Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission

We need to find the number of audio channels that can be accommodated in the available bandwidth of an optical communication system.

First, we find the frequency of the optical source. The wavelength is $$\lambda = 1000$$ nm $$= 10^{-6}$$ m. The frequency is:

$$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{10^{-6}} = 3 \times 10^{14} \text{ Hz}$$

Next, we determine the available channel bandwidth. Only 2% of the source frequency is available:

$$\text{Available bandwidth} = 0.02 \times 3 \times 10^{14} = 6 \times 10^{12} \text{ Hz}$$

Finally, we calculate the number of channels. Each audio signal requires a bandwidth of 8 kHz $$= 8 \times 10^3$$ Hz. Therefore:

$$\text{Number of channels} = \frac{6 \times 10^{12}}{8 \times 10^3} = 0.75 \times 10^{9} = 75 \times 10^7$$

Hence, the correct answer is Option B ($$75 \times 10^7$$).