Q. Consider an optical communication system operating at $\lambda\sim 800\ nm$. Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting audio signals requiring a bandwidth of $8\ kHz$
A) $4.8 \times 10^{8}$
B) 48
C) $6.2 \times 10^{8}$
D) $4.8 \times 10^{5}$
Explanation
(a) Optical source frequency $f = \frac{c}{\lambda}$
$= 3 \times 10^{8}/(800 \times 10^{- 9}) = 3.8 \times 10^{14}\ Hz$
Bandwidth of channel (1% of above) $= 3.8 \times 10^{12}Hz$
Number of channels = (Total bandwidth of channel)/(Bandwidth needed per channel)
$= (3.8 \times 10^{12})(8 \times 10^{3}) \sim 4.8 \times 10^{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.