An electromagnetic wave of frequency 2 text{ | vedclass

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(D) In vacuum,the wavelength is given by $\lambda_0 = \frac{c}{f} = \frac{3 	imes 10^8 	ext{ m/s}}{2 	imes 10^6 	ext{ Hz}} = 150 	ext{ m}$.In a n

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decreases by $100 	ext{ m}$

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(D) In vacuum,the wavelength is given by $\lambda_0 = \frac{c}{f} = \frac{3 	imes 10^8 	ext{ m/s}}{2 	imes 10^6 	ext{ Hz}} = 150 	ext{ m}$.In a non-magnetic medium,the relative permeability $\mu_r = 1$. The speed of the wave in the medium is $v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} = \frac{c}{\sqrt{9 	imes 1}} = \frac{c}{3}$.Substituting the value of $c$,we get $v = \frac{3 	imes 10^8}{3} = 1 	imes 10^8 	ext{ m/s}$.The frequency $f$ remains constant when the wave enters a different medium. Therefore,the wavelength in the medium is $\lambda = \frac{v}{f} = \frac{10^8 	ext{ m/s}}{2 	imes 10^6 	ext{ Hz}} = 50 	ext{ m}$.The change in wavelength is $\Delta \lambda = \lambda_0 - \lambda = 150 	ext{ m} - 50 	ext{ m} = 100 	ext{ m}$.Thus,the wavelength decreases by $100 	ext{ m}$.

An electromagnetic wave of frequency $2 \text{ MHz}$ propagates from vacuum to a non-magnetic medium of relative permittivity $9$. Then its wavelength

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