How to determine the direction of the E-field of an EM wave

AI Thread Summary
To determine the direction of the electric field amplitude in an electromagnetic wave hitting a boundary between water and air, it's essential to understand the relationship between the wave vector, magnetic field, and electric field. The magnetic field has only a y-component, and both the electric and magnetic fields are perpendicular to each other and to the wave vector. The correct application of the right-hand rule is crucial for finding the direction of the electric field, as it dictates the orientation of the vectors involved. A geometric representation can aid in visualizing the relationships between the vectors, ensuring that the cross product correctly points in the direction of the wave vector. Clarifying the order of the cross product and redrawing the vectors may resolve confusion regarding their orientations.
Luk

Homework Statement


Consider a wave vector which hits a plain boundary between water and air. The wave vector hits the boundary with an angle α1 measured from the vertical axis. The magnetic field amplitude has a y-component only. Also, notice: The z-axis is the horizontal axis, the x-axis is the vertical axis and the y-axis is the axis that goes into the paper. The question is, in what direction does the electrical field amplitude point?
The wave vector has components $$ \vec{k} = \begin{array}{c}\sin(\alpha_1)\\0\\-\cos(\alpha_1)\end{array} $$

Homework Equations


Well I know that the magnetic and the electrical field are both perpendicular to each other and to the wave vector. I also know that $$\vec{H} = 1/\eta * \vec{e_k} \times \vec{E} $$

The Attempt at a Solution


I was thinking that, maybe, this equation holds: $$ \vec{E} = \eta * \vec{e_k} \times \vec{H} $$
But I'm not sure about the correct order within the cross product. So it could also be this: $$ \vec{E} = \eta * \vec{H} \times \vec{e_k} $$
I simply don't know.
 
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The order of the cross product is such that ##\vec{E}\times\vec{H}## points in the direction of ##\hat{e}_k##.
 
right. Well since H points in the positive y-direction, is the cross product then $$ \begin{array}{c}-\sin(\alpha_e)\\0\\-\cos(\alpha_e)\end{array} $$ ?
 
Luk said:
... is the cross product then ...
Which cross product is this? Do you mean the electric field? It is best to make a drawing showing the three principal axes, the k-vector and the H-vector. Add to this drawing the E-vector in such a way that ##\vec{E}\times\vec{H}## points in the direction of ##\hat{e}_k##. Then look at the drawing and find the components of the E-vector.
 
Ok. Well, here is my gemoetric solution and how I choose the cross product. I hope it is readable:
.. wait. How can I upload a photo ?

allright, should be uploaded. You find the images at the end of this page
 
IMG_0081.JPG
IMG_0082.JPG
IMG_0083.JPG
 
So, what's confusing me is: This answer appears to be correct. But as you can see on the last page, the cross product of E x H does NOT yield a vector in k-direction, but in negative k-direction. So, WTF ?
 
It seems you have applied the right hand rule incorrectly. If you use this definition of the rhr and your sketch, you should have index finger along E, middle finger along H then the thumb along k is opposite to what you have drawn. Now the directions of H and k are given, so you need to redraw your E-vector.

Right_hand_rule.png
 
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