A) \[4.8\times {{10}^{8}}\]
B) 48
C) \[6.2\times {{10}^{8}}\]
D) \[4.8\times {{10}^{5}}\]
Correct Answer: A
Solution :
[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) \[\Rightarrow \]Number of channels for audio signal \[=(3.8\times {{10}^{12}})/(8\times {{10}^{3}})=4.8\times {{10}^{8}}\]You need to login to perform this action.
You will be redirected in
3 sec