When light is incident on a boundary where the index of refraction is changing, some of the light is reflected and some is transmitted. Consider the situation below, which could represent a thin layer of oil between air and water.
Figure8.7.1.Thin film interference.
Here light ray (1) is the original light beam incident on the top interface. It splits into a reflected ray (2) and a transmitted ray (3). At the bottom interface, the transmitted ray then splits into another reflected ray (4) and transmitted ray (6). Lastly, ray 4 interacts again with the top surface and some of it is transmitted (5). There are more reflections and transmissions, but the first two are the most dominant due to the intensity decreasing at every interface. Now light rays (2) and (5), which can be seen by an eye in the original medium, interference with each other. They are coherent since they originated from the same source and there is a path length difference \((\Delta D)\) between the two.
Principle8.7.2.Phase Shifts for Reflected Light.
When light reflects off a material with a higher index of refraction, the reflected wave has a phase shift of \(\pi\text{.}\) In contrast, if light reflects off a material with a lower index of refraction, there is no \(\pi\) phase shift. Transmitted waves never have a phase shift.
Additional Detail8.7.3.Where does the phase shift come from?
An analogy to the phase shift for reflected light is a wave pulse interacting with a medium of a higher density (slower wave speed). As seen in Figure 7.6.5, once the wave starts to interact with the higher density string, the reflected wave is inverted, shifting the wave by half of a cycle \((\pi)\text{.}\) In contrast, as seen in Figure 7.6.4, once the wave starts to interact with the lower density string, the reflected wave is not inverted. The transmitted pulse does not invert in either case.
ExercisesActivities
Explanation8.7.1.Path Length Difference.
Assume the thickness of the thin film is \(t\) and the incident ray is nearly normal to the surface. Determine the path length difference, \(\Delta D\) (the extra distance traveled by ray (5) compared to ray (2).
Solution.
At near normal incidence the \(\Delta D\) is equal to twice the thickness \(t\) of the film, \(\Delta D = 2t\text{,}\) because ray (5) travels first down then back up.
Explanation8.7.2.Oil on Water.
Suppose you have the air \(\rightarrow\) oil \(\rightarrow\) water system described above. The index of refraction for oil is \(n_{oil} = 1.44\) and the index of refraction for water is \(n_{water} = 1.33\text{.}\) For each of the two reflections, identify whether or not there is a phase shift.
Solution.
Oil has a higher index of refraction than air, so ray (2) has a \(\pi\) phase shift. When ray (3) reflects off the bottom surface, it is in oil and bouncing off water. Since \(n_{water}\) is smaller than \(n_{oil}\) there is no phase shift in reflected ray (4) or in (5) since it is transmitted.
ReferencesReferences
[1]
Content provided by BoxSand: https://boxsand.physics.oregonstate.edu/welcome.
[2]
Figure 8.7.1 modified from OpenStax, https://openstax.org/books/college-physics-2e/pages/27-7-thin-film-interference
