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(i) The intensity of a light pulse trave...

(i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation `I = I_0 e^(-alphax)`, where `I_0` is the intensity at ` x = 0 ` and `alpha` is the attenuation constant.
Show that the intensity reduces by 75 percent after a distance of `(ln 4)/(alpha)`
(ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation
dB `= 10log_10 (I//I_0).` What is the attenuation in `dB//km` for an optical fibre in which the intensity falls by 50 percent over a distance of 50 km?

Text Solution

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Given, `I = I_0e^(-alphax) or I/I_0 = e^(-alphax) , log_e.I/I_0 = -alphax or log_e. I_0/I = alphax`
When `x = 1/alpha log_e4, then log_e .I_0/I = alpha xx 1/alphalog_e 4 or I_0/I = 4 or I/I_0 = 1/4`
% decrease in intensity `= ((I_0-I)/I_0) xx 100 = (1-I/I_0) xx 100 = (1-1/4) xx 100 = 75%`
(ii) Let `alpha` be the attenuation in `dB//km`. If x is the distance travelled by signal, the `10 log_10. I/I_0 = - alpha x`
Given `I/I_0 = 1/2 , x = 50 km :. 10 log_(10) (1/2) = -alpha xx 50 or 10log_(10) 2 = 50 alpha`
or `log_(10) 2 = 5 alpha or alpha = 1/5 log_(10) 2 = 0.3010/5 = 0.0602 dB//km`.
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