Question

The attenuation in an optical fiber is 4.2 dB/km. What fraction of its initial intensity remains after

traveling 3km and 5km​

Answer 1

The intensity received is 5% for 3 km and 0.79% for 5 km.

Here, according to the given information, we are given that,

The attenuation in an optical fiber is 4.2 dB/km.

This means that the coefficient of attenuation is $\alpha$ = 4.2 dB/km.

Now, we need to find the fraction of its initial intensity remains after

traveling 3km and 5km.

For the case of 3 kms, we have,

Let L be the length of the optical fiber. Here, L is equal to 3 km.

We need to find the fractional intensity that is $\frac{P_{0} }{P_{i} }$.

Then, the attenuation constant is given as,

$-\frac{\alpha L}{10} =log_{10} \frac{P_{0} }{P_{i}}$

Or, we can write this as,

$\frac{P_{0} }{P_{i}} =10^{\frac{-\alpha L}{10} }$

Then, putting the values, we get,

$\frac{P_{0} }{P_{i}} =10^{\frac{-(4.2).3}{10} } = 0.05$

Then, the intensity received is 5%

Similarly, in the case of 5 km, we get,

Let L be the length of the optical fiber. Here, L is equal to 5 km.

We need to find the fractional intensity that is $\frac{P_{0} }{P_{i} }$.

Then, the attenuation constant is given as,

$-\frac{\alpha L}{10} =log_{10} \frac{P_{0} }{P_{i}}$

Or, we can write this as,

$\frac{P_{0} }{P_{i}} =10^{\frac{-\alpha L}{10} }$

Then, putting the values, we get,

$\frac{P_{0} }{P_{i}} =10^{\frac{-(4.2).5}{10} } = 0.0079$

Then, the intensity received is 0.79%

Hence, the intensity received is 5% for 3 km and 0.79% for 5 km.

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