Mean optical power launched into an 8 km fib | vedclass

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Question

(A) The attenuation in an optical fibre is given by the formula: $	ext{Attenuation} = 10 \log_{10} \left( \frac{P_{	ext{in}}}{P_{	ext{out}}} ight

Vedclass · Accepted Answer

$14.77$

Vedclass · Answer

(A) The attenuation in an optical fibre is given by the formula: $	ext{Attenuation} = 10 \log_{10} \left( \frac{P_{	ext{in}}}{P_{	ext{out}}} ight) \ dB$. Given: $P_{	ext{in}} = 120 \ \mu W$ and $P_{	ext{out}} = 4 \ \mu W$. Substituting the values: $	ext{Attenuation} = 10 \log_{10} \left( \frac{120}{4} ight) = 10 \log_{10} (30)$. Using the given value $\log_{10} 30 = 1.477$: $	ext{Attenuation} = 10 	imes 1.477 = 14.77 \ dB$.

Mean optical power launched into an $8 \ km$ fibre is $120 \ \mu W$ and mean output power is $4 \ \mu W$,then the overall attenuation is ....... $dB$ (Given $\log 30 = 1.477$)

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