Question

In a communication system operating at wavelength 800nm, 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 $c = 3 \times {10^8}m{s^{ - 1}}$,$h = 6.6 \times {10^{ - 34}}Js$)
a. $3.75 \times {10^6}$
b. $4.87 \times {10^5}$
c. $3.86 \times {10^6}$
d. $6.25 \times {10^5}$

Answer 1

The given problem is from Communication System. It is based on the relation between frequency and wavelength. So, first we should calculate the frequency then the number of channels can be calculated with the help of available frequency and total bandwidth that will be transmitted.

Complete step by step answer:
According to the question, the given wavelength of the communication system is 800nm. So with the help of wavelength we can calculate the source frequency using this formula-
$f = \dfrac{c}{\lambda }$
Where, f= Frequency (Hz)
c= Speed of light ( $3 \times {10^8}m{s^{ - 1}}$)
$\lambda$= Wavelength
So, we put the given value in this formula to calculate source frequency-
$\lambda = 800nm = 800 \times {10^{ - 9}}m$
$f = \dfrac{{3 \times {{10}^8}}}{{800 \times {{10}^{ - 9}}}} \\\ \Rightarrow f = 3.57 \times {10^{14}}Hz \\\$
The total source frequency is $3.75 \times {10^{14}}Hz$

The condition in question is given as only one percent of the source frequency is available as signal bandwidth. So we should calculate the frequency for signal bandwidth using source frequency.
Available frequency= 1% of Source frequency
Available frequency= 1% of $3.75 \times {10^{14}}Hz$
Available frequency= $3.75 \times {10^{14}}Hz$
We know that $1MHz = {10^6}Hz$. So the available frequency in MHz is given by-
Available frequency= $\dfrac{{3.75 \times {{10}^{12}}}}{{{{10}^6}}}$
Available frequency is = $3.73 \times {10^6}MHz$
The number of channels accommodated for transmitting TV signals of bandwidth 6 MHz are given by-
$\text{No. of channels}= \dfrac{\text{Available frequency}}{\text{Total bandwidth}}$
No. of channels=$\dfrac{{3.73 \times {{10}^6}}}{6} = 6.25 \times {10^5}$

Hence, the correct answer is option (D).

Note: Sometimes we don’t focus on the conversion of Hz into MHz for available frequencies that make the result wrong. So always use the same unit for the same quantity in a single formula to calculate right values.