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Consider the problem shown in the "wave_incidence_1.png" attached image. An electromagnetic wave is travelling towards an interface between its current medium and a new medium, which has a refractive index [itex]n_2 \neq n_1[/itex]. The interface is represented by the [itex](x,y)[/itex] plane.
The electric field [itex]\mathbf{E}[/itex] is parallel to that plane and it is directed outwards with respect to the screen, like the [itex]x[/itex] axis. It is orthogonal to the plane of incidence, which is the [itex](y,z)[/itex] plane. The magnetic field instead lays in the plane of incidence, but for this problem only the [itex]H_y[/itex] component has to be considered. [itex]E_x, H_y[/itex] constitute a field which only has tangential components to the interface plane [itex](x,y)[/itex].
It is also possible to represent this problem as a transmission line with characteristic impedance
[itex]\eta^{(1)} = \displaystyle \frac{E_x}{H_y} = \displaystyle \frac{\displaystyle \sqrt{\frac{\mu_0}{\epsilon_1}}}{\cos \theta_i}[/itex]
(so computed only with [itex]E_x, H_y[/itex]) terminated over a load which represents the second medium, as shown in the "wave_incidence_equivalent_line.png" attached image. The load impedance can be computed as
[itex]\eta^{(2)} = \displaystyle \frac{\displaystyle \sqrt{\frac{\mu_0}{\epsilon_2}}}{\cos \theta_t}[/itex]
So, one can obtain the reflection coefficient [itex]\Gamma[/itex] for the field, by computing it for the line:
[itex]\Gamma = \displaystyle \frac{\eta^{(2)}  \eta^{(1)}}{\eta^{(2)} + \eta^{(1)}}[/itex]
In [itex]z = 0[/itex] we have
[itex]E_0 e^{j k_y^{(1)} y} + \Gamma E_0 e^{j k_y^{(1)} y} = T E_0 e^{j k_y^{(2)} y}[/itex]
and so the entire problem is solved with the transmissionline approach (and the equality [itex]k_y^{(1)} = k_y^{(2)}[/itex] will be implied).
But why just the tangential components of the fields to the interface should be considered? [itex]H_z[/itex] in fact is not taken into account. I know that the tangential fields must be continuous over an interface between two dielectrics; but the orthogonal fields also have their boundary conditions!
The electric field [itex]\mathbf{E}[/itex] is parallel to that plane and it is directed outwards with respect to the screen, like the [itex]x[/itex] axis. It is orthogonal to the plane of incidence, which is the [itex](y,z)[/itex] plane. The magnetic field instead lays in the plane of incidence, but for this problem only the [itex]H_y[/itex] component has to be considered. [itex]E_x, H_y[/itex] constitute a field which only has tangential components to the interface plane [itex](x,y)[/itex].
It is also possible to represent this problem as a transmission line with characteristic impedance
[itex]\eta^{(1)} = \displaystyle \frac{E_x}{H_y} = \displaystyle \frac{\displaystyle \sqrt{\frac{\mu_0}{\epsilon_1}}}{\cos \theta_i}[/itex]
(so computed only with [itex]E_x, H_y[/itex]) terminated over a load which represents the second medium, as shown in the "wave_incidence_equivalent_line.png" attached image. The load impedance can be computed as
[itex]\eta^{(2)} = \displaystyle \frac{\displaystyle \sqrt{\frac{\mu_0}{\epsilon_2}}}{\cos \theta_t}[/itex]
So, one can obtain the reflection coefficient [itex]\Gamma[/itex] for the field, by computing it for the line:
[itex]\Gamma = \displaystyle \frac{\eta^{(2)}  \eta^{(1)}}{\eta^{(2)} + \eta^{(1)}}[/itex]
In [itex]z = 0[/itex] we have
[itex]E_0 e^{j k_y^{(1)} y} + \Gamma E_0 e^{j k_y^{(1)} y} = T E_0 e^{j k_y^{(2)} y}[/itex]
and so the entire problem is solved with the transmissionline approach (and the equality [itex]k_y^{(1)} = k_y^{(2)}[/itex] will be implied).
But why just the tangential components of the fields to the interface should be considered? [itex]H_z[/itex] in fact is not taken into account. I know that the tangential fields must be continuous over an interface between two dielectrics; but the orthogonal fields also have their boundary conditions!
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