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 $c = 3 \times 10^8 m/s,h = 6.6 \times 10^{-34} J-s)$

Solution:

$f = \frac{3 \times10^{8}}{8 \times10^{-7}} = \frac{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 $= \frac{3.75 \times10^{6}}{6} = 6.25 \times10^{5} $
$\therefore$ correct answer is (4)

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