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An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\varepsilon=4.0$ Then:
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Verified Answer
The correct answer is:
Wavelength is halved and frequency remains unchanged.
Velocity of electromagnetic wave in vacuum.
$$
C=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=V_{\mathrm{vacum}}
$$
Velocity of electromagnetic wave in medium
$$
V_{m e d}=\frac{1}{\sqrt{\mu_0 \mu_0 \varepsilon_0 c_{n r}}}=\frac{\mathrm{C}}{\sqrt{\mu_r \varepsilon_r}}
$$
$\mu_0$ and $\varepsilon_0$ is in vacuum and $\mu \mathrm{r} \varepsilon_0$ is in medium.
For dielectric medium $r=1$
$$
\begin{aligned}
& \therefore V_{\text {med }}=\frac{C}{\sqrt{\varepsilon_r}} \text { Given } \varepsilon_0=4 \\
& \therefore V_{\text {med }}=\frac{C}{\sqrt{4}}=\frac{\mathrm{C}}{2}
\end{aligned}
$$
$\therefore$ Wavelength of the wave in medium $\lambda_{\text {med }}=\frac{V_{\text {med }}}{V}=\frac{C}{2 V}=$ $\frac{\lambda_{\text {vacum }}}{2}$
$$
C=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=V_{\mathrm{vacum}}
$$
Velocity of electromagnetic wave in medium
$$
V_{m e d}=\frac{1}{\sqrt{\mu_0 \mu_0 \varepsilon_0 c_{n r}}}=\frac{\mathrm{C}}{\sqrt{\mu_r \varepsilon_r}}
$$
$\mu_0$ and $\varepsilon_0$ is in vacuum and $\mu \mathrm{r} \varepsilon_0$ is in medium.
For dielectric medium $r=1$
$$
\begin{aligned}
& \therefore V_{\text {med }}=\frac{C}{\sqrt{\varepsilon_r}} \text { Given } \varepsilon_0=4 \\
& \therefore V_{\text {med }}=\frac{C}{\sqrt{4}}=\frac{\mathrm{C}}{2}
\end{aligned}
$$
$\therefore$ Wavelength of the wave in medium $\lambda_{\text {med }}=\frac{V_{\text {med }}}{V}=\frac{C}{2 V}=$ $\frac{\lambda_{\text {vacum }}}{2}$
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