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The input power to an optical fiber is $2 \mathrm{~mW}$ while the power measured at the output end is $2 \mu \mathrm{W}$. If the fiber attenuation is $0.5 \mathrm{~dB} / \mathrm{km}$, calculate the le
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written 17 months ago by
prajapatijaimin
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Solution:
$ \begin{aligned} P(0) & =2 \text { mwatt }=2 \times 10^{-3} \text { watt } \\\\ P(z) & =2 \mu \mathrm{watt}=2 \times 10^{-6} \text { watt } \\\\ \alpha & =0.5 \mathrm{~dB} / \mathrm{km}\\ \end{aligned} $
$ \alpha=10 \times \frac{1}{z}\left[\frac{p(0)}{p(z)}\right]\\ $
$ \begin{aligned} & 0.5=\frac{1}{\mathrm{z}} \mathrm{x} 3 \\\\ & …
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