Question

An electromagnetic wave of frequency $\upsilon =3MHz$ passes from vacuum into a dielectric medium with relative permittivity ${{\varepsilon }_{r}}=4$ . Then:
(a) Wavelength is halved and frequency remains unchanged
(b) Wavelength is doubled and frequency remains becomes half
(c) Wavelength is doubled and frequency remains unchanged
(d) Wavelength and frequency both remain unchanged

Answer 1

As the electromagnetic wave passes from one vacuum to a dielectric, it changes its medium. So, in order to comment on the parameters like wavelength and frequency, we will first calculate the speed of the electromagnetic wave in the new medium and then use it to calculate the refractive index of the new medium.

Complete answer:
Since the electromagnetic wave is moving from one medium to another without any loss in its energy, then it is simply a case of refraction and we also know that.
So, by using the property of refraction we can say that the frequency of the EM wave will remain constant throughout.
Hence, the frequency of the EM wave will not change.
Now, let the wavelength of the wave before entering the dielectric be ${{\lambda }_{1}}$ and the wavelength after entering the dielectric be ${{\lambda }_{2}}$.
Now, we know the relation between ${{\lambda }_{1}}$ and ${{\lambda }_{2}}$ can be given as:
$\Rightarrow \dfrac{{{\lambda }_{2}}}{{{\mu }_{air}}}=\dfrac{{{\lambda }_{1}}}{{{\mu }_{dielectric}}}$
Here, ${{\mu }_{air}}$ is equal to 1.
And, ${{\mu }_{dielectric}}$ can be found out using the relation:
$\Rightarrow {{\mu }_{dielectric}}=\sqrt{{{\mu }_{r}}{{\varepsilon }_{r}}}$
Here,
$\Rightarrow {{\mu }_{r}}=1$ [as it is not mentioned in the problem, we will take the standard value]
$\Rightarrow {{\varepsilon }_{r}}=4$
Putting these values in the above equation, we get:
$\begin{aligned} & \Rightarrow {{\mu }_{dielectric}}=\sqrt{1\times 4} \\\ & \Rightarrow {{\mu }_{dielectric}}=2 \\\ \end{aligned}$
Therefore, the wavelength after reflection can be calculated as follows:
$\begin{aligned} & \Rightarrow \dfrac{{{\lambda }_{2}}}{1}=\dfrac{{{\lambda }_{1}}}{2} \\\ & \Rightarrow {{\lambda }_{2}}=\dfrac{{{\lambda }_{1}}}{2} \\\ \end{aligned}$
Hence, the wavelength is halved after refraction.
Thus, the final conclusion is, the wavelength is halved and frequency remains unchanged.

Hence, option (a) is the correct option.

Note:
We should know the change in parameters of an electromagnetic wave upon entering a dielectric medium, like the frequency remains unchanged wavelength and speed of the EM wave changes. Upon entering a dielectric, the phase constant of an EM wave can also change.