Presentation on theme: "Thin Film Interference"— Presentation transcript:
1Thin Film Interference AP Physics BMontwood High SchoolR. Casao
2You 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.
3Light shining on the upper surface of a thin film with thickness t is partly reflected at the upper surface (path abc).Light transmittedthrough the uppersurface is partlyreflected at the lowersurface (path abdef).The two reflectedwaves come togetherat point P on theretina of the eye.
4Depending on the phase relationship, they may interfere constructively or destructively. Different colors havedifferent wavelengths,so the interferencemay be constructivefor some colors anddestructive for others,which is why we seecolored rings orfringes.
5The complex shapes of the colored rings in the photo result from differences in the thickness of thefilm.The bottom figure showsa thin transparent film ofuniform thickness L andindex of refraction n2illuminated by bright lightof wavelength λ from a distantpoint source.
6Assume 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 istrikes the left frontsurface of the film at point aand undergoes both reflectionand refraction there.
7The 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 itundergoes bothreflection and refraction.The light refracted at point c,represented by ray r2 isintercepted by the observer’s eye.
8If 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 isilluminated.If there is some intermediatephase difference, there areintermediate interference andbrightness.
9The 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 thefilm.Because is about zero, weapproximate the path lengthdifference between the wavesof r1 and r2 as 2·L.
10This approach is not possible for two reasons: 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, andreflections areinvolved, which canchange the phase.The phase difference betweentwo waves can change if one orboth are reflected.
11Reflection 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).
12When a pulse traveling slowly along the denser string reaches the interface with the lighter string, the pulse is partially transmitted and partiallyreflected, 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.
13Light 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.
14When 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 oflesser index ofrefraction (withgreater speed).
15The wave that is reflected at the interface undergoes a phase shift of π rad, or half a wavelength (½·λ).
16Light 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.
17If 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.
18Reflection Phase Shift Summary:ReflectionReflection Phase ShiftOff lower index of refractionOff higher index of refraction½ wavelengthRemember as “higher means half”
19Equations for Thin-Film Interference Three ways in which the phase difference between two waves can change:By reflectionBy the waves traveling along paths of different lengthsBy the waves traveling through media of different indexes of refractionWhen light reflects from athin film, producing thewaves of rays r1 and r2shown, all three ways areinvolved.
20Examine 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) havingthe lower of the two indexes ofrefraction; so the reflected waveis not shifted in phase by thereflection, and neither is theportion of it that exits the film asray r2.
21As 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 beexactly in phase so that they producefully constructiveinterference, the path length2·L must cause an additionalphase difference of 0.5, 1.5, 2.5, …wavelengths.
22Only 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 thewavelength λn2 of the light in themedium containing pathlength 2·L (in the mediumwith index of refraction n2)
23Rewrite 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 noadditional phase difference or a phasedifference of 1, 2, 3, …wavelengths. Only then willthe net phase difference be anodd number of half-wavelengths.
24For a dark film, we must have: (for out-of-phase waves)where the wavelength is the wavelength λn2 in the medium containing 2·L.
25Remembering 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 ofthe incident light in a vacuum (air).
26Substituting intoand 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)
27For 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.
29Summary: 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, …
30If 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, …
31Summary: Interference in Thin Films Normal incidence:Constructive reflection, no phase shift2·t = m·λ, m = 0, 1, 2, 3, ...Destructive reflection2·t = (m+½) ·λ, m = 0, 1, 2, 3, ...λ: Light wavelength in the filmλο: Light wavelength in airλ = λο/n
32Phase Shift at Interface When na<nb, a phase shift of π, or ½·λ, occurs.Constructive reflection: 2·L = (m+½)·λ, m=0, 1, 2, 3…Destructive reflection: 2·L = m·λ, m=0, 1, 2, 3...
33If 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:1. Constructive interference (no relative phase shift):2·L = m·λ2. Destructive interference (no relative phase shift) (no relative phase shift):2·L = (m + ½)·λ
34If the film has thickness L, the light is at normal incidence and has wavelength λ in the film; if one of the two waves has a half-cycle reflection phase shift:1. Constructive interference (half-cycle phase shift):2·L = (m + ½)·λ2. Destructive interference (half-cycle phase shift) (no relative phase shift):2·L = m·λ
35Interference in Thin Films Equation1 phase reversal0 or 2 phase reversals2·L = (m + ½)·lconstructivedestructive2·L = m·l