# Phase change in reflection

1. Oct 1, 2011

### McLaren Rulez

Hi,

When there is reflection, we generally use the phase shift upon reflection to be $\pi$. Where does this $\pi$ come from or is it arbitrary? I ask because I came across an optics book which describes beam splitters (a mirror is of course a beam splitter with reflectivity, R=1 and transmittivity, T=0) and as long as we have

$$e^{ikx} -> \sqrt{T}e^{ikx} + \sqrt{R}e^{i\theta}e^{iky}$$
$$e^{iky} -> \sqrt{T}e^{iky} + \sqrt{R}e^{i\theta'}e^{iky}$$

and $\theta+\theta'=\pi$

it is perfectly valid to choose any phase shift for the reflected beam. So why is $\pi$ everywhere? In particular, if we talk about thin film interference, the fact that it is $\pi$ seems to be very important.

And on the same note, reflected light will experience a 180 degree phase change when it reflects from a medium of higher index of refraction and no phase change when it reflects from a medium of smaller index. Why is this so?

Thank you!

2. Oct 1, 2011

### DiracRules

Just answering the first question (why $\pi$)

Consider a polarized wave directed perpendicularly towards a plane surface (not to have sines and cosines to mess things up). Be $\mathbf{k}$ the wave vector.
Since the directions of $\mathbf{E},\mathbf{B},\mathbf{k}$ (not their magnitude) are linked by $\mathbf{E}\wedge\mathbf{B}=\mathbf{k}$, and since the magnetic field does not change in direction on the interface of a medium if the latter has no densities of current on it ($div\mathbf{H}=0$), when $\mathbf{k}$ is reflected, to maintain the vectorial relationship, the phase of $\mathbf{E}$ has to be shifted from a $\pi$ factor.

I can't answer now (and it's quite disturbing, since I studied it last year) why passing from a medium with higher refractive index to one with a lower on you do not have reflection.

3. Oct 1, 2011

### JeffKoch

I don't have Born and Wolfe in front of me, I'd have to dig it up to explain since it's been many years since I looked at this, but the pi shift is *not* universal at all, particularly for thin films. It's the limit for reflection from a conducting metal surface, as I recall, but it does not apply to thin dielectric films or stacks of films - you have to do more work to calculate the phase shift. I waded into this in some detail while designing an interferometer for a company I worked for, turned out - to my great advantage - that reflection off a non-polarizing beamsplitter was giving me a 40-degree phase shift compared with the transmitted beam coming from the other direction, which was enough for electronic quad counter circuits to treat them as sines and cosines, which turned the system into an interferometer encoder without the need for 1/4 waveplates, polarization splitting, etc.

4. Oct 3, 2011

### DiracRules

Ok, I studied better this topic in today optical physics, so now I can answer to you properly :D
Some conventions: $n=\frac{n_2}{n_1}$, where n1 is the first medium and n2 the second, so that you have internal reflection for n<1;
$E_{0,i},E_{0,r},E_{0,t}$ are the amplitudes of the incoming, reflected and transmitted waves.

From Fresnel equation linking the amplitudes of the incoming, the reflected and the transmitted waves, you get the Fresnel coefficient for reflection and transmission:
$r=\frac{E_{0,r}}{E_{0,i}}=\frac{\cos(\theta_i)-\sqrt{n^2-\sin^2(\theta_i)}}{\cos(\theta_i)+\sqrt{n^2-\sin^2(\theta_i)}}$ Actually this expression is for a particular type of waves - the ones with the electric field perpendicular to the incident plane, but the qualitative considerations fit with all the other types since they do not differ too much from this expression.

Now, we have to consider 2 cases: n>1 (external reflection) and n<1 (internal reflection).
1) n>1
If you plot r, you'll see that $\forall\theta_i,\,\, r<0$
This means that $E_{0,r}=rE_{0,i}=-|r|E_{0,i}=|r|E_{0,i}\cdot e^{i\pi}$, where I transformed $-1=e^{i\pi}$
If we consider now the global equation for E, we get

$\vec{E_r}=E_{0,r}e^{i\{\vec{k}\cdot\vec{r}-\omega\cdot t\}} =|r|E_{0,i}e^{i\{\vec{k}\cdot\vec{r}-\omega\cdot t+\pi\}}$

As you can see, the reflected wave has a pi phase shift.

2) n<1
For internal reflection things are a bit different: for $\theta_i\in[0,\theta_{critical}]$ ($\theta_{critical}$ is the angle when complete internal reflection occurs, and satisfies $\sin\theta_{critical}=n$), the phase shift is 0, while for higher $\theta_i$ the phase shift depends on $\theta_i$. If $\phi$ is the phase shift,
$\tan\frac{\phi}{2}=\frac{\sqrt{\sin^2(\theta_i)-n^2}}{\cos(\theta_i)}$

This is the analytical explanation. Dunno if it's too complicate :D Hope you are satisfied :D

5. Oct 3, 2011

### sophiecentaur

Whenever there is a discontinuity, you get a reflection. Beyond the Critical angle, it's Total, too.

6. Oct 3, 2011

### McLaren Rulez

Thank you DiracRules for writing it out fully for me. Much appreciated!

Two quick questions:

1) How come in thin film interference, we always assume the light reflecting off the back end of the film undergoes zero phase shift? That seems wrong.

2) For a beam splitter in an optical circuit, how come we have a liberty to choose any $\theta$ and $\theta'$ as long as $\theta+\theta'=\pi$. Your explanation suggests that the phase shift is always defined based on incidence angle and the refractive indices so how come this freedom exists?

Last edited: Oct 3, 2011
7. Oct 3, 2011

### sophiecentaur

The reflectivity formula depends upon the difference between the two refractive indices. In one case the result is positive and in the other the result is negative. So a phase change for reflection into the more dense and no phase change for reflection going out of the more dense.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/reflco.html" [Broken]

Last edited by a moderator: May 5, 2017