A) 1 million
B) 50 million
C) 1 thousand
D) 50 thousand
Correct Answer: A
Solution :
[a] Optical source frequency \[v=\frac{c}{\lambda }\] \[\Rightarrow \,\,\,\,\,\,\,\,v=\frac{3\times {{10}^{8}}m{{s}^{-1}}}{1200\times {{10}^{6}}m}=2.5\times {{10}^{14}}Hz\] Bandwidth of channel (\[2%\]of the source frequency) \[=5\times {{10}^{12}}Hz\] Number of channels \[=\frac{Total\text{ }bandwidth}{Bandwidth\text{ }needed\text{ }per\text{ }channel}\] \[=\frac{5\times {{10}^{12}}Hz}{5\times {{10}^{6}}Hz}={{10}^{6}}=1\] million.
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