# Q. In a communication system operating at wavelength 800 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 MHz are (Take velocity of light $c = 3 \times 10^8 m/s,h = 6.6 \times 10^{-34} J-s)$

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Solution:

## $f = \dfrac{3 \times10^{8}}{8 \times10^{-7}} = \dfrac{30}{8} \times10^{14} Hz$ $= 3.75 \times 10^{14} Hz$ $1 \%$ of $f =0.0375 \times10^{14} Hz$ $= 3.75 \times10^{12} Hz = 3.75 \times10^{6}$ number of channels $= \dfrac{3.75 \times10^{6}}{6} = 6.25 \times10^{5}$ $\therefore$ correct answer is (4)

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