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In a communication system operating at w...

In a communication system operating at wavelength 800 nm, only one percent of source frequency is a available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light `c=3xx10^(8)m//s, h=6.6xx10^(-34)J-s`)

A

`6.25 xx 10^(5)`

B

`4.87 xx 10^(5)`

C

`3.75 xx 10^(6)`

D

`3.86 xx 10^(6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`lambda = 800 nm`
`f = (3 xx 10^8)/(800 xx 10^(-9)) = (3000)/(800) xx 10^(14) Hz`
Now 1% = `(3000)/(800) xx 10^(14) xx 1`
Number of channels = `(3/8 xx 10^(13))/(6 xx 10^6) = 3/48 xx 10^7 = (300)/48 xx 10^5 = 6.25 xx 10^(5) = 6.25 xx 10^(5)`.
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