# EM Radiation with Oblique Insidence of Dielectric Boundary

So in my university course, we have studied the EM radiation infalling perpendicularly onto a dielectric boundary. I have now tried to go beyond and work out how the relationships look for oblique incidence, but I am running into trouble as some of my results are not consistent with eachother. I would be really happy if somebody could point out what I am doing wrong.

Throughout this analysis I assume $\mu = 1$.

Consider a dielectric boundary in the xy-plane, separating air at z>0 and a dielectric with relative permittivity $\epsilon$ at z<0. Take the incoming radiation to have angle $\theta$ to the surface normal and be polarised in the xz-plane, ie $$\vec{E_0} = (-E_0\cos \theta, 0, -E_0 \sin \theta) e^{-iwt}$$ and $$\vec{H_0} = (0,H_0,) e^{-iwt}$$

Now we have the continuity conditions for a dielectric boundary:

$D_{perpendicular}, B_{perpendicular} =$ cont.

$E_{parallel}, H_{paralell} =$ cont.

Hence inside the plasma (I believe) we should have:

$$\vec{E_p} = (-E_0\cos \theta, 0, \frac{-E_0}{\epsilon} \sin \theta) e^{-iwt}$$ and $$\vec{H_p} = (0,H_0,) e^{-iwt}$$

This gives the ratios of the fields as: $$|\frac{E_p}{H_p}| = \frac{E_0}{H_0} \sqrt{(\cos\theta)^2+(\frac{\sin\theta}{\epsilon})^2} = Z_0\sqrt{(\cos\theta)^2+(\frac{\sin\theta}{\epsilon})^2}$$

However, we also know that $$|\frac{E}{H}| = \frac{Z_0}{n}$$ where $n = \sqrt{\epsilon}$

This is clearly not consistent with the above expression involving the angles, worryingly enough not even for perpendicular incidence $\theta = 0$.

Where do I go wrong in my line of thought?