14.9 Thin Film Interference

Thin-film interference happens when light reflects from the two surfaces of a thin layer (thickness comparable to the light’s wavelength) and those two reflected waves interfere. Key points from the AP CED: part of the light is reflected and part transmitted at each boundary (14.9.A.1); a 180° phase flip occurs when reflection is from a medium with higher n, none when from lower n (14.9.A.2); refraction doesn’t change phase (14.9.A.3). The interference (constructive or destructive) depends on the optical path difference 2nt (n = film index, t = thickness), the wavelength in the film, any phase shifts on reflection, and angle of incidence (14.9.A.4). At normal incidence you often use 2nt = mλ (constructive, accounting for phase flips) or 2nt = (m + 1/2)λ (destructive)—quantitative work on the AP is limited to normal incidence. Practical examples: soap bubbles and oil films (color from varying t) and quarter-wave antireflection coatings (t = λ/4 in coating, n_coating between air and substrate) (14.9.A.5). For a quick CED-aligned review, see the topic study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze); for more unit context (https://library.fiveable.me/ap-physics-2-revised/unit-14) and tons of practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Soap-bubble colors come from thin-film interference. Light hits the thin soap layer and part reflects at the top surface and part transmits, reflects off the inner surface, then exits. Those two reflected waves interfere—sometimes adding (constructive) and sometimes canceling (destructive) for different wavelengths. Phase-change rules matter: reflection from air (n ≈ 1) into the higher-n soap film gives a 180° phase flip, while reflection from the film back into air does not, so the effective condition for bright/dark wavelengths depends on that extra half-cycle plus the film’s optical path 2n t (n = film index, t = thickness). Because t varies across the bubble and is comparable to visible wavelengths, different places favor different colors, producing the shifting rainbow patterns you see. This is exactly the thin-film behavior described in the CED (14.9.A; phase change on reflection and optical path difference). For a quick review, see the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and try practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

A 180° (π rad) phase flip on reflection happens whenever light reflects from a boundary into a medium with a higher index of refraction than the medium it’s coming from (CED 14.9.A.2.i). So at an air (n≈1) → glass (n>1) surface the reflected ray gets a half-wavelength phase shift; at a glass → air surface it does not (CED 14.9.A.2.ii). Remember: refraction (transmission) does NOT change phase (CED 14.9.A.3). In thin films, check each interface separately: the top-surface reflection gets a π shift if n_film > n_incident; the bottom-surface reflection gets a π shift if n_substrate > n_film. Those relative shifts combine with the extra path (2·film thickness) to determine constructive or destructive interference (CED 14.9.A.4). For the common quarter-wave antireflection coating: choose n_coating between air and substrate and thickness = λ/(4n_coating) so the two reflected rays cancel (CED 14.9.A.5.iii). For a quick review, see the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and more practice at (https://library.fiveable.me/practice/ap-physics-2-revised).

Constructive interference in a thin film means the two reflected rays add to make a brighter (stronger) reflected wave; destructive means they cancel and the reflected light is reduced or gone. For normal incidence (what the AP exam limits quantitative work to), whether they add or cancel depends on two things from the CED: (1) optical path difference 2 n t (where n is film index and t its thickness) and (2) any phase shifts on reflection (180° phase flip occurs when light reflects from a medium with higher n; no flip when reflecting from lower n). Quick rules you can use: - If both reflected rays experience the same net phase change, constructive occurs when 2 n t = m λ (m integer), destructive when 2 n t = (m + 1/2) λ. - If one reflected ray gets an extra 180° flip (net difference of π), the conditions swap: constructive when 2 n t = (m + 1/2) λ, destructive when 2 n t = m λ. Practical note: antireflection coatings use a quarter-wave (t = λ/(4n)) film with n between air and glass to force destructive interference of reflections (CED 14.9.A.5 and 14.9.A.4). For more practice and a clear study guide, see the Fiveable thin-film page (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and many practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Antireflection coatings use thin-film interference to make the reflected waves cancel. A thin layer (coating) with index n_coat is put on the glass (n_glass). For light at normal incidence, you choose the coating thickness t = λ0/(4 n_coat) (λ0 is the wavelength in air). Light reflected from the top surface and light that enters the coating, reflects off the coating–glass interface, then exits travel an extra 2t = λ0/2 (in air-equivalent), so those two reflected rays are half a wavelength out of phase. Combined with the phase-change rules (reflection from a higher n gives a 180° shift), the two reflected rays end up out of phase and destructively interfere—greatly reducing reflected light and increasing transmission. In practice n_coat is chosen between air and glass (ideally ≈ sqrt(n_air·n_glass)) and coatings are optimized for particular wavelengths. For AP exam work, remember the quarter-wave result and apply phase-change rules at normal incidence (CED Topic 14.9; see the Fiveable thin-film study guide: https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze). For more practice, check the AP Physics 2 practice bank (https://library.fiveable.me/practice/ap-physics-2-revised).

Thickness matters because it sets the optical path difference between the ray reflected from the top surface and the ray that travels through the film, reflects from the bottom surface, and comes back out. If that extra distance (2 · n · t for normal incidence, where n is the film’s index and t its thickness) equals an integer number of wavelengths (after accounting for any 180° phase shifts on reflection), the two reflected waves add (constructive interference); if it equals a half-integer number of wavelengths they cancel (destructive interference). Thin-film effects are strongest when t is comparable to the light’s wavelength (CED 14.9.A.4). That’s why soap bubbles show colors (different t → different wavelengths reinforced) and why anti-reflection coatings work: a quarter-wave coating (t = λ/4 in the coating, with index between air and substrate) produces destructive interference of reflections (CED 14.9.A.5.iii). For AP-style problems remember phase changes on reflection and that quantitative work is limited to normal incidence (CED boundary). For a clear review and practice questions check the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and more unit resources (https://library.fiveable.me/ap-physics-2-revised/unit-14).

You see rainbows because of thin-film interference: sunlight reflects off the top and bottom surfaces of the oil layer and those two reflected waves add together. Two things set whether a given color is bright or dark: (1) any phase flips on reflection (a 180° flip happens when light reflects off a medium with larger index of refraction) and (2) the extra distance the wave travels inside the film (optical path = 2 n_film t). For normal incidence, constructive interference for wavelength λ in air happens when 2 n_film t = m λ (or m+½ λ) depending on whether the net reflection phase shift is 0 or π. Because the oil’s thickness varies across the puddle, different spots meet the constructive condition for different visible wavelengths, so you get spatially varying colors. This is exactly the thin-film behavior described in the CED (phase change on reflection, optical path difference, constructive/destructive interference). For a quick review, check the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze). For more practice problems across Unit 14, see (https://library.fiveable.me/practice/ap-physics-2-revised).

When light hits a thin film from air it does three things: some is reflected at the top surface, some is transmitted into the film (and later partly reflected from the bottom surface), and some can be absorbed (14.9.A.1). Each reflected ray can pick up a phase shift of 180° if it reflects from a boundary to a higher index (n increases) and no phase shift if it reflects to a lower index (14.9.A.2). Refraction itself doesn’t change the wave’s phase (14.9.A.3). The two reflected rays recombine and interfere—their interference (constructive or destructive) depends on film thickness, wavelength in the film (=λ/n), any phase shifts, and angle of incidence (14.9.A.4). That’s why soap bubbles and oil films show color and why quarter-wave (λ/4 in the coating) antireflection layers work (14.9.A.5). For AP exam problems, quantitative thin-film work is limited to normal incidence (boundary statement). For a compact review, see the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) or the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-14) and try practice questions (https://library.fiveable.me/practice/ap-physics-2-revised).

Quick rule: when light reflects off a boundary with a higher index of refraction (n increases), the reflected wave gets a 180° phase flip (half-wavelength). When it reflects off a boundary with a lower index (n decreases), there’s no phase flip. Refraction (transmission) doesn’t add a phase flip. Apply that to thin films (normal incidence, AP scope): there are two reflected rays—one from the top surface (air → film) and one from the bottom surface (film → substrate). Check the n’s for each reflection: - If top reflection is air (n1) → film (n2) and n2>n1, top reflection flips. - For bottom reflection, compare film (n2) → substrate (n3). If n3>n2 that reflection flips, otherwise it doesn’t. Interference depends on both the optical path difference (2 n2 t) and whether 0, 1, or 2 half-wavelength flips occur. For normal incidence: - If the two reflected rays have a net extra half-wavelength from phase flips, constructive interference occurs when 2 n t = (m + 1/2) λ (m integer). If they have no net flip, constructive when 2 n t = m λ. (Destructive is the opposite.) Want worked examples and AP-style practice? Check the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze), the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-14), and the practice set (https://library.fiveable.me/practice/ap-physics-2-revised).

For a simple single-layer antireflection coating at normal incidence (AP CED boundary), choose a coating index n_coat between air (n≈1) and the substrate (n_substrate). The reflected waves from the top and bottom of the thin film will destructively interfere when the optical path difference equals half a wavelength: 2 n_coat t = λ_coating/2. Since λ_coating = λ_vacuum / n_coat, this gives the quarter-wave result: t = λ_vacuum / (4 n_coat). Also note the phase-change rule: a 180° shift occurs on reflection from a higher-n medium. For the usual case (n_air < n_coat < n_substrate) both reflections get a π phase shift (they cancel), so the quarter-wave thickness produces destructive interference of reflected light. AP questions restrict analysis to normal incidence—use this formula on the exam. For review and extra practice see the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and lots of practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

You see different colors on a soap bubble because the soap film’s thickness changes across the surface, so different wavelengths of light interfere constructively or destructively at different places. Light reflecting from the top film surface (air → film, n increases) gets a 180° phase flip; light that transmits, reflects off the inner surface (film → air, n decreases) does not. The two reflected rays recombine with an optical path difference ≈ 2 n t (film thickness t and film index n) plus any phase flips. For normal incidence, constructive interference for a particular wavelength happens when the total phase difference matches an integer multiple of 2π (effectively 2 n t ≈ (m + 1/2)λ because of the one 180° flip), so different t favor different λ. Changing viewing angle also changes effective path length, so colors shift with angle. This is thin-film interference (see CED 14.9.A: phase change on reflection, optical path difference, constructive/destructive interference). For a focused review, check the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze). For more unit review and practice, see Unit 14 (https://library.fiveable.me/ap-physics-2-revised/unit-14) and Fiveable’s practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Thin-film interference depends on the optical path difference between the ray reflected from the top surface and the ray that travels through the film, reflects from the bottom surface, and comes back out. For normal incidence (what the AP exam limits you to), the extra path = 2 n t, where n is the film’s index and t its thickness. Whether that path gives constructive or destructive interference also depends on phase shifts on reflection: - If the two reflected rays pick up the same net phase shift, constructive: 2 n t = m λ (m = 0,1,2...), destructive: 2 n t = (m + 1/2) λ. - If one ray gets a 180° (π) phase flip and the other doesn’t (common for a film with n_f > air but < substrate), the conditions swap: constructive: 2 n t = (m + 1/2) λ, destructive: 2 n t = m λ. A common antireflection trick: make t = λ/(4n) (quarter-wave) so reflections destructively interfere for that λ. Thin-film color effects (soap bubbles, oil) come from thickness varying across the film so different wavelengths satisfy these conditions at different places. For the AP CED guidance and review, check the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-14). For practice, use Fiveable’s problem set (https://library.fiveable.me/practice/ap-physics-2-revised).

The index of refraction matters two ways in thin-film interference: it controls phase shifts on reflection and it changes the wavelength (optical path) inside the film. If light reflects from a boundary into a medium with higher n, that reflected ray picks up a 180° phase flip; no flip happens if it reflects into a lower-n medium (CED 14.9.A.2). Also the wavelength in the film is λfilm = λvac/n, so the extra distance the second ray travels inside the film gives an optical path difference 2 n t (for normal incidence) that sets whether the two reflected rays interfere constructively or destructively (CED 14.9.A.4). That’s why soap-bubble colors vary with thickness and n, and why a quarter-wave antireflection coating uses t = λ/(4ncoating) with ncoating between air and substrate to force destructive interference (CED 14.9.A.5.iii). For AP problems they limit quantitative work to normal incidence—use phase-flip rules plus 2 n t to decide bright/dark. For a quick review, see the AP Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and more unit resources (https://library.fiveable.me/ap-physics-2-revised/unit-14). For extra practice, try the AP Physics 2 problem set (https://library.fiveable.me/practice/ap-physics-2-revised).

Because two reflected waves combine, they can cancel when they’re exactly out of phase. In a thin film you get one ray reflected at the top surface and another that goes into the film, reflects at the bottom surface, and comes back out. The relative phase between those two depends on: - phase shifts on reflection (180° if reflecting off a higher n, none if off a lower n)—EK 14.9.A.2, and - the extra distance traveled inside the film (optical path = 2 n t)—EK 14.9.A.4. For normal incidence (the AP boundary), destructive interference occurs when the optical path difference plus any reflection phase shifts make the two waves differ by an odd multiple of half a wavelength. Practically that’s why a quarter-wave coating (t = λ/(4n)) can eliminate reflection: it makes the two reflected rays 180° out of phase (EK 14.9.A.5.iii). For more practice and review, see the Topic 14.9 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and the unit page (https://library.fiveable.me/ap-physics-2-revised/unit-14).

Thin-film interference shows up all over real optics beyond bubbles. Common examples you should know for the CED: oil slicks and thin oil films on water (iridescent colors from varying film thickness), and antireflection coatings on eyeglasses, camera lenses, and smartphone screens—those use quarter-wave coatings and index choices to create destructive interference of reflected light (CED 14.9.A.5.ii–iii). Other real-world uses: dielectric (Bragg) mirrors and bandpass/edge filters in lasers and telescopes, iridescent insect/peacock wing colors (biological structural coloration), interference-based metrology (Newton’s rings) for measuring tiny thickness changes, and specialized thin-film sensors and decorative iridescent paints. For AP exam focus, be ready to link these examples to phase changes on reflection, optical path differences, and how thickness ~ wavelength controls constructive vs destructive interference (CED 14.9.A). Review the Topic 14.9 study guide here (https://library.fiveable.me/ap-physics-2-revised/unit-6/9-thin-film-interference/study-guide/OyOfqxqTZ5XbL7ze) and practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).