What is the worst case acceptance value for attenuation when testing 2000 m (6562 ft) of 50/125 multimode fiber at a wavelength of 1300 nm when the fiber has 2 connector pairs and 3 splices?

Question

Quizplus · Accepted Answer

The attenuation values for connectors and splices are:

- Connector pair: 0.75 dB (0.15 dB x 2)
- Splice: 0.30 dB x 3 = 0.90 dB

The worst case attenuation value calculation is as follows:

- Attenuation of the fiber: 3.5 dB/Km x 2 Km = 7 dB
- Attenuation for connectors: 0.75 dB x 2 = 1.5 dB
- Attenuation for splices: 0.9 dB x 3 = 2.7 dB
- Total attenuation: 7 dB + 1.5 dB + 2.7 dB = 11.2 dB
- Attenuation per km: 11.2 dB / 2 km = 5.6 dB/km

The worst case acceptance value for attenuation can be calculated by multiplying the attenuation per km with the length of the fiber and adding the attenuation for connectors and splices. Therefore, the worst case acceptance value for attenuation is:

- Attenuation: 5.6 dB/km x 2 km + 1.5 dB + 2.7 dB = 5.4 dB

Therefore, the best choice is C, with a worst-case acceptance value for attenuation of 5.4 dB.

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