Explanation

(d) The frequency 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
$v^{'} = 2.5 \times 10^{14}\ \times 2\%$
Or $v^{'} = 2.5 \times 10^{14}\ \times \frac{2}{100}$
Or $v^{'} = 5 \times 10^{12}\ Hz$
$\therefore Member\ of\ channels = \frac{v^{'}}{bond\ width}$
Or $No\ of\ channels = \frac{{5 \times 10}^{12}}{{5 \times 10}^{6}} = 10^{6} = 1\ million$