In a communication system operating at wavelength $800 \mathrm{~nm}$, only one per cent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of bandwidth $6 \mathrm{MHz}$ is

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Aicharya · Accepted Answer

Given that, wavelength, $\lambda=800 \mathrm{~nm}$
speed of light, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Then, frequency, $f=c / \lambda$
Substituting the values, we get
Source frequency, $f=\frac{3 \times 10^8}{800 \times 10^{-9}} \mathrm{~Hz}$
$=3.75 \times 10^{14} \mathrm{~Hz}$
It is said in the question that only $1 \%$ of source frequency is available as signal bandwidth.
$\therefore$ Total available signal bandwidth
$\begin{aligned}
& =1 \% \text { of } 3.75 \times 10^{14} \\
& =\frac{1}{100} \times 3.75 \times 10^{14} \mathrm{~Hz}=3.75 \times 10^{12} \mathrm{~Hz}
\end{aligned}$
Given that, the frequency bandwidth of one signal
$=6 \mathrm{MHz}=6 \times 10^6 \mathrm{~Hz}$
Hence, the number of channels
$\begin{aligned}
& =\frac{\text { Total available signal bandwidth }}{\text { Bandwidth of one signal }} \\
& =\frac{3.75 \times 10^{12}}{6 \times 10^6}=\frac{1}{16} \times 10^7
\end{aligned}$

In a communication system operating at wavelength $800 \mathrm{~nm}$, only one per cent of source frequency…

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