Question

1. Explain why for (non-magnetic materials) there is a Brewster angle for TM waves, but not for TE waves. (Hint: use Snell's Law and the expressions for「E and「M-) Will the situation change when you have a magnetic material (ie, when 14-# 1)?

Answer 1

For non-magnetic material, Brewster’s effect exists only for TM waves, and not for TE waves. because for a plane electromagnetic wave incident on the plane boundary between medium 1 and medium 2, the amplitude reflectivities of TE and TM waves are given by the Fresnel formulae:

= sin (θt − θi) / sin (θt + θi) ............ Equ 1

TE

and = tan (θt − θi) / tan (θt + θi) ............Equ 2

「TM

where θi and θt are the angles of incidence and transmission respectively.

The numerators in Eq. 1 and Eq. 2 cannot vanish because θi is not equal to θt. However,
「TM
can vanish because tan (θi+θt) diverges to infinity when (θi+θt) is equal to π/2 but the reverse is not possible and
TE
can diverge to infinity.

On the other hand, with regard to TE waves, each dipole is perpendicular to the plane of incidence and emits waves isotropically in the plane. Therefore, no special angles exist for TE waves, so no TE waves exist.

For Magnetic material, $\mu _{r}$ not equals to 1, Brewster angle can be observed for TE waves in the case of such magnetic materials.

We assume that medium 1 is a vacuum and medium 2 is a medium with εr and $\mu _{r}$ . The amplitude reflectivities for TE waves and TM waves are expressed as follows:

= (Zr cos θi − cos θt) / (Zr cos θi + cos θt)

TE

and = (cos θi − Zr cos θt) / (cos θi + Zr cos θt)

「TM

where Zr = $\sqrt{}$(µr/εr) is the normalized wave impedance of medium 2. The incident angle θi and the transmitted angle θt are related by Snell’s law sin θi/ sin θt = n with n = √ εr *√ µr. The no-reflection conditions,
TE
= 0 or
「TM
= 0, can be written as follows:

0 ≤ $sin^{2}$ θi = (
an
)/($\alpha ^{2} - 1$) ≤ 1

where α = µr for TE waves and α = εr for TM waves. So, the Brewster angle is possible for TE mode.