A radio can tune to any station in the 6,MHz | vedclass

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(B) Given: Frequency $f_{1} = 6\,MHz = 6 	imes 10^{6}\,Hz$ and $f_{2} = 10\,MHz = 10 	imes 10^{6}\,Hz$.Speed of light $c = 3 	imes 10^{8}\,m/s$.The

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(B) Given: Frequency $f_{1} = 6\,MHz = 6 	imes 10^{6}\,Hz$ and $f_{2} = 10\,MHz = 10 	imes 10^{6}\,Hz$.Speed of light $c = 3 	imes 10^{8}\,m/s$.The wavelength $\lambda$ is given by $\lambda = \frac{c}{f}$.For $f_{1} = 6 	imes 10^{6}\,Hz$,$\lambda_{1} = \frac{3 	imes 10^{8}}{6 	imes 10^{6}} = 50\,m$.For $f_{2} = 10 	imes 10^{6}\,Hz$,$\lambda_{2} = \frac{3 	imes 10^{8}}{10 	imes 10^{6}} = 30\,m$.The wavelength bandwidth is the difference between the two wavelengths: $\Delta\lambda = \lambda_{1} - \lambda_{2} = 50\,m - 30\,m = 20\,m$.

$A$ radio can tune to any station in the $6\,MHz$ to $10\,MHz$ band. The value of the corresponding wavelength bandwidth will be $....\,m$.

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