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An electromagnetic wave of frequency $2 \mathrm{MHz}$ propagates from vacuum to a non-magnetic medium of relative permittivity 9 . Then its' wavelength
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decreases by 100 m
In vacuum wavelength,
$\lambda_0=\frac{c}{f_0}=\frac{3 \times 10^8}{2 \times 10^6}=1.5 \times 10^2=150 \mathrm{~m}$
In medium, speed of wave is
$v=\frac{\mathcal{c}}{\sqrt{\varepsilon_r \mu_r}} \approx \frac{\mathcal{c}}{\sqrt{\varepsilon_r}}$
(as medium is non-magnetic)
$\therefore \quad v=\frac{\mathcal{c}}{\sqrt{9}}=\frac{\mathcal{c}}{3}=1 \times 10^8 \mathrm{~ms}^{-1}$
Frequency remaining same, wavelength in medium is $\lambda=\frac{v}{f_0}=\frac{10^8}{2 \times 10^6}=50 \mathrm{~m}$
$\therefore$ Wavelength decreases by $100 \mathrm{~m}$.
$\lambda_0=\frac{c}{f_0}=\frac{3 \times 10^8}{2 \times 10^6}=1.5 \times 10^2=150 \mathrm{~m}$
In medium, speed of wave is
$v=\frac{\mathcal{c}}{\sqrt{\varepsilon_r \mu_r}} \approx \frac{\mathcal{c}}{\sqrt{\varepsilon_r}}$
(as medium is non-magnetic)
$\therefore \quad v=\frac{\mathcal{c}}{\sqrt{9}}=\frac{\mathcal{c}}{3}=1 \times 10^8 \mathrm{~ms}^{-1}$
Frequency remaining same, wavelength in medium is $\lambda=\frac{v}{f_0}=\frac{10^8}{2 \times 10^6}=50 \mathrm{~m}$
$\therefore$ Wavelength decreases by $100 \mathrm{~m}$.
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