Question

The frequency of electromagnetic waves in free space is 2 MHz. When it passes through a region of relative permittivity ${\varepsilon _r} = 4.0$, then its wavelength _________ & frequency _________.
A. Becomes double, becomes half
B. Becomes double, remains constant
C. Becomes half, becomes double
D. Becomes half, remains constant.

Answer 1

Hint: Before we proceed into the problem, it is important to know the definitions of permittivity and permeability.
Permittivity is the measure of the electric polarizability of a dielectric and Permeability is the measure of resistance of the material against the formation of magnetic field lines through the materials.
Now that we have the definitions of the permittivity and permeability here, we can solve this problem by establishing the relation between these two terms and the speed of light.

Complete step-by-step answer:
The dielectric is defined as the electric insulator that can be polarised by applying an electric field. That means that whenever a dielectric medium is placed in an external electric field, the electric charges do not flow through the material like it happens in a normal electric conductor, but the charge carries a slight shift from their equilibrium positions. This results in dielectric polarization. In other words, the positive and negative charges re-align themselves along the direction of the electric field.
The permittivity is a measure of this particular property.
This means that the materials which have a higher permittivity polarizes to a greater extent in the presence of an external electric field compared to a material with lower permittivity.
The field created due to the polarisation –
$D = \varepsilon E$
The permittivity of free space, also called vacuum permittivity, is the ratio of $\dfrac{D}{E}$in space.
The value is –
${\varepsilon _0} = 8.854 \times {10^{ - 12}}F/m$
The permeability is the measure of resistance of the material against the formation of magnetic field lines through the materials. The absolute value of the permeability is –
${\mu _0} = 4\pi \times {10^{ - 7}}H/m$
If we multiply these two quantities, we get –
$
  {\varepsilon _0}{\mu _0} = 8.854 \times {10^{ - 12}} \times 4\pi \times {10^{ - 7}} = 111.26 \times {10^{ - 19}} \\
  Inverting, \\
  \dfrac{1}{{{\varepsilon _0}{\mu _0}}} = \dfrac{1}{{111.26 \times {{10}^{ - 19}}}} = 8.98 \times {10^{16}} \\
 $
This value is almost equal to the square of the speed of light.
Hence, we have –
$\dfrac{1}{{{\varepsilon _0}{\mu _0}}} = {c^2}$
$ \to c = \dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$
Since the electromagnetic wave passes through a dielectric of relative permittivity, ${\varepsilon _r} = 4.0$
The actual permittivity, $\varepsilon = {\varepsilon _0}{\varepsilon _r} = 4{\varepsilon _0}$
Velocity of the light in the medium, $v = \dfrac{1}{{\sqrt {\varepsilon \mu } }}$
Since, there is no change in magnetic field in the dielectric, $\mu = {\mu _0}$
Substituting, we get –
$v = \dfrac{1}{{\sqrt {\varepsilon \mu } }}$
\[
  v = \dfrac{1}{{\sqrt {4{\varepsilon _0}{\mu _0}} }} \\
   \to v = \dfrac{1}{{2\sqrt {{\varepsilon _0}{\mu _0}} }} \\
   \to v = \dfrac{c}{2}\because c = \dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} \\
 \]
Here, we see that the velocity of light reduces by half.
Velocity of light is the product of its frequency and wavelength. When the light propagates, the frequency never changes in the path since it is dependent on the starting point of the light. Only the wavelength is variable. Since, we see that the velocity of light reduces by 2 and the frequency is constant, it is clear that the wavelength reduces by 2.
Thus, frequency remains constant and wavelength reduces by half.

Hence, the correct option is Option D.

Note: The concept of permittivity is applicable in the theory of capacitance.
Capacitor is an electronic device that is used to store charges. It consists of two plates separated by a dielectric medium like air, paper, ceramic etc.
In such cases, the capacitance depends on the total permittivity of the dielectric medium.
Capacitance,
$C = {\varepsilon _0}\dfrac{A}{d}$