Question:
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated 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)
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated 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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Source frequency = c/lambda; usable bandwidth = 1% of it; number of channels = usable bandwidth / 6 MHz.
Updated On: Nov 5, 2026
- \(3.75 \times 10^{6}\)
- \(3.86 \times 10^{6}\)
- \(6.25 \times 10^{5}\)
- \(4.87 \times 10^{5}\)
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The Correct Option is C
Solution and Explanation
Source frequency \(\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{800 \times 10^{-9}} = 3.75 \times 10^{14}\) Hz.
Available signal bandwidth \(= 1\% \text{ of } \nu = 3.75 \times 10^{12}\) Hz.
Number of channels \(= \frac{3.75 \times 10^{12}}{6 \times 10^{6}} = 6.25 \times 10^{5}\).
Answer: (3) \(6.25 \times 10^{5}\)
Available signal bandwidth \(= 1\% \text{ of } \nu = 3.75 \times 10^{12}\) Hz.
Number of channels \(= \frac{3.75 \times 10^{12}}{6 \times 10^{6}} = 6.25 \times 10^{5}\).
Answer: (3) \(6.25 \times 10^{5}\)
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