Question on phasors at boundary of oblique incidence.

[yungman]

A plane wave travels in ##\hat k_I=\hat x \sin\theta_I+\hat z \cos \theta_I## direction hitting a boundary formed by xy plane ( z=0). The incidence wave is in the plane of incident formed by xz plane where y=0.

We let ##\tilde E_I(\vec k_I)= \hat x E_{I_x}+\hat y E_{I_y}+\hat z E_{I_z} =E_{0I}e^{-j\vec k_I\cdot \vec r}##. This means ##\tilde E_{0I}## has to have x, y and z components ##\Rightarrow\;\tilde E_{0I}=\hat x E_{0I_x}+\hat y E_{0I_y}+\hat z E_{0I_z}##

But at ##z=0##, ##E_{0I_z}## has to be zero!
If we let ##E_{0I_z}=0##, then it won't work for the vector where z is not 0! how do I resolve this? Only way I can think of is ##E_{0I_z}## is a function of z and it's zero at z=0. Am I right?

thanks

 

[eljose79]

for the help!Hello,

Thank you for your question. It seems like you are trying to understand the behavior of a plane wave hitting a boundary formed by the xy plane. In this situation, the incident wave is in the plane of incidence, which is formed by the xz plane where y=0.

You are correct in thinking that the electric field components of the incident wave, denoted by ##\tilde E_I(\vec k_I)##, must have x, y, and z components. This is because the incident wave is a vector quantity and it has to have components in all three directions.

Now, when the incident wave hits the boundary formed by the xy plane at z=0, the z component of the electric field, ##E_{0I_z}##, has to be zero. This is because the boundary does not allow any electric field in the z direction. However, this does not mean that ##E_{0I_z}## has to be zero for all values of z. It can still have a non-zero value for z ≠ 0.

To resolve this, you can think of ##E_{0I_z}## as a function of z, as you have suggested. This means that the z component of the incident wave's electric field is dependent on the value of z. At z=0, it is zero, but for other values of z, it can have a non-zero value.

I hope this helps to clarify your understanding. Let me know if you have any further questions.