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A low-loss fiber has an average loss of 3 dB/km at 900 nm. Compute the length over which, a) Power decreases by 50% b) Power decreases by 75%.
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written 17 months ago by
prajapatijaimin
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Solution:
α = 3 dB/km
$ \frac{p(0)}{p(z)}=50 \%=0.5\\ $
a) Power decreases by 50 %:
$ \begin{aligned}\\ & {\left[\frac{200 \mu \mathrm{W}}{\mathrm{P}(\mathrm{z})}\right]=10^{2.4}} \\\\ & 3=10 \cdot \frac{1}{\mathrm{z}} \log [0.5]\\ \end{aligned}\\ $
z = 1 km… Ans.
$ \text { b) } \frac{\mathrm{p}(0)}{\mathrm{p}(\mathrm{z})}=25 \%=0.25 \text { Since power decrease by } 75%: …
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