Question

The attenuation of light in an optical fiber is estimated at 2.2dB/km. What  Fractional initial intensity remains after 2 km & 6 km?

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Answer 1

Explanation:

For multimode fiber, the loss is about 3 dB per km for 850 nm sources, 1 dB per km for 1300 nm. (3.5 and 1.5 dB/km max per EIA/TIA 568) This roughly translates into a loss of 0.1 dB per 100 feet (30 m) for 850 nm, 0.1 dB per 300 feet(100 m) for 1300 nm.

Answer 2

Answer:

0.363 and 0.048 are the respective fractional intensities that are left after 2 and 6 kilometers of travelled light.

Explanation:

As the value of coefficient of attenuation,

$\alpha =-\frac{10}{L} log(\frac{P_{out} }{P_{in} } )$   --------(1)

and the given value of α=2.2db/km

Part1:

Fractional Intensity after light has transverses 2Km.

L=2Km

Putting the values of L and α in the equation (1) ,

$2.2 =-\frac{10}{2} log_{10} (\frac{P_{out} }{P_{in} } )\\log_{10} (\frac{P_{out} }{P_{in} } )=-0.44\\\frac{P_{out} }{P_{in} }=10^{-0.44} \\\frac{P_{out} }{P_{in} } =0.363$

Hence, 0.363 is the fractional intensity of light left after 2 Kilometers.

Part2:

Fractional Intensity after light has transverses 6Km.

L=6Km

Putting the values of L and α in the equation (1) ,

$2.2 =-\frac{10}{6} log_{10} (\frac{P_{out} }{P_{in} } )\\log_{10} (\frac{P_{out} }{P_{in} } )=1.32\\\frac{P_{out} }{P_{in} }=10^{-1.32} \\\frac{P_{out} }{P_{in} } =0.048$

Hence, 0.048 is the fractional intensity of light left after 6 Kilometers.