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Question

(A) The frequency of the optical source is given by $f = c / \lambda = (3 	imes 10^{8} \, m/s) / (800 	imes 10^{-9} \, m) = 3.75 	imes 10^{14} \, H

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$4.8 	imes 10^{8}$

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(A) The frequency of the optical source is given by $f = c / \lambda = (3 	imes 10^{8} \, m/s) / (800 	imes 10^{-9} \, m) = 3.75 	imes 10^{14} \, Hz$.The available bandwidth for communication is $1\%$ of the source frequency:$\Delta f = 0.01 	imes 3.75 	imes 10^{14} \, Hz = 3.75 	imes 10^{12} \, Hz$.The number of channels that can be accommodated for a bandwidth of $8 \, kHz$ $(8 	imes 10^{3} \, Hz)$ is:$N = \Delta f / 	ext{bandwidth per channel} = (3.75 	imes 10^{12} \, Hz) / (8 	imes 10^{3} \, Hz) \approx 4.6875 	imes 10^{8} \approx 4.8 	imes 10^{8}$.

Consider an optical communication system operating at a wavelength of $\lambda \sim 800 \, nm$. If only $1\%$ of the optical source frequency is available for optical communication,how many channels can be accommodated for a bandwidth of $8 \, kHz$?

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