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An electromagnetic wave of frequency 3.0 MHz passes from vacuum into a non-magnetic medium with permittivity, $\epsilon=16 \epsilon_0$. Where, $\epsilon_0$ is the free space permittivity. The change in wavelength is
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The correct answer is:
$-75 \mathrm{~m}$
Given, frequency of EM wave,
$\begin{aligned}
f & =3 \mathrm{MHz} \\
& =3 \times 10^6 \mathrm{~Hz}
\end{aligned}$
and permittivity of non-magnetic medium,
$\varepsilon=16 \varepsilon_0$
Wavelength of EM wave,
$\begin{aligned}
& \lambda=\frac{c}{f}=\frac{3 \times 10^8}{3 \times 10^6} \\
& \lambda=100 \mathrm{~m}
\end{aligned}$
$\lambda=100 \mathrm{~m}$
Velocity of electromagnetic (EM) into non-magnetic material,
$\begin{aligned}
v & =\frac{c}{\sqrt{\varepsilon_{\varphi}}}=\frac{3 \times 10^8}{\sqrt{16}} \quad\left[\because \varepsilon_{\varphi}=\frac{\varepsilon}{\varepsilon_0}=16\right] \\
& =\frac{3}{4} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\therefore$ Wavelength, $\lambda^{\prime}=\frac{v}{f}=\frac{\frac{3}{4} \times 10^8}{3 \times 10^6}=25 \mathrm{~m}$
$\therefore$ Change in wavelength $=\lambda^{\prime}-\lambda$
$=25-100=-75 \mathrm{~m}$
$\begin{aligned}
f & =3 \mathrm{MHz} \\
& =3 \times 10^6 \mathrm{~Hz}
\end{aligned}$
and permittivity of non-magnetic medium,
$\varepsilon=16 \varepsilon_0$
Wavelength of EM wave,
$\begin{aligned}
& \lambda=\frac{c}{f}=\frac{3 \times 10^8}{3 \times 10^6} \\
& \lambda=100 \mathrm{~m}
\end{aligned}$
$\lambda=100 \mathrm{~m}$
Velocity of electromagnetic (EM) into non-magnetic material,
$\begin{aligned}
v & =\frac{c}{\sqrt{\varepsilon_{\varphi}}}=\frac{3 \times 10^8}{\sqrt{16}} \quad\left[\because \varepsilon_{\varphi}=\frac{\varepsilon}{\varepsilon_0}=16\right] \\
& =\frac{3}{4} \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}$
$\therefore$ Wavelength, $\lambda^{\prime}=\frac{v}{f}=\frac{\frac{3}{4} \times 10^8}{3 \times 10^6}=25 \mathrm{~m}$
$\therefore$ Change in wavelength $=\lambda^{\prime}-\lambda$
$=25-100=-75 \mathrm{~m}$
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