# Change of phase in total internal reflection

1. Aug 17, 2010

### htg

I would like to know if there is change of phase during total internal reflection. (In particular I would like to know if I can cover copper with several microns of glass and have it reflect 200 MHz EM waves coming from water (epsilon(H2O)=80, epsilon(glass)=4)).

2. Aug 17, 2010

### sophiecentaur

The thin layer (compared with the wavelength involved) of glass would not have any significant effect, I think. The copper would dominate and it would be a case of reflection by a metal surface.

3. Aug 17, 2010

### htg

It it not clear to me. The layer of copper is also thin, but it reflects close to 100% of tha wave energy. This is why I am interested in the phase change in the case of reflection from the water/glass interface.

4. Aug 17, 2010

### sophiecentaur

But it can't significantly alter the phase as its so thin - the transit time must be only a few picoseconds.

5. Aug 17, 2010

### htg

Somehow comparably thin layer of Cu changes the phase by about Pi radians during reflection.

6. Aug 17, 2010

### sophiecentaur

Ah yes but the skin depth is minute for copper because of it high conductivity. Glass is a dielectric and things are different because the wave can penetrate. The evanescent wave 'behind' the interface doesn't have a chance to form before it hits the copper.

7. Aug 17, 2010

### htg

I hope you are right. By the way, what happens to phase during total internal reflection when we have a thick enough layer of dielectric?

8. Aug 18, 2010

### DrDu

200 MHz corresponds still to a wavelength on the order of 1 meter in glass.
To see an effect of the glass on the radio waves it should be at least as thick.
In copper the situation is different. The dielectric constant is very large and negative so that the penetration depth is only, say, some micro meters.

9. Aug 18, 2010

### Born2bwire

As I recall, the reflection coefficient is +1 for total internal reflection at a dielectric to dielectric interface at the critical angle. But of course there is a 180 degree phase shift for one of the components (the magnetic field I think) due to the change in the direction. Beyond the critical angle, the reflection coefficient becomes complex and you introduce a phase shift to both components I believe.

But yeah, having a thin layer of dielectric on the copper is not going to do anything. The resulting reflection coefficient is

$$R = \frac{R_{12}+R_{23}e^{2ik_zd}}{1+R_{12}R_{23}e^{2ik_zd}}$$

For a thin dielectric the exponential is approximately 1 and R_{23} = -1 for a PEC and thus the total reflection coefficient is approximately -1. So no real difference.

We can also recast it as

$$R = R_{12} + \frac{T_{12}R_{23}T_{21}e^{2ik_zd}}{1-R_{21}R_{23}e^{2ik_zd}}$$

If we have total internal reflection, then R_{21} = 1, T_{21} = 2 and as before R_{23} = -1. Thus,

$$R = R_{12} - T_{12} = -1$$

We can easily calculate this exactly for a water interface of 1 mm over copper with an incident angle of 45 degrees. The total reflection coefficient is -1 - 2.40081795943819e-009i.

Last edited: Aug 18, 2010
10. Aug 18, 2010

### htg

What is e-009i ? (The decimal point of the exponent is not there)

11. Aug 18, 2010

10^(-9)*i.