## Reflection And Refraction

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# Reflection and Refraction of Waves by the ionosphere

The mechanism of reflection and refraction of radio waves by the ionosphere is very much a function of frequency.**Case I. Reflection of Low Frequencies.**In this case the wavelength is considered to be sufficiently long that there is a great change in the ionization density in the course of a wavelength. The layer then may be considered a reflecting surface.

For this type of reflection, the reflection coefficient of the medium will depend upon the frequency, polarization, and angle of incidence of the wave. The reflection curves will be those for reflection from a perfect dielectric that has a refractive index of less than unity. For angles of incidence greater than a certain critical angle (which depends upon the refractive index), there will be complete reflection of the signal for both polarization of the wave. For angles less than the critical (that is, closer to the normal), the reflection coefficient will be less than unity and will depend on the angle of incidence.

**Case II: Reflection or Refraction at High Frequencies**

Assuming the permeability of the ionosphere to be μr = 1, the phase velocity of the wave u = c /√ε

_{r}... (1)

c is the velocity of light in a vacuum.

εr depends upon the electron density in the distance of a wavelength is smaller, the change in phase velocity will also be small.

The wave penetrates the lower edge of the ionosphere without reflection. But within the ionosphere the wave travels

a path that is curved away from the region of greater electron density (smaller index of refraction). At any point along the path, the angle Φ between the path and the normal is given by Snell’s law of refraction.

sin Φi = nsin Φ

or sin Φ = sin Φi/n ... (2)

n is the index of refraction at the point where Φ is observed, and Φi is the angle of incidence (measured from the normal to the ionosphere layer).

Now, n = √ε

_{r}... (3)

At great heights, the collision frequency v is small and ω

^{2}>> v

^{2}.

Then the expression for ε

_{r}is ε

_{n}= (1 – Ne

^{2}/ε

_{0}mω

^{2}) ... (4)

For an electron, e = 1.59 x 10-19 coulombs, m = 9 x 10-31 kg, so that (4) becomes

εr = (1 – 81N/f

^{2}) ... (5)

N is the number of electrons per cubic meter and f is the frequency in Hertz.

From Eq. (3) the refractive index is

n = √1 – 81N/f

^{2}... (6)

The refractive index decreases as the wave penetrates into regions of greater electron density and the angle of refraction increases correspondingly. When n has decreased to the point where n = sin Φi, the angle of refraction Φ will be 90o and the wave will be travelling horizontally. The highest point reached by the wave is therefore that point at which the electron density N satisfies the relation.

√1 – 81N/f

^{2}= sin Φi

Or N’ f

^{2}cos

^{2}Φi /81 ... (7)

If the electron density at some level in a layer is sufficiently great to satisfy relation (7), the wave will be returned to earth from that level. If the maximum electron density in a layer is less than that required by (7), the wave will penetrate the layer (though it may be reflected back from a higher layer for which N is greater).

The largest electron density required for reflection occurs when the angle of incidence Φi is zero, that is, for vertical incidence. For any give layer the highest frequency that will be reflected back for vertical incidence will be

f

_{cr}= √81N

_{max }... (8)

Here Nmax is the maximum ionization density (electrons per cubic meter) and fcr is the critical frequency for the layer.

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