Q 12-08-193JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
In a communication system operating at wavelength $800\ \text{nm}$, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width $6\ \text{MHz}$ are (Take velocity of light $c = 3\times10^8\ \text{m/s}$, $h = 6.6\times10^{-34}\ \text{J s}$)
Answer: (A) $6.25\times10^5$
Source frequency $f = \dfrac c\lambda = \dfrac{3\times10^8}{800\times10^{-9}} = 3.75\times10^{14}\ \text{Hz}$.
Available bandwidth $= 1\%$ of this $= 3.75\times10^{12}\ \text{Hz}$.
$$N = \frac{3.75\times10^{12}}{6\times10^6} = 6.25\times10^5$$
Solution by Sreeraj P, M.Sc Physics