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Question:

In a communication system operating at wavelength 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 are (Take velocity of light c=3×10⁸m/s)

3.75×10⁶

3.86×10⁶

4.87×10⁵

6.25×10⁵

Solution:

The correct option is D
6.25×10⁵
f=c/λ = (3×10⁸)/(800×10⁻⁹) = 3.75×10¹⁴ Hz
Signal bandwidth = 1% of source frequency = 0.01 × 3.75×10¹⁴ Hz = 3.75×10¹² Hz
Number of channels = (Signal bandwidth) / (bandwidth of each TV signal) = (3.75×10¹² Hz) / (6×10⁶ Hz) = 6.25×10⁵