# Normal incidence over conductors

|The following image shows the different wave components (incidence, reflection and transmission) that we are going to study in this post:

## Reflection and Transmission Coefficients

The **reflection coefficient** from medium n1 to medium n2 (refraction indexes) can be expressed as the ration between the electric field reflection component and the electric field incidence component.

The **transmission coefficient** also depends on the refraction indexes and it is the ration between the transmission field component and the incidence field component:

If there is **normal incidence**, the following relation is met:

The **standing wave ratio** is the relation between the maximum and the minimum values of the electric field:

## Characteristics of a Conductor Medium

The following expressions are characteristics of a conductor medium:

**Intrinsic impedance (Ohms):**

**Phase velocity (meter/second):** **Propagation constant (radian/meter): **

In this case, the** attenuation constant and the phase constant** are equivalent (Neper/meter):

**Wavelength (meters):**

## Perfect Conductor

In a perfect conductor, we have:

The field components of the **incident wave** are:

The field components of the **reflected wave** are:

Therefore, the total **components of the standing wave** are:

## Incidence Exercise:

A plane wave with linear polarization arrives to a real conductor plane with a finite high conductivity. The current density which is induced in the conductor will stop being strictly superficial and will penetrate in some way in the conductor.

- Is there any electric field in the conductor?
- What is the value of the wave impedance in the conductor? Use the following values:
*μ =*μ0 = 4π x 10-7 H/m, σ = 5,0 x 105 1/Ωm, and calculate it for*f = 90 MHz and for f = 10*GHz.

## 1.

The field inside a perfect conductor is zero. In our case, σ is finite (the conductor is real as supposed to ideal), so the field won’t be zero.

## 2.

The impedance for an uniform plane wave is expressed as:

where

Therefore, the permittivity won’t be always a real number. By using the previous formulas, we have:

For f=10 GHz For f=90 GHz

## Conclusions

Why did we study polarization in the previous post and now we study the incidence? There are important applications where these two concepts are developed:

- The electric field of the signals from
**AM radio stations**is linearly polarized, with the electric field perpendicular to the Earth. Thus,the receiver antenna needs to be in vertical position to achieve maximum reception. - The electric field of
**TV signals**is linearly polarized with the horizontal direction. Therefore, the receiver antennas need to be in horizontal position in the roof of the houses and buildings. - The wave transmitted by the
**FM radio stations**are usually circularly polarized.

And these are just some of the simplest applications. Do you know any other applications where polarization and incidence is important? Leave a comment below! 🙂

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