I have seen a lot about when light undergoes a phase change when it is reflected. But does it undergo a phase change when refracted and if so why and if not why not?
[Physics] Does light change phase on refraction
boundary conditionselectromagnetic-radiationmaxwell-equationsopticsrefraction
Related Solutions
Great question!
You might have learned that the amplitude of compression and the amplitude of particle displacements are not synonymous. In fact, the maximum amplitude of pressure and the maximum amplitude of particle displacements are out of phase for $\pi/2$. And twice $\pi/2$ (one for original, and one for the reflected wave) accounts for the missing $\pi$ in the phase change of particle displacement.
Imagine, that rarefaction travels towards the wall, which is on the right side. On the moment the wave strikes the wall, maximum displacement is left of rarefaction, that is $\pi/2$ behind it. The same is true for the reflected wave, that is, maximum displacement is again left of the rarefaction, only the direction of the wave is opposite, so maximum displacement amplitude is $\pi/2$ in front of rarefaction.
Thus, the phase of particle displacement changes phase for $\pi$, while the phase of pressure does not change at all at rigid surface.
It actually depends on what kind of beam spitter you have.
I'll give a general treatment and shows that the conclusions of both Emilio Pisanty and Steven Sagona are basically correct, corresponding to different specific beam splitters, which are all common in the laboratory. For simplicity we don't consider loss in this answer.
First of all, the definition of "phase shift" in this specific answer is chosen as the relative phase between reflected and transmitted light from the same port. More specifically, we let the transmission coefficient to be real number $t$, and the reflection coefficient will thus carries the information about relative phase shift $r e^{i\theta_\alpha}$, where $\alpha=1,2$ representing the light coming from beam splitter port 1 or 2 (see picture below).
Instead of assuming symmetrical phase shift, we allow any possible phase shift, as we are discussing general beam splitter which is not necessarily symmetric.
Then we write the physics procedure happening in beam splitter into the following matrix form: \begin{eqnarray} \begin{pmatrix} E_3 \\ E_4 \end{pmatrix} = \underbrace{ \begin{pmatrix} t & r e^{i\theta_2} \\ r e^{i\theta_1} & t \end{pmatrix}}_{ M} \begin{pmatrix} E_1\\ E_2 \end{pmatrix} \end{eqnarray}
The conservation of energy requires $|E_3|^2 + |E_4|^2 = |E_1|^2 + |E_2|^2$, which is equivalent to the mathematical statement that the beam splitter matrix is unitary. That gives us \begin{eqnarray} M^\dagger M = \begin{pmatrix}r^2 + t^2 & rt(e^{i\theta_2}+e^{-i\theta_1}) \\ rt(e^{i\theta_1}+e^{-i\theta_2}) & r^2 + t^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \end{eqnarray}
It can be seen that $r^2 + t^2 =1$ is automatically true as we deal with lossless beam splitter here. The rest of the above matrix equation gives us \begin{eqnarray} e^{i\theta_2}+e^{-i\theta_1} \equiv 2 e^{i(\theta_2-\theta_1)/2} \cos \frac{\theta_1+\theta_2}{2} =0 \end{eqnarray} from which we conclude that \begin{equation} \theta_1 + \theta_2 = \pi \end{equation}
If we have a beam splitter with symmetric phase shifts, $\theta_1 =\theta_2 = \pi /2$, then \begin{equation} M= \begin{pmatrix} t & ir \\ ir & t \end{pmatrix} \end{equation} This is consistent with Steven Sagona's argument based on the same assumption.
If we have a beam splitter that is not only symmetric in phase shifts ($\theta_1 = \theta_2=\pi/2$), but also symmetric in reflection and transmission ($r=t=1/\sqrt 2$), then we have \begin{equation} M=\frac{1}{\sqrt 2} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \end{equation} Then we basically arrive at Emilio Pisanty's second equation.
If we have a beam splitter that is not symmetric in phase shifts ($\theta_1 = 0, \theta_2=\pi$), which is also very common in the lab, then we have \begin{equation} M=\frac{1}{\sqrt 2} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \end{equation} Remember I have been putting the relative phase factor to reflection instead of transmission so that we have $\{t, r e^{i \theta_\alpha}\}$. Now, we can also put the phase factor to transmission so that we have $\{t e^{-i\theta_\alpha}, r\}$, without losing any physics. What we got after is Emilio Pisanty's first equation, namely \begin{equation} M=\frac{1}{\sqrt 2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}
Best Answer
The key word here is continuity.
The continuity boundary conditions for the electromagnetic field vectors sets these phenomenons. The tangential components of $\vec{E}$ and $\vec{H}$ must be continuous across an interface - the only way that they can not be is if there is a surface current flowing (which cannot happen in dielectrics). Likewise, the normal components of the displacement and induction must be continuous across the boundary (the only way that they cannot is if nett charge gathers on the surface - again - not possible in a dielectric.
On the incidence side there are two fields - the incident and reflected one. The sum of these two must have the same phase as the transmitted wave, by the continuity conditions. If you write down boundary conditions for tangential and normal components, then you find that the only way that continuity can be upheld is if the incident and transmitted waves are in phase, whilst the reflected wave may be either exactly in or out of phase, depending on whether the waves are crossing an increase or decrease in refractive index.
Continuity of boundary conditions holds for different physical reasons for different waves, but the conditions are often similar. We don't often find fields in nature with a sudden, discontinuous change in them.