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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Electromagnetic Waves

NEET 2013 · Physics · Electromagnetic Waves

An electromagnetic wave of frequency v=3.0 MHz passes from vacuum...

An electromagnetic wave of frequency v=3.0MHzv=3.0 \mathrm{MHz} passes from vacuum into a dielectric medium with relative permittivity ε=4.0\varepsilon=4.0 Then:

Options

  1. AWavelength is double and frequency become half.
  2. BWavelength is halved and frequency remains unchanged.
  3. CWavelength and frequency both remain unchanged.
  4. DWavelength is doubled and frequency unchanged.

Correct answer

B. Wavelength is halved and frequency remains unchanged.

Step-by-step solution

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}=1μ0μ0ε0cnr=Cμrεrμ0andε0\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 μrε0\mu \mathrm{r} \varepsilon_0 is in medium. For dielectric medium r=1r=1∴Vmed =Cεr Given ε0=4∴Vmed =C4=C2\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 λmed =Vmed V=C2V=λvacum 2\lambda_{\text{med }}=\frac{V_{\text{med }}}{V}=\frac{C}{2 V}=\frac{\lambda_{\text{vacum }}}{2}
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