Question

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 accomodated for transmitting $\text{TV}$ signals of band width $6\text{MHz}$ are
(Take velocity of light $c=3\times {10}^8\text{m/s},h=6.6\times {10}^{-34}\text{J-s}$)

• A. $3.75\times {10}^6$
• B. $3.86\times {10}^6$
• C. $4.87\times {10}^5$
• D. $6.25\times {10}^5$

Answer 1

The correct option is D $6.25\times {10}^5$

As we know, number of channels accomodated for transmission
$=\frac{\text{total bandwidth of channel}}{\text{bandwidth needed per channel}}....\left(1\right)$

Source frequency, $f=\frac{V}{\lambda }=\frac{3\times {10}^8}{8\times {10}^{-7}}=\frac{30}{8}\times {10}^{14}\text{Hz}=3.75\times {10}^{14}\text{Hz}$

Available frequency$=1\%\text{of source frequency}\Rightarrow \frac{1}{100}\times 3.75\times {10}^{14}\text{Hz}$
$=3.75\times {10}^{12}\text{Hz}=3.75\times {10}^6\text{MHz}$
From equation $1,$
Number of channels accomodated for transmission $=\frac{3.75\times {10}^6}{6}=6.25\times {10}^5$