I've got quite easy question (and my attempt to solve it). So there is an optical fiber with Dispersion index: D = -100ps/nm•km, length: L = 10km. The question is: which wavelength will achieve faster speed and the end of the optical fibre: λ_1 = 1000nm or λ_2 = 1001nm? In my opinion if the speed of the wave is dependent on the refractive index which will be the same in the same fibre then the speed of both wavelengths will have the same speed. Is it true? Does dispersion have any impact here?

  • 7
    \$\begingroup\$ First, remind me, what does "dispersion" mean? \$\endgroup\$ – The Photon Nov 23 '16 at 15:50
  • \$\begingroup\$ It relates wavelength to refraction index (i think) \$\endgroup\$ – Makoto Nov 23 '16 at 15:52
  • 3
    \$\begingroup\$ This is obviously homework because the numbers are so neat. But The Photon gave you a good clue, and the sign of the dispersion is important. \$\endgroup\$ – Brian Drummond Nov 23 '16 at 16:18
  • \$\begingroup\$ @Makoto um, well, yeah. But I'd say that is kind of the macroscopic view at the effect that happens on the phase level (hint! hint!) \$\endgroup\$ – Marcus Müller Nov 23 '16 at 16:27
  • 2
    \$\begingroup\$ I'm voting to close this question as off-topic because this is clearly physics \$\endgroup\$ – Scott Seidman Nov 24 '16 at 14:34

First, when you talk about the "speed" of a signal in optical fiber, that's ambiguous. You should be clear about whether you're interested in the latency (the time it takes a signal to travel from one end of the fiber to the other) or the bit rate. In this case, it seems most likely you're interested in the latency, or propagation delay.

In my opinion if the speed of the wave is dependent on the refractive index which will be the same in the same fibre then the speed of both wavelengths will have the same speed. Is it true?

No. This is not true. The index of refraction of a material varies (at least slightly) depending on the wavelength of the light being considered.

In addition, in a dielectric waveguide like optical fiber, as the wavelength changes a different proportion of the signal power travels in the core and in the cladding, leading to (at least small) changes in the effective index of the fiber.

In fact, dispersion can be either negative or positive (also called anomolous and normal dispersion), depending on the wavelength and the design of the fiber. We can also engineer the dispersion properties of the fiber in some cases to optimize the fiber for different applications.

But all of that is irrelevant to answering the question, because the total effect is summarized in the dispersion parameter.

When you specify the dispersion as you did, D = -100ps/nm•km, you're saying we already know the effect of all those variations, and that effect is that the propagation delay through 1 km of fiber changes by -100 ps for every nanometer of change in the wavelength of the signal light.

So you don't need to worry about the physical mechanism. You just need to apply the definition of the dispersion parameter to decide whether a longer or shorter wavelength travels faster through this fiber.

  • 1
    \$\begingroup\$ +1 Because "The Photon" answered the optics question. Also it's a good answer. \$\endgroup\$ – ambitiose_sed_ineptum Nov 23 '16 at 17:13
  • 3
    \$\begingroup\$ @ambitiose_sed_ineptum, There is a reason for my nickname. \$\endgroup\$ – The Photon Nov 23 '16 at 17:16
  • \$\begingroup\$ @ThePhoton So to sum up and get an exact answer for my question, with D = -100ps/nm•km while increasing the wavelength the delay will be decreasing (it will decrease 100ps while adding 1nm to the wavelenght). On the other hand if D = 100ps/nm•km then the delay will be increasing while increasing the wavelength? So when D = -100 the time to travel to the end of the fibre optic will be greater for 1000nm then 1001. \$\endgroup\$ – mikro098 Nov 23 '16 at 17:21
  • \$\begingroup\$ @codddeer123, yes. \$\endgroup\$ – The Photon Nov 23 '16 at 17:23
  • 1
    \$\begingroup\$ @Naz, that wouldn't work. TIR requires the core region to have higher index than the cladding. What is actually done is the core and cladding are both glass, and the core region has some chemical dopant to slightly change the index (by a fraction of a percent). \$\endgroup\$ – The Photon Nov 23 '16 at 17:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.