Thin films, optical structures and photonic uses

Reflections can combine

Light reflected from different interfaces can interfere. Relative phase depends on optical path length, including thickness, refractive response and angle. A thin film can therefore alter reflection without requiring a new bulk absorption mechanism.

Imagine putting a transparent layer on a transparent substrate. Some incident light returns from the first boundary; some enters the layer, reflects from the second boundary and returns through the first. The detector receives their combined electric field. Adding the two reflected powers separately would miss interference: the instantaneous fields add before the time-averaged power is calculated.

Phase describes position within an oscillation. Two same-frequency field components with a stable relative phase can reinforce when their oscillations agree and oppose when they are half a cycle apart. A half-cycle change is π radians, or 180°; a full cycle is 2π radians. Complete cancellation also requires the relevant returning field amplitudes to match. A phase condition alone does not set their sizes.

Use vacuum wavelength λ₀ to label the light. In a transparent scalar medium of phase index n, wavelength is λ₀/n. The index is dimensionless; multiplying a physical distance by n gives its optical path length for this simple propagation model. Wavelength, refractive index and dispersion provides the broader phase/refraction context, while Reflection, absorption and transmission supplies the separate power and collection account. The needed film quantities are defined here.

Count reflection changes as well as the extra travel

Begin at normal incidence on flat, parallel interfaces: propagation is perpendicular to both faces. Let the incident region, film and substrate have positive real indices n₀, n_f and n_s. At a lower-to-higher-index reflection the electric field reverses sign, a π phase change in this regime. A higher-to-lower-index reflection does not reverse it. These rules are not a general phase prescription for absorbing media, total internal reflection or every polarisation at an oblique interface.

The first returning contribution reflects at the incident-region/film boundary. The second makes a round trip through physical film thickness d, so its additional propagation optical path at normal incidence is 2 n_f d. Its propagation phase relative to the first is φ = 2π(2 n_f d/λ₀) = 4πn_f d/λ₀. Use the same length units for d and λ₀. The boundary reflection changes must also be included; the longer path alone does not tell us reinforcement versus cancellation.

For an original fictional example, set n₀ = 1.00, n_f = 1.50, n_s = 2.25 and λ₀ = 600 nm. Both relevant reflections go towards a higher index, so both acquire the same π change. Their difference from reflection is zero modulo 2π. Choosing d = 100 nm gives 2 n_f d = 300 nm = λ₀/2, so φ = π and the two contributions oppose. Equivalently d = λ₀/(4 n_f), a quarter of the in-film wavelength, not a quarter of λ₀.

The first figure separates these path accounts for reading. A real plane wave has repeated returns inside the film as well, not just two isolated rays. The next section includes those returns in the reflected fraction. If the substrate instead had index below the film, one of the two reflection sign changes would differ and the same quarter-wave travel could reinforce rather than cancel.

Phase matching and amplitude matching are different requirements

Use the following bounded model to quantify the example. The film and substrate are homogeneous, isotropic, nonmagnetic and lossless, with abrupt flat parallel interfaces. Illumination is coherent and normally incident, and the film has uniform thickness. The substrate extends onward without a returning rear-surface beam. This is a single coating on that substrate, not an entire finite window or lens. All indices and thicknesses below are hypothetical teaching inputs, not measured material specifications.

For a normal-incidence electric-field amplitude written on one fixed transverse axis, define the signed interface coefficients a = (n₀ − n_f)/(n₀ + n_f) and b = (n_f − n_s)/(n_f + n_s). A negative coefficient includes the reflection’s π sign change. The reflected power fraction of the complete film, after summing its repeated internal returns, is R = [a² + b² + 2ab cos φ]/[1 + (ab)² + 2ab cos φ]. Here R is dimensionless and cos uses φ in radians. You may use this stated response formula without deriving a multilayer matrix; a and b are interface coefficients, not two independently measured outgoing beam amplitudes.

For n₀ = 1.00, n_f = 1.50 and n_s = 2.25, a = b = −0.200. At the 100 nm quarter-wave thickness and 600 nm wavelength, cos φ = −1. The numerator is (a − b)² = 0, while the denominator is (1 − ab)² = 0.9216. Thus this ideal complete-film model gives R = 0. The zero is not evidence that the incident power was absorbed: this lossless boundary sends all of it forward into the substrate.

The zero requires a = b as well as the opposing phase. Solving that equality gives n_f² = n₀ n_s, so n_f = √(n₀ n_s) for positive indices. The square root here means the positive value whose square is the product. With a different supplied film index 1.60, its quarter-wave thickness at 600 nm is 93.75 nm, but the index match is absent. The quarter-wave expression R = [(n₀ n_s − n_f²)/(n₀ n_s + n_f²)]² gives R ≈ 0.00415368, or 0.415368%. Correct phase has reduced reflection, but has not made it zero.

As another check, setting d = 0 removes the film’s propagation region and the formula returns the bare incident-region/substrate value R_bare = [(n₀ − n_s)/(n₀ + n_s)]² = 25/169 ≈ 0.147929, or 14.7929%. Adding a film has changed a wave boundary problem; its index name alone does not prescribe that change.

One reflection minimum does not determine the whole spectrum

Keep n₀ = 1.00, n_f = 1.50 and n_s = 2.25 constant over the following fictional spectral interval. Real indices can disperse; holding them constant is an explicit teaching approximation. Compare films 100 and 300 nm thick. At λ₀ = 600 nm their round-trip phases are π and 3π. Both are an odd number of half-cycles and both give R = 0 in this matched-index model.

They do not have the same response elsewhere. At 450 nm the 100 nm film has φ = 4π/3 and R ≈ 4.15973%; the 300 nm film has φ = 4π and R = 14.7929%, the bare value. At 900 nm their reflected fractions are again 4.15973% and 14.7929%, but for different phase orders. The table and graph use the complete response formula, not an assumption that the first two reflected powers simply add.

A requirement of R ≤ 5% at the three specified wavelengths 450, 600 and 900 nm is met by the 100 nm film at all three, while the 300 nm film meets it only at 600 nm. That is a three-point requirement, not proof from those observations alone that every intermediate wavelength passes. For the stipulated continuous model, the plotted curve supplies additional predictions; a real broadband claim needs relevant spectral measurements and their uncertainty.

This also exposes an inverse problem. A measured dip near 600 nm alone does not identify 100 nm as the unique thickness, even when the indices are known: an odd higher-quarter-wave thickness can share that dip. Measurements away from the dip, independent thickness evidence and appropriate index data can distinguish alternatives. A colour impression or one low-reflection point is not a composition or complete-structure measurement.

Performance has a range

A coating designed for one wavelength and incidence may behave differently elsewhere. Loss, thickness variation and roughness matter. A photonic structure’s effective response belongs to its geometry and operating range, not solely to the constituent material name.

In the matched-index normal-incidence model, the first quarter-wave minimum follows λ_min = 4 n_f d. If the thickness is 90, 100 or 110 nm at unchanged n_f = 1.50, the first minima move to 540, 600 or 660 nm. At the fixed 600 nm evaluation, both 90 and 110 nm films give about 0.423059% reflection. Equal departures on the two sides of this particular minimum have equal R because the cosine in the formula is symmetric about φ = π.

Suppose a thickness record supplies d anywhere from 90 to 110 nm, with every other model input fixed. The phase at 600 nm then lies from 0.9π to 1.1π. Across that interval the largest reflected fraction is at its endpoints, so 0 ≤ R ≤ 0.00423059. This meets an invented R ≤ 0.5% test at that one wavelength within this ideal model. It is a bounded calculation from supplied thickness limits, not a manufacturing capability, uncertainty confidence level or real coating tolerance.

Angle changes the phase comparison too. For incidence θ₀ from the normal, refraction gives n₀ sin θ₀ = n_f sin θ_f. Comparing returned waves on a common outgoing phase plane gives propagation phase φ = 4π n_f d cos θ_f/λ₀. It is not found by merely replacing d with the longer zigzag distance d/cos θ_f: the waves being compared also enter and leave at different surface positions. At oblique incidence the field reflection amplitudes depend on polarisation, so the normal-incidence formula and its zero must not be transferred unchanged.

Surface roughness can redirect light, variations of d can make different regions respond differently, and absorbing constituents change the balance. Illumination bandwidth, angular spread and coherence can also weaken the visibility of a simple interference pattern. A good fit at one position and angle does not establish the response of a curved, stressed or nonuniform component.

Repeated optical structure can select and confine waves

Photonic structures use spatial variations of refractive response to control light. A repeated high/low-index layer sequence can arrange many interface returns to reinforce reflection over a range, rather than suppress it as the single film did at its chosen wavelength. This is a different boundary structure; putting an average index into the one-film formula does not calculate the stack.

For a separate geometric example at normal incidence and λ₀ = 600 nm, quarter-wave layers with n_H = 2.00 and n_L = 1.50 have physical thicknesses 75.0 and 100 nm. One high/low period is 175 nm; three such pairs contain six layers and total 525 nm. Equal optical thickness does not mean equal physical thickness. These dimensions specify a hypothetical arrangement, not its achieved reflectance, usable bandwidth or fabrication recipe.

Such repeated-index reflectors can provide mirrors in optical cavities. A cavity supports fields repeatedly returned between reflecting regions; the phase accumulated between those regions helps select resonant wavelengths. A transmission filter can instead be arranged to favour a selected band through its interference conditions. Layer sequence, termination, surrounding media and loss must be retained when comparing either function.

A photonic crystal more generally has a periodic optical structure. Its permitted wave patterns depend on that period, the geometry within it, direction and polarisation; a forbidden frequency range for one class of waves is not automatically a gap for all directions and polarisations. A finite sample’s reflected/transmitted spectrum also depends on its boundaries. The core lesson here is to connect a named function to those architectural conditions, not to assign universal performance from the constituent material or from an unqualified band-gap label.

Test the structure and the optical boundary together

For a coating claim, record the layer sequence, measured thickness and uniformity, actual substrate and surrounding media, wavelength range, incidence distribution, polarisation, surface condition and specimen temperature or processing state where relevant. Compare reflection and transmission under a common incident-power reference and collection geometry. A narrow detector can miss scattered light; missing specular reflection is not automatically transmitted power or absorption.

The model’s zero reflection assumes negligible absorption and no uncounted outgoing route. On a real absorbing substrate, less returned power can coexist with more absorbed power rather than useful transmission. A finite rear surface adds further returns; a thick incoherent substrate requires a different treatment from one coherent cavity. Test those alternatives before interpreting a fitted layer thickness as an established physical state.

A meaningful application comparison therefore names its target band, allowed angles and desired output, then checks the actual structure and response over that domain. Do not certify a real window, laser mirror, thermal limit or service life from the ideal zero or a few matching points. The original narrow-band example asks whether the observed behaviour meets the stated design range, not whether every coating should work for every colour.

Worked examples

Try the reasoning

1. Can equal chemical composition but different film thickness change colour?

   Show the answer and reasoning

   Yes.

   Optical path and interference change.

2. Does a successful flat sample establish every curved-component response?

   Show the answer and reasoning

   No.

   Angles, thickness uniformity and geometry can differ.

3. Keep the fictional matched indices, but target the first normal-incidence quarter-wave minimum at 800 nm. What physical thickness is required? Is 800/4 = 200 nm correct?

   Show the answer and reasoning

   d = 800/(4 × 1.50) = 133.333 nm. A 200 nm layer uses a quarter of the vacuum wavelength instead of the in-film wavelength.

   The required optical thickness is n_f d = λ₀/4 = 200 nm, so physical d is smaller by n_f. This is an ideal design input, not a real deposition specification.

4. At normal incidence let n₀ = 1.00, n_f = 1.50 and n_s = 1.00. A quarter-wave film gives φ = π. Are the two relevant reflection changes the same, and does that travel condition produce destructive reflection in this arrangement?

   Show the answer and reasoning

   No. The front reflection reverses sign and the film-to-substrate reflection does not. Quarter-wave travel now reinforces the returning field contributions.

   The propagation half-cycle and the additional reflection half-cycle together give a full relative cycle. The quarter-wave slogan has changed meaning because the surrounding-index order changed. Amplitudes and all returns are still needed for a quantitative R.

5. A spreadsheet reports zero reflection for every wavelength after setting d = 0 while keeping a film-index entry of 1.50. What should the complete one-film model give for n₀ = 1.00 and n_s = 2.25?

   Show the answer and reasoning

   The bare-interface fraction 25/169 ≈ 0.147929, or 14.7929%, rather than zero.

   With φ = 0 the response becomes (a + b)²/(1 + ab)². This reduces to the direct incident-region/substrate reflection. A zero-thickness entry has not created an amplitude-matched quarter-wave coating.

6. For the original 100 nm film, change external incidence to 30.0° from n₀ = 1.00. Keep the supplied indices constant. At what vacuum wavelength is the propagation phase π? Does that establish zero oblique reflection?

   Show the answer and reasoning

   sin θ_f = 1/3, so cos θ_f = √(8/9). The phase-π wavelength is 600√(8/9) ≈ 565.685 nm. It does not establish zero reflected power.

   Use λ₀ = 4 n_f d cos θ_f, not a longer-path substitution by 1/cos θ_f. At oblique incidence the interface amplitudes are polarisation dependent and the normal-incidence amplitude match does not automatically survive.

7. For the same indices, a uniform 110 nm film is evaluated at 600 nm. State its first minimum wavelength and model reflected percentage. Has a higher minimum wavelength proved a changed chemistry?

   Show the answer and reasoning

   The first minimum is 660 nm; at 600 nm the model gives about 0.423059% reflection. No changed chemistry is established.

   Thickness alone has changed n_f d while the material indices were stipulated unchanged. In a real inverse measurement, index and other structural changes can also shift a spectral feature, so additional evidence is needed.

8. At 600 nm, a hypothetical repeated structure uses n_H = 2.00 and n_L = 1.50 quarter-wave layers. Find each thickness and the total physical thickness of three H/L pairs. Can the single-film R formula establish its reflection?

   Show the answer and reasoning

   H is 75.0 nm, L is 100 nm, one pair is 175 nm and three pairs total 525 nm. The single-film formula does not calculate that six-layer structure.

   Each layer has the same optical thickness 150 nm, not the same physical thickness. The many interfaces, their sequence and boundaries must be retained in an appropriate multilayer calculation or measurement.

9. A coated sample sends less light to a specular reflection detector, but no other outgoing directions or transmission have been measured. Does that prove useful forward transmission increased by the missing amount?

   Show the answer and reasoning

   No. Absorption and uncollected scattering remain possible, as do other changes in the measurement boundary.

   The ideal lossless model and complete collection permitted R + T = 1 for its stated boundary. That conclusion is not licensed by a narrow reflection reading on an uncharacterised real specimen.

10. The 100 nm coating meets R ≤ 5% at 450, 600 and 900 nm in the table. A report calls those three observations proof of compliance at every wavelength from 450 to 900 nm. Is that inference sufficient?

    Show the answer and reasoning

    No. The three-point requirement and an entire-band requirement are different claims.

    The stipulated continuous model can predict intermediate values, but real compliance requires relevant spectral evidence and uncertainties over the claimed domain. Do not silently turn discrete observations into a complete measured spectrum.

What to carry forward

• Optical architecture controls phase relationships.
• State bandwidth, angle and losses.
• Add coherent fields before forming their combined reflected power; reflection sign changes matter as well as extra travel.
• An opposing phase does not guarantee matched amplitudes, and repeated internal returns belong in a quantitative film response.
• A single spectral dip can fit more than one thickness or structure; measurements away from it can discriminate alternatives.
• A film, repeated-index mirror, cavity and finite component have different boundaries; one constituent index or one formula does not describe them all.
• Bounded hypothetical calculations are useful predictions under their assumptions, not fabrication tolerances, complete spectra or certified component performance.