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Question

The wave impedance of a medium is equal to:

The correct answer is

the wave impedance of free-space divided by the refractive index of the medium

The wave impedance of a medium is a fundamental property that describes the ratio of the electric field strength to the magnetic field strength of an electromagnetic wave propagating through that medium. It is also known as the intrinsic impedance or characteristic impedance of the medium. Understanding this concept is crucial for studying how electromagnetic waves behave in different materials.

Wave Impedance Explained

The wave impedance, often denoted by \(Z\), for a uniform plane wave propagating in a linear, homogeneous, and isotropic medium is given by the formula:

\[ Z = \sqrt{\frac{\mu}{\epsilon}} \]

  • Here, \(\mu\) represents the magnetic permeability of the medium. It measures how much the medium supports the formation of a magnetic field within itself.
  • And \(\epsilon\) represents the electric permittivity of the medium. It measures how much the medium stores electric potential energy in an electric field.

Free-Space Wave Impedance

For free-space (a vacuum), the magnetic permeability is denoted as \(\mu_0\) and the electric permittivity as \(\epsilon_0\). The wave impedance of free-space, denoted as \(Z_0\), is a constant value and is given by:

\[ Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \]

The approximate value of \(Z_0\) is 377 ohms (or more precisely, \(120\pi\) ohms).

Refractive Index of a Medium

The refractive index (\(n\)) of a medium describes how fast light (or more generally, an electromagnetic wave) travels through it compared to its speed in a vacuum. It is defined as the ratio of the speed of light in free-space (\(c\)) to the speed of light in the medium (\(v\)):

\[ n = \frac{c}{v} \]

We know that the speed of light in a medium is \(v = \frac{1}{\sqrt{\mu\epsilon}}\) and in free-space is \(c = \frac{1}{\sqrt{\mu_0\epsilon_0}}\). Therefore, the refractive index can also be expressed as:

\[ n = \frac{\frac{1}{\sqrt{\mu_0\epsilon_0}}}{\frac{1}{\sqrt{\mu\epsilon}}} = \sqrt{\frac{\mu\epsilon}{\mu_0\epsilon_0}} \]

For most non-magnetic dielectric materials, the magnetic permeability \(\mu\) is very close to \(\mu_0\). In such cases, we can approximate \(\mu \approx \mu_0\), which simplifies the refractive index formula to:

\[ n \approx \sqrt{\frac{\epsilon}{\epsilon_0}} = \sqrt{\epsilon_r} \]

Here, \(\epsilon_r\) is the relative permittivity of the medium.

Relating Wave Impedance, Free-Space Impedance, and Refractive Index

Let's establish the relationship between the wave impedance of a medium (\(Z\)), the wave impedance of free-space (\(Z_0\)), and the refractive index (\(n\)).

We have:

  • Wave impedance of the medium: \(Z = \sqrt{\frac{\mu}{\epsilon}}\)
  • Wave impedance of free-space: \(Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}\)

Consider the ratio \(\frac{Z_0}{Z}\):

\[ \frac{Z_0}{Z} = \frac{\sqrt{\frac{\mu_0}{\epsilon_0}}}{\sqrt{\frac{\mu}{\epsilon}}} = \sqrt{\frac{\mu_0}{\epsilon_0} \cdot \frac{\epsilon}{\mu}} = \sqrt{\frac{\mu_0\epsilon}{\mu\epsilon_0}} \]

As discussed, for most non-magnetic materials, we can assume \(\mu \approx \mu_0\). Under this common assumption, the expression simplifies significantly:

\[ \frac{Z_0}{Z} = \sqrt{\frac{\mu_0\epsilon}{\mu_0\epsilon_0}} = \sqrt{\frac{\epsilon}{\epsilon_0}} \]

We also know that for non-magnetic materials, the refractive index \(n = \sqrt{\frac{\epsilon}{\epsilon_0}}\). Therefore, we can substitute \(n\) into the equation:

\[ \frac{Z_0}{Z} = n \]

Rearranging this equation to solve for the wave impedance of the medium (\(Z\)), we get:

\[ Z = \frac{Z_0}{n} \]

Conclusion on Wave Impedance

This derivation shows that the wave impedance of a medium is equal to the wave impedance of free-space divided by the refractive index of the medium, especially for non-magnetic materials which is a common assumption in many applications concerning electromagnetic wave propagation.

Therefore, the correct relationship is that the wave impedance of a medium is equal to:

  • the wave impedance of free-space divided by the refractive index of the medium.
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Important Questions from Electromagnetic Wave Propagation

  1. For sky waves, following statements are given:

    (A) n > 1, this shows 81 \(\rm\frac{N}{f^2}\) positive

    (B) n > 1, show 81 \(\rm\frac{N}{f^2}\)  Negative

    (C) n < 1 shows 81 \(\rm\frac{N}{f^2}\)  < 1

    (D) v g x v p= c 2

    (E) n = 0 shows 81 \(\rm\frac{N}{f^2}\)  = 1, f = f c

    Choose the correct answer from the options given below:

  2. If the Polarization vector is given as N and the Direction of propagation is given as K then which one of the following relations is correct?

  3. The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:

  4. Bending of light wave as it passes between material of different optical density

  5. The average radiation on a surface by a normally electromagnetic wave is _______ in case of complete absorption by the surface.

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