In a communication system operating at wavelength 800 $\mathrm{nm}$, only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width $6 \mathrm{MHz}$ are (Take velocity of light $\left.\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, \mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$
Correct Option: , 3
(3) Frequency, $\mathrm{f}=\frac{V}{\lambda}=\frac{3 \times 10^{8}}{8 \times 10^{-7}}=\frac{30}{8} \times 10^{14} \mathrm{~Hz}$
$=3.75 \times 10^{14} \mathrm{~Hz}$
$1 \%$ of f $=0.0375 \times 10^{14} \mathrm{~Hz}$
$=3.75 \times 10^{12} \mathrm{~Hz}=3.75 \times 10^{6} \mathrm{MHz}$
As we know, number of channels accomodated for transmission $=$
$\frac{\text { total bandwidth of Channel }}{\text { bandwidth needed perchannel }}=\frac{3.75 \times 10^{6}}{6}=6.25 \times 10^{5}$
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