# Optics: how can I see 180 degree relative phase shift between internally and externally reflected light?

Guys this is an example from chapter 4.3 of Hecht's Optics.

Top picture shows a light beam moving through a glass medium

The second picture shows that when the glass medium is cut into two halves (across the blue surface), the atoms on the blue surface oscillates thereby reflecting part of the light.

The arrow coming out of the rightmost blue surface is termed "external reflection", and the arrow coming out of the leftmost blue surface is termed "internal reflection".

Two questions:

Why doesn't the internal reflection reflect to the right?

The book also claims that there is a 180 relative phase shift between internally and externally reflected light. But both reflected light beams are traveling to the left so there is no phase shift.

Can someone help me understand this better?

• As soon as you type "Two questions" take that as a strong hint that your question should be more focused. Try to limit posts to one single well defined question if possible. – DanielSank Jan 7 '15 at 4:41
• @IllegalImmigrant The internal reflection yes reflects also to the right as follows: the beam comes from the left, and the part $T_{l,1}$ that is transmitted at the leftmost cut flies to the rightmost cut. Here, again, part is reflected $R_r$, part is transmitted. The reflected part $R_{r,1}$ flies back to the leftmost cut, where the process repeats, i.e. part is reflected $R_{l,2}$. This part flies again to the right where part of it is transmitted $T_{r,2}$. So, we can have also a reflected part to the right, but much weakened. – Sofia Jan 7 '15 at 5:17
• Hi Sofia, so the left most arrow denoting the internal reflection should really point right, is that correct? What is wrong with this textbook....? – Rodrigo Amarante Jan 7 '15 at 5:18
• @IllegalImmigrant The process is that at every interface between two media, we have reflection and refraction. In the case when the beam falls vertically on this surface, we use the word transmission instead of refraction. – Sofia Jan 7 '15 at 5:19
• @IllegalImmigrant : I see in your picture two green arrows. The first reflected one is reflected by the 1st medium-air interface. It should go to the left, as the picture indicates. – Sofia Jan 7 '15 at 5:21

Why doesn't the internal reflection reflect to the right?

Reflection to the right in your diagram is called "transmission."

The book also claims that there is a 180 relative phase shift between internally and externally reflected light. But both reflected light beams are traveling to the left so there is no phase shift.

Your drawing doesn't show the phase of the light. Remember that light is an oscillating electromagnetic field subject to boundary conditions. At a dielectric boundary the electric field component parallel to the surface must be continuous, because there's no charge geometry that would explain a discontinuity; the electric field component perpendicular to the surface will be discontinuous, because there is electric field normal to the surface from bound surface charges in the polarized material. This difference between the parallel and perpendicular E-field boundary conditions is the reason we have Snell's Law in dielectrics.

You have light normal to the surface, so you must have $$E_\text{incident} + E_\text{reflected} = E_\text{transmitted}$$ at both interfaces. However the material in the dielectric is polarized, with $$(\epsilon-\epsilon_0)E = P$$ For the interface with internal reflection, the polarization $P$ contributes to both the incident and reflected waves. However for the interface with external reflection the polarization contributes only to the transmitted wave. I think this gives you a sign difference in the contribution of $P$ to the reflected wave, but I'm too sleepy to remember the details — I'm pretty sure it has something to do with a wave equation for $D=\epsilon E$.

The way you measure a phase difference, of course, is with an interference experiment.

Here's another way you can think of it. Your phase difference is with respect to the incident light; there'll be another phase difference because there's a path length difference between the internally- and externally-reflected light. You can reduce this path length difference by bringing your two surfaces closer together. In the limit where the two surfaces are separated by much less than the wavelength of light and are exactly parallel, the internally- and externally-reflected rays should be exactly out of phase and interfere destructively. But the limit where the two surfaces have zero separation is the same as the limit where there's no interface at all and the dielectric is continuous — in that case there is also no reflection, only transmission.