Thin Film Interference

AP PhysicsMontwood High SchoolR. Casao

•You often see bright bands of color when light reflects from a thin layer of oil floating on water or from a soap bubble as a result of interference.

•Light waves are reflected from the front and back surfaces of the thin film, and constructive interference between the two reflected waves with different wavelengths occurs in different places for different wavelengths.

•Light shining on the upper surface of a thin film with thickness t is partly reflected at the upper surface (path abc).

•Light transmitted through the upper surface is partly reflected at the lower surface (path abdef).•The two reflected waves come together at point P on the retina of the eye.

•Depending on the phase relationship, they may interfere constructively or destructively.

•Different colors have different wavelengths, so the interference may be constructive for some colors and destructive for others, which is why we see colored rings or fringes.

•The complex shapes of the colored rings in the photo result from differences

in the thickness of the film. •The bottom figure shows a thin transparent film of uniform thickness L and index of refraction n2

illuminated by bright light of wavelength λ from a distant point source.

•Assume that air lies on both sides of the film; so n1 = n3.

•Assume that the incident light ray is almost perpendicular to the film ( ).

•We are interested in whether the film is bright or dark to an observer viewing it almost perpendicularly.

•Incident light ray i strikes the left front surface of the film at point a and undergoes both reflection and refraction there.

•The reflected ray r1 is intercepted by the observer’s eye.

•The refracted light crosses the film to point b on the back surface, where it undergoes both reflection and refraction.

•The light reflected at b crosses back through the film to point c, where it

undergoes both reflection and refraction.•The light refracted at point c, represented by ray r2 is intercepted by the observer’s eye.

•If the light waves of rays r1 and r2 are exactly in phase at the eye, they produce an interference maximum and region ac on the film is bright to the observer.

•It rays r1 and r2 are out of phase at the eye, they produce in interference minimum and region ac is dark to the observer,

even though it is illuminated.•If there is some intermediate phase difference, there are intermediate interference and brightness.

•The key to what the observer sees is the phase difference between the waves of rays r1 and r2.

•Both rays are derived from the same ray i, but the path involved in producing r2 involves light traveling twice across the film (a to b, and then b to c), whereas the path involved in producing r1 involves no travel through the

film.•Because is about zero, we approximate the path length difference between the waves of r1 and r2 as 2·L.

•To find the phase difference between the waves, we cannot just find the number of wavelengths λ that is equivalent to a path difference of 2·L.

•This approach is not possible for two reasons:▫the path length difference occurs in a medium

other than air, and▫reflections are involved, which can change the phase.

•The phase difference between two waves can change if one or both are reflected.

Reflection Phase Shifts•Refraction at an interface never causes a

phase change – but reflection can, depending on the indexes of refraction on the two sides of the interface.

•Consider what happens when reflection causes a phase change using the example of pulses on a denser string (along which pulse travel is slow) and a lighter string (along which pulse travel is fast).

•When a pulse traveling slowly along the denser string reaches the interface with the lighter string, the pulse is partially transmitted and partially

reflected, with no change in orientation.•For light, this corresponds to the incident

wave traveling in the medium of greater index of refraction (recall that greater n means slower speed).

•The wave that is reflected at the interface does not undergo a change in phase; its reflection phase shift is zero.

• Light traveling from a more dense medium to a less dense medium is reflected from the interface with no phase change.

• A crest is reflected as a crest and a trough is reflected as a trough.

•When a pulse traveling fast along a lighter string reaches the interface with a denser string, the pulse is again partially transmitted and partially reflected.

•The transmitted pulse has the same orientation as the incident pulse, but now the reflected pulse in inverted.

•For a sinusoidal wave, the inversion involves a phase change of π rad, or half a wavelength (½·λ).

•For light, the situation corresponds to the incident wave traveling in the medium of

lesser index of refraction (with greater speed).

•The wave that is reflected at the interface undergoes a phase shift of π rad, or half a wavelength (½·λ).

•Light traveling from a less dense medium to a more dense medium is reflected from the interface with a phase change of π rad or ½·λ.•A crest is reflected as a trough and a trough is reflected as a crest.

•If the index of refraction for both media is the same, then the incident and transmitted waves have the same speed and there is no reflection from the interface.

•Summary:Reflection Reflection Phase Shift

Off lower index of refraction

Off higher index of refraction ½ wavelengthRemember as “higher means half”

Equations for Thin-Film Interference•Three ways in which the phase difference

between two waves can change:▫By reflection▫By the waves traveling along paths of different

lengths▫By the waves traveling through media of

different indexes of refraction•When light reflects from a thin film, producing the waves of rays r1 and r2 shown, all three ways are involved.

•Examine the two reflections in the figure.▫At point a on the front interface, the incident

wave in air reflects from the medium having the higher of the two indexes of refraction; so the reflected ray r1 has its phase shifted by ½·λ.

▫At point b on the back interface, the incident wave reflects from the medium (air) having

the lower of the two indexes of refraction; so the reflected wave is not shifted in phase by the reflection, and neither is the portion of it that exits the film as ray r2.

•As a result of the reflection phase shifts, the waves r1 and r2 have a phase difference of ½·λ and are exactly out of phase.

•Now consider the path length difference 2·L that occurs because the waves of ray r2 crosses the film twice.

•If the waves of r1 and r2 are to be exactly in phase so that they produce fully constructive interference, the path length 2·L must cause an additional phase difference of 0.5, 1.5, 2.5, … wavelengths.

•Only then will the net phase difference be an integer number of wavelengths.

•For a bright film, we must have:

(for in-phase waves)

•The wavelength we need here is the wavelength λn2 of the light in the medium containing path length 2·L (in the medium with index of refraction n2)

2 2odd numberL

•Rewrite the previous equation:

(for in-phase waves)

•If the waves are to be exactly out of phase so that there is fully destructive interference, the path length 2·L must cause either no

additional phase difference or a phase difference of 1, 2, 3, … wavelengths. Only then will the net phase difference be an odd number of half-wavelengths.

22 2 nodd numberL

•For a dark film, we must have:

(for out-of-phase waves)

where the wavelength is the wavelength λn2 in the medium containing 2·L.

2 integerL

22 integer nL

•Remembering that the greater the index of refraction of a medium, the smaller the wavelength of light in that medium to rewrite the wavelength of ray r2 inside the film:

where λ is the wavelength of the incident light in a vacuum (air).•

22

n n

•Substituting into

and replacing “odd number/2”:

for m = 0, 1, 2, . . . (maxima for bright film in air)

•Similarly, with m replacing “integer”:

for m = 0, 1, 2, . . . (minima for dark film in air)

22 2 nodd numberL

2

12 2L mn

22 L m

n

•For a given film thickness L, these equations tell us the wavelengths of light for which the film appears bright and dark, respectively, one wavelength for each value of m.

•Intermediate thicknesses give intermediate brightnesses.

•These equations also tell us the thicknesses of the films that appear bright and dark in that light, respectively, one thickness for each value of m.

•Intermediate thicknesses give intermediate brightnesses.

•Summary: if the film has thickness L, the light is at normal incidence and has wavelength λ in the film:▫If neither or both of the reflected waves from

the two surfaces have a half-cycle reflection phase shift, the conditions for constructive and destructive interference are:

▫Constructive (no relative phase shift): 2·L = m·λ where m = 0, 1, 2, …

▫Destructive (no relative phase shift):2·L = (m + ½)·λ where m = 0, 1, 2, …

▫If one of the two waves has a half-cycle reflection phase shift, the conditions for constructive and destructive interference are reversed:

▫Constructive (half-cycle relative phase shift):2·L = (m + ½)·λ where m = 0, 1, 2, …

▫Destructive (half-cycle relative phase shift):2·L = m·λ where m = 0, 1, 2, …