Discussion
It is desired to send a sequence of computer screen images over an optical fiber. The screen is 480 x 640 pixels, each pixel being 24 bits. There are 60 screen images per second. How much bandwidth is needed, and how many microns of wavelenth are needed for this band at 1.30 microns?
Answer: C
Correct answer: 1.5 × 10−6 microns
The screen needs
\[ 480\times 640\times 24\times 60 \approx 4.42\times 10^{8}\ \text{bps} \]
and the spectral width for that bandwidth at \( \lambda = 1\ \mu\text{m} \) is
\[ \Delta\lambda = \frac{\lambda^{2}\Delta f}{c} = \frac{(10^{-6})^{2}\times 4.42\times 10^{8}}{3\times 10^{8}} \approx 1.5\times 10^{-12}\ \text{m} \]
which is \( 1.5\times 10^{-6} \) microns - a minute slice of the fibre's available spectrum.
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