## Polarization angles for normal transmission and reflection

Posted by Peeter Joot on January 22, 2014

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## Question: Polarization angles for normal transmission and reflection ([1] pr 9.14)

For normal incidence, without assuming that the reflected and transmitted waves have the same polarization as the incident wave, prove that this must be so.

## Answer

Working with coordinates as illustrated in fig. 1.1, the incident wave can be assumed to have the form

Assuming a polarization for the reflected wave, we have

And finally assuming a polarization for the transmitted wave, we have

With no components of any of the or waves in the directions the boundary value conditions at require the equality of the and components of

With , those components are

Equality of eq. 1.0.5b, and eq. 1.0.5c require

or . It turns out that all of these solutions correspond to the same physical waves. Let’s look at each in turn

- . The system eq. 1.0.5.5 is reduced to
with solution

- . The system eq. 1.0.5.5 is reduced to
with solution

Effectively the sign for the magnitude of the transmitted and reflected phasors is toggled, but the polarization vectors are also negated, with , and . The resulting and are unchanged relative to those of the solution above.

- . The system eq. 1.0.5.5 is reduced to
with solution

Effectively the sign for the magnitude of the transmitted phasor is toggled. The polarization vectors in this case are , and , so the transmitted phasor magnitude change of sign does not change relative to that of the solution above.

- . The system eq. 1.0.5.5 is reduced to
with solution

This time, the sign for the magnitude of the reflected phasor is toggled. The polarization vectors in this case are , and . In this final variation the reflected phasor magnitude change of sign does not change relative to that of the solution.

We see that there is only one solution for the polarization angle of the transmitted and reflected waves relative to the incident wave. Although we fixed the incident polarization with along , the polarization of the incident wave is maintained regardless of TE or TM labeling in this example, since our system is symmetric with respect to rotation.

# References

[1] D.J. Griffith. *Introduction to Electrodynamics*. Prentice-Hall, 1981.

## Antonio said

Very nice Peeter. Probably the normal incidence is the only case where you need to prove that there are no other independent solutions, as only 1 solution was provided, for the oblique incident case, two independent solutions are derived and therefore it is probably unecessary to prove that there is not another one.