SATHEE: Modern Physics 7 Question 28

Modern Physics 7 Question 28

29. In communication system, only one percent frequency of signal of wavelength $800 n m$ can be used as bandwidth. How many channal of $6 M H z$ bandwidth can be broadcast this? $(c = 3 \times 10^{8} m / s, h = 6.6 \times 10^{- 34} J - s)$

(Main 2019, 9 Jan II)

(a) $3.75 \times 10^{6}$

(b) $3.86 \times 10^{6}$

(d) $4.87 \times 10^{5}$

Show Answer

Solution:

Here,

Signal wavelength, $λ = 800 n - m = 8 \times 10^{- 7} m$

Frequency of source is

$As, \begin{aligned} f & = \frac{c}{λ} = \frac{3 \times 10^{8}}{8 \times 10^{- 7}} \\ = 3.75 \times 10^{14} H z \end{aligned}$

$∴$ Total bandwidth used for communication

$\begin{aligned} = 1 & = 3.75 \times 10^{12} H z \end{aligned}$

So, number of channel for signals $= \underset{―}{total bandwidth available for communication}$

bandwidth of TV signal

$= \frac{3.75 \times 10^{12}}{6 \times 10^{6}} = 0.625 \times 10^{6} = 6.25 \times 10^{5}$

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Modern Physics 7 Question 28

29. In communication system, only one percent frequency of signal of wavelength 800nm can be used as bandwidth. How many channal of 6MHz bandwidth can be broadcast this? (c=3×108m/s,h=6.6×10−34J−s)

Solution:

bandwidth of TV signal

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29. In communication system, only one percent frequency of signal of wavelength $800 n m$ can be used as bandwidth. How many channal of $6 M H z$ bandwidth can be broadcast this? $(c = 3 \times 10^{8} m / s, h = 6.6 \times 10^{- 34} J - s)$