shine with bulb on fiber frontface. Talking about standard single mode telecommunication fiber, core diameter (e.g. this part of fiber, where is the light guided) is around 9 micrometers. If you have multimode, core will be like 50 micrometers, so this is the diameter you need to have in focus. Structure of single multimode fiber optical cable is concentric like this: core (50 um) - cladding (250um) - primary protection (1 mm) - secondary protection (rest of diameter). If you have fiber with larger diameter of core and cladding, it is not called "fiber" but "light guide", used for example in endoscopy. Because fibers are guiding light due to total internal reflection effect, crucial characteristic is numerical aperture of fiber (NA), which defines the broadness of acceptance "cone" where you can achieve total internal reflection. Obviously, to confine maximum light into fiber you need to match the cone of focusing element with acceptance cone of the fiber. Common approach is to use microscope objective with similar NA as fiber and micrometric mount to launch the light. If you need to confine broad spectra, you can use GRIN objective with abberation compensation, but ussually you need just to pick objective for the strongest wavelength and you are fine.

Answer 2

When you try to focus light into a fiber, you have to match the fiber mode to the spatial mode of the light beam for best coupling. For example, with a microscope objective lens, a single mode fiber and a good TEM00 (Gaussian mode) laser, you can expect a maximum coupling efficiency of about 60% if you are tuning with a 3-stage X-Y-Z translation micrometer stage. It can be very very very painful to do (speaking from experience).

With multi-mode fibers, things get a bit easier and you can achieve higher coupling efficiency. I have not worked with incoherent light, but I'd imagine that the efficiency cannot be higher that with coherent light. Can you give us a few more details about your experimental arrangement? Pictures would help quite a bit.

A useful read. Also check this wiki-article that talks about NA and acceptance angle.

Answer 3

"focal point", "image point" - careful with the terms.

All you need to worry about is where the light is focused in the smallest space. Yes, that should be pretty bright. Yes, you could use a business card to detect it - just figure out where the image is smallest.

Now, how to capture more light. Look at this image:

http://upload.wikimedia.org/wikipedia/commons/7/71/Lens3.svg

As the object moves further away from the lens and to the left, the image also moves to the left, closer to the lens. When S1 = infinity, then S2 = f, and that's the closest the image will get to the lens, and the image is smallest at that point. However, because the source is so far away, the image may be small, but it's pretty faint.

So yes, the answer is: use a more convergent lens (a stronger lens) because then the source could be closer to it. Also, a bigger lens might help. There's a trade-off how far to place the source; if it's too close, you capture a lot of light, but the image is large; if it's too far, the image is small, but you don't capture much light. At S1 = 2f, the image is about the same size as the source.

It might help to enclose the source in some sort of box, the inside should be either bright white (good), or reflective (better), or reflective with a shape that focuses light back into the source (best). The size of the lens becomes somewhat less important in this setup.

This is a hard problem, don't be surprised if there are no clearly defined answers to it. Unless the source is equal to or smaller than the fiber entrance, then it's somewhat easier - put a reflective sphere around the source, poke a hole, put a ball lens on the hole and couple the sphere with the entrance of the fiber.

http://www.edmundoptics.com/technical-support/optics/understanding-ball-lenses/

A tiny LED chip is easy to couple with a fiber aperture. A large fuzzy lamp is much harder.

Answer 4

I don't know what's wrong with using a white card to find the focus. The focus is where you see the image of the lamp filament. You cannot focus the filament to a smaller image size than the actual filament itself without running up against the laws of thermodynamics.

EDIT: OK, this answer is really wrong (see comments below). Leaving it up because I don't believe in covering my tracks.

EDIT 2: I'm going to try this one more time. What is probably true is that no system of lenses is going to cram more light into the fibre than you would get just by butting up the end of the fibre directly against the lamp. Any takers?

EDIT 3: Second answer also wrong. Let's try this: Express the capture area of the lens as a fraction of the total emission angle of the source. That is the same fraction which gives you the minimum (optimum) image size as compared to the size of the source.

(Also, re-reading the original question, I think I now understand the point about image point versus focal point: the focal point is what you get with parallel rays, i.e. a source at infinity. If the source is close, so you don't have parallel rays, the focus moves correspondingly. The focus is literally where you see the image on a white paper, and the calculated focal length of 220 mm is irrelevant.)

EDIT 4: My third edit assumes that your source and target are in fixed locations, and you have an infinite number of lenses with different focal lengths to play with. I will now, finally I hope, try to address the actual optimization of the system which you actually have: the big Fresnel lens with the 220-mm focal length. First let's imagine the "symmetric" case, where the source and the target are 880 mm apart, and the lens is right in the middle. When you put your white piece of paper at the target location, you should see an image size exactly the size of the actual filament. You can see this should be so because of the symmetry of the ray tracing patterns. Let us examine what happens if you try longer or shorter combinations.

First imagine what happens if you put your eyeball directly at the image point and look at the lens. (Imagine your eyeball to be the size of a geometric point.) If your eyeball is placed within the actual filament image, then when you look at the lens you will see the entire surace illumainated with the fiery color of a hot filament; if your eyeball moves outside the image zone, you see merely ambient light, or darkness as it may be. Now let us move our target backwards, away from the lens. Which way do we have to move the lens to adjust the image to be on target? Answer: if we move the lens in the opposite direction, towards the target, we will find a new focus. It is easy to see this because if we move our target all the way back to infinity, the lens must be 220-mm away from the source. (Remember we said we were starting at optimum conditions with the lens 440-mm from the source.) If we put our eyeball exactly right on target, we still see the lens filled with that fiery color; but we are so far away that the lens looks very small, and therefore the total amount of light it delivers is small. The image is actually quite large, but very faint.

But wait: if instead of moving the lens towards the source, if we had instead followed the target, we would also find a solution at infinity where the lens is 220 mm from the image point and the light bulb is at infinity! If we put our eyeball right at the image point, we once again see the bright fiery color of the hot filament; but this time, it does not fill the whole size of the lens, which is close to us and looks quite large. The image is now quite bright, (like when you focus sunlight) but it is absolutely tiny, so if we move our eyeball off target in the slightest, we see merely ambient light, or darkness. Again, the total amount of light collected is small.

To summarize: the closest proximity of target to source at which the lens focuses an image is 4x the focal length; at this position, the image size is the same as the actual size of the source. There are no closer solutions. For every longer case of source-to-target distance, there are two solutions for lens position; in one case, the image is larger and fainter, and in the other case, the image is smaller and brighter.

Where is your maximum light collection? As you bring the lens closer to the source, you are intercepting a bigger solid angle, so you are collecting more light; but the image size at the focus is also getting bigger (and fainter), so eventually your are not able to concentrate all the availble light into your target. I'm guessing the maximum light collection occurs when your image size is the same diameter as your optical fiber, because that's the closest lens-to-source proximity and hence the largest solid-angle interception without "spillage" at the point of collection.