In a communication system operating at wavelength 800 ~nm , only one per cent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz is

Step-by-step solution

Given that, wavelength, =800 ~nm speed of light, c=3 10^8 ~m / s Then, frequency, f=c / Substituting the values, we get Source frequency, f= 3 10^8 800 10⁻⁹ ~Hz =3.75 10¹⁴ ~Hz It is said in the question that only 1 % of source frequency is available as signal bandwidth. Total available signal bandwidth aligned & =1 % of 3.75 10¹⁴ & = 1 100 3.75 10¹⁴ ~Hz =3.75 10¹² ~Hz aligned Given that, the frequency bandwidth of one signal =6 MHz =6 10^6 ~Hz Hence, the number of channels aligned & = Total available signal bandwid