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 accommodated for transmitting $ T V $ signals of band width $ 6 \mathrm{MHz} $ are (Take velocity of light $ c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, h=6.6 \times 10^{-34} J_{-s} $ )

Question

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 accommodated for transmitting $ T V $ signals of band width $ 6 \mathrm{MHz} $ are (Take velocity of light $ c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, h=6.6 \times 10^{-34} J_{-s} $ ), $ 6.25 \times 10^{5} $, $ 4.87 \times 10^{5} $, $ 3.75 \times 10^{6} $, $ 3.86 \times 10^{6} $

MARKS App · Accepted Answer

Available signal width = 1 100 × c λ = 1 100 × 3 × 10 8 800 × 10 - 9 H z = 3.75 × 10 12 H z Number of available channels = 3.75 × 10 12 6 × 10 6 = 6.25 × 10 5

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