In a communication system operating at a wavelength of $800\,nm,$ only one percent of the source frequency is available as signal bandwidth. The number of channels accommodated for transmitting $TV$ signals of bandwidth $6\,MHz$ is (Take velocity of light $c = 3 \times 10^8\,m/s$)

  • A
    $3.75 \times 10^6$
  • B
    $3.86 \times 10^6$
  • C
    $6.25 \times 10^5$
  • D
    $4.87 \times 10^5$

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