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(B) The frequency of the optical source is given by $f = \frac{c}{\lambda}$.Substituting the values $c = 3 	imes 10^8 	ext{ m/s}$ and $\lambda = 1.3

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$1.15 	imes 10^{10}$

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(B) The frequency of the optical source is given by $f = \frac{c}{\lambda}$.Substituting the values $c = 3 	imes 10^8 	ext{ m/s}$ and $\lambda = 1.3 	imes 10^{-6} 	ext{ m}$:$f = \frac{3 	imes 10^8}{1.3 	imes 10^{-6}} \approx 2.3 	imes 10^{14} 	ext{ Hz}$.The number of subscribers (channels) is the ratio of the total frequency to the bandwidth required per channel:$	ext{Number of subscribers} = \frac{f}{	ext{bandwidth per channel}} = \frac{2.3 	imes 10^{14}}{20 	imes 10^3}$.$= 0.115 	imes 10^{11} = 1.15 	imes 10^{10}$.

An optical fibre communication system works on a wavelength of $1.3 \mu m$. The number of subscribers it can feed if a channel requires $20 \text{ kHz}$ is:

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