Verified Solution • 2025–26 Session moderate

QJEEM19Ph9aa1: JEE Main 2019 Physics - Class 12 JEEM19Ph9aa1 Solution: In a communication system operating at wavelength 800 nm, only…

In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwith. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 ×\times 108m/s, h = 6.6 ×\times 10 −- 34 J-s)

Multiple Choice Options

  • A 3.75 ×\times 106
  • B 3.86 ×\times 106
  • C 6.25 ×\times 105
  • D 4.87 ×\times 105

Step-by-Step Verified Solution

Given,

λ\lambda = 800 nm

∴\therefore   f = 3×108800×10−9{{3 \times {{10}^8}} \over {800 \times {{10}^{ - 9}}}}

= 3.75 ×\times 1014 Hz

Available frequency for signal bandwith

= 1% of F

= 3.75 ×\times 1014 ×\times 1100{1 \over {100}}

= 3.75 ×\times 1012 Hz

One TV signal needs = 6 MHz band width

∴\therefore   Total number of channel possible

= 3.75×10126×106{{3.75 \times {{10}^{12}}} \over {6 \times {{10}^6}}}

= 6.25 ×\times 105

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