I'm trying to understand how the (power) loss in optical fibre works and I'm having some trouble.

enter image description here

This is a classic 50/50 fused fibre Y coupler. The output on port 2 is the same as port 3 (so half the power of port 1 plus the losses).

The losses considered here are the coupling loss (same for each port) and the radiated loss at the point of the split. We want to measure the coupling loss.

So my first idea would be to go from port 2 and measure the loss. However, while the radiated loss might be smaller, it seems unlikely that it would completely bypass it. Going from port 3 would bring the same result.

So is there any way I can measure the coupling loss in this configuration?

  • \$\begingroup\$ I haven't worked with commerical fiber, but with laser tables. I would tackle this with an interference technique. \$\endgroup\$ – crasic Jan 19 '16 at 9:14
  • \$\begingroup\$ I don't understand what it means to "go from" a port. But to eliminate the effect of radiated loss, can you not use an ordinary fibre of the same material and dimensions but without the Y coupler? Coupling loss should occur predominantly at the ends of the fibre. I don't think it makes that much sense to distinguish between coupling loss and radiated loss at the Y itself, although you can try to measure the reflected power coming back out of port 1 if you're interested in that. \$\endgroup\$ – Oleksandr R. Jan 19 '16 at 13:15
  • \$\begingroup\$ @OleksandrR. I meant that instead of using port 1 as the input, we'd use port 2. So the power would flow from port 2 to port 1. I do agree that there is little point but it's a question my professor asked that in class. Your suggestion is that the reflected power we get in port 1 depends only on the coupling loss? \$\endgroup\$ – meneldal Jan 19 '16 at 23:52
  • \$\begingroup\$ If the power is not radiated (absorbed somewhere in the fibre cladding), and not transmitted to the output, then where else does it have to go apart from being reflected? You can't easily measure the radiated power (maybe by measuring the temperature rise of the fibre?), but you can subtract the transmitted and reflected powers from the input power and get it that way. I don't see any other loss mechanisms for this device. I expect the losses from reflection at the air/glass interface at the ports will far exceed the reflection loss at the Y. You will have to compare against a standard fibre. \$\endgroup\$ – Oleksandr R. Jan 20 '16 at 0:25
  • 1
    \$\begingroup\$ I don't know why you think there is no reflection. There is a glass transition at the junction in both directions. Since its an optical beam splitter. You need to mea sure reflected power and see how much of your missing power it fills up, assume the remainder is radiated. Since there is a change in n, a portion of the light will always be reflected at the beam split even on the pass through direction \$\endgroup\$ – crasic Jan 20 '16 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.