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A fiber of length 10cm is illuminated wi...

A fiber of length `10`cm is illuminated with light from an light emitting diode (LED) which is turned on and off repeatedely for equal amount of time. The speed of the pulse of light are `2.00xx10^(8)m//s` and `2.1xx10^(8)m//s` in fiber. Maximum frequency of LED os that pulse arrive without overlapping is `60X(KHz)`. Calculate `X`.

Text Solution

Verified by Experts

The correct Answer is:
7
`Deltat=(10km)/(2xx10^(8))-(10km)/(2.1xx10^(8))`
`=(10xx10^(3))/(10^(8))[(2.1-2)/4.2]=1/(10^(4))xx1/((42))`
`f=42xx10^(4)=420Khz=60xxKhzimpliesX=7`
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