Question

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

• A. 2 million
• B. 10 million
• C. 0.1 million
• D. 1 million

Answer 1

Answer: (D)

1 million

Answer 2

: Optical source frequency, $\upsilon = \frac{c}{\lambda}$

Or $\upsilon = \frac{3 \times 10^{8}ms^{- 1}}{1200 \times 10^{- 9}m} = 2.5 \times 10^{14}Hz$

Bandwidth of channel= 2% of the source frequency

$= \frac{2}{100} \times 2.5 \times 10^{14}Hz = 5 \times 10^{12}Hz$

Number of channels

$\frac{TotalbandwidthofChannel}{Bandwidthneededperchannel}$

$= \frac{5 \times 10^{12}Hz}{5 \times 10^{6}Hz} = 10^{6} = 1million$