A) 2 million
B) 10 million
C) 0.1 million
D) 1 million
E) 0.5 million
Correct Answer: D
Solution :
The frequency for optical communication \[v=\frac{c}{\lambda }\] or \[v=\frac{3\times {{10}^{8}}}{1200\times {{10}^{-9}}}\] \[v=25\times {{10}^{13}}Hz\] But only 2% of the source frequency is available for TV transmission. \[\therefore \] \[v=2.5\times {{10}^{14}}\times 2%\] or \[v=2.5\times {{10}^{14}}\times \frac{2}{100}\] or \[v=5\times {{10}^{12}}Hz\] \[\therefore \]Number of channels\[=\frac{v}{bandwidth}\] or number of channels\[=\frac{5\times {{10}^{12}}}{5\,MHz}\] or number of channels\[=\frac{5\times {{10}^{12}}}{5\times {{10}^{6}}}\] \[={{10}^{6}}\] = 1 million
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