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?


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.

  • $\begingroup$ So, if I had understood it correctly this means that there could be phase differences between the incident and transmitted waves if the interface is a conductor. $\endgroup$ – budding physicist Dec 20 '19 at 23:56

Whenever light (or any wave in general) goes from one medium to another, some of the energy of the wave is 'reflected' back through the first medium (at the same angle as the incident wave) and some of the energy (may be) refracted (bent) through the second medium.

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When light goes from a low refractive index medium to a high refractive index medium (such as air to water), the reflection undergoes a 180 degree phase change.

Conversely, when light goes from a high refractive index medium to a low refractive index medium (such as water to air) it DOES NOT undergo a phase change.

This is a general property of waves. See: Phase shift of 180 degrees on reflection from optically denser medium

The 'refracted' wave does not undergo a phase change either way.

  • $\begingroup$ Sorry I am a bit confused by your answer, does light undergo a phase change when it is refracted?? $\endgroup$ – user43487 Dec 6 '14 at 8:22
  • $\begingroup$ When a wave undergoes a change in refractive index, some the wave is reflected and some of the wave (may be) refracted. No phase change for the 'refracted' wave. But the reflected portion can undergo a phase change, if the wave is going from a lower to higher refractive index. $\endgroup$ – theo Dec 6 '14 at 8:29

Yes. The "Phase Shift" is:

δ = 2πΔ/λ

with Δ = Optical Path Difference (OPD) = (n2 - n1) × t

where n2 is the refractive index of a sample medium (e.g. glass, air, etc.) and n1 is the refractive index of the surrounding medium (e.g. air)

t is the thickness of the sample medium and λ is the wavelength of the light

  • $\begingroup$ I am sorry but I was wondering the consistency between this answer and the answers given above. Do you mean to say there is phase change due to refraction when the medium is optically thick? $\endgroup$ – budding physicist Dec 20 '19 at 23:16

The phase change is what guides the process. 180 degrees phase change is reflection. Other phase changes are refraction.


Consider the problem of thin film interference:

Thin film interference

n1,n2: refractive indices of medium 1,2 respectively

Let $n1<n2$ : at the interface between the two media we can calculate the reflectance vector of light as follows: r = (n1-n2)/(n1+n2) This tells you what portion of the incident light is reflected.

Consider the interface between n1,n2. Let us calculate 'r' here. Say n1 is air and n2 is oil.

$n1=1$, $n2=1.5$(approx.)

As light moves from air to oil, r = (1-1.5)/(1+1.5) = -0.2.

The "-" sign indicates that there is a reversal in the Electric field vector, implying a phase shift of 180 degrees. If light moved from n2 to n1, there would be no sign change and 'r' would turn out to be positive.

Therefore, there is a phase change of 180 degrees upon reflection from a denser medium and no change upon reflection from a rarer medium.

There is no phase change associated with refraction. This is because the Electric flux needs to be equal on either side of the interface. This can only be satisfied if the refracted ray has the same phase as the incident ray. What changes in refraction is the speed of light as it approaches n2 due to its different refractive index. This change in speed is compensated by a reduction in the wavelength of light so that there is no change in the frequency.

Hope this makes it clear.