Question

Only $2\%$ of the optical source frequency is the available channel bandwidth for an optical communicating system operating at $1000nm$. If an audio signal requires a bandwidth of $8kHz$, 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 $1000nm=\frac{3\times 10}{1000\times 10^{-9}}$
$\Rightarrow 3\times 10^{14}Hz$ available for channel band width
$=\frac{2}{100}\times 3\times 10^{14}$
$\Rightarrow 6\times 10^{12}Hz$
Bandwidth for 1 channel $=8000Hz$
$\therefore$ No. of channel
$=\frac{6\times 10^{12}}{8\times 10^3}$
$\Rightarrow \frac{600}{8}\times 10^7=75\times 10^7$