(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?

Explanation

(a) Given, the intensity of a light pulse `I=I_(0)e^(-alphax)`,
where, `I_(0)` is the intensity at x = 0 and `alpha` is constant.
According to the question, I = 25% of `I_(0)=(25)/(100).I_(0)=(I_(0))/(4)`
Using the formula mentioned in the question,
`I=I_(0)e^(-alphax)`
`(I_(0))/(4)=I_(0)e^(-alphax)`
or `(1)/(4) = e^(-alphax)`
Taking log on both sides, we get
`ln1-ln4=-alphaxlne" "(becauselne=1)`
`-ln4=-alphax`
`x=(ln4)/(alpha)`
Therefore, at distance `x=(ln4)/(alpha)`, the intensity is reduced to 75% of initial intensity. (b) Let `alpha` be the attenuation in dB/km. If x is the distance travelled by signal, then
`10log_(10)((I)/(I_(0)))=-alphax " ....(i)"`
where, `I_(0)` is the intensity initially.
According to the question, I = 50 % of `I_(0)=(I_(0))/(2)` and x = 50 km
Putting the value of x in Eq. (i), we get
`10log_(10).(I_(0))/(2I_(0))=-alphaxx50`
`10[log1-log2]=-50 alpha`
`(10xx0.3010)/(50)=alpha`
`therefore` The attenuation for an optical fibre
`alpha = 0.0602` dB/km