An electromagnetic wave of frequency 5.0 x 1014Hz, is incident on the surface of water of refractive index \(\frac{4}{3}\). Taking the speed of the wave in ...

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The formula for calculating the wavelength of a wave is given by: \begin{equation*} \lambda = \frac{v}{f} \end{equation*} where - \(\lambda\) is the wavelength of the wave - v is the velocity of the wave in the medium - f is the frequency of the wave. The velocity of the electromagnetic wave in air is given as 3.0 x 10⁸ms^-1. The refractive index of water is given as \(\frac{4}{3}\). The refractive index is the ratio of the speed of light in a vacuum to the speed of light in a medium. Hence the speed of the wave in water is given as: \begin{equation*} v_{water} = \frac{v_{air}}{n} = \frac{3.0\times10^8}{4/3} = 2.25\times10^8ms^{-1} \end{equation*} Substituting the values of v and f in the formula above, we have: \begin{equation*} \lambda = \frac{v}{f} = \frac{2.25\times10^8}{5.0\times10^{14}} = 4.5\times10^{-7}m \end{equation*} Therefore, the wavelength of the electromagnetic wave in water is 4.5 x 10^-7m. Hence the correct option is: 4.5 x 10^-7m.