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In a communication system operating at wavelength 800nm, only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6MHz are (Take velocity of light c= 3 \times10^{8}m/s, h = 6.6\times 10^{-34}Js)

 

  • Option 1)

    4.87 \times 10^{5}

  • Option 2)

    3.86 \times 10^{6}

  • Option 3)

    6.25\times 10^{5}

  • Option 4)

    3.75 \times 10^{6}

Answers (1)

best_answer

 

Band width -

It is frequency range over which an equipment operate or the portion of the spectrum occupied by the signal.

-

f=\frac{3\times 10^{8}}{8\times 10^{-7}}

     =\frac{30}{8}\times 10^{14} Hz

     =3.75 \times 10^{14} Hz

1% of f =3.75 \times 10^{6} MHz

Number of channel = \frac{3.75 \times 10^{6}}{6}

                              = 6.25 \times 10^{25}

 

 


Option 1)

4.87 \times 10^{5}

Option 2)

3.86 \times 10^{6}

Option 3)

6.25\times 10^{5}

Option 4)

3.75 \times 10^{6}

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