# EM wave at boundary

I have a simple question about reflecting EM waves at dielectric boundaries. To best illustrate my question, consider normal incidence. The incident wave has the wavevector k positive, and the reflected has k negative. Since B = k x E , and k has changed sign, B must also change sign. This is my problem - why cant E change sign instead? This would satisfy the necessary equations. It also makes intuitive sense, at normal incidence E and B are both parallel to the plane, and flipping either would preserve the handedness. Why is B special?

Meir Achuz
Homework Helper
Gold Member
It depends on the dielectric constant epsilon. If epsilon is greater than 1, which is the usual case, then the reflected E will not flip. For epsilon less than 1, E would flip.
For instance, for a wave leaving a dielectric into air, the reflected E does flip.

Thanks clem. Do you know how to show this from the boundary conditions?

k is the propagation vector, it represents the direction of the wave. When the wave reflects, the direction is reversed, hence k changes sign.
If you change the sign of E instead of that of k, you have a wave traveling in the same direction as initially.
Clem, can you help me understand what role permittivity has to play in the direction of E? I don't know at all about it..

ksac I think you have misunderstood my question. Given that k changes sign, E or B (but not both) could change sign to satisfy the B = k x E condition. A further condition is required to constrain which one will flip.

Born2bwire
Gold Member
Thanks clem. Do you know how to show this from the boundary conditions?

The Fresnel reflection coefficients have already done this for you.

The Fresnel reflection coefficients are derived from the boundary conditions - they assume what I am trying to prove.

Meir Achuz
Homework Helper
Gold Member
It is done in all textbooks.
In a dielectric, the magnitudes of E and B are related by B=sqrt{epsilon mu} E.
From Maxwell's equations, both E and B tangential are continuous at the interface.
This means (with mu =1) that E_1+E_1'=sqrt{epsilon}E_2
and E_1-E_1'=E_2.
Solve for the reflected E_1'.

This is exactly the problem! Why do you choose to put the negative sign on the second equation rather than the first? From what I can see you are choosing to flip the electric field in this case.

Born2bwire