Q154 / 160 · TS EAMCET 2019 (03 May Shift 1) · Physics
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
- Correct answerA$-75 \mathrm{~m}$
- B$+75 \mathrm{~m}$
- C$-50 \mathrm{~m}$
- D$+50 \mathrm{~m}$
Solution
Given, frequency of EM wave,
$\begin{aligned}<br />f & =3 \mathrm{MHz} \\<br />& =3 \times 10^6 \mathrm{~Hz}<br />\end{aligned}$
and permittivity of non-magnetic medium,
$\varepsilon=16 \varepsilon_0$
Wavelength of EM wave,
$\begin{aligned}<br />& \lambda=\frac{c}{f}=\frac{3 \times 10^8}{3 \times 10^6} \\<br />& \lambda=100 \mathrm{~m}<br />\end{aligned}$
$\lambda=100 \mathrm{~m}$
Velocity of electromagnetic (EM) into non-magnetic material,
$\begin{aligned}<br />v & =\frac{c}{\sqrt{\varepsilon_{\varphi}}}=\frac{3 \times 10^8}{\sqrt{16}} \quad\left[\because \varepsilon_{\varphi}=\frac{\varepsilon}{\varepsilon_0}=16\right] \\<br />& =\frac{3}{4} \times 10^8 \mathrm{~m} / \mathrm{s}<br />\end{aligned}$
$\therefore$ Wavelength, $\lambda^{\prime}=\frac{v}{f}=\frac{\frac{3}{4} \times 10^8}{3 \times 10^6}=25 \mathrm{~m}$
$\therefore$ Change in wavelength $=\lambda^{\prime}-\lambda$
$=25-100=-75 \mathrm{~m}$
$\begin{aligned}<br />f & =3 \mathrm{MHz} \\<br />& =3 \times 10^6 \mathrm{~Hz}<br />\end{aligned}$
and permittivity of non-magnetic medium,
$\varepsilon=16 \varepsilon_0$
Wavelength of EM wave,
$\begin{aligned}<br />& \lambda=\frac{c}{f}=\frac{3 \times 10^8}{3 \times 10^6} \\<br />& \lambda=100 \mathrm{~m}<br />\end{aligned}$
$\lambda=100 \mathrm{~m}$
Velocity of electromagnetic (EM) into non-magnetic material,
$\begin{aligned}<br />v & =\frac{c}{\sqrt{\varepsilon_{\varphi}}}=\frac{3 \times 10^8}{\sqrt{16}} \quad\left[\because \varepsilon_{\varphi}=\frac{\varepsilon}{\varepsilon_0}=16\right] \\<br />& =\frac{3}{4} \times 10^8 \mathrm{~m} / \mathrm{s}<br />\end{aligned}$
$\therefore$ Wavelength, $\lambda^{\prime}=\frac{v}{f}=\frac{\frac{3}{4} \times 10^8}{3 \times 10^6}=25 \mathrm{~m}$
$\therefore$ Change in wavelength $=\lambda^{\prime}-\lambda$
$=25-100=-75 \mathrm{~m}$
Hence, the correct answer is option: 'A'.