# reflected electric and magnetic fields component

The question states that the direction of propagation of E is +x and +z plane. The goal here is to find the reflected component of E and H. I know that the propagation changes sign once it hits the boundary. but what i don't understand in part a) why z component changes sign not x? why not both? Also in part b) I cannot visualize how the the sin and cosine component works here, also why these are added to reflected component of H ? can anyone draw this out for me please and explain it? Thanks a) Find the reflected electric field in phasor form. b) Find the reflected magnetic field in phasor form • Try drawing a picture - if E propagates in x and z and then is relected by "something", we don't know what that something is or its angle. Hopefully I'm being stupid and someone with more knowledge can answer! – Andy aka Oct 31 '15 at 23:33

I took your picture from your other question and I edited to make it match to your problem. As you can see the wave progresses in the [3,0,4] direction. This is the argument of the complex exponential. Remember than ax+by+cz=K (K is a constant) would be a plane and the vector perpendicular to it is [a,b,c].

Vector $\vec \beta _i$ is therefore [3,0,4]. You can see from the picture that the reflected $\vec \beta _r$ is [3,0,-4]. It reflects on the z=0 plane (and that's why z changes sign).

In the other hand, the cos and sin components allow to write the vector from its polar coordinates.

If you look at $\vec H _i$ it is easy to verify that its angle with $\beta _z$ is $\theta_i$ therefore its coordinates are $[-H_i \cos(\theta_i) , 0, H_i sin(\theta_i)]$.

Because $\Gamma$ is negative we have a direction change for the reflected E field. Also E, H and propagation direction must follow the 3 fingers rule, before and after the reflection, as they do in the figure.

For $\vec H _r$ you can see from the figure that its coordinates will be $[-H_r \cos(\theta_r) , 0, -H_r sin(\theta_r)]$ as in the solution.