A) 2 million
B) 10 million
C) \[0.1\] million
D) 1 million
Correct Answer: D
Solution :
| Optical source frequency \[\text{v=}\frac{c}{\lambda }\] \[\Rightarrow \,\,\,\text{v=}\frac{3\times {{10}^{8}}m{{s}^{-1}}}{1200\times {{10}^{-9}}m}=2.5\times {{10}^{14}}Hz\] |
| Bandwidth of channel (\[2%\]of the source frequency) \[=5\times {{10}^{12}}Hz\] |
| Number of channels \[=\frac{\text{Total band}\,\text{width}}{\text{ }\!\!~\!\!\text{ Bandwidth needed per channel}}\]\[=\frac{5\times {{10}^{12}}Hz}{5\times {{10}^{16}}Hz}={{10}^{6}}=1\,million\] |
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