# 180 degree out of phase reflection?

1. Jan 4, 2010

### stringbean

I was just reading my book when it said that the fact that newtons rings were always dark in the center where the lens touches the glass plate is proof that reflections are 180 degrees out of phase with the incident light, but after further thinking, I think that logic doesn't make complete sense. It says there are other experiments that prove reflections are 180 degrees out of phase so I would like it if someone would tell me what they are. The problem with the logic is that if an incident wave reflects first at one surface and then at the other like in thin films, even if the surfaces are closer than a wavelength, they would both be out of phase with the incident light and would therefore be in phase with each other if the surfaces are close enough. The book implies that they should now be out of phase to match the experiment but according to my logic, the reflected waves, should still be in phase with each other. I would also wonder in the experiment if any reflection is taking place in the center anyway. Are some reflections perfectly in phase and other reflections perfectly out of phase with the incident light? Are there other experiments that prove perfectly out of phase reflections?

2. Jan 4, 2010

### Stonebridge

Standing wave patterns on a string are produced by the reflected wave, 180 deg out of phase, interacting with the incident wave.

3. Jan 4, 2010

### Staff: Mentor

There are two reflected waves to consider: The reflection at the lens/air interface and the reflection at the air/glass interface. Only the second one gets a 180° phase shift, thus the two reflected waves are out of phase with respect to each other.

When light reflects from a medium of higher index of refraction there is a 180° phase shift; When light reflects from a medium of lower index of refraction there is no phase shift. This is also what happens in thin film interference.

4. Jan 4, 2010

### stringbean

Thanks for your input. I understand now, but I have other questions. We could only deduce or reason from this experiment that one of the reflections is out of phase. How do we prove which one? Also is there other experiments that prove which reflections are out of phase with the incident light?

5. Jan 4, 2010

### MrFrankie

I'll post here a question that i've posted in another thread. The topic is the same.
If you have only unidirectional wave and monochromatic light, and parallel film surfaces, you should observe the same overlapping on the whole wavefront. Let's suppose that the interference is completely destructive... the energy reflected is zero. So... where does the energy go? The fraction of light transmitted is the same, so it doesn't depend on the reflected waves interference.

6. Jan 4, 2010

### Bob S

Here is another example of the reflection of a TEM signal normally incident at a boundary.
Consider a TEM RF signal in a coaxial cable, like RG-58 or other coax. The ratio of the H (azimuthal magnetic) and E (radial electric) is conveniently specified by an impedance Z = E/H = 50 ohms. This TEM signal propagates indefinitely in the coax until it encounters an impedance mismatch (similar to a change in the index of refraction).
1) If the cable is open at the end (very high impedance termination), there can be no current, so the H' of the reflected wave is -H. Thus for the reflected signal E'=E since there is no transmitted signal. Either E or H has to be inverted because the Poynting vector, S = E x H, has changed direction. So the voltage at the end of the coax is doubled.
2) If the cable is shorted at the end, there can be no voltage, so for the reflected signal E'=-E. Thus for the reflected wave H'=H, and the current at the end has doubled.

So in one case, The E field is inverted (going from higher to lower impedance), and in the other case, the E field is not inverted (going from a lower impedance to a higher impedance).

Bob S

7. Jan 4, 2010

### Staff: Mentor

That light reflected from a medium of higher index of refraction gets a 180° phase change is a consequence of Maxwell's equations and the boundary conditions at the interface between the two media. But you're right--this experiment only shows that one of the reflected waves has been phase shifted, but not which one. I'll see if I can think of an experiment that could tell which wave gets shifted.

8. Jan 4, 2010

### Staff: Mentor

The energy goes into the transmitted wave.
Not true. For example, by properly coating a lens with a film of the right thickness and index of refraction the two reflected waves can be made to destructively interfere, thus reducing the reflected energy and increasing the amount of light transmitted. (This is called an 'anti-reflection coating'.)

9. Jan 4, 2010

### MrFrankie

Thank you for the answer, but i'm still not sure.
The point is that the amount of light transmitted is due to the transmission or reflection coefficients, a factor on the amplitude on the wave. Still the phenomenon of interference is due to the phase of the coherent overlapping waves.
If you change the thickness of the film you only change the optic path, therefore the phase, not the reflection coefficient and so the relative amount of transmitted amplitude.
And more, between the surfaces of the film you don't have overlapping, so the density of energy inside should be the same, changing only because of the larger volume, not because of the phase.
What do you think about it?