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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)` |
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Answer» Correct Answer - 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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