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The intensity of a light pulse travellin...

The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation `I=I_0 e^(-alpha x)`, where `I_0` is the intensity at x = 0 and `alpha` is the attenuation constant. What is the distance travelled by the wave, when the intensity reduces by 75% of its initial intensity?

A

`(log_e4)/alpha`

B

`(log_e2)/alpha`

C

`alpha/(log_e 4)`

D

`alpha/(log_e 2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`I=I_0e^(-alpha x)` is the intensity of the light pulse.
When the initial intensity is reduced by 75 % , the intensity becomes (25%) `I=I_0/4`
`because I=I_0 e^(-alpha x ) therefore I_0/4 = I_0e^(-alpha x ) therefore 1/4 = e^(-alpha x )`
Taking logs on both sides , we get
`therefore log_e 1-log_e 4=-alphax`
`therefore log_e4=alphax`
`therefore x=(log_e4)/alpha`
`therefore` The intensity is reduced by 75% at a distance of `(log_e 4)/alpha `
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