Question

How do you calculate the wavelength of electromagnetic waves?

Answer 1

Hint: Before we proceed into understanding of the electromagnetic wave, 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.

Complete 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 –
$\Rightarrow {\varepsilon _0} = 8.854 \times {10^{ - 12}}F{m^{ - 1}}$
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 –
$\Rightarrow {\mu _0} = 4\pi \times {10^{ - 7}}H{m^{ - 1}}$
If we multiply these two quantities, we get –
$\Rightarrow {\varepsilon _0}{\mu _0} = 8.854 \times {10^{ - 12}} \times 4\pi \times {10^{ - 7}} = 111.26 \times {10^{ - 19}}$
Inverting,
$\Rightarrow \dfrac{1}{{{\varepsilon _0}{\mu _0}}} = \dfrac{1}{{111.26 \times {{10}^{ - 19}}}} = 8.98 \times {10^{16}}$
This value is almost equal to the square of the speed of light.
Hence, we have –
$\Rightarrow \dfrac{1}{{{\varepsilon _0}{\mu _0}}} = {c^2}$
$ \Rightarrow c = \sqrt {\dfrac{1}{{{\varepsilon _0}{\mu _0}}}} $
The velocity of the wave is equal to the product of frequency $f$and wavelength of the wave $\lambda $. Hence, we have –
$\Rightarrow c = f\lambda $
Substituting in the above, we have –
$\Rightarrow f\lambda = \sqrt {\dfrac{1}{{{\varepsilon _0}{\mu _0}}}} $
$ \Rightarrow \lambda = \dfrac{1}{f}\sqrt {\dfrac{1}{{{\varepsilon _0}{\mu _0}}}} $
Therefore, the expression to calculate the wavelength of an electromagnetic wave is derived as above.

Note: The electromagnetic wave consists of transverse electric field waves and transverse magnetic field waves travelling at right angles to each other, such that the ratio of the amplitudes of magnetic field and electric field is equal to the velocity of the wave, which is the standard value of velocity of light in vacuum.
If ${H_0}$ and ${E_0}$ are the amplitudes of the magnetic field and electric field of an electromagnetic wave, we have –
$\dfrac{{{H_0}}}{{{E_0}}} = \dfrac{1}{c}$
Also, wavelength can be expressed as –
$\dfrac{{{H_0}}}{{{E_0}}} = \dfrac{1}{{f\lambda }}$
$\lambda = \dfrac{{{E_0}}}{{f{H_0}}}$
The wavelength can also be expressed in the above equation.