Question

An electromagnetic wave of frequency $v$ = $3.0$ $MHz$ passes from vacuum into a dielectric medium with relative permittivity $\epsilon$ = $4$. Then,

• A. Wavelength is doubled and frequency becomes half.
• B. Wavelength is halved and frequency remains unchanged.
• C. Wavelength and frequency both remains unchanged.
• D. Wavelength is doubled and frequency remains unchanged.

Answer 1

Answer: (B)

Step 1: Given that:

The frequency of electromagnetic wave= $3.0MHz$ =$3\times {10}^{6}Hz$

Relative permittivity of the medium(${\epsilon }_r$)= $4$

Step 2: Determination of frequency of the EM wave when passes from vacuum to dielectric medium:

1. The frequency of the electromagnetic wave does not change with the change in medium.
2. ​The wavelength and speed of the wave change with the change in the medium.

Thus, the frequency of the EM wave when passes from vacuum to dielectric medium will not change.

Step 3: Determination of wavelength of EM wave:

The velocity of the EM wave in vacuum is given by,

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

If n is the frequency of the wave and $\lambda$ is the wavelength of the wave, then

$c=n\lambda$

Thus,

$c=\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}=\text{n}\lambda$

$n=\frac{c}{\lambda }$ ...............(1)

Where $\lambda$ is the wavelength of EM wave in vacuum.

Now, the Velocity of the electromagnetic wave in the medium is given by;

$v_{medium}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0{\mu }_r{\epsilon }_r}}$

Since, $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

Therefore,

$v_{medium}=\frac{c}{\sqrt{{\mu }_r{ϵ}_r}}$

Where, ${\epsilon }_r$ and ${\mu }_r$ are the relative permittivity and relative permeability of the medium.

For dielectric medium, ${\mu }_r=1$ ,

Thus,

$v_{medium}=\frac{c}{\sqrt{{\epsilon }_r}}$

Now,

$v_{medium}=\frac{c}{\sqrt{4}}$

$v_{medium}=\frac{c}{2}$

Since, frequency does not change, thus,

$v_{medium}=\text{n}\times {\lambda }_{medium}$

$\frac{c}{2}=\text{n}\times {\lambda }_{medium}$

$\Rightarrow \text{ν}=\frac{c}{2{\lambda }_{medium}}$ ............(2)

From (1) and (2) we get;

$\frac{c}{\lambda }=\frac{c}{2\lambda {}_{medium}}$

${\lambda }_{medium}=\frac{\lambda }{2}$

Thus,

Option B) Wavelength is halved and frequency remains unchanged the correct option.