Answer : The wavelength of the signal is [tex]3.23\times 10^8nm[/tex]. This is not a visible light.
Solution : Given,
Frequency = 927.9 MHz = [tex]927.9\times 10^6Hz[/tex] [tex](1MHz=10^6Hz)[/tex]
Speed of light = [tex]3\times 10^8m/s[/tex]
Formula used :
[tex]\nu=\frac{c}{\lambda}\\\lambda=\frac{c}{\nu}[/tex]
where,
[tex]\nu[/tex] = frequency
[tex]\lambda[/tex] = wavelength
c = speed of light
Now put all the given values in this formula, we get the wavelength of the signal.
[tex]\lambda=\frac{3\times 10^8m/s}{927.9\times 10^6Hz}=0.323m=0.323\times 10^9nm=3.23\times 10^8nm[/tex] [tex](1m=10^9nm)[/tex]
This is not a visible light because the range of visible light is (380 to 700 nm).
Therefore, the wavelength of the signal is [tex]3.23\times 10^8nm[/tex]. This is not a visible light.
