Does impedance matching eliminate reflection at non-normal incidence?

[a b]

Hello everybody,
I've often read that if a light ray travels from a medium 1 to a medium 2 and the two media have the same impedance, there is no reflected light at all, only transmitted light.
So i made some calculations assuming to have two media with two different refractive indexes but same impedance :I think there's no problem with it, if I have correctly understood the equations:

impedance Z1= sqrt( (mu_zero*mu_r1)/(epsilon_zero*epsilon_r1))
impedance Z2= sqrt( (mu_zero*mu_r2)/(epsilon_zero*epsilon_r2))
refractive index n1= sqrt(mu_r1*epsilon_r1)
refractive index n2= sqrt(mu_r2*epsilon_r2)

so i must choose:
mu_r1= k*mu_r2
epsilon_r1=k*epsilon_r2
with k a real number

I've chosen:

mu_r1=2
epsilon_r1=2
mu_r2=1
epsilon_r2=1

with these parameters, using the fresnel equations it turns out to me that if the incoming light ray is not perpendicular to the surface of separation between the two media, there may be some reflection; if t he angle is large enough there seems to be even total internal reflection.
I'm wrong with my calculations or it makes sense?
Maybe the rule I've read (impedance matching= no reflection) is valid only if the incoming ray is perpendicular to the separation surface?

Thank you

P.S: English is not my first language, excuse me if I don't write well.

 

[chrisbaird]

If the indices of refraction are different, there will be reflection. If the impedances are different, there will be reflection. I think the correct rule of thumb is "if the two media have the same impedance and magnetic effects are neglected, there will be no reflection." Magnetic effects are much rarer than dielectric effects in everyday materials (mu_r ≈ 1 much more often than epsilon_r ≈ 1) so that often in literature the magnetic effects are assumed to be negligible unless otherwise stated.

 

[Claude Bile]

chrisbaird is right.

Fresnel equations (where the refractive index matching condition comes from) require the permeability to be unity.

Claude.