Question

Choose the correct answer from the alternatives given.
An electromagnetic wave of frequency $\nu =3MHz$ passes from vacuum into a dielectric medium with permittivity $\epsilon =4$. Then

• A. wavelength and frequency both become half.
• B. wavelength is doubled and frequency remains unchanged.
• C. wavelength and frequency both remain unchanged.
• D. wavelength is halved and frequency remains unchanged.

Answer 1

The correct option is D wavelength is halved and frequency remains unchanged.

Given : frequency $v=3MHz=3\times {10}^{6}Hz$, relative permitivity ${\epsilon }_r=4$

Here the frequency of electromagnetic wave remains unchanged but the wavelength of electromagnetic wave changes when it passes from one medium to another.

The refractive index is the square root of permeability and permittivity product.

For formula,

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}⟹c\propto \frac{1}{\sqrt{{\epsilon }_0}}$

Similarly,

$v\propto \frac{1}{\sqrt{\epsilon }}$

Therefore,

$\frac{c}{v}=\sqrt{\frac{\epsilon }{{\epsilon }_0}}=\sqrt{\frac{4}{1}}=\mathrm{2........}\left(i\right)$

But

$\frac{c}{v}=\frac{\nu \lambda }{\nu {\lambda }^{\prime }}⟹\frac{c}{v}=\frac{\lambda }{{\lambda }^{\prime }}⟹2=\frac{\lambda }{{\lambda }^{\prime }}⟹{\lambda }^{\prime }=\frac{\lambda }{2}$

Hence wavelength is halved.