Question

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

• A. $3.75\times {10}^6$
• B. $4.87\times {10}^5$
• C. $3.86\times {10}^6$
• D. $6.25\times {10}^5$

Answer 1

The correct option is D $6.25\times {10}^5$

$f=\frac{3\times {10}^8}{8\times {10}^{-7}}=\frac{30}{8}\times {10}^{14}Hz$
$=3.75\times {10}^{14}Hz$
1% of $f=0.0375\times {10}^{14}Hz$
$=3.75\times {10}^{12}Hz=3.75\times {10}^{6}MHz$
number of channels = $\frac{3.75\times {10}^6}{6}=6.25\times {10}^5$
$\therefore$ correct answer is (4)