Engineering Optical Fibres: Wavelength for Zero net Dispersion

[Master1022]

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:
$$D = a \lambda^2 + b\lambda + C \text{ps/km}$$
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:

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

 

[Charles Link]

You need at least one of the coefficients of the dispersion function $D$ to be negative if you are going to have a positive root $\lambda$.

 

[Master1022]

You need at least one of the coefficients of the dispersion function $D$ to be negative if you are going to have a positive root $\lambda$.

Ah yes, you are right! Sorry, I think I forgot the -ve sign on $b$.

 

[Charles Link]

Ah yes, you are right! Sorry, I think I forgot the -ve sign on $b$.

I think "c" also might need a minus sign, or you get imaginary roots. Please check your arithmetic=putting in minus signs on "b" and "c", I get roots of $\lambda=+100$ nm, and $\lambda=-50$ nm.

This one doesn't even need the quadratic formula=it factors: multiplying by 100 we get $(\lambda-100) (\lambda+50)=0$.