Answer to Question #138977 in Optics for Yashwanth

Question

3. The photo detector output in a cutback attenuation set up is 3.5V at the far end of the fiber. After cutting the fiber 2m at the near end, 5km from the far end, photo detector output read was 4.2 V. What is the attenuation of fiber in dB/km?

Answer 1

As per the given question,

$(V_1)=3.5V$

$L_1=2m$

$L_2=5km= 5000m$

$V_2=4.2V$

$\Rightarrow \alpha_{dB}=[\frac{10}{(L1-L2)}] \times log_{10}(\frac{V_2}{V_1})$

Now, substituting the values in the above,

$\Rightarrow \alpha_{dB}=[\frac{10}{(2-5000)}] \times log_{10}(\frac{4.2}{3.5})$

$\Rightarrow \alpha_{dB}=[\frac{10}{(4998)}] \times log_{10}(1.2)$

$\Rightarrow \alpha_{dB}=-[\frac{10}{(4998)}] \times 0.791$

$\Rightarrow \alpha_{dB}=-0.002\times 0.791$

$\Rightarrow \alpha_{dB}=-0.001582dB/m$

$\Rightarrow \alpha_{dB}=-1.582dB/km$