# Behaviour at an interface plane wave

I have this example diagram that was given in one of my lectures and I am just going through what the equation given actually mean and calculating some results from the equation. Which are the angle of incidence and the type of polarisation the incident wave is.

So here is the diagram with equation given:

So starting with the first part of the equation:

$$30\left(0.866i-0.5k\right)$$

What I understand of this is that this represent the direction and magnitude of the plane wave, so if a particle was put in front of this plane then, it would oscillate in the $$30\left(0.866i-0.5k\right)$$ direction.

So for the next part of the equation

$$e^{i\left(0.5x+0.866z\right)}$$

From this I gather that as the wave propagates parallel to the plane which is indicated by the $$x$$ and $$z$$ the we are looking at what is called a p polarisation, but I assume that if say the $$z$$ or $$x$$ component was replaced with a $$y$$ then we would be looking at a s polarisation due to the wave propagating perpendicular to the plane?

So to find the angle of incidence I did the following:

I took the experiential part to of the plane wave equation:

$$e^{\left(k.r\right)}$$

and then looking at the diagram from the propagation vector for incidence $$k_i$$ I get the following.

$$k_i=\left|k_i\right|\sin\left(90-\theta _i\right)\hat{z}+\left|k_i\right|\cos\left(90-\theta _i\right)\hat{x}$$

which simplify to

$$k_i=\left|k_i\right|\sin\left(\theta _i\right)\hat{x}+\left|k_i\right|\cos\left(\theta _i\right)\hat{z}$$

Now r vector is given by

$$r=x\hat{x}+z\hat{z}$$

using the dot product relationship I get the following for $$k \cdot r$$

$$\left|k_i\right|\sin\left(\theta _i\right)x+\left|k_i\right|\cos\left(\theta \:_i\right)z$$

comparing this to the equation given I make

$$\left|k_i\right|=1$$

$$\theta _i=\tan^{-1}\left(\frac{0.5}{0.866}\right)=30°.$$

Have I understood the equation correctly?

From this I gather that as the wave propagates parallel to the plane which is indicated by the $$x$$ and $$z$$ the we are looking at what is called a p polarisation,

Yes.

but I assume that if say the $$z$$ or $$x$$ component was replaced with a $$y$$ then we would be looking at a s polarisation due to the wave propagating perpendicular to the plane?

No. The $$s$$ polarization corresponds to a pure $$y$$ polarization (in the conventions you've set up). If you just replaced $$x$$ or $$z$$ in your $$p$$-polarized wave for $$y$$, i.e. if you had a polarization along $$0.5\hat{\mathbf j}+0.866\hat{\mathbf k}$$ or $$0.5\hat{\mathbf i}+0.866\hat{\mathbf j}$$, then the result would not be a viable plane wave, as the polarization would not be orthogonal to the wave vector.

Other than that, though,

Have I understood the equation correctly?

Yes.