14.9 Thin-Film Interference

1. ASome is reflected and some enters the glass, where a small part of it is absorbed. Correct
   At a boundary between media with different indices of refraction, the light divides: part is reflected and part is transmitted into the glass. Glass is not perfectly transparent, so a small part of the transmitted light is absorbed as it travels through the thick block; the reflected, transmitted and absorbed parts together account for all of the incident light.
2. BNone of it is reflected; all of it enters the glass, which is clear rather than shiny.
   A student who thinks only shiny surfaces reflect light picks this. Clear glass reflects part of the light at its surface (about 4% at normal incidence), which is why you can see faint reflections in a window.
3. CAll of it is reflected or all is transmitted, with the angle deciding which one.
   A student who pictures light at a boundary as going one way or the other picks this. The light divides at the surface at any angle of incidence; all-or-nothing reflection happens only in total internal reflection, which needs light traveling toward a smaller index, not from air into glass.
4. DSome is reflected and the rest crosses the block, since clear glass absorbs no light.
   A student who thinks only dark materials absorb light picks this. Clear glass absorbs a small fraction of the light passing through it, more for a thicker block; a thick slab of window glass looks green when viewed through its edge.

CED 14.9.A.1 · Read this in Fix

1. AThe index of refraction of the second medium, from which the light reflects
   A student who treats the phase change as a property of the reflecting material picks this. The same material can give either result: light in air reflecting from glass changes phase, but light in diamond reflecting from glass does not. The index of the first medium is needed for the comparison.
2. BWhich of the two media has the greater mass density, the first or the second
   A student who reads 'denser medium' as a medium of greater mass density picks this. The comparison is between indices of refraction: oil is less dense than water but can have the greater index.
3. CHow the index of refraction of the second medium compares with that of the first Correct
   The phase change on reflection depends on the relative indices of the two media: light reflected from a medium of greater index changes phase by 180°, and light reflected from a medium of smaller index does not. Both indices are needed to make the comparison.
4. DThe thickness of the second medium and the wavelength of the light used
   A student who mixes up the phase change on reflection with the phase difference from the extra path in a film picks this. Thickness and wavelength matter for the path difference; the phase change on reflection depends only on the two indices.

CED 14.9.A.2 · Read this in Fix

1. AIts phase shifts by 180° at the boundary, as glass has the greater index.
   A student who applies the reflection rule to transmitted light picks this. The 180° change happens to the light reflected from the glass; the light transmitted into the glass has no phase change.
2. BIts frequency decreases in the glass, because it travels more slowly there.
   A student who links a smaller speed to a smaller frequency picks this. The frequency is set by the source and is the same in air and in glass; the smaller speed goes with a shorter wavelength, λ/n.
3. CIts phase is continuous across the boundary, with no 180° shift. Correct
   Refraction does not change the phase of a wave: the transmitted wave continues from the incident wave without a sudden shift. Its speed and wavelength change in the glass, but its frequency does not. A 180° phase change can happen only to the reflected light.
4. DIts wavelength is unchanged in the glass, because its color does not change.
   A student who ties each color to one fixed wavelength picks this. In the glass the light is slower at the same frequency, so its wavelength is shorter, λ/n; it has the same color again when it leaves.

Working Refraction leaves the phase continuous (no 180° shift); f is unchanged, v = c/n and λ_n = λ/n decrease.

CED 14.9.A.3 · Read this in Fix

1. AThe bubble's wall is comparable in thickness to a wavelength of light; the pane is thousands of times thicker. Correct
   Thin-film interference occurs when the film's thickness is comparable to the wavelength of the light. A bubble's wall is a few hundred nanometers thick, like a wavelength of visible light, while a 3 mm pane is thousands of wavelengths thick, so it shows no thin-film colors.
2. BThe soap contains dyes that absorb some of the colors in sunlight, but clear window glass contains no such dyes.
   A student who thinks thin-film colors come from dyes picks this. A bubble made from clear soap solution still shows colors, and the colors change as the wall thins; they come from interference of light reflected from the wall's two surfaces.
3. CThe curved bubble spreads white light into its colors like a prism, which a flat pane cannot do.
   A student who attributes the colors to dispersion picks this. The colors come from interference of light reflected from the two surfaces of the bubble's wall, and flat soap films in a wire loop show them just as well.
4. DClear glass transmits all the light striking it, so the pane reflects no light that could interfere.
   A student who thinks clear materials reflect no light picks this. Each surface of a window pane reflects a few percent of the light; you can see reflections in a window. The pane shows no colors because it is far too thick.

CED 14.9.A.4 · Read this in Fix

1. AIt is destroyed where the two reflected waves interfere destructively.
   A student who thinks destructive interference destroys energy picks this. Energy cannot be destroyed; where the reflected waves cancel, the energy goes into the transmitted light instead.
2. BIt is absorbed by the coating, which warms up slightly as a result.
   A student who thinks antireflection coatings work like tinted glass picks this. The coating is transparent and absorbs very little; it reduces reflection by interference, so the lens transmits more light, not less.
3. CIt is scattered off to the sides, so it does not come back to the viewer.
   A student who confuses an antireflection coating with a matte finish picks this. Scattering would still reflect the light, only in many directions; the coating reduces the total reflected light, and the rest is transmitted.
4. DIt is transmitted into the lens, so more green light reaches the camera's sensor. Correct
   Energy is conserved. Destructive interference of the two reflected waves means that less energy is reflected, and the coating is transparent, so that energy is transmitted through the lens instead.

CED 14.9.A.5.ii · Read this in Fix

1. ARay 2 changes phase by 180°, but ray 1 does not.
   A student who has the rule the wrong way round, with the 180° change on reflection from a smaller index, picks this. Ray 1 reflects from the oil, of greater index than air, so it is the one that changes phase; ray 2 reflects from water, of smaller index than oil, so it does not.
2. BBoth rays change phase by 180° as each reflects.
   A student who compares mass densities picks this: oil is denser than air, and water is denser than oil. The rule uses indices of refraction, and water (n = 1.33) has a smaller index than oil (n = 1.45), so ray 2 has no phase change.
3. CNeither of the rays changes phase on reflection.
   A student who thinks reflection never changes the phase of light picks this. Ray 1 reflects from the oil, which has a greater index than air, so it changes phase by 180°.
4. DRay 1 changes phase by 180°, but ray 2 does not. Correct
   Ray 1, traveling in air (n = 1.00), reflects from oil (n = 1.45), which has the greater index, so it changes phase by 180°. Ray 2, traveling in oil, reflects from water (n = 1.33), which has the smaller index, so it has no phase change.

Working Ray 1: air (1.00) → oil (1.45), reflects from the greater index: 180° change. Ray 2: oil (1.45) → water (1.33), reflects from the smaller index: no change.

CED 14.9.A.2.i · Read this in Fix

1. ADark, since only ray 1 changes phase by 180° on reflecting
   A student who puts the 180° change on reflection from the smaller index picks this. The prediction is right but the reasoning is not: ray 1 reflects from air, of smaller index than glass, so it keeps its phase; it is ray 2, reflecting from glass, that changes phase.
2. BDark, since only ray 2 changes phase by 180° on reflecting Correct
   Ray 1 travels in glass and reflects from air, which has the smaller index, so it keeps its phase. Ray 2 travels in air and reflects from glass, which has the greater index, so it changes phase by 180°. Near P the wedge is so thin that the path difference is almost zero, so the two waves are 180° out of phase and cancel: the region near P is dark.
3. CBright, since both rays change phase by 180° on reflecting
   A student who thinks every reflection changes the phase picks this. Ray 1 travels in glass and reflects from air, of smaller index, so it keeps its phase. Only ray 2 changes phase, so the waves near P are 180° out of phase.
4. DBright, since both rays travel equal distances very near P
   A student who uses the path difference alone picks this. The path difference near P is indeed almost zero, but ray 2 alone changes phase by 180° on reflection, so the two waves cancel and the region near P is dark.

Working Ray 1: glass (1.52) → air (1.00), reflects from the smaller index: no change. Ray 2: air (1.00) → glass (1.52), reflects from the greater index: 180° change. Near P the wedge thickness t → 0, so the path difference 2t → 0 and the net phase difference is 180°: destructive interference, dark.

CED 14.9.A.2.ii · Read this in Fix

1. A4 units
   A student who adds the amplitudes whatever the phases picks this, 3 + 1 = 4. That is right only for waves in phase; here each crest of wave 1 meets a trough of wave 2, so the amplitudes subtract.
2. B2 units Correct
   The two waves combine into one wave whose disturbance at each instant is the sum of theirs. Every crest of wave 1 (+3) occurs with a trough of wave 2 (−1), so the resulting wave has crests of 3 − 1 = 2 units: amplitude 2 units.
3. C3 units
   A student who thinks the stronger wave passes on unchanged picks this. The waves do not compete; they add, and wave 2's trough reduces each crest of wave 1 from 3 to 2 units.
4. D0 units
   A student who thinks destructive interference always cancels completely picks this. Complete cancellation needs equal amplitudes; here the amplitudes are 3 and 1, so a wave of amplitude 2 remains.

Working The waves have the same period and are 180° out of phase (each crest of wave 1 coincides with a trough of wave 2). Resulting amplitude = 3 − 1 = 2 units.

CED 14.9.A.4.i · Read this in Fix

1. A306 nm
   A student who ignores the phase changes on reflection and sets 2nt = λ picks this: 2(1.33)(115 nm) = 306 nm. One reflection changes phase by 180°, so 2nt = λ/2 for the strongest reflection, giving 612 nm.
2. B460 nm
   A student who compares the extra path with the wavelength in air, 2t = λ/2, picks this: λ = 4t = 460 nm. The extra path is inside the soap, so it must be compared with λ/n, which gives λ = 4nt = 612 nm.
3. C346 nm
   A student who converts between wavelengths the wrong way, dividing by n, picks this: 4(115 nm)/1.33 = 346 nm. The wavelength in the film is λ/n, so the wavelength in air is n times the film wavelength: λ = 4nt = 612 nm.
4. D612 nm Correct
   Only the reflection at the top surface (air to soap, greater index) changes phase by 180°, so constructive interference needs 2t = (m + ½)λ/n. The longest wavelength has m = 0: λ = 4nt = 4(1.33)(115 nm) = 612 nm.

Working Top surface: air → soap (n increases), 180° phase change. Bottom surface: soap → air (n decreases), none. Net 180° from reflection, so constructive interference requires 2t = (m + ½)λ_n with λ_n = λ/n. Longest wavelength: m = 0, so 2t = λ/(2n) and λ = 4nt = 4(1.33)(115 nm) = 612 nm.

CED 14.9.A.4.ii · Read this in Fix

1. A250 nm
   A student who takes the extra path to be t rather than 2t picks this, setting t = λ/n = 600 nm/2.40 = 250 nm. The wave reflected from the bottom crosses the layer twice, so 2t = λ/n and t = 125 nm.
2. B300 nm
   A student who compares the extra path with the wavelength in air picks this, setting 2t = 600 nm. Inside the layer the wavelength is 600 nm/2.40 = 250 nm, so 2t = 250 nm and t = 125 nm.
3. C125 nm Correct
   Only the top reflection (air to layer, greater index) changes phase by 180°; the bottom reflection (layer to glass, smaller index) does not. With a net 180° from reflection, destructive interference needs a path difference of a whole number of wavelengths in the layer: 2t = λ/n for the smallest nonzero thickness, so t = 600 nm/(2 × 2.40) = 125 nm.
4. D720 nm
   A student who takes the wavelength in the layer to be n times the wavelength in air (1440 nm) picks this. Light is slower in the layer at the same frequency, so its wavelength there is shorter, 600 nm/2.40 = 250 nm, and t = 125 nm.

Working Top surface: air (1.00) → layer (2.40), n increases: 180° change. Bottom surface: layer (2.40) → glass (1.50), n decreases: no change. Net 180° from reflection, so destructive interference requires 2t = mλ_n with λ_n = λ/n = 600 nm/2.40 = 250 nm. Smallest nonzero thickness: m = 1, t = 250 nm/2 = 125 nm.

CED 14.9.A.4.ii · Read this in Fix

1. AThe oil refracts each color by a different amount, sending each a different way.
   A student who explains thin-film colors by dispersion picks this. For a flat film, light of every color reflected from the two surfaces leaves at the angle of reflection; the colors come from interference, which changes with angle.
2. BThe path difference between the two reflected waves changes with the viewing angle. Correct
   The amount of constructive or destructive interference depends on the angle at which the light crosses the film as well as on the thickness, wavelength and phase changes. At a low angle the path difference between the two reflected waves is different, so a different wavelength interferes constructively and the color changes.
3. CAt a low angle, all of the light reflects from the top surface and none enters the oil.
   A student who thinks the angle decides whether light is all reflected or all transmitted picks this. At any angle, part of the light reflects from the top surface and part enters the oil and reflects from the bottom, so the two reflected waves still interfere; the color changes because their path difference changes with the angle.
4. DThe oil contains pigments, and each pigment shows a different color from each direction.
   A student who thinks the film's colors come from pigments picks this. The oil film is essentially clear; its colors come from interference of the waves reflected from its two surfaces.

CED 14.9.A.4.ii · Read this in Fix

1. AThe film's thickness changes with height, so the wavelength reflected most strongly changes too. Correct
   The side view shows that the film gets thicker from top to bottom. At each height, the wavelengths that interfere constructively depend on the thickness there, so each height reflects a different color, and the bands run horizontally, along lines of equal thickness.
2. BThe film acts like a prism, spreading the colors of white light out from its top down to its bottom.
   A student who explains thin-film colors by dispersion picks this. The film's colors come from interference of light reflected from its two surfaces; the side view shows the thickness changing with height, which is what sets the color.
3. CDye in the soap drains downward, so each height absorbs a different part of the white light.
   A student who thinks the colors come from dyes picks this. Clear soap solution gives the same bands; the diagram's side view shows the thickness changing with height, and the thickness sets which wavelengths interfere constructively.
4. DThicker parts of the film reflect more light, so each band is brighter than the band above it.
   A student who thinks thicker films reflect more light picks this. The front view shows repeating bands of different colors, not a steady brightening downward; reflection rises and falls with thickness because of interference.

CED 14.9.A.5.i · Read this in Fix

1. A109 nm Correct
   The index increases at both surfaces (1.00 to 1.38 to 1.50), so both reflected rays change phase by 180° and those changes cancel. Destructive interference then needs a path difference of half a wavelength in the coating: 2t = λ/(2n), so t = λ/(4n) = 600 nm/(4 × 1.38) = 109 nm.
2. B100 nm
   A student who uses the index of the glass instead of the coating picks this: 600 nm/(4 × 1.50) = 100 nm. The extra path is traveled in the coating, so its index, 1.38, sets the wavelength: t = 600 nm/(4 × 1.38) = 109 nm.
3. C150 nm
   A student who uses the wavelength in air picks this: 600 nm/4 = 150 nm. Inside the coating the wavelength is 600 nm/1.38 = 435 nm, and a quarter of that is 109 nm.
4. D217 nm
   A student who takes the extra path to be t rather than 2t picks this, setting t = λ/(2n) = 600 nm/(2 × 1.38) = 217 nm. The wave reflected from the glass crosses the coating twice, so 2t = λ/(2n) and t = 109 nm.

Working Air (1.00) → coating (1.38): 180° change. Coating (1.38) → glass (1.50): 180° change. The two cancel, so destructive interference requires 2t = (m + ½)λ_c with λ_c = λ/n_c. Minimum thickness (m = 0): t = λ/(4n_c) = 600 nm/(4 × 1.38) = 109 nm.

CED 14.9.A.5.iii · Read this in Fix

1. AThe coating absorbs green light, so the light that it reflects contains no green light.
   A student who thinks the coating works by absorbing light picks this. The coating is transparent; the green light that is not reflected is transmitted into the lens.
2. BReflection is canceled best for green, so some red and blue light is still reflected. Correct
   The coating's thickness gives exactly half a wavelength of path difference only for green light. For red and blue light the two reflected waves are not exactly 180° out of phase, so they cancel only partly, and a mixture of red and blue, which looks purple, is reflected.
3. CThe coating spreads white light into its colors like a prism and reflects the purple part.
   A student who explains colors from thin layers by dispersion picks this. The coating is flat and thin; the color comes from interference, which cancels the reflection of some wavelengths more than others.
4. DThe coating scatters green light off to the sides, so less green reaches the eye.
   A student who thinks the coating works like a matte finish picks this. The coating is smooth; less green light is reflected because the two reflected green waves interfere destructively.

CED 14.9.A.5.ii · Read this in Fix

1. ABrighter at 100 nm, where the film has less pigment to absorb the green
   A student who thinks a film's color and brightness come from pigments absorbing light picks this. Soap film has no pigment and absorbs almost no light at either thickness; the difference comes from interference of the two reflected waves, which are in phase at 100 nm and out of phase at 200 nm.
2. BBrighter at 200 nm, where the path difference is one wavelength in the film
   A student who uses the path difference alone picks this. The path difference at 200 nm is one wavelength in the film, but one reflection changes phase by 180°, so the waves there are out of phase and the reflection is weak.
3. CBrighter at 200 nm, where the thicker film reflects more of the light
   A student who thinks a thicker film reflects more light picks this. Brightness depends on the phase difference between the two reflected waves; at 200 nm they are out of phase and the reflection is weak.
4. DBrighter at 100 nm, where the two reflected waves arrive in phase Correct
   In the film λ/n = 532 nm/1.33 = 400 nm. At 100 nm the path difference 2t = 200 nm is half a wavelength, which with the single 180° change at the top surface puts the waves in phase: strong reflection. At 200 nm, 2t = 400 nm is one whole wavelength, so the single 180° change leaves the waves out of phase: weak reflection.

Working λ_film = 532 nm/1.33 = 400 nm. One 180° change (top surface only). t = 100 nm: 2t = 200 nm = λ_film/2, plus the 180° change → in phase (constructive). t = 200 nm: 2t = 400 nm = λ_film, plus the 180° change → 180° out of phase (destructive).

CED 14.9.A.4.ii · Read this in Fix

1. Aλ/2
   A student who compares the extra path 2t with the wavelength in air sets 2t = λ and picks this. The extra path is traveled inside the coating, where the wavelength is λ/n_c.
2. Bλ/(2n_c) Correct
   Both reflections go from lower to higher index, so both change phase by 180° and the phase changes cancel. Constructive interference then needs the extra path 2t to be a whole number of wavelengths in the coating, λ/n_c; the smallest nonzero thickness is λ/(2n_c).
3. Cn_cλ/2
   A student who takes the wavelength in the coating to be n_cλ sets 2t = n_cλ and picks this. Light is slower in the coating at the same frequency, so its wavelength there is λ/n_c.
4. Dλ/n_c
   A student who takes the extra distance traveled by the wave reflected from the bottom surface to be t, instead of 2t, sets t = λ/n_c and picks this. That wave crosses the coating twice, down and back up.

Working Top surface: air (1) to coating (n_c > 1), 180° phase change. Bottom surface: coating (n_c) to glass (n_g > n_c), 180° phase change. The two changes cancel, so only the path difference 2t counts, measured in wavelengths in the coating, λ/n_c. Constructive: 2t = mλ/n_c; smallest nonzero thickness (m = 1): t = λ/(2n_c).

CED 14.9.A.4.ii · Read this in Fix

1. A0.87t_X
   A student who takes the wavelength in a coating to be n_c times the wavelength in air makes t proportional to n_cλ and picks this. The wavelength in the coating is λ/n_c, so a coating of greater index is thinner, not thicker.
2. B0.75t_X
   A student who uses the wavelength in air, t = λ/4, scales only by 450/600 and picks this. The quarter wavelength is measured inside the coating, so the coating index must be included.
3. C0.60t_X
   A student who finds the wavelength in each coating with the index of the glass beneath it scales by (450/600)(1.52/1.90) and picks this. The light travels through the coating, so the coating's index, not the glass's, sets its wavelength there.
4. D0.65t_X Correct
   The minimum thickness is a quarter of the wavelength in the coating, t = λ/(4n_c), so t is proportional to λ/n_c. Changing λ by 450/600 and n_c by 1.60/1.38 gives t_Y = (0.750)(1.38/1.60)t_X = 0.65t_X.

Working Simplest coating: thickness one-quarter of the wavelength in the coating, t = λ/(4n_c) (both reflections change phase by 180°, since 1 < n_c < n_g). t_Y/t_X = (λ_Y/λ_X)(n_cX/n_cY) = (450/600)(1.38/1.60) = 0.750 × 0.8625 = 0.647, so t_Y = 0.65t_X. (t_X = 109 nm, t_Y = 70.3 nm.) The glass index does not enter.

CED 14.9.A.5.iii · Read this in Fix

1. A0.500λ
   A student who compares the extra path inside the film with the wavelength in air sets 2Δt = λ and picks this. Inside the oil the wavelength is λ/1.50, and it is this wavelength that the extra path must change by between neighboring dark bands.
2. B0.750λ
   A student who takes the wavelength in the oil as n times the wavelength in air sets 2Δt = 1.50λ and picks this. Light travels more slowly in the oil at the same frequency, so its wavelength there is shorter: λ/1.50.
3. C0.333λ Correct
   Reflection at the top (air to oil) changes the phase by 180° and reflection at the bottom (oil to water, a smaller index) does not, so dark bands are where 2t = mλ/1.50. Neighboring dark bands differ by one in m, so 2Δt = λ/1.50 and Δt = λ/3.00 = 0.333λ.
4. D0.667λ
   A student who takes the extra distance traveled by the light reflected from the bottom surface as t, not 2t, sets Δt = λ/1.50 and picks this. That light crosses the film twice, down and back up, so the extra distance is 2t and Δt = λ/(2 × 1.50).

Working Top surface: air (1.00) to oil (1.50), reflection from a greater index: 180° phase change. Bottom surface: oil (1.50) to water (1.33), reflection from a smaller index: no phase change. Net 180°, so dark bands are where 2t = mλ_oil, with λ_oil = λ/1.50. Neighboring dark bands differ by 1 in m, so 2Δt = λ/1.50 and Δt = λ/(2 × 1.50) = λ/3.00 = 0.333λ. (The phase changes set which thicknesses are dark, but not the spacing between neighboring dark bands.)

CED 14.9.A.5.i · Read this in Fix