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(C) The frequency of the source is given by $f = \frac{c}{\lambda}$.Substituting the values,$f = \frac{3 	imes 10^8\,m/s}{800 	imes 10^{-9}\,m} = \f

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$6.25 	imes 10^5$

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(C) The frequency of the source is given by $f = \frac{c}{\lambda}$.Substituting the values,$f = \frac{3 	imes 10^8\,m/s}{800 	imes 10^{-9}\,m} = \frac{3 	imes 10^8}{8 	imes 10^{-7}} = 0.375 	imes 10^{15}\,Hz = 3.75 	imes 10^{14}\,Hz$.The available signal bandwidth is $1\%$ of the source frequency,so $	ext{Bandwidth} = 0.01 	imes f = 0.01 	imes 3.75 	imes 10^{14}\,Hz = 3.75 	imes 10^{12}\,Hz$.The number of channels $n$ that can be accommodated for $TV$ signals of bandwidth $6\,MHz$ $(6 	imes 10^6\,Hz)$ is given by $n = \frac{	ext{Total Bandwidth}}{	ext{Bandwidth per channel}}$.$n = \frac{3.75 	imes 10^{12}}{6 	imes 10^6} = 0.625 	imes 10^6 = 6.25 	imes 10^5$.

In a communication system operating at a wavelength of $800\,nm,$ only one percent of the source frequency is available as signal bandwidth. The number of channels accommodated for transmitting $TV$ signals of bandwidth $6\,MHz$ is (Take velocity of light $c = 3 \times 10^8\,m/s$)

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