In optical communication systems operating at 1200nm , only 2 % of...

The frequency of optical communication is given by,

$v = \dfrac{c}{\lambda} or$v = \dfrac{3 \times {10}^{8}}{1200 \times {10}^{- 9}}$<br /><br />$v = 25 \times {10}^{13}Hz$<br /><br />But only$2 %$of the source frequency is available for TV transmission.<br /><br />${v}^{′} = 2 .5 \times {10}^{14} \times 2%$<br /><br />${v}^{′} = 2 .5 \times {10}^{14} \times \dfrac{2}{100}$<br /><br />${v}^{′} = 5 \times {10}^{12}Hz$<br /><br />$∴ Numberofchannels = \dfrac{{v}^{′}}{bandwidth}$<br /><br />Noofchannels = \dfrac{{5 \times 10}^{12}}{{5 \times 10}^{6}} = {10}^{6}$