# Power loss in optical fibers

1.27 Power loss in optical fibers and the use of dB. An optical fiber is rated as having a loss of 4 dB/km. Given an input light power density of 10$\, \mathrm{mW}/\mathrm{mm}^2$, what is the light power density at the end of a fiber 6 km long?

## Solution

The input power density $\mathbf{P_{in}}$ is: $10 \, \mathrm{mW}/\mathrm{mm}^2$

The total attenuation along the line:

$$4 \, \mathrm{dB/km} \times 6 \, \mathrm{km} = 24 \, \mathrm{dB}$$

We need to calculate the input power density in dB:

$$\mathbf{P_{in}} = 10 \, \mathrm{mW}/\mathrm{mm}^2 \rightarrow P_{in} = 10 \log_{10} \left(\frac{10 \, \mathrm{mW}/\mathrm{mm}^2}{1 \, \mathrm{mW}/\mathrm{mm}^2}\right) = 10 \, \mathrm{dB}$$

Note that we have arbitrarily defined as reference (the denominator) $1 \, \mathrm{mW}/\mathrm{mm}^2$. We need to keep this in mind when converting back to power density.

The output power density in dB:

$$P_{out} = 10 \, \mathrm{dB} - 24 \, \mathrm{dB} = - 14 \, \mathrm{dB}$$

Now we convert this back to power density:

$$-14 \, \mathrm{dB} = 10 \log_{10} \left( \frac{\mathbf{P_{out}}}{1 \, \mathrm{mW}/\mathrm{mm}^2} \right)$$$$10^{-1.4} = \frac{\mathbf{P_{out}}}{1 \, \mathrm{mW}/\mathrm{mm}^2}$$$$\therefore \; \mathbf{P_{out}} = 0.0398 \, \mathrm{mW}/\mathrm{mm}^2$$$$\mathbf{P_{out}} = 39.8 \, \mathrm{\mu W}/\mathrm{mm}^2$$

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