Optical Communication Bandwidth Calculation

In optical communication systems, understanding the relationship between spectral width and bandwidth is crucial. Spectral width refers to the range of wavelengths present in a light source, while bandwidth refers to the range of frequencies. These two quantities are inversely related, meaning a smaller spectral width corresponds to a larger bandwidth, and vice-versa.

Silica Optical Fiber Parameters

We are provided with the following parameters for the silica optical fiber:

We also know the speed of light in vacuum, which is $c = 3 \times 10^8$ m/s.

Wavelength to Frequency Conversion Principle

The relationship between the speed of light ($v$), frequency ($f$), and wavelength ($\lambda$) in a medium is given by the formula:

$$f = \frac{v}{\lambda}$$

Where $v$ is the speed of light in the specific medium. Since the light is propagating through a silica optical fiber, we must first calculate the speed of light within this fiber using its refractive index. The speed of light in a medium is given by:

$$v = \frac{c}{n}$$

Substituting the values:

$$v = \frac{3 \times 10^8 \text{ m/s}}{1.5} = 2 \times 10^8 \text{ m/s}$$

Bandwidth from Spectral Width Derivation

To find the bandwidth ($\Delta f$) corresponding to a given spectral width ($\Delta\lambda$), we can differentiate the frequency-wavelength relationship with respect to wavelength:

$$f = \frac{v}{\lambda}$$

Taking the derivative of $f$ with respect to $\lambda$:

$$\frac{df}{d\lambda} = -\frac{v}{\lambda^2}$$

For small changes, the magnitude of the bandwidth ($\Delta f$) can be approximated as:

$$\Delta f = \left| \frac{df}{d\lambda} \right| \Delta\lambda = \frac{v}{\lambda^2} \Delta\lambda$$

Numerical Calculation of Bandwidth

Let's convert the given wavelengths from nanometers (nm) to meters (m):

• Spectral width ($\Delta\lambda$) = 1 nm = $1 \times 10^{-9}$ m
• Central Wavelength ($\lambda$) = 1300 nm = $1300 \times 10^{-9}$ m = $1.3 \times 10^{-6}$ m

Now, substitute the values into the derived formula:

$$\Delta f = \frac{2 \times 10^8 \text{ m/s}}{(1.3 \times 10^{-6} \text{ m})^2} \times (1 \times 10^{-9} \text{ m})$$

First, calculate the square of the wavelength:

$$(1.3 \times 10^{-6} \text{ m})^2 = 1.69 \times 10^{-12} \text{ m}^2$$

Now, substitute this back into the equation for $\Delta f$:

$$\Delta f = \frac{2 \times 10^8}{1.69 \times 10^{-12}} \times 10^{-9}$$

$$\Delta f = \frac{2}{1.69} \times 10^{8 + 12 - 9}$$

$$\Delta f = \frac{2}{1.69} \times 10^{11} \text{ Hz}$$

$$\Delta f \approx 1.1834 \times 10^{11} \text{ Hz}$$

To convert this frequency to Gigahertz (GHz), we divide by $10^9$ (since 1 GHz = $10^9$ Hz):

$$\Delta f \approx 1.1834 \times 10^{11} \times 10^{-9} \text{ GHz}$$

$$\Delta f \approx 118.34 \text{ GHz}$$

Rounding this value to the nearest option, we get approximately 120 GHz.