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Master1022

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- Homework Statement
- An optical fibre transmission system uses a step-index multimode optical fibre which has a core refractive index of 1.49 and a cladding refractive index of 1.48. The fibre is also subject to material dispersion which is a function of wavelength ## \lambda ##. The material dispersion coefficient D is given by:

[tex] D = a \lambda^2 + b\lambda + C \text{ps/km} [/tex]

where ## a = 0.01 ##, ##b = 0.50 ## and ## c = 50## and ## \lambda ## is the wavelength in nanometres. Estimate the wavelength at which the fibre has zero net dispersion.

- Relevant Equations
- D = 0

Hi,

I was working on the problem below:

An optical fibre transmission system uses a step-index multimode optical fibre which has a core refractive index of 1.49 and a cladding refractive index of 1.48. The fibre is also subject to material dispersion which is a function of wavelength ## \lambda ##. The material dispersion coefficient D is given by:

[tex] D = a \lambda^2 + b\lambda + C \text{ps/km} [/tex]

where ## a = 0.01 ##, ##b = 0.50 ## and ## c = 50## and ##\lambda ## is the wavelength in nanometres. Estimate the wavelength at which the fibre has zero net dispersion.

Do we just let D = 0 and solve the quadratic equation? Is is that simple...

I was slightly confused as I know ## D = -\frac{\lambda}{c} \frac{d^2 n}{d \lambda^2} ## and perhaps there was a trick that we needed to use this.

Nonetheless, the first method yields the answers 4999 nm and 1 nm. Not sure how to choose between them, but I think the higher value might be correct as it is closer to the order of magnitude that we saw in a graph in the lectures (order of microns).

Any help or guidance would be appreciated

I was working on the problem below:

**Question:**An optical fibre transmission system uses a step-index multimode optical fibre which has a core refractive index of 1.49 and a cladding refractive index of 1.48. The fibre is also subject to material dispersion which is a function of wavelength ## \lambda ##. The material dispersion coefficient D is given by:

[tex] D = a \lambda^2 + b\lambda + C \text{ps/km} [/tex]

where ## a = 0.01 ##, ##b = 0.50 ## and ## c = 50## and ##\lambda ## is the wavelength in nanometres. Estimate the wavelength at which the fibre has zero net dispersion.

**Attempt:**Do we just let D = 0 and solve the quadratic equation? Is is that simple...

I was slightly confused as I know ## D = -\frac{\lambda}{c} \frac{d^2 n}{d \lambda^2} ## and perhaps there was a trick that we needed to use this.

Nonetheless, the first method yields the answers 4999 nm and 1 nm. Not sure how to choose between them, but I think the higher value might be correct as it is closer to the order of magnitude that we saw in a graph in the lectures (order of microns).

Any help or guidance would be appreciated