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The intensity of a light pulse travelling along a communication channel decreases exponentially with distance $x$ according to the relation $I = {I_0}{e^{ - \alpha x}}$ , where $I_0$ is the intensity at $x = 0$ and $\alpha $ is the attenuation constant. The attenuation in $dB/km$ for an optical fibre in which the intensity falls by $50$ percent over a distance of $50\ km$ is
$0.3010$
$0.0602$
$0.1505$
$0.1204$
Solution
If attenuation of a signal is expressed in $\mathrm{dB}$
$10 \log _{10}\left(\mathrm{I} / \mathrm{I}_{0}\right)=-\alpha \mathrm{x}$
($\alpha$ is the attenuation in $\mathrm{dB} / \mathrm{km}$ )
or $\quad \frac{\mathrm{I}}{\mathrm{I}_{0}}=\frac{\mathrm{I}}{2}$
Thus, $10 \log _{10}\left(\frac{1}{2}\right)=-\alpha \mathrm{X}$
or $10 \log _{10} 2=50 \alpha$
or $\quad \alpha=\frac{\log _{10} 2}{5}=\frac{0.3010}{3}=0.0602 \mathrm{\,dB} / \mathrm{km}$
