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(A) The frequency of the optical source is given by $f = \frac{c}{\lambda}$.Substituting the values: $f = \frac{3 	imes 10^{8} \, m/s}{800 	imes 10^

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$4.8 	imes 10^{8}$

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(A) The frequency of the optical source is given by $f = \frac{c}{\lambda}$.Substituting the values: $f = \frac{3 	imes 10^{8} \, m/s}{800 	imes 10^{-9} \, m} = 3.75 	imes 10^{14} \, Hz$.The available channel bandwidth is $1\%$ of the optical source frequency:$	ext{Bandwidth} = 0.01 	imes 3.75 	imes 10^{14} \, Hz = 3.75 	imes 10^{12} \, Hz$.The number of channels that can be accommodated for audio signals requiring $8 \, kHz$ $(8 	imes 10^{3} \, Hz)$ is:$	ext{Number of channels} = \frac{	ext{Total Bandwidth}}{	ext{Bandwidth per channel}} = \frac{3.75 	imes 10^{12}}{8 	imes 10^{3}} = 0.46875 	imes 10^{9} = 4.6875 	imes 10^{8}$.Rounding to the nearest provided option,we get $4.8 	imes 10^{8}$.

Consider an optical communication system operating at $800 \, nm$. Suppose,only $1\%$ of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting audio signals requiring a bandwidth of $8 \, kHz$?

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