Modern Physics Ques 275
- 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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Answer:
Correct Answer: 275.(c)
Solution:
Formula:
- 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
$=\frac{\text { total bandwidth available for communication }}{\text { bandwidth of TV signal}}$
$=\frac{3.75 \times 10^{12}}{6 \times 10^{6}}=0.625 \times 10^{6}=6.25 \times 10^{5}$
BETA
