Question

Only $2 \%$ of the optical source frequency is the available channel bandwidth for an optical communicating system operating at $1000 \,nm$. If an audio signal requires a bandwidth of $8 \,kHz$, how many channels can be accommodated for transmission :

• A. $375 \times 10^{7}$
• B. $75 \times 10^{7}$
• C. $375 \times 10^{8}$
• D. $75 \times 10^{9}$

Answer 1

Answer: (B)

Frequency at $1000\, nm =\frac{3 \times 10}{1000 \times 10^{-9}} $
$\Rightarrow 3 \times 10^{14} Hz$ available for channel band width
$=\frac{2}{100} \times 3 \times 10^{14}$
$ \Rightarrow 6 \times 10^{12} Hz$
Bandwidth for 1 channel $=8000\, Hz$
$\therefore$ No. of channel
$=\frac{6 \times 10^{12}}{8 \times 10^{3}}$
$ \Rightarrow \frac{600}{8} \times 10^{7}=75 \times 10^{7}$