Modern Physics 7 Question 28
29. In communication system, only one percent frequency of signal of wavelength $800 nm$ can be used as bandwidth. How many channal of $6 MHz$ bandwidth can be broadcast this? $\left(c=3 \times 10^{8} m / s, h=6.6 \times 10^{-34} J-s\right)$
(Main 2019, 9 Jan II)
(a) $3.75 \times 10^{6}$
(b) $3.86 \times 10^{6}$
(c) $6.25 \times 10^{5}$
(d) $4.87 \times 10^{5}$
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Solution:
- Here,
Signal wavelength, $\lambda=800 n-m=8 \times 10^{-7} m$
Frequency of source is
$$ \text { As, } \quad \begin{aligned} f & =\frac{c}{\lambda}=\frac{3 \times 10^{8}}{8 \times 10^{-7}} \\ & =3.75 \times 10^{14} Hz \end{aligned} $$
$\therefore$ Total bandwidth used for communication
$$ \begin{aligned} & =1 % \text { of } 3.75 \times 10^{14} \\ & =3.75 \times 10^{12} Hz \end{aligned} $$
So, number of channel for signals $=\underline{\text { total bandwidth available for communication }}$
bandwidth of TV signal
$=\frac{3.75 \times 10^{12}}{6 \times 10^{6}}=0.625 \times 10^{6}=6.25 \times 10^{5}$
