72.

In optical communication system operating at 1200 nm, only 2% of the source frequency is available for TV transmission having a bandwidth of 5 MHz. The number of TV channels that can be transmitted is

• 2 million

• 10 million

• 0.1 million

• 1 million

The frequency for optical communication

          $$\begin{gathered} \mathrm{v} = \frac{\mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{or} \mathrm{v} = \frac{3 \times {10}^8}{1200 \times {10}^{-9}} \\ \mathrm{v} = 25 \times {10}^{13} \mathrm{Hz} \end{gathered}$$

But only 2% of the source frequency is available for TV transmission.

∴    v' = 2.5 × 10¹⁴ × 2 %

or   v' = 2.5 × 10¹⁴ × $\frac{2}{100}$

or   v' = 5 × 10¹² Hz

∴  Number of channels = $\frac{\mathrm{v}\text{'}}{\mathrm{bandwidth}}$

or  number of channels = $\frac{5 \times {10}^{12}}{5 \mathrm{MHz}}$

or  number of channels = $\frac{5 \times {10}^{12}}{5 \times {10}^6}$

                                  = 10⁶

                                  = 1 million