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(B) The frequency of the optical source is given by $f = \frac{c}{\lambda} = \frac{3 	imes 10^{8} \,m/s}{1000 	imes 10^{-9} \,m} = 3 	imes 10^{14}

Vedclass · Accepted Answer

$75 	imes 10^{7}$

Vedclass · Answer

(B) The frequency of the optical source is given by $f = \frac{c}{\lambda} = \frac{3 	imes 10^{8} \,m/s}{1000 	imes 10^{-9} \,m} = 3 	imes 10^{14} \,Hz$. The available channel bandwidth is $2 \%$ of this frequency: $	ext{Bandwidth} = \frac{2}{100} 	imes 3 	imes 10^{14} \,Hz = 6 	imes 10^{12} \,Hz$. The bandwidth required for one audio signal channel is $8 \,kHz = 8 	imes 10^{3} \,Hz$. The number of channels that can be accommodated is given by the ratio of the total available bandwidth to the bandwidth per channel: $	ext{Number of channels} = \frac{6 	imes 10^{12} \,Hz}{8 	imes 10^{3} \,Hz} = 0.75 	imes 10^{9} = 75 	imes 10^{7}$.

Column $A$	Column $B$
$A$. $D$-layer	$I$. $250-400 \ km$
$B$. $E$-layer	$II$. $170-190 \ km$
$C$. $F_1$-layer	$III$. $95-120 \ km$
$D$. $F_2$-layer	$IV$. $65-75 \ km$

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

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